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Proposition 6.1.6. Let \( \chi \) be a character of \( G \) and let \( n \geq 1 \) be an integer. Set \( \chi \left( \mathfrak{p}\right) = 0 \) if \( \mathfrak{p} \) divides the conductor of \( \chi \) . Then \( {a}_{1}\left( \chi \right) = 1 \), and for \( n > 1 \) , if \( n = {p}_{1}^{{m}_{1}}\ldots {p}_{k}^{{m}_{k}} \) is the prime factorization of \( n \), then\n\n\[ \n{a}_{n}\left( \chi \right) = {a}_{{p}_{1}^{{m}_{1}}}\left( \chi \right) \ldots {a}_{{p}_{k}^{{m}_{k}}}\left( \chi \right) ,\n\]\n\nwhere the coefficients \( {a}_{{p}^{m}}\left( \chi \right) \) are given by one of the following formulas:\n\n(1) if \( p \) is inert, \( {a}_{{p}^{m}}\left( \chi \right) = 0 \) if \( m \) is odd and \( {a}_{{p}^{m}}\left( \chi \right) = \chi {\left( p\right) }^{m/2} \) if \( m \) is even;\n\n(2) if \( p{\mathbb{Z}}_{K} = {\mathfrak{p}}^{2} \) is ramified, \( {a}_{{p}^{m}}\left( \chi \right) = \chi {\left( \mathfrak{p}\right) }^{m} \) ;\n\n(3) if \( p{\mathbb{Z}}_{K} = {\mathfrak{{pp}}}^{\prime } \) splits, \( {a}_{{p}^{m}}\left( \chi \right) = \left( {m + 1}\right) \chi {\left( \mathfrak{p}\right) }^{m} \) if \( \chi \left( \mathfrak{p}\right) = \chi \left( {\mathfrak{p}}^{\prime }\right) \) and otherwise\n\n\[ \n{a}_{{p}^{m}}\left( \chi \right) = \frac{\chi {\left( \mathfrak{p}\right) }^{m + 1} - \chi {\left( {\mathfrak{p}}^{\prime }\right) }^{m + 1}}{\chi \left( \mathfrak{p}\right) - \chi \left( {\mathfrak{p}}^{\prime }\right) }.\n\]
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Proof. The first assertion is a translation of the fact that the map \( n \mapsto \) \( {a}_{n}\left( \chi \right) \) is multiplicative. The formulas for \( n = {p}^{m} \) are clear when \( p \) is inert or ramified and are easily proved by induction when \( p \) is split.
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No
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Proposition 6.2.5. Keep the above notation. If \( \gamma = a + {b\omega } \in {\mathbb{Z}}_{K} \) satisfies \( \left| {{\tau }_{2}\left( \gamma \right) - \beta }\right| \leq \varepsilon \) and \( \left| {{\tau }_{1}\left( \gamma \right) }\right| \leq B \), then \( q\left( {a, b,1}\right) \leq 3{B}^{2} \) . Conversely, assume in addition that \( \varepsilon < 1/\left( {3\left( {B + 1}\right) \left( {\sqrt{D} + 1}\right) }\right) \) . Then if \( q\left( {x, y, z}\right) \leq 3{B}^{2} \) and \( \left( {x, y, z}\right) \neq \left( {0,0,0}\right) \), we have \( z = \pm 1 \) and \( \gamma = z\left( {x + {y\omega }}\right) \) is a solution to the slightly weaker problem \( \left| {{\tau }_{2}\left( \gamma \right) - \beta }\right| \leq \varepsilon \sqrt{3} \) and \( \left| {{\tau }_{1}\left( \gamma \right) }\right| \leq B\sqrt{3} \) .
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Proof. The first statement is clear. Conversely, assume that \( q\left( {x, y, z}\right) \leq \) \( 3{B}^{2} \) and that \( \left( {x, y, z}\right) \neq \left( {0,0,0}\right) \) . It is clear that \( \left| z\right| \leq 1 \), and Exercise 4 (which is an excellent exercise on the properties of continued fractions) shows that if we assume the given inequality for \( \varepsilon \), then \( z \neq 0 \), so \( z = \pm 1 \) and the rest of the proposition follows.
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No
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Proposition 6.3.1. Let \( \mathfrak{a} \) and \( \mathfrak{c} \) be fractional ideals of \( K \), and let \( \operatorname{Art}\left( \mathfrak{c}\right) \) be the element of \( \operatorname{Gal}\left( {K\left( 1\right) /K}\right) \) corresponding to the ideal \( \mathfrak{c} \) by the Artin reciprocity map (since \( K\left( 1\right) /K \) is unramified, there are no ramification conditions on \( \mathfrak{c} \) ). Then \( j\left( \mathfrak{a}\right) \in {\mathbb{Z}}_{K\left( 1\right) } \) (in particular, it is an algebraic integer) and \[ j{\left( \mathfrak{a}\right) }^{\operatorname{Art}\left( \mathfrak{c}\right) } = j\left( {\mathfrak{a}{\mathfrak{c}}^{-1}}\right) . \]
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Since we know that \( K\left( 1\right) = K\left( {j\left( {\mathbb{Z}}_{K}\right) }\right) \), it follows from this proposition that \[ \alpha = {\operatorname{Tr}}_{K\left( 1\right) /K{\left( 1\right) }^{C}}\left( {j\left( {\mathbb{Z}}_{K}\right) }\right) = \mathop{\sum }\limits_{{\overline{\mathfrak{c}} \in \overline{C}}}j\left( {\mathfrak{c}}^{-1}\right) \in K{\left( 1\right) }^{C}. \] It can be shown (see [Sch3]) that \( \alpha \) does not belong to any subfield of \( K{\left( 1\right) }^{C} \) , in other words, that for any \( \mathfrak{b} \notin C \) we have \[ \mathop{\sum }\limits_{{\overline{\mathfrak{c}} \in \bar{C}}}j\left( {{\mathfrak{b}}^{-1}{\mathfrak{c}}^{-1}}\right) \neq \mathop{\sum }\limits_{{\overline{\mathfrak{c}} \in \bar{C}}}j\left( {\mathfrak{c}}^{-1}\right) . \] It follows that \( K{\left( 1\right) }^{C} = K\left( \alpha \right) \), hence the problem of the construction of \( K{\left( 1\right) }^{C} \) is in principle solved.
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No
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Proposition 6.3.3. Let \( {\Gamma }^{0}\left( {pq}\right) \) be the group of matrices \( \gamma = \left( \begin{array}{ll} a & b \\ c & d \end{array}\right) \in \) \( {\mathrm{{PSL}}}_{2}\left( \mathbb{Z}\right) \) such that \( {pq} \mid b \) .\n\n(1) For any \( \gamma \in {\Gamma }^{0}\left( {pq}\right) \), we have \( {g}_{p, q}\left( {\gamma \left( \tau \right) }\right) = {v}_{g}\left( \gamma \right) {g}_{p, q}\left( \tau \right) \), where the multiplier system \( {v}_{g} \) is given by\n\n\[ \n{v}_{g}\left( \gamma \right) = \exp \left( {-\frac{2i\pi }{24}\left( {p - 1}\right) \left( {q - 1}\right) \left( {{cd}\left( {1 - {a}^{2}}\right) - {ac} + 3\left( {a - 1}\right) + a\frac{b}{pq}}\right) }\right) \n\]\n\n(using the same normalization for \( \gamma \) as above).\n\n(2) If \( e \) is an integer such that \( {24} \mid e\left( {p - 1}\right) \left( {q - 1}\right) \), the function \( {g}_{p, q}^{e}\left( \tau \right) \) is invariant under \( {\Gamma }^{0}\left( {pq}\right) \) .
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Proof. Left to the reader (Exercise 10).
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No
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Proposition 6.3.4. Let \( \mathfrak{p} \) and \( \mathfrak{q} \) be two primitive ideals of respective norms \( p \) and \( q \) such that the product \( \mathfrak{{pq}} \) is also primitive. If\n\n\[ \mathfrak{p}\mathfrak{q} = {pq}\mathbb{Z} \oplus \frac{-w + \sqrt{D}}{2}\mathbb{Z} \]\n\nthen \( \mathfrak{p} = p\mathbb{Z} \oplus \left( {\left( {-w + \sqrt{D}}\right) /2}\right) \mathbb{Z} \) and \( \mathfrak{q} = q\mathbb{Z} \oplus \left( {\left( {-w + \sqrt{D}}\right) /2}\right) \mathbb{Z} \) .
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Proof. Since \( \mathfrak{{pq}} \subset \mathfrak{p} \), we have \( w \equiv u\left( {{\;\operatorname{mod}\;2}p}\right) \) and similarly for \( \mathfrak{q} \), proving the proposition.
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No
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Corollary 6.3.5. Let \( \mathfrak{a},\mathfrak{p},\mathfrak{q} \) be three primitive ideals such that apq is a primitive ideal. Let \( a = \mathcal{N}\left( \mathfrak{a}\right), p = \mathcal{N}\left( \mathfrak{p}\right), q = \mathcal{N}\left( \mathfrak{q}\right) \), and assume that \( e \) is a positive integer such that \( {24} \mid e\left( {p - 1}\right) \left( {q - 1}\right) \) . (1) There exists an oriented basis \( \left( {{\omega }_{1},{\omega }_{2}}\right) \) of a such that \( \left( {{\omega }_{1}, p{\omega }_{2}}\right) \) is a basis of ap, \( \left( {{\omega }_{1}, q{\omega }_{2}}\right) \) is a basis of aq, and \( \left( {{\omega }_{1},{pq}{\omega }_{2}}\right) \) is a basis of apq. (2) The quantity \( {g}_{p, q, e}\left( {{\omega }_{1}/{\omega }_{2}}\right) \) is independent of the choice of oriented basis satisfying (1).
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Proof. For (1), we write apq \( = \operatorname{apq}\mathbb{Z} \oplus \left( {\left( {-w + \sqrt{D}}\right) /2}\right) \mathbb{Z} \) . It follows from the proposition that \( {\omega }_{1} = \left( {-w + \sqrt{D}}\right) /2 \) and \( {\omega }_{2} = a \) is a suitable basis. For (2), we note that \( \left( {{\omega }_{1}^{\prime },{\omega }_{2}^{\prime }}\right) \) is another suitable basis of \( \mathfrak{a} \) if and only if there exists \( \gamma = \left( \begin{matrix} a & b \\ c & d \end{matrix}\right) \in \widetilde{{\mathrm{{PSL}}}_{2}}\left( \mathbb{Z}\right) \) such that \( a{\omega }_{1} + b{\omega }_{2} \in \mathfrak{{apq}} \) and \( {pq}\left( {c{\omega }_{1} + }\right. \) \( \left. {d{\omega }_{2}}\right) \in \) apq. This is equivalent to the single condition \[ b{\omega }_{2} \in \mathfrak{{apq}} = {\omega }_{1}\mathbb{Z} \oplus {pq}{\omega }_{2}\mathbb{Z}, \] hence to \( {pq} \mid b \), so \( \gamma \in {\Gamma }^{0}\left( {pq}\right) \) . Therefore, \( {g}_{p, q, e}\left( {{\omega }_{1}^{\prime }/{\omega }_{2}^{\prime }}\right) = {g}_{p, q, e}\left( {{\omega }_{1}/{\omega }_{2}}\right) \) since \( {g}_{p, q, e} \) is invariant under \( {\Gamma }^{0}\left( {pq}\right) \), as was to be proved.
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Yes
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Theorem 6.3.7 (Schertz). Let \( {\left( {\mathfrak{a}}_{i}\right) }_{1 \leq i \leq h\left( K\right) } \) be a system of representatives of the ideal classes of \( K = \mathbb{Q}\left( \sqrt{D}\right) \), chosen to be primitive. Let \( \mathfrak{p} \) and \( \mathfrak{q} \) be ideals of \( K \) of norm \( p \) and \( q \), respectively. Assume that\n\n(1) the ideals \( \mathfrak{p} \) and \( \mathfrak{q} \) are primitive ideals that are nonprincipal;\n\n(2) if both classes of \( \mathfrak{p} \) and \( \mathfrak{q} \) are of order 2 in the class group, these classes are equal;\n\n(3) for all \( i,{\mathfrak{{pqa}}}_{i} \) is a primitive ideal;\n\n(4) \( e \) is a positive integer chosen such that \( {24} \mid e\left( {p - 1}\right) \left( {q - 1}\right) \) .\n\nSet\n\n\[ \n{P}_{\mathfrak{p},\mathfrak{q}, e}\left( X\right) = \mathop{\prod }\limits_{{1 \leq i \leq h\left( K\right) }}\left( {X - {g}_{\mathfrak{p},\mathfrak{q}, e}\left( {\mathfrak{a}}_{i}\right) }\right) = \mathop{\prod }\limits_{{1 \leq i \leq h\left( K\right) }}\left( {X - {\left( \frac{\eta \left( {{\tau }_{i}/p}\right) \eta \left( {{\tau }_{i}/q}\right) }{\eta \left( {{\tau }_{i}/{pq}}\right) \eta \left( {\tau }_{i}\right) }\right) }^{e}}\right) , \n\]\nwhere \( {\mathfrak{a}}_{i}\mathfrak{p}\mathfrak{q} = {a}_{i}\left( {{pq}\mathbb{Z} + {\tau }_{i}\mathbb{Z}}\right) \).\n\nThen \( {P}_{\mathfrak{p},\mathfrak{q}, e}\left( X\right) \in \mathbb{Z}\left\lbrack X\right\rbrack \), it is irreducible in \( \mathbb{Z}\left\lbrack X\right\rbrack \) and in \( K\left\lbrack X\right\rbrack \), its constant term is equal to \( \pm 1 \), and the field obtained by adjoining to \( K \) a root of \( {P}_{\mathfrak{p},\mathfrak{q}, e} \) is the Hilbert class field \( K\left( 1\right) \) of \( K \) .
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We refer to [Sch1] for the proof. The statement (and the proof) given by Schertz is slightly incorrect because of the omission of condition (2) above on classes of order 2 . An example is \( K = \mathbb{Q}\left( \sqrt{-{30}}\right) \), which has class group isomorphic to \( {\left( \mathbb{Z}/2\mathbb{Z}\right) }^{2} \), with \( \mathfrak{p} \) an ideal above \( {11},\mathfrak{q} \) an ideal above 37, and \( e = 1 \) . The polynomial \( {P}_{\mathfrak{p},\mathfrak{q}, e} \) found in this case is the square of an irreducible polynomial in \( \mathbb{Z}\left\lbrack X\right\rbrack \) . However, it is easy to correct the statement and proof as above, as was remarked by the author and Schertz himself (see [Sch4]).
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Yes
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Proposition 6.3.11. Let \( L \) be a complex lattice.\n\n(1) There exists a unique meromorphic function \( \zeta \left( {z, L}\right) \), called the Weierstrass \( \zeta \) -function, such that \( {\zeta }^{\prime }\left( {z, L}\right) = - \wp \left( {z, L}\right) \) and such that \( \zeta \left( {z, L}\right) \) is an odd function.
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Proof. Since the proofs are easy, we leave some details to the reader (Exercise 13).\n\nIt is clear that the series defining \( \zeta \left( {z, L}\right) \) converges uniformly on any compact subset not containing points of \( L \) (this is the case as soon as the general term goes to zero faster than \( 1/{\left| \omega \right| }^{\alpha } \) for any \( \alpha > 2 \) ). Thus the series defines a meromorphic function on \( \mathbb{C} \) with poles at points of \( L \), and by differentiating termwise it is clear that \( \zeta {\left( z, L\right) }^{\prime } = - \wp \left( {z, L}\right) \) . In addition,\n\n\[ - \zeta \left( {-z, L}\right) = {z}^{-1} + {z}^{2}\mathop{\sum }\limits_{{\omega \in {L}^{ * }}}{\left( {\omega }^{2}\left( z + \omega \right) \right) }^{-1} = {z}^{-1} + {z}^{2}\mathop{\sum }\limits_{{\omega \in {L}^{ * }}}{\left( {\omega }^{2}\left( z - \omega \right) \right) }^{-1}, \]\n\nso \( \zeta \left( {z, L}\right) \) is an odd function, thus proving (1) and (2), since clearly the property of being odd makes \( \zeta \) unique among all antiderivatives of \( - \wp \) .
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No
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Corollary 6.3.12. For any \( \left( \begin{array}{ll} a & b \\ c & d \end{array}\right) \in {\operatorname{PSL}}_{2}\left( \mathbb{Z}\right) \) and \( \tau \in \mathcal{H} \), we have\n\n\[ \n{E}_{2}\left( \frac{{a\tau } + b}{{c\tau } + d}\right) = {\left( c\tau + d\right) }^{2}{E}_{2}\left( \tau \right) + \frac{{12c}\left( {{c\tau } + d}\right) }{2i\pi }.\n\]
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Proof. Set \( {\omega }_{1}^{\prime } = a{\omega }_{1} + b{\omega }_{2},{\omega }_{2}^{\prime } = c{\omega }_{1} + d{\omega }_{2} \), and \( {\tau }^{\prime } = {\omega }_{1}^{\prime }/{\omega }_{2}^{\prime } \) . By assumption, \( \left( {{\omega }_{1}^{\prime },{\omega }_{2}^{\prime }}\right) \) is still an oriented basis of \( L \), so Proposition 6.3.11 applied to this basis gives in particular\n\n\[ \n\zeta \left( {z + {\omega }_{2}^{\prime }, L}\right) = \zeta \left( {z, L}\right) + \frac{{\pi }^{2}}{3{\omega }_{2}^{\prime }}{E}_{2}\left( {\tau }^{\prime }\right) = \zeta \left( {z, L}\right) + \frac{{\pi }^{2}}{3{\omega }_{2}\left( {{c\tau } + d}\right) }{E}_{2}\left( \frac{{a\tau } + b}{{c\tau } + d}\right) \;. \n\]\n\nOn the other hand, the same proposition applied to the basis \( \left( {{\omega }_{1},{\omega }_{2}}\right) \) gives\n\n\[ \n\zeta \left( {z + {\omega }_{2}^{\prime }, L}\right) = \zeta \left( {z, L}\right) + c{\eta }_{1} + d{\eta }_{2} = \zeta \left( {z, L}\right) + \frac{{\pi }^{2}}{3{\omega }_{2}}\left( {{c\tau } + d}\right) {E}_{2}\left( \tau \right) - c\frac{2i\pi }{{\omega }_{2}}, \n\]\n\nwhich gives the corollary by identification.
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Yes
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Proposition 6.3.14. Let \( L \) be a complex lattice.\n\n(1) There exists a unique holomorphic function \( \sigma \left( {z, L}\right) \), called the Weierstrass \( \sigma \) -function, such that \( {\sigma }^{\prime }\left( {z, L}\right) /\sigma \left( {z, L}\right) = \zeta \left( {z, L}\right) \) and \( \mathop{\lim }\limits_{{z \rightarrow 0}}\sigma \left( {z, L}\right) /z = 1.\n\n(2) The function \( \sigma \left( {z, L}\right) \) is an odd function having simple zeros exactly at all points of \( L \) . More precisely, we have the expansion\n\n\[ \sigma \left( {z, L}\right) = z\mathop{\prod }\limits_{{\omega \in {L}^{ * }}}\left( {1 - \frac{z}{\omega }}\right) {e}^{\frac{z}{\omega } + \frac{{z}^{2}}{2{\omega }^{2}}}. \]
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Proof. Once again, we leave some details to the reader (Exercise 16). The general term of the given product expansion tends to 1 as \( {\left| \omega \right| }^{-3} \), hence the product converges uniformly on any compact subset of \( \mathbb{C} \) and so defines a holomorphic function. By definition, its logarithmic derivative is equal to \( \zeta \left( {z, L}\right) \), and we have \( \mathop{\lim }\limits_{{z \rightarrow 0}}\sigma \left( {z, L}\right) /z = 1 \), a condition that also ensures uniqueness of \( \sigma \left( {z, L}\right) \) . Since \( L \) is symmetrical with respect to the origin, \( \sigma \left( {z, L}\right) \) is an odd function, proving (1) and (2).
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No
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Corollary 6.3.16.\n\n\[ \eta \left( \tau \right) = {q}^{1/{24}}\left( {1 + \mathop{\sum }\limits_{{k \geq 1}}{\left( -1\right) }^{k}\left( {{q}^{k\left( {{3k} - 1}\right) /2} + {q}^{k\left( {{3k} + 1}\right) /2}}\right) }\right) . \]
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Proof. This follows from the Jacobi identity by replacing \( \left( {u, q}\right) \) by \( \left( {q,{q}^{3}}\right) \) and rearranging terms. The details are left to the reader (Exercise 17). Note that this is the identity we used in Algorithm 6.3.2.
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No
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Corollary 6.3.17.\n\n\[ \n{\eta }^{3}\left( \tau \right) = {q}^{1/8}\mathop{\sum }\limits_{{k \geq 0}}{\left( -1\right) }^{k}\left( {{2k} + 1}\right) {q}^{k\left( {k + 1}\right) /2}.\n\]
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Proof. This follows from the Jacobi identity by making \( u \rightarrow 1 \) .
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No
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Corollary 6.3.18. With the same notation as above, we have the formula\n\n\\[ \n\\sigma \\left( {z, L}\\right) = \\frac{{\\omega }_{2}}{2i\\pi }{e}^{{\\eta }_{2}{z}^{2}/\\left( {2{\\omega }_{2}}\\right) }\\frac{\\mathop{\\sum }\\limits_{{k \\geq 0}}{\\left( -1\\right) }^{k}\\left( {{u}^{k + 1/2} - {u}^{-\\left( {k + 1/2}\\right) }}\\right) {q}^{k\\left( {k + 1}\\right) /2}}{{q}^{-1/8}\\eta {\\left( \\tau \\right) }^{3}}.\n\\]
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## Proof. Clear from Proposition 6.3.14 (4).
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No
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Proposition 6.3.20. Let \( L \) be a complex lattice and \( a \notin L \) . Then\n\n\[ \wp \left( {z, L}\right) - \wp \left( {a, L}\right) = - \frac{\sigma \left( {z - a, L}\right) \sigma \left( {z + a, L}\right) }{\sigma {\left( a, L\right) }^{2}\sigma {\left( z, L\right) }^{2}}. \]
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Proof. Using Proposition 6.3.14, it is easy to check that the ratio of the left- and right-hand sides is an elliptic function with no zero or poles, hence it is constant by Liouville’s theorem. Making \( z \rightarrow 0 \) and using the expansions of \( \wp \) and \( \sigma \) around 0 gives the result.
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No
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Proposition 6.3.21. Let \( \omega = m{\omega }_{1} + n{\omega }_{2} \in L \) with \( m \) and \( n \) in \( \mathbb{Z} \), and let \( z = {x}_{1}{\omega }_{1} + {x}_{2}{\omega }_{2} \) with \( {x}_{1} \) and \( {x}_{2} \) in \( \mathbb{R} \), as above. Then\n\n\[ \n{\phi }^{ * }\left( {z + \omega, L}\right) = s\left( \omega \right) {e}^{{2i\pi }\left( {n{x}_{1} - m{x}_{2}}\right) }{\phi }^{ * }\left( {z, L}\right) ,\n\]\n\nwhere \( s\left( \omega \right) = + 1 \) if \( \omega /2 \in L, s\left( \omega \right) = - 1 \) otherwise. In addition, the function \( {\phi }^{ * } \) is bounded on \( \mathbb{C} \) .
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Proof. The proof of the transformation formula follows immediately from the corresponding formula for the \( \sigma \) -function seen above. It follows in particular that \( \left| {{\phi }^{ * }\left( {z + \omega, L}\right) }\right| = \left| {{\phi }^{ * }\left( {z, L}\right) }\right| \), and since \( {\phi }^{ * } \) is a continuous (although nonholomorphic) function, \( {\phi }^{ * }\left( {z, L}\right) \) is bounded in any compact set, hence in any fundamental parallelogram of the form \( \left( {a, a + {\omega }_{2}, a + {\omega }_{1} + {\omega }_{2}, a + {\omega }_{1}}\right) \) . It follows that it is bounded on all of \( \mathbb{C} \) . In fact, it is not difficult to give explicit bounds if desired, using the fact that we can reduce to \( \operatorname{Im}\left( {{\omega }_{1}/{\omega }_{2}}\right) \geq \sqrt{3}/2 \) and to \( z/{\omega }_{2} = x + {iy} \) with \( \left| x\right| \leq 1/2 \) and \( \left| y\right| \leq 1/2 \) (Exercise 22).
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Yes
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Theorem 6.3.24. Let \( K \) be an imaginary quadratic field of discriminant \( D \) , let \( h \) be its class number, and let \( \mathfrak{f} \) be a conductor of \( K \) different from \( {\mathbb{Z}}_{K} \) (see Proposition 3.3.20 for all the possible conductors). Denote as usual by \( {\zeta }_{m} \) a primitive mth root of unity, and for any prime number \( \ell \) dividing \( D \) , denote by \( {\mathfrak{p}}_{\ell } \) the unique ramified prime ideal above \( \ell \) .
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Proof. The special cases \( D = - 3,\mathfrak{f} = {\mathfrak{p}}_{3}^{3} \) and \( D = - 4,\mathfrak{f} = 4{\mathbb{Z}}_{K} \) are easily treated directly, so we exclude these cases. We first prove a lemma.\n\nLemma 6.3.25. For any integer \( m \), denote by \( \mathfrak{f}\left( m\right) \) the conductor of the Abelian extension \( K\left( {\zeta }_{m}\right) /K \).\n\n(1) We have \( \mathfrak{f}\left( m\right) \mid m{\mathbb{Z}}_{K} \).\n\n(2) If \( K = \mathbb{Q}\left( \sqrt{D}\right) \) with \( D \equiv 8\left( {\;\operatorname{mod}\;{16}}\right) \), then
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No
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Lemma 6.3.25. For any integer \( m \), denote by \( \mathfrak{f}\left( m\right) \) the conductor of the Abelian extension \( K\left( {\zeta }_{m}\right) /K \). (1) We have \( \mathfrak{f}\left( m\right) \mid m{\mathbb{Z}}_{K} \). (2) If \( K = \mathbb{Q}\left( \sqrt{D}\right) \) with \( D \equiv 8\left( {\;\operatorname{mod}\;{16}}\right) \), then \( \mathfrak{f}\left( 4\right) = 2{\mathbb{Z}}_{K} \) ; while if \( D \equiv {12} \) \( {\;\operatorname{mod}\;{16}} \), then \( \mathfrak{f}\left( 4\right) = {\mathbb{Z}}_{K} \).
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Proof. (1) is nothing else but Proposition 3.5.5. For (2), we can, for example, use Hecke's Theorem 10.2.9 from which we borrow the notation. Indeed, since \( \left\lbrack {K\left( i\right) : K}\right\rbrack = 2 \), we have \( \mathfrak{f}\left( 4\right) = \mathfrak{d}\left( {K\left( i\right) /K}\right) \). Denote by \( \mathfrak{p} \) the unique prime ideal above 2, so that \( z\left( {\mathfrak{p},2}\right) = 5 \). It is easily checked by an explicit calculation that the congruence \( {x}^{2} \equiv - 1\left( {{\;\operatorname{mod}\;4}{\mathbb{Z}}_{K}}\right) \) is not soluble if \( D \equiv 8 \) (mod 16) and that it is soluble if \( D \equiv {12}\left( {\;\operatorname{mod}\;{16}}\right) \), while of course \( {x}^{2} \equiv - 1 \) \( \left( {{\;\operatorname{mod}\;2}{\mathbb{Z}}_{K}}\right) \) is always soluble. Since the largest value of \( a < z\left( {\mathfrak{p},2}\right) \) that occurs in Hecke’s theorem is odd, we deduce that \( a = 3 \) if \( D \equiv 8\left( {\;\operatorname{mod}\;{16}}\right) \) and \( a \geq 4 \) if \( D \equiv {12}\left( {\;\operatorname{mod}\;{16}}\right) \), proving the lemma.
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Yes
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Lemma 6.3.26. Let \( f \) be an integer such that \( f \nmid {12} \) .\n\n(1) If \( D \) is any integer, there exists \( t \) such that \( \left( {{t}^{2} - D,{2f}}\right) = 1 \) and \( f \nmid {2t} \) .
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Proof. Let \( {E}_{f} \) be the set of \( t \) such that \( 0 \leq t < {2f} \) and \( \left( {{t}^{2} - D,{2f}}\right) = 1 \) . This condition means that \( t ≢ D\left( {\;\operatorname{mod}\;2}\right) \) and that for each prime \( p > 2 \) dividing \( f, t \) must not be congruent to the square roots of \( D \) modulo \( p \) if they exist. It follows that\n\n\[ \left| {E}_{f}\right| = f\mathop{\prod }\limits_{{p \mid f, p > 2}}\left( {1 - \frac{1 + \left( \frac{D}{p}\right) }{p}}\right) . \]\n\nIf we write \( f = \mathop{\prod }\limits_{{p \mid f}}{p}^{{v}_{p}} \), we thus have\n\n\[ \left| {E}_{f}\right| \geq {2}^{{v}_{2}}\mathop{\prod }\limits_{{p \mid f, p > 2}}{p}^{{v}_{p} - 1}\left( {p - 2}\right) . \]\n\nOn the other hand, let \( {F}_{f} \) be the set of \( t \) such that \( 0 \leq t < {2f} \) and \( f \mid {2t} \) . Clearly, \( \left| {F}_{f}\right| = 2 \) if \( f \) is odd and \( \left| {F}_{f}\right| = 4 \) if \( f \) is even.\n\nIf there exists a prime \( p \geq 5 \) such that \( p \mid f \), then \( \left| {E}_{f}\right| \geq 3 \cdot {2}^{{v}_{2}} \), hence \( \left| {E}_{f}\right| > 2 \) if \( {v}_{2} = 0 \) and \( \left| {E}_{f}\right| > 4 \) if \( {v}_{2} \geq 1 \) ; hence \( \left| {{E}_{f} \smallsetminus {F}_{f}}\right| > 0 \) . A similar reasoning shows that the same conclusion still holds if \( {v}_{3} \geq 2 \) or if \( {v}_{2} \geq 3 \) . Thus the only \( f \) for which (1) may be false are numbers of the form \( f = {2}^{{v}_{2}}{3}^{{v}_{3}} \) with \( 0 \leq {v}_{2} \leq 2 \) and \( 0 \leq {v}_{3} \leq 1 \) ; in other words, the divisors of 12 . It is easy to check that the conclusion of (1) (and of (2)) is false if \( f \) is a divisor of 12 .
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Yes
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Theorem 7.1.5. Let \( L/K \) be a relative extension of degree \( n \), and let \( h\left( K\right) \) be the class number of \( K \). (1) If \( \left( {n, h\left( K\right) }\right) = 1 \) (for example, if \( K \) has class number 1), the natural map \( {\psi }_{N, i} \) from \( C{l}_{N}\left( {L/K}\right) \) to \( C{l}_{i}\left( {L/K}\right) \) defined above is an isomorphism, and \( {\mathrm{{Cl}}}_{N}\left( K\right) = {\mathrm{{Cl}}}_{i}\left( K\right) = \{ 1\} .
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Proof. It is clear that statements (3) and (4) imply statement (2), which implies statement (1). Furthermore, the formulas \( {\mathcal{N}}_{L/K}\left( {{i}_{L/K}\left( \overline{\mathfrak{a}}\right) }\right) = {\overline{\mathfrak{a}}}^{n} \) and \( {\overline{\mathfrak{a}}}^{h\left( K\right) } = \overline{1} \) in \( {Cl}\left( K\right) \) immediately imply (5). Thus we need only to prove (3) and (4). Recall from Proposition 4.1.17 that in the category of finite Abelian groups, taking \( p \) -Sylow subgroups is an exact functor. Applying this to the two exact sequences (1) and (2), which define the two notions of class groups, we obtain the following two exact sequences:\n\n\[ 1 \rightarrow {\operatorname{Cl}}_{i}{\left( K\right) }_{p} \rightarrow \operatorname{Cl}{\left( K\right) }_{p}\overset{{i}_{L/K, p}}{ \rightarrow }\operatorname{Cl}{\left( L\right) }_{p} \rightarrow {\operatorname{Cl}}_{i}{\left( L/K\right) }_{p} \rightarrow 1 \]\n\nand\n\n\[ 1 \rightarrow {\operatorname{Cl}}_{N}{\left( L/K\right) }_{p} \rightarrow \operatorname{Cl}{\left( L\right) }_{p}\overset{{\mathcal{N}}_{L/K, p}}{ \rightarrow }\operatorname{Cl}{\left( K\right) }_{p} \rightarrow {\operatorname{Cl}}_{N}{\left( K\right) }_{p} \rightarrow 1. \]\n\n(3)\n\nAssume first that \( p \) is a prime such that \( p \nmid h\left( K\right) \), in other words that \( {Cl}{\left( K\right) }_{p} = \{ 1\} \) . The two exact sequences above imply immediately that \( C{l}_{i}{\left( K\right) }_{p} = C{l}_{N}{\left( K\right) }_{p} = \{ 1\} \) and that \( C{l}_{i}{\left( L/K\right) }_{p} \simeq {Cl}{\left( L\right) }_{p} \simeq C{l}_{N}\left( {L/K}\right) , \) proving (3).
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Yes
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Proposition 7.2.3. (1) We have the following exact sequence\n\n\[ 1 \rightarrow U\left( K\right) \overset{{i}_{L/K}}{ \rightarrow }U\left( L\right) \overset{j}{ \rightarrow }{U}_{i}\left( {L/K}\right) \overset{\pi }{ \rightarrow }C{l}_{i}\left( K\right) \rightarrow 1 \]
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Proof. (1). If \( \alpha \in U\left( L\right) \), then \( \alpha {\mathbb{Z}}_{L} = {\mathbb{Z}}_{L} \), so \( \alpha {\mathbb{Z}}_{K}{\mathbb{Z}}_{L} = {\mathbb{Z}}_{L} \) . Hence \( j\left( \alpha \right) \in \) \( {U}_{i}\left( {L/K}\right) \) is well-defined and is a group homomorphism. If \( \overline{\left( \mathfrak{a},\alpha \right) } = \overline{\left( \mathfrak{b},\beta \right) } \) in \( {U}_{i}\left( {L/K}\right) \), there exists \( \gamma \in {K}^{ * } \) such that \( \mathfrak{b} = \gamma \mathfrak{a} \) and \( \beta = \alpha /\gamma \), and furthermore \( \alpha \mathfrak{a}{\mathbb{Z}}_{L} = {\mathbb{Z}}_{L} \) . It follows that \( \mathfrak{a} \) and \( \mathfrak{b} \) are in the same ideal class in \( {Cl}\left( K\right) \) and that the class of \( \mathfrak{a} \) is in the capitulating class group, so the map \( \pi \) is well-defined and is a group homomorphism.\n\nLet us show exactness. Exactness at \( U\left( K\right) \) follows from the fact that \( {i}_{L/K} \) is injective on elements (recall that it was not injective on ideal classes). The kernel of \( j \) is the set of \( \alpha \in U\left( L\right) \) such that \( \overline{\left( {\mathbb{Z}}_{K},\alpha \right) } \) is the unit element of the group \( {U}_{i}\left( {L/K}\right) \), which is the class of \( \overline{\left( {\mathbb{Z}}_{K},1\right) } \) . By definition of the equivalence relation, this means that there exists \( u \in {K}^{ * } \) such that \( {\mathbb{Z}}_{K} = u{\mathbb{Z}}_{K} \) and \( \alpha = u \cdot 1 \), in other words that \( \alpha = u \) is a unit of \( {\mathbb{Z}}_{K} \), so the kernel of \( i \) is the group \( U\left( K\right) \), thus showing exactness at \( U\left( L\right) \) .\n\nThe kernel of \( \pi \) is the set of \( \overline{\left( \mathfrak{a},\alpha \right) } \) such that \( \mathfrak{a} \) is a principal ideal in \( {\mathbb{Z}}_{K} \) , hence such that there exists \( \gamma \in {K}^{ * } \) with \( \mathfrak{a} = \gamma {\mathbb{Z}}_{K} \), and also \( \alpha \mathfrak{a}{\mathbb{Z}}_{L} = {\mathbb{Z}}_{L} \) . Hence, \( \overline{\left( \mathfrak{a},\alpha \right) } = \overline{\left( {\mathbb{Z}}_{K},\alpha \gamma \right) } \), and \( {\mathbb{Z}}_{L} = \alpha \mathfrak{a}{\mathbb{Z}}_{L} = {\alpha \gamma }{\mathbb{Z}}_{L} \), so \( {\alpha \gamma } \in U\left( L\right) \) and the kernel of \( \pi \) is thus equal to \( j\left( {U\left( L\right) }\right) \), proving exactness at \( {U}_{i}\left( {L/K}\right) \) .\n\nFinally, we must show that \( \pi \) is surjective. But if \( \overline{\mathfrak{a}} \) is an ideal class in \( C{l}_{i}\left( K\right) \), there exists \( \alpha \in {L}^{ * } \) such that \( \mathfrak{a}{\mathbb{Z}}_{L} = \alpha {\mathbb{Z}}_{L} \), and it is clear that \( \overline{\left( \mathfrak{a},1/\alpha \right) } \in {U}_{i}\left( {L/K}\right) \) and satisfies \( \pi \left( \overline{\left( \mathfrak{a},1/\alpha \right) }\right) = \overline{\mathfrak{a}} \), thus finishing the proof of (1).
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Yes
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Corollary 7.2.4. Assume that \( \\left( {n, h\\left( K\\right) }\\right) = 1 \) . Then the group \( {U}_{i}\\left( {L/K}\\right) \) is isomorphic to the quotient group \( U\\left( L\\right) /{i}_{L/K}\\left( {U\\left( K\\right) }\\right) \) .
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Proof. By Theorem 7.1.5 (1), when \( \\left( {n, h\\left( K\\right) }\\right) = 1 \) we have \( C{l}_{i}\\left( K\\right) = \\{ 1\\} \) , and so the corollary follows immediately from Proposition 7.2.3.
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Yes
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Corollary 7.2.5. There exists a six-term exact sequence \[ 1 \rightarrow U\left( K\right) \overset{{i}_{L/K}}{ \rightarrow }U\left( L\right) \overset{j}{ \rightarrow }{U}_{i}\left( {L/K}\right) \] \[ \rightarrow {Cl}\left( K\right) \overset{{i}_{L/K}}{ \rightarrow }{Cl}\left( L\right) \rightarrow C{l}_{i}\left( {L/K}\right) \rightarrow 1. \]
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Proof. This is a trivial consequence of exact sequences (1) and (4).
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No
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Proposition 7.2.6. Let \( A \) be an \( m \times n \) integer matrix of rank \( n \), hence with \( m \geq n \). There exist two unimodular matrices \( U \) and \( V \) of size \( n \times n \) and \( m \times m \), respectively, such that \[ B = {VAU} = \left( \begin{matrix} {d}_{1} & 0 & \ldots & 0 \\ 0 & {d}_{2} & \ddots & \vdots \\ \vdots & \ddots & \ddots & 0 \\ 0 & \ldots & 0 & {d}_{n} \\ 0 & \ldots & \ldots & 0 \\ \vdots & \vdots & \vdots & \vdots \\ 0 & \ldots & \ldots & 0 \end{matrix}\right) \] with \( {d}_{i - 1} \mid {d}_{i} \) for \( i > 1 \).
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Proof. As usual, denote by \( {A}^{t} \) the transpose of \( A \), which is an \( n \times m \) matrix of rank \( n \). By the theorem on the HNF ([Coh0, Theorem 2.4.4]), there exists a unimodular \( m \times m \) matrix \( {V}_{1} \) and an \( n \times n \) matrix \( H \) in HNF such that \( {A}^{t}{V}_{1} = \left( {0 \mid H}\right) \). Calling \( {V}_{2} \) the matrix obtained by putting the first \( m - n \) columns of \( {V}_{1} \) after the last \( n \) columns, the matrix \( {V}_{2} \) is still unimodular and \( {A}^{t}{V}_{2} = \left( {H \mid 0}\right) \), hence \( {V}_{2}^{t}A = \left( \frac{{H}^{t}}{0}\right) \). Let \( {V}_{3} \) and \( U \) be \( n \times n \) unimodular matrices such that \( {V}_{3}{H}^{t}U = D = \operatorname{diag}\left( {d}_{i}\right) \) is in SNF. If we set \[ \begin{array}{r} V = \left( \begin{matrix} {V}_{3} & 0 \\ 0 & {I}_{m - n} \end{matrix}\right) {V}_{2}^{t} \end{array}, \] then \( {VAU} = \left( \frac{D}{0}\right) = B \) satisfies the conditions of the proposition. Note of course that the above proof gives an algorithm to compute \( U, V \), and \( B \).
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Yes
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Corollary 7.2.8. Keep the notation of the above proposition, denote as usual by \( \mu \left( L\right) \) the group of roots of unity of \( L \), and set \( w\left( {L/K}\right) = w\left( L\right) /w\left( K\right) \) . Then:\n\n(1)\n\n\[ \frac{U\left( L\right) }{\mu \left( L\right) \cdot {i}_{L/K}\left( {U\left( K\right) }\right) } = {\bigoplus }_{1 \leq i \leq r\left( K\right) }\left( {\mathbb{Z}/{d}_{i}\mathbb{Z}}\right) \overline{{\eta }_{i}} \oplus {\bigoplus }_{r\left( K\right) < i \leq r\left( L\right) }\mathbb{Z}\overline{{\eta }_{i}}, \] \n\nwhere \( \bar{\eta } \) denotes the class of \( \eta \) modulo \( \mathbf{\mu }\left( L\right) \cdot {i}_{L/K}\left( {U\left( K\right) }\right) \) .
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Proof. Statement (1) follows immediately from Proposition 7.2.7. Note that because of the presence of the integers \( {e}_{i} \), we cannot give such a clean formula for \( U\left( L\right) /{i}_{L/K}\left( {U\left( K\right) }\right) \) .\n\nBy (1), \( \operatorname{diag}\left( {d}_{i}\right) \) is the Smith normal form of the torsion submodule of \( U\left( L\right) /\left( {\mathbf{\mu }\left( L\right) \cdot {i}_{L/K}\left( {U\left( K\right) }\right) }\right) \) . Hence the uniqueness of the \( {d}_{i} \) follows from the uniqueness of the SNF, proving (2).\n\nBy definition of \( {d}_{i} \) and \( {e}_{i} \), we have \( {\varepsilon }_{i} = {\zeta }^{{e}_{i}}{\eta }_{i}^{{d}_{i}} \), where the \( \left( {\varepsilon }_{i}\right) \) form a system of fundamental units of \( K \) . Taking the norm from \( L \) to \( K \), we obtain\n\n\[ {\varepsilon }_{i}^{n} = {\zeta }^{k{e}_{i}}{\mathcal{N}}_{L/K}{\left( {\eta }_{i}\right) }^{{d}_{i}} \] \n\nfor some integer \( k \) such that \( {\mathcal{N}}_{L/K}\left( \zeta \right) = {\zeta }^{k} \in K \) . Since the \( {\varepsilon }_{i} \) form a system of fundamental units and \( {\zeta }^{w\left( {L/K}\right) } \) generates \( \mathbf{\mu }\left( K\right) \), there exist integers \( {x}_{j} \) such that\n\n\[ {\mathcal{N}}_{L/K}\left( {\eta }_{i}\right) = {\zeta }^{{x}_{0}w\left( {L/K}\right) }\mathop{\prod }\limits_{j}{\varepsilon }_{j}^{{x}_{j}}. \] \n\nSince the \( {\varepsilon }_{j} \) are independent, if we replace in the formula for \( {\varepsilon }_{i}^{n} \), we obtain \( {x}_{j} = 0 \) for \( j \neq i \) and \( j \neq 0,{d}_{i}{x}_{i} = n \), and \( k{e}_{i} + {d}_{i}{x}_{0}w\left( {L/K}\right) \equiv 0{\;(\operatorname{mod}\;{w\left( L\right) })}. \) In particular, we see that \( {d}_{i} \mid n \), proving (3).\n\nSince \( {\mathcal{N}}_{L/K}\left( \zeta \right) \) is a root of unity of \( K \), there exists an integer \( m \) such that \( {\mathcal{N}}_{L/K}\left( \zeta \right) = {\zeta }^{w\left( {L/K}\right) m} \) . On the other hand, since \( {\zeta }^{w\left( {L/K}\right) } \in K \), we have \( {\mathcal{N}}_{L/K}\left( {\zeta }^{w\left( {L/K}\right) }\right) = {\zeta }^{w\left( {L/K}\right) n} \), hence\n\n\[ {\zeta }^{w\left( {L/K}\right) \left( {w\left( {L/K}\right) m - n}\right) } = 1 \] \n\nso \( w\left( L\right) \mid w\left( {L/K}\right) \left( {w\left( {L/K}\right) m - n}\right) \) or, equivalently, \( w\left( K\right) \mid w\left( {L/K}\right) m - n \) . It follows in particular that \( \left( {w\left( K\right), w\left( {L/K}\right) }\right) \mid n \), proving (4).
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Yes
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Proposition 7.2.11. (1) We have the following exact sequences:\n\n\[ 1 \rightarrow {\\operatorname{Cl}}_{N}\\left( K\\right) \rightarrow {U}_{N}\\left( {L/K}\\right) \rightarrow U\\left( L\\right) \\overset{{\\mathcal{N}}_{L/K}}{ \\rightarrow }\\frac{U\\left( K\\right) }{\\mu \\left( K\\right) } \rightarrow {U}_{N}\\left( K\\right) \rightarrow 1 \]\n\nand\n\n\[ 1 \rightarrow {\\operatorname{Cl}}_{N}\\left( {L/K}\\right) \rightarrow \\operatorname{Cl}\\left( L\\right) \\overset{{\\mathcal{N}}_{L/K}}{ \\rightarrow }\\operatorname{Cl}\\left( K\\right) \]\n\n\[ \\rightarrow {U}_{N}\\left( {L/K}\\right) \rightarrow U\\left( L\\right) \\overset{{\\mathcal{N}}_{L/K}}{ \\rightarrow }\\frac{U\\left( K\\right) }{\\mu \\left( K\\right) } \rightarrow {U}_{N}\\left( K\\right) \rightarrow 1. \]
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Proof. Note that the map from \( C{l}_{N}\\left( K\\right) \) to \( {U}_{N}\\left( {L/K}\\right) \) is the map sending \( \\overline{\\mathfrak{a}} \) to \( \\left( {\\overline{\\mathfrak{a}},1}\\right) \), and the map from \( {U}_{N}\\left( {L/K}\\right) \) to \( U\\left( L\\right) \) is the map sending \( \\left( {\\overline{\\mathfrak{a}},\\alpha }\\right) \) to \( \\alpha \), so the exact sequences of (1) immediately follow from the definitions. For (2), we note that as a quotient of \( U\\left( K\\right) ,{U}_{N}\\left( K\\right) \) is a finitely generated Abelian group, and if \( u \\in U\\left( K\\right) \) we have \( {\\mathcal{N}}_{L/K}\\left( {{i}_{L/K}\\left( u\\right) }\\right) = {u}^{n} \), hence \( {u}^{n} \\in \) \( {\\mathcal{N}}_{L/K}\\left( {U\\left( L\\right) }\\right) \), so the exponent of \( {U}_{N}\\left( K\\right) \) divides \( n \), and in particular since it is finitely generated, it is finite. (3) follows from the finiteness of \( C{l}_{N}\\left( K\\right) \) , \( \\mathbf{\\mu }\\left( K\\right) ,{U}_{N}\\left( K\\right) \), and the exact sequence (7).
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Yes
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Proposition 7.4.4. We have a canonical isomorphism\n\n\[ \nC{l}_{S}\left( K\right) \simeq {Cl}\left( K\right) / < \overline{{\mathfrak{p}}_{i}}{ > }_{{\mathfrak{p}}_{i} \in S}, \n\] \n\nwhere \( < \overline{{\mathfrak{p}}_{i}} > \) denotes the subgroup of \( {Cl}\left( K\right) \) generated by ideal classes of the prime ideals in \( S \) .
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Proof. Let \( \bar{I} \) be an ideal class in \( C{l}_{S}\left( K\right) \), and define \( f\left( \bar{I}\right) = \overline{I \cap {\mathbb{Z}}_{K}} \) in \( {Cl}\left( K\right) / < \overline{{\mathfrak{p}}_{i}} > \) . The map \( f \) is well-defined and is a group homomorphism. Assume that \( f\left( \bar{I}\right) = 1 \) . This means that \( I \cap {\mathbb{Z}}_{K} = \alpha \mathop{\prod }\limits_{i}{\mathfrak{p}}_{i}^{{x}_{i}} \) for some \( \alpha \in {K}^{ * } \) . Multiplying by \( {\mathbb{Z}}_{K, S} \) and using Proposition 7.4.2 and \( \mathfrak{p}{\mathbb{Z}}_{K, S} = {\mathbb{Z}}_{K, S} \) for \( \mathfrak{p} \in S \), we obtain \( I = \alpha {\mathbb{Z}}_{K, S} \), so \( \bar{I} \) is trivial, hence \( f \) is injective. Finally, if \( \overline{\mathfrak{a}} \) is some ideal class in \( {Cl}\left( K\right) / < \overline{{\mathfrak{p}}_{i}} > \), by Corollary 1.2.11 we can choose as representative an integral ideal \( \mathfrak{a} \) coprime to the product of all prime ideals of \( S \), and then by Proposition 7.4.2, we have \( f\left( \overline{\mathfrak{a}{\mathbb{Z}}_{K, S}}\right) = \overline{\mathfrak{a}} \), so \( f \) is surjective.
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Yes
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Corollary 7.4.5. There exists \( {S}_{1} \) such that for any \( S \supset {S}_{1} \), the ring \( {\mathbb{Z}}_{K, S} \) is a principal ideal domain.
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Proof. Let \( \left( \overline{{\mathfrak{a}}_{i}}\right) \) be generators of \( {Cl}\left( K\right) \), and let \( {S}_{1} \) be a set containing all Archimedean places and all prime ideals \( \mathfrak{p} \) such that \( {v}_{\mathfrak{p}}\left( {\mathfrak{a}}_{i}\right) \neq 0 \) for some \( i \) . This set is finite, and the proposition implies that if \( S \supset {S}_{1} \) then \( C{l}_{S}\left( K\right) \) is trivial, so that \( {\mathbb{Z}}_{K, S} \) is a principal ideal domain.
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Yes
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Proposition 7.4.7. Let \( {Cl}\left( K\right) = \left( {B,{D}_{B}}\right) \) with \( B = \left( \overline{{\mathfrak{b}}_{i}}\right) \) be the \( {SNF} \) of \( {Cl}\left( K\right) \), where \( {\mathfrak{b}}_{i} \) are ideals and \( {D}_{B} = \operatorname{diag}\left( {b}_{i}\right) \). Let\n\n\[ U\left( K\right) = \left( {\mathbb{Z}/w\left( K\right) \mathbb{Z}}\right) {\varepsilon }_{0} \oplus {\bigoplus }_{1 \leq i \leq r}\mathbb{Z}{\varepsilon }_{i} \]\n\nbe the unit group of \( K \) in SNF. Let \( {\beta }_{i} \in K \) be such that \( {\mathfrak{b}}_{i}^{{b}_{i}} = \left( {1/{\beta }_{i}}\right) {\mathbb{Z}}_{K} \). For each prime ideal \( {\mathfrak{p}}_{j} \in S \), write\n\n\[ {\mathfrak{p}}_{j} = {\alpha }_{j}\mathop{\prod }\limits_{i}{\mathfrak{b}}_{i}^{{p}_{i, j}} \]\n\nwith \( {\alpha }_{j} \in {K}^{ * } \) and \( {p}_{i, j} \in \mathbb{Z} \), and let \( P = \left( {p}_{i, j}\right) \). Finally, let \( U = \left( \begin{array}{ll} {U}_{1} & {U}_{2} \\ {U}_{3} & {U}_{4} \end{array}\right) \) be the unimodular matrix such that \( \left( {P \mid {D}_{B}}\right) U = \left( {0 \mid H}\right) \), where \( H \) is the HNF of \( \left( {P \mid {D}_{B}}\right) \). If\n\n\[ \left\lbrack {{\gamma }_{1},\ldots ,{\gamma }_{s}}\right\rbrack = \left\lbrack {{\alpha }_{1},\ldots ,{\alpha }_{s},{\beta }_{1},\ldots ,{\beta }_{m}}\right\rbrack \left( \frac{{U}_{1}}{{U}_{3}}\right) \]\n\nin the usual multiplicative sense used in Chapter 4, then\n\n\[ {U}_{S}\left( K\right) = \left( {\mathbb{Z}/w\left( K\right) \mathbb{Z}}\right) {\varepsilon }_{0} \oplus {\bigoplus }_{1 \leq i \leq r}\mathbb{Z}{\varepsilon }_{i} \oplus {\bigoplus }_{1 \leq i \leq s}\mathbb{Z}{\gamma }_{i}. \]\n\nIn particular, the torsion subgroup of \( {U}_{S}\left( K\right) \) is equal to that of \( U\left( K\right) \), and the rank of \( {U}_{S}\left( K\right) \) is equal to \( r + s = {r}_{1} + {r}_{2} - 1 + \left| {S}_{0}\right| \), where \( {S}_{0} \) denotes the set of prime ideals belonging to \( S \) (this can also be written \( \left| S\right| - 1 \), since there are \( {r}_{1} + {r}_{2} \) Archimedean places in \( S \) ).
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Proof. Let \( {S}_{0} \) be the row vector of the prime ideals belonging to \( S \) and let \( {B}^{\prime } = \left\lbrack {{\mathfrak{b}}_{1},\ldots ,{\mathfrak{b}}_{m}}\right\rbrack \) be the row vector of the ideals \( {\mathfrak{b}}_{i} \). We use again matrix notation as we did in Chapter 4. I first claim that\n\n\[ {S}_{0}{U}_{1} = \left\lbrack {{\gamma }_{1}{\mathbb{Z}}_{K},\ldots ,{\gamma }_{m}{\mathbb{Z}}_{K}}\right\rbrack \]\n\nIndeed, by definition, we have\n\n\[ {B}^{\prime }\left( {P \mid {D}_{B}}\right) = \left( {\left( {{\alpha }_{j}^{-1}{\mathfrak{p}}_{j}}\right) \mid \left( {{\beta }_{j}^{-1}{\mathbb{Z}}_{K}}\right) }\right) .\n\nHence, multiplying on the right by \( \left( \frac{{U}_{1}}{{U}_{3}}\right) \), we obtain\n\n\[ \left\lbrack {{\mathbb{Z}}_{K},\ldots ,{\mathbb{Z}}_{K}}\right\rbrack = \left( {\left( {\alpha }_{j}^{-1}\right) \mid \left( {\beta }_{j}^{-1}\right) }\right) \left( \frac{{U}_{1}}{{U}_{3}}\right) \cdot {S}_{0}{U}_{1} = \left( {\gamma }_{j}^{-1}\right) {S}_{0}{U}_{1}, \]\n\nproving my claim.\n\nNow,\n\n\[ \alpha \in {U}_{S}\left( K\right) \Leftrightarrow \forall \mathfrak{p} \notin S,{v}_{\mathfrak{p}}\left( \alpha \right) = 0 \Leftrightarrow \exists X \in {\mathbb{Z}}^{{S}_{0}},\alpha {\mathbb{Z}}_{K} = {S}_{0}X. \]\n\nTaking ideal classes, we see that \( \overline{{S}_{0}}X = \mathbf{1} \) in \( {Cl}\left( K\right) \). But, by definition,\n\n\[ \overline{{S}_{0}}X = 1 \Leftrightarrow {BPX} = 1 \Leftrightarrow \exists Y \in {\mathbb{Z}}^{m},{PX} = {D}_{B}Y \]\n\n\[ \Leftrightarrow \left( \frac{X}{-Y}\right) \in \operatorname{Ker}\left( {P \mid {D}_{B}}\right) \]\n\nwhere of course Ker denotes the integer kernel. If \( U = \left( \begin{array}{ll} {U}_{1} & {U}_{2} \\ {U}_{3} & {U}_{4} \end{array}\right) \) is the unimodular matrix such that \( \left( {P \mid {D}_{B}}\right) U = \left( {0 \mid H}\right) \), where \( H \) is the
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No
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Proposition 7.5.1. The exponent of the quotient group\n\n\[ \n\\left( {{\\mathcal{N}}_{L/K}\\left( {L}^{ * }\\right) \\cap {U}_{S}\\left( K\\right) }\\right) /{\\mathcal{N}}_{L/K}\\left( {{U}_{S}\\left( L\\right) }\\right) \n\]\n\ndivides \( \\left\\lbrack {L : K}\\right\\rbrack \) .
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Proof. Indeed, if \( a \\in {U}_{S}\\left( K\\right) \\subset {K}^{ * } \), then\n\n\[ \n{a}^{\\left\\lbrack L : K\\right\\rbrack } = {\\mathcal{N}}_{L/K}\\left( a\\right) \\in {\\mathcal{N}}_{L/K}\\left( {{U}_{S}\\left( L\\right) }\\right) \n\]
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Yes
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Proposition 7.5.3. Assume that \( {S}_{0} \) is suitable for the extension \( L/K \), let \( a \in {K}^{ * } \), and call \( {S}_{a} \) the set of prime ideals \( \mathfrak{p} \) of \( K \) such that \( {v}_{\mathfrak{p}}\left( a\right) \neq 0 \) . Then the equation \( {\mathcal{N}}_{L/K}\left( x\right) = a \) is soluble with \( x \in L \) if and only if it is soluble with \( x \in {U}_{L,{S}_{0} \cup {S}_{a}} \) .
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Proof. Indeed, if \( S = {S}_{0} \cup {S}_{a} \), then \( a \in {U}_{S}\left( K\right) \) by definition, and since \( {S}_{0} \) is suitable and \( S \supset {S}_{0} \), if the equation \( {\mathcal{N}}_{L/K}\left( x\right) = a \) is soluble, we have \( a \in {\mathcal{N}}_{L/K}\left( {{U}_{S}\left( L\right) }\right) .
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Yes
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Theorem 7.5.4. Let \( L/K \) be a Galois extension, and let \( {S}_{0} \) be a set of prime ideals of \( K \) such that \( C{l}_{i}\left( {L/K}\right) \) can be generated by the classes of ideals divisible only by prime ideals of \( L \) above the ideals of \( {S}_{0} \) . Then \( {S}_{0} \) is suitable for the extension \( L/K \) in the sense of Definition 7.5.2: in other words, for all \( S \supset {S}_{0} \), we have \( {\mathcal{N}}_{L/K}\left( {{U}_{S}\left( L\right) }\right) = {\mathcal{N}}_{L/K}\left( {L}^{ * }\right) \cap {U}_{S}\left( K\right) \) . In addition, we also have \( {\mathcal{N}}_{L/K}\left( {\mathbb{Z}}_{L, S}\right) = {\mathcal{N}}_{L/K}\left( {L}^{ * }\right) \cap {\mathbb{Z}}_{K, S} \) .
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Proof of Theorem 7.5.4. The inclusions \( {\mathcal{N}}_{L/K}\left( {{U}_{S}\left( L\right) }\right) \subset {\mathcal{N}}_{L/K}\left( {L}^{ * }\right) \cap \) \( {U}_{S}\left( K\right) \) and \( {\mathcal{N}}_{L/K}\left( {\mathbb{Z}}_{L, S}\right) \subset {\mathcal{N}}_{L/K}\left( {L}^{ * }\right) \cap {\mathbb{Z}}_{K, S} \) are trivial. Conversely, let \( a \in {\mathcal{N}}_{L/K}\left( {L}^{ * }\right) \cap {\mathbb{Z}}_{K, S} \), and let \( x, y \) in \( L \) be such that \( {\mathcal{N}}_{L/K}\left( {x/y}\right) = a \) . We may, of course, assume that \( x \) and \( y \) are in \( {\mathbb{Z}}_{K, S} \) (in fact, in \( {\mathbb{Z}}_{K} \) if desired). By definition, we have
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Yes
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Lemma 7.5.5. Let \( L/K \) be a Galois extension of number fields, let \( S \) be a finite set of prime ideals of \( K \), and let \( I \) and \( J \) be \( S \)-integral ideals of \( L \) such that \( {\mathcal{N}}_{L/K}\left( I\right) = \mathfrak{a}{\mathcal{N}}_{L/K}\left( J\right) \) for some \( S \)-integral ideal \( \mathfrak{a} \) of \( K \). Assume that for \( 1 \leq i \leq k \) there exist prime ideals \( {\mathfrak{P}}_{i} \) of \( L \) such that \( \mathop{\prod }\limits_{{1 \leq i \leq k}}{\mathfrak{P}}_{i} \mid J \) as ideals of \( {\mathbb{Z}}_{L, S} \). Then for each \( i \leq k \) there exists \( {\sigma }_{i} \in \operatorname{Gal}\left( {L/K}\right) \) such that \( \mathop{\prod }\limits_{{1 \leq i \leq k}}{\sigma }_{i}\left( {\mathfrak{P}}_{i}\right) \mid I \) as ideals of \( {\mathbb{Z}}_{L, S} \).
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Proof. We prove the lemma by induction on \( k \), the case \( k = 0 \) being trivial. Assume first \( k = 1 \), and let \( \mathfrak{P} \) be a prime ideal dividing \( J \). Since \( {\mathcal{N}}_{L/K}\left( \mathfrak{P}\right) \) divides \( {\mathcal{N}}_{L/K}\left( J\right) \), it also divides \( {\mathcal{N}}_{L/K}\left( I\right) \) since \( \mathfrak{a} \) is \( S \)-integral. Since the extension is Galois, by the remark made at the end of Section 2.2.5, we have for any ideal \( I,{\mathcal{N}}_{L/K}\left( I\right) = \mathop{\prod }\limits_{{1 \leq i \leq n}}{\sigma }_{i}\left( I\right) \); hence, in particular, there exists \( \sigma \in \operatorname{Gal}\left( {L/K}\right) \) such that \( \mathfrak{P} \mid \sigma \left( I\right) \), and hence \( {\sigma }^{-1}\left( \mathfrak{P}\right) \mid I \), as claimed.\n\nAssume now that the lemma is true for some \( k \geq 1 \), and assume that \( \mathop{\prod }\limits_{{1 \leq i \leq k + 1}}{\mathfrak{P}}_{i} \) divides \( J \). In particular, the product of the first \( k \) primes divides \( J \); hence by our induction hypothesis there exist \( {\sigma }_{i} \in \operatorname{Gal}\left( {L/K}\right) \) such that \( \mathop{\prod }\limits_{{1 \leq i \leq k}}{\sigma }_{i}\left( {\mathfrak{P}}_{i}\right) \) divides \( I \). It follows that\n\n\[{\mathcal{N}}_{L/K}\left( {I/\mathop{\prod }\limits_{{1 \leq i \leq k}}{\sigma }_{i}\left( {\mathfrak{P}}_{i}\right) }\right) = \mathfrak{a}{\mathcal{N}}_{L/K}\left( {J/\mathop{\prod }\limits_{{1 \leq i \leq k}}{\mathfrak{P}}_{i}}\right) .\n\]\n\nWe conclude by applying the case \( k = 1 \) proved above.
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Yes
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Proposition 7.5.6. Let \( L/K \) be a Galois extension, and let \( r \) be an integer such that \( \operatorname{Gal}\left( {L/K}\right) \) can be generated by \( r \) elements. In addition, for any finite set \( S \) of prime ideals of \( K \), denote by \( C{l}_{i, S}\left( {L/K}\right) \) the quotient of \( C{l}_{i}\left( {L/K}\right) \) by the group generated by the classes of ideals divisible only by prime ideals of \( L \) above the ideals of \( S \) . Then for all \( S \) the quotient group \( \left( {{\mathcal{N}}_{L/K}\left( {L}^{ * }\right) \cap {U}_{S}\left( K\right) }\right) /{\mathcal{N}}_{L/K}\left( {{U}_{S}\left( L\right) }\right) \) is a subquotient (in other words, a quotient of a subgroup) of \( C{l}_{i, S}{\left( L/K\right) }^{r} \) .
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I refer to [Sim1] for the proof.
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No
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Corollary 7.5.7. Let \( L/K \) be a Galois extension, and let \( {S}_{0} \) be a set of prime ideals such that \( \left| {C{l}_{i,{S}_{0}}\left( {L/K}\right) }\right| \) is coprime to \( \left\lbrack {L : K}\right\rbrack \) . Then \( {S}_{0} \) is suitable for the extension \( L/K \) .
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Proof. This follows immediately from the above proposition and Proposition 7.5.1. This corollary is a slight strengthening of part of Theorem 7.5.4, which asserts only that \( {S}_{0} \) is suitable if \( \left| {C{l}_{i,{S}_{0}}\left( {L/K}\right) }\right| = 1 \) .
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Yes
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Proposition 7.5.8. Let \( S \) be a finite set of primes of \( K \) . The exponent of \( \left( {{\mathcal{N}}_{L/K}\left( {L}^{ * }\right) \cap {U}_{S}\left( K\right) }\right) /{\mathcal{N}}_{L/K}\left( {{U}_{S}\left( L\right) }\right) \) divides the GCD \( \left( {n,\left| H\right| \cdot \left| {C{l}_{i, S}\left( {N/K}\right) }\right| }\right) \) .
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Proof. By Proposition 7.5.1, we already know that this exponent divides \( n \) . Set \( h = \left| {C{l}_{i, S}\left( {N/K}\right) }\right| \), and let \( a \in {\mathcal{N}}_{L/K}\left( {L}^{ * }\right) \cap {U}_{S}\left( K\right) \), so that we may write \( a = {\mathcal{N}}_{L/K}\left( x\right) \) for some \( x \in {L}^{ * } \) . We have\n\n\[ {a}^{\left| H\right| } = {\mathcal{N}}_{L/K}\left( {x}^{\left| H\right| }\right) = {\mathcal{N}}_{L/K}\left( {{\mathcal{N}}_{N/L}\left( x\right) }\right) = {\mathcal{N}}_{N/K}\left( x\right) ; \]\n\nhence \( {a}^{\left| H\right| } \in {\mathcal{N}}_{N/K}\left( {N}^{ * }\right) \cap {U}_{K, S} \) . Applying Proposition 7.5.6 to the Galois extension \( N/K \), we deduce in particular that the exponent of the quotient group \( \left( {{\mathcal{N}}_{N/K}\left( {N}^{ * }\right) \cap {U}_{K, S}}\right) /{\mathcal{N}}_{N/K}\left( {U}_{N, S}\right) \) divides \( h \) ; hence there exists \( s \in \) \( {U}_{N, S} \) such that\n\n\[ {a}^{h\left| H\right| } = {\mathcal{N}}_{N/K}\left( s\right) = {\mathcal{N}}_{L/K}\left( {{\mathcal{N}}_{N/L}\left( s\right) }\right) \]\n\nso \( {a}^{h\left| H\right| } \in {\mathcal{N}}_{L/K}\left( {{U}_{S}\left( L\right) }\right) \), as claimed.
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Yes
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Theorem 7.5.10. Keep the above notation, and let \( {S}_{0} \) be a finite set of primes of \( K \) containing all the prime ideals of \( K \) ramified in \( L/K \) . Assume that \( \left| {C{l}_{i,{S}_{0}}\left( {N/K}\right) }\right| \) is coprime to \( n \) and that for all cyclic subgroups \( C \) of \( G = \operatorname{Gal}\left( {N/K}\right) \) of prime power order \( {p}^{a} \) with \( p\left| {\left( {n,\left| H\right| }\right) ,\left| {C{l}_{i,{S}_{0}}\left( {{N}^{C}/K}\right) }\right| }\right| \) is coprime to \( \left( {n,\left| H\right| }\right) \) . Then \( {S}_{0} \) is suitable for \( L/K \) .
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We refer to [Sim1] for the (quite technical) proof. Please note the condition that \( {S}_{0} \) must contain all the ramified prime ideals, which did not occur in the previous results. It is not difficult to give examples showing that this condition (or a similar one) is necessary (see [Sim1]). Note also that, at least in the known proof, it is necessary to consider the cyclic subgroups of \( G \), and not only of \( H \) .
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No
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Proposition 8.1.1. Let \( \gamma = \left( \begin{array}{ll} A & B \\ C & D \end{array}\right) \) . Then\n\n\[ \operatorname{disc}\left( {F \circ \gamma }\right) = {\left( AD - BC\right) }^{n\left( {n - 1}\right) }\operatorname{disc}\left( F\right) . \]
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Proof. Let \( \left( {{\alpha }_{i} : {\beta }_{i}}\right) \) be the roots of \( F \) chosen as above so that\n\n\[ F\left( {X, Y}\right) = \mathop{\prod }\limits_{{1 \leq i \leq n}}\left( {{\beta }_{i}x - {\alpha }_{i}y}\right) . \]\n\nThen\n\n\[ F \circ \gamma \left( {X, Y}\right) = \mathop{\prod }\limits_{{1 \leq i \leq n}}\left( {{\beta }_{i}\left( {{Ax} + {By}}\right) - {\alpha }_{i}\left( {{Cx} + {Dy}}\right) }\right) = \mathop{\prod }\limits_{{1 \leq i \leq n}}\left( {{\beta }_{i}^{\prime }x - {\alpha }_{i}^{\prime }y}\right) ,\]\n\nwith\n\n\[ \left( \begin{matrix} {\alpha }_{i}^{\prime } \\ {\beta }_{i}^{\prime } \end{matrix}\right) = \left( \begin{matrix} D & - B \\ - C & A \end{matrix}\right) \left( \begin{matrix} {\alpha }_{i} \\ {\beta }_{i} \end{matrix}\right) . \]\n\nHence,\n\n\[ \operatorname{disc}\left( {F \circ \gamma }\right) = \mathop{\prod }\limits_{{1 \leq i < j \leq n}}{\left( {\alpha }_{i}^{\prime }{\beta }_{j}^{\prime } - {\alpha }_{j}^{\prime }{\beta }_{i}^{\prime }\right) }^{2} \]\n\n\[ = \mathop{\prod }\limits_{{1 \leq i < j \leq n}}{\left( \left( AD - BC\right) \left( {\alpha }_{i}{\beta }_{j} - {\alpha }_{j}{\beta }_{i}\right) \right) }^{2} \]\n\n\[ = {\left( AD - BC\right) }^{n\left( {n - 1}\right) }\operatorname{disc}\left( F\right) . \]
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Yes
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Proposition 8.1.3. Let \( {f}_{1} \) and \( {f}_{2} \) be two covariants on \( {\Phi }_{n} \) of degree \( {d}_{1} \) , \( {d}_{2} \), weight \( {w}_{1},{w}_{2} \), and with values in \( {\Phi }_{{m}_{1}} \) and \( {\Phi }_{{m}_{2}} \), respectively. For any nonnegative integer \( h \) and \( F \in {\Phi }_{n} \), set\n\n\[ \n{\phi }_{h}\left( {{f}_{1},{f}_{2}}\right) \left( F\right) = \mathop{\sum }\limits_{{j = 0}}^{h}{\left( -1\right) }^{j}\left( \begin{array}{l} h \\ j \end{array}\right) \frac{{\partial }^{h}}{\partial {X}^{h - j}\partial {Y}^{j}}{f}_{1}\left( F\right) \frac{{\partial }^{h}}{\partial {X}^{j}\partial {Y}^{h - j}}{f}_{2}\left( F\right) .\n\]\n\nThen \( {\phi }_{h}\left( {{f}_{1},{f}_{2}}\right) \) is a covariant on \( {\Phi }_{n} \) of degree \( {d}_{1} + {d}_{2} \), weight \( {w}_{1} + {w}_{2} + h \) , with values in \( {\Phi }_{{m}_{1} + {m}_{2} - {2h}} \) .
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Proof. To simplify notation, write \( {\partial }_{X} \) for \( \partial /\partial X \) and \( {\partial }_{Y} \) for \( \partial /\partial Y \) . The only operators that occur are \( {\partial }_{X} \) and \( {\partial }_{Y} \), which commute, and the multiplication operator, which does not commute with \( {\partial }_{X} \) or \( {\partial }_{Y} \) . Let \( \gamma = \left( \begin{array}{ll} A & B \\ C & D \end{array}\right) \in \) \( {\mathrm{{GL}}}_{2}\left( K\right) \) and set \( G = {\phi }_{h}\left( {{f}_{1},{f}_{2}}\right) \left( {F \circ \gamma }\right) \) . Then\n\n\[ \nG = \mathop{\sum }\limits_{{j = 0}}^{h}{\left( -1\right) }^{j}\left( \begin{array}{l} h \\ j \end{array}\right) {\left( A{\partial }_{X} + C{\partial }_{Y}\right) }^{h - j}{\left( B{\partial }_{X} + D{\partial }_{Y}\right) }^{j}{f}_{1}\left( {F \circ \gamma }\right)\n\]\n\n\[ \n\; \cdot {\left( A{\partial }_{X} + C{\partial }_{Y}\right) }^{j}{\left( B{\partial }_{X} + D{\partial }_{Y}\right) }^{h - j}{f}_{2}\left( {F \circ \gamma }\right)\n\]\n\n\[ \n= {\left( AD - BC\right) }^{{w}_{1} + {w}_{2}}( - \left( {B{\partial }_{X} + D{\partial }_{Y}}\right) {f}_{1}\left( F\right) \circ \gamma \left( {A{\partial }_{X} + C{\partial }_{Y}}\right) {f}_{2}\left( F\right)\n\]\n\n\[ \n\; + {\left( A{\partial }_{X} + C{\partial }_{Y}){f}_{1}\left( F\right) \left( B{\partial }_{X} + D{\partial }_{Y}\right) {f}_{2}\left( F\right) \circ \gamma \right) }^{h}\n\]\n\n\[ \n= {\left( AD - BC\right) }^{{w}_{1} + {w}_{2} + h}{\left( {\partial }_{X}{f}_{1}\left( F\right) {\partial }_{Y}{f}_{2}\left( F\right) - {\partial }_{Y}{f}_{1}\left( F\right) {\partial }_{X}{f}_{2}\left( F\right) \right) }^{h} \circ \gamma\n\]\n\n\[ \n= {\left( AD - BC\right) }^{{w}_{1} + {w}_{2} + h}{\phi }_{h}\left( {{f}_{1},{f}_{2}}\right) \left( F\right) \circ \gamma .\n\]\n\nIn addition, it is clear that each term of \( {\phi }_{h}\left( {{f}_{1},{f}_{2}}\right) \left( F\right) \) is in \( {\Phi }_{{m}_{1} + {m}_{2} - {2h}} \) and is of degree \( {d}_{1} + {d}_{2} \) in the coefficients of the form, proving the proposition.
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Yes
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Proposition 8.1.5. Let \( F \) be an integral form and \( \gamma \in {\mathrm{{GL}}}_{2}\left( \mathbb{Z}\right) \) . Then \( F \circ \gamma \) is irreducible if and only if \( F \) is irreducible, and \( F \circ \gamma \) is primitive if and only \( F \) is primitive.
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Proof. This immediately follows from the fact that the action of \( {\mathrm{{GL}}}_{2}\left( \mathbb{Z}\right) \) is reversible, that is, \( F = \left( {F \circ \gamma }\right) \circ {\gamma }^{-1} \) .
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Yes
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Proposition 8.2.1. Let \( \mathcal{B} = \left( {1,\alpha ,\beta }\right) \) be an integral basis of a cubic number field \( K \) as above. For \( x \) and \( y \) elements of \( \mathbb{Q} \), set\n\n\[ \n{F}_{\mathcal{B}}\left( {x, y}\right) = \frac{\mathop{\prod }\limits_{{1 \leq i < j \leq 3}}\left( {\left( {{\beta }_{i} - {\beta }_{j}}\right) x - \left( {{\alpha }_{i} - {\alpha }_{j}}\right) y}\right) }{\mathop{\sum }\limits_{{1 \leq i < j \leq 3}}{\left( -1\right) }^{i - j}\left( {{\alpha }_{i}{\beta }_{j} - {\alpha }_{j}{\beta }_{i}}\right) }.\n\]\n\n(1) For \( x \) and \( y \) in \( \mathbb{Q} \), we have\n\n\[ \n{F}_{\mathcal{B}}\left( {x, y}\right) = \pm \sqrt{\frac{\operatorname{disc}\left( {{\beta x} - {\alpha y}}\right) }{d\left( K\right) }} = \pm \frac{{\mathcal{N}}_{K/\mathbb{Q}}\left( {\left( {\beta - {\beta }^{\prime }}\right) x - \left( {\alpha - {\alpha }^{\prime }}\right) y}\right) }{\sqrt{d\left( K\right) }}.\n\]
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Proof. (1) follows directly from the definitions, after noting that \( d\left( K\right) \) is the square of the determinant of the matrix\n\n\[ \n\left( \begin{matrix} 1 & {\alpha }_{1} & {\beta }_{1} \\ 1 & {\alpha }_{2} & {\beta }_{2} \\ 1 & {\alpha }_{3} & {\beta }_{3} \end{matrix}\right) \;. \n\]\n\nIf we take one of the expressions on the right-hand side of the equalities in (1) as the definition of \( {F}_{\mathcal{B}} \), there is a sign ambiguity. However, this will not matter in the sequel since \( {F}_{\mathcal{B}}\left( {-x, - y}\right) = - {F}_{\mathcal{B}}\left( {x, y}\right) \) so that \( {F}_{\mathcal{B}} \) and \( - {F}_{\mathcal{B}} \) are \( {\mathrm{{SL}}}_{2}\left( \mathbb{Z}\right) \) -equivalent.
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Yes
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Proposition 8.2.2. We have \( {\phi }_{\Phi \mathcal{C}} \circ {\phi }_{\mathcal{C}\Phi } = 1 \) ; hence \( {\phi }_{\mathcal{C}\Phi } \) is injective and \( {\phi }_{\Phi \mathcal{C}} \) is surjective.
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Proof. Let \( K \) be a cubic field and let \( \left( {1,\alpha ,\beta }\right) \) be an integral basis. We have (for example) \( {K}_{{F}_{K}} = \mathbb{Q}\left( {\left( {{\alpha }_{2} - {\alpha }_{3}}\right) /\left( {{\beta }_{2} - {\beta }_{3}}\right) }\right) \subset {K}^{g} \) . If \( K \) is a cyclic cubic field, then \( {K}_{{F}_{K}} = K \) . Otherwise, \( {K}^{g} \) is a number field of degree 6, with Galois group isomorphic to \( {S}_{3} \), and \( \left( {{\alpha }_{2} - {\alpha }_{3}}\right) /\left( {{\beta }_{2} - {\beta }_{3}}\right) \) is fixed by the transposition (23) of order 2, hence belongs to a cubic subfield, and so \( {K}_{{F}_{K}} \) is isomorphic to \( K \) . In fact, by choosing the numbering such that \( \alpha = {\alpha }_{1} \) and \( \beta = {\beta }_{1},{K}_{{F}_{K}} \) is even equal to \( K \), not only conjugate to it. Note also that we cannot have \( \left( {{\alpha }_{2} - {\alpha }_{3}}\right) /\left( {{\beta }_{2} - {\beta }_{3}}\right) \in \mathbb{Q} \), because \( {F}_{K} \) would then be reducible.
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No
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(1) \( p \mid \operatorname{disc}\left( F\right) \) if and only if \( F \) has at least a double root in \( \overline{{\mathbb{F}}_{p}} \), and if this is the case, all the roots of \( F \) are in fact in \( {\mathbb{F}}_{p} \) itself.
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Proof. (1). Assume that \( p \mid \operatorname{disc}\left( F\right) \) . We know that any nonzero polynomial in one variable over \( {\mathbb{F}}_{p} \) can be written as \( \mathop{\prod }\limits_{{i > 1}}{A}_{i}^{i} \), where \( {A}_{i} \in {\mathbb{F}}_{p}\left\lbrack X\right\rbrack \) are pairwise coprime and squarefree polynomials, and essentially in a unique manner (up to multiplication of each \( {A}_{i} \) by suitable constants). This result can be homogenized and transformed into an identical one for homogeneous polynomials in two variables.\n\nSince \( F \) is primitive, it is nonzero modulo \( p \) . Since \( p \mid \operatorname{disc}\left( F\right) \), by definition of the discriminant this means that \( F \) has at least a double root in \( {\mathbb{P}}_{1}\left( \overline{{\mathbb{F}}_{p}}\right) \) . In other words, in the decomposition above there exists \( i > 1 \) such that \( {A}_{i} \) is not equal to a constant. Since \( F \) is of degree 3, this means that \( F = {A}_{3}^{3} \) or else \( F = {A}_{1}{A}_{2}^{2} \) with \( {A}_{1},{A}_{2} \), and \( {A}_{3} \) of degree 1 . It follows in particular that all the roots of \( F \) modulo \( p \) are in \( {\mathbb{F}}_{p} \) itself.
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Yes
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Corollary 8.3.4. Let \( F = \left( {a, b, c, d}\right) \) be a primitive form, and let \( {H}_{F} = \) \( \left( {P, Q, R}\right) \) be its Hessian. Then \( F \notin {U}_{2} \) if and only if \( \operatorname{disc}\left( F\right) \equiv 0\left( {\;\operatorname{mod}\;{16}}\right) \) or \( \operatorname{disc}\left( F\right) \equiv 4\left( {\;\operatorname{mod}\;{16}}\right) \) and \( P \) or \( R \) is odd.
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Proof. The proof is trivial and is left to the reader (Exercise 7).
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No
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Corollary 8.3.6. Let \( F \) be a primitive cubic form and \( p \) be a prime. Then \( F \notin {U}_{p} \) if and only if \( F \) has at least a double root \( \left( {\gamma : \delta }\right) \) modulo \( p \), and \( F\left( {\gamma ,\delta }\right) \equiv 0\left( {\;\operatorname{mod}\;{p}^{2}}\right) \) .
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Proof. If \( F \notin {U}_{p} \), then in particular \( F \notin {V}_{p} \), hence \( p \mid f \), and so by Proposition 8.3.5, either \( F\left( {\gamma ,\delta }\right) \equiv 0\left( {\;\operatorname{mod}\;{p}^{2}}\right) \) if \( F \) has a double root, or \( \left( {F, p}\right) = \left( {1}^{3}\right) \) , but then by definition of \( {U}_{p} \), we again have \( F\left( {\gamma ,\delta }\right) \equiv 0\left( {\;\operatorname{mod}\;{p}^{2}}\right) \) . Conversely, if \( F \) has at least a double root \( \left( {\gamma : \delta }\right) \) modulo \( p \), and \( F\left( {\gamma ,\delta }\right) \equiv 0\left( {\;\operatorname{mod}\;{p}^{2}}\right) \) , then if it is a triple root, by definition \( F \notin {U}_{p} \) . If it is only a double root, by Proposition 8.3.5 we have \( p \mid f \), so \( F \notin {V}_{p} \), and hence \( F \notin {U}_{p} \) . When \( \left( {\gamma ,\delta }\right) \) is at least a double root modulo \( p \), it is easily checked that the condition \( F\left( {\gamma ,\delta }\right) \equiv 0\left( {\;\operatorname{mod}\;{p}^{2}}\right) \) depends only on \( \gamma \) and \( \delta \) modulo \( p \) .
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Yes
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Lemma 8.4.3. Assume that \( F \) is a primitive cubic form and \( p \) a prime such that \( \left( {F, p}\right) = \left( {1}^{3}\right) \) . Then \( F \in {U}_{p} \) if and only if there exists \( \left( {u, v}\right) \in {\mathbb{Z}}^{2} \) such that \( F\left( {u, v}\right) = {ep} \) with \( p \nmid e \) .
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Let us prove the lemma. By lifting the condition \( \left( {F, p}\right) = \left( {1}^{3}\right) \) to \( \mathbb{Z} \), we can write\n\n\[ F\left( {x, y}\right) = \lambda {\left( \delta x - \gamma y\right) }^{3} + {pG}\left( {x, y}\right) ,\]\n\nwith \( \lambda ,\gamma ,\delta \) in \( \mathbb{Z} \) and \( G \) an integral cubic form. Then \( F \in {U}_{p} \) if and only if \( p \nmid G\left( {\gamma ,\delta }\right) \) . Thus if \( F \in {U}_{p} \), we have \( F\left( {\gamma ,\delta }\right) = {ep} \) with \( e = G\left( {\gamma ,\delta }\right) ≢ 0 \) \( \left( {\;\operatorname{mod}\;p}\right) \) .\n\nConversely, assume that there exists \( \left( {u, v}\right) \in {\mathbb{Z}}^{2} \) such that \( F\left( {u, v}\right) = {ep} \) with \( p \nmid e \) . Then \( F\left( {u, v}\right) \equiv 0\left( {\;\operatorname{mod}\;p}\right) \), hence \( \lambda \left( {{\delta u} - {\gamma v}}\right) \equiv 0\left( {\;\operatorname{mod}\;p}\right) \) . Since \( F \) is primitive, \( p \nmid \lambda \) and hence \( {\delta u} - {\gamma v} \equiv 0\left( {\;\operatorname{mod}\;p}\right) \) . Again, since \( F \) is primitive, \( \gamma \) and \( \delta \) cannot both be divisible by \( p \), from which it follows that there exists \( \mu \) such that \( \gamma \equiv {\mu u}\left( {\;\operatorname{mod}\;p}\right) \) and \( \delta \equiv {\mu v}\left( {\;\operatorname{mod}\;p}\right) \) . Also, since \( p \nmid e \), we have \( p \nmid \mu \) . But then \( {ep} = F\left( {u, v}\right) \equiv {pG}\left( {u, v}\right) \left( {\;\operatorname{mod}\;{p}^{3}}\right) \), hence \( G\left( {u, v}\right) \equiv e \) \( \left( {\;\operatorname{mod}\;{p}^{2}}\right) \), and so \( G\left( {\gamma ,\delta }\right) \equiv {\mu }^{3}G\left( {u, v}\right) \equiv {\mu }^{3}e ≢ 0\left( {\;\operatorname{mod}\;p}\right) \), so \( F \in {U}_{p} \), proving the lemma.
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Yes
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Lemma 8.4.5. Let \( F \) be any form in \( \Phi \) (in other words, primitive and irreducible). Then there exists a number field \( K \) such that \( F \) is rationally equivalent to \( {F}_{K} \) .
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Proof. In some algebraic closure of \( \mathbb{Q} \), write \( F = a\left( {x - {\lambda y}}\right) \left( {x - {\lambda }^{\prime }y}\right) (x - \) \( \left. {{\lambda }^{\prime \prime }y}\right) \) . Since \( F \) is irreducible, \( \lambda \) is a cubic irrationality, and we will take \( K = \) \( \mathbb{Q}\left( \lambda \right) \) . Write \( {F}_{K} = {a}_{K}\left( {x - {\nu y}}\right) \left( {x - {\nu }^{\prime }y}\right) \left( {x - {\nu }^{\prime \prime }y}\right) \) so that \( \nu \in K \) (we saw above that this is always possible). Since \( K \) is a \( \mathbb{Q} \) -vector space of dimension 3, there exist four integers \( k, l, m \), and \( n \) not all zero such that \( l + {k\lambda } - {n\nu } - {m\lambda \nu } = 0 \) . Taking conjugates, we obtain the same equality with \( {}^{\prime } \) and \( {}^{\prime \prime } \) . Using \( \lambda = \) \( \left( {{\nu n} - l}\right) /\left( {k - {\nu m}}\right) \), we obtain\n\n\[ \n{F}_{K}\left( {{kx} + {ly},{mx} + {ny}}\right) = \rho \left( {x - {\lambda y}}\right) \left( {x - {\lambda }^{\prime }y}\right) \left( {x - {\lambda }^{\prime \prime }y}\right) = \left( {\rho /a}\right) F \n\]\n\nwith \( \rho = {a}_{K}\mathcal{N}\left( {k - {\nu m}}\right) \in \mathbb{Q} \) . Furthermore, the determinant \( {kn} - {lm} \) is nonzero since otherwise either \( \lambda \) or \( \nu \) would be in \( \mathbb{Q} \) . Thus, \( {F}_{K} \) is rationally equivalent to \( F \) .
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Yes
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Proposition 8.4.8. Let \( K \) be a cubic number field, and as before write \( d\left( K\right) = {d}_{k}{f}^{2} \), where \( {d}_{k} \) is a fundamental discriminant. Let \( {F}_{K} \) be the cubic form associated to \( K \) by the Davenport-Heilbronn map, and let \( {H}_{K} = \) \( \left( {P, Q, R}\right) \) be its Hessian. Finally, set \( {f}_{H} = \gcd \left( {P, Q, R}\right) \) . Then\n\n(1) \( {f}_{H} = f \) or \( {f}_{H} = {3f} \) and apart from powers of \( 3, f \) and \( {f}_{H} \) are squarefree,\n\n(2) if \( {f}_{H} = {3f} \), then \( 3 \mid f \) ,\n\n(3) if \( 3\parallel f \), then \( {f}_{H} = {3f} \) ,\n\n(4) in particular, \( f \mid {f}_{H}, f \) and \( {f}_{H} \) have the same prime divisors, and we can have \( \left( {{v}_{3}\left( f\right) ,{v}_{3}\left( {f}_{H}\right) }\right) \) only equal to \( \left( {0,0}\right) ,\left( {1,2}\right) ,\left( {2,2}\right) \), and \( \left( {2,3}\right) \) .
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Proof. By Proposition 8.3.3 (2), we have \( p \mid {f}_{H} \) if and only if \( \left( {F, p}\right) = \left( {1}^{3}\right) \) , and by Proposition 8.2.3 (3) this is true if and only if \( p \) is totally ramified, hence by Proposition 8.4.1 (1), if and only if \( p \mid f \) . Hence \( f \) and \( {f}_{H} \) have the same prime divisors. Let \( p \) be such a prime divisor. By Davenport-Heilbronn’s theorem, \( {F}_{K} \in U \) . Hence by Proposition 8.3.3 (3) we have \( {p}^{3} \nmid d\left( K\right) \) if \( p \neq 3 \) , so if \( p \neq 3 \) , we have \( {p}^{2} \nmid f \) and \( p \nmid {d}_{k} \), and so up to powers of \( 3, f \) is squarefree. Since \( {f}_{H}^{2} \mid {3d}\left( K\right) = {d}_{k}{f}^{2} \), we have \( {v}_{p}\left( {f}_{H}^{2}\right) \leq 2 \), and thus up to powers of 3, \( {f}_{H} \) is also squarefree.\n\nAssume now that \( p = 3 \) and that \( p \) divides \( f \) (hence also \( {f}_{H} \) ). By Proposition 8.3.3 (3) we have \( {v}_{3}\left( f\right) \leq 2 \) and since \( {f}_{H}^{2} \mid {3d}\left( K\right) \), we have \( {v}_{3}\left( {f}_{H}\right) \leq 3 \) . Furthermore, if \( {v}_{3}\left( f\right) = 1 \), we have \( {v}_{3}\left( {f}_{H}^{2}\right) \leq {3}^{4} \), and so \( {v}_{3}\left( {f}_{H}\right) \leq 2 \) . Finally, since \( 3 \mid \left( {P, Q, R}\right) \), using the explicit formulas in terms of the coefficients of the form \( {F}_{K} \) we see that \( 3\left| {b\text{and}3}\right| c \), which implies \( 9 \mid \left( {P, Q, R}\right) \), hence \( {v}_{3}\left( {f}_{H}\right) \geq 2 \) . This proves all the assertions of the proposition.
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Yes
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Lemma 8.5.2. Let \( H = \left( {P, Q, R}\right) \) and \( {H}^{\prime } = \left( {{P}^{\prime },{Q}^{\prime },{R}^{\prime }}\right) \) be two reduced, definite, integral, binary quadratic forms such that there exists \( M \in {\mathrm{{GL}}}_{2}\left( \mathbb{Z}\right) \) with \( {H}^{\prime } = H \circ M \) . Then, either \( {H}^{\prime } = H \) and \( M \in \operatorname{Aut}\left( H\right) \), or \( {H}^{\prime } = {H}^{-1} \) and \( M \in \operatorname{Aut}\left( H\right) \sigma \) . Furthermore, the only elements of \( \operatorname{Aut}\left( H\right) \) are \( \pm {I}_{2} \), except in the following special cases that can occur simultaneously:
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Proof. Since \( H \) and \( {H}^{\prime } \) are equivalent, they have the same discriminant and represent the same integers. To say that \( H \) is reduced implies that \( P \) is the minimum of \( H \) on \( {\mathbb{Z}}^{2} - \{ \left( {0,0}\right) \} \) and that \( R \) is the second minimum; hence \( P = {P}^{\prime } \) and \( R = {R}^{\prime } \) . Equality of discriminants implies \( Q = \pm {Q}^{\prime } \) . Hence \( {H}^{\prime } = H \) or \( {H}^{\prime } = {H}^{-1} = H \circ \sigma \), so we need only to compute \( \operatorname{Aut}\left( H\right) \) . Let \( T = \left( \begin{array}{ll} 1 & 1 \\ 0 & 1 \end{array}\right) \) and \( S = \left( \begin{matrix} 0 & - 1 \\ 1 & 0 \end{matrix}\right) \) be the usual generators of the modular group \( {\operatorname{PSL}}_{2}\left( \mathbb{Z}\right) \), and let \( \mathcal{F} \) be the usual compact fundamental domain for the modular group in the upper half-plane \( \mathcal{H} \) . Let \( M = \left( \begin{array}{ll} a & b \\ c & d \end{array}\right) \) be an automorphism of \( H = \left( {P, Q, R}\right) \) . If \( M \in {\mathrm{{SL}}}_{2}\left( \mathbb{Z}\right) \), then it fixes \( \tau \in \mathcal{F} \) . Thus, by the usual theory, it is either \( \pm I, \pm S \) (if \( H = \left( {P,0, P}\right) \) ), \( \pm {ST} \) or \( \pm {\left( ST\right) }^{2} \) (if \( H = \left( {P, P, P}\right) \) ), \( \pm {TS} \) or \( \pm {\left( TS\right) }^{2} \) (if \( H = \left( {P, - P, P}\right) \) ). If \( M \notin {\mathrm{{SL}}}_{2}\left( \mathbb{Z}\right) \), then \( M \) swaps the two complex roots \( \tau \) and \( \bar{\tau } \) of \( H\left( {x,1}\right) \) , so that \[ {a\tau } + b = {c\tau }\bar{\tau } + d\bar{\tau } \] Taking imaginary parts, we get \( a = - d \), and taking real parts and using \( \tau + \bar{\tau } = - Q/P \) and \( \tau \bar{\tau } = R/P \), we obtain \( {bP} = {aQ} + {cR} \) . Finally, the determinant condition gives \( {a}^{2} + {bc} = 1 \) . One easily checks that the three conditions \( a = - d,{a}^{2} + {bc} = 1 \), and \( {bP} = {aQ} + {cR} \) are necessary and sufficient conditions for \( M \) to be an automorphism of \( H \) . Thus, \( H\left( {a, c}\right) = P{a}^{2} + {Qac} + R{c}^{2} = P{a}^{2} + {Pbc} = P \), and \( H\left( {a, c}\right) \geq \) \( \left( {P - Q + R}\right) \min \left( {{a}^{2},{c}^{2}}\right) \), and since \( H \) is reduced we have \( \left| Q\right| \leq P \leq R \) . We thus obtain the following.
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Yes
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Proposition 8.5.5. Let \( F = \left( {a, b, c, d}\right) \) be a reduced form such that \( 0 < \) \( \operatorname{disc}\left( F\right) \leq X \) . We have the following inequalities.\n\n(1)\n\n\[ 1 \leq a \leq \frac{2}{3\sqrt{3}}{X}^{1/4}. \]\n\n(2) If \( a \leq {X}^{1/4}/3 \), we have\n\n\[ 0 \leq b \leq \frac{3a}{2} + \sqrt{\sqrt{X} - \frac{{27}{a}^{2}}{4}} \]\n\nwhile if \( {X}^{1/4}/3 < a \leq 2{X}^{1/4}/\left( {3\sqrt{3}}\right) \), we have\n\n\[ \frac{3a}{2} - \sqrt{\sqrt{X} - \frac{{27}{a}^{2}}{4}} \leq b \leq \frac{3a}{2} + \sqrt{\sqrt{X} - \frac{{27}{a}^{2}}{4}}. \]\n\n(3) If \( a > {X}^{1/4}/3 \) or \( a \leq {X}^{1/4}/3 \) and \( b \geq - {3a}/2 + \sqrt{\sqrt{X} - {27}{a}^{2}/4} \), we have\n\n\[ \frac{{b}^{2} - \sqrt{X}}{3a} \leq c \leq b - {3a} \]\n\nwhile if \( a \leq {X}^{1/4}/3 \) and \( b \leq - {3a}/2 + \sqrt{\sqrt{X} - {27}{a}^{2}/4} \), we have\n\n\[ \frac{{b}^{2} - {P}_{2}}{3a} \leq c \leq b - {3a} \]\n\nwhere \( {P}_{2} \) is the unique positive solution of the equation\n\n\[ 4{P}_{2}^{3} - {\left( 3a + 2b\right) }^{2}{P}_{2}^{2} - {27}{a}^{2}X = 0. \]
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Proof. Let \( H = \left( {P, Q, R}\right) \) be the Hessian of \( F \), and set \( \Delta = \operatorname{disc}\left( F\right) \) so that \( {4PR} - {Q}^{2} = {3\Delta } \) . Since \( F \) is reduced, we have\n\n\[ {3X} \geq {3\Delta } \geq {4PR} - {P}^{2} \geq 3{P}^{2}. \]\n\nHowever, it is easily checked that\n\n\[ P{b}^{2} - {3Qab} + {9R}{a}^{2} - {P}^{2} = 0. \]\n\nThis is a quadratic equation in \( b \), which therefore must have a nonnegative discriminant. As its discriminant is equal to\n\n\[ 9{a}^{2}\left( {{Q}^{2} - {4PR}}\right) + 4{P}^{3} = 4{P}^{3} - {27}{a}^{2}\Delta \]\n\nwe thus have\n\n\[ {a}^{2} \leq \frac{4{P}^{3}}{27\Delta } \leq \frac{4P}{27} \leq \frac{4\sqrt{X}}{27} \]\n\nproving the inequality for \( a \) .\n\nFor \( b \) and \( c \), we note that \( P = {b}^{2} - {3ac} \leq \sqrt{\Delta } \leq \sqrt{X} \) ; hence the lower bound \( c \geq \left( {{b}^{2} - \sqrt{X}}\right) /\left( {3a}\right) \) is clear. Furthermore, the inequality \( Q \leq P \) gives \( {bc} - {9ad} \leq {b}^{2} - {3ac} \), hence \( {9ad} \geq \left( {b + {3a}}\right) c - {b}^{2} \) . The inequality \( P \leq R \) thus gives\n\n\[ {b}^{2} - {3ac} \leq {c}^{2} - {3bd} \leq {c}^{2} - \frac{b}{3a}\left( {\left( {b + {3a}}\right) c - {b}^{2}}\right) , \]\n\nor in other words the quadratic inequality\n\n\[ {c}^{2} + c\left( {{3a} - \frac{{b}^{2}}{3a} - b}\right) + \frac{{b}^{3}}{3a} - {b}^{2} \geq 0. \]\n\nThe roots of the polynomial in \( c \) are \( {b}^{2}/\left( {3a}\right) \) and \( b - {3a} \), and since \( b - {3a} \leq \) \( {b}^{2}/\left( {3a}\right) \) (the corresponding quadratic equation having a negative discriminant), we have \( c \leq b - {3a} \) or \( c \geq {b}^{2}/\left( {3a}\right) \) . The latter is impossible, however, since it would imply that \( P \leq 0 \) . Thus \( c \leq b - {3a} \), proving the inequalities \( \left( {{b}^{2} - \sqrt{X}}\right) /\left( {3a}\right) \leq c \leq b - {3a} \) . In particular, it implies that \( {b}^{2} - {3ab} + 9{a}^{2} - \sqrt{X} \leq 0 \), hence \( b \) lies between the roots of this quadratic equation; in other words,\n\n\[ \frac{3a}{2} - \sqrt{\sqrt{X} - \frac{{27}{a}^{2}}{4}} \leq b \leq \frac{3a}{2} + \sqrt{\sqrt{X} - \frac{{27}{a}^{2}}{4}}, \]\n\nas claimed. It is immediately checked that this lower bound for \( b \) is sharper than the trivial lower bound \( b \geq 0 \) if and only if \( a > {X}^{1/4}/3 \) .\n\nThe upper bounds for \( a \) and \( b \) are sharp, since they are reached for \( P = \) \( Q = R \) .\n\nTo finish the proof of the proposition, we must prove the other lower bound for \( c \) . By Proposition 8.3.2, we have \( {3aR} - {bQ} + {cP} = 0 \) . Since \( R = \) \( \left( {{3\Delta } + {Q}^{2}}\right) /\left( {4P}\right) \), this gives\n\n\[ {9a\Delta } + {3a}{Q}^{2} - {4bPQ} + {4c}{P}^{2} = 0 \]\n\nand solving this quadratic equation in \( Q \) gives\n\n\[ Q = \frac{{2bP} + \varepsilon \sqrt{4{P}^{3} - {27}{a}^{2}\Delta }}{3a} \]\n\nfor some
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Yes
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Proposition 8.5.7. Let \( K \) be a totally real cubic number field, \( {F}_{K} \) the unique reduced form associated to \( K \), and \( {H}_{K} \) its Hessian. Then\n\n(1) \( K \) is cyclic (that is, \( d\left( K\right) = {f}^{2} \) ) if and only if \( {H}_{K} = {f}_{H}\left( {1, \pm 1,1}\right) \).\n\n(2) \( d\left( K\right) = 5{f}^{2} \) if and only if \( {H}_{K} = {f}_{H}\left( {1, \pm 1,4}\right) \) or \( {H}_{K} = {f}_{H}\left( {2, \pm 1,2}\right) \).\n\n(3) \( d\left( K\right) = 8{f}^{2} \) if and only if \( {H}_{K} = {f}_{H}\left( {1,0,6}\right) \) or \( {H}_{K} = {f}_{H}\left( {2,0,3}\right) \).\n\n(4) \( d\left( K\right) = {12}{f}^{2} \) if and only if \( {H}_{K} = {f}_{H}\left( {1,0,9}\right) ,{H}_{K} = {f}_{H}\left( {2, \pm 2,5}\right) \), or \( {H}_{K} = {f}_{H}\left( {1,0,1}\right) \n\n(5) Let \( \Delta > 0 \) be a fundamental discriminant. Then \( d\left( K\right) = \Delta {f}^{2} \) if and only if \( {H}_{K} \) is a multiple of a primitive, reduced, positive definite, quadratic form of discriminant \( - {3\Delta }\left( {\text{case}\;{f}_{H} = f}\right) \) or \( - \Delta /3( \) case \( \;{f}_{H} = {3f}\; \) and \( 3 \mid f) \) .
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Proof. Using Proposition 8.4.8, proving this is just a matter of listing reduced quadratic forms. We leave it to the reader (see Exercise 8).
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No
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Lemma 8.6.2. Let \( F = \left( {a, b, c, d}\right) \) be a complex cubic form. Then \( F \) is reduced if and only if\n\n\[ \n{d}^{2} - {a}^{2} + {ac} - {bd} > 0 \n\]\n\n\[ \n- {\left( a - b\right) }^{2} - {ac} < {ad} - {bc} < {\left( a + b\right) }^{2} + {ac} \n\]\n\n\[ \na > 0,\;b \geq 0,\;d \neq 0,\text{ and }\;d > 0\text{ if }b = 0.\n\]
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Proof. The condition \( B < A \) is equivalent to \( {a\theta } + b < a \) and hence to \( \theta < \left( {a - b}\right) /a \) since \( a > 0 \) . Since \( F \) has only one real root (and again \( a > 0 \) ), this is equivalent to \( F\left( {a - b, a}\right) > 0 \), which gives \( - {\left( a - d\right) }^{2} - {ac} < {ad} - {bc} \) . Similarly, the condition \( - B < A \) is equivalent to \( - {a\theta } - b < a \), hence to \( \theta > - \left( {a + b}\right) /a \), so \( F\left( {-\left( {a + b}\right), a}\right) < 0 \), which gives \( {ad} - {bc} < {\left( a + b\right) }^{2} + {ac} \) .\n\nFinally, the condition \( A < C \) is equivalent to \( a{\theta }^{2} + {b\theta } + \left( {c - a}\right) > 0 \), and this is equivalent to \( \mathcal{R}\left( {{F}_{1}, G}\right) > 0 \), where \( \mathcal{R} \) is the resultant, \( {F}_{1}\left( X\right) = F\left( {X,1}\right) \) and \( G\left( X\right) = a{X}^{2} + {bX} + \left( {c - a}\right) \) . An immediate computation shows that \( \mathcal{R}\left( {{F}_{1}, G}\right) = {a}^{3}\left( {{d}^{2} - {a}^{2} + {ac} - {bd}}\right) \), and this gives the last unproved condition.
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Yes
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Proposition 8.6.3. (1) Two equivalent, irreducible, reduced, complex cubic forms are equal.
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Proof. (1). Let \( {F}^{\prime } = F \circ M \), where \( F \) and \( {F}^{\prime } \) are reduced and \( M \in \) \( {\mathrm{{GL}}}_{2}\left( \mathbb{Z}\right) \) . Then by the formula proved above, there exists \( \lambda > 0 \) such that \( {H}_{{F}^{\prime }} = \lambda {H}_{F} \circ M \) . As before, we deduce from the inequalities \( \left| B\right| < A < C \) and \( \left| {B}^{\prime }\right| < {A}^{\prime } < {C}^{\prime } \) that \( {H}_{{F}^{\prime }} = \lambda {H}_{F} \) and hence that \( M \) is an automorphism of \( {H}_{F} \) . The proof then terminates as in the real case, except that there are no special cases to consider since the forms are irreducible.
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Yes
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Proposition 8.6.4. Let \( F = \left( {a, b, c, d}\right) \) be a reduced form such that \( - X \leq \) \( \operatorname{disc}\left( F\right) < 0 \) . We have the following inequalities:\n\n(1)\n\n\[ 1 \leq a \leq \frac{2{X}^{1/4}}{{3}^{3/4}} \]\n\n(2)\n\n\[ 0 \leq b \leq \frac{3a}{2} + \sqrt{{\left( \frac{X}{3}\right) }^{1/2} - \frac{3{a}^{2}}{4}} \]\n\n(3)\n\n\[ 1 - b \leq c \leq {\left( \frac{X}{4a}\right) }^{1/3} + U\left( {a, b}\right) \]\n\nwhere \( U\left( {a, b}\right) = {b}^{2}/\left( {3a}\right) \) if \( b \leq {3a}/2 \), while \( U\left( {a, b}\right) = b - {3a}/4 \) if \( b > {3a}/2 \) .
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Proof. Set \( \Delta = \left| {\operatorname{disc}\left( F\right) }\right| \) and \( {3D} = {4AC} - {B}^{2} \) . The inequalities \( \left| B\right| < \) \( A < C \) imply as usual \( {A}^{2} < D \) or, equivalently, \( {a}^{2} < D \) . In addition, by a computation made above, we have \( \Delta = {3D}\left( {A{\theta }^{2} + {B\theta } + C}\right) \) . Solving this as a quadratic equation in \( \theta \), we obtain\n\n\[ {2a\theta } = - B \pm \sqrt{{4a}{\left( \frac{\Delta }{3D}\right) }^{1/2} - {3D}}. \]\n\nSince the expression under the square root is nonnegative, we obtain \( {16}{a}^{2}\mathit{Δ} \geq \) \( {27}{D}^{3} \geq {27}{a}^{6} \), proving (1) since \( \Delta \leq X \) .\n\nFrom the expression for \( {2a\theta } \), we also obtain\n\n\[ b = B - {a\theta } = \frac{3B}{2} \pm \sqrt{a{\left( \frac{\Delta }{3D}\right) }^{1/2} - \frac{3D}{4}}. \]\n\nSince the expression under the square root is a decreasing function of \( D \) and since \( D \geq {a}^{2} \) and \( \left| B\right| \leq A = a \), we obtain\n\n\[ b \leq \frac{3a}{2} + \sqrt{{\left( \frac{\Delta }{3}\right) }^{1/2} - \frac{3{a}^{2}}{4}} \]\n\nproving the inequality for \( b \) .\n\nWe have \( c = C - {\theta B} > A - \left| \theta \right| A = a - \left| {a\theta }\right| \) . But \( \left| {a\theta }\right| = \left| {b + {a\theta } - b}\right| \leq \) \( b + \left| B\right| \leq b + A = a + b \) ; hence \( c > - b \) so \( c \geq 1 - b \), as claimed, since \( b \) and \( c \) are integers.\n\nFor the upper bound, we check that\n\n\[ {4ac} = - 3{B}^{2} + {4bB} + {3D}. \]\n\nThis is a quadratic in \( B \) whose derivative is positive for \( B < {2b}/3 \) and negative for \( B > {2b}/3 \) . Since we must have \( - a < B < a \), it follows that the maximum of this quadratic is attained for \( B = {2b}/3 \) when \( {2b}/3 \leq a \) and for \( B = a \) when \( {2b}/3 > a \) . The upper bound for \( c \) follows immediately.
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Yes
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Lemma 9.2.1. Let \( L/K \) be an Abelian extension of number fields of conductor \( \mathfrak{m} \), let \( {n}_{L} = \left\lbrack {L : \mathbb{Q}}\right\rbrack \) and \( {n}_{K} = \left\lbrack {K : \mathbb{Q}}\right\rbrack \), and assume that the root discriminant satisfies \( {\left| d\left( L\right) \right| }^{1/{n}_{L}} \leq B \) for some number \( B \) . Then\n\n\[ \mathcal{N}\left( \mathfrak{m}\right) d{\left( K\right) }^{2} \leq {B}^{2{n}_{K}} \]\n\nIn particular, the number of possible base fields \( K \) and moduli \( \mathfrak{m} \) is finite.
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Proof. By Theorem 3.5.11, we have\n\n\[ {\left| d\left( L\right) \right| }^{1/{h}_{\mathfrak{m}, C}} = \left| {d\left( K\right) }\right| \frac{\mathcal{N}\left( \mathfrak{m}\right) }{\mathop{\prod }\limits_{{\mathfrak{p} \mid \mathfrak{m}}}\mathcal{N}{\left( \mathfrak{p}\right) }^{\mathop{\sum }\limits_{{1 \leq k \leq {v}_{\mathfrak{p}}\left( \mathfrak{m}\right) }}{h}_{\mathfrak{m}/{\mathfrak{p}}^{k}, C}/{h}_{\mathfrak{m}, C}}}.\n\nSince \( \mathfrak{m} \) is assumed to be the conductor, we have \( {h}_{\mathfrak{m}/{\mathfrak{p}}^{k}, C} < {h}_{\mathfrak{m}, C} \) for \( k \geq 1 \), and in particular \( {h}_{\mathfrak{m}/{\mathfrak{p}}^{k}, C} \leq {h}_{\mathfrak{m}, C}/2 \) since it is a divisor. From this and the above formula giving \( d\left( L\right) \), we obtain\n\n\[ \mathcal{N}{\left( \mathfrak{m}\right) }^{1/2}\left| {d\left( K\right) }\right| \leq {\left| d\left( L\right) \right| }^{1/{h}_{\mathfrak{m}, C}} = {\left| d\left( L\right) \right| }^{{n}_{K}/{n}_{L}} \leq {B}^{{n}_{K}}, \]\n\nfrom which the result follows.
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Yes
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Lemma 9.2.2. As usual, denote by \( {V}_{2}\left( K\right) \) the group of 2-virtual units of \( K \) (see Definition 5.2.4), and let \( {I}_{s} \) be the group of squarefree integral ideals of \( K \) whose ideal class is a square. Then\n\n(1) The map \( \phi \), which sends the class modulo \( {V}_{2}\left( K\right) \) of an element \( \alpha \in {K}^{ * } \) to the squarefree part of the ideal \( \alpha {\mathbb{Z}}_{K} \), is an isomorphism from \( {K}^{ * }/{V}_{2}\left( K\right) \) to \( {I}_{s} \) . (2) There is a natural exact sequence\n\n\[ 1 \rightarrow \frac{{V}_{2}\left( K\right) }{{K}^{*2}} \rightarrow \frac{{K}^{ * }}{{K}^{*2}} \rightarrow {I}_{s} \rightarrow 1 \]
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Proof. (1). We first show that the map \( \phi \) is well-defined. Indeed, we can write in a unique way \( \alpha {\mathbb{Z}}_{K} = \mathfrak{a}{\mathfrak{q}}^{2} \) with \( \mathfrak{a} \) integral and squarefree, and the ideal class of \( \mathfrak{a} \) is equal to that of \( {\mathfrak{q}}^{-2} \), so is a square, so \( \phi \left( \alpha \right) \in {I}_{s} \) . In addition, if \( {\alpha }^{\prime } = {\alpha \beta } \) with \( \beta \in {V}_{2}\left( K\right) \) is equivalent to \( \alpha \) modulo \( {V}_{2}\left( K\right) \), then since \( \beta \) is a virtual unit there exists an ideal \( \mathfrak{b} \) such that \( \beta {\mathbb{Z}}_{K} = {\mathfrak{b}}^{2} \), so that \( {\alpha }^{\prime }{\mathbb{Z}}_{K} = \mathfrak{a}{\left( \mathfrak{{qb}}\right) }^{2} \), hence \( \mathfrak{a} \) is also equal to the image of \( {\alpha }^{\prime } \) .\n\nIn addition, \( \phi \) is injective since \( \phi \left( \alpha \right) = \phi \left( {\alpha }^{\prime }\right) \) implies that \( \alpha /{\alpha }^{\prime } = {\left( q/{q}^{\prime }\right) }^{2} \) for some ideals \( q \) and \( {q}^{\prime } \), which means exactly that \( \alpha /{\alpha }^{\prime } \in {V}_{2}\left( K\right) \) .\n\nFinally, \( \phi \) is surjective since if \( \mathfrak{a} \in {I}_{s} \) then \( \mathfrak{a} = \alpha {\mathfrak{b}}^{2} \) for some ideal \( \mathfrak{b} \), so \( \alpha {\mathbb{Z}}_{K} = \mathfrak{a}{\mathfrak{q}}^{2} \) with \( \mathfrak{q} = {\mathfrak{b}}^{-1} \), finishing the proof of (1).\n\n(2) follows from (1) and the natural exact sequence\n\n\[ 1 \rightarrow \frac{{V}_{2}\left( K\right) }{{K}^{*2}} \rightarrow \frac{{K}^{ * }}{{K}^{*2}} \rightarrow \frac{{K}^{ * }}{{V}_{2}\left( K\right) } \rightarrow 1. \]
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Yes
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Theorem 9.2.6. Let \( K \) be a number field, \( L \) an extension of \( K \) of degree \( n \) , and let \( {L}_{2} \) be the Galois closure of \( L/K \) in some algebraic closure of \( K \) . Assume that \( \operatorname{Gal}\left( {{L}_{2}/K}\right) \) is isomorphic to the dihedral group \( {D}_{n} \), and that \( n \geq 3 \) with \( n \) odd. Finally, let \( {K}_{2} \) be the unique quadratic subextension of \( {L}_{2}/K \), and let \( \left( {\mathfrak{m}, C}\right) \) be the conductor of the Abelian extension \( {L}_{2}/{K}_{2} \) (see the diagram below). Then for each \( d \mid n \) there exists an integral ideal \( {\mathfrak{a}}_{d} \) of \( K \) (not only of \( {K}_{2} \) ) such that the following holds.\n\n(1) The conductor \( \mathfrak{m} \) of \( {L}_{2}/{K}_{2} \) is obtained by extending \( {\mathfrak{a}}_{n} \) to \( {K}_{2} \), in other words, \( {\mathfrak{m}}_{0} = {\mathfrak{a}}_{n}{\mathbb{Z}}_{{K}_{2}} \) and \( {\mathfrak{m}}_{\infty } = \varnothing \) .\n\n(2) More generally, the conductor of the unique subextension \( {L}_{2, d}/{K}_{2} \) of degree \( d \) of \( {L}_{2}/{K}_{2} \) is obtained by extending \( {\mathfrak{a}}_{d} \) to \( {K}_{2} \) . In particular, \( {\mathfrak{a}}_{1} = {\mathbb{Z}}_{K} \) and \( {d}_{1}\left| {d}_{2}\right| n \) implies that \( {\mathfrak{a}}_{{d}_{1}} \mid {\mathfrak{a}}_{{d}_{2}} \) .\n\n(3) If \( \tau \) is a generator of \( \operatorname{Gal}\left( {{K}_{2}/K}\right) \) and if we set\n\n\[ C{l}_{\mathrm{m}}\left( {K}_{2}\right) /\bar{C} = \langle \bar{I}\rangle \]\n\nfor some ideal \( I \) of \( {K}_{2} \), then \( \tau \left( \bar{I}\right) = \overline{{I}^{-1}} \) .\n\n(4) The relative discriminant ideal \( \mathfrak{d}\left( {L/K}\right) \) is given by\n\n\[ \mathfrak{d}\left( {L/K}\right) = \mathfrak{d}{\left( {K}_{2}/K\right) }^{\left( {n - 1}\right) /2}\mathop{\prod }\limits_{{d \mid n}}{\mathfrak{a}}_{d}^{\phi \left( d\right) }. \]\n\n(5) Let \( \left( {{r}_{1},{r}_{2}}\right) \) be the signature of \( K \) and \( \left( {{r}_{1}^{\prime },{r}_{2}^{\prime }}\right) \) the signature of \( {K}_{2} \) . Then the signature of \( {L}_{2} \) is equal to \( \left( {n{r}_{1}^{\prime }, n{r}_{2}^{\prime }}\right) \) and the signature of \( L \) is equal to \( \left( {{r}_{1} + \left( {\left( {n - 1}\right) /2}\right) {r}_{1}^{\prime },{r}_{2} + \left( {\left( {n - 1}\right) /2}\right) {r}_{2}^{\prime }}\right) \) .\n\n(6) Let \( \mathfrak{p} \) be a prime ideal of \( K \) . Then we have the following properties:\n\na) \( \mathfrak{p} \mid {\mathfrak{a}}_{n} \) if and only if \( \mathfrak{p} \) is totally ramified in \( L/K \) ;\n\nb) \( \mathfrak{p} \mid \left( {{\mathfrak{a}}_{d},\mathfrak{d}\left( {{K}_{2}/K}\right) }\right) \) implies \( \mathfrak{p} \mid d \) ;\n\nc) \( {\mathfrak{p}}^{2} \mid {\mathfrak{a}}_{d} \) implies \( \mathfrak{p} \mid d \) .
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See Section 10.1.5 for a proof of this theorem in the case \( n \) prime.
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No
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Lemma 9.3.6. (1) If the \( {x}_{j} \) for \( 1 \leq j \leq n \) are nonnegative real numbers and \( k \) is a real number such that \( k \geq 1 \), then \[ \mathop{\sum }\limits_{{1 \leq j \leq n}}{x}_{j}^{k} \leq {\left( \mathop{\sum }\limits_{{1 \leq j \leq n}}{x}_{j}\right) }^{k} \]
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Proof. For (1), we could say that for all \( p \geq 1 \), the \( {L}^{p} \) norm is less than or equal to the \( {L}^{1} \) norm. For a simpler proof, let us show the statement by induction on \( n \) . It is trivially true for \( n \leq 1 \) . Set \[ f\left( {x}_{n}\right) = {\left( \mathop{\sum }\limits_{{1 \leq j \leq n}}{x}_{j}\right) }^{k} - \mathop{\sum }\limits_{{1 \leq j \leq n}}{x}_{j}^{k} \] Then \[ {f}^{\prime }\left( {x}_{n}\right) = k\left( {{\left( \mathop{\sum }\limits_{{1 \leq j \leq n}}{x}_{j}\right) }^{k - 1} - {x}_{n}^{k - 1}}\right) , \] and since \( k \geq 1 \) this is nonnegative. It follows that \( f\left( {x}_{n}\right) \) is a nondecreasing function of \( {x}_{n} \), hence that \( f\left( {x}_{n}\right) \geq f\left( 0\right) \geq 0 \) by our induction hypothesis, proving (1).
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Yes
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Proposition 9.3.7. If \( {r}_{2} = 0 \), then we have the following inequalities:\n\n(1) \( \left| {a}_{n}\right| < {s}_{2}^{n/2}/{n}^{n/2} \) ;\n\n(2) \( {s}_{2} \geq n + 1 \) or, equivalently, \( {a}_{2} \leq \left( {{a}_{1}^{2} - 1 - n}\right) /2 \) ;\n\n(3) for \( k \) even, \( {s}_{k} > n{\left| {a}_{n}\right| }^{k/n} \) .
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Proof. Since \( {T}_{2} = {s}_{2} \), the above remark and the arithmetic-geometric mean inequality used above gives the bound \( \left| {a}_{n}\right| < {s}_{2}^{n/2}/{n}^{n/2} \), the inequality being strict since the arithmetic-geometric mean inequality becomes an equality only if all components are equal, which is not possible in the totally real case. Since \( \left| {a}_{n}\right| \geq 1 \), this same inequality can also be used backwards to say that \( {s}_{2} \geq n + 1 \) or, equivalently, that \( {a}_{2} \leq \left( {{a}_{1}^{2} - 1 - n}\right) /2 \) . Similarly for \( k \) even, the arithmetic-geometric mean inequality gives the stated inequality.
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Yes
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Proposition 9.3.8. If \( {r}_{2} = 0 \), then for \( 1 \leq k \leq n - 1 \) we have the inequality\n\n\[ \n{a}_{k - 1}{a}_{k + 1} \leq \frac{k\left( {n - k}\right) }{\left( {k + 1}\right) \left( {n - k + 1}\right) }{a}_{k}^{2}.\n\]
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Proof. Write as usual \( P\left( X\right) = \mathop{\prod }\limits_{j}\left( {X - {x}_{j}}\right) \) with the \( {x}_{j} \) real by assumption, and let \( x \) be a real number different from the \( {x}_{j} \) . Then\n\n\[ \n\frac{{P}^{\prime }\left( x\right) }{P\left( x\right) } = \mathop{\sum }\limits_{j}\frac{1}{x - {x}_{j}}\n\]\n\nhence\n\[ \n\frac{P\left( x\right) {P}^{\prime \prime }\left( x\right) - {P}^{\prime }{\left( x\right) }^{2}}{P{\left( x\right) }^{2}} = - \mathop{\sum }\limits_{j}\frac{1}{{\left( x - {x}_{j}\right) }^{2}}.\n\]\n\nBy Schwartz's inequality we have\n\n\[ \n{\left( \mathop{\sum }\limits_{j}\frac{1}{x - {x}_{j}}\right) }^{2} \leq n\mathop{\sum }\limits_{j}\frac{1}{{\left( x - {x}_{j}\right) }^{2}}\n\]\n\nhence\n\[ \n\frac{P\left( x\right) {P}^{\prime \prime }\left( x\right) - {P}^{\prime }{\left( x\right) }^{2}}{P{\left( x\right) }^{2}} \leq - \frac{1}{n}{\left( \frac{{P}^{\prime }\left( x\right) }{P\left( x\right) }\right) }^{2},\n\]\n\nfrom which it follows that\n\n\[ \nP\left( x\right) {P}^{\prime \prime }\left( x\right) \leq \frac{n - 1}{n}{P}^{\prime }{\left( x\right) }^{2}.\n\]\n\nNow since \( P\left( X\right) \) is totally real, it follows that all the derivatives of \( P\left( X\right) \) are also totally real (there must always be a root of \( {P}^{\prime }\left( X\right) \) between two roots of \( P\left( X\right) \) ), hence we may apply this result to the \( \left( {n - k + 1}\right) \) st derivative of \( P\left( x\right) \) . Using the formula \( {P}^{\left( n - k\right) }\left( 0\right) = {\left( -1\right) }^{n - k}\left( {n - k}\right) !{a}_{k} \) and simplifying, we obtain the inequality of the proposition.
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Yes
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Proposition 9.3.9. Let \( {g}_{1},\ldots ,{g}_{m} \) and \( f \) be \( {C}^{1} \) functions in \( {\mathbb{R}}^{n} \), let \( A \) be the subset of \( {\mathbb{R}}^{n} \) defined by the equations \( {g}_{k}\left( \mathbf{x}\right) = 0 \) for \( 1 \leq k \leq m \), and let \( \mathbf{x} \) be a local extremum of the function \( f \) on \( A \) . Then the vectors \( {f}^{\prime }\left( \mathbf{x}\right) \) and \( {g}_{k}^{\prime }\left( \mathbf{x}\right) \) for \( 1 \leq k \leq m \) are linearly dependent; in other words, there exist \( {\lambda }_{k} \in \mathbb{R} \) and \( {\lambda }_{0} \in \) \( \mathbb{R} \) not all equal to zero such that if we set \( g\left( \mathbf{x}\right) = {\lambda }_{0}f\left( \mathbf{x}\right) + \mathop{\sum }\limits_{{1 \leq k \leq m}}{\lambda }_{k}{g}_{k}\left( \mathbf{x}\right) \), then \( \frac{\partial g}{\partial {x}_{j}}\left( \mathbf{x}\right) = 0 \) for all \( j \) with \( 1 \leq j \leq n \) .
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Thus, we may apply this proposition to \( A = {G}_{3} \) and \( A = {G}_{4} \), but not to \( G \) itself since there is an inequality in the definition of \( G \), and this is the reason for which we have had to introduce the auxiliary sets \( {G}_{3} \) and \( {G}_{4} \) . Note also that, as in the one variable case, the condition is necessary but not sufficient for \( \mathbf{x} \) to be an extremum (consider \( f\left( x\right) = {x}^{3} \) at \( x = 0 \) ).
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No
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Lemma 9.3.11. Let \( {R}_{m, k}\left( x\right) \) be the above function with \( 1 \leq m \leq n - 1 \) .\n\n(1) For \( x > 0 \), the function \( {R}_{m, k}\left( x\right) \) has a unique minimum at 1, it decreases for \( x < 1 \), and it increases for \( x > 1 \) .
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The proof of the lemma is straightforward. We have\n\n\[ {R}_{m, k}^{\prime }\left( x\right) = \frac{{km}\left( {n - m}\right) }{2}\left( {{x}^{{km}/2 - 1} - {x}^{k\left( {m - n}\right) /2 - 1}}\right) ,\]\n\nso \( {R}_{m, k}^{\prime }\left( x\right) = 0 \) if and only if \( x = 1 \), and it is negative for \( x < 1 \) and positive for \( x > 1 \), proving (1).
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Yes
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Proposition 9.4.2. Assume that \( P\left( X\right) \) has signature \( \left( {0,2}\right) \), keep the notation of Section 9.3.4, and let \( \mathbf{x} = \left( {{x}_{1},{x}_{2},{x}_{3},{x}_{4}}\right) \) be a local extremum on \( {G}_{3} \) or \( {G}_{4} \) of the function\n\n\[ \n{s}_{3}\left( \mathbf{x}\right) = 2\left( {{x}_{1}^{3} + {x}_{2}^{3}}\right) - 6\left( {{x}_{1}{x}_{3}^{2} + {x}_{2}{x}_{4}^{2}}\right) .\n\]\n\n(1) If \( \mathbf{x} \in {G}_{3} \smallsetminus {G}_{4} \), then either \( {x}_{1} = {x}_{2},{x}_{3} = 0 \), or \( {x}_{4} = 0 \) .
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Proof. With the notation of Section 9.3.4, we have \( {g}_{1}\left( \mathbf{x}\right) = 2\left( {{x}_{1} + {x}_{2}}\right) - {a}_{1} \) , \( {g}_{2}\left( \mathbf{x}\right) = 2\left( {{x}_{1}^{2} + {x}_{2}^{2} - {x}_{3}^{2} - {x}_{4}^{2}}\right) - {s}_{2},{g}_{3}\left( \mathbf{x}\right) = \left( {{x}_{1}^{2} + {x}_{2}^{2}}\right) \left( {{x}_{3}^{2} + {x}_{4}^{2}}\right) - {a}_{4},{g}_{4}\left( \mathbf{x}\right) = \) \( 2\left( {{x}_{1}^{2} + {x}_{2}^{2} + {x}_{3}^{2} + {x}_{4}^{2}}\right) - {t}_{2} \), and \( {s}_{3}\left( \mathbf{x}\right) = 2\left( {{x}_{1}^{3} + {x}_{2}^{3}}\right) - 6\left( {{x}_{1}{x}_{3}^{2} + {x}_{2}{x}_{4}^{2}}\right) \) .\n\nFor (1), assume that \( \mathbf{x} \in {G}_{3} \smallsetminus {G}_{4} \) is a local extremum of \( {s}_{3}\left( \mathbf{x}\right) \) . Then by Proposition 9.3.9, the \( 4 \times 4 \) matrix whose columns are \( {g}_{1}^{\prime }\left( \mathbf{x}\right) ,{g}_{2}^{\prime }\left( \mathbf{x}\right) ,{g}_{3}^{\prime }\left( \mathbf{x}\right) \) , and \( {s}_{3}^{\prime }\left( \mathbf{x}\right) \) must have rank at most equal to 3 ; in other words, its determinant \( D \) must be equal to 0 . A computation shows that\n\n\[ \nD = - {192}{x}_{3}{x}_{4}{\left( {x}_{1} - {x}_{2}\right) }^{2}\left( {{x}_{1}^{2} + {x}_{2}^{2} + {x}_{3}^{2} + {x}_{4}^{2}}\right) .\n\]\n\nThus \( D = 0 \) if and only if \( {x}_{1} = {x}_{2} \) or
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Yes
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(1) If \( \mathbf{x} \in {G}_{3} \smallsetminus {G}_{4} \), then either \( {x}_{1} = {x}_{2} \) or \( {x}_{4} = 0 \) .
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Proof. With the notation of Section 9.3.4, we have \( {g}_{1}\left( \mathbf{x}\right) = {x}_{1} + {x}_{2} + 2{x}_{3} - {a}_{1},{g}_{2}\left( \mathbf{x}\right) = {x}_{1}^{2} + {x}_{2}^{2} + 2{x}_{3}^{2} - 2{x}_{4}^{2} - {s}_{2},{g}_{3}\left( \mathbf{x}\right) = {x}_{1}{x}_{2}\left( {{x}_{3}^{2} + {x}_{4}^{2}}\right) - {a}_{4}, \) \( {g}_{4}\left( \mathbf{x}\right) = {x}_{1}^{2} + {x}_{2}^{2} + 2{x}_{3}^{2} + 2{x}_{4}^{2} - {t}_{2} \), and \( {s}_{3}\left( \mathbf{x}\right) = {x}_{1}^{3} + {x}_{2}^{3} + 2{x}_{3}^{3} - 6{x}_{3}{x}_{4}^{2} \) . For (1), assume that \( \mathbf{x} \in {G}_{3} \smallsetminus {G}_{4} \) is a local extremum of \( {s}_{3}\left( \mathbf{x}\right) \) . Then by Proposition 9.3.9, the \( 4 \times 4 \) matrix whose columns are \( {g}_{1}^{\prime }\left( \mathbf{x}\right) ,{g}_{2}^{\prime }\left( \mathbf{x}\right) ,{g}_{3}^{\prime }\left( \mathbf{x}\right) \) , and \( {s}_{3}^{\prime }\left( \mathbf{x}\right) \) must have rank at most equal to 3 ; in other words, its determinant \( D \) must be equal to 0 . A computation (using the fact that \( {g}_{3}\left( \mathbf{x}\right) = 0 \) ) shows that \[ D = {24}{x}_{4}\left( {{x}_{1} - {x}_{2}}\right) \left( {{\left( {x}_{1} - {x}_{3}\right) }^{2} + {x}_{4}^{2}}\right) \left( {{\left( {x}_{2} - {x}_{3}\right) }^{2} + {x}_{4}^{2}}\right) . \] Thus \( D = 0 \) if and only if \( {x}_{1} = {x}_{2} \) or \( {x}_{4} = 0 \) (note that if one of the last two factors is equal to 0, we also have \( {x}_{4} = 0 \) ), and in both cases w
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Yes
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Proposition 9.4.4. Assume that \( P\left( X\right) \) has signature \( \left( {4,0}\right) \), keep the notation of Section 9.3.4, set\n\n\[ Q\left( X\right) = 3{X}^{4} - 2{a}_{1}{X}^{3} + {a}_{2}{X}^{2} - {a}_{4} = X{P}^{\prime }\left( X\right) - P\left( X\right) ,\]\n\n\[ S\left( X\right) = {12}{X}^{3} - 9{a}_{1}{X}^{2} + 6{a}_{2}X + {a}_{1}^{3} - 3{a}_{1}{a}_{2},\]\n\nand let \( \mathbf{x} = \left( {{x}_{1},{x}_{2},{x}_{3},{x}_{4}}\right) \) be a local extremum on \( {G}_{3} \) of the function\n\n\[ {s}_{3}\left( \mathbf{x}\right) = {x}_{1}^{3} + {x}_{2}^{3} + {x}_{3}^{3} + {x}_{4}^{3}. \]\n\nThen at least two of the \( {x}_{i} \) are equal. If, for example, \( {x}_{3} = {x}_{4} \), then \( {x}_{3} = {x}_{4} \) is a real root of \( S\left( X\right) = 0,{x}_{1} \) and \( {x}_{2} \) are the two real roots of\n\n\[ {X}^{2} - \left( {{a}_{1} - 2{x}_{3}}\right) X + 3{x}_{3}^{2} - 2{a}_{1}{x}_{3} + {a}_{2} = 0 \]\n\nand \( {s}_{3}\left( \mathbf{x}\right) = S\left( {x}_{3}\right) \) .
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Proof. This signature is much simpler than the other two. Indeed, we have \( {g}_{1}\left( \mathbf{x}\right) = {x}_{1} + {x}_{2} + {x}_{3} + {x}_{4} - {a}_{1},{g}_{2}\left( \mathbf{x}\right) = {x}_{1}^{2} + {x}_{2}^{2} + {x}_{3}^{2} + {x}_{4}^{2} - {s}_{2} \) , \( {g}_{3}\left( \mathbf{x}\right) = {x}_{1}{x}_{2}{x}_{3}{x}_{4} - {a}_{4} \), and \( {s}_{3}\left( \mathbf{x}\right) = {x}_{1}^{3} + {x}_{2}^{3} + {x}_{3}^{3} + {x}_{4}^{3} \) . If follows that the \( 4 \times 4 \) matrix that we must consider according to Proposition 9.3.9 is, up to trivial factors, a Vandermonde matrix in the \( {x}_{i} \), so its determinant vanishes if and only if two of the \( {x}_{i} \) are equal. Replacing and solving for the \( {x}_{i} \) and \( {s}_{3}\left( \mathbf{x}\right) \) gives the proposition.
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Yes
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Lemma 10.1.1. Let \( R \) be a Dedekind domain, \( M \) a finitely generated, torsion-free \( R \) -module, \( N \) a submodule of \( M \) of finite index, and \( \mathfrak{a} = \) Ann \( \left( {M, N}\right) \) and \( \mathfrak{b} = \left\lbrack {M : N}\right\rbrack \) be as above. Finally, let \( \mathfrak{p} \) be a (nonzero) prime ideal of \( R \) . The following conditions are equivalent:\n\n(1) \( \mathfrak{p} \nmid \mathfrak{a} \) ;\n\n(2) \( \mathfrak{p} \nmid \mathfrak{b} \) ;\n\n(3) \( \mathfrak{p}M + N = M \) ;\n\n(4) the injection from \( N \) to \( M \) induces an isomorphism from \( N/\mathfrak{p}N \) to \( M/\mathfrak{p}M \)\n\n(5) \( \mathfrak{p}M \cap N = \mathfrak{p}N \) .\n\nIf, in addition, \( N \) is free with basis \( \left( {\omega }_{i}\right) \), these conditions are also equivalent to:\n\n(6) the classes modulo \( \mathfrak{p}M \) of the \( {\omega }_{i} \) form an \( R/\mathfrak{p} \) -basis of \( M/\mathfrak{p}M \) .
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Proof. Since \( \mathfrak{a}\left| \mathfrak{b}\right| {\mathfrak{a}}^{n} \), as already stated \( \mathfrak{a} \) and \( \mathfrak{b} \) have the same prime ideal divisors, so the equivalence of (1) and (2) is trivial.\n\n\( \left( 2\right) \Rightarrow \left( 3\right) \) By \( \left( 2\right) \), we have \( \mathfrak{b} + \mathfrak{p} = R \), hence multiplying by \( M \) and using \( \mathfrak{b}M \subset \mathfrak{a}M \subset N \), we obtain \( M \subset \mathfrak{p}M + N \) . The other inclusion being obvious, this proves (3).\n\n\( \left( 3\right) \Rightarrow \left( 4\right) \) . The natural map \( i \) from \( N/\mathfrak{p}N \) to \( M/\mathfrak{p}M \) induced by the injection from \( N \) to \( M \) is well-defined. By (3), it is surjective. On the other hand, both \( M/\mathfrak{p}M \) and \( N/\mathfrak{p}N \) are \( R/\mathfrak{p} \) -vector spaces of the same dimension \( n \) ; hence the map \( i \) is an isomorphism.\n\n\( \left( 4\right) \Rightarrow \left( 5\right) \) . This simply expresses the fact that the kernel of \( i \) is trivial.\n\n\( \left( 5\right) \Rightarrow \left( 1\right) \) . Assume by contradiction that \( \mathfrak{p} \mid \mathfrak{a} \), in other words that \( \mathfrak{a} \subset \mathfrak{p} \) . This implies that \( \mathfrak{a}M \subset \mathfrak{p}M \) . On the other hand, by definition of \( \mathfrak{a} \) we have \( \mathfrak{a}M \subset N \), so \( \mathfrak{a}M \subset \mathfrak{p}M \cap N = \mathfrak{p}N \) by (5). It follows that \( {\mathfrak{{ap}}}^{-1}M \subset N \), hence that \( {\mathfrak{{ap}}}^{-1} \subset \mathfrak{a} \), a contradiction, thus proving (1). We have thus proved that conditions (1) to (5) are equivalent.\n\nAssume now that \( N \) is free and that \( \left( {\omega }_{i}\right) \) is a basis of \( N \) .\n\n(3) \( \Leftrightarrow \) (6). It is clear that \( M = N + \mathfrak{p}M \) is equivalent to the statement that the classes modulo \( \mathfrak{p}M \) of the \( {\omega }_{i} \) generate \( M/\mathfrak{p}M \) . Since this is an \( R/\mathfrak{p} \) - vector space of dimension \( n \), and since there are \( n \) generating elements \( {\omega }_{i} \) , this proves the equivalence of (3) and (6) and hence the lemma.
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Yes
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Lemma 10.1.2. Let \( L/K \) be a finite extension of number fields of degree \( n \) , and let \( \mathfrak{p} \) be a prime ideal of \( K \) that is totally ramified in \( L \), so that \( \mathfrak{p}{\mathbb{Z}}_{L} = {\mathfrak{P}}^{n} \) . Let \( \pi \in \mathfrak{P} \smallsetminus {\mathfrak{P}}^{2} \) . Then \( {\mathbb{Z}}_{K}\left\lbrack \pi \right\rbrack \) is a \( \mathfrak{p} \) -maximal order of \( {\mathbb{Z}}_{L} \) .
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Proof. Since \( \mathfrak{P} \) is totally ramified, \( {\mathbb{Z}}_{L}/{\mathfrak{P}}^{k} \) is a \( {\mathbb{Z}}_{K}/\mathfrak{p} \) -vector space (of dimension \( k \) ) for all \( k \leq n \) . We will show by induction that for all \( k \leq n \), the classes modulo \( {\mathfrak{P}}^{k} \) of the \( {\pi }^{i} \) for \( 0 \leq i < k \) generate \( {\mathbb{Z}}_{L}/{\mathfrak{P}}^{k} \) . Applied to \( k = n \) , this will show that the classes modulo \( \mathfrak{p}{\mathbb{Z}}_{L} \) of the \( {\pi }^{i} \) for \( 0 \leq i < n \) generate \( {\mathbb{Z}}_{L}/\mathfrak{p}{\mathbb{Z}}_{L} \), hence form a basis, and the lemma follows from the equivalence \( \left( 2\right) \Leftrightarrow \left( 6\right) \) of Lemma 10.1.1.\n\nSince \( {\mathbb{Z}}_{L}/\mathfrak{P} \) is a \( {\mathbb{Z}}_{K}/\mathfrak{p} \) -vector space of dimension 1, it is generated by the class of 1, so our claim is true for \( k = 1 \) . Assume that it is true for some \( k < n \), and let \( x \in {\mathbb{Z}}_{L} \) . Thus, there exist \( {x}_{i} \in {\mathbb{Z}}_{K} \) such that \( x \equiv \mathop{\sum }\limits_{{0 \leq i < k}}{x}_{i}{\pi }^{i} \) (mod \( {\mathfrak{P}}^{k} \) ). On the other hand, since \( \pi \in \mathfrak{P} \smallsetminus {\mathfrak{P}}^{2} \), multiplication by \( {\pi }^{\bar{k}} \) induces\n\nan isomorphism from \( {\mathbb{Z}}_{K}/\mathfrak{p} \simeq {\mathbb{Z}}_{L}/\mathfrak{P} \) to \( {\mathfrak{P}}^{k}/{\mathfrak{P}}^{k + 1} \) . Since \( x - \mathop{\sum }\limits_{{0 \leq i < k}}{x}_{i}{\pi }^{i} \in \) \( {\mathfrak{P}}^{k} \), this implies that there exists \( {x}_{k} \in {\mathbb{Z}}_{K} \) such that\n\n\[ x - \mathop{\sum }\limits_{{0 \leq i < k}}{x}_{i}{\pi }^{i} \equiv {x}_{k}{\pi }^{k}\left( {\;\operatorname{mod}\;{\mathfrak{P}}^{k + 1}}\right) ,\]\n\nthus proving our induction hypothesis and hence the lemma.
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Yes
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Proposition 10.1.3. Let \( L/K \) be a normal extension of number fields as above.\n\n(1) The ideals \( {\mathfrak{P}}_{i} \) are permuted transitively by the Galois group \( G \) : in other words, for every pair \( \left( {i, j}\right) \) there exists a (not necessarily unique) \( {\sigma }_{i, j} \in G \) such that \( \sigma \left( {\mathfrak{P}}_{i}\right) = {\mathfrak{P}}_{j} \) .
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Proof. (1) Fix \( {\mathfrak{P}}_{i} \), and assume that for some \( j,{\mathfrak{P}}_{j} \) is not of the form \( \sigma \left( {\mathfrak{P}}_{i}\right) \) for \( \sigma \in G \) . By the weak approximation theorem (Proposition 1.2.3), we can find \( x \in {\mathfrak{P}}_{j} \) such that \( x \notin \sigma \left( {\mathfrak{P}}_{i}\right) \) for all \( \sigma \in G \) . If we let \( a = {\mathcal{N}}_{L/K}\left( x\right) = \) \( \mathop{\prod }\limits_{{\sigma \in G}}\sigma \left( x\right) \), we have \( a \in {\mathbb{Z}}_{K} \cap {\mathfrak{P}}_{j} = \mathfrak{p} \) . On the other hand, for all \( \sigma \in G \) we have \( \sigma \left( x\right) \notin {\mathfrak{P}}_{i} \) (otherwise, \( x \in {\sigma }^{-1}\left( {\mathfrak{P}}_{i}\right) \) ), and since \( {\mathfrak{P}}_{i} \) is a prime ideal we have \( a \notin {\mathfrak{P}}_{i} \) and in particular \( a \notin \mathfrak{p} \), which is absurd.
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Yes
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(1) The map \( s \) defined above is a surjective homomorphism from \( D\left( {\mathfrak{P}/\mathfrak{p}}\right) \) to \( \operatorname{Gal}\left( {\left( {{\mathbb{Z}}_{L}/\mathfrak{P}}\right) /\left( {{\mathbb{Z}}_{K}/\mathfrak{p}}\right) }\right) \) whose kernel is equal to \( I\left( {\mathfrak{P}/\mathfrak{p}}\right) \) . In other words, \( s \) induces an isomorphism from \( D\left( {\mathfrak{P}/\mathfrak{p}}\right) /I\left( {\mathfrak{P}/\mathfrak{p}}\right) \) to \( \operatorname{Gal}\left( {\left( {{\mathbb{Z}}_{L}/\mathfrak{P}}\right) /\left( {{\mathbb{Z}}_{K}/\mathfrak{p}}\right) }\right) \) .
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Proof. (1) The finite field extension \( \left( {{\mathbb{Z}}_{L}/\mathfrak{P}}\right) /\left( {{\mathbb{Z}}_{K}/\mathfrak{p}}\right) \) is Galois (that is, normal and separable, and even cyclic). Hence, in particular, the primitive element theorem applies, so we can find \( \bar{\theta } \in {\mathbb{Z}}_{L}/\mathfrak{P} \) such that \( {\mathbb{Z}}_{L}/\mathfrak{P} = \) \( \left( {{\mathbb{Z}}_{K}/\mathfrak{p}}\right) \left( \bar{\theta }\right) \) . By the weak approximation theorem, we can find \( \theta \in {\mathbb{Z}}_{L} \) such that \( \theta \equiv \bar{\theta }\left( {\;\operatorname{mod}\;\mathfrak{P}}\right) \) and \( \theta \in \sigma \left( \mathfrak{P}\right) \) for all \( \sigma \notin D\left( {\mathfrak{P}/\mathfrak{p}}\right) \) . This is possible since by definition of \( D\left( {\mathfrak{P}/\mathfrak{p}}\right) \) all the \( \sigma \left( \mathfrak{P}\right) \) for \( \sigma \notin D\left( {\mathfrak{P}/\mathfrak{p}}\right) \) are distinct from \( \mathfrak{P} \) . Set\n\n\[ P\left( X\right) = \mathop{\prod }\limits_{{\sigma \in G}}\left( {X - \sigma \left( \theta \right) }\right) \]\n\nand let \( \bar{P}\left( X\right) \) be the polynomial in \( \left( {{\mathbb{Z}}_{L}/\mathfrak{P}}\right) \left\lbrack X\right\rbrack \) obtained by reducing the coefficients of \( P \) modulo \( \mathfrak{P} \) . Since \( \sigma \left( \theta \right) \in \mathfrak{P} \) for all \( \sigma \notin D\left( {\mathfrak{P}/\mathfrak{p}}\right) \), it follows that for \( m = \left| G\right| - \left| {D\left( {\mathfrak{P}/\mathfrak{p}}\right) }\right| \) we have\n\n\[ \bar{P}\left( X\right) = {X}^{m}\overline{{P}_{1}}\left( X\right) \;\text{ with }\;\overline{{P}_{1}}\left( X\right) = \mathop{\prod }\limits_{{\sigma \in D\left( {\mathfrak{P}/\mathfrak{p}}\right) }}\left( {X - \sigma \left( \theta \right) }\right) . \]\n\nBy Galois theory \( \bar{P}\left( X\right) \in \left( {{\mathbb{Z}}_{K}/\mathfrak{p}}\right) \left\lbrack X\right\rbrack \), hence \( \overline{{P}_{1}}\left( X\right) \in \left( {{\mathbb{Z}}_{K}/\mathfrak{p}}\right) \left\lbrack X\right\rbrack \), so \( \bar{\theta } \) is a root of \( \overline{{P}_{1}}\left( X\right) \), hence its minimal polynomial is a divisor of \( \overline{{P}_{1}} \), so the conjugates of \( \bar{\theta } \) are among the \( \sigma \left( \bar{\theta }\right) \) for \( \sigma \in D\left( {\mathfrak{P}/\mathfrak{p}}\right) \) . This means exactly that the homomorphism \( s \) is surjective.\n\nThe kernel of \( s \) is the set of \( \sigma \in D\left( {\mathfrak{P}/\mathfrak{p}}\right) \) such that \( s\left( \sigma \right) \) is the identity on \( {\mathbb{Z}}_{L}/\mathfrak{P} \), in other words such that \( \sigma \left( x\right) \equiv x\left( {\;\operatorname{mod}\;\mathfrak{P}}\right) \) for all \( x \in {\mathbb{Z}}_{L} \), and by definition this is the inertia group \( I\left( {\mathfrak{P}/\mathfrak{p}}\right) \), finishing the proof of (1).
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Yes
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Corollary 10.1.6. We have\n\n\\[ \n\\left| {D\\left( {\\mathfrak{P}/\\mathfrak{p}}\\right) }\\right| = e\\left( {\\mathfrak{P}/\\mathfrak{p}}\\right) f\\left( {\\mathfrak{P}/\\mathfrak{p}}\\right) \\;\\text{ and }\\;\\left| {I\\left( {\\mathfrak{P}/\\mathfrak{p}}\\right) }\\right| = e\\left( {\\mathfrak{P}/\\mathfrak{p}}\\right) .\n\\]
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Proof. Since there are \( g \) ideals \( {\\mathfrak{P}}_{i} \) above \( \\mathfrak{p} \) and the \( D\\left( {{\\mathfrak{P}}_{i}/\\mathfrak{p}}\\right) \) are conjugate groups, and hence have the same cardinality, we have \( D\\left( {\\mathfrak{P}/\\mathfrak{p}}\\right) = n/g = \) \( e\\left( {\\mathfrak{P}/\\mathfrak{p}}\\right) f\\left( {\\mathfrak{P}/p}\\right) \) . The cardinality of \( I\\left( {\\mathfrak{P}/\\mathfrak{p}}\\right) \) then follows from the proposition.
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Yes
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Corollary 10.1.7. If \( L/K \) is a normal extension of number fields which is not cyclic, no prime ideal of \( K \) is inert in \( L/K \) .
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Proof. Indeed, if \( \mathfrak{p} \) is inert, then \( D\left( {\mathfrak{P}/\mathfrak{p}}\right) = \operatorname{Gal}\left( {L/K}\right) \) and \( I\left( {\mathfrak{P}/\mathfrak{p}}\right) = \{ 1\} \) , hence Proposition 10.1.5 shows that \( \operatorname{Gal}\left( {L/K}\right) = D\left( {\mathfrak{P}/\mathfrak{p}}\right) /I\left( {\mathfrak{P}/p}\right) \) is cyclic. Note that the converse is true: if \( L/K \) is a cyclic extension, there exist inert primes in \( L/K \), even a positive density. This follows immediately from the Chebotarev density theorem (see [Lan3]).
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Yes
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Lemma 10.1.8. Recall that \( D = {G}_{-1} \) and \( I = {G}_{0} \). (1) We have the inclusions \[ G \supset {G}_{-1} \supset {G}_{0} \supset {G}_{1}\cdots \] (2) There exists \( n \) such that \( {G}_{k} = \left\{ {1}_{G}\right\} \) for all \( k \geq n \) .
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Proof. (1) is trivial. Furthermore, if \( \sigma \in {G}_{k} \) for all \( k \), then for all \( x \in {\mathbb{Z}}_{L} \) and for all \( k,\sigma \left( x\right) \equiv x\left( {\;\operatorname{mod}\;{\mathfrak{P}}^{k + 1}}\right) \), so \( \sigma \left( x\right) = x \) and hence \( \sigma = {1}_{G} \). Since \( G \) is finite, it follows that \( {G}_{k} = \left\{ {1}_{G}\right\} \) for \( k \) sufficiently large.
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Yes
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Lemma 10.1.9. For all \( k \geq 0,{G}_{k} \) is a normal subgroup of \( D = {G}_{-1} \) (not of \( G \) itself, however).
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Proof. Let \( \sigma \in {G}_{k} \) and \( \tau \in D \) . For all \( x \in {\mathbb{Z}}_{L} \), we have\n\n\[ \sigma \left( {{\tau }^{-1}\left( x\right) }\right) \equiv {\tau }^{-1}\left( x\right) \left( {\;\operatorname{mod}\;{\mathfrak{P}}^{k + 1}}\right) ; \]\n\nhence\n\n\[ \tau \left( {\sigma \left( {{\tau }^{-1}\left( x\right) }\right) }\right) \equiv x\left( {{\;\operatorname{mod}\;\tau }{\left( \mathfrak{P}\right) }^{k + 1}}\right) ,\]\n\nand since \( \tau \in D \), we have \( \tau \left( \mathfrak{P}\right) = \mathfrak{P} \), so the result follows.
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Yes
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Lemma 10.1.10. Keep the above notation. Then for all \( k \geq 0 \) we have \( {G}_{k}\left( {\mathfrak{P}/{\mathfrak{P}}_{I}}\right) = {G}_{k}\left( {\mathfrak{P}/\mathfrak{p}}\right) \) and we also have \( {v}_{\mathfrak{P}}\left( {\mathfrak{D}\left( {L/K}\right) }\right) = {v}_{\mathfrak{P}}\left( {\mathfrak{D}\left( {L/{L}^{I}}\right) }\right) \) , where as usual \( \mathfrak{D}\left( {L/K}\right) \) denotes the relative different (see Definition 2.3.16).
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Proof. Since \( I = {G}_{0} \), it is clear that for \( k \geq 0,\sigma \in {G}_{k}\left( {\mathfrak{P}/\mathfrak{p}}\right) \) if and only if \( \sigma \in {G}_{k}\left( {\mathfrak{P}/{\mathfrak{P}}_{I}}\right) \), so the groups are equal. To prove the second statement, we note that the extension \( {L}^{I}/K \) is unramified at \( {\mathfrak{P}}_{I} \), hence \( {\mathfrak{P}}_{I} \) and hence also \( \mathfrak{P} \) do not divide the relative different \( \mathfrak{D}\left( {{L}^{I}/K}\right) \) . By the transitivity formula for the different (see Proposition 2.3.17), it follows that \( {v}_{\mathfrak{P}}\left( {\mathfrak{D}\left( {L/K}\right) }\right) = {v}_{\mathfrak{P}}\left( {\mathfrak{D}\left( {L/{L}^{I}}\right) }\right) \), as was to be proved.
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Yes
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Lemma 10.1.11. Assume that \( \mathfrak{p} = {\mathfrak{P}}^{e} \) is totally ramified in \( L/K \) and let \( \pi \in \mathfrak{P} \smallsetminus {\mathfrak{P}}^{2} \) be a uniformizer at \( \mathfrak{P} \). If \( \sigma \in G = \operatorname{Gal}\left( {L/K}\right) \), then \( \sigma \in {G}_{k} \) if and only if \( \sigma \left( \pi \right) \equiv \pi \left( {\;\operatorname{mod}\;{\mathfrak{P}}^{k + 1}}\right) \).
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Proof. The condition is evidently necessary. Conversely, if it is satisfied, then \( \sigma \left( x\right) \equiv x\left( {\;\operatorname{mod}\;{\mathfrak{P}}^{k + 1}}\right) \) for all \( x \in {\mathbb{Z}}_{K}\left\lbrack \pi \right\rbrack \). Let \( y \) be an arbitrary element of \( {\mathbb{Z}}_{L} \). By Lemma 10.1.2 we know that \( {\mathbb{Z}}_{K}\left\lbrack \pi \right\rbrack \) is a \( \mathfrak{p} \)-maximal order, so it follows that the index-ideal \( \mathfrak{b} = \left\lbrack {{\mathbb{Z}}_{L} : {\mathbb{Z}}_{K}\left\lbrack \pi \right\rbrack }\right\rbrack \) is not divisible by \( \mathfrak{p} \). By the weak approximation theorem we can find \( d \in {\mathbb{Z}}_{K} \) such that \( {v}_{\mathfrak{p}}\left( d\right) = 0 \) and \( {v}_{\mathfrak{q}}\left( d\right) \geq \) \( {v}_{\mathfrak{q}}\left( \mathfrak{b}\right) \) for \( \mathfrak{q} \mid \mathfrak{b} \). Thus \( {dy} \in \mathfrak{b}y \subset {\mathbb{Z}}_{K}\left\lbrack \pi \right\rbrack \), hence \( \sigma \left( {dy}\right) \equiv {dy}\left( {{\;\operatorname{mod}\;{\mathfrak{P}}^{k + 1}}\text{. Since}}\right) \) \( d \in {\mathbb{Z}}_{K} \) we have \( \sigma \left( d\right) = d \), and since \( d \) is coprime to \( \mathfrak{p} \) and hence to \( \mathfrak{P} \), we deduce that \( \sigma \left( y\right) \equiv y\left( {\;\operatorname{mod}\;{\mathfrak{P}}^{k + 1}}\right) \), as desired.
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Yes
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Proposition 10.1.13. Let \( \mathfrak{p} \) be a prime ideal of \( K \) which is not necessarily totally ramified in \( L/K \), let \( \pi \) be a uniformizer of \( \mathfrak{P} \), and let \( \sigma \in {G}_{0} = I \).\n\n(1) We have \( {v}_{\mathfrak{P}}\left( {\sigma \left( \pi \right) /\pi }\right) = 0 \).\n\n(2) For \( k \geq 1,\sigma \in {G}_{k} \Leftrightarrow \sigma \left( \pi \right) /\pi \equiv 1\left( {{\;\operatorname{mod}\;{}^{ * }}{\mathfrak{P}}^{k}}\right) \).\n\nNote that since \( \sigma \left( \pi \right) /\pi \) is not an algebraic integer in general, the congruence is multiplicative, in other words (2) reads \( \sigma \in {G}_{k} \Leftrightarrow {v}_{\mathfrak{P}}\left( {\sigma \left( \pi \right) /\pi - 1}\right) \geq \) \( k \) .
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Proof. Since \( \pi \) is a uniformizer of \( \mathfrak{P} \), we have\n\n\[ {v}_{\mathfrak{P}}\left( {\sigma \left( \pi \right) /\pi }\right) = {v}_{{\sigma }^{-1}\left( \mathfrak{P}\right) }\left( \pi \right) - {v}_{\mathfrak{P}}\left( \pi \right) = 0 \]\n\nsince \( \sigma \) is in the inertia group, hence a fortiori in the decomposition group, proving (1).\n\nFor (2), by Lemma 10.1.10, we have \( {G}_{k}\left( {\mathfrak{P}/{\mathfrak{P}}_{I}}\right) = {G}_{k}\left( {\mathfrak{P}/\mathfrak{p}}\right) \). Since we assume that \( \sigma \in {G}_{0} = I \), this implies that we can replace the extension \( L/K \) by the extension \( L/{L}^{I} \) without changing the ramification groups, and in this extension \( {\mathfrak{P}}_{I} \) is totally ramified. We clearly have\n\n\[ {v}_{\mathfrak{P}}\left( {\sigma \left( \pi \right) /\pi - 1}\right) = {v}_{\mathfrak{P}}\left( {\sigma \left( \pi \right) - \pi }\right) - 1 = {i}_{G}\left( \sigma \right) - 1 \]\n\nhence the definition shows that \( \sigma \in {G}_{k} \) if and only if \( {i}_{G}\left( \sigma \right) \geq k + 1 \) if and only if \( \sigma \left( \pi \right) /\pi \equiv 1\left( {{\;\operatorname{mod}\; * }{\mathfrak{P}}^{k}}\right) \), as was to be proved.
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Yes
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Proposition 10.1.14. As above, let \( \pi \) be a uniformizer of \( \mathfrak{P} \) . (1) The map that sends \( \sigma \in {G}_{0} \) to \( \sigma \left( \pi \right) /\pi \) induces an injection \( {\theta }_{0} \) from \( {G}_{0}/{G}_{1} \) to \( {\left( {\mathbb{Z}}_{L}/\mathfrak{P}\right) }^{ * } \) . (2) For \( k \geq 1 \) the map that sends \( \sigma \in {G}_{k} \) to \( \sigma \left( \pi \right) /\pi - 1 \) induces an injection \( {\theta }_{k} \) from \( {G}_{k}/{G}_{k + 1} \) to \( {\mathfrak{P}}^{k}/{\mathfrak{P}}^{k + 1} \) . (3) The maps \( {\theta }_{k} \) do not depend on the choice of the uniformizer \( \pi \) .
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Proof. First note that if \( {v}_{\mathfrak{P}}\left( \alpha \right) \geq k \), by Lemma 1.2.31 we may write \( \alpha = x/d \) with \( {v}_{\mathfrak{P}}\left( d\right) = 0 \), hence with \( {v}_{\mathfrak{P}}\left( x\right) \geq k \) . It follows that we can send \( \alpha \) to \( {\mathfrak{P}}^{k}/{\mathfrak{P}}^{k + 1} \) by sending it to the class of \( x{d}^{-1} \), where \( {d}^{-1} \) is the inverse of \( d \) modulo \( {\mathfrak{P}}^{k + 1} \), and this clearly does not depend on the chosen representation of \( \alpha \). On the other hand, proposition 10.1.13 shows that for \( k \geq 1,\sigma \left( \pi \right) /\pi \equiv 1 \) (mod * \( {\mathfrak{P}}^{k} \) ) if and only if \( \sigma \in {G}_{k} \), showing both that for \( k \geq 1 \) the maps \( {\theta }_{k} \) are well-defined and that they are injective. For \( k = 0 \), a similar reasoning shows that the map \( {\theta }_{0} \) is well-defined, and \( \sigma \left( \pi \right) /\pi \equiv 1\left( {{\;\operatorname{mod}\; * }\mathfrak{P}}\right) \) is equivalent to \( \sigma \in {G}_{1} \) by Proposition 10.1.13, so the map is also injective in this case. We leave the proof of (3) to the reader.
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No
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Corollary 10.1.15. Let \( p \) be the prime number below \( \mathfrak{p} \) and \( \mathfrak{P} \) or, equivalently, the characteristic of the residue fields \( {\mathbb{Z}}_{K}/\mathfrak{p} \) and \( {\mathbb{Z}}_{L}/\mathfrak{P} \) . Then we have the following.\n\n(1) The group \( {G}_{0}/{G}_{1} \) is a cyclic group of order prime to \( p \), and in fact \( \left| {{G}_{0}/{G}_{1}}\right| = e/\left( {{p}^{\infty }, e}\right) \), the prime to \( p \) part of the ramification index \( e = \) \( e\left( {\mathfrak{P}/\mathfrak{p}}\right) \) .\n\n(2) For all \( k \geq 1,{G}_{k}/{G}_{k + 1} \) is an Abelian group isomorphic to a product of copies of \( \mathbb{Z}/p\mathbb{Z} \) (in other words, an elementary p-group). In particular, \( {G}_{1} \) is a (not necessarily Abelian) p-group, and \( \left| {G}_{1}\right| = \left( {{p}^{\infty }, e}\right) \) .
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Proof. Since \( {\left( {\mathbb{Z}}_{L}/\mathfrak{P}\right) }^{ * } \) is the multiplicative group of a finite field with \( {p}^{f\left( {\mathfrak{P}/p}\right) } \) elements, it is a cyclic group of order \( {p}^{f\left( {\mathfrak{P}/p}\right) } - 1 \) . Proposition 10.1.14 thus implies that \( {G}_{0}/{G}_{1} \) is a subgroup of this group, hence is a cyclic group of order dividing \( {p}^{f\left( {\mathfrak{P}/p}\right) } - 1 \), and in particular of order prime to \( p \) . Since we know that \( \left| {G}_{0}\right| = e = e\left( {\mathfrak{P}/\mathfrak{p}}\right) \) and we will see in (2) that \( {G}_{1} \) is a \( p \) -group,(1) follows.\n\nSimilarly, Proposition 10.1.14 shows that for \( k \geq 1,{G}_{k}/{G}_{k + 1} \) is isomorphic to a subgroup of the additive group \( {\mathfrak{P}}^{k}/{\mathfrak{P}}^{k + 1} \), which is (noncanonically) isomorphic to \( {\mathbb{Z}}_{L}/\mathfrak{P} \) as we saw in Section 4.2.3. Since this is the additive group of a finite field, it is an \( {\mathbb{F}}_{p} \) -vector space and hence an elementary \( p \) -group, finishing the proof of the corollary.
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Yes
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Proposition 10.1.16. Let \( \sigma \in {G}_{0} \) and let \( \tau \in {G}_{k} \) for some \( k \geq 1 \) . Then in \( {\mathfrak{P}}^{k}/{\mathfrak{P}}^{k + 1} \) we have\n\n\[{\theta }_{k}\left( {{\sigma \tau }{\sigma }^{-1}}\right) = {\theta }_{0}{\left( \sigma \right) }^{k}{\theta }_{k}\left( \tau \right) .\]
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Proof. Note first that this formula makes sense, since \( {\theta }_{0}{\left( \sigma \right) }^{k} \in {\left( {\mathbb{Z}}_{L}/\mathfrak{P}\right) }^{ * } \) , which operates multiplicatively on the \( {\mathbb{Z}}_{L}/\mathfrak{P} \) -vector space \( {\mathfrak{P}}^{k}/{\mathfrak{P}}^{k + 1} \) .\n\nSet \( {\pi }_{1} = {\sigma }^{-1}\left( \pi \right) \) . Since \( \sigma \) is in the decomposition group, \( {\pi }_{1} \) is also a uniformizer of \( \mathfrak{P} \) . Thus by definition, if we set \( a = \tau \left( {\pi }_{1}\right) /{\pi }_{1} - 1 \), we have \( {v}_{\mathfrak{P}}\left( a\right) \geq k \) and the class of \( a \) modulo \( {\mathfrak{P}}^{k + 1} \) is equal to \( {\theta }_{k}\left( \tau \right) \) . On the other hand, we have\n\n\[ \sigma \left( a\right) = \sigma \left( {\tau \left( {\pi }_{1}\right) }\right) /\sigma \left( {\pi }_{1}\right) - 1 = {\sigma \tau }{\sigma }^{-1}\left( \pi \right) /\pi - 1, \]\n\nso \( {\theta }_{k}\left( {{\sigma \tau }{\sigma }^{-1}}\right) \) is the class of \( \sigma \left( a\right) \) modulo \( {\mathfrak{P}}^{k + 1} \) .\n\nFinally, if we set \( b = a/{\pi }^{k} \), we have \( {v}_{\mathfrak{P}}\left( b\right) \geq 0 \), and since \( \sigma \in {G}_{0} \) we have \( \sigma \left( b\right) \equiv b\left( {{\;\operatorname{mod}\; * }\mathfrak{P}}\right) \), so\n\n\[ \sigma \left( a\right) = {\left( \sigma \left( \pi \right) /\pi \right) }^{k}\sigma \left( b\right) {\pi }^{k} \equiv {\theta }_{0}{\left( \sigma \right) }^{k}b{\pi }^{k} \]\n\n\[ \equiv {\theta }_{0}{\left( \sigma \right) }^{k}a \equiv {\theta }_{0}{\left( \sigma \right) }^{k}{\theta }_{k}\left( \tau \right) {\;(\operatorname{mod}\;{{}^{ * }{\mathfrak{P}}^{k + 1}})}\text{,} \]\n\nproving the proposition.
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Yes
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Corollary 10.1.17. Let \( \sigma \in {G}_{0} \) and let \( \tau \in {G}_{k} \) for some \( k \geq 1 \) . Then \( {\sigma \tau }{\sigma }^{-1}{\tau }^{-1} \in {G}_{k + 1} \) if and only if either \( \tau \in {G}_{k + 1} \) or \( {\sigma }^{k} \in {G}_{1} \) .
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Indeed, by the above proposition\n\n\[ \n{\sigma \tau }{\sigma }^{-1}{\tau }^{-1} \in {G}_{k + 1} \Leftrightarrow {\sigma \tau }{\sigma }^{-1} \in \tau {G}_{k + 1} \n\]\n\n\[ \n\Leftrightarrow {\theta }_{k}\left( {{\sigma \tau }{\sigma }^{-1}}\right) = {\theta }_{k}\left( \tau \right) \n\]\n\n\[ \n\Leftrightarrow {\theta }_{0}{\left( \sigma \right) }^{k}{\theta }_{k}\left( \tau \right) = {\theta }_{k}\left( \tau \right) \n\]\n\n\[ \n\Leftrightarrow {\theta }_{k}\left( \tau \right) \left( {{\theta }_{0}{\left( \sigma \right) }^{k} - 1}\right) = 0 \n\]\n\n\[ \n\Leftrightarrow {\theta }_{k}\left( \tau \right) = 0\text{ or }{\theta }_{0}\left( {\sigma }^{k}\right) = 1 \n\]\n\n\[ \n\Leftrightarrow \tau \in {G}_{k + 1}\text{ or }{\sigma }^{k} \in {G}_{1} \n\]\n\nproving the corollary.
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Yes
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Corollary 10.1.18. Let \( {e}_{0} = e/\left( {{p}^{\infty }, e}\right) \) be the order of \( {G}_{0}/{G}_{1} \) . (1) If \( {G}_{0} \) is an Abelian group, then for \( k \geq 0,{G}_{k} \neq {G}_{k + 1} \) implies \( {e}_{0} \mid k \) .
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Proof. Let \( \sigma \in {G}_{0} \) be such that the image of \( \sigma \) in \( {G}_{0}/{G}_{1} \) generates the cyclic group \( {G}_{0}/{G}_{1} \) . (1). Assume that \( {e}_{0} \nmid k \), and in particular \( k \geq 1 \) . This means that \( {\sigma }^{k} \notin {G}_{1} \) . It follows that for any \( \tau \in {G}_{k},{\sigma \tau }{\sigma }^{-1}{\tau }^{-1} \in {G}_{k + 1} \) if and only if \( \tau \in {G}_{k + 1} \) . Since \( {G}_{0} \) is Abelian, \( {\sigma \tau }{\sigma }^{-1}{\tau }^{-1} = {1}_{G} \), so \( \tau \in {G}_{k + 1} \) for all \( \tau \in {G}_{k} \), and hence \( {G}_{k} = {G}_{k + 1} \) .
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Yes
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Lemma 10.1.20. Let \( \pi \) be as above, and let \( f \) be the minimal polynomial of \( \pi \) in \( {\mathbb{Z}}_{K}\left\lbrack X\right\rbrack \) . Then \[ {\mathbb{Z}}_{K}{\left\lbrack \pi \right\rbrack }^{ * } = \frac{1}{{f}^{\prime }\left( \pi \right) }{\mathbb{Z}}_{K}\left\lbrack \pi \right\rbrack \]
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Proof. Set \( {a}_{i, j} = {\operatorname{Tr}}_{L/K}\left( {{\pi }^{i}{\pi }^{j}/{f}^{\prime }\left( \pi \right) }\right) \), and let \( A \) be the \( n \times n \) matrix whose entries are the \( {a}_{i, j} \). For any \( x \in L \) we can write \( x = \mathop{\sum }\limits_{i}{x}_{i}{\pi }^{i}/{f}^{\prime }\left( \pi \right) \) with \( {x}_{i} \) in \( K \) but not necessarily in \( {\mathbb{Z}}_{K} \). If \( X \) is the column vector of the \( {x}_{i} \), the condition \( x \in {\mathbb{Z}}_{K}{\left\lbrack \pi \right\rbrack }^{ * } \) is equivalent to \( {AX} \in {\mathbb{Z}}_{K}^{n} \), hence to prove the lemma we must prove that \( A \in {\mathrm{{GL}}}_{n}\left( {\mathbb{Z}}_{K}\right) \) - in other words, that \( A \) has entries in \( {\mathbb{Z}}_{K} \) and determinant equal to a unit.\n\nTo do this, we first note the rational function identity \[ \frac{1}{f\left( X\right) } = \mathop{\sum }\limits_{{k = 1}}^{n}\frac{1}{{f}^{\prime }\left( {\alpha }_{k}\right) \left( {X - {\alpha }_{k}}\right) } \] where the \( {\alpha }_{k} \) are the roots of \( f\left( X\right) \) in \( \overline{\mathbb{Q}} \) (see Exercise 5).\n\nSince \( f\left( X\right) \) is a monic polynomial of degree \( n \), if we expand this identity in powers of \( 1/X \), all the coefficients of \( {\left( 1/X\right) }^{i} \) for \( i < n \) vanish and the coefficient of \( {\left( 1/X\right) }^{n} \) is equal to 1 . Looking at the right-hand side, this gives \[ \mathop{\sum }\limits_{{1 \leq k \leq n}}\frac{{\alpha }_{k}^{i}}{{f}^{\prime }\left( {\alpha }_{k}\right) } = \left\{ \begin{array}{ll} 0 & \text{ if }i \leq n - 2 \\ 1 & \text{ if }i = n - 1 \end{array}\right. \] in other words, \( {\operatorname{Tr}}_{L/K}\left( {{\pi }^{i}/{f}^{\prime }\left( \pi \right) }\right) = 0 \) for \( i \leq n - 2 \) and \( {\operatorname{Tr}}_{L/K}\left( {{\pi }^{n - 1}/{f}^{\prime }\left( \pi \right) }\right) = \) 1.\n\nLet us now prove that \( A \in {\mathrm{{GL}}}_{n}\left( {\mathbb{Z}}_{K}\right) \). Thanks to the above formula, we already see that \( {a}_{i, j} = {\operatorname{Tr}}_{L/K}\left( {{\pi }^{i + j}/{f}^{\prime }\left( \pi \right) }\right) \) is equal to 0 for \( i + j \leq n - 2 \) and is equal to 1 for \( i + j = n - 1 \). For \( i + j \geq n \) we have \[ {a}_{i, j} = {\operatorname{Tr}}_{L/K}\left( {{\pi }^{n}{\pi }^{i + j - n}/{f}^{\prime }\left( \pi \right) }\right) . \] Since \( f \) is monic, \( {\pi }^{n} \) is equal to a \( {\mathbb{Z}}_{K} \) -linear combination of the \( {\pi }^{k} \) for \( k < n \) , and this shows by induction that \( {a}_{i, j} \in {\mathbb{Z}}_{K} \) for all \( i \) and \( j \). Finally, up to reversal of rows or columns \( A \) is a triangular matrix, and clearly \( \det \left( A\right) = \) \( {\left( -1\right) }^{n\left( {n - 1}\right) /2} \) is indeed a unit, proving our claim and the lemma.
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Yes
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Corollary 10.1.21. With the above notation,\n\n\[ \n{v}_{\mathfrak{P}}\left( {\mathfrak{D}\left( {L/K}\right) }\right) = {v}_{\mathfrak{P}}\left( {{f}^{\prime }\left( \pi \right) }\right) \n\]
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Proof. As in Section 10.1.1, set\n\n\[ \n\mathfrak{a} = \operatorname{Ann}\left( {{\mathbb{Z}}_{L},{\mathbb{Z}}_{K}\left\lbrack \pi \right\rbrack }\right) = \left\{ {x \in {\mathbb{Z}}_{K}/x{\mathbb{Z}}_{L} \subset {\mathbb{Z}}_{K}\left\lbrack \pi \right\rbrack }\right\} \n\]\n\nSince \( {\mathbb{Z}}_{K}\left\lbrack \pi \right\rbrack \) is \( \mathfrak{p} \) -maximal, we know that \( \mathfrak{p} \nmid \mathfrak{a} \).\n\nOn the other hand,\n\n\[ \nx \in \mathfrak{a} \Leftrightarrow x{\mathbb{Z}}_{L} \subset {\mathbb{Z}}_{K}\left\lbrack \pi \right\rbrack \; \Leftrightarrow x{\mathbb{Z}}_{L}/{f}^{\prime }\left( \pi \right) \subset {\mathbb{Z}}_{K}{\left\lbrack \pi \right\rbrack }^{ * } \n\]\n\n\[ \n\Leftrightarrow {\operatorname{Tr}}_{L/K}\left( {x{\mathbb{Z}}_{L}/{f}^{\prime }\left( \pi \right) }\right) \subset {\mathbb{Z}}_{K} \Leftrightarrow x/{f}^{\prime }\left( \pi \right) \in \mathfrak{D}{\left( L/K\right) }^{-1} \n\]\n\n\[ \n\Leftrightarrow x \in {f}^{\prime }\left( \pi \right) \mathfrak{D}{\left( L/K\right) }^{-1}. \n\]\n\nIt follows that \( \mathfrak{D}\left( {L/K}\right) = {f}^{\prime }\left( \pi \right) {\mathfrak{a}}^{-1} \), and since \( \mathfrak{p} \nmid \mathfrak{a} \), the result follows.
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Yes
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Theorem 10.1.22. Let \( L/K \) be a normal extension of number fields, set \( G = \operatorname{Gal}\left( {L/K}\right) \), let \( \mathfrak{P} \) be a prime ideal of \( L \), and let \( \mathfrak{p} \) be the prime ideal of \( K \) below \( \mathfrak{P} \) . The valuation at \( \mathfrak{P} \) of the relative different is given by the formula\n\n\[ \n{v}_{\mathfrak{P}}\left( {\mathfrak{D}\left( {L/K}\right) }\right) = \mathop{\sum }\limits_{{k \geq 0}}\left( {\left| {G}_{k}\right| - 1}\right) .\n\]
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Proof. By Lemma 10.1.10, if we set \( {K}^{\prime } = {L}^{I} \), the ideal \( \mathfrak{P} \) has the same ramification groups (for \( k \geq 0 \) ) in the extension \( L/{K}^{\prime } \) as in the extension \( L/K \), and \( {v}_{\mathfrak{P}}\left( {\mathfrak{D}\left( {L/{K}^{\prime }}\right) }\right) = {v}_{\mathfrak{P}}\left( {\mathfrak{D}\left( {L/K}\right) }\right) \) . We may thus assume that \( \mathfrak{p} \) is totally ramified in the extension \( L/K \), so that we can apply the results proved in that case.\n\nThus, let \( \pi \in \mathfrak{P} \smallsetminus {\mathfrak{P}}^{2} \) with minimal monic polynomial \( f\left( X\right) \in {\mathbb{Z}}_{K}\left\lbrack X\right\rbrack \) , so that we know that \( {\mathbb{Z}}_{K}\left\lbrack \pi \right\rbrack \) is a p-maximal order in \( {\mathbb{Z}}_{L} \) (by Lemma 10.1.2) and that \( {v}_{\mathfrak{P}}\left( {\mathfrak{D}\left( {L/K}\right) }\right) = {v}_{\mathfrak{P}}\left( {{f}^{\prime }\left( \pi \right) }\right) \) (by Corollary 10.1.21). Since \( f\left( \pi \right) = \) \( \mathop{\prod }\limits_{{\sigma \in G}}\left( {X - \sigma \left( \pi \right) }\right) \), we have\n\n\[ \n{f}^{\prime }\left( \pi \right) = \mathop{\prod }\limits_{{\sigma \in G\smallsetminus \left\{ {1}_{G}\right\} }}\left( {\pi - \sigma \left( \pi \right) }\right)\n\]\n\nso that\n\n\[ \n{v}_{\mathfrak{P}}\left( {\mathfrak{D}\left( {L/K}\right) }\right) = {v}_{\mathfrak{P}}\left( {{f}^{\prime }\left( \pi \right) }\right) = \mathop{\sum }\limits_{{\sigma \in G\smallsetminus \{ {1}_{G}\} }}{i}_{G}\left( \sigma \right)\n\]\n\nby definition of the function \( {i}_{G}\left( \sigma \right) \) .\n\nTo finish the proof, set \( {g}_{k} = \left| {G}_{k}\right| - 1 \) and let \( n \) be such that \( {G}_{n} = \left\{ {1}_{G}\right\} \) . If \( \sigma \in {G}_{k - 1} \smallsetminus {G}_{k} \), we have by definition \( {i}_{G}\left( \sigma \right) = k \) . It follows that\n\n\[ \n\mathop{\sum }\limits_{{\sigma \in G \smallsetminus \left\{ {1}_{G}\right\} }}{i}_{G}\left( \sigma \right) = \mathop{\sum }\limits_{{k = 0}}^{n}k\left( {{g}_{k - 1} - {g}_{k}}\right) = \mathop{\sum }\limits_{{k = 0}}^{n}{g}_{k}\left( {k + 1 - k}\right) = \mathop{\sum }\limits_{{k = 0}}^{n}{g}_{k},\n\]\n\nas was to be proved. Note that in general the intermediate result\n\n\[ \n{v}_{\mathfrak{P}}\left( {\mathfrak{D}\left( {L/K}\right) }\right) = \mathop{\sum }\limits_{{\sigma \in G\smallsetminus \left\{ {1}_{G}\right\} }}{i}_{G}\left( \sigma \right)\n\]\n\nthat we have found is only valid when \( \mathfrak{p} \) is totally ramified in \( L/K \) .
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Yes
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Proposition 10.1.23. Let \( L/K \) be a cyclic extension of number fields of prime degree \( \ell \), let \( \mathfrak{p} \) be a prime ideal of \( K \) above \( \ell \), and denote as usual by \( {G}_{k} \) the ramification groups of \( \mathfrak{P}/\mathfrak{p} \) for any prime ideal \( \mathfrak{P} \) of \( L \) above \( \mathfrak{p} \) . If \( {k}_{0} \) is the largest \( k \) such that \( {G}_{k} \neq \{ 1\} \), then \( {k}_{0} \leq \lfloor \ell e\left( {\mathfrak{p}/\ell }\right) /\left( {\ell - 1}\right) \rfloor .
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Proof. If \( \mathfrak{p} \) is unramified in \( L/K \), we have \( {G}_{0} = \{ 1\} \), so the inequality is trivial. Thus, since \( \ell \) is prime, we may assume that \( \mathfrak{p} \) is totally ramified in \( L/K \), so that \( \mathfrak{p}{\mathbb{Z}}_{L} = {\mathfrak{P}}^{\ell } \) . For simplicity set \( e = e\left( {\mathfrak{p}/\ell }\right) = {v}_{\mathfrak{p}}\left( \ell \right) \) . By Theorem 10.1.22, since \( \left| {G}_{k}\right| = \ell \) for \( k \leq {k}_{0} \) and \( \left| {G}_{k}\right| = 1 \) for \( k > {k}_{0} \), we have \( {v}_{\mathfrak{P}}\left( {\mathfrak{D}\left( {L/K}\right) }\right) = \left( {{k}_{0} + 1}\right) \left( {\ell - 1}\right) \) . Hence\n\n\[ \n{k}_{0} > \lfloor \ell e/\left( {\ell - 1}\right) \rfloor \Rightarrow {k}_{0} > \ell e/\left( {\ell - 1}\right) \Rightarrow \left( {{k}_{0} + 1}\right) \left( {\ell - 1}\right) > \ell e + \ell - 1 \n\]\n\n\[ \n\Rightarrow \left( {{k}_{0} + 1}\right) \left( {\ell - 1}\right) \geq \ell \left( {e + 1}\right) \Rightarrow {\mathfrak{P}}^{\ell \left( {e + 1}\right) } \mid \mathfrak{D}\left( {L/K}\right) \n\]\n\n\[ \n\Rightarrow {\mathfrak{p}}^{e + 1} \mid \mathfrak{D}\left( {L/K}\right) \Rightarrow {\mathfrak{p}}^{-\left( {e + 1}\right) }{\mathbb{Z}}_{L} \subset {\mathfrak{D}}^{-1}\left( {L/K}\right) \n\]\n\n\[ \n\Rightarrow {\operatorname{Tr}}_{L/K}\left( {{\mathfrak{p}}^{-\left( {e + 1}\right) }{\mathbb{Z}}_{L}}\right) \subset {\mathbb{Z}}_{K} \n\]\n\n\[ \n\Rightarrow {\mathfrak{p}}^{-\left( {e + 1}\right) }{\operatorname{Tr}}_{L/K}\left( {\mathbb{Z}}_{L}\right) \subset {\mathbb{Z}}_{K} \Rightarrow {\operatorname{Tr}}_{L/K}\left( {\mathbb{Z}}_{L}\right) \subset {\mathfrak{p}}^{e + 1} \n\]\n\n\[ \n\Rightarrow \ell = {\operatorname{Tr}}_{L/K}\left( 1\right) \in {\mathfrak{p}}^{e + 1} \Rightarrow e = {v}_{\mathfrak{p}}\left( \ell \right) \geq e + 1, \n\]\n\nand this contradiction proves the proposition.
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Yes
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Corollary 10.1.24. Let \( L/K \) be a cyclic extension of number fields of prime degree \( \ell \) and let \( \mathfrak{p} \) be a prime ideal of \( K \) above \( \ell \) . If \( \mathfrak{f} \) denotes the conductor of \( L/K \), then \( {v}_{\mathfrak{p}}\left( \mathfrak{f}\right) \leq \lfloor \ell e\left( {\mathfrak{p}/\ell }\right) /\left( {\ell - 1}\right) \rfloor + 1 \), and this upper bound is best possible.
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Proof. Indeed, we may assume that \( \mathfrak{p} \mid \mathfrak{f} \), so \( \mathfrak{p} \) is ramified, hence totally ramified in \( L/K \) . Thus, using the same notation as that of the proposition, we have \( {v}_{\mathfrak{p}}\left( {\mathfrak{d}\left( {L/K}\right) }\right) = {v}_{\mathfrak{P}}\left( {\mathfrak{D}\left( {L/K}\right) }\right) = \left( {\ell - 1}\right) \left( {{k}_{0} + 1}\right) \) . On the other hand, by Corollary 3.5.12 we have \( \mathfrak{d}\left( {L/K}\right) = {\mathfrak{f}}^{\ell - 1} \), so \( {v}_{\mathfrak{p}}\left( \mathfrak{f}\right) = {k}_{0} + 1 \), and the upper bound thus follows from Proposition 10.1.23. The fact that it is the best possible is an immediate consequence of Hecke's Theorem 10.2.9 (1), which we will prove later.
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No
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Proposition 10.1.26. Keep the above hypotheses and notation, and let \( \mathfrak{p} \) be a prime ideal of \( K \) . (1) The prime ideal \( \mathfrak{p} \) cannot be inert in \( N/K \) .
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Proof. Note first that if \( g \) is the number of prime ideals of \( N \) above \( \mathfrak{p} \) and if \( {\mathfrak{P}}_{N} \) is one of them, we have \( e\left( {{\mathfrak{P}}_{N}/\mathfrak{p}}\right) f\left( {{\mathfrak{P}}_{N}/\mathfrak{p}}\right) g = \left\lbrack {N : K}\right\rbrack = 2\ell \), hence the possibilities listed in the proposition are exhaustive. (1). This follows immediately from Corollary 10.1.7 since \( {D}_{\ell } \) is not a cyclic group.
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Yes
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Lemma 10.1.27. Keep the above hypotheses and notation. We have\n\n\[ \n{\mathcal{N}}_{L/K}\left( {\mathfrak{d}\left( {N/L}\right) }\right) = \mathfrak{d}\left( {{K}_{2}/K}\right) \n\]
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Proof. We are going to show that for every prime ideal \( \mathfrak{p} \) of \( K \), the \( \mathfrak{p} \) - adic valuations of both sides are equal. Thus, let \( \mathfrak{p} \) be a prime ideal of \( K \) , and assume first that \( \mathfrak{p} \nmid 2 \) . If \( \mathfrak{p} \mid \mathfrak{d}\left( {{K}_{2}/K}\right) \), then \( \mathfrak{p}{\mathbb{Z}}_{{K}_{2}} = {\mathfrak{P}}_{{K}_{2}}^{2} \), hence we are necessarily in cases (6) or (9) of the above proposition, since case (5) cannot occur. In case (9), \( \mathfrak{p} \) is totally ramified everywhere, and since \( \mathfrak{p} \mid \ell \) and \( \ell \neq 2 \), it is easily seen by applying Proposition 3.3.21 that \( {v}_{\mathfrak{p}}\left( {\mathfrak{d}\left( {N/L}\right) }\right) = \) \( {v}_{\mathfrak{p}}\left( {\mathfrak{d}\left( {{K}_{2}/K}\right) }\right) = 1 \)\n\nIn case (6), we have \( \mathfrak{p}{\mathbb{Z}}_{L} = {\mathfrak{P}}_{L,1}{\mathfrak{P}}_{L,2}^{2}\ldots {\mathfrak{P}}_{L,\left( {\ell + 1}\right) /2}^{2},{\mathfrak{P}}_{L,1} \) is ramified in \( N/L \), and the \( {\mathfrak{P}}_{L, i} \) are split in \( N/L \) for \( i > 1 \) . It follows that \( {\mathfrak{P}}_{L, i} \nmid \mathfrak{d}\left( {N/L}\right) \) for \( i > 1 \) . By Proposition 3.3.21 once again, we know that \( {v}_{\mathfrak{p}}\left( {\mathfrak{d}\left( {{K}_{2}/K}\right) }\right) = \) \( {v}_{{\mathfrak{P}}_{L,1}}\left( {\mathfrak{d}\left( {N/L}\right) }\right) = 1 \), and since \( {\mathfrak{P}}_{L,1} \) is of degree 1 above \( \mathfrak{p} \) and is the only prime ideal of \( L \) above \( \mathfrak{p} \) ramified in \( N/L \), we have\n\n\[ \n{v}_{\mathfrak{p}}\left( {\mathfrak{d}\left( {{K}_{2}/K}\right) }\right) = {v}_{\mathfrak{p}}\left( {{\mathcal{N}}_{L/K}\left( {\mathfrak{d}\left( {N/L}\right) }\right) }\right) = 1. \n\]\n\nConversely, assume that \( \mathfrak{p} \nmid \mathfrak{d}\left( {{K}_{2}/K}\right) \) . Then for any prime ideal \( {\mathfrak{P}}_{N} \) of \( N \) above \( \mathfrak{p}, e\left( {{\mathfrak{P}}_{N}/\mathfrak{p}}\right) \) is odd (it must be equal to 1 or \( \ell \) ), hence no prime ideal of \( L \) above \( \mathfrak{p} \) can ramify in \( N/L \) by transitivity of ramification indices. Thus \( {v}_{\mathfrak{p}}\left( {{\mathcal{N}}_{L/K}\left( {\mathfrak{d}\left( {N/L}\right) }\right) }\right) = {v}_{\mathfrak{p}}\left( {\mathfrak{d}\left( {{K}_{2}/K}\right) }\right) = 0 \) . We have thus proved that if \( \mathfrak{p} \nmid 2 \) (the \
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Yes
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Proposition 10.1.28. Keep the above hypotheses and notation. In particular, recall that \( \mathfrak{f} \) is an ideal of \( K \) such that \( \mathfrak{f}{\mathbb{Z}}_{{K}_{2}} \) is the conductor of \( N/{K}_{2} \) . Let \( \mathfrak{p} \) be a prime ideal of \( K \) .\n\n(1) We have\n\n\[ \mathfrak{d}\left( {L/K}\right) = \mathfrak{d}{\left( {K}_{2}/K\right) }^{\left( {\ell - 1}\right) /2}{\mathfrak{f}}^{\ell - 1} = {\left( \mathfrak{d}\left( {K}_{2}/K\right) {\mathfrak{f}}^{2}\right) }^{\left( {\ell - 1}\right) /2}. \]
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Proof. (1). By Theorem 2.5.1 we know that\n\n\[ \mathfrak{d}\left( {N/K}\right) = \mathfrak{d}{\left( L/K\right) }^{2}{\mathcal{N}}_{L/K}\left( {\mathfrak{d}\left( {N/L}\right) }\right) = \mathfrak{d}{\left( {K}_{2}/K\right) }^{\ell }{\mathcal{N}}_{{K}_{2}/K}\left( {\mathfrak{d}\left( {N/{K}_{2}}\right) }\right) \]\n\n\[ = \mathfrak{d}{\left( {K}_{2}/K\right) }^{\ell }{\mathcal{N}}_{{K}_{2}/K}\left( {\mathfrak{f}}_{2}^{\ell - 1}\right) = \mathfrak{d}{\left( {K}_{2}/K\right) }^{\ell }{\mathfrak{f}}^{2\left( {\ell - 1}\right) }, \]\n\nwhere we have used Corollary 3.5.12 and Proposition 10.1.25. By Lemma 10.1.27, we deduce that\n\n\[ \mathfrak{d}{\left( L/K\right) }^{2} = \mathfrak{d}{\left( {K}_{2}/K\right) }^{\ell - 1}{\mathfrak{f}}^{2\left( {\ell - 1}\right) }, \]\n\nproving (1).
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Yes
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Lemma 10.2.1 (Dirichlet's Character Independence Theorem). Let \( G \) be a group, let \( L \) be a field, and let \( {\chi }_{1},\ldots ,{\chi }_{m} \) be distinct characters of \( G \) with values in \( {L}^{ * } \) . The characters \( {\chi }_{i} \) are \( L \) -linearly independent, in other words a relation \( \mathop{\sum }\limits_{{1 \leq i \leq m}}{a}_{i}{\chi }_{i} = 0 \) for \( {a}_{i} \in L \) implies that \( {a}_{i} = 0 \) for all \( i \) .
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Proof. Assume that the characters are \( L \) -linearly dependent. Choose a dependence relation of minimal length, so that, up to reordering of the \( {\chi }_{i} \) ,\n\n\[ \forall h \in G\;\mathop{\sum }\limits_{{1 \leq i \leq n}}{a}_{i}{\chi }_{i}\left( h\right) = 0 \]\n\n(1)\n\nwith \( n \) minimal. For any \( g \in G \), we have for all \( h,\mathop{\sum }\limits_{{1 < i < n}}{a}_{i}{\chi }_{i}\left( {gh}\right) = 0 \) . Multiplying relation (1) by \( {\chi }_{1}\left( g\right) \) and subtracting, we obtain that for all \( g \) and \( h \) in \( G \) we have\n\n\[ \mathop{\sum }\limits_{{1 \leq i \leq n}}{a}_{i}\left( {{\chi }_{i}\left( g\right) - {\chi }_{1}\left( g\right) }\right) {\chi }_{i}\left( h\right) \]\n\nand since the first coefficient vanishes, this is a relation of length \( n - 1 \) between the characters. By the minimality of \( n \), this must be the trivial relation, and\n\nagain by minimality the \( {a}_{i} \) are nonzero, hence \( {\chi }_{i}\left( g\right) = {\chi }_{1}\left( g\right) \) for all \( i \leq n \) and all \( g \in G \) . Since the characters are distinct, this implies \( n = 1 \), hence \( {\chi }_{1} = 0 \) , which is absurd.
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Yes
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Corollary 10.2.2. Let \( K \) and \( L \) be fields, and let \( {\sigma }_{1},\ldots ,{\sigma }_{m} \) be distinct homomorphisms from \( K \) to \( L \) . Then the \( {\sigma }_{i} \) are \( L \) -linearly independent.
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Proof. Simply apply the preceding lemma to \( G = {K}^{ * } \) and to \( {\chi }_{i} \) equal to the restriction of \( {\sigma }_{i} \) to \( {K}^{ * } \) .
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No
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Lemma 10.2.3 (Noether’s Theorem). Let \( L/K \) be a normal extension with Galois group \( G \), and let \( \phi \) be a map from \( G \) to \( {L}^{ * } \) . We will say that \( \phi \) satisfies the cocycle condition if for all \( g, h \) in \( G \) we have\n\n\[ \phi \left( {gh}\right) = \phi \left( g\right) \cdot g\left( {\phi \left( h\right) }\right) . \]\n\nThen \( \phi \) satisfies the cocycle condition if and only if there exists \( \alpha \in {L}^{ * } \) such\n\nthat\n\[ \forall g \in G,\phi \left( g\right) = \frac{\alpha }{g\left( \alpha \right) }.\]
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Proof. If \( \phi \left( g\right) = \alpha /g\left( \alpha \right) \), we have\n\n\[ \phi \left( g\right) \cdot g\left( {\phi \left( h\right) }\right) = \frac{\alpha }{g\left( \alpha \right) }g\left( \frac{\alpha }{h\left( \alpha \right) }\right) = \frac{\alpha }{g\left( {h\left( \alpha \right) }\right) } = \phi \left( {gh}\right) ,\]\n\nso \( \phi \) satisfies the cocycle condition. Conversely, assume that \( \phi \) satisfies the cocycle condition. For \( x \in L \), set\n\n\[ \sigma \left( x\right) = \mathop{\sum }\limits_{{h \in G}}\phi \left( h\right) h\left( x\right) \]\n\nThen \( \sigma \) is an additive map from \( L \) to \( L \) . Applying Lemma 10.2.2 to the distinct homomorphisms \( h \in G \), we deduce that \( \sigma \) is not identically zero (recall that \( \phi \left( h\right) \neq 0 \) for all \( h \) by assumption). Hence, let \( x \in L \) such that \( \alpha = \sigma \left( x\right) \neq 0 \) . We have\n\n\[ g\left( \alpha \right) = g\left( {\mathop{\sum }\limits_{{h \in G}}\phi \left( h\right) h\left( x\right) }\right) = \mathop{\sum }\limits_{{h \in G}}g\left( {\phi \left( h\right) }\right) {gh}\left( x\right) ; \]\n\nhence by the cocycle condition\n\n\[ g\left( \alpha \right) = \phi {\left( g\right) }^{-1}\mathop{\sum }\limits_{{h \in G}}\phi \left( {gh}\right) {gh}\left( x\right) = \phi {\left( g\right) }^{-1}\mathop{\sum }\limits_{{h \in G}}\phi \left( h\right) h\left( x\right) = \phi {\left( g\right) }^{-1}\alpha ,\]\n\nproving the lemma.
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Yes
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Lemma 10.2.4 (Hilbert’s Theorem 90). Let \( L/K \) be a cyclic extension with Galois group \( G \) generated by an element \( \sigma \) . Then \( \alpha \in L \) is an element of relative norm equal to 1 if and only if there exists \( \beta \in L \) such that \( \alpha = \beta /\sigma \left( \beta \right) \) .
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Proof. Clearly \( {\mathcal{N}}_{L/K}\left( {\sigma \left( \beta \right) }\right) = {\mathcal{N}}_{L/K}\left( \beta \right) \), hence the relative norm of \( \beta /\sigma \left( \beta \right) \) is equal to 1 . Conversely, assume that \( {\mathcal{N}}_{L/K}\left( \alpha \right) = 1 \) . Let \( n = \left| G\right| \) , and for \( 0 \leq i < n \), set\n\n\[ \phi \left( {\sigma }^{i}\right) = \mathop{\prod }\limits_{{0 \leq k < i}}{\sigma }^{k}\left( \alpha \right) \]\n\nI claim that \( \phi \) satisfies the cocycle condition. Indeed,\n\n\[ \phi \left( {\sigma }^{i}\right) {\sigma }^{i}\left( {\phi \left( {\sigma }^{j}\right) }\right) = \mathop{\prod }\limits_{{0 \leq k < i}}{\sigma }^{k}\left( \alpha \right) \mathop{\prod }\limits_{{i \leq k < i + j}}{\sigma }^{k}\left( \alpha \right) = \mathop{\prod }\limits_{{0 \leq k < i + j}}{\sigma }^{k}\left( \alpha \right) . \]\n\nHence if \( i + j < n \), this is equal to \( \phi \left( {\sigma }^{i + j}\right) \), while if \( i + j \geq n \), this is equal to\n\n\[ \mathop{\prod }\limits_{{0 \leq k < n}}{\sigma }^{k}\left( \alpha \right) \mathop{\prod }\limits_{{0 \leq k < i + j - n}}{\sigma }^{k}\left( \alpha \right) = {\mathcal{N}}_{L/K}\left( \alpha \right) \phi \left( {\sigma }^{i + j}\right) = \phi \left( {\sigma }^{i + j}\right) \]\n\nonce again, since \( {\mathcal{N}}_{L/K}\left( \alpha \right) = 1 \) . Thus \( \phi \) satisfies the cocycle condition. Hence by Lemma 10.2.3, there exists \( \beta \in {L}^{ * } \) such that \( \phi \left( {\sigma }^{i}\right) = \beta /{\sigma }^{i}\left( \beta \right) \) for all \( i \) , and in particular \( \alpha = \phi \left( \sigma \right) = \beta /\sigma \left( \beta \right) \), as desired. Note that by choosing \( \gamma = {\sigma }^{n - 1}\left( \beta \right) \), we would get \( \alpha = \sigma \left( \gamma \right) /\gamma \) .
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Yes
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Lemma 10.2.6. Any homomorphism from \( G \) to \( {\mu }_{n} \) is of the form\n\n\[ \sigma \mapsto < \sigma ,\bar{b} > \]\n\nfor some \( \bar{b} \in B \) .
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To prove the lemma, let \( \phi \) be a homomorphism from \( G \) to \( {\mathbf{\mu }}_{n} \) . Recall that \( {\mathbf{\mu }}_{n} \subset K \), hence that any element of \( G = \operatorname{Gal}\left( {L/K}\right) \) fixes \( {\mathbf{\mu }}_{n} \) pointwise. Thus, for all \( \sigma \) and \( \tau \) in \( G \) we have\n\n\[ \phi \left( {\sigma \tau }\right) = \phi \left( \sigma \right) \phi \left( \tau \right) = \phi \left( \sigma \right) \sigma \left( {\phi \left( \tau \right) }\right) . \]\n\nThus the map \( \phi \) considered as a map from \( G \) to \( {L}^{ * } \) satisfies the conditions of Noether’s theorem (Lemma 10.2.3); therefore there exists \( \alpha \in {L}^{ * } \) such that \( \phi \left( \sigma \right) = \sigma \left( \alpha \right) /\alpha \) for all \( \sigma \in G \) . Since we also have \( \phi {\left( \sigma \right) }^{n} = 1 \), we obtain \( \sigma {\left( \alpha \right) }^{n} = {\alpha }^{n} \) for all \( \sigma \in G \) . Hence by Galois theory \( {\alpha }^{n} \in {K}^{ * } \), and so \( {\alpha }^{n} \in \) \( {L}^{*n} \cap {K}^{ * } \) . It is clear that \( \bar{b} = \overline{{\alpha }^{n}} \) is such that \( \phi \left( \sigma \right) = < \sigma ,\bar{b} > \), proving the lemma and hence the theorem.
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Yes
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Corollary 10.2.7. Let \( K \) be a number field and \( n \geq 1 \) be an integer such that \( {\zeta }_{n} \in K \) .\n\n(1) An extension \( L/K \) is a cyclic extension of degree \( n \) if and only if there exists \( \alpha \in {K}^{ * } \) such that \( \bar{\alpha } \) is exactly of order \( n \) in \( {K}^{ * }/{K}^{*n} \) and such that \( L = K\left( \sqrt[n]{\alpha }\right) .\n\n(2) The cyclic extensions \( {L}_{1} = K\left( \sqrt[n]{{\alpha }_{1}}\right) \) and \( {L}_{2} = K\left( \sqrt[n]{{\alpha }_{2}}\right) \) are \( K \) - isomorphic if and only if there exists an integer \( j \) coprime to \( n \) and \( \gamma \in {K}^{ * } \), such that \( {\alpha }_{2} = {\alpha }_{1}^{j}{\gamma }^{n} \).
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Proof. (1) Let \( L/K \) be a cyclic extension of degree \( n \) . By Theorem 10.2.5, there exists a subgroup \( B \) of \( {K}^{ * }/{K}^{*n} \) such that \( L = {K}_{B} \) and \( B \simeq \) \( \operatorname{Gal}\left( {L/K}\right) \simeq \mathbb{Z}/n\mathbb{Z} \) . If \( \bar{\alpha } \) is a generator of \( B \), it is clear that \( {K}_{B} = K\left( \sqrt[n]{\alpha }\right) \) . Conversely, if \( L = K\left( \sqrt[n]{\alpha }\right) \) with \( \alpha \in {K}^{ * } \), then \( L/K \) is a cyclic extension of degree \( n \) if and only if \( \bar{\alpha } \) generates a subgroup of order \( n \) of \( {K}^{ * }/{K}^{*n} \) .\n\n(2) Let \( \phi \) be a \( K \) -isomorphism from \( {L}_{1} \) to \( {L}_{2} \), and let \( {B}_{1} \) and \( {B}_{2} \) be the subgroups of \( {K}^{ * }/{K}^{*n} \) corresponding to \( {L}_{1} \) and \( {L}_{2} \), respectively. If \( z = \) \( \phi \left( \sqrt[n]{{\alpha }_{1}}\right) \), we thus have \( {z}^{n} \in {L}_{2}^{*n} \cap {K}^{ * } \), hence \( \overline{{z}^{n}} \in {B}_{2} \), and since \( \phi \) is a \( K \) - isomorphism, we have \( \overline{{z}^{n}} = \overline{\phi \left( {\alpha }_{1}\right) } = \overline{{\alpha }_{1}} \in {B}_{2} \), and similarly \( \overline{{\alpha }_{2}} \in {B}_{1} \) . Since \( \overline{{\alpha }_{i}} \) is a generator of \( {B}_{i} \), it follows that \( {\alpha }_{2} = {\alpha }_{1}^{j}{\gamma }^{n} \) and \( {\alpha }_{1} = {\alpha }_{2}^{k}{\delta }^{n} \) . Hence \( {\alpha }_{1}^{{kj} - 1} \in {K}^{*n} \), and since \( {\alpha }_{1} \) is exactly of order \( n \) in \( {K}^{ * }/{K}^{*n} \), this implies that \( {kj} \equiv 1\left( {\;\operatorname{mod}\;n}\right) \), hence \( j \) is coprime to \( n \), as claimed.
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Yes
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Theorem 10.2.9. Let \( K \) be a number field, \( \ell \) a prime number such that \( {\zeta }_{\ell } \in K \), and \( L = K\left( \sqrt[\ell ]{\alpha }\right) \), where \( \alpha \in {K}^{ * } \smallsetminus {K}^{*\ell } \) . If \( \mathfrak{p} \) is a prime ideal of \( {\mathbb{Z}}_{K} \) , we set\n\n\[ e\left( {\mathfrak{p}/\ell }\right) = {v}_{\mathfrak{p}}\left( \ell \right) = \left( {\ell - 1}\right) {v}_{\mathfrak{p}}\left( {1 - {\zeta }_{\ell }}\right) \]\n\nso that \( e\left( {\mathfrak{p}/\ell }\right) \) is the absolute ramification index of \( \mathfrak{p} \) if \( \mathfrak{p} \) is above \( \ell \) and 0 otherwise, and we also set\n\n\[ z\left( {\mathfrak{p},\ell }\right) = \ell \frac{e\left( {\mathfrak{p}/\ell }\right) }{\ell - 1} + 1 = \ell {v}_{\mathfrak{p}}\left( {1 - {\zeta }_{\ell }}\right) + 1. \]\n\n(1) Assume that \( \ell \nmid {v}_{\mathfrak{p}}\left( \alpha \right) \) . Then \( \mathfrak{p} \) is totally ramified in \( L/K \) and\n\n\[ {v}_{\mathfrak{p}}\left( {\mathfrak{d}\left( {L/K}\right) }\right) = \ell - 1 + \ell e\left( {\mathfrak{p}/\ell }\right) = \left( {\ell - 1}\right) z\left( {\mathfrak{p},\ell }\right) . \]\n\nIn particular, \( {v}_{\mathfrak{p}}\left( {\mathfrak{d}\left( {L/K}\right) }\right) = \ell - 1 \) if \( \ell \nmid {v}_{\mathfrak{p}}\left( \alpha \right) \) and \( \mathfrak{p} \nmid \ell \) .
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Proof. To simplify notation, write \( \zeta \) for \( {\zeta }_{\ell } \) . For (1), assume that \( \ell \nmid {v}_{\mathfrak{p}}\left( \alpha \right) \) . Let \( \pi \) be a uniformizer of \( \mathfrak{p} \) in \( K \) . There exist integers \( x \) and \( y \) such that \( x\ell + y{v}_{\mathfrak{p}}\left( \alpha \right) = 1 \) . Thus, if \( \beta = {\alpha }^{y}{\pi }^{x\ell } \) we have \( {v}_{\mathfrak{p}}\left( \beta \right) = 1 \) . Furthermore, since \( \left( {y,\ell }\right) = 1 \), by Corollary 10.2.7 we have \( K\left( \sqrt[\ell ]{\beta }\right) = K\left( \sqrt[\ell ]{\alpha }\right) \) . Thus, replacing \( \alpha \) by \( \beta \), we may assume that
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Yes
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Lemma 10.2.10. Assume that \( {v}_{\mathfrak{p}}\left( \alpha \right) = 0 \) and \( \mathfrak{p} \mid \ell \) . The congruence \( {x}^{\ell } \equiv \alpha \) \( \left( {\;\operatorname{mod}\;{\mathfrak{p}}^{k}}\right) \) has a solution in \( K \) for \( k = z\left( {\mathfrak{p},\ell }\right) \) if and only if it has a solution for all \( k \geq z\left( {\mathfrak{p},\ell }\right) \) .
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Proof. Assume that the congruence has a solution \( x \) for some \( k \geq z\left( {\mathfrak{p},\ell }\right) \) . We apply a Newton-Hensel iteration: in other words, we set\n\n\[ \n{x}_{1} = x + y\;\text{ with }\;y = - \frac{{x}^{\ell } - \alpha }{\ell {x}^{\ell - 1}}.\n\]\n\nThen \( {v}_{\mathfrak{p}}\left( y\right) \geq k - e > 0 \) . On the other hand,\n\n\[ \n{x}_{1}^{\ell } - \alpha = {x}^{\ell } - \alpha + \ell {x}^{\ell - 1}y + \mathop{\sum }\limits_{{j \geq 2}}\left( \begin{array}{l} \ell \\ j \end{array}\right) {x}^{\ell - j}{y}^{j} = \mathop{\sum }\limits_{{2 \leq j \leq \ell - 1}}\left( \begin{array}{l} \ell \\ j \end{array}\right) {x}^{\ell - j}{y}^{j} + {y}^{\ell }.\n\]\n\nBy our assumption on \( k \), for \( 2 \leq j \leq \ell - 1 \) we have\n\n\[ \n{v}_{\mathfrak{p}}\left( {\left( \begin{array}{l} \ell \\ j \end{array}\right) {x}^{\ell - j}{y}^{j}}\right) = e + j{v}_{\mathfrak{p}}\left( y\right) \geq e + 2\left( {k - e}\right) = {2k} - e \geq k + 1.\n\]\n\nAlso, \( {v}_{\mathfrak{p}}\left( {y}^{\ell }\right) = \ell {v}_{\mathfrak{p}}\left( y\right) \geq \ell \left( {k - e}\right) \), and the inequality \( \ell \left( {k - e}\right) \geq k + 1 \) is equivalent to \( k \geq 1/\left( {\ell - 1}\right) + e\ell /\left( {\ell - 1}\right) \), hence to \( k \geq z\left( {\mathfrak{p},\ell }\right) \), which is true by assumption. Thus all the terms have a \( \mathfrak{p} \) -adic valuation greater than or equal to \( k + 1 \) ; hence \( {x}_{1}^{\ell } \equiv \alpha \left( {\;\operatorname{mod}\;{\mathfrak{p}}^{k + 1}}\right) \), proving the lemma.
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Yes
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Lemma 10.2.11. With this notation, we have \( a \geq 1 \) and \( \ell \nmid a \) .
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Proof. This follows immediately from Proposition 10.2.13, which we will prove in the next section.
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No
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Corollary 10.2.12. Let \( K \) be a number field, \( \ell \) a prime number such that \( {\zeta }_{\ell } \in K \), and \( L = K\left( \sqrt[\ell ]{\alpha }\right) \), where \( \alpha \in {K}^{ * } \smallsetminus {K}^{*\ell } \) . Let \( \mathfrak{p} \) be a prime ideal of \( {\mathbb{Z}}_{K} \) . Then we have the following results.\n\n(1) The ideal \( \mathfrak{p} \) is unramified in \( L/K \) if and only if \( \ell \mid {v}_{\mathfrak{p}}\left( \alpha \right) \), and in addition, either \( \mathfrak{p} \nmid \ell \) or \( \mathfrak{p} \mid \ell \) and the congruence\n\n\[ \n{x}^{\ell } \equiv \alpha \left( {\;\operatorname{mod}\;{\mathfrak{p}}^{z\left( {\mathfrak{p},\ell }\right) - 1 + {v}_{\mathfrak{p}}\left( \alpha \right) }}\right)\n\]\n\nhas a solution in \( K \) .\n\n(2) The ideal \( \mathfrak{p} \) is ramified in \( L/K \) if and only if either \( \ell \nmid {v}_{\mathfrak{p}}\left( \alpha \right) \) or if \( \ell \mid {v}_{\mathfrak{p}}\left( \alpha \right) \) and the congruence (1) has no solution in \( K \) .\n\n(3) We always have \( \left( {\ell - 1}\right) \mid {v}_{\mathfrak{p}}\left( {\mathfrak{d}\left( {L/K}\right) }\right) \), and \( {v}_{\mathfrak{p}}\left( {\mathfrak{d}\left( {L/K}\right) }\right) = \ell - 1 \) if and only if \( \ell \nmid {v}_{\mathfrak{p}}\left( \alpha \right) \) and \( \mathfrak{p} \nmid \ell \) .
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Proof. This is an immediate consequence of Theorem 10.2.9.
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No
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Proposition 10.2.13. Keep the above notation and hypotheses, and let \( \pi \) be a uniformizer of \( \mathfrak{p} \) . Let \( k \) be an integer such that \( 1 \leq k \leq z\left( {\mathfrak{p},\ell }\right) - 1 = \) \( \ell e\left( {\mathfrak{p}/\ell }\right) /\left( {\ell - 1}\right) \), and let \( {x}_{k - 1} \in {\mathbb{Z}}_{K} \) be such that \( {x}_{k - 1}^{\ell } \equiv \alpha \left( {\;\operatorname{mod}\;{\mathfrak{p}}^{k - 1}}\right) \) . The congruence \( {x}_{k}^{\ell } \equiv \alpha \left( {\;\operatorname{mod}\;{\mathfrak{p}}^{k}}\right) \) is soluble if and only if one of the following two conditions holds:\n\n(1) \( {v}_{\mathfrak{p}}\left( {{x}_{k - 1}^{\ell } - \alpha }\right) \geq k \), in which case we can take \( {x}_{k} = {x}_{k - 1} \) ;\n\n(2) \( {v}_{\mathfrak{p}}\left( {{x}_{k - 1}^{\ell } - \alpha }\right) = k - 1 \) and \( k \equiv 1\left( {\;\operatorname{mod}\;\ell }\right) \), in which case we can take\n\n\( {x}_{k} = {x}_{k - 1} + {\pi }^{\left( {k - 1}\right) /\ell }y \), where \( y \) is a solution of the congruence \( {y}^{\ell } \equiv \)\n\n\( \left( {\alpha - {x}_{k - 1}^{\ell }}\right) /{\pi }^{k - 1}{\;(\operatorname{mod}\;{}^{ * }\mathfrak{p})}. \)
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Proof. Assume first that the congruence modulo \( {\mathfrak{p}}^{k} \) is satisfied. Since \( {\mathbb{Z}}_{K}/\mathfrak{p} \) is a perfect field of characteristic \( \ell \), the map \( x \mapsto {x}^{\ell } \) is a bijection (hence an injection) from \( {\mathbb{Z}}_{K}/\mathfrak{p} \) to itself, so if we write \( {x}_{k} = {x}_{k - 1} + u \), we know that \( u \in \mathfrak{p} \) . If \( u = 0 \) - in other words, if \( {v}_{\mathfrak{p}}\left( {{x}_{k - 1}^{\ell } - \alpha }\right) \geq k \) - we are in case (1). Thus, assume that \( {v}_{\mathfrak{p}}\left( {{x}_{k - 1}^{\ell } - \alpha }\right) = k - 1 \) .\n\nBy the binomial theorem we can write\n\n\[ \n{x}_{k}^{\ell } - \alpha = {x}_{k - 1}^{\ell } - \alpha + s + {u}^{\ell }\n\]\n\nwith\n\n\[ \ns = \mathop{\sum }\limits_{{1 \leq j \leq \ell - 1}}\left( \begin{array}{l} \ell \\ j \end{array}\right) {x}_{k - 1}^{\ell - j}{u}^{j}\n\]\n\nand in particular \( {v}_{\mathfrak{p}}\left( s\right) = e\left( {\mathfrak{p}/\ell }\right) + {v}_{\mathfrak{p}}\left( u\right) \) .\n\nIf \( {v}_{\mathfrak{p}}\left( u\right) \geq e\left( {\mathfrak{p}/\ell }\right) /\left( {\ell - 1}\right) \), we have \( {v}_{\mathfrak{p}}\left( {u}^{\ell }\right) \geq z\left( {\mathfrak{p},\ell }\right) - 1 \) and \( {v}_{\mathfrak{p}}\left( s\right) \geq \) \( z\left( {\mathfrak{p},\ell }\right) - 1 \), hence since \( k \leq z\left( {\mathfrak{p},\ell }\right) - 1 \) we have \( {v}_{\mathfrak{p}}\left( {{x}_{k}^{\ell } - \alpha }\right) = {v}_{\mathfrak{p}}\left( {{x}_{k - 1}^{\ell } - }\right. \) \( \alpha ) = k - 1 \), which is a contradiction. Thus if we are not in case (1), we have \( 1 \leq {v}_{\mathfrak{p}}\left( u\right) < e\left( {\mathfrak{p}/\ell }\right) /\left( {\ell - 1}\right) \) . In this case we have \( {v}_{\mathfrak{p}}\left( {s + {u}^{\ell }}\right) = \ell {v}_{\mathfrak{p}}\left( u\right) \) ; hence a necessary condition for the solubility of the congruence modulo \( {\mathfrak{p}}^{k} \) is that \( \ell {v}_{\mathfrak{p}}\left( u\right) = {v}_{\mathfrak{p}}\left( {{x}_{k - 1}^{\ell } - \alpha }\right) = k - 1 \), that is, \( k \equiv 1\left( {\;\operatorname{mod}\;\ell }\right) \), in which case we must choose \( u \) such that \( {v}_{\mathfrak{p}}\left( u\right) = \left( {k - 1}\right) /\ell \) . Since \( k \equiv 1\left( {\;\operatorname{mod}\;\ell }\right) \) and \( k \leq z\left( {\mathfrak{p},\ell }\right) - 1 = \ell e\left( {\mathfrak{p}/\ell }\right) /\left( {\ell - 1}\right) \), we have \( k \leq \ell e\left( {\mathfrak{p}/\ell }\right) /\left( {\ell - 1}\right) - \ell + 1 \) ; hence \( e\left( {\mathfrak{p}/\ell }\right) \geq \left( {\ell - 1}\right) \left( {\left( {k - 1}\right) /\ell }\right) + \ell - 1 \) . Thus\n\n\[ \n{v}_{\mathfrak{p}}\left( s\right) = e\left( {\mathfrak{p}/\ell }\right) + {v}_{\mathfr
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Yes
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(1) The function \( \gamma \left( s\right) \) is a meromorphic function of \( s \) .
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Proof. It is clear from the definition that \( \gamma \left( s\right) \) is a meromorphic function whose poles are the complex numbers \( s \) such that \( {a}_{i}s + {b}_{i} = - k \) for some nonnegative integer \( k \) -in other words, the numbers \( s = - \left( {{b}_{i} + k}\right) /{a}_{i} \) . Since the \( {a}_{i} \) are positive real numbers, we have \( \operatorname{Re}\left( s\right) = - \left( {\operatorname{Re}\left( {b}_{i}\right) + k}\right) /{a}_{i} \), so we can take \( {\sigma }_{0} = \mathop{\max }\limits_{i}\left( {-\operatorname{Re}\left( {b}_{i}\right) /{a}_{i}}\right) \) . In addition, for each \( i \) there are only a finite number of values of the integer \( k \) such that \( {\sigma }_{1} < - \left( {\operatorname{Re}\left( {b}_{i}\right) + k}\right) /{a}_{i} < {\sigma }_{2} \), so this proves (1), (2), and (3).
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Yes
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