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Proposition 2.4.13. Let \( \mathbb{F} = {\mathbb{F}}_{q} \) be some finite field and let \( n \geq 1 \) .\n\n(1) In \( \mathbb{F}\left\lbrack X\right\rbrack \) we have the decomposition into irreducibles\n\n\[{\Omega }_{{q}^{n}}\left( X\right) = {X}^{{q}^{n}} - X = \mathop{\prod }\limits_{\substack{{P\text{ monic irred... | Proof. (1). Since \( {\mathbb{F}}_{q} \) is perfect, irreducible polynomials have only simple roots in \( \overline{{\mathbb{F}}_{q}} \) (see Proposition 3.1.1). Monic irreducible polynomials being pairwise coprime, it follows that both sides of the equation are polynomials with only simple roots in \( \overline{{\math... | Yes |
Corollary 2.4.14. Let \( p\\left( n\\right) \) be the number of monic irreducible polynomials of degree \( n \) in \( {\\mathbb{F}}_{q}\\left\\lbrack X\\right\\rbrack \) . Then \( p\\left( n\\right) \\geq 1 \) and \( p\\left( n\\right) \) is given by the explicit formula\n\n\[ p\\left( n\\right) = \\frac{1}{n}\\mathop{... | Proof. If \( x \\in {\\mathbb{F}}_{{q}^{n}} \) generates \( {\\mathbb{F}}_{{q}^{n}} \) over \( {\\mathbb{F}}_{q} \), then the degree of the minimal polynomial of \( x \) is equal to \( n \) ; hence \( p\\left( n\\right) \\geq 1 \) . Taking degrees in the decomposition of \( {X}^{{q}^{n}} - X \) given in the above propo... | Yes |
Lemma 2.5.1. (1) For \( 0 \leq m < q - 1 \) we have \( \mathop{\sum }\limits_{{a \in {\mathbb{F}}_{q}}}{a}^{m} = 0 \) . (2) For any \( m \in \mathbb{Z} \) we have \[ \mathop{\sum }\limits_{{a \in {\mathbb{F}}_{q}^{ * }}}{a}^{m} = \left\{ \begin{array}{ll} 0 & \text{ when }\left( {q - 1}\right) \nmid m, \\ - 1 & \text{ ... | Proof. The result is clear for \( m = 0 \) since \( p \mid q \), so assume \( 0 < m < q - 1 \) . It is clear that the map \( a \mapsto {a}^{m} \) is a group homomorphism \( \chi \) from \( {\mathbb{F}}_{a}^{ * } \) to \( {\mathbb{F}}_{q}^{ * } \) . Furthermore, by Corollary 2.4.3 we know that \( {\mathbb{F}}_{q}^{ * } ... | No |
Theorem 2.5.2 (Chevalley-Warning). (1) Let \( {\left( {P}_{i}\left( \underline{X}\right) \right) }_{1 \leq i \leq r} \) be a family of \( r \) polynomials in \( {\mathbb{F}}_{q}\left\lbrack {{X}_{1},\ldots ,{X}_{n}}\right\rbrack \) of respective total degrees \( {d}_{i} \), and let\n\n\[V = \left\{ {\left( {{a}_{1},\ld... | Proof. Define \( P\left( \underline{X}\right) = \mathop{\prod }\limits_{{1 \leq i \leq r}}\left( {1 - {P}_{i}{\left( \underline{X}\right) }^{q - 1}}\right) \) . Since \( {a}^{q - 1} = 1 \) when \( a \in \) \( {\mathbb{F}}_{q}^{ * } \), it is clear that if \( \underline{A} = \left( {{a}_{1},\ldots ,{a}_{n}}\right) \in V... | Yes |
Proposition 2.5.4. Let \( b \in {\mathbb{F}}_{q} \) be fixed. The map \[ {\psi }_{b} : \;x \mapsto {\zeta }_{p}^{{\operatorname{Tr}}_{{\mathbb{F}}_{q}/{\mathbb{F}}_{p}}\left( {bx}\right) } \] is an additive character of \( {\mathbb{F}}_{q} \). Furthermore, the map \( b \mapsto {\psi }_{b} \) is a group isomorphism from... | Proof. The first statement is clear by linearity of the trace and the fact that we are in characteristic \( p \). For the second statement, we note that the map \( b \mapsto {\psi }_{b} \) is clearly a group homomorphism from \( {\mathbb{F}}_{q} \) to \( \widehat{{\mathbb{F}}_{q}} \), which are two groups with the same... | Yes |
Lemma 2.5.6. For \( b \neq 0 \) and any a we have \( \tau \left( {\chi ,{\psi }_{ab}}\right) = \chi {\left( b\right) }^{-1}\tau \left( {\chi ,{\psi }_{a}}\right) \) , and in particular \( \tau \left( {\chi ,{\psi }_{b}}\right) = \chi {\left( b\right) }^{-1}\tau \left( {\chi ,{\psi }_{1}}\right) \) . | Proof. This is the analogue of Proposition 2.1.39 and is proved in the same way:\n\n\[ \tau \left( {\chi ,{\psi }_{ab}}\right) = \mathop{\sum }\limits_{{x \in {\mathbb{F}}_{q}^{ * }}}\chi \left( x\right) {\psi }_{a}\left( {xb}\right) = \mathop{\sum }\limits_{{y \in {\mathbb{F}}_{q}^{ * }}}\chi \left( {y{b}^{-1}}\right)... | Yes |
Lemma 2.5.8. Let \( \psi \) be a nontrivial additive character.\n\n(1) We have \( \tau \left( {\varepsilon ,\psi }\right) = - 1 \) .\n\n(2) For any character \( \chi \) we have\n\n\[ \tau \left( {{\chi }^{-1},\psi }\right) = \chi \left( {-1}\right) \overline{\tau \left( {\chi ,\psi }\right) }.\]\n\n(3) If \( b \in {\ma... | Proof. We have \( \tau \left( {\varepsilon ,\psi }\right) = - \psi \left( 0\right) + \mathop{\sum }\limits_{{a \in {\mathbb{F}}_{q}}}\psi \left( a\right) = - 1 \) by Proposition 2.1.20, proving (1). For (2), note that \( 1 = \psi \left( 0\right) = \psi \left( x\right) \psi \left( {-x}\right) \) ; hence \( \psi \left( {... | Yes |
Proposition 2.5.9. Let \( \psi \) be a nontrivial additive character.\n\n(1) If \( \chi \) is a nontrivial multiplicative character then \( \left| {\tau \left( {\chi ,\psi }\right) }\right| = {q}^{1/2} \). | Proof. Setting \( z = x{y}^{-1} \), we have\n\n\[ \n{\left| \tau \left( \chi ,\psi \right) \right| }^{2} = \tau \left( {\chi ,\psi }\right) \overline{\tau \left( {\chi ,\psi }\right) } = \mathop{\sum }\limits_{{x, y \in {\mathbb{F}}_{q}^{ * }}}\chi \left( x\right) \bar{\chi }\left( y\right) \psi \left( x\right) \bar{\p... | Yes |
Corollary 2.5.10. Let \( \psi \) be a nontrivial additive character and let \( b \in {\mathbb{F}}_{q}^{ * } \) . Then \[ \left| {\mathop{\sum }\limits_{{x \in {\mathbb{F}}_{q}}}\psi \left( {b{x}^{m}}\right) }\right| \leq \left( {\gcd \left( {m, q - 1}\right) - 1}\right) {q}^{1/2} \leq \left( {m - 1}\right) {q}^{1/2}. \... | Proof. Set \( d = \gcd \left( {m, q - 1}\right) \) . By Corollary 2.4.4 we have \[ \mathop{\sum }\limits_{{x \in {\mathbb{F}}_{q}}}\psi \left( {b{x}^{m}}\right) = 1 + d\mathop{\sum }\limits_{{y \in {\mathbb{F}}_{q}^{*d}}}\psi \left( {by}\right) . \] By the orthogonality relations in the group \( {\mathbb{F}}_{q}^{ * }/... | Yes |
Proposition 2.5.11. Let \( a, b \), and \( c \) be nonzero elements of \( {\mathbb{F}}_{q} \), let \( m \geq {\mathbb{Z}}_{ \geq 1} \) , and set \( d = \gcd \left( {m, q - 1}\right) \) . (1) The number \( N \) of solutions \( \left( {x, y, z}\right) \in {\mathbb{F}}_{q}^{3} \) of the equation \( a{x}^{m} + b{y}^{m} + c... | Proof. By orthogonality, denoting as usual by \( {\psi }_{0} \) the trivial additive character, we have \[ {qN} = \mathop{\sum }\limits_{{\left( {x, y, z}\right) \in {\mathbb{F}}_{q}^{3}}}\mathop{\sum }\limits_{\psi }\psi \left( {a{x}^{m} + b{y}^{m} + c{z}^{m}}\right) \] \[ = {q}^{3} + \mathop{\sum }\limits_{{\psi \neq... | Yes |
Lemma 2.5.13. For \( a \neq 0 \) we have\n\n\[ \n{J}_{k}\left( {{\chi }_{1},\ldots ,{\chi }_{k};a}\right) = \left( {{\chi }_{1}\cdots {\chi }_{k}}\right) \left( a\right) {J}_{k}\left( {{\chi }_{1},\ldots ,{\chi }_{k}}\right) ,\n\]\n\nwhile (abbreviating as above \( {J}_{k}\left( {{\chi }_{1},\ldots ,{\chi }_{k};0}\righ... | Proof. The formula for \( a \neq 0 \) is clear by setting \( {y}_{k} = {x}_{k}/a \), so assume \( a = 0 \) . If all the \( {\chi }_{j} \) are equal to \( \varepsilon \) then \( {J}_{k}\left( 0\right) \) is equal to the number of \( \left( {{x}_{1},\ldots ,{x}_{k}}\right) \in {\mathbb{F}}_{q}^{k} \) such that \( {x}_{1}... | Yes |
Proposition 2.5.14. Let \( \psi \) be a nontrivial additive character and let \( {\chi }_{1} \) , \( \ldots ,{\chi }_{k} \) be multiplicative characters of \( {\mathbb{F}}_{q} \) . Denote by \( t \) the number of such \( {\chi }_{i} \) equal to the trivial character \( \varepsilon \) . (1) If \( t = k \) then \( {J}_{k... | Proof. (1) and (2). For \( k = 1 \) the result is trivial, so assume that \( k \geq 2 \) . We can write \[ {J}_{k}\left( {{\chi }_{1,\ldots ,{\chi }_{k}}\right) = \mathop{\sum }\limits_{{{x}_{1},\ldots ,{x}_{k - 1} \in {\mathbb{F}}_{q}}}{\chi }_{1}\left( {x}_{1}\right) \cdots {\chi }_{k - 1}\left( {x}_{k - 1}\right) {\... | Yes |
Corollary 2.5.15. If \( \chi \) is a character of order dividing \( m \) then \( \tau {\left( \chi ,\psi \right) }^{m} \in \) \( \mathbb{Q}\left( {\zeta }_{m}\right) \) . | Proof. This is trivial if \( \chi \) or \( \psi \) is a trivial character. Otherwise we apply the last formula of the above proposition to \( k = m \) and \( {\chi }_{i} = \chi \) for all \( i \), and we deduce the result since evidently \( {J}_{k}\left( {\chi ,\ldots ,\chi }\right) \in \mathbb{Q}\left( {\zeta }_{m}\ri... | No |
Corollary 2.5.16. As above, denote by \( t \) the number of indices \( i \) such that \( {\chi }_{i} = \varepsilon \) . Then \[ {J}_{k}\left( 0\right) = \left\{ \begin{array}{ll} {q}^{k - 1} & \text{ if }t = k, \\ 0 & \text{ if }1 \leq t \leq k - 1\text{ or if }{\chi }_{1}\cdots {\chi }_{k} \neq \varepsilon , \\ \left(... | Proof. By Lemma 2.5.13, the result is clear for \( t = k \) and when \( {\chi }_{1}\cdots {\chi }_{k} \neq \) \( \varepsilon \) . Otherwise, we may assume by symmetry that \( {\chi }_{k} \neq {\chi }_{0} \), and the lemma gives \( {J}_{k}\left( 0\right) = {\chi }_{k}\left( {-1}\right) \left( {q - 1}\right) {J}_{k - 1}\... | Yes |
Proposition 2.5.18. For any \( n \geq 2 \) denote as usual by \( {\zeta }_{n} \) a primitive nth root of unity.\n\n(1) Assume that \( q \) is odd. If \( \chi \) is a character of order \( n > 2 \) and \( \rho \) is the character of order 2 we have the identity\n\n\[ \chi \left( 4\right) J\left( {\chi ,\chi }\right) = J... | Proof. (1) is an easy exercise left to the reader (Exercise 40). | No |
Corollary 2.5.19. (1) Assume that \( q \) is odd. If \( \chi \) is a character of order \( n > 2 \) then\n\n\[ J\left( {\chi ,\chi }\right) \equiv - {\chi }^{-1}\left( 4\right) q\left( {{\;\operatorname{mod}\;2}\left( {1 - {\zeta }_{n}}\right) }\right) . \] | Proof. By (1) and (2) of the proposition we have\n\n\[ \chi \left( 4\right) J\left( {\chi ,\chi }\right) = J\left( {\chi ,\rho }\right) \equiv - q\left( {{\;\operatorname{mod}\;2}\left( {1 - i}\right) }\right) \] \n\nsince \( {\zeta }_{2} = - 1 \) . | No |
Proposition 2.5.20. (1) Let \( q \) be a prime power such that \( q \equiv 1\left( {\;\operatorname{mod}\;4}\right) \), let \( \chi \) be one of the two characters of order 4 on \( {\mathbb{F}}_{q}^{ * } \), and write \( J\left( {\chi ,\chi }\right) = a + {bi} \). Then \( {a}^{2} + {b}^{2} = q,2 \mid b \), and \( a \eq... | Proof. (1). Since the group of characters of \( {\mathbb{F}}_{q}^{ * } \) is isomorphic to \( {\mathbb{F}}_{q}^{ * } \), hence is cyclic of order \( q - 1 \), for any \( n \mid q - 1 \) there exists a character of order exactly equal to \( n \) . Thus, when \( q \equiv 1\left( {\;\operatorname{mod}\;4}\right) \) let \(... | Yes |
Lemma 2.5.21. Let \( d \mid \left( {q - 1}\right) \), and denote by \( {G}_{d} \) the group of multiplicative characters \( \chi \) on \( {\mathbb{F}}_{q} \) such that \( {\chi }^{d} = \varepsilon \) . Then \( \left| {G}_{d}\right| = d \), and if \( y \in {\mathbb{F}}_{q} \) we have\n\n\[ \mathop{\sum }\limits_{{\chi \... | Proof. The group \( {G}_{d} \) is canonically isomorphic to the group of characters of the abelian group \( {\mathbb{F}}_{q}^{ * }/{\mathbb{F}}_{q}^{*d} \) ; hence the lemma is an immediate consequence of the orthogonality of characters (Proposition 2.1.20).\n\nNote that the result for \( y = 0 \) comes from the conven... | Yes |
Theorem 2.5.22. (1) For any nontrivial additive character \( \psi \) we have\n\n\[ N\left( q\right) = {q}^{k - 1} + \left( {1 - \frac{1}{q}}\right) \mathop{\sum }\limits_{\substack{{{\chi }_{i} \in {G}_{{d}_{i}}\smallsetminus \{ \varepsilon \} } \\ {{\chi }_{1}\cdots {\chi }_{k} = \varepsilon } }}\tau \left( {{\chi }_{... | Proof. (1). By the above remark we have\n\n\[ N\left( q\right) = \mathop{\sum }\limits_{\substack{{{y}_{i} \in {\mathbb{F}}_{q}} \\ {{y}_{1} + \cdots + {y}_{k} = 0} }}\mathop{\prod }\limits_{{1 \leq i \leq k}}\mathop{\sum }\limits_{{\chi \in {G}_{{d}_{i}}}}\chi \left( {{y}_{i}/{a}_{i}}\right) \]\n\n\[ = \mathop{\sum }\... | Yes |
Corollary 2.5.23. Let \( a, b \), and \( c \) be nonzero elements of \( {\mathbb{F}}_{q} \), let \( m \geq {\mathbb{Z}}_{ \geq 1} \) , and set \( d = \gcd \left( {m, q - 1}\right) \) . The number \( M\left( q\right) \) of projective solutions in \( {\mathbb{P}}^{2}\left( {\mathbb{F}}_{q}\right) \) to the equation \( a{... | \[ \left| {M\left( q\right) - \left( {q + 1}\right) }\right| \leq \left( {d - 1}\right) \left( {d - 2}\right) {q}^{1/2}. \] | Yes |
Corollary 2.5.24. Assume that \( {m}_{i} = m \) for all \( i \) and that the \( {a}_{i} \) are all nonzero, and as usual set \( d = \gcd \left( {m, q - 1}\right) \) . The number \( M\left( q\right) \) of projective solutions in \( {\mathbb{P}}^{k - 1}\left( {\mathbb{F}}_{q}\right) \) to the equation \( {a}_{1}{x}_{1}^{... | Proof. This immediately follows from Theorem 2.5.22 and Proposition 2.5.14, using as above the fact that \( M\left( q\right) = \left( {N\left( q\right) - 1}\right) /\left( {q - 1}\right) \), and \( {\chi }_{k}^{-1}\left( {-{a}_{k}}\right) = \mathop{\prod }\limits_{{1 \leq i \leq k - 1}}{\chi }_{i}\left( {-{a}_{k}}\righ... | Yes |
Corollary 2.5.25. Assume that \( q \equiv 1\left( {\;\operatorname{mod}\;6}\right) \), and let \( \chi \) be a character of order 3 on \( {\mathbb{F}}_{q} \) . The number \( M\left( q\right) \) of projective solutions in \( {\mathbb{P}}^{2}\left( {\mathbb{F}}_{q}\right) \) to the equation \( {x}^{3} + {y}^{3} + {z}^{3}... | Proof. This immediately follows from the above corollary and Proposition 2.5.20 and is left to the reader (Exercise 43 (1)). | No |
Corollary 2.5.27. With the above notation and assumptions, we have the formula\n\n\[ \n{N}_{C}\left( {q}^{n}\right) = {q}^{n} + 1 - \mathop{\sum }\limits_{{1 \leq j \leq {2g}}}{\alpha }_{j}^{n}. \]\n\nIn particular, the number \( {N}_{C}\left( q\right) \) of projective points on a curve of genus \( g \) defined over \(... | Proof. Expanding into power series \( \log \left( {{\zeta }_{C}\left( T\right) }\right) \) and replacing by the formula given by Weil's theorem immediately gives the first result. The second follows from \( \left| {\alpha }_{j}\right| = {q}^{1/2} \) . | Yes |
Theorem 2.5.28. There exist constants \( c\left( {n, d, r}\right) \) such that if \( V \) is an algebraic variety defined over \( {\mathbb{F}}_{q} \) of dimension \( r \) and degree \( d \) in \( n \) -dimensional projective space, then the number \( N \) of points of \( V \) defined over \( {\mathbb{F}}_{q} \) satisfi... | Since we have seen that the genus of a general curve of degree \( d \) is bounded by \( \left( {d - 1}\right) \left( {d - 2}\right) /2 \), we see that the above theorem exactly generalizes (without specifying \( c\left( {n, d, r}\right) ) \) Corollary 2.5.27. | No |
Proposition 3.1.1. Let \( K \) be a perfect field and \( \alpha \) an element that is algebraic over \( K \) . Then the minimal polynomial of \( \alpha \) in \( K\left\lbrack X\right\rbrack \) is separable; in other words, it is coprime to its derivative, or equivalently, it has no multiple roots in \( \bar{K} \) . | Proof. First, it is easy to check and left to the reader that over any field \( K \) a polynomial is coprime to its derivative if and only if it has no multiple roots in \( \bar{K} \) . Now assume that \( K \) is perfect, and let \( A \in K\left\lbrack X\right\rbrack \) be the minimal polynomial of \( \alpha \) . Since... | No |
Proposition 3.1.3. Let \( K \) be a perfect field, and let \( L/K \) be an extension of degree \( n \) .\n\n(1) Any embedding of \( K \) into \( \bar{K} \) extends to exactly \( {nK} \) -embeddings of \( L \) into\n\n(2) There exist at most \( {nK} \) -automorphisms of \( L \), which are the \( K \) -embeddings \( \sig... | Proof. By the primitive element theorem (which is true because we have assumed \( K \) to be perfect) we can write \( L = K\left( \alpha \right) \) for some \( \alpha \in L \) . Let \( A\left( X\right) \in K\left\lbrack X\right\rbrack \) be the minimal polynomial of \( \alpha \), which is therefore of degree \( n \) (a... | Yes |
Proposition 3.1.5. Let \( K \) be a perfect field and \( L/K \) a finite extension. The following three properties are equivalent:\n\n(1) \( L \) is closed under conjugation over \( K \) (i.e., if \( \alpha \in L \), every conjugate of \( \alpha \) also belongs to \( L \) ).\n\n(2) There exist exactly \( n = \left\lbra... | Proof. By the proof of the above proposition, if \( L \) is closed under conjugation then all the roots of the minimal polynomial \( A\left( X\right) \) of \( \alpha \) belong to \( L \) ; hence there are indeed \( {nK} \) -automorphisms, so (1) implies (2). The equivalence of (2) and (3) is clear from Proposition 3.1.... | Yes |
Proposition 3.1.7. If \( L = K\left( {{\alpha }_{1},\ldots ,{\alpha }_{k}}\right) \) and \( L \) contains the conjugates of all the \( {\alpha }_{i} \), then \( L/K \) is Galois. | Proof. Any element \( x \in L \) has the form \( x = U\left( {{\alpha }_{1},\ldots ,{\alpha }_{k}}\right) \), where \( U \) has coefficients in \( K \) . If \( \sigma \) is a \( K \) -embedding of \( L \) into \( \bar{K} \), then the coefficients of \( U \) are fixed by \( \sigma \) ; hence\n\n\[ \sigma \left( x\right)... | Yes |
Corollary 3.1.8. If \( L/K \) is a finite extension, there exists a finite extension \( N \) of \( L \) that is Galois over \( K \), and any such \( N \) will also be Galois over \( L \) . | Proof. Write \( L = K\left( \alpha \right) \), and let \( {\alpha }_{1},\ldots ,{\alpha }_{k} \) be the conjugates of \( \alpha \) in \( \bar{K} \) . Then by the above proposition, \( N = K\left( {{\alpha }_{1},\ldots ,{\alpha }_{k}}\right) \) is a finite Galois extension of \( K \) . Furthermore, if \( \sigma \) is an... | Yes |
Proposition 3.1.9. Let \( L/K \) be a Galois extension with Galois group \( G \) and let \( H \) be a subgroup of \( G \) . Then \( {L}^{H} = K \) if and only if \( H = G \). | Proof. We have clearly \( {L}^{H} \supset K \) for all \( H \) . Choose first \( H = G \), assume that \( x \in {L}^{G} \), and set \( {K}_{1} = K\left( x\right) \) . Then \( L/{K}_{1} \) is a field extension, and by assumption every \( \sigma \in G \) is a \( {K}_{1} \) -automorphism of \( L \) . It follows from Propo... | Yes |
Theorem 3.1.10 (Fundamental theorem of Galois theory). Let \( K \) be a perfect field, let \( L/K \) be a finite Galois extension, and set \( G = \operatorname{Gal}\left( {L/K}\right) \). There exists a one-to-one reverse-ordering correspondence between on the one hand subfields \( {L}_{1} \) of \( L \) containing \( K... | Proof. Let \( {L}_{1} \) be a subfield of \( L \) containing \( K \). Since \( L/K \) is Galois, any \( K \) -embedding of \( L \) into \( \bar{K} \) is a \( K \) -automorphism; hence any \( {L}_{1} \) -embedding of \( L \) into \( \overline{{L}_{1}} = \bar{K} \) is an automorphism, hence an \( {L}_{1} \) -automorphism... | Yes |
Theorem 3.1.11. Assume that \( L/K \) is Galois, and let \( M/K \) be any finite extension. Then the extension \( {LM}/M \) is also Galois, and \( \operatorname{Gal}\left( {{LM}/M}\right) \) can be considered as a subgroup of \( \operatorname{Gal}\left( {L/K}\right) \) by restriction of automorphisms. Furthermore, we h... | Proof. Write \( L = K\left( \alpha \right) \) for some \( \alpha \in L \) . Clearly \( {LM} = M\left( \alpha \right) \), and since the conjugates of \( \alpha \) belong to \( L \), hence to \( {LM} \), Proposition 3.1.7 implies that \( {LM}/M \) is Galois.\n\nThe restriction of an \( M \) -automorphism of \( {ML} \) to... | Yes |
Proposition 3.1.12. Let \( K \) be a number field and let \( \Omega \) be an algebraically closed field containing \( K \) . If \( L/K \) is an extension of degree \( n \), there exist exactly \( {nK} \) -embeddings of \( L \) into \( \Omega \), i.e., injective field homomorphisms from \( L \) to \( \Omega \) that leav... | Proof. The proof is essentially the same as that of Proposition 3.1.3 and is left to the reader (Exercise 2). | No |
Proposition 3.1.13. Let \( K \) be a number field of degree \( n \) . The number of embeddings of \( K \) into \( \mathbb{C} \) whose image is not contained in \( \mathbb{R} \) (we will simply say nonreal embeddings, or sometimes complex embeddings) is even. | Proof. If \( \sigma \) is such an embedding and if \( c \) denotes complex conjugation, it is clear that \( c \circ \sigma \) is also such an embedding, and we have \( c \circ \sigma \neq \sigma \) since otherwise the image of \( \sigma \) would be contained in \( \mathbb{R} \) . | Yes |
Proposition 3.1.15. Let \( \sigma \) be a continuous field homomorphism from \( \mathbb{R} \) or \( \mathbb{C} \) into \( \mathbb{C} \) . Then either \( \sigma \) is the identity or it is complex conjugation. | Proof. We have \( \sigma \left( 1\right) = 1 \) ; hence for any positive integer \( n \), by induction we have \( \sigma \left( n\right) = n \), hence \( \sigma \left( {-n}\right) = - n \), so that \( \sigma \left( n\right) = n \) for all \( n \in \mathbb{Z} \), and \( \sigma \left( {p/q}\right) = \sigma \left( p\right... | Yes |
Proposition 3.1.18. Let \( K \) be a number field of degree \( n \) . The map \( \left( {x, y}\right) \mapsto \) \( {\operatorname{Tr}}_{K/\mathbb{Q}}\left( {xy}\right) \) defines a nondegenerate \( \mathbb{Q} \) -bilinear form on \( K \times K \) with values in \( \mathbb{Q} \) . | Proof. The above proposition shows immediately that this map is a \( \mathbb{Q} \) - bilinear form. If \( x \) is such that \( {\operatorname{Tr}}_{K/\mathbb{Q}}\left( {xy}\right) = 0 \) for all \( y \in K \), then if \( x \neq 0 \) we can choose \( y = 1/x \), so that \( 0 = {\operatorname{Tr}}_{K/\mathbb{Q}}\left( 1\... | Yes |
Lemma 3.1.19 (Noether). Let \( L/K \) be a Galois 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( {\p... | 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(... | No |
Lemma 3.1.21. Any homomorphism from \( G \) to \( {\mathbf{\mu }}_{n} \) has the form\n\n\[ \sigma \mapsto \langle \sigma ,\bar{b}\rangle \]\n\nfor some \( \bar{b} \in B \) . | 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 h... | Yes |
Corollary 3.1.22. Let \( K \) be a commutative field and \( n \geq 1 \) an integer not divisible by the characteristic of \( K \) such that \( {\zeta }_{n} \in K \). (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... | Proof. (1) Let \( L/K \) be a cyclic extension of degree \( n \). By Theorem 3.1.20, 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 cle... | Yes |
Proposition 3.1.24. Let \( K \) be a perfect field of characteristic 2. Quadratic extensions \( L \) of \( K \) have the form \( K\left( \theta \right) \), where \( \theta \) is a root of an Artin-Schreier polynomial \( {X}^{2} - X - a \) for some \( a \in K \) not of the form \( {x}^{2} - x \) for \( x \in K \) . Furt... | Proof. Since \( K \) is perfect, the extension is separable, and hence by the primitive element theorem, \( L = K\left( \theta \right) \), where \( \theta \) is a root of \( {X}^{2} - {bX} - c = 0 \) for some \( b \) and \( c \) in \( K \) . Since \( K \) is perfect we cannot have \( b = 0 \), otherwise \( {X}^{2} - c ... | Yes |
Lemma 3.2.1 (Dedekind independence). Let \( G \) be a group, \( L \) a field, and let \( {\sigma }_{1},\ldots ,{\sigma }_{m} \) be distinct group homomorphisms from \( G \) to \( {L}^{ * } \) . Then they are \( L \) -linearly independent. In other words, if there exist \( {a}_{i} \in L \) and a relation \( \mathop{\sum... | Proof. Assume that there exists a nontrivial relation. Choose such a relation of minimal length, so that, up to reordering of the \( {\sigma }_{i} \) ,\n\n\[ \forall h \in G\;\mathop{\sum }\limits_{{1 \leq i \leq k}}{a}_{i}{\sigma }_{i}\left( h\right) = 0 \]\n\n(1)\n\nwith \( k \) minimal. For any \( g \in G \), we hav... | Yes |
Corollary 3.2.2. Let \( K \) and \( L \) be fields, and let \( {\sigma }_{1},\ldots ,{\sigma }_{m} \) be distinct field homomorphisms from \( K \) to \( L \) . Then the \( {\sigma }_{i} \) are \( L \) -linearly independent in the vector space of linear maps from \( K \) to \( L \) . | Proof. Clear by applying the above lemma to \( G = {K}^{ * } \) . | No |
Corollary 3.2.3. Let \( E/F \) be a finite extension of commutative fields of degree \( n \) . The elements of \( \operatorname{Gal}\left( {E/F}\right) \) form an \( E \) -basis of the space \( {\mathcal{L}}_{F}\left( E\right) \) of \( F \) -linear maps from \( E \) to \( E \) . In other words, we have the direct sum d... | Proof. Indeed, \( {\mathcal{L}}_{F}\left( E\right) \) is an \( F \) -vector space of dimension \( {n}^{2} \), hence an \( E \) -vector space of dimension \( n \), so any family of \( {nE} \) -linearly independent elements form an \( E \) -basis. | No |
Proposition 3.2.4 (Hilbert’s Theorem 90). Let \( E/F \) be a finite cyclic extension of commutative fields of degree \( n \) and let \( \sigma \) be a generator of the Galois group \( G = \operatorname{Gal}\left( {E/F}\right) \). If \( \alpha \in {E}^{ * } \) is such that \( \mathop{\prod }\limits_{{0 \leq j < n}}{\sig... | Proof. Consider the map \( \phi \) from \( G \) to \( {E}^{ * } \) defined for \( k \geq 0 \) by \( \phi \left( {\sigma }^{k}\right) = \) \( \mathop{\prod }\limits_{{0 \leq j < k}}{\sigma }^{j}\left( \alpha \right) \). Since \( \mathop{\prod }\limits_{{0 \leq j < n}}{\sigma }^{j}\left( \alpha \right) \) and \( \sigma \... | Yes |
Theorem 3.2.5. Let \( F \) be a commutative field, let \( \sigma \) be an endomorphism of a finite-dimensional \( F \) -vector space \( E \), let \( P \) be its minimal polynomial over \( F \), and let \( s \) be the degree of \( P \) . There exists \( \theta \in E \) such that the minimal polynomial of the restriction... | Proof. First, let \( A \) and \( B \) be coprime polynomials in \( F\left\lbrack X\right\rbrack \) . We leave as an easy exercise for the reader (Exercise 10) to show that\n\n\[ \operatorname{Ker}\left( {\left( {AB}\right) \left( \sigma \right) }\right) = \operatorname{Ker}\left( {A\left( \sigma \right) }\right) \oplus... | No |
Theorem 3.2.6. Let \( E/F \) be a finite cyclic extension of commutative fields of degree \( n \) and let \( G = \operatorname{Gal}\left( {E/F}\right) \) be its Galois group. There exists an element \( \theta \in E \) whose conjugates under the action of \( G \) form an \( F \) -basis of \( E \), in other words \( E \)... | Proof. Let \( \sigma \) be a generator of \( G \) (in the case of finite fields, we have seen that we can choose for \( \sigma \) the Frobenius automorphism corresponding to \( F \) ). Since the \( n \) elements \( {\sigma }^{i} \) for \( 0 \leq i \leq n - 1 \) are distinct, they are \( F \) -linearly independent by Co... | Yes |
Proposition 3.2.8. A polynomial \( P \in F\left\lbrack {{X}_{1},\ldots ,{X}_{n}}\right\rbrack \) is additive if and only if it has the form \( \mathop{\sum }\limits_{{1 \leq i \leq n}}{P}_{i}\left( {X}_{i}\right) \), where the \( {P}_{i}\left( {X}_{i}\right) \) are additive polynomials in one variable. | Proof. We can write\n\n\[ \left( {{X}_{1},\ldots ,{X}_{n}}\right) = \mathop{\sum }\limits_{{1 \leq i \leq n}}\left( {0,\ldots ,{X}_{i},\ldots ,0}\right) ,\]\n\nand so by definition of additive polynomials we have\n\n\[ P\left( {{X}_{1},\ldots ,{X}_{n}}\right) = \mathop{\sum }\limits_{{1 \leq i \leq n}}{P}_{i}\left( {X}... | Yes |
Theorem 3.2.10 (Artin). If the \( {\sigma }_{i} \) are algebraically dependent as above, we can choose \( P \) to be a (nonzero) additive polynomial. | Proof. For simplicity, we will write \( \sum \left( x\right) \) instead of \( \left( {{\sigma }_{1}\left( x\right) ,\ldots ,{\sigma }_{n}\left( x\right) }\right) \) . Let \( P\left( {{X}_{1},\ldots ,{X}_{n}}\right) \) be the polynomial of lowest possible total degree that is nonzero and is such that \( P\left( {\sum \l... | Yes |
Theorem 3.2.11. As above, let \( F \) be an infinite field, and let \( {\sigma }_{1},\ldots ,{\sigma }_{n} \) be distinct elements of a finite group of automorphisms of \( F \) . Then the \( {\sigma }_{i} \) are algebraically independent over \( F \) . | Proof. By Theorem 3.2.10, there exists a nonzero additive polynomial \( P \) such that \( P\left( {\sum \left( x\right) }\right) = 0 \) for all \( x \) . If \( F \) has characteristic 0, by Theorem 3.2.8 such a polynomial is simply a linear form, and hence the result follows from linear independence (Corollary 3.2.2). ... | Yes |
Theorem 3.2.12. Let \( E/F \) be a finite Galois extension of commutative fields of degree \( n \), and let \( G = \operatorname{Gal}\left( {E/F}\right) \) be its Galois group. There exists an element \( \theta \in E \) whose conjugates under the action of \( G \) form an \( F \) -basis of \( E \), in other words \( E ... | Proof. If \( F \) is a finite field, then \( E/F \) is a cyclic extension, so that the result is Theorem 3.2.6. We can therefore assume that \( F \) is infinite.\n\nLet \( {\sigma }_{1},\ldots ,{\sigma }_{n} \) be the (distinct) elements of \( G \), numbered so that \( {\sigma }_{1} \) is the identity. We can write\n\n... | Yes |
Proposition 3.3.2 (Gauss’s lemma). If \( A \) and \( B \) are two nonzero polynomials in \( \mathbb{Z}\left\lbrack X\right\rbrack \), we have \( c\left( {AB}\right) = c\left( A\right) c\left( B\right) \) . | Proof. Let us say that a polynomial \( A \in \mathbb{Z}\left\lbrack X\right\rbrack \) is primitive if its content is equal to 1 . Since \( A = c\left( A\right) {A}_{1} \) with \( {A}_{1} \) primitive, it is clear that the proposition is equivalent to the statement that the product of two primitive polynomials \( A \) a... | Yes |
Corollary 3.3.3. Let \( C \in \mathbb{Z}\left\lbrack X\right\rbrack \) be a monic polynomial and assume that \( A \in \mathbb{Q}\left\lbrack X\right\rbrack \) is a monic polynomial such that \( A \mid C \) in \( \mathbb{Q}\left\lbrack X\right\rbrack \) . Then in fact \( A \in \mathbb{Z}\left\lbrack X\right\rbrack \) . | Proof. Write \( C = {AB} \) with \( B \in \mathbb{Q}\left\lbrack X\right\rbrack \) . Let \( {d}_{A} \) (respectively \( {d}_{B} \) ) be the smallest integer such that \( {d}_{A}A \) (respectively \( {d}_{B}B \) ) is in \( \mathbb{Z}\left\lbrack X\right\rbrack \), in other words, the LCM of the denominators of the coeff... | Yes |
Proposition 3.3.4. Let \( \alpha \) be an algebraic number. The following four properties are equivalent.\n\n(1) The number \( \alpha \) is a root of a monic polynomial with coefficients in \( \mathbb{Z} \) .\n\n(2) The minimal monic polynomial of \( \alpha \) has coefficients in \( \mathbb{Z} \) .\n\n(3) The ring \( \... | Proof. (1) \( \Rightarrow \) (2): Assume that \( P\left( \alpha \right) = 0 \) with \( P \in \mathbb{Z}\left\lbrack X\right\rbrack \) monic, and let \( T \) be the minimal monic polynomial of \( \alpha \) . By definition, \( T \) divides \( P \) in \( \mathbb{Q}\left\lbrack X\right\rbrack \), and \( T \) is monic, so w... | Yes |
Proposition 3.3.6. If \( \alpha \) and \( \beta \) are algebraic integers, then so are \( \alpha + \beta \) and \( {\alpha \beta } \) . In other words, algebraic integers belonging to a fixed algebraic closure of \( \mathbb{Q} \) form a ring. | Proof. Consider \( R = \mathbb{Z}\left\lbrack {\alpha ,\beta }\right\rbrack \), the ring of polynomials in \( \alpha \) and \( \beta \) . Since \( \alpha \) and \( \beta \) are algebraic integers, of respective degree \( m \) and \( n \), say, it is clear that the \( {\left( {\alpha }^{i}{\beta }^{j}\right) }_{0 \leq i... | Yes |
Proposition 3.3.7. Let \( P\left( X\right) \) be a monic polynomial whose coefficients are algebraic integers, and let \( \alpha \) be such that \( P\left( \alpha \right) = 0 \) . Then \( \alpha \) is an algebraic integer. | Proof. Write \( P\left( X\right) = {X}^{n} + \mathop{\sum }\limits_{{1 \leq i \leq n - 1}}{\beta }_{i}{X}^{i} \) . Since the \( {\beta }_{i} \) are algebraic integers, it follows that \( \mathbb{Z}\left\lbrack {{\beta }_{1},\ldots ,{\beta }_{n - 1}}\right\rbrack \) is a finitely generated \( \mathbb{Z} \) -module. Let ... | Yes |
Proposition 3.3.8 (Dedekind). Let \( \alpha \) be an algebraic number and let \( T \in \) \( \mathbb{Z}\left\lbrack X\right\rbrack \) be a nonzero polynomial such that \( T\left( \alpha \right) = 0 \) . Write \( T\left( X\right) = {a}_{n}{X}^{n} + \) \( {a}_{n - 1}{X}^{n - 1} + \cdots + {a}_{1}X + {a}_{0} \), let \( {\... | Proof. If we define \( {\omega }_{j} \) for all \( j \geq 1 \) by the formula of the proposition, it is clear by definition that \( {\omega }_{j} = 0 \) for \( j \geq n \) . We can thus consider \( R \) as the \( \mathbb{Z} \) -module generated by all the \( {\omega }_{j} \) for \( j \geq 0 \) . Now it is clear that fo... | Yes |
Proposition 3.3.9 (Kronecker). Let \( T\left( X\right) \in \mathbb{Z}\left\lbrack X\right\rbrack \) be a monic polynomial with integer coefficients. Assume that all the roots of \( T \) in \( \mathbb{C} \) have modulus equal to 1 . Then all the roots of \( T \) are roots of unity. | Proof. Without loss of generality we may assume that \( T\left( X\right) \) is irreducible. Write \( T\left( X\right) = \mathop{\prod }\limits_{{1 \leq i \leq n}}\left( {X - {\alpha }_{i}}\right) \) with \( {\alpha }_{i} \in \mathbb{C} \), and for any \( k \geq 1 \), consider the polynomial \( {T}_{k}\left( X\right) = ... | Yes |
Proposition 3.3.11. Let \( K \) be a number field of degree \( n \) . The ring \( {\mathbb{Z}}_{K} \) is a free \( \mathbb{Z} \) -module of rank \( n \) . | Proof. Let \( {\alpha }_{1},\ldots ,{\alpha }_{n} \) be a \( \mathbb{Q} \) -basis of \( K \) . Multiplying if necessary each \( {\alpha }_{i} \) by some nonzero element of \( \mathbb{Z} \), we may assume that \( {\alpha }_{i} \in {\mathbb{Z}}_{K} \) for all \( i \) . Let \( \Lambda \) be the free \( \mathbb{Z} \) -modu... | Yes |
Proposition 3.3.13. Let \( I \) be a fractional ideal of \( L \). (1) There exist elements \( {\omega }_{i} \in I \) and fractional ideals \( {\mathfrak{a}}_{i} \) of \( K \) for \( 1 \leq i \leq n \) such that \( I = {\bigoplus }_{1 \leq i \leq n}{\mathfrak{a}}_{i}{\omega }_{i} \) in an evident sense. (2) The fraction... | Proof. Left to the reader (Exercise 16). | No |
Theorem 3.3.17. Let \( I \) be a fractional ideal of \( K \) . There exists a factorization\n\n\[ I = \mathop{\prod }\limits_{{i = 1}}^{g}{\mathfrak{p}}_{i}^{{v}_{i}} \]\n\nwhere the \( {\mathfrak{p}}_{i} \) are distinct prime ideals and \( {v}_{i} \in \mathbb{Z} \smallsetminus \{ 0\} \), and this factorization is uniq... | This theorem is one of the most important consequences of the fact that \( {\mathbb{Z}}_{K} \) is a Dedekind domain. | No |
Theorem 3.3.18 (Finiteness of the class group). Define two fractional ideals \( \mathfrak{a} \) and \( \mathfrak{b} \) to be equivalent if there exists \( \alpha \in {K}^{ * } \) such that \( \mathfrak{b} = \alpha \mathfrak{a} \) . This equivalence relation is compatible with the multiplicative group structure of ideal... | The group of ideal classes of \( K \) is denoted by \( {Cl}\left( K\right) \), and the class number, in other words \( \left| {{Cl}\left( K\right) }\right| \), is denoted by \( h\left( K\right) \) . Standard group theory implies that for any ideal \( \mathfrak{a} \) the ideal \( {\mathfrak{a}}^{h\left( K\right) } \) ha... | Yes |
Proposition 3.3.21. Let \( M/L \) and \( L/K \) be extensions of number fields, \( \mathfrak{p} \) an ideal of \( K,{\mathfrak{P}}_{L} \) an ideal of \( L \) above \( \mathfrak{p} \), and \( {\mathfrak{P}}_{M} \) an ideal of \( M \) above \( {\mathfrak{P}}_{L} \) . We have the transitivity relations:\n\n\[ e\left( {{\m... | Proof. Left to the reader (Exercise 18). | No |
Lemma 3.3.24. Let \( L = K\left( \theta \right) \), where \( \theta \) is an algebraic integer, and denote by \( f = \left\lbrack {{\mathbb{Z}}_{L} : {\mathbb{Z}}_{K}\left\lbrack \theta \right\rbrack }\right\rbrack \) the index of \( {\mathbb{Z}}_{K}\left\lbrack \theta \right\rbrack \) in \( {\mathbb{Z}}_{L} \) . Let \... | Proof. Clearly \( \mathfrak{p} \nmid y \) ; otherwise, \( \mathfrak{P} \mid x \), and hence \( \mathfrak{p} \mid x \) and \( \mathfrak{p} \mid y \), contradicting the fact that the ideal \( x{\mathbb{Z}}_{K} + y{\mathbb{Z}}_{K} \) is coprime to \( \mathfrak{p} \) . Assume that \( \mathfrak{p} \nmid f \), and let \( {y}... | Yes |
Proposition 3.3.25. Let \( L/K \) be a Galois extension, let \( G = \operatorname{Gal}\left( {L/K}\right) \) , and let \( \mathfrak{p} \) be a prime ideal of \( {\mathbb{Z}}_{K} \) . The decomposition of \( \mathfrak{p} \) in \( L \) is given by \( \mathfrak{p}{\mathbb{Z}}_{L} = \mathop{\prod }\limits_{{1 \leq i \leq g... | Proof. Left to the reader (Exercise 18). | No |
Proposition 3.3.27. We have \( \left| {D\left( {\mathfrak{P}/p}\right) }\right| = e\left( {\mathfrak{P}/\mathfrak{p}}\right) f\left( {\mathfrak{P}/\mathfrak{p}}\right) ,\left| {I\left( {\mathfrak{P}/\mathfrak{p}}\right) }\right| = \) \( e\left( {\mathfrak{P}/\mathfrak{p}}\right) \), and \( D\left( {\mathfrak{P}/\mathfr... | Proof. Left to the reader (Exercise 18). | No |
Lemma 3.3.28. Let \( {K}_{1}/K \) and \( {K}_{2}/K \) be two extensions of number fields, and let \( \mathfrak{p} \) be a prime ideal of \( {\mathbb{Z}}_{K} \). Denote by \( \mathfrak{P} \) a prime ideal of the compositum \( {K}_{1}{K}_{2} \) above \( \mathfrak{p} \), and by \( {\mathfrak{p}}_{1} \) and \( {\mathfrak{p... | Proof. Let \( \sigma \) be a \( {K}_{2} \) -embedding of \( {K}_{1}{K}_{2} \) into \( \mathbb{C} \), so that \( \sigma \) is the identity on \( {K}_{2} \), and \( {\left. \sigma \right| }_{{K}_{1}} \) is a \( K \) -embedding of \( {K}_{1} \) into \( \mathbb{C} \). Since \( {K}_{1}/K \) is Galois, \( \sigma \left( {K}_{... | Yes |
Lemma 3.3.30. Let \( {K}_{1}/K \) and \( {K}_{2}/K \) be two extensions of number fields, and let \( \mathfrak{p} \) be a prime ideal of \( {\mathbb{Z}}_{K} \) . Denote by \( \mathfrak{P} \) a prime ideal of the compositum \( {K}_{1}{K}_{2} \) above \( \mathfrak{p} \), and by \( {\mathfrak{p}}_{1} \) and \( {\mathfrak{... | Proof. Denote by \( N \) the Galois closure of \( {K}_{1}{K}_{2}/K \), set \( {H}_{i} = \operatorname{Gal}\left( {N/{K}_{i}}\right) \) for \( i = 1,2 \), so that \( \operatorname{Gal}\left( {N/{K}_{1}{K}_{2}}\right) = {H}_{1} \cap {H}_{2} \), and let \( {\mathfrak{P}}_{N} \) be a prime ideal of \( N \) above \( \mathfr... | Yes |
Proposition 3.4.1. Let \( K = \mathbb{Q}\left( \sqrt{d}\right) \) be a quadratic field with \( d \) squarefree, and let \( {\mathbb{Z}}_{K} \) be its ring of integers. \( A\mathbb{Z} \) -basis of \( {\mathbb{Z}}_{K} \) is given by \( \left( {1,\omega }\right) \), where \( \omega = \sqrt{d} \) when \( d \equiv 2 \) or 3... | Proof. Although the proof is simple, it is not completely trivial, and the reader who has never seen it should try it for himself before reading on. Let \( \alpha = a + b\sqrt{d} \in {\mathbb{Z}}_{K} \) with \( a, b \) in \( \mathbb{Q} \) . The characteristic polynomial of \( \alpha \) is equal to \( {\left( X - a\righ... | No |
Proposition 3.4.3. Let \( K = \mathbb{Q}\left( \sqrt{D}\right) \) be a quadratic field of discriminant \( D \), set \( \omega = \left( {D + \sqrt{D}}\right) /2 \), and let \( p \) be a prime number. The prime ideal decomposition of \( p{\mathbb{Z}}_{K} \) is given as follows.\n\n(1) If \( p \mid D \) then \( p \) is ra... | Proof. Since \( {\mathbb{Z}}_{K} = \mathbb{Z}\left\lbrack \omega \right\rbrack \), this follows immediately from Proposition 3.3.22 and is left to the reader (Exercise 21). | No |
Proposition 3.4.5. We have\n\n\[ \frac{h\left( D\right) R\left( D\right) }{w\left( D\right) } = \frac{{\left| D\right| }^{1/2}}{{c}_{\operatorname{sign}\left( D\right) }}\mathop{\sum }\limits_{{n \geq 1}}\frac{\left( \frac{D}{n}\right) }{n} = \frac{{\left| D\right| }^{1/2}}{{c}_{\operatorname{sign}\left( D\right) }}L\l... | We will study in detail \( L \) -series in Chapter 10, and in particular we will see that \( L\left( {\left( \frac{D}{}\right) ,1}\right) = O\left( {\log \left( \left| D\right| \right) }\right) \) . On the other hand, it is immediate to see from the definition that \( R\left( D\right) > \log \left( D\right) /2 + o\left... | No |
Proposition 3.5.2. (1) We have\n\n\[ \n{\Phi }_{n}\left( X\right) = \mathop{\prod }\limits_{{d \mid n}}{\left( {X}^{d} - 1\right) }^{\mu \left( {n/d}\right) }, \n\]\n\nwhere \( \mu \) is the Möbius function (see Definition 10.1.4). | Proof. (1). Let us show that the right-hand side satisfies the defining property of \( {\Phi }_{n} \) : it is equal to \( X - 1 \) for \( n = 1 \), and otherwise\n\n\[ \n\mathop{\prod }\limits_{{d \mid n}}\mathop{\prod }\limits_{{e \mid d}}{\left( {X}^{e} - 1\right) }^{\mu \left( {d/e}\right) } = \mathop{\prod }\limits... | Yes |
Lemma 3.5.3. Let \( m \in {\mathbb{Z}}_{ \geq 1} \) . For all \( d \mid m \) we have \( {\Phi }_{m}\left( X\right) \mid {\Phi }_{m/d}\left( {X}^{d}\right) \) . | Proof. For any \( a \) we have \( {X}^{d} - {a}^{d} = \mathop{\prod }\limits_{{k{\;\operatorname{mod}\;d}}}\left( {X - {\zeta }_{d}^{k}a}\right) \), so applying this to \( a = {\zeta }_{m}^{i} \), we deduce that\n\n\[ \n{X}^{d} - {\zeta }_{m/d}^{i} = \mathop{\prod }\limits_{{k{\;\operatorname{mod}\;d}}}\left( {X - {\ze... | Yes |
Proposition 3.5.4. For all \( n > 1 \) we have\n\n\[ \n{\Phi }_{n}\left( 1\right) = \mathop{\prod }\limits_{{a \in {\left( \mathbb{Z}/n\mathbb{Z}\right) }^{ * }}}\left( {1 - {\zeta }_{n}^{a}}\right) = \left\{ \begin{array}{ll} p & \text{ if }n = {p}^{k}, \\ 1 & \text{ otherwise. } \end{array}\right.\n\] | Proof. By definition of the Möbius function (Definition 10.1.4), for \( n > 1 \) we have \( \mathop{\sum }\limits_{{d \mid n}}\mu \left( d\right) = \mathop{\sum }\limits_{{d \mid n}}\mu \left( {n/d}\right) = 0 \) . It follows from Proposition 3.5.2 (1) that \( {\Phi }_{n}\left( X\right) = \mathop{\prod }\limits_{{d \mi... | Yes |
Proposition 3.5.5. (1) Let \( a \) and \( b \) be integers coprime to \( n \) . The elements \( {C}_{a, b} = \left( {1 - {\zeta }_{n}^{a}}\right) /\left( {1 - {\zeta }_{n}^{b}}\right) \) are units, called cyclotomic units. | Proof. (1). Since \( a \) and \( b \) are coprime to \( n \) there exists \( c \) such that \( a \equiv {bc} \) \( \left( {\;\operatorname{mod}\;n}\right) \) . It follows that\n\n\[ \n{C}_{a, b} = \frac{1 - {\zeta }_{n}^{bc}}{1 - {\zeta }_{n}^{b}} = \mathop{\sum }\limits_{{0 \leq j < c}}{\zeta }_{n}^{bj} \n\]\n\nis an ... | Yes |
Proposition 3.5.8. Let \( 1 \leq m < n \) . Then \( R\left( {{\Phi }_{m},{\Phi }_{n}}\right) = 1 \), unless \( m \mid n \) and \( n/m = {p}^{a} \) is a power of a prime \( p \), in which case \( R\left( {{\Phi }_{m},{\Phi }_{n}}\right) = {p}^{\phi \left( m\right) } \) . | Proof. Assume first that \( \gcd \left( {m, n}\right) = 1 \) and that \( m > 1 \) . Thus \( {\Phi }_{m}\left( X\right) \mid \) \( \left( {{X}^{m} - 1}\right) /\left( {X - 1}\right) \), so by multiplicativity of the resultant \( R\left( {{\Phi }_{m},{\Phi }_{n}}\right) \mid R\left( {{X}^{m} - }\right. \) \( \left. {1,{\... | Yes |
Proposition 3.5.10. (1) The polynomial \( {\Phi }_{n}\left( X\right) \) is irreducible in \( \mathbb{Q}\left\lbrack X\right\rbrack \) ; in other words, \( {\Phi }_{n}\left( X\right) \) is the minimal polynomial of \( {\zeta }_{n} \) over \( \mathbb{Q} \), and \( \left\lbrack {\mathbb{Q}\left( {\zeta }_{n}\right) }\righ... | Proof. (I thank J.-F. Jaulent for the following proof of this very classical result.) Let \( P \) be an irreducible factor of \( {\Phi }_{n} \) in \( \mathbb{Q}\left\lbrack X\right\rbrack \), let \( \zeta \) be a root of \( P \) , and let \( p \) be a prime number such that \( p \nmid n \) . I claim that \( {\zeta }^{p... | Yes |
Corollary 3.5.11. There is a one-to-one correspondence between subfields of \( \mathbb{Q}\left( {\zeta }_{n}\right) \) and subgroups of \( {\left( \mathbb{Z}/n\mathbb{Z}\right) }^{ * } \) . | Proof. This is simply Galois theory. | No |
Corollary 3.5.12. The subgroup of roots of unity of \( \mathbb{Q}\left( {\zeta }_{n}\right) \) is the group of \( \pm {\zeta }_{n}^{i} \) for \( 0 \leq i < n \), or equivalently, the subgroup of order \( {2n} \) generated by \( {\zeta }_{2n} = - {\zeta }_{n} \) when \( n \) is odd, or the subgroup of order \( n \) gene... | Proof. Let \( {\zeta }_{m} \) be an \( m \) th root of unity in \( \mathbb{Q}\left( {\zeta }_{n}\right) \) . We thus have \( \mathbb{Q}\left( {\zeta }_{m}\right) \subset \) \( \mathbb{Q}\left( {\zeta }_{n}\right) \), hence by the proposition \( \phi \left( m\right) \mid \phi \left( n\right) \) . Since \( \phi \left( m\... | Yes |
Theorem 3.5.13 (Kronecker-Weber). Any Abelian extension \( K \) of \( \mathbb{Q} \) is a subfield of some cyclotomic field \( \mathbb{Q}\left( {\zeta }_{n}\right) \) . | Proof. The reader will find a proof of an essential special case of this theorem in Exercises 28 and 29, from which it is not difficult to deduce the general statement. | No |
Proposition 3.5.14. Let \( k = \mathbb{Q}\left( \sqrt{D}\right) \) be a quadratic field of discriminant \( D \), and set \( m = \left| D\right| \) . Then \( k \) is a subfield of \( \mathbb{Q}\left( {\zeta }_{m}\right) \) . | Proof. Set \[ \tau = \mathop{\sum }\limits_{{x{\;\operatorname{mod}\;m}}}\left( \frac{D}{x}\right) {\zeta }_{m}^{x} \] where \( \left( \frac{D}{x}\right) \) is the Legendre-Kronecker symbol (see Definition 2.2.5). Since it is a primitive character, Corollary 2.1.47 shows that \( {\tau }^{2} = \operatorname{sign}\left( ... | Yes |
Proposition 3.5.15. Assume that \( n = {p}^{k} \) is a prime power. The ideal \( (1 - \) \( \zeta ){\mathbb{Z}}_{K} \) is the unique prime ideal of \( K \) above \( p \), and \( p \) is totally ramified in \( K \) . | Proof. By Proposition 3.5.5, there exists a unit \( u \in K \) such that \( (1 - \) \( \zeta {)}^{\phi \left( {p}^{k}\right) } = {up} \) . It follows that if we set \( \mathfrak{p} = \left( {1 - \zeta }\right) {\mathbb{Z}}_{K} \) we have the ideal equality \( {\mathfrak{p}}^{\phi \left( k\right) } = p{\mathbb{Z}}_{K} \... | Yes |
Proposition 3.5.17. Set \( \zeta = {\zeta }_{{p}^{k}} \) .\n\n(1) The ring of integers of \( K = \mathbb{Q}\left( \zeta \right) \) is equal to \( \mathbb{Z}\left\lbrack \zeta \right\rbrack \) . | Proof. Since \( \zeta \in {\mathbb{Z}}_{K} \) we clearly have \( \mathbb{Z}\left\lbrack \zeta \right\rbrack \subset {\mathbb{Z}}_{K} \), so we must show the reverse inclusion. Since evidently \( K = \mathbb{Q}\left( \zeta \right) = \mathbb{Q}\left( {1 - \zeta }\right) \), by Proposition 3.5.10 the \( {\left( 1 - \zeta ... | Yes |
Lemma 3.5.19. Let \( \alpha \in K \) be such that \( \alpha /\iota \left( \alpha \right) \in {\mathbb{Z}}_{K} \) . Then \( \alpha /\iota \left( \alpha \right) \) is in fact a root of unity. In particular, if \( u \in U\left( K\right) \) then \( u/\iota \left( u\right) \) is a root of unity. | Proof. Since complex conjugation belongs to \( \operatorname{Gal}\left( {K/\mathbb{Q}}\right) \), which is abelian, it commutes with all elements of \( \operatorname{Gal}\left( {K/\mathbb{Q}}\right) \) . Thus if \( \sigma \in \operatorname{Gal}\left( {K/\mathbb{Q}}\right) \) we have \( \sigma \left( {\alpha /\iota \lef... | Yes |
Proposition 3.5.20. If \( p \neq 2 \) we have \( U\left( K\right) = \left\langle {\zeta }_{{p}^{k}}\right\rangle U\left( {K}^{ + }\right) \) . | Proof. The right-hand side is clearly a subgroup of the left-hand side. Thus let \( u \in U\left( K\right) \), and write for simplicity \( \zeta \) instead of \( {\zeta }_{{p}^{k}} \) . By the above lemma \( u/\iota \left( u\right) \) is a root of unity. The group of roots of unity in \( \mathbb{Q}\left( \zeta \right) ... | Yes |
Proposition 3.5.21. If \( p \geq 3 \) is prime, the map sending an ideal \( \mathfrak{a} \) of \( {\mathbb{Z}}_{{K}^{ + }} \) to \( \mathfrak{a}{\mathbb{Z}}_{K} \) induces an injective homomorphism from \( {Cl}\left( {K}^{ + }\right) \) to \( {Cl}\left( K\right) \) . In particular, \( {h}_{{p}^{k}}^{ + } \) divides \( ... | Proof. This map evidently induces a group homomorphism from \( {Cl}\left( {K}^{ + }\right) \) to \( {Cl}\left( K\right) \), and we must show that it is injective. Thus assume that \( \mathfrak{a} \) is an ideal of \( {\mathbb{Z}}_{{K}^{ + }} \) (which we may assume integral) such that \( \mathfrak{a}{\mathbb{Z}}_{K} = ... | Yes |
Proposition 3.5.22. Let \( K \) be a totally complex field, \( {K}^{ + } \) its maximal totally real subfield, and assume that \( \left\lbrack {K : {K}^{ + }}\right\rbrack = 2 \) . If we denote by \( h \) (respectively \( \left. {h}^{ + }\right) \) the class number of \( K \) (respectively of \( {K}^{ + } \) ), then \(... | Proof. We give a proof that assumes the basic definitions and properties of the Hilbert class field. Denote by \( H \) the Hilbert class field of \( {K}^{ + } \), so that \( H/{K}^{ + } \) is an everywhere unramified Abelian extension, and \( \left\lbrack {H : {K}^{ + }}\right\rbrack = {h}^{ + } \) . Consider the field... | Yes |
Lemma 3.6.1. Let \( {\mu }_{q - 1} = \left\{ {{\zeta }_{q - 1}^{j},0 \leq j \leq q - 2}\right\} \subset L \) be the group of \( \left( {q - 1}\right) \) st roots of unity in \( L \) . The map \( {u}_{\mathfrak{P}} \) sending \( x \in {\mu }_{q - 1} \) to its class modulo \( \mathfrak{P} \) in \( {\mathbb{F}}_{q}^{ * } ... | Proof. It is clear that \( {u}_{\mathfrak{P}} \) is a group homomorphism, and since \( {\mu }_{q - 1} \) and \( {\mathbb{F}}_{q}^{ * } \) both have \( q - 1 \) elements it is sufficient to show that the kernel of \( {u}_{\mathfrak{P}} \) is equal to 1 . Now\n\n\[ \mathop{\prod }\limits_{{1 \leq k \leq q - 2}}\left( {x ... | Yes |
For all \( \sigma \in \operatorname{Gal}\left( {L/\mathbb{Q}}\right) \) we have\n\n\[{\omega }_{\sigma \left( \mathfrak{P}\right) } = \sigma \circ {\omega }_{\mathfrak{P}} \circ {\sigma }^{-1}.\] | Since \( {\omega }_{\mathfrak{P}}\left( {{\sigma }^{-1}\left( x\right) }\right) \) is a \( \left( {q - 1}\right) \) st root of unity, so is \( \sigma \left( {{\omega }_{\mathfrak{P}}\left( {{\sigma }^{-1}\left( x\right) }\right) }\right) \) . Furthermore, \( {\omega }_{\mathfrak{P}}\left( {{\sigma }^{-1}\left( x\right)... | Yes |
Proposition 3.6.4. If \( a \) and \( b \) are integers such that \( 1 \leq a, b \leq q - 2 \) we have \( J\left( {{\omega }^{-a},{\omega }^{-b}}\right) \equiv 0\left( {\;\operatorname{mod}\;\mathfrak{P}}\right) \) if \( a + b \geq q \), and otherwise | Proof. For any \( x \in {\mathbb{F}}_{q}^{ * } = {\left( {\mathbb{Z}}_{L}/\mathfrak{P}\right) }^{ * } \) we let \( {x}^{\prime } \in {\mu }_{q - 1} \subset {\mathbb{Z}}_{L} \) be such that \( {x}^{\prime } \equiv x\left( {\;\operatorname{mod}\;\mathfrak{P}}\right) \), so that \( \omega \left( x\right) = {x}^{\prime } \... | Yes |
(1) For all \( r \in \mathbb{Z} \) we have\n\n\[ s\left( r\right) = \left( {p - 1}\right) \mathop{\sum }\limits_{{0 \leq i < f}}\left\{ \frac{{p}^{i}r}{q - 1}\right\} . \] | Proof. (1). Both sides of the formula are periodic of period dividing \( q - 1 \) ; hence we may assume that \( 0 \leq r < q - 1 \), so that \( r = \mathop{\sum }\limits_{{0 \leq j < f}}{r}_{j}{p}^{j} \) with \( 0 \leq {r}_{j} \leq p - 1 \) . For \( 0 \leq i \leq f - 1 \) we have\n\n\[ {p}^{i}r = \mathop{\sum }\limits_... | Yes |
Corollary 3.6.8. We have \( {v}_{\mathfrak{P}}\left( {\tau \left( {\omega }^{-r}\right) }\right) = s\left( r\right) \) . | Proof. By definition \( t\left( r\right) \) is coprime to \( p \) hence is invertible modulo \( \mathfrak{P} \), so by the theorem \( {v}_{\mathfrak{P}}\left( {\tau \left( {\omega }^{-r}\right) }\right) = s\left( r\right) {v}_{\mathfrak{P}}\left( {\zeta - 1}\right) = s\left( r\right) e\left( {\mathfrak{P}/\mathfrak{p}}... | Yes |
Proposition 3.6.10. Let \( m \mid \left( {q - 1}\right) \), set \( d = \left( {q - 1}\right) /m \), and recall that \( {L}_{m} = \mathbb{Q}\left( {{\zeta }_{m},{\zeta }_{p}}\right) \) and that \( {\mathfrak{P}}_{m} \) is the prime ideal of \( {L}_{m} \) below \( \mathfrak{P} \) . Then \[ \tau \left( {\omega }^{-{rd}}\r... | Proof. First note that the values of \( {\omega }^{-{rd}} \) are in \( {K}_{m} \), so that \( \tau \left( {\omega }^{-{rd}}\right) \in {L}_{m} \) . Since \( {\mathfrak{P}}_{m} \) is a prime ideal of \( {L}_{m} \) above \( \mathfrak{p} \), by Galois theory all the prime ideals of \( {L}_{m} \) above \( \mathfrak{p} \) h... | Yes |
Corollary 3.6.11. Let \( m \mid \left( {q - 1}\right) \), and recall that \( {K}_{m} = \mathbb{Q}\left( {\zeta }_{m}\right) \) and that \( {\mathfrak{p}}_{m} \) is the prime ideal of \( {K}_{m} \) below \( \mathfrak{P} \) . Then\n\n\[ \tau {\left( {\omega }^{-d}\right) }^{m}{\mathbb{Z}}_{{K}_{m}} = \mathop{\prod }\limi... | Proof. By Corollary 2.5.15 we know that \( \tau {\left( {\omega }^{-d}\right) }^{m} \in {K}_{m} \) . Since \( e\left( {{\mathfrak{P}}_{m}/{\mathfrak{p}}_{m}}\right) = \) \( e\left( {\mathfrak{p}/p}\right) = p - 1 \) we have\n\n\[ {v}_{{\mathfrak{p}}_{m}}\left( {\tau {\left( {\omega }^{-d}\right) }^{m}}\right) = \left( ... | Yes |
Proposition 3.6.13. Set\n\n\[ \Theta = \mathop{\sum }\limits_{{t \in {\left( \mathbb{Z}/m\mathbb{Z}\right) }^{ * }}}\left\{ \frac{t}{m}\right\} {\sigma }_{t}^{-1} = \frac{1}{m}\mathop{\sum }\limits_{\substack{{1 \leq t \leq m - 1} \\ {\gcd \left( {t, m}\right) = 1} }}t{\sigma }_{t}^{-1}. \]\n\nWith the same notation as... | Proof. Let \( T \) be a system of representatives of \( {\left( \mathbb{Z}/m\mathbb{Z}\right) }^{ * } \) modulo the cyclic subgroup \( \langle p\rangle \) generated by \( p \) . Proposition 3.6.11 can be restated by saying that \( \tau {\left( {\omega }^{-d}\right) }^{m}{\mathbb{Z}}_{{K}_{m}} = {\mathfrak{p}}_{m}^{m\th... | Yes |
Corollary 3.6.14. Let \( \mathbb{Q}\left( {\zeta }_{m}\right) \) be a cyclotomic field and let \( \Theta \) be defined as above. For all fractional ideals \( \mathfrak{a} \) of \( \mathbb{Q}\left( {\zeta }_{m}\right) \) the ideal \( {\mathfrak{a}}^{m\Theta } \) is a principal ideal. | Proof. We know that in any ideal class there exists an integral ideal co-prime to any fixed ideal, in particular to \( m \) . In other words, there exists an element \( \alpha \) such that \( \mathfrak{b} = \alpha \mathfrak{a} \) is an integral ideal coprime to \( m \) . If \( {\mathfrak{p}}_{m} \) is a prime ideal div... | Yes |
Lemma 3.6.16. We have \( {\Theta }_{b} \in {I}_{s}\left( m\right) \) . More precisely, we have\n\n\[ \n{\Theta }_{b} = - \mathop{\sum }\limits_{{1 \leq t \leq m - 1,\gcd \left( {t, m}\right) = 1}}\left\lfloor \frac{bt}{m}\right\rfloor {\sigma }_{t}^{-1}.\n\] | Proof. Since evidently \( {\Theta }_{b} \in \mathbb{Z}\left\lbrack G\right\rbrack \Theta \), we must show that \( {\Theta }_{b} \in \mathbb{Z}\left\lbrack G\right\rbrack \) . Setting \( u = {b}^{-1}t \) we have\n\n\[ \n{\sigma }_{b}\Theta = \mathop{\sum }\limits_{{t \in {\left( \mathbb{Z}/m\mathbb{Z}\right) }^{ * }}}\{... | Yes |
Lemma 3.6.17. The ideal \( {I}_{s}\left( m\right) \) is generated by the \( {\Theta }_{b} \) as a \( \mathbb{Z} \) -module (hence also as an ideal). More precisely, it is generated over \( \mathbb{Z} \) by \( {\Theta }_{m + 1} \) and the \( {\Theta }_{b} \) for \( 1 \leq b \leq m \) with \( \gcd \left( {b, m}\right) = ... | Proof. By definition an element \( \delta \in {I}_{s}\left( m\right) \) has the form \( \delta = {\gamma \Theta } \), where \( {\gamma \Theta } \in \mathbb{Z}\left\lbrack G\right\rbrack \) and \( \gamma \in \mathbb{Z}\left\lbrack G\right\rbrack \) . If we write \( \gamma = \mathop{\sum }\limits_{{t \in {\left( \mathbb{... | Yes |
Lemma 3.6.18. Recall that \( d = \left( {q - 1}\right) /m \) . For all \( b \) coprime to \( m \) we have \( \tau {\left( {\omega }^{-d}\right) }^{{\sigma }_{b} - b} \in {K}_{m}. \) | Proof. Since \( m \) is coprime to \( p,\operatorname{Gal}\left( {{L}_{m}/{K}_{m}}\right) \) is the cyclic group of order \( p - 1 \) formed by the maps \( {\alpha }_{k} \) such that \( {\alpha }_{k}\left( {\zeta }_{p}\right) = {\zeta }_{p}^{k} \) and \( {\alpha }_{k}\left( {\zeta }_{m}\right) = {\zeta }_{m} \) for \( ... | Yes |
Theorem 3.6.19 (Stickelberger). The Stickelberger ideal \( {I}_{s}\left( m\right) \) annihilates the class group of \( {K}_{m} = \mathbb{Q}\left( {\zeta }_{m}\right) \) ; in other words, for any \( \gamma \in {I}_{s}\left( m\right) \) and any fractional ideal \( \mathfrak{a} \) of \( {K}_{m} \) the ideal \( {\mathfrak{... | Proof. As in the proof of Corollary 3.6.14, it is sufficient to prove that \( {\mathfrak{p}}_{m}^{\gamma } \) is a principal ideal for any \( \gamma \in {I}_{s}\left( m\right) \) and any prime ideal \( {\mathfrak{p}}_{m} \) coprime to \( m \) . Raising the equality of Proposition 3.6.13 to the power \( {\sigma }_{b} - ... | Yes |
Proposition 3.6.20. Let \( m ≢ 2\left( {\;\operatorname{mod}\;4}\right) \) and denote by \( {K}_{m}^{ + } = \mathbb{Q}\left( {{\zeta }_{m} + {\zeta }_{m}^{-1}}\right) \) the maximal totally real subfield of \( {K}_{m} \) . Then for any ideal \( \mathfrak{a} \) of \( {K}_{m}^{ + } \) and any \( b \) coprime to \( m \) w... | Proof. Recall that we denote complex conjugation by \( \iota \), and that \( \iota {\sigma }_{t} = \) \( {\sigma }_{m - t} \), so that when \( \mathfrak{a} \subset {K}_{m}^{ + } \) we have \( {\sigma }_{m - t}\left( {\mathfrak{a}{\mathbb{Z}}_{{K}_{m}}}\right) = {\sigma }_{t}\left( {\mathfrak{a}{\mathbb{Z}}_{{K}_{m}}}\r... | Yes |
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