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Proposition 9.9.7. Let \( q \) be prime \( q \equiv 1\\left( 4\\right) \) . Then \( {\\chi }_{\\pi }\\left( q\\right) = {\\chi }_{q}\\left( \\pi \\right) \) . | Proof. Since \( q \equiv 1\\left( 4\\right) \)\n\n\\[ \ng{\\left( {\\chi }_{\\pi }\\right) }^{q} \equiv \sum {\\chi }_{\\pi }{\\left( j\\right) }^{q}{\\zeta }^{qj} \equiv \sum {\\chi }_{\\pi }\\left( j\\right) {\\zeta }^{qj} \equiv {\\bar{\\chi }}_{\\pi }\\left( q\\right) g\\left( {\\chi }_{\\pi }\\right) \\left( q\\ri... | Yes |
Lemma 1. \( {\chi }_{\pi }\left( q\right) = 1 \) iff \( {x}^{4} \equiv q\left( p\right) \) has a solution with \( x \in \mathbb{Z} \) . | Proof. By Proposition 9.8.1 the integers \( 0,1,2,\ldots, p - 1 \) form a complete set of residues for the residue classes of \( \mathbb{Z}\left\lbrack i\right\rbrack \) modulo \( \pi \) . Thus \( {\chi }_{\pi }\left( q\right) = 1 \) iff \( {x}^{4} \equiv q\left( \pi \right) \) has a solution with \( x \in \mathbb{Z} \... | Yes |
Lemma 2. If \( {\psi }_{p}\left( q\right) = 1 \) then \( {\chi }_{\pi }\left( q\right) = \pm 1 \) . | Proof. Since \( {q}^{\left( {p - 1}\right) /2} \equiv 1\left( p\right) \) it follows that \( {\chi }_{\pi }^{2}\left( q\right) \equiv {\left( {q}^{\left( {p - 1}\right) /4}\right) }^{2} \equiv {q}^{\left( {p - 1}\right) /2} \equiv \) \( 1\left( \pi \right) \) . Thus \( {\chi }_{\pi }^{2}\left( q\right) = 1 \) . Thus, a... | Yes |
Proposition 9.10.1. Let \( \pi \) be the primary irreducible dividing \( p \) . Then\n\n\[ g{\left( {\chi }_{\pi }\right) }^{2} = - {\left( -1\right) }^{\left( {p - 1}\right) /4}\sqrt{p}\pi \]\n\nwhere \( \sqrt{p} \) denotes the positive square root. | Proof. By Proposition 9.9.4 and Theorem 1, Chapter 8 we have\n\n\[ J\left( {{\chi }_{\pi },{\chi }_{\pi }}\right) = - {\chi }_{\pi }\left( {-1}\right) \pi = \frac{g{\left( {\chi }_{\pi }\right) }^{2}}{g\left( {\psi }_{p}\right) }.\]\n\nThe proposition follows from Theorem 1, Chapter 6 and the observation that \( {\chi ... | Yes |
Proposition 9.10.2. If \( \pi \) is a primary irreducible dividing \( p \) then \( {\chi }_{\pi }\left( q\right) {\chi }_{\lambda }\left( p\right) \equiv \) \( {\pi }^{\left( {q - 1}\right) /2}\left( q\right) \) . | Proof. We have, in the ring of all algebraic integers,\n\n\[ g{\left( {\chi }_{\pi }\right) }^{q} = {\left( \sum {\chi }_{\pi }\left( j\right) {\zeta }^{j}\right) }^{q} \]\n\n\[ \equiv \sum {\chi }_{\pi }\left( j\right) {\zeta }^{qj}\left( q\right) \]\n\n\[ \equiv {\chi }_{\pi }\left( {q}^{-1}\right) g\left( {\chi }_{\... | Yes |
Proposition 9.10.3. \( {\pi }^{\left( {q - 1}\right) /2} \equiv {\psi }_{q}\left( d\right) {\psi }_{q}\left( {{ad} - {bc}}\right) \left( q\right) \). | Proof. Since \( {d\pi } \equiv {ad} - {bc}\left( \lambda \right) \) one has\n\n\[ \n{\left( d\pi \right) }^{\left( {q - 1}\right) /2} \equiv {\left( ad - bc\right) }^{\left( {q - 1}\right) /2}\left( \lambda \right) .\n\]\n\nThus\n\n\[ \n{\psi }_{q}\left( d\right) {\pi }^{\left( {q - 1}\right) /2} \equiv {\psi }_{q}\lef... | Yes |
Lemma 3. \( {\psi }_{q}\left( {{ad} - {bc}}\right) = {\psi }_{q}\left( {{ad} + {bc}}\right) \). | Proof. Since \( {c}^{2} \equiv - {d}^{2}\left( q\right) \) one has\n\n\[ \n{\psi }_{q}\left( {{ad} - {bc}}\right) {\psi }_{q}\left( {{ad} + {bc}}\right) = {\psi }_{q}\left( {{a}^{2}{d}^{2} - {b}^{2}{c}^{2}}\right) = {\psi }_{q}\left( {{d}^{2}p}\right) = {\psi }_{q}\left( p\right) = 1. \n\] | Yes |
Lemma 4. If \( q = {c}^{2} + {d}^{2}, c > 0, c \equiv 1\\left( 2\\right) \) then \( {\\psi }_{q}\\left( d\\right) = {\\left( -1\\right) }^{\\left( {q - 1}\\right) /4} \) . | Proof. Let \( {\\psi }_{\\mathrm{c}} \) denote the Jacobi symbol. Then by Proposition 5.2.2 one has \( {\\psi }_{q}\\left( c\\right) = {\\psi }_{c}\\left( q\\right) = {\\psi }_{c}\\left( {d}^{2}\\right) = 1 \) . (Cf. Exercise 26, Chapter 5). But \( {c}^{2} \equiv - {d}^{2}\\left( q\\right) \) implies \( {c}^{\\left( {q... | No |
Lemma 1. \( {\zeta }_{{2}^{n}} \) is constructible, \( n = 1,2,\ldots \) . | Proof. Since \( {\left( {\zeta }_{{2}^{n}}\right) }^{2} = {\zeta }_{{2}^{n - 1}} \) the result follows by induction \( \left( {\zeta }_{2}\right. \) is certainly constructible!). | No |
Lemma 2. \[ \mathop{\sum }\limits_{\chi }\chi \left( t\right) = \left\{ \begin{array}{ll} 1, & \text{ if }t = 0 \\ p - 1, & \text{ if }t = 1 \\ 0, & \text{ if }t \neq 0,1 \end{array}\right. \] the sum being over all characters of \( {F}_{p}^{ * } \) . | Proof. If \( \chi = \varepsilon \), the trivial character then \( \varepsilon \left( 0\right) = 1 \) . Thus the result holds for \( t = 0 \) . It is true when \( t = 1 \) by Proposition 8.1.3 while the remaining case is the corollary to Proposition 8.1.3. | No |
Theorem 4. If \( p \) is a Fermat prime then \( {\zeta }_{p} \) is constructible. | Proof. If \( g\left( \chi \right) = \mathop{\sum }\limits_{{t = 0}}^{{p - 1}}\chi \left( t\right) {\zeta }_{p}^{t} \) is the Gauss sum associated with \( \chi \) then\n\n\[ \mathop{\sum }\limits_{\chi }g\left( \chi \right) = \mathop{\sum }\limits_{{t = 0}}^{{p - 1}}\left( {\mathop{\sum }\limits_{\chi }\chi \left( t\rig... | Yes |
Lemma 1. \( G = g\left( {\chi }_{\pi }\right) + \overline{g\left( {\chi }_{\pi }\right) } \) . | Proof. If \( \zeta = {e}^{{2\pi i}/p} \) then since \( G \) is real, and \( - 1 = {\left( -1\right) }^{3} \)\n\n\[ G = \mathop{\sum }\limits_{{t = 0}}^{{p - 1}}{\zeta }^{{t}^{3}} = \mathop{\sum }\limits_{{t = 0}}^{{p - 1}}{\zeta }^{t}\left( {1 + {\chi }_{\pi }\left( t\right) + {\chi }_{\pi }\left( {t}^{2}\right) }\righ... | Yes |
Lemma 2. \( G \) is a real root of \( {x}^{3} - {3px} - \left( {{2a} - b}\right) p = 0 \) . | Proof. By Lemma 1, writing \( \chi \) for \( {\chi }_{\pi } \), \[ {G}^{3} = g{\left( \chi \right) }^{3} + \overline{g{\left( \chi \right) }^{3}} + {3g}\left( \chi \right) \overline{g\left( \chi \right) }\left( {g\left( \chi \right) + \overline{g\left( \chi \right) }}\right) \] \[ = {p\pi } + p\bar{\pi } + {3pG} \] \[ ... | Yes |
Lemma 1. Let \( \bar{H} \) be the set of \( \left\lbrack x\right\rbrack \in {P}^{n}\left( F\right) \) such that \( {x}_{0} = 0 \) . Then \( \phi \) maps \( {P}^{n}\left( F\right) - \bar{H} \) to \( {A}^{n}\left( F\right) \) and this map is one to one and onto. | Proof. If \( \phi \left( \left\lbrack x\right\rbrack \right) = \phi \left( \left\lbrack y\right\rbrack \right) \), then \( {x}_{i}/{x}_{0} = {y}_{i}/{y}_{0} \) for \( i = 0,1,\ldots, n \) . Let \( \gamma = {y}_{0}/{x}_{0} \) . Then \( \gamma {x}_{i} = {y}_{i} \) for \( i = 0,1,\ldots, n \) and so \( \left\lbrack x\righ... | Yes |
Lemma 2. \( \widetilde{f}\left( y\right) \) is a homogeneous polynomial of degree equal to \( \deg f \) . Moreover, \( f\left( {1,{y}_{1},{y}_{2},\ldots ,{y}_{n}}\right) = f\left( {{y}_{1},{y}_{2},\ldots ,{y}_{n}}\right) \) . | Proof. Set \( d = \deg f \) and consider a monomial \( {x}_{1}^{{i}_{1}}{x}_{2}^{{i}_{2}}\cdots {x}_{n}^{{i}_{n}} \) of degree \( l \leq d \) . Then \( {y}_{0}^{d}{\left( {y}_{1}/{y}_{0}\right) }^{{i}_{1}}\cdots {\left( {y}_{n}/{y}_{0}\right) }^{{i}_{n}} = {y}_{0}^{d - l}{y}_{1}^{{i}_{1}}{y}_{2}^{{i}_{2}}\cdots {y}_{n}... | Yes |
Lemma 1. Let \( f\left( {{x}_{1},{x}_{2},\ldots ,{x}_{n}}\right) \) be a polynomial that is of degree less than \( q \) in each of its variables. Then if \( f \) vanishes on all of \( {A}^{n}\left( F\right) \), it is the zero polynomial. | Proof. The proof is by induction on \( n \) . If \( n = 1, f\left( x\right) \) is a polynomial in one variable of degree less than \( q \) with \( q \) distinct roots, namely, all the elements of \( F \) . Thus \( f \) is identically zero.\n\nSuppose that we have proved the result for \( n - 1 \) and consider\n\n\[ f\l... | Yes |
Lemma 2. Each polynomial \( f\left( x\right) \in F\left\lbrack {{x}_{1},\ldots ,{x}_{n}}\right\rbrack \) is equivalent to a reduced polynomial. | Proof. Consider the case of one variable. Clearly \( {x}^{q} \sim x \) . If \( m > 0 \) is an integer, let \( l \) be the least positive integer such that \( {x}^{m} \sim {x}^{l} \) . We claim that \( l < q \) . If not, \( l = {qs} + r \) with \( 0 \leq r < q \) and \( s \neq 0 \) . Then \( {x}^{l} = {\left( {x}^{q}\ri... | Yes |
Lemma 3. Let \( {i}_{1},{i}_{2},\ldots ,{i}_{n} \) be nonnegative integers. Then unless each \( {i}_{j} \) is nonzero and divisible by \( q - 1 \) we have\n\n\[ \mathop{\sum }\limits_{{a \in {A}^{n}\left( F\right) }}{a}_{1}^{{i}_{1}}{a}_{2}^{{i}_{2}}\cdots {a}_{n}^{{i}_{n}} = 0. \]\n | Proof. Suppose first that \( n = 1 \) . If \( i = 0 \), then \( \mathop{\sum }\limits_{{a \in F}}{a}^{0} = q = 0 \) in \( F \) . Suppose \( i \neq 0.{F}^{ * } \) is cyclic. Let \( b \) be a generator. If \( q - 1 \nmid i \), then\n\n\[ \mathop{\sum }\limits_{{a \in F}}{a}^{i} = \mathop{\sum }\limits_{{k = 0}}^{{q - 2}}... | Yes |
Proposition 10.3.1. If \( \alpha ,\beta \in F \) and \( a \in {F}_{p} \), then\n\n(a) \( \operatorname{tr}\left( \alpha \right) \in {F}_{p} \).\n\n(b) \( \operatorname{tr}\left( {\alpha + \beta }\right) = \operatorname{tr}\left( \alpha \right) + \operatorname{tr}\left( \beta \right) \).\n\n(c) \( \operatorname{tr}\left... | Proof.\n\n(a) We have\n\n\[{\left( \alpha + {\alpha }^{p} + \cdots + {\alpha }^{{p}^{n - 1}}\right) }^{p} = {\alpha }^{p} + {\alpha }^{{p}^{2}} + \cdots + {\alpha }^{{p}^{n - 1}} + {\alpha }^{{p}^{n}}.\]\n\nSince \( {\alpha }^{{p}^{n}} = {\alpha }^{q} = \alpha \) we see that \( \operatorname{tr}{\left( \alpha \right) }... | Yes |
Proposition 10.3.2. The function \( \psi \) has the following properties:\n\n(a) \( \psi \left( {\alpha + \beta }\right) = \psi \left( \alpha \right) \psi \left( \beta \right) \) .\n\n(b) There is an \( \alpha \in F \) such that \( \psi \left( \alpha \right) \neq 1 \) .\n\n(c) \( \mathop{\sum }\limits_{{a \in F}}\psi \... | Proof.\n\n(a) \( \psi \left( {\alpha + \beta }\right) = {\zeta }_{p}^{\operatorname{tr}\left( {\alpha + \beta }\right) } = {\zeta }_{p}^{\operatorname{tr}\left( \alpha \right) + \operatorname{tr}\left( \beta \right) } = {\zeta }_{p}^{\operatorname{tr}\left( \alpha \right) }{\zeta }_{p}^{\operatorname{tr}\left( \beta \r... | Yes |
Proposition 10.3.3. Let \( \alpha, x, y \in F \) . Then\n\n\[ \frac{1}{q}\mathop{\sum }\limits_{{x \in F}}\psi \left( {\alpha \left( {x - y}\right) }\right) = \delta \left( {x, y}\right) \]\n\nwhere \( \delta \left( {x, y}\right) = 1 \) if \( x = y \) and zero otherwise. | Proof. If \( x = y \), then \( \mathop{\sum }\limits_{{\alpha \in F}}\psi \left( {\alpha \left( {x - y}\right) }\right) = \mathop{\sum }\limits_{{\alpha \in F}}\psi \left( 0\right) = q \) .\n\nIf \( x \neq y \), then \( x - y \neq 0 \) and \( \alpha \left( {x - y}\right) \) ranges over all of \( F \) as \( x \) ranges ... | Yes |
Theorem 2. Suppose that \( F \) is a field with \( q \) elements and \( q \equiv 1\\left( m\\right) \) . The homogeneous equation \( {a}_{0}{y}_{0}^{m} + {a}_{1}{y}_{1}^{m} + \\cdots + {a}_{n}{y}_{n}^{m} = 0,{a}_{0},{a}_{1},\\ldots ,{a}_{n} \\in {F}^{ * } \), defines a hypersurface in \( {P}^{n}\\left( F\\right) \) . T... | Proof. The number of points \( N \) on the hypersurface in \( {A}^{n + 1}\\left( F\\right) \) defined by \( {a}_{0}{y}_{0}^{m} + {a}_{1}{y}_{1}^{m} + \\cdots + {a}_{n}{y}_{n}^{m} = 0 \) is given by\n\n\[ \n{q}^{n} + \\mathop{\\sum }\\limits_{{{\\chi }_{0},{\\chi }_{1}\\ldots .,{\\chi }_{n}}}{\\chi }_{0}\\left( {a}_{0}^... | Yes |
Proposition 11.1.1. The zeta function is rational iff there exist complex numbers \( {\alpha }_{i} \) and \( {\beta }_{j} \) such that\n\n\[ \n{N}_{s} = \mathop{\sum }\limits_{j}{\beta }_{j}^{s} - \mathop{\sum }\limits_{i}{\alpha }_{i}^{s} \n\] | Proof. Suppose that the zeta function is rational. Then by the above remarks\n\n\[ \nZ\left( u\right) = \frac{\mathop{\prod }\limits_{i}\left( {1 - {\alpha }_{i}u}\right) }{\mathop{\prod }\limits_{j}\left( {1 - {\beta }_{j}u}\right) } \n\]\n\nwith \( {\alpha }_{i},{\beta }_{j} \in \mathbb{C} \) . Taking the logarithmic... | Yes |
Proposition 11.1.2. \( E \) and \( {E}^{\prime } \) are isomorphic over \( F \) ; i.e., there exists a map \( \sigma : E \rightarrow {E}^{\prime } \) such that\n\n(a) \( \sigma \) is one to one and onto.\n\n(b) \( \sigma \left( a\right) = a \) for all \( a \in F \) .\n\n(c) \( \sigma \left( {\alpha + \beta }\right) = \... | Proof. We shall show that both \( E \) and \( {E}^{\prime } \) are isomorphic over \( F \) to \( F\left\lbrack x\right\rbrack /\left( {f\left( x\right) }\right) \) for some irreducible polynomial \( f\left( x\right) \in F\left\lbrack x\right\rbrack \) .\n\nTo begin with there is an \( {\alpha }^{\prime } \in {E}^{\prim... | Yes |
Proposition 11.1.3. \( {Z}_{v}\left( u\right) = \mathop{\prod }\limits_{\mathfrak{P}}\left( {1/\left( {1 - {u}^{\deg \mathfrak{P}}}\right) }\right) \) . | Proof. The right-hand side is clearly\n\n\[ \mathop{\prod }\limits_{{n = 1}}^{\infty }{\left( \frac{1}{1 - {u}^{n}}\right) }^{{a}_{n}} \]\n\nThe logarithmic derivative of this expression is\n\n\[ \frac{1}{u}\mathop{\sum }\limits_{{n = 1}}^{\infty }\frac{n{a}_{n}{u}^{n}}{1 - {u}^{n}}. \]\n\nExpanding the denominator int... | Yes |
Proposition 11.2.2. If \( \alpha ,\beta \in E \) and \( a \in F \), then\n\n(a) \( {N}_{E/F}\left( \alpha \right) \in F \) .\n\n(b) \( {N}_{E/F}\left( {\alpha \beta }\right) = {N}_{E/F}\left( \alpha \right) {N}_{E/F}\left( \beta \right) \) .\n\n(c) \( {N}_{E/F}\left( {ax}\right) = {a}^{s}{N}_{{E}_{l}F}\left( \alpha \ri... | Proof. The proof of Proposition 11.2.1 is exactly analogous to that of Proposition 10.3.1 and will be omitted.\n\nTo prove Proposition 11.2.2 notice that\n\n\[ \n{N}_{E/F}{\left( \alpha \right) }^{q} = {\left( \alpha \cdot {\alpha }^{q} \cdot \cdots \cdot {\alpha }^{{q}^{s - 1}}\right) }^{q} = {\alpha }^{q} \cdot {\alp... | Yes |
Proposition 11.2.3. Let \( F \subset E \subset K \) be three finite fields and \( \alpha \in K \) . Then\n\n(a) \( {\operatorname{tr}}_{K/F}\left( \alpha \right) = {\operatorname{tr}}_{E/F}\left( {{\operatorname{tr}}_{K/E}\left( \alpha \right) }\right) \) . | Proof. We shall prove only property (a). The proof of property (b) is similar.\n\nLet \( d = \left\lbrack {E : F}\right\rbrack, m = \left\lbrack {K : E}\right\rbrack \), and \( n = \left\lbrack {K : F}\right\rbrack \) . As we have pointed out above, \( n = {dm} \).\n\nThe number of elements in \( E \) is \( {q}_{1} = {... | Yes |
Proposition 11.2.4. Write \( f\left( x\right) = {x}^{d} - {c}_{1}{x}^{d - 1} + \cdots + {\left( -1\right) }^{d}{c}_{d} \). Then\n\n(a) \( f\left( x\right) = \left( {x - \alpha }\right) \left( {x - {\alpha }^{q}}\right) \cdots \left( {x - {\alpha }^{{q}^{d - 1}}}\right) \).\n\n(b) \( {\operatorname{tr}}_{K/F}\left( \alp... | Proof. Since the coefficients of \( f \) satisfy \( {a}^{q} = a \) we have\n\n\[ 0 = f{\left( \alpha \right) }^{q} = f\left( {\alpha }^{q}\right) \]\n\nThus \( {\alpha }^{q} \) is a root of \( f \). Similarly,\n\n\[ 0 = f{\left( {\alpha }^{q}\right) }^{q} = f\left( {\alpha }^{{q}^{2}}\right) . \]\n\nThus \( {\alpha }^{... | Yes |
Theorem 2. Let \( {a}_{0},{a}_{1},\ldots ,{a}_{n} \in {F}^{ * } \), where \( F \) is a finite field with \( q \) elements, and \( q \equiv 1\left( m\right) \). Let \( f\left( {{x}_{0},\ldots ,{x}_{n}}\right) = {a}_{0}{x}_{0}^{m} + {a}_{1}{x}_{1}^{m} + \cdots + {a}_{n}{x}_{n}^{m} \). Then the zeta function \( {Z}_{f}\le... | \[ \frac{P{\left( u\right) }^{{\left( -1\right) }^{n}}}{\left( {1 - u}\right) \left( {1 - {qu}}\right) \cdots \left( {1 - {q}^{n - 1}u}\right) }, \] where \( P\left( u\right) \) is the polynomial \[ \mathop{\prod }\limits_{{{\chi }_{0},{\chi }_{1},\ldots ,{\chi }_{n}}}\left( {1 - {\left( -1\right) }^{n + 1}\frac{1}{q}{... | Yes |
Lemma 1. \( \lambda \left( {fg}\right) = \lambda \left( f\right) \lambda \left( g\right) \) for all monic \( f, g \in F\left\lbrack x\right\rbrack \) . | Proof. If \( g\left( x\right) = {x}^{m} - {b}_{1}{x}^{m - 1} + \cdots + {\left( -1\right) }^{m}{b}_{m} \), then \( f\left( x\right) g\left( x\right) = {x}^{n + m} - \) \( \left( {{b}_{1} + {c}_{1}}\right) {x}^{n + m - 1} + \cdots + {\left( -1\right) }^{n + m}{b}_{m}{c}_{n} \) . Thus \( \lambda \left( {fg}\right) = \psi... | Yes |
Lemma 2. Let \( \\alpha \\in {F}_{s} \) and \( f\\left( x\\right) \) be the monic irreducible polynomial for \( \\alpha \) over \( F \) . Then \[ \\lambda {\\left( f\\right) }^{{s}_{i}d} = {\\chi }^{\\prime }\\left( \\alpha \\right) {\\psi }^{\\prime }\\left( \\alpha \\right) ,\\;\\text{ where }d = \\deg f. \] | Proof. This result follows easily from Proposition 11.2.4. Namely, if \( f\\left( x\\right) = \) \( {x}^{d} - {c}_{1}{x}^{d - 1} + \\cdots + {\\left( -1\\right) }^{d}{c}_{d} \), then \[ {\\operatorname{tr}}_{{F}_{s}/F}\\left( \\alpha \\right) = \\frac{s}{d}{c}_{1}\\;\\text{ and }\\;{N}_{{F}_{s}/F}\\left( \\alpha \\righ... | Yes |
Lemma 3. \( g\left( {\chi }^{\prime }\right) = \sum \left( {\deg f}\right) \lambda {\left( f\right) }^{{s}_{i}\deg f} \), where the sum is over all monic irreducible polynomials of \( F\left\lbrack x\right\rbrack \) with degree dividing \( s \) . | Proof. According to Theorem 1 of Chapter 7-generalized to \( F \) as base field \( - {x}^{{q}^{s}} - x \) is the product of all monic irreducible polynomials of degree dividing \( s \) . It follows that every such irreducible polynomial has all its roots in \( {F}_{s} \) and conversely that every element in \( {F}_{s} ... | Yes |
Proposition 12.1.1. If \( \Delta \left( {{\alpha }_{1},\ldots ,{\alpha }_{n}}\right) \neq 0 \) then \( {\alpha }_{1},\ldots ,{\alpha }_{n} \) is a basis for \( L/K \) . If \( L/K \) is separable and \( {\alpha }_{1},\ldots ,{\alpha }_{n} \) is a basis for \( L/K \) then \( \Delta \left( {{\alpha }_{1},\ldots ,{\alpha }... | Proof. Suppose \( {\alpha }_{1},\ldots ,{\alpha }_{n} \) are linearly dependent. Then there exist \( {a}_{1},\ldots ,{a}_{n} \in K \), not all zero, such that \( \sum {a}_{i}{\alpha }_{i} = 0 \) . Multiply this equation by \( {\alpha }_{j} \) and take the trace. One finds\n\n\[ \mathop{\sum }\limits_{i}{a}_{i}t\left( {... | Yes |
Proposition 12.1.2. Suppose \( {\alpha }_{1},\ldots ,{\alpha }_{n} \) and \( {\beta }_{1},\ldots ,{\beta }_{n} \) are bases for \( L/K \) . Let \( {\alpha }_{i} = \mathop{\sum }\limits_{j}{a}_{ij}{\beta }_{j},{a}_{ij} \in K \) . Then \( \Delta \left( {{\alpha }_{1},\ldots ,{\alpha }_{n}}\right) = \det {\left( {a}_{ij}\... | Proof. Take the trace of both sides of the identity \( {\alpha }_{i}{\alpha }_{k} = \mathop{\sum }\limits_{j}\mathop{\sum }\limits_{l}{a}_{ij}{a}_{kl}{\beta }_{j}{\beta }_{l} \) . Let \( A = \left( {t\left( {{\alpha }_{i}{\alpha }_{j}}\right) }\right), B = \left( {t\left( {{\beta }_{j}{\beta }_{l}}\right) }\right) \), ... | Yes |
Proposition 12.1.3. For \( {\alpha }_{1},{\alpha }_{2},\ldots ,{\alpha }_{n} \in L \) and \( L/K \) separable we have\n\n\[ \Delta \left( {{\alpha }_{1},\ldots ,{\alpha }_{n}}\right) = \det {\left( {\alpha }_{i}^{\left( j\right) }\right) }^{2}. \] | Proof. \( t\left( {{\alpha }_{i}{\alpha }_{j}}\right) = {\alpha }_{i}^{\left( 1\right) }{\alpha }_{j}^{\left( 1\right) } + {\alpha }_{i}^{\left( 2\right) }{\alpha }_{j}^{\left( 2\right) } + \cdots + {\alpha }_{i}^{\left( n\right) }{\alpha }_{j}^{\left( n\right) } \) . Let \( A = \left( {t\left( {{\alpha }_{i}{\alpha }_... | Yes |
Proposition 12.1.4. Suppose \( 1,\beta ,\ldots ,{\beta }^{n - 1} \) are in \( L \) and linearly independent over \( K \) . Let \( f\left( x\right) \in K\left\lbrack x\right\rbrack \) be the minimal polynomial for \( \beta \) over \( K \) . If \( L/K \) is separable then\n\n\[ \n\Delta \left( {1,\beta ,\ldots ,{\beta }^... | Proof. The matrix \( \left( {\left( {\beta }^{\left( j\right) }\right) }^{i}\right) \) where \( j = 1,\ldots, n \) and \( i = 0,\ldots, n - 1 \) is of Vandermonde type and so its determinant is\n\n\[ \n\mathop{\prod }\limits_{{i < j}}\left( {{\beta }^{\left( j\right) } - {\beta }^{\left( i\right) }}\right) \n\]\n\nThus... | Yes |
Lemma 1. Suppose \( \beta \in F \) . There is a \( b \in \mathbb{Z}, b \neq 0 \), such that \( {b\beta } \in D \) . | Proof. \( \beta \) satisfies an equation \( {a}_{0}{\beta }^{n} + {a}_{1}{\beta }^{n - 1} + \cdots + {a}_{n} = 0 \) with the \( {a}_{i} \in \mathbb{Z},{a}_{0} \neq 0 \) . Multiply both sides by \( {a}_{0}^{n - 1} \) and notice that \( {\left( {a}_{0}\beta \right) }^{n} + {a}_{1}{\left( {a}_{0}\beta \right) }^{n - 1} \)... | Yes |
Proposition 12.2.1. Every ideal A of D contains a basis for \( F \) over \( \mathbb{Q} \) . | Proof. Let \( {\beta }_{1},\ldots ,{\beta }_{n} \) be a basis for \( F \) over \( \mathbb{Q} \) . By the preceding lemma there is a \( b \in \mathbb{Z}, b \neq 0 \), such that \( b{\beta }_{1},\ldots, b{\beta }_{n} \in D \) . Choose \( \alpha \in A,\alpha \neq 0 \) . Then the elements \( b{\beta }_{1}\alpha ,\ldots, b{... | Yes |
Proposition 12.2.2. Let \( A \) be an ideal in \( D \) and suppose \( {\alpha }_{1},\ldots ,{\alpha }_{n} \in A \) is a basis for \( F/\mathbb{Q} \) with \( \left| {\Delta \left( {{\alpha }_{1},\ldots ,{\alpha }_{n}}\right) }\right| \) minimal. Then \( A = \mathbb{Z}{\alpha }_{1} + \mathbb{Z}{\alpha }_{2} + \cdots + \m... | Proof. Since the absolute value of the discriminant of a basis in \( A \) is a positive integer, there is such a basis with \( \left| {\Delta \left( {{\alpha }_{1},\ldots ,{\alpha }_{n}}\right) }\right| \) minimal.\n\nSuppose \( \alpha \in A \) and write \( \alpha = {\gamma }_{1}{\alpha }_{1} + {\gamma }_{2}{\alpha }_{... | Yes |
Lemma 2. If \( A \subset D \) is an ideal then \( A \cap \mathbb{Z} \neq 0 \) . | Proof. Let \( \alpha \in A,\alpha \neq 0 \) . There exist \( {a}_{i} \in \mathbb{Z} \) such that \( {\alpha }^{m} + {a}_{1}{\alpha }^{m - 1} + \cdots \) \( + {a}_{m} = 0 \) . Since we are working in a field we may assume \( {a}_{m} \neq 0 \) . But then, \( 0 \neq {a}_{m} \in A \cap \mathbb{Z} \) . | Yes |
Proposition 12.2.3. For any ideal \( A, D/A \) is finite. | Proof. By the lemma there is an \( a \in A \cap \mathbb{Z}, a \neq 0 \) . Let \( \left( a\right) \) be the principal ideal generated by \( a \) in \( D \) . Since \( D/\left( a\right) \) maps onto \( D/A \) it is enough to show \( D/\left( a\right) \) is finite. In fact we will show it has precisely \( {a}^{n} \) eleme... | Yes |
Corollary 1. D is a Noetherian ring, i.e., every ascending chain of ideals \( {A}_{1} \subset {A}_{2} \subset {A}_{3} \subset \cdots \) terminates. In other words, there is an \( N > 0 \) such that \( {A}_{m} = {A}_{m + 1} \) for all \( m \geq N \) . | Proof. Since \( D/{A}_{1} \) is finite there are only finitely many ideals containing \( {A}_{1} \) . | No |
Corollary 2. Every prime ideal of \( D \) is maximal. | Proof. If \( P \) is a prime ideal then \( D/P \) is a finite integral domain. Such a ring is necessarily a field (see Exercise 19). Thus \( D/P \) is a field and so \( P \) is maximal. | No |
Lemma 3. Let \( A \subset D \) be an ideal. If \( \beta \in F \) is such that \( {\beta A} \subset A \) then \( \beta \in D \) . | Proof. By Proposition 12.2.2 \( A \) is a finitely generated \( \mathbb{Z} \) module so the result follows from Proposition 6.1.4. | No |
Proposition 12.2.4. Let \( A, B \subset D \) be ideals and suppose \( \omega \in D \) is such that \( \left( \omega \right) A = {BA} \) . Then \( \left( \omega \right) = B \) . | Proof. If \( \beta \in B \) we see \( \left( {\beta /\omega }\right) A \subset A \) so by Lemma \( 3,\beta /\omega \in D \) . It follows that \( B \subset \left( \omega \right) \) and so \( {\omega }^{-1}B \subset D \) is an ideal. Since \( A = {\omega }^{-1}{BA} \), Lemma 4 shows \( {\omega }^{-1}B = D \) and so \( B ... | Yes |
Lemma 5. There exists a positive integer \( M \) depending only on \( F \) with the following property. Given \( \alpha ,\beta \in D,\beta \neq 0 \), there is an integer \( t,1 \leq t \leq M \) , and an element \( \omega \in D \) such that \( \left| {N\left( {{t\alpha } - {\omega \beta }}\right) }\right| < \left| {N\le... | Proof. We first reformulate the statement slightly. Let \( \gamma = \alpha /\beta \in F \) . Then it is sufficient to show that for all \( \gamma \in F \) there is an \( M \) such that \( \left| {N\left( {{t\gamma } - \omega }\right) }\right| < 1 \) for some \( 1 \leq t \leq M \) and \( \omega \in D \) .\n\nLet \( {\om... | Yes |
Theorem 1. The class number of \( F \) is finite. | Proof. Let \( A \) be an ideal in \( D \) . For \( \alpha \in A,\alpha \neq 0,\left| {N\left( \alpha \right) }\right| \) is a positive integer. Choose \( \beta \in A,\beta \neq 0 \), so that \( \left| {N\left( \beta \right) }\right| \) is minimal. For any \( \alpha \in A \) there is a \( t \) , \( 1 \leq t \leq M \), s... | Yes |
Proposition 12.2.5. For any ideal \( A \subset D \) there is an integer \( k,1 \leq k \leq {h}_{F} \) , such that \( {A}^{k} \) is principal. | Proof. Consider the set of ideals \( \left\{ {{A}^{i} \mid 1 \leq i \leq {h}_{F} + 1}\right\} \) . At least two of these ideals must lie in the same class, say \( {A}^{i} \sim {A}^{j} \) with \( i < j \) . There exist \( \alpha ,\beta \in D \) such that \( \left( \alpha \right) {A}^{i} = \left( \beta \right) {A}^{j} \)... | Yes |
Proposition 12.2.7. If \( A \) and \( B \) are ideals, such that \( B \supset A \), then there is an ideal \( C \) such that \( A = {BC} \) . | Proof. As above there is a \( k > 0 \) such that \( {B}^{k} = \left( \beta \right) \). Now, since \( A \subset B \) we have \( {B}^{k - 1}A \subset {B}^{k} = \left( \beta \right) \) so \( C = \left( {1/\beta }\right) {B}^{k - 1}A \subset D \) is an ideal. Thus, \( {BC} = \left( {1/\beta }\right) {B}^{k}A = \left( {1/\b... | Yes |
Proposition 12.2.8. Every ideal in D can be written as a product of prime ideals. | Proof. Let \( A \) be a proper ideal. Since \( D/A \) is finite, \( A \) is contained in a maximal ideal \( {P}_{1} \) (using Zorn’s lemma one can show that in an arbitrary commutative ring with identity a proper ideal is contained in a maximal ideal). By the last proposition \( A = {P}_{1}{B}_{1} \) for some ideal \( ... | Yes |
Proposition 12.2.9. Let \( P \) be a prime ideal and \( A \) and \( B \) ideals. Then\n\n(i) \( {\operatorname{ord}}_{P}P = 1 \)\n\n(ii) If \( {P}^{\prime } \neq P \) is prime \( {\operatorname{ord}}_{P}{P}^{\prime } = 0 \)\n\n(iii) \( {\operatorname{ord}}_{P}{AB} = {\operatorname{ord}}_{P}A + {\operatorname{ord}}_{P}B... | Proof. The first assertion is clear. As for (ii) assume ord \( {}_{P}{P}^{\prime } > 0 \) . Then \( P \supset {P}^{\prime } \) . Since prime ideals are maximal \( P = {P}^{\prime } \) contradicting the assumption.\n\nLet \( t = {\operatorname{ord}}_{P}A \) and \( s = {\operatorname{ord}}_{P}B \) . By Proposition 12.2.7... | Yes |
Proposition 12.3.1. Let \( R \) be a commutative ring with identity. Suppose \( {A}_{1} \) , \( {A}_{2},\ldots ,{A}_{g} \) are ideals such that \( {A}_{i} + {A}_{j} = R \) for \( i \neq j \) . Let \( A = {A}_{1}{A}_{2}\cdots {A}_{g} \) . Then\n\n\[ R/A \approx R/{A}_{1} \oplus R/{A}_{2} \oplus \cdots \oplus R/{A}_{q}. ... | Proof. Let \( {\psi }_{i} \) be the natural map from \( R \) to \( R/{A}_{i} \) and define \( \psi : R \rightarrow \) \( R/{A}_{1} \oplus \cdots \oplus R/{A}_{g} \) by \( \psi \left( \gamma \right) = \left( {{\psi }_{1}\left( \gamma \right) ,{\psi }_{2}\left( \gamma \right) ,\ldots ,{\psi }_{g}\left( \gamma \right) }\r... | Yes |
Proposition 12.3.2. Let \( P \subset D \) be a prime ideal and let \( {p}^{f} \) be the number of elements in \( D/P \) . The number of elements in \( D/{P}^{e} \) is \( {p}^{ef} \) . | Proof. The assertion is true for \( e = 1 \) . If \( e > 1 \) then \( D/{P}^{e} \) has \( {P}^{e - 1}/{P}^{e} \) as a subgroup and the quotient is isomorphic to \( D/{P}^{e - 1} \) (second law of isomorphism). If we can show \( {P}^{e - 1}/{P}^{e} \) has \( {p}^{f} \) elements then the result will follow by induction.\... | Yes |
Proposition 12.3.3. Let \( p \in \mathbb{Z} \) be a prime number. Suppose \( {P}_{i} \) and \( {P}_{j} \) are prime ideals of \( D \) containing \( p \) . Then there is a \( \sigma \in G \) such that \( \sigma {P}_{i} = {P}_{j} \) . | Proof. Suppose there is a prime ideal \( {P}_{0} \) containing \( p \) and not in the set \( \left\{ {\sigma {P}_{i} \mid \sigma \in G}\right\} \) . By Proposition 12.3.1 we can find an \( \alpha \in D \) such that \( \alpha \equiv 0\left( {P}_{0}\right) \) and \( \alpha \equiv 1\left( {\sigma {P}_{i}}\right) \) for al... | Yes |
Proposition 13.1.1. If \( d \equiv 2,3\left( 4\right) \) then \( D = \mathbb{Z} + \mathbb{Z}\sqrt{d} \) . If \( d \equiv 1\left( 4\right) \) then \( D = \mathbb{Z} + \mathbb{Z}\left( {\left( {-1 + \sqrt{d}}\right) /2}\right) \) . | Proof. Suppose \( \gamma = r + s\sqrt{d}, r, s \in \mathbb{Q} \) . Then \( \gamma \in D \) iff \( {2r} \) and \( {r}^{2} - {s}^{2}d \in \mathbb{Z} \) . Since \( {2r} \in \mathbb{Z} \) it follows from the second condition that \( 4{s}^{2}d \in \mathbb{Z} \) . Since \( d \) is square-free it follows that \( {2s} \in \mat... | Yes |
Proposition 13.1.2. Let \( {\delta }_{F} \) denote the discriminant of \( F \) . If \( d \equiv 2,3\left( 4\right) \) then \( {\delta }_{F} = {4d} \) . If \( d \equiv 1\left( 4\right) \) then \( {\delta }_{F} = d \) . | Proof. If \( d \equiv 2,3\left( 4\right) \) set \( {\omega }_{1} = 1 \) and \( {\omega }_{2} = \sqrt{d} \) . Then \[ \left( {t\left( {{\omega }_{i}{\omega }_{j}}\right) }\right) = \left( \begin{matrix} 2 & 0 \\ 0 & {2d} \end{matrix}\right) \] Thus \( {\delta }_{F} = \det \left( {t\left( {{\omega }_{i}{\omega }_{j}}\rig... | Yes |
Proposition 13.1.3. Suppose \( p \) is odd.\n\n(i) If \( p \nmid {\delta }_{F} \) and \( {x}^{2} \equiv d\left( p\right) \) is solvable in \( \mathbb{Z} \) then \( \left( p\right) = P{P}^{\prime }, P \neq {P}^{\prime } \).\n\n(ii) If \( p \nmid {\delta }_{F} \) and \( {x}^{2} \equiv d\left( p\right) \) is not solvable ... | Proof. In case (i) suppose \( {a}^{2} \equiv d\left( p\right) \) with \( a \in \mathbb{Z} \) . We claim that \( \left( p\right) = \) \( \left( {p, a + \sqrt{d}}\right) \left( {p, a - \sqrt{d}}\right) \) . In fact, \( \left( {p, a + \sqrt{d}}\right) \left( {p, a - \sqrt{d}}\right) = \left( p\right) (p, a + \sqrt{d} \) ,... | Yes |
Proposition 13.1.4. Suppose \( p = 2 \) .\n\n(i) If \( 2 \times {\delta }_{F} \) and \( d \equiv 1\\left( 8\\right) \) then \( \\left( 2\\right) = P{P}^{\\prime } \) and \( P \neq {P}^{\\prime } \).\n\n(ii) If \( 2 \times {\delta }_{F} \) and \( d \equiv 5\\left( 8\\right) \) then \( \\left( 2\\right) = P \).\n\n(iii) ... | Proof. If \( d \equiv 1\\left( 8\\right) \) we claim that \( \\left( 2\\right) = \\left( {2,\\left( {1 + \\sqrt{d}}\\right) /2}\\right) \\left( {2,\\left( {1 - \\sqrt{d}}\\right) /2}\\right) \). In fact \( \\left( {2,\\left( {1 + \\sqrt{d}}\\right) /2}\\right) \\left( {2,\\left( {1 - \\sqrt{d}}\\right) /2}\\right) = \\... | Yes |
Proposition 13.1.5. If \( d < 0 \) and square free then\n\n(a) \( {U}_{-1} = \{ 1, i, - 1, - i\} \).\n\n(b) \( {U}_{-3} = \left\{ {\pm 1, \pm \omega , \pm {\omega }^{2}}\right\} \), where \( \omega = \left( {-1 + \sqrt{-3}}\right) /2 \).\n\n(c) \( {U}_{d} = \{ 1, - 1\} \) for \( d < - 3 \), or \( d = - 2 \). | Proof. If \( d \equiv 2 \) or 3 (4) then any unit may be written in the form \( x + \sqrt{d}y \) , \( x, y \in \mathbb{Z} \) . Thus \( N\left( \alpha \right) = \pm 1 \) is equivalent to \( {x}^{2} + \left| d\right| {y}^{2} = 1 \) . If \( d = - 1 \) we obtain (a). If \( \left| d\right| > 1 \) then clearly \( {U}_{d} = \... | Yes |
Proposition 13.1.6. If D is the ring of integers in \( \mathbb{Q}\left( \sqrt{d}\right), d > 0 \) then there exists a unit \( u > 1 \) such that every unit is of the form \( \pm {u}^{m}, m \in \mathbb{Z} \) . | Proof. By Proposition 17.5.2 there exist positive nonzero integers \( x, y \) such that \( {x}^{2} - d{y}^{2} = + 1 \) . Thus \( x + \sqrt{d}y = u \) is a unit in \( D, u > 1 \) . Let \( M \) be a fixed real number, \( M > u \) . By Exercise 4, Chapter 12 there are at most a finite number of \( \alpha \in D \) with \( ... | Yes |
Proposition 13.2.1. Let \( G \) be the Galois group of \( F/\mathbb{Q} \) . There is a monomorphism \( \theta : G \rightarrow U\left( {\mathbb{Z}/m\mathbb{Z}}\right) \) such that for \( \sigma \in G \)\n\n\[ \sigma {\zeta }_{m} = {\zeta }_{m}^{\theta \left( \sigma \right) } \] | Proof. Since \( {\zeta }_{m}^{m} = 1 \) we have \( {\left( \sigma {\zeta }_{m}\right) }^{m} = 1 \) . Thus \( \sigma {\zeta }_{m} = {\zeta }_{m}^{\theta \left( \sigma \right) } \) where \( \theta \left( \sigma \right) \) is an integer modulo \( m \) . If \( \tau = {\sigma }^{-1} \) then \( {\zeta }_{m} = {\tau \sigma }{... | Yes |
Proposition 13.2.2. \( {x}^{m} - 1 = \mathop{\prod }\limits_{{d/m}}{\Phi }_{d}\left( x\right) \) . | Proof.\n\n\[ \n{x}^{m} - 1 = \mathop{\prod }\limits_{{i = 0}}^{{m - 1}}\left( {x - {\zeta }_{m}^{i}}\right) = \mathop{\prod }\limits_{{d/m}}\mathop{\prod }\limits_{{\left( {i, m}\right) = d}}\left( {x - {\zeta }_{m}^{i}}\right) .\n\] \n\nWe claim \( \mathop{\prod }\limits_{{\left( {i, m}\right) = d}}\left( {x - {\zeta ... | Yes |
Proposition 13.2.3. Suppose \( p \) is a rational prime and \( p \times m \) . Let \( P \) be a prime ideal in \( D \) containing \( p \) . Then the cosets of \( 1,\zeta ,{\zeta }^{2},\ldots ,{\zeta }^{m - 1} \) in \( D/P \) are all distinct. If \( f \) denotes the degree of \( P \) then \( {p}^{f} \equiv 1\left( m\rig... | Proof. For \( w \in D \) let \( \bar{w} \) denote its coset in \( D/P \) .\n\nDivide both sides of \( {x}^{m} - 1 = \prod \left( {x - {\zeta }^{i}}\right) \) by \( x - 1 \) . We find\n\n\[ 1 + x + \cdots + {x}^{m - 1} = \mathop{\prod }\limits_{{i = 1}}^{{m - 1}}\left( {x - {\zeta }^{i}}\right) . \]\n\nLet \( x = 1 \) i... | Yes |
Theorem 1. The mth cyclotomic polynomial, \( {\Phi }_{m}\left( x\right) \), is irreducible in \( \mathbb{Z}\left\lbrack x\right\rbrack \) . | Proof. Let \( f\left( x\right) \in \mathbb{Z}\left\lbrack x\right\rbrack \) be the monic irreducible polynomial for \( \zeta \) . The fact that \( f\left( x\right) \) has coefficients in \( \mathbb{Z} \) follows from the fact that \( \zeta \) is an algebraic integer (Exercise 16, Chapter 6). If \( p \nmid m \) is a pri... | No |
Corollary 2. The map \( \theta \) of Proposition 13.2.1 is an isomorphism of \( G \) onto \( U\left( {\mathbb{Z}/m\mathbb{Z}}\right) \) . | Proof. Both \( G \) and \( U\left( {\mathbb{Z}/m\mathbb{Z}}\right) \) have \( \phi \left( m\right) \) elements. Since \( \theta \) is one-to-one it must be onto. | Yes |
Lemma 1. Let \( F/\mathbb{Q} \) be an algebraic number field of degree \( n \) . Let \( D \subset F \) be the ring of integers and \( {\alpha }_{1},{\alpha }_{2},\ldots ,{\alpha }_{n} \in D \) a field basis for \( F/\mathbb{Q} \) . Let \( \Delta = \Delta \left( {{\alpha }_{1}\text{,}}\right. \) \( \left. {{\alpha }_{2}... | Proof. Let \( w \in D \) . We have \( w = \sum {r}_{i}{\alpha }_{i} \) with \( {r}_{i} \in \mathbb{Q} \) . Multiply both sides by \( {\alpha }_{j} \) and take the trace. We find \( t\left( {w{\alpha }_{j}}\right) = \sum {r}_{i}t\left( {{\alpha }_{i}{\alpha }_{j}}\right) \) . The elements \( t\left( {w{\alpha }_{j}}\rig... | Yes |
Lemma 2. The discriminant \( \Delta = \Delta \left( {1,\zeta ,\ldots ,{\zeta }^{\phi \left( m\right) - 1}}\right) \) divides \( {m}^{\phi \left( m\right) } \) . | Proof. Differentiate both sides of \( {x}^{m} - 1 = {\Phi }_{m}\left( x\right) g\left( x\right) \) . We find \( m{x}^{m - 1} = \) \( {\Phi }_{m}^{\prime }\left( x\right) g\left( x\right) + {\Phi }_{m}\left( x\right) {g}^{\prime }\left( x\right) \) . Substitute \( x = \zeta \) . The result is \( m{\zeta }^{m - 1} = {\Ph... | Yes |
Proposition 13.2.4. Let \( p \in \mathbb{Z} \) be a prime such that \( p \nmid m \) . Let \( w \in D \) the ring of integers in \( \mathbb{Q}\left( \zeta \right) \) . There is an element \( \sum {a}_{i}{\zeta }^{i} \in \mathbb{Z}\left\lbrack \zeta \right\rbrack \) such that \( w \equiv \sum {a}_{i}{\zeta }^{i}\left( p\... | Proof. Let \( \Delta = \Delta \left( {1,\zeta ,\ldots ,{\zeta }^{\phi \left( m\right) - 1}}\right) \) . By Lemma \( 2, p \nmid \Delta \) . Thus there is a \( {\Delta }^{\prime } \in \mathbb{Z} \) such that \( {\Delta }^{\prime }\Delta \equiv 1\left( p\right) \) . Thus \( w \equiv {\Delta }^{\prime }{\Delta w}\left( p\r... | No |
Proposition 13.2.5. If \( p \) is a prime and \( p \nmid m \) the every prime ideal \( P \) in \( D \) containing \( p \) is unramified. | Proof. Assume \( P \) is ramified. Then \( \left( p\right) \subset {P}^{2} \) . Let \( w \) be an element of \( P \) not in \( {P}^{2} \) . By the above corollary \( {w}^{{p}^{n}} \equiv w\left( p\right) \) and so \( {w}^{{p}^{n}} \equiv w\left( {P}^{2}\right) \) . Since \( {p}^{n} \geq 2 \) it follows that \( w \in {P... | Yes |
Proposition 13.2.6. For all \( w \in D \) we have \( {\sigma }_{p}w \equiv {w}^{p}\left( p\right) \) . | Proof. By Proposition 13.2.4 we have \( w \equiv \sum {a}_{i}{\zeta }^{i}\left( p\right) \) . Apply \( {\sigma }_{p} \) to both sides. We find that \( {\sigma }_{p}w \equiv \sum {a}_{i}{\zeta }^{pi}\left( p\right) \) . Since the \( {a}_{i} \in \mathbb{Z} \) we have \( \sum {a}_{i}{\zeta }^{pi} \equiv \sum {a}_{i}^{p}{\... | Yes |
Theorem 2. Let \( p \) be a prime, \( p \times m \) . Let \( f \) be the smallest positive integer such that \( {p}^{f} \equiv 1\left( m\right) \) . Then in \( D \subset \mathbb{Q}\left( \zeta \right) \) we have \[ \left( p\right) = {P}_{1}{P}_{2}\cdots {P}_{g} \] where each \( {P}_{i} \) has degree \( f \) and \( g = ... | Proof. We first observe that it follows directly from the definition that \( f \) is the order of the automorphism \( {\sigma }_{p} \) . Now, \( {p}^{{f}_{1}} = \left| {D/{P}_{1}}\right| \) where \( {f}_{1} \) is the degree of \( {P}_{1} \) . Since \( D/{P}_{1} \) is a finite field we have \( {w}^{{p}^{{r}_{1}}} \equiv... | Yes |
Proposition 13.2.7. Let \( l \) be a prime in \( \mathbb{Z} \) . Then, in \( \mathbb{Q}\left( {\zeta }_{l}\right) \) , \( l \) ramifies completely. More precisely, let \( L = \left( {1 - {\zeta }_{l}}\right) \) . Then \( L \) is a prime ideal and \( \left( l\right) = {L}^{l - 1} \) Moreover \( L \) has degree 1. | Proof. As in the proof of Proposition 13.2.3 we have \( l = \prod \left( {1 - {\zeta }_{l}^{i}}\right) \) where the product is over \( 1 \leq i \leq l - 1 \) .\n\nLet \( {u}_{i} = \left( {1 - {\zeta }^{i}}\right) /\left( {1 - \zeta }\right) = 1 + \zeta + \cdots + {\zeta }^{i - 1} \) . We claim that \( {u}_{i} \) is a u... | Yes |
Proposition 13.2.8. Let \( P \) be a prime ideal in \( \mathbb{Q}\left( {\zeta }_{m}\right) \) and set \( P \cap \mathbb{Z} = p\mathbb{Z} \) . If \( p \) is odd then \( P \) is ramified iff \( p \mid m \) . If \( p = 2 \) then \( P \) is ramified iff \( 4 \mid m \) . | Proof. By Proposition 13.2.5 we know that \( p \nmid m \) implies \( P \) is unramified.\n\nSuppose \( p \) is odd and \( p \mid m \) . Then \( \mathbb{Q}\left( {\zeta }_{p}\right) \subset \mathbb{Q}\left( {\zeta }_{m}\right) \) . Let \( {D}_{p} \) and \( {D}_{m} \) be the rings of integers in \( \mathbb{Q}\left( {\zet... | Yes |
Lemma 3. If \( \left( {m, n}\right) = 1 \) then \( \mathbb{Q}\left( {{\zeta }_{m},{\zeta }_{n}}\right) = \mathbb{Q}\left( {\zeta }_{mn}\right) \) . | Proof. Since \( {\zeta }_{mn}^{m} = {\zeta }_{n} \) and \( {\zeta }_{mn}^{n} = {\zeta }_{m} \) we have \( \mathbb{Q}\left( {{\zeta }_{m},{\zeta }_{n}}\right) \subset \mathbb{Q}\left( {\zeta }_{mn}\right) \). On the other hand, since \( \left( {m, n}\right) = 1 \) there exist integers \( u \) and \( v \) such that \( {u... | Yes |
Proposition 13.2.9. Let \( p \) be a prime such that \( p \times m \) . Let \( D \) be the ring of integers in \( \mathbb{Q}\left( {{\zeta }_{p},{\zeta }_{m}}\right) \) . Then\n\n\[ \n{pD} = {\left( {P}_{1}{P}_{2}\cdots {P}_{g}\right) }^{p - 1},\n\]\n\nwhere the \( {P}_{i} \) are distinct prime ideals of degree \( f \)... | Proof. Since \( \mathbb{Q}\left( {\zeta }_{p}\right) \subset \mathbb{Q}\left( {{\zeta }_{p},{\zeta }_{m}}\right) \) we see, as in the proof of the last proposition, that all the ramification indices of primes in \( D \) containing \( p \) are divisible by \( p - 1 \) . Thus\n\n\[ \n{pD} = {\left( {P}_{1}{P}_{2}\cdots {... | Yes |
Proposition 13.2.10. If \( l \) is prime then \( D = \mathbb{Z}\left\lbrack {\zeta }_{l}\right\rbrack \) . | Proof. Clearly \( \mathbb{Z}\left\lbrack {\zeta }_{l}\right\rbrack \subset D \) . If \( \alpha \in D \) there exist \( {a}_{0},{a}_{1},\ldots ,{a}_{l - 2} \) rational numbers such that \( \alpha = {a}_{0} + {a}_{1}\zeta + \cdots + {a}_{l - 2}{\zeta }^{l - 2} \) . We show first of all that \( l{a}_{i} \in \mathbb{Z}, i ... | Yes |
Proposition 14.1.1. If \( A, B \subset D \) are ideals, then \( N\left( {AB}\right) = N\left( A\right) N\left( B\right) \) . | Proof. If \( A \) and \( B \) are relatively prime, then \( D/{AB} \approx D/A \oplus D/B \) so the assertion is clear in this case.\n\nLet \( A = {P}_{1}^{{a}_{1}}{P}_{2}^{{a}_{2}}\cdots {P}_{t}^{{a}_{t}} \) be the prime decomposition of \( A \) . We claim \( N\left( A\right) = {\left( N\left( {P}_{1}\right) \right) }... | Yes |
Proposition 14.1.2. Suppose \( K/\mathbb{Q} \) is a Galois extension with group \( G \) . Then\n\n\[ \mathop{\prod }\limits_{{\sigma \in G}}\sigma \left( A\right) = \left( {N\left( A\right) }\right) \] | Proof. Since both sides are multiplicative in \( A \) it suffices to prove the result when \( A \) is a prime ideal \( P \) .\n\nLet \( {P}_{1},{P}_{2},\ldots ,{P}_{g} \) be the distinct prime ideals in the set \( \{ \sigma \left( P\right) \mid \sigma \in G\} \) . Then \( \left| G\right| = g\left| {G\left( P\right) }\r... | Yes |
Proposition 14.1.3. Let \( K/\mathbb{Q} \) be Galois with group \( G \) . Let \( \alpha \in D \) and let \( A = \left( \alpha \right) \) be the principal ideal generated by \( \alpha \) . Let \( {N\alpha } \) be the norm of \( \alpha \) . Then \( N\left( A\right) = \) \( \left| {N\left( \alpha \right) }\right| \). | Proof. \( \left( {N\left( A\right) }\right) = \prod \sigma \left( A\right) = \prod \sigma \left( \left( \alpha \right) \right) = \prod \left( {\sigma \alpha }\right) = \left( {\prod \sigma \left( \alpha \right) }\right) = \left( {N\left( \alpha \right) }\right) \) . Thus \( N\left( A\right) \) and \( N\left( \alpha \ri... | Yes |
Proposition 14.2.1. Let \( \alpha \in {D}_{m},\alpha \notin P \) . There is an integer \( i \), unique modulo \( m \) , such that\n\n\[ \n{\alpha }^{\left( {q - 1}\right) /m} \equiv {\zeta }_{m}^{i}\left( P\right) \n\] | Proof. Since the multiplicative group of \( {D}_{m}/P \) has \( q - 1 \) elements we have \( {\alpha }^{q - 1} \equiv 1\left( P\right) \) . Thus\n\n\[ \n\mathop{\prod }\limits_{{i = 0}}^{{m - 1}}\left( {{\alpha }^{\left( {q - 1}\right) /m} - {\zeta }_{m}^{i}}\right) \equiv 0\left( P\right) . \n\]\n\nSince \( P \) is a ... | Yes |
(a) \( {\left( \alpha /P\right) }_{m} = 1 \) iff \( {x}^{m} \equiv \alpha \left( P\right) \) is solvable in \( {D}_{m} \) . | Proof. Since the result has been proven earlier for \( m = 2,3 \), and 4 we may safely leave the details to the reader. | No |
Proposition 14.2.3. Suppose \( A \) and \( B \) are ideals prime to \( \left( m\right) \) . Then\n\n(a) \( {\left( \alpha \beta /A\right) }_{m} = {\left( \alpha /A\right) }_{m}{\left( \beta /A\right) }_{m} \).\n\n(b) \( {\left( \alpha /AB\right) }_{m} = {\left( \alpha /A\right) }_{m}{\left( \alpha /B\right) }_{m} \).\n... | Proof. All three assertions are straightforward to prove using the last proposition and the above definition. We remark that the converse of part (c) is not true. | No |
Proposition 14.2.4. Let \( A \) be an ideal prime to \( m \) and \( \sigma \in G \) . Then\n\n\[{\left( \frac{\alpha }{A}\right) }_{m}^{\sigma } = {\left( \frac{{\alpha }^{\sigma }}{{A}^{\sigma }}\right) }_{m}\] | Proof. Since both sides of the asserted equality are multiplicative in \( A \) it will be enough to check the case where \( A = P \) is a prime ideal. By definition\n\n\[{\alpha }^{\left( {{NP} - 1}\right) /m} \equiv {\left( \frac{\alpha }{P}\right) }_{m}\left( P\right)\]\n\nApplying \( \sigma \) to this congruence we ... | Yes |
Theorem 1 (The Eisenstein Reciprocity Law). Let \( l \) be an odd prime, \( a \in \mathbb{Z} \) prime to \( l \), and \( \alpha \in {D}_{l} \) a primary element. Suppose moreover that \( \alpha \) and a are prime to each other. Then\n\n\[{\left( \frac{\alpha }{a}\right) }_{l} = {\left( \frac{a}{\alpha }\right) }_{l}\] | The proof of this elegant theorem will be given in Section 5. It is a consequence of the Stickelberger relation which will be stated in the next section and proven in Section 4. Since this process is long, and somewhat involved, the reader may wish to skip to the last part of the chapter, Section 6, where three interes... | No |
(c) \( \Phi \left( P\right) \in \mathbb{Q}\left( {\zeta }_{m}\right) \) . | Proof. (a) has already been discussed. (b) follows in the same way as when \( F \) is the prime field. (c) follows from Proposition 8.3.3 which is stated over \( \mathbb{Z}/p\mathbb{Z} \) but generalizes easily to \( F \) . We will give another proof of (c) based on Galois theory. Consider the diagram of fields The Gal... | Yes |
Lemma 1. Let \( p > 1 \) be a positive integer. Every positive integer can be written uniquely in the form \( \mathop{\sum }\limits_{{i = 0}}^{n}{a}_{i}{p}^{i} \) where \( 0 \leq {a}_{i} < p \) . | Proof. Let \( a \) be a positive integer. There is a unique nonnegative integer \( n \) such that \( {p}^{n} \leq a < {p}^{n + 1} \) . By the division algorithm we have \( a = {a}_{n}{p}^{n} + r \) where \( 0 \leq r < {p}^{n} \) . The number \( {a}_{n} \) is less than \( p \) since otherwise \( a \geq {p}^{n + 1} \) . ... | Yes |
Lemma 2. \( S\left( a\right) = \left( {p - 1}\right) \mathop{\sum }\limits_{{i = 0}}^{{f - 1}}\left\langle {{p}^{i}a/\left( {q - 1}\right) }\right\rangle \) . | Proof. Both sides are unchanged if a multiple of \( q - 1 \) is added to \( a \) . Thus we may assume \( 1 \leq a < q - 1 \) .\n\nWrite \( a = {a}_{0} + {a}_{1}p + \cdots + {a}_{f - 1}{p}^{f - 1} \) where \( 0 \leq {a}_{i} < p \) . Since \( {p}^{f} = \) \( q \equiv 1\left( {q - 1}\right) \) we have\n\n\[ a = {a}_{0} + ... | Yes |
Lemma 3. \( \mathop{\sum }\limits_{{a = 1}}^{{q - 2}}S\left( a\right) = \left( {f\left( {p - 1}\right) \left( {q - 2}\right) }\right) /2 \) . | Proof. Write \( a = {a}_{0} + {a}_{1}p + \cdots + {a}_{f - 1}{p}^{f - 1} \) with \( 0 \leq {a}_{i} < p \) . Notice that \( q - 1 = \left( {p - 1}\right) + \left( {p - 1}\right) p + \cdots + \left( {p - 1}\right) {p}^{f - 1} \) . It follows that \( q - 1 - a = \left( {p - 1 - {a}_{0}}\right) + \left( {p - 1 - {a}_{1}}\r... | Yes |
Lemma 5. \( {D}_{m}/P \approx {D}_{q - 1}/\mathfrak{P} \) . | Proof. There is a natural monomorphism from \( {D}_{m}/P \) to \( {D}_{q - 1}/\mathfrak{P} \) . To show this is an isomorphism it suffices to show both fields have the same number of elements. By Theorem 2 of Chapter 13 we have \( \left| {{D}_{q - 1}/\mathfrak{P}}\right| = {p}^{{f}^{\prime }} \) where \( {f}^{\prime } ... | Yes |
Lemma 7. \( \omega \left( {\bar{\zeta }}_{q - 1}^{i}\right) = {\zeta }_{q - 1}^{i} \) . | Proof. Immediate from the definition. | No |
Theorem 3. \( {\operatorname{ord}}_{\mathcal{P}}\left( {g}_{a}\right) = S\left( a\right) \), where \( 1 \leq a < q \) . | Proof. To begin with we show that \( {\operatorname{ord}}_{\mathcal{P}}\left( {g}_{1}\right) = 1 \) . Recall\n\n\[ \n{g}_{1} = \mathop{\sum }\limits_{{t \in \mathbb{F}}}\omega {\left( t\right) }^{-1}{\zeta }_{p}^{\operatorname{tr}\left( t\right) }\n\]\n\nUsing Lemma 7 we will convert this into a sum over the powers of ... | Yes |
Lemma 8. \( {\operatorname{ord}}_{{P}_{t}}\left( {\Phi \left( P\right) }\right) = \left( {m/\left( {p - 1}\right) }\right) S\left( {t\left( {\left( {q - 1}\right) /m}\right) }\right) \) . | Proof. It follows quickly from the definitions that\n\n\[ \n{\operatorname{ord}}_{{P}_{t}}\left( {\Phi \left( P\right) }\right) = {\operatorname{ord}}_{P}\left( {\Phi {\left( P\right) }^{{\sigma }_{t}}}\right) \n\] \n\nChoose an integer \( {t}^{\prime } \) such that \( {t}^{\prime } \equiv t\left( m\right) \) and \( {t... | Yes |
Lemma 1. The only roots of unity in \( \mathbb{Q}\left( {\zeta }_{m}\right) \) are \( \pm {\zeta }_{m}^{i}i = 1,2,\ldots, m \) . | Proof. In the proof of Theorem 1 we only need this result when \( m \) is an odd prime. We will leave the proof for general \( m \) as an exercise and assume \( m = l \), an odd prime.\n\nSuppose \( {\zeta }_{n} \in \mathbb{Q}\left( {\zeta }_{l}\right) \) . If \( 4 \mid n \) then \( \sqrt{-1} \in \mathbb{Q}\left( {\zet... | No |
Lemma 2. Let \( K/\mathbb{Q} \) be an algebraic number field and let \( {\sigma }_{1},{\sigma }_{2},\ldots ,{\sigma }_{n} \) be the \( n = \left\lbrack {K : \mathbb{Q}}\right\rbrack \) isomorphisms of \( K \) into \( \mathbb{C} \) . If \( \alpha \in K \) is such that \( \left| {\alpha }^{{\sigma }_{i}}\right| \leq 1 \)... | Proof. \( \alpha \) is a root of\n\n\[ f\left( x\right) = \mathop{\prod }\limits_{{i = 1}}^{n}\left( {x - {\alpha }^{{\sigma }_{i}}}\right) \in \mathbb{Z}\left\lbrack x\right\rbrack .\n\]\n\nThe hypothesis of the lemma implies that the coefficient of \( {x}^{m} \) in \( f\left( x\right) \) is an integer bounded by the ... | Yes |
Lemma 3. Suppose \( A \subset {D}_{m} \) is an ideal prime to \( m \) and let \( \sigma \) be an automorphism of \( \mathbb{Q}\left( {\zeta }_{m}\right) /\mathbb{Q} \) . Then\n\n\[ \Phi {\left( A\right) }^{\sigma } = \Phi \left( {A}^{\sigma }\right) \] | Proof. To see this it is convenient to write \( g\left( P\right) \) in the following form\n\n\[ g\left( P\right) = \sum {\left( \frac{\alpha }{P}\right) }_{m}^{-1}{\zeta }_{p}^{\operatorname{tr}\left( \widetilde{\alpha }\right) } \]\n\nwhere the sum is over a set of representatives for the cosets of \( {D}_{m}/P \) .\n... | Yes |
Lemma 4. For \( \alpha \in {D}_{m},{\left| {\alpha }^{\gamma }\right| }^{2} = {\left| N\alpha \right| }^{m} \). | Proof. The automorphism \( {\sigma }_{-1} \) is complex conjugation on \( \mathbb{Q}\left( {\zeta }_{m}\right) \) since it takes \( {\zeta }_{m} \) to \( {\zeta }_{m}^{-1} = {\zeta }_{m} \). Thus\n\n\[ \n{\left| {\alpha }^{\gamma }\right| }^{2} = {\alpha }^{\gamma }{\alpha }^{{\gamma \sigma } - 1} = {\alpha }^{\gamma \... | Yes |
Proposition 14.5.2. Let \( \alpha \in {D}_{m},\alpha \) prime to \( m \) . Then \( \Phi \left( \alpha \right) = \varepsilon \left( \alpha \right) {\alpha }^{\gamma } \) where \( \varepsilon \left( \alpha \right) = \pm {\zeta }_{m}^{i} \) for some \( i \) . | Proof. In the light of part (d) of the last proposition it is enough to prove the assertion about \( \varepsilon \left( \alpha \right) \) . We have \( {\left| \Phi \left( \alpha \right) \right| }^{2} = {\left( N\left( \alpha \right) \right) }^{m} \) by Proposition 14.5.1 and \( {\left| {\alpha }^{\gamma }\right| }^{2} ... | Yes |
Proposition 14.5.3. Suppose \( P,{P}^{\prime } \subset {D}_{m} \) are prime ideals both prime to \( m \) . Suppose further that \( {NP} \) and \( N{P}^{\prime } \) are relatively prime. Then\n\n\[{\left( \frac{\Phi \left( P\right) }{{P}^{\prime }}\right) }_{m} = {\left( \frac{N{P}^{\prime }}{P}\right) }_{m}.\] | Proof. Let \( {q}^{\prime } = {p}^{\prime {f}^{\prime }} = N{P}^{\prime } \) . Recall \( {q}^{\prime } \equiv 1\left( m\right) \) . The following congruences are taken modulo \( {p}^{\prime } \) in \( {D}_{m} \)\n\n\[g{\left( P\right) }^{{q}^{\prime }} \equiv \sum {\chi }_{P}{\left( t\right) }^{{q}^{\prime }}\psi {\lef... | Yes |
Corollary 1. Suppose \( A, B \subset {D}_{m} \) are ideals prime to \( m \) and that \( {NA} \) and \( {NB} \) are prime to each other. Then\n\n\[ \n{\left( \frac{NB}{A}\right) }_{m} = {\left( \frac{\Phi \left( A\right) }{B}\right) }_{m}.\n\] | Proof. As usual, the corollary follows from the proposition by multi-plicativity. | No |
Corollary 2. Suppose \( A \) and \( B \) are as in Corollary 1 and moreover that \( A = \left( \alpha \right) \) is principal. Then \[ {\left( \frac{\varepsilon \left( \alpha \right) }{B}\right) }_{m}{\left( \frac{\alpha }{NB}\right) }_{m} = {\left( \frac{NB}{\alpha }\right) }_{m}. \] | Proof. To begin with \[ {\left( \frac{\Phi \left( \alpha \right) }{B}\right) }_{m} = {\left( \frac{\varepsilon \left( \alpha \right) }{B}\right) }_{m}{\left( \frac{{\alpha }^{\gamma }}{B}\right) }_{m}. \] Notice that \( {\left( {\alpha }^{t{\sigma }_{t}^{-1}}/B\right) }_{m} = {\left( {\alpha }^{{\sigma }_{t}^{-1}/B}\ri... | Yes |
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