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Lemma 5. If \( A \subset {D}_{l} \) is an ideal prime to \( l \), then \( \Phi \left( A\right) \equiv \pm 1\left( l\right) \) . | Proof. It is enough to show that \( \Phi \left( P\right) \equiv - 1\left( l\right) \) where \( P \subset {D}_{l} \) is a prime ideal prime to \( l \) . Well,\n\n\[ \Phi \left( P\right) = g{\left( P\right) }^{l} \equiv \mathop{\sum }\limits_{t}{\chi }_{P}{\left( t\right) }^{l}\psi {\left( t\right) }^{l}\left( l\right) \... | Yes |
Lemma 6. If \( \alpha \in D \) is primary, then \( \varepsilon \left( \alpha \right) = \pm 1 \) . | Proof. Since \( \left( {1 - {\zeta }_{l}}\right) \) is the unique prime above \( l \) in \( {D}_{l} \) we have \( {\left( 1 - {\zeta }_{l}\right) }^{\sigma } = \left( {1 - {\zeta }_{l}}\right) \) for all \( \sigma \in G \) . It follows that \( {\left( 1 - {\zeta }_{l}\right) }^{\gamma } \subset \left( {1 - {\zeta }_{l}... | Yes |
Proposition 14.5.4. If \( \alpha \in {D}_{l} \) is primary, and \( B \) is an ideal prime to \( l \), and \( {NB} \) is prime to \( \alpha \), then\n\n\[{\left( \frac{\alpha }{NB}\right) }_{l} = {\left( \frac{NB}{\alpha }\right) }_{l}\] | Proof. By Corollary 2 to Proposition 14.5.3 we need only show \( {\left( \varepsilon \left( \alpha \right) /B\right) }_{l} = 1 \) .\n\nSince \( \alpha \) is primary \( \varepsilon \left( \alpha \right) = \pm 1 \) by the above lemma. Since \( l \) is odd, \( {\left( \pm 1\right) }^{l} = \) \( \pm 1 \) and we are done. | No |
Theorem 4. Suppose \( a \in \mathbb{Z} \) and that \( l \times a \) where \( l \) is an odd prime. If \( {x}^{l} \equiv a\left( p\right) \) is solvable for all but finitely many primes \( p \) then \( a = {b}^{l} \) . | Proof. We can restate the theorem as follows. If \( a \) is not an \( l \) th power then there are infinitely many primes \( p \) such that \( {x}^{l} \equiv a\left( p\right) \) is not solvable.\n\nAssume \( a \) is not an \( l \) th power in \( \mathbb{Z} \) . Let \( a{D}_{l} = {P}_{1}^{{a}_{1}}{P}_{2}^{{a}_{2}}\cdots... | Yes |
Lemma 1. Suppose \( i \neq j \) and \( 0 \leq i, j < l \) . Then \( x + {\zeta }^{i}y \) and \( x + {\zeta }^{j}y \) are relatively prime in \( {D}_{l} \) . | Proof. Suppose \( A \subset {D}_{l} \) is an ideal containing \( x + {\zeta }^{i}y \) and \( x + {\zeta }^{j}y \) . Then \( \left( {{\zeta }^{j} - {\zeta }^{i}}\right) x \) and \( \left( {{\zeta }^{j} - {\zeta }^{i}}\right) y \) are in \( A \) . Since \( x \) and \( y \) are relatively prime it follows that \( {\zeta }... | Yes |
Lemma 2. \( \mathfrak{P}\widetilde{D} = \prod {P}^{{\sigma }_{s}} \) where \( P \) is a prime ideal of \( \widetilde{D}, P \cap D = \mathfrak{P} \), and \( s \) runs over the nonzero squares modulo \( l \) . | Proof. The set of \( {\sigma }_{s} \) in the statement of the lemma form the Galois group of \( \mathbb{Q}\left( {\zeta }_{l}\right) \) over \( \mathbb{Q}\left( \sqrt{-l}\right) \) . Since \( p\widetilde{D} = {P}^{{\sum }_{t}^{1 - 1}{\sigma }_{t}} \) and \( {\sigma }_{n}\left( \mathfrak{P}\right) = \overline{\mathfrak{... | Yes |
Lemma 1. Expand \( t/\left( {{e}^{t} - 1}\right) \) in a power series about the origin as follows \( t/\left( {{e}^{t} - 1}\right) = \mathop{\sum }\limits_{{m = 0}}^{\infty }{b}_{m}\left( {{t}^{m}/m!}\right) \) . Then for all \( m,{b}_{m} = {B}_{m} \) . | Proof. Multiply both sides by \( {e}^{t} - 1 \) to obtain\n\n\[ t = \mathop{\sum }\limits_{{n = 1}}^{\infty }\frac{{t}^{n}}{n!}\mathop{\sum }\limits_{{m = 0}}^{\infty }{b}_{m}\frac{{t}^{m}}{m!} \]\n\nEquating coefficients of \( {t}^{m + 1} \) gives \( 1 = {b}_{0} \) for \( m = 0 \) and\n\n\[ \mathop{\sum }\limits_{{k =... | Yes |
For \( m \geq 1 \) the sums \( {S}_{m}\left( n\right) \) satisfy\n\n\[ \left( {m + 1}\right) {S}_{m}\left( n\right) = \mathop{\sum }\limits_{{k = 0}}^{m}\left( \begin{matrix} m + 1 \\ k \end{matrix}\right) {B}_{k}{n}^{m + 1 - k}. \] | In \( {e}^{kt} = \mathop{\sum }\limits_{{m = 0}}^{\infty }{k}^{m}\left( {{t}^{m}/m!}\right) \) substitute \( k = 0,1,2,\ldots, n - 1 \) and add. This results in\n\n\[ 1 + {e}^{t} + {e}^{2t} + \cdots + {e}^{\left( {n - 1}\right) t} = \mathop{\sum }\limits_{{m = 0}}^{\infty }{S}_{m}\left( n\right) \frac{{t}^{m}}{m!}. \]\... | Yes |
Theorem 2. For \( m \) a positive integer\n\n\[ \n{2\zeta }\left( {2m}\right) = {\left( -1\right) }^{m + 1}\frac{{\left( 2\pi \right) }^{2m}}{\left( {2m}\right) !}{B}_{2m} \n\] | Proof. The proof of this result requires a fact from classical analysis. Namely, we need the partial fraction expansion for cot \( x \) .\n\n\[ \n\cot x = \frac{1}{x} - 2\mathop{\sum }\limits_{{n = 1}}^{\infty }\frac{x}{{n}^{2}{\pi }^{2} - {x}^{2}}. \n\]\n\n(4)\n\nThere are several ways to derive this expansion. Perhap... | Yes |
Theorem 4. Let \( p \) be a regular prime. Then \( {x}^{p} + {y}^{p} = {z}^{p} \) has no solution in positive integers. | Actually Kummer proved that Fermat’s conjecture is true if \( p \) does not divide the class number of \( \mathbb{Q}\left( {\zeta }_{p}\right) \) . In other words the criterion is that for any nonprincipal ideal \( A \) in \( Z\left\lbrack {\zeta }_{p}\right\rbrack ,{A}^{p} \) is not principal. This condition is equiva... | No |
Lemma 1. Let \( p \) be a prime number and \( k \geq 1 \) an integer. Then\n\n(a) \( {p}^{k}/\left( {k + 1}\right) \) is \( p \) -integral.\n\n(b) \( {p}^{k}/\left( {k + 1}\right) \equiv 0\left( p\right) \) if \( k \geq 2 \) .\n\n(c) \( {p}^{k - 2}/\left( {k + 1}\right) \) is p-integral if \( k \geq 3 \) and \( p \geq ... | Proof. To prove (a) we show that \( k + 1 \leq {p}^{k} \) for \( k \geq 1 \) . If \( k = 1 \) the result is true. If \( k + 1 \leq {p}^{k} \) then \( k + 2 \leq {p}^{k} + 1 < 2{p}^{k} \leq {p}^{k + 1} \) . Now write \( k + 1 = \) \( {p}^{a}q \) where \( \left( {q, p}\right) = 1 \) . Then \( {p}^{k}/\left( {k + 1}\right... | No |
Proposition 15.2.1. Let \( p \) be a prime and \( m \geq 1 \) an integer. Then \( p{B}_{m} \) is \( p \) - integral. If \( m \geq 2 \) is even then \( p{B}_{m} \equiv {S}_{m}\left( p\right) \left( p\right) . | Proof. The first assertion states that if \( p \) divides the denominator of \( {B}_{m} \) then \( {p}^{2} \) does not. First of all, \( p{B}_{1} = - p/2 \) which is indeed \( p \) -integral for all \( p \) . We proceed by induction.\n\nSuppose \( m > 1 \) . Applying Equation (10) with \( n = p \) we see that, since \(... | Yes |
Lemma 2. Let \( p \) be a prime. Then if \( p - 1 \times m,{S}_{m}\left( p\right) \equiv 0\left( p\right) \) . If \( p - 1 \mid m \) then \( {S}_{m}\left( p\right) \equiv - 1\left( p\right) \) . | Proof. Let \( g \) be a primitive root modulo \( p \) . Then\n\n\[ \n{S}_{m}\left( p\right) = {1}^{m} + {2}^{m} + \cdots + {\left( p - 1\right) }^{m} \]\n\n\[ \n\equiv {1}^{m} + {g}^{m} + {g}^{2m} + \cdots + {g}^{\left( {p - 2}\right) m}\left( p\right) . \]\n\nThus \( \left( {{g}^{m} - 1}\right) {S}_{m}\left( p\right) ... | Yes |
Proposition 15.2.2. If \( m \) is even, \( m \geq 2 \) then for all \( n \geq 1 \) we have\n\n\[ \n{V}_{m}{S}_{m}\left( n\right) \equiv {U}_{m}n\left( {n}^{2}\right) \n\] | Proof. Consider the terms in Equation (10) for \( k \geq 1 \) and fixed \( n \)\n\n\[ \n\left( \begin{matrix} m \\ k \end{matrix}\right) \left( {{B}_{m - k}\frac{{n}^{k - 1}}{k + 1}}\right) {n}^{2} = {A}_{k}^{m}{n}^{2} \n\]\n\n(12)\n\nWe will show that for \( p \mid n \) and \( p \neq 2,3{\operatorname{ord}}_{p}\left( ... | Yes |
Proposition 15.2.3. Let \( m \geq 2 \) be even and define \( {U}_{m} \) and \( {V}_{m} \) as in the last proposition. Suppose \( a \) and \( n \) are positive integers with \( \left( {a, n}\right) = 1 \) . Then\n\n\[ \left( {{a}^{m} - 1}\right) {U}_{m} \equiv m{a}^{m - 1}{V}_{m}\mathop{\sum }\limits_{{j = 1}}^{{n - 1}}... | Proof. For \( 1 \leq j < n \) write \( {ja} = {q}_{j}n + {r}_{j} \) where \( 0 \leq {r}_{j} < n \) . Then \( \left\lbrack {{ja}/n}\right\rbrack = \) \( {q}_{j} \) and since \( \left( {a, n}\right) = 1 \) the two sets \( \{ 1,2,3,\ldots, n - 1\} \) and \( \left\{ {{r}_{1},{r}_{2},\ldots ,{r}_{n - 1}}\right\} \) are iden... | Yes |
Proposition 15.2.4. If \( p - 1 \times m \) then \( {B}_{m}/m \) is a p-integer. | Proof. By Theorem \( 3,{B}_{m} \) is a \( p \) -integer. Write \( m = {p}^{t}{m}_{0} \) where \( p \nmid {m}_{0} \) . In the Voronoi congruence, Equation (14), put \( n = {p}^{t} \) . Then \( \left( {{a}^{m} - 1}\right) {U}_{m} \equiv \) \( 0\left( {p}^{t}\right) \) . Choose \( a \) to be a primitive root modulo \( p \... | Yes |
Theorem 6. The set of irregular primes is infinite. | Proof. Let \( \left\{ {{p}_{1},\ldots ,{p}_{s}}\right\} \) be a set of irregular primes. We will find an irregular prime not in this set.\n\nLet \( k \geq 2 \) be even and set \( n = k\left( {{p}_{1} - 1}\right) \cdots \left( {{p}_{s} - 1}\right) \) . If the set is empty choose \( n = k \) . By Proposition 15.1.1, part... | Yes |
Lemma 1. \( \mathcal{A} \) is the direct product of the \( {\mathcal{A}}_{i} \) . In other words, \( \mathcal{A} = {\mathcal{A}}_{1}{\mathcal{A}}_{2}\cdots {\mathcal{A}}_{l - 1} \) and \( {\mathcal{A}}_{i} \cap \mathop{\prod }\limits_{{j \neq i}}{\mathcal{A}}_{j} = e \) (the identity class) for \( i = 1,2,\cdots, l - 1... | Proof. For each \( i \) with \( 1 \leq i \leq l - 1 \) we define elements \( {\varepsilon }_{i} \in \mathbb{Z}/l\mathbb{Z}\left\lbrack G\right\rbrack \) by the formula\n\n\[ \n{\varepsilon }_{i} = - \mathop{\sum }\limits_{{t = 1}}^{{l - 1}}{\bar{t}}^{-i}{\sigma }_{t} \n\]\n\nReplacing \( t \) by \( {ts} \) in the formu... | Yes |
Theorem 7 (J. Herbrand). Let \( i \) be an odd integer \( 1 \leq i < l \) and define \( j \) by \( i + j = l \) . Then \( {\mathcal{A}}_{1} = \left( e\right) \) . If \( i \geq 3 \) and \( l \nmid {B}_{j} \) then \( {\mathcal{A}}_{i} = \left( e\right) \) . | Proof. Let \( A \in {\mathcal{A}}_{1} \) . Then, by Stickelberger’s relation \[ e = {A}^{{\sum t}{\sigma }_{t}^{-1}} = {A}^{{\sum t}{i}^{-1}} = {A}^{l - 1} = {A}^{-1}. \] This shows \( {\mathcal{A}}_{1} = \left( e\right) \) as asserted. Now suppose \( i \) is odd and \( 3 \leq i \leq l - 2 \) . Let \( A \in {\mathcal{A... | Yes |
Lemma 3. Let \( b \) be an integer prime to \( m \) . Determine \( {b}^{\prime } \) by the conditions \( {b}^{\prime } \equiv b\left( m\right) \) and \( {b}^{\prime } \equiv 1\left( p\right) \) . Let \( {\sigma }_{{b}^{\prime }} \) be the corresponding automorphism of \( \mathbb{Q}\left( {\zeta }_{pm}\right) \) . Then\... | Proof. The automorphisms of \( \mathbb{Q}\left( {\zeta }_{pm}\right) \) which leave \( {\zeta }_{m} \) fixed are of the form \( {\sigma }_{c} \) where \( \left( {c,{pm}}\right) = 1 \) and \( c \equiv 1\left( m\right) \) . Let\n\n\[ {\Omega }_{b}\left( P\right) = g{\left( P\right) }^{{\sigma }_{b} \cdot - b}. \]\n\nWe w... | Yes |
Proposition 16.1.1. For \( s > 1 \)\n\n\[ \zeta \left( s\right) = \mathop{\prod }\limits_{p}{\left( 1 - {p}^{-s}\right) }^{-1}, \]\n\nwhere the product is over all primes \( p > 0 \) . | Proof. For \( s > 1,{p}^{-s} < 1 \), so we have \( {\left( 1 - {p}^{-s}\right) }^{-1} = \mathop{\sum }\limits_{{m = 0}}^{\infty }{p}^{-{ms}} \) . By the theorem of unique factorization\n\n\[ \mathop{\prod }\limits_{{p \leq N}}{\left( 1 - {p}^{-s}\right) }^{-1} = \mathop{\sum }\limits_{{n \leq N}}{n}^{-s} + {R}_{N}\left... | Yes |
Proposition 16.1.2. Assume \( s > 1 \) . Then\n\n\[ \mathop{\lim }\limits_{{s \rightarrow 1}}\left( {s - 1}\right) \zeta \left( s\right) = 1 \] | Proof. For fixed \( s,{t}^{-s} \) is a monotone decreasing function of \( t \) . Thus,\n\n\[ {\left( n + 1\right) }^{-s} < {\int }_{n}^{n + 1}{t}^{-s}{dt} < {n}^{-s}. \]\n\nSumming from \( n = 1 \) to \( \infty \) ,\n\n\[ \zeta \left( s\right) - 1 < {\int }_{1}^{\infty }{t}^{-s}{dt} < \zeta \left( s\right) \]\n\nThe va... | Yes |
Proposition 16.1.3. \( \ln \zeta \left( s\right) = \mathop{\sum }\limits_{p}{p}^{-s} + R\left( s\right) \) where \( R\left( s\right) \) remains bounded as \( s \rightarrow 1 \) . | Proof. We use the formula \( - \ln \left( {1 - x}\right) = x + {x}^{2}/2 + {x}^{3}/3 + \cdots \) which is valid for \( - 1 < x < 1 \) .\n\nBy Proposition 16.1.1 we have\n\n\[ \zeta \left( s\right) = \mathop{\prod }\limits_{{p \leq N}}{\left( 1 - {p}^{-s}\right) }^{-1}{\lambda }_{N}\left( s\right) \]\n\nwhere \( {\lambd... | Yes |
Proposition 16.1.4. Let \( \mathcal{P} \) be a set of positive prime numbers. Then\n\n(a) If \( \mathcal{P} \) is finite, then \( d\left( \mathcal{P}\right) = 0 \) .\n\n(b) If \( \mathcal{P} \) consists of all but finitely many positive primes, then \( d\left( \mathcal{P}\right) = 1 \) .\n\n(c) If \( \mathcal{P} = {\ma... | Proof. Parts (a) and (c) are clear from the definition of Dirichlet density. Part (b) follows quickly from the corollary to Proposition 16.1.2 and Proposition 16.1.3. | No |
Proposition 16.3.1. Let \( A \) be a finite abelian group. If \( \chi ,\psi \in \widehat{A} \) and \( a, b \in A \), then\n\n(i) \( \mathop{\sum }\limits_{{a \in A}}\chi \left( a\right) \overline{\psi \left( a\right) } = {n\delta }\left( {\chi ,\psi }\right) \) where \( \delta \left( {\chi ,\chi }\right) = 1 \) and \( ... | Proof. Since \( \mathop{\sum }\limits_{{a \in A}}\chi \left( a\right) \overline{\psi \left( a\right) } = \mathop{\sum }\limits_{{a \in A}}\chi {\psi }^{-1}\left( a\right) \) it will be enough to show (i) that we can prove \( \mathop{\sum }\limits_{{a \in A}}\chi \left( a\right) = n \) if \( \chi = {\chi }_{0} \) and \(... | Yes |
If \( {\chi }_{0} \) denotes the trivial character modulo \( m \), then \( \mathop{\lim }\limits_{{s \rightarrow 1}} \) \( G\left( {s,{\chi }_{0}}\right) /\ln {\left( s - 1\right) }^{-1} = 1 \) . | The first assertion is easy. \( L\left( {s,{\chi }_{0}}\right) \) is a real valued function of positive real numbers. We have seen \( L\left( {s,{\chi }_{0}}\right) = \mathop{\prod }\limits_{{p \mid m}}\left( {1 - {p}^{-s}}\right) \zeta \left( s\right) \) . It follows that \( G\left( {s,{\chi }_{0}}\right) = \mathop{\s... | Yes |
Lemma 1. Suppose \( \left\{ {a}_{n}\right\} \) and \( \left\{ {b}_{n}\right\} \) for \( n = 1,2,3,\ldots \) are sequences of complex numbers such that \( \mathop{\sum }\limits_{{n = 1}}^{\infty }{a}_{n}{b}_{n} \) converges. Let \( {A}_{n} = {a}_{1} + {a}_{2} + \cdots + {a}_{n} \) and suppose \( {A}_{n}{b}_{n} \rightarr... | Proof. Let \( {S}_{N} = \mathop{\sum }\limits_{{n = 1}}^{N}{a}_{n}{b}_{n} \) . Set \( {A}_{0} = 0 \) . Then\n\n\[ {S}_{N} = \mathop{\sum }\limits_{{n = 1}}^{N}\left( {{A}_{n} - {A}_{n - 1}}\right) {b}_{n} = \mathop{\sum }\limits_{{n = 1}}^{N}{A}_{n}{b}_{n} - \mathop{\sum }\limits_{{n = 1}}^{N}{A}_{n - 1}{b}_{n} \]\n\n\... | Yes |
Proposition 16.5.1. \( \zeta \left( s\right) - {\left( s - 1\right) }^{-1} \) can be continued to an analytic function on the region \( \{ s \in \mathbb{C} \mid \sigma > 0\} \) . | Proof. Assume \( \sigma > 1 \) . Then, by the lemma\n\n\[ \zeta \left( s\right) = \mathop{\sum }\limits_{{n = 1}}^{\infty }{n}^{-s} = \mathop{\sum }\limits_{{n = 1}}^{\infty }n\left( {{n}^{-s} - {\left( n + 1\right) }^{-s}}\right) . \]\n\nFor a real number \( x \) recall that \( \left\lbrack x\right\rbrack \) is the gr... | Yes |
Lemma 2. Let \( \chi \) be a nontrivial character modulo \( m \) . For all \( N > 0 \) we have \( \left| {\mathop{\sum }\limits_{{n = 0}}^{N}\chi \left( n\right) }\right| \leq \phi \left( m\right) . | Proof. Write \( N = {qm} + r \) where \( 0 \leq r < m \) . Since \( \chi \left( {n + m}\right) = \chi \left( n\right) \) for all \( n \) we see\n\n\[ \mathop{\sum }\limits_{{n = 1}}^{N}\chi \left( n\right) = q\left( {\mathop{\sum }\limits_{{n = 0}}^{{m - 1}}\chi \left( n\right) }\right) + \mathop{\sum }\limits_{{n = 0}... | Yes |
Proposition 16.5.2. Let \( \chi \) be a nontrivial Dirichlet character modulo \( m \) . Then, \( L\left( {s,\chi }\right) \) can be continued to an analytic function in the region \( \{ s \in \mathbb{C} \mid \sigma > 0\} \) . | Proof. Define \( S\left( x\right) = \mathop{\sum }\limits_{{n \leq x}}\chi \left( n\right) \).\n\nBy Lemma 1 we have for \( \sigma > 1 \) ,\n\n\[ L\left( {s,\chi }\right) = \mathop{\sum }\limits_{{n = 1}}^{\infty }S\left( n\right) \left( {{n}^{-s} - {\left( n + 1\right) }^{-s}}\right) \]\n\n\[ = s\mathop{\sum }\limits_... | Yes |
Proposition 16.5.3. Let \( F\left( s\right) = \mathop{\prod }\limits_{\chi }L\left( {s,\chi }\right) \) where the product is over all Dirichlet characters modulo \( m \) . Then, for \( s \) real and \( s > 1 \) we have \( F\left( s\right) \geq 1 \) . | Proof. Assume \( s \) is real and \( s > 1 \) . Recall that\n\n\[ G\left( {s,\chi }\right) = \mathop{\sum }\limits_{p}\mathop{\sum }\limits_{{k = 1}}^{\infty }\frac{1}{k}\chi \left( {p}^{k}\right) {p}^{-{ks}}. \]\n\nSumming over \( \chi \) and using Proposition 16.3.2, part (ii), we find\n\n\[ \mathop{\sum }\limits_{\c... | Yes |
Proposition 16.5.4. If \( \chi \) is a nontrivial complex character modulo \( m \), then \( L\left( {1,\chi }\right) \neq 0 \) . | Proof. From the series defining \( L\left( {s,\chi }\right) \) we see that for \( s \) real, \( s > 1,\overline{L\left( {s,\chi }\right) } = \) \( L\left( {s,\bar{\chi }}\right) \) . Letting \( s \) tend towards 1 it follows that \( L\left( {1,\chi }\right) = 0 \) implies \( L\left( {1,\bar{\chi }}\right) = 0 \) . Assu... | Yes |
Lemma 3. Suppose \( f \) is a nonnegative, multiplicative function on \( {\mathbb{Z}}^{ + } \), i.e., for all \( m, n > 0 \) with \( \left( {m, n}\right) = 1,{f}^{\prime }\left( {mn}\right) = f\left( m\right) f\left( n\right) \) . Assume there is a constant \( c \) such that \( f\left( {p}^{k}\right) < c \) for all pri... | Proof. Fix \( s > 1 \) . Let \( a\left( p\right) = \mathop{\sum }\limits_{{k = 1}}^{\infty }f\left( {p}^{k}\right) {p}^{-{ks}} \) . Then \( a\left( p\right) < c{p}^{-s}\mathop{\sum }\limits_{{k = 0}}^{\infty }{p}^{-{ks}} = \) \( c{p}^{-s}{\left( 1 - {p}^{-s}\right) }^{-1} \), and so \( a\left( p\right) < {2c}{p}^{-s} \... | Yes |
Proposition 16.6.1. Let \( k \) be a positive integer. Then, \( \zeta \left( 0\right) = - \frac{1}{2} \) and for \( k > 1,\zeta \left( {1 - k}\right) = - {B}_{k}/k \) where \( {B}_{k} \) is the \( k \) th Bernoulli number. | Proof. In Equation (viii) substitute \( s = 1 - k \) . The result is \( {\zeta }^{ * }\left( {1 - k}\right) = \) \( {\left( -1\right) }^{k}{\int }_{0}^{\infty }{R}_{k}\left( t\right) {dt} \) . Since \( {R}_{k}\left( t\right) = \left( {d/{dt}}\right) {R}_{k - 1}\left( t\right) \) we deduce \( \left( {1 - {2}^{k}}\right)... | Yes |
Lemma 1. \( t{F}_{\chi }\left( {e}^{-t}\right) = \mathop{\sum }\limits_{{n = 0}}^{\infty }{\left( -1\right) }^{n}\left( {{B}_{n,\chi }/n!}\right) {t}^{n} \) . | Proof. Simply substitute \( - t \) for \( t \) in Equation (xi). | No |
Proposition 16.6.2. Let \( k \) be a positive integer. Then \( L\left( {1 - k,\chi }\right) = - {B}_{k,\chi }/k \) . | Proof. In Equation (x) substitute \( s = 1 - k \) . The result is \( \left( {1 - {2}^{k}}\right) L\left( {1 - k,\chi }\right) \) \( = {\left( -1\right) }^{k}{\int }_{0}^{\infty }{R}_{\chi, k}\left( t\right) {dt} \) . Since \( {R}_{\chi, k}\left( t\right) = \left( {d/{dt}}\right) {R}_{\chi, k - 1}\left( t\right) \) it f... | Yes |
Proposition 17.3.1. Let \( a, b, c \) be nonzero integers, square free, pairwise relatively prime and not all positive nor all negative. Then (17) has a nontrivial integral solution iff the following conditions are satisfied\n\n(i) \( - {abRc} \) .\n\n(ii) \( - {acRb} \) .\n\n(iii) \( - {bcRa} \) . | It is convenient to prove this result in the following equivalent form.\n\nProposition 17.3.2. Let | No |
Proposition 17.4.1. If \( p \) is an odd prime such that \( {2p} + 1 = q \) is also prime then (24) has no integral solution with \( p \smallsetminus {xyz} \) | Proof. Assume on the contrary that such a solution exists and suppose that \( \left( {x, y, z}\right) = 1 \) . Write\n\n\[- {x}^{p} = \left( {y + z}\right) \left( {{z}^{p - 1} - {z}^{p - 2}y + \cdots + {y}^{p - 1}}\right) .\]\n\n(25)\n\nThe two factors on the right are relatively prime. For clearly \( p \smallsetminus ... | Yes |
Proposition 17.5.1. If \( \xi \) is irrational then there are infinitely many rational numbers \( x/y,\left( {x, y}\right) = 1 \) such that \( \left| {x/y - \xi }\right| < 1/{y}^{2} \) . | Proof. Partition the half-open interval \( \lbrack 0,1) \) by\n\n\[ \lbrack 0,1) = \left\lbrack {0,\frac{1}{n}}\right) \cup \left\lbrack {\frac{1}{n},\frac{2}{n}}\right) \cup \cdots \cup \left\lbrack {\frac{n - 1}{n},1}\right) . \]\n\nIf \( \left\lbrack \alpha \right\rbrack \) denotes, as usual, the largest integer les... | Yes |
Lemma 1. If \( d \) is a positive square-free integer then there is a constant \( M \) such that \( \left| {{x}^{2} - d{y}^{2}}\right| < M \) has infinitely many integral solutions. | Proof. Write \( {x}^{2} - d{y}^{2} = \left( {x + \sqrt{d}y}\right) \left( {x - \sqrt{d}y}\right) \) . By Proposition 17.5.1 there exist infinitely many pairs of relatively prime integers \( \left( {x, y}\right), y > 0 \) satisfying \( \left| {x - \sqrt{dy}}\right| < 1/y \) . It follows that\n\n\[ \left| {x + \sqrt{d}y}... | Yes |
Corollary 1. The equation \( {x}^{2} + {y}^{2} = n, n > 0 \) has an integral solution iff ord, \( n \) is even for every prime \( p \equiv 3\left( 4\right) \) . When that is the case the number of solutions is \( \mathop{\prod }\limits_{{p \equiv 1\left( 4\right) }}\left( {1 + {\operatorname{ord}}_{p}n}\right) \) . | Proof. Since \( \chi \left( n\right) \) is multiplicative it follows by Exercise 10, Chapter 2 that \( \mathop{\sum }\limits_{{d \mid n}}\chi \left( d\right) \) is multiplicative. If \( p \equiv 1\left( 4\right) \) then \( \mathop{\sum }\limits_{{d \mid {p}^{n}}}\chi \left( d\right) = n + 1 \) while if \( p \equiv 3\le... | No |
Corollary 2. Let \( m \) be a positive odd integer. The number of integral solutions \( \left( {x, y}\right), x > 0, y > 0 \) to \( {x}^{2} + {y}^{2} = {2m} \) is \( \mathop{\sum }\limits_{{d \mid m}}\chi \left( d\right) \) . | Proof. Since \( {2m} \equiv 2\left( 4\right), y \) is positive. On the other hand \( \chi \left( {2d}\right) = 0 \) for any divisor \( {2d} \) of \( {2m} \) . | No |
Lemma 1. If \( p \) is prime the congruence \( {x}^{2} + {y}^{2} + 1 \equiv 0\left( p\right) \) has a solution in integers \( x, y \) . | Proof. Denote by \( S \) the set of squares modulo \( p \) . Then \( S \) and \( \{ - 1 - x \mid x \in S\} \) \( = {S}^{\prime } \) each have \( \left( {p + 1}\right) /2 \) elements. Thus \( S \) and \( {S}^{\prime } \) are not disjoint and the result follows. | Yes |
Lemma 2. Suppose for a prime \( p \) there is an integer \( m,1 < m < p \) such that \( {mp} \) is the sum of four squares. Then there is an \( n,0 < n < m \) such that \( {np} \) is the sum of four squares. | Proof. Write\n\n\[ {mp} = {x}_{1}^{2} + {x}_{2}^{2} + {x}_{3}^{2} + {x}_{4}^{2}. \]\n\n(40)\n\nLet \( {x}_{i} \equiv {y}_{i}\left( m\right) \) with \( - m/2 < {y}_{i} \leq m/2 \) . Then \( {y}_{1}^{2} + {y}_{2}^{2} + {y}_{3}^{2} + {y}_{4}^{2} \equiv 0\left( m\right) \) so that there is an integer \( r \geq 0 \) such th... | Yes |
Proposition 17.7.1. Any positive integer is the sum of four squares. | Proof. This follows immediately from Lemmas 1 and 2 and Exercise 28. | No |
Proposition 17.7.2. Let \( n \) be a positive integer \( n \equiv 4\\left( 8\\right) \) . The number of integral solutions \( \\left( {x, y, z, w}\\right), x, y, z, w \) positive and odd to the equation\n\n\[ \n{x}^{2} + {y}^{2} + {z}^{2} + {w}^{2} = n \n\]\n\nis the sum of the positive odd divisors of \( n \) . | The proof of the proposition is divided into several lemmas. Let \( N \) denote the number of integral solutions \( \\left( {x, y, z, w}\\right) \) to (43) with \( x, y, z, w \) positive and odd. Since \( n \equiv 4\\left( 8\\right) \) we may write \( n = {2m}, m \equiv 2\\left( 4\\right) \) . | No |
Lemma 3. \( N \) is the number of solutions \( \left( {x, y, z, u, v}\right) \) to the system of Diophantine equations\n\n\[ \n{x}^{2} + {y}^{2} = {2u} \]\n\n\[ \n{z}^{2} + {w}^{2} = {2v} \]\n\n(44)\n\n\[ \nu + v = m, \]\n\nwith \( x, y, z, u, v \) odd and positive. | Proof. This is left as a simple exercise. | No |
Lemma 4. \( N = \sum \chi \left( {de}\right) = \sum {\left( -1\right) }^{\left( {{de} - 1}\right) /2} = \sum {\left( -1\right) }^{\left( {d - e}\right) /2} \) the sum over all solutions \( \left( {d, e, t, s}\right) \) in positive odd integers to \( {ds} + {et} = m \) . | Proof. By Lemma 3 and Corollary 2 of Proposition 17.6.1 we see easily that\n\n\[ N = \mathop{\sum }\limits_{\substack{{u, v} \\ {u + v = m} }}\left( {\mathop{\sum }\limits_{\substack{{d \mid u} \\ {e \mid v} }}\chi \left( d\right) \chi \left( e\right) }\right) \]\n\n(45)\n\nWrite \( u = {ds}, v = {et} \) so that the te... | Yes |
Lemma 5. Given \( \left( {d, e, t, s}\right) \in S \) there is a unique \( n \in {\mathbb{Z}}^{ + } \) such that \( {\psi }_{n}\left( {d, e, t, s}\right) \in S \) . | Proof. One sees immediately using (46) that \( {d}^{\prime },{e}^{\prime },{t}^{\prime },{s}^{\prime } \) are odd, \( {d}^{\prime } > {e}^{\prime },{d}^{\prime } > 0 \) , \( {e}^{\prime } > 0 \) . Furthermore the conditions \( {s}^{\prime } > 0,{t}^{\prime } > 0 \) are equivalent to, by (46), \( e/\left( {d - e}\right)... | Yes |
Lemma 6. \( \Phi \) is a bijection. | Proof. We will show that \( {\Phi }^{2} \) is the identity map. For if \( \left( {d, e, t, s}\right) \in S \) then\n\n\[ \n{\Phi }^{2}\left( {d, e, t, s}\right) = \Phi \left( {{\left( {A}_{n}\left( \begin{array}{l} t \\ s \end{array}\right) \right) }^{ * }, \cdot {\left( {A}_{n}^{-1}\left( \begin{array}{l} d \\ e \end{... | Yes |
Lemma 1. The equation \( {x}^{3} + {y}^{3} = u{z}^{3} \) , \( u \) a unit in \( \mathbb{Z}\left\lbrack \omega \right\rbrack \) has no solution with \( x, y, z \in \mathbb{Z}\left\lbrack \omega \right\rbrack ,\lambda \smallsetminus {xyz} \) . | Proof. Note that since \( \lambda \) is irreducible the condition \( \lambda \smallsetminus {xyz} \) is equivalent to \( \lambda \smallsetminus x,\lambda \smallsetminus y,\lambda \smallsetminus z \) . If \( x \in \mathbb{Z}\left\lbrack \omega \right\rbrack, x \equiv 1\left( \lambda \right) \) then \( {x}^{3} \equiv 1\l... | Yes |
Lemma 2. If \( {x}^{3} + {y}^{3} = u{z}^{3} \) for \( x, y, z \in \mathbb{Z}\left\lbrack \omega \right\rbrack ,\lambda \smallsetminus {xy},\lambda \mid z \) then \( {\lambda }^{2} \mid z \) . | Proof. Reduction of (51) modulo \( {\lambda }^{4} \) gives\n\n\[ \pm 1 \pm 1 \equiv u{z}^{3}\left( {\lambda }^{4}\right) \]\n\nIf \( 0 \equiv u{z}^{3}\left( {\lambda }^{4}\right) \) then 3 or \( {\mathrm{d}}_{\lambda }z \geq 4 \) so that \( {\operatorname{ord}}_{\lambda }z \geq 2 \) . If \( \pm 2 \equiv u{z}^{3}\left( ... | Yes |
Lemma 1. If \( u \) is a unit in \( \mathbb{Z}\left\lbrack \zeta \right\rbrack \) then \( {\zeta }^{s}u \) is real for some rational integer \( s \) . | Proof. Observe first of all that complex conjugation is an automorphism of \( \mathbb{Q}\left( \zeta \right) \) since \( \bar{\zeta } = {\zeta }^{l - 1} \) . Thus if \( u \) is a unit then \( \bar{u} \) is a unit and \( \tau = u/\bar{u} \in \mathbb{Z}\left\lbrack \zeta \right\rbrack \) . Furthermore if \( \rho \) is an... | Yes |
Proposition 17.11.1. If \( l \) is a regular prime then the diophantine equation\n\n\[ \n{x}^{l} + {y}^{l} = {z}^{l} \n\]\n\nhas no solution in rational integers \( x, y, z \) with \( l + {xyz} \) . | The proof of this proposition will be presented in several lemmas. We begin by factoring the left-hand side of (74)\n\n\[ \n{x}^{l} + {y}^{l} = \left( {x + y}\right) \left( {x + {\zeta y}}\right) \cdots \left( {x + {\zeta }^{l - 1}y}\right) . \n\]\n\nRecall that two ideals \( \mathfrak{A} \) and \( \mathfrak{B} \) are ... | Yes |
Lemma 2. The ideals \( \left( {x + {\zeta }^{i}y}\right) \) and \( \left( {x + {\zeta }^{j}y}\right) \) are relatively prime if \( i ≢ j\left( l\right) \) . | This lemma has already been proven in Section 6, Chapter 14. | No |
Lemma 3. There exist \( u,\beta \in \mathbb{Z}\left\lbrack \zeta \right\rbrack, u \) is a real unit such that \( x + {\zeta y} = {\zeta }^{s}{u\beta } \) , where \( s \in \mathbb{Z} \) and \( \beta \equiv n\left( l\right) \) for some \( n \in \mathbb{Z} \) . | Proof. Using Lemma 2, Corollary in Section 6, Chapter 14, and the fact that the right-hand side of (74) is an \( l \) th power we see that \( \left( {x + {\zeta y}}\right) = {\mathfrak{A}}^{l} \) for some ideal \( \mathfrak{A} \) . Since \( l\backprime h \) it follows that \( \mathfrak{A} \) is principal. Thus \( x + {... | Yes |
Lemma 4. \( x + {\zeta y} - {\zeta }^{2s}x - {\zeta }^{{2s} - 1}y \in l\mathbb{Z}\left\lbrack \zeta \right\rbrack \) . | By Proposition 6.4.1. \( 1,\zeta ,{\zeta }^{2},\ldots ,{\zeta }^{l - 2} \) are linearly independent over \( \mathbb{Q} \) . Furthermore we may assume \( l > 3 \) (by Section 8) and \( 0 \leq s \leq l - 1 \) . The proof of Proposition 17.11.1 will be completed by deriving a contradiction from the relation of Lemma 4. By... | Yes |
If \( \alpha \) is a real algebraic number of degree \( n, n \geq 2 \) then there is a constant \( c > 0 \) such that for any rational number \( p/q, q > 0 \) | Proof. It is clearly enough to assume \( \left| {\alpha - p/q}\right| \leq 1 \) . By the mean value theorem \( \left| {f\left( {p/q}\right) }\right| = \left| {f\left( \alpha \right) - f\left( {p/q}\right) }\right| \leq \left| {\alpha - p/q}\right| A \) where \( f\left( x\right) \in \mathbb{Z}\left\lbrack x\right\rbrack... | Yes |
Theorem 4. Suppose \( p \neq 2 \) or 3 and \( p \nmid D \) . Consider the elliptic curve \( {y}^{2} = {x}^{3} + D \) over \( {\mathbb{F}}_{p} \) . If \( p \equiv 2\\left( 3\\right) \) then \( {N}_{p} = p + 1 \) . If \( p \equiv 1\\left( 3\\right) \) let \( p = \\pi \\bar{\\pi } \) with \( \\pi \\in \\mathbb{Z}\\left\\l... | As an example consider the curve \( {y}^{2} = {x}^{3} + 1 \) over \( {\\mathbb{F}}_{13} \) . We find \( {13} = \\left( {-1 + {3\\omega }}\\right) \\left( {-1 + 3{\\omega }^{2}}\\right) \) and \( - 1 + {3\\omega } \\equiv 2\\left( 3\\right) \) . To apply the formula in the theorem we must know \( {\\left( 4/ - 1 + 3\\om... | Yes |
Theorem 5. Suppose \( p \neq 2 \) and \( p \nmid D \) . Consider the elliptic curve \( {y}^{2} = {x}^{3} - {Dx} \) over \( {\mathbb{F}}_{p} \) . If \( p \equiv 3\left( 4\right) \) then \( {N}_{p} = p + 1 \) . If \( p \equiv 1\left( 4\right) \) let \( p = \pi \bar{\pi } \) with \( \pi \in \mathbb{Z}\left\lbrack i\right\... | As an example, consider \( {y}^{2} = {x}^{3} - x \) over \( {\mathbb{F}}_{13} \) . One sees\n\n\[ \n{13} = \left( {3 + {2i}}\right) \left( {3 - {2i}}\right)\n\]\n\nand \( 3 + {2i} \equiv 1\left( {2 + {2i}}\right) \) . The formula of the theorem tells us that \( {N}_{13} = \) \( {13} + 1 - \left( {3 + {2i}}\right) - \le... | Yes |
Proposition 18.5.1. Let \( \chi \) be an algebraic Hecke character of weight \( m \) . Then if \( \left( {A, M}\right) = \left( 1\right) ,\left| {\chi \left( A\right) }\right| = N{A}^{m/2} \) . | Proof. Let \( {I}_{M} \) be the set of ideals in \( \mathcal{O} \) which are relatively prime to \( M \) . We put an equivalence relation on \( {I}_{M} \) as follows; if \( A, B \in {I}_{M} \) we say \( A \sim B \) if there exist \( \alpha ,\beta \in \mathcal{O} \) such that \( \alpha ,\beta \equiv 1\left( M\right) \) ... | Yes |
Theorem 7. Let \( E \) be the elliptic curve defined by \( {y}^{2} = {x}^{3} - {Dx} \) with \( D \in \mathbb{Z} \). The character \( \chi \) defined above is an algebraic Hecke character of weight 1 for the modulus \( \left( {8D}\right) \). Moreover, \( L\left( {E, s}\right) = L\left( {s,\chi }\right) \). | Proof. Assume to begin with that \( p \equiv 3\left( 4\right) \) and \( p \nmid {2D} \). By Theorem 5, \( {N}_{p} = p + 1 \) so that \( {a}_{p} = 0 \). Let \( P = \left( p\right) \). Then \( {NP} = {p}^{2} \) and \( \chi \left( P\right) = - p \). Thus\n\n\[ 1 - {a}_{p}{p}^{-s} + {p}^{1 - {2s}} = 1 + {p}^{1 - {2s}} = 1 ... | Yes |
Lemma 1. Suppose \( p \) is an odd prime and \( p \equiv 2\\left( 3\\right) \) . Then \( {\\left( 4D/p\\right) }_{6} = 1 \) . | Proof. It follows from the hypotheses that \( p + 1 \) is divisible by 6 . We know \( {\\left( 4D\\right) }^{p - 1} \equiv 1\\left( p\\right) \) . Raising both sides of this congruence to the \( \\left( {\\left( {p + 1}\\right) /6}\\right) \) th power gives the result. | Yes |
Lemma 2. Suppose \( \alpha \in \mathbb{Z}\left\lbrack \omega \right\rbrack \) and \( \left( {\alpha ,{2D}}\right) = \left( 1\right) \) . Define \( {\left( D/\alpha \right) }_{2} \) to be \( {\left( D/\alpha \right) }_{6}^{3} \) . Then \( {\left( D/\alpha \right) }_{2} = \left( {D/{N\alpha }}\right) \), where this last ... | Proof. Both \( {\left( D/\alpha \right) }_{2} \) and \( \left( {D/{N\alpha }}\right) \) are multiplicative in \( \alpha \) . Thus it is enough to check that they are equal when \( \alpha = \pi \), a prime element.\n\nSuppose \( \pi = p \neq 2 \), a rational prime with \( p \equiv 2 \) (3). Then \( {Np} = {p}^{2} \) and... | Yes |
Theorem 8. Let \( E \) be the elliptic curve over \( \mathbb{Q} \) defined by \( {y}^{2} = {x}^{3} + D, D \in \mathbb{Z} \). The character \( \chi \) defined above is an algebraic Hecke character of weight 1 for the modulus (12D). Moreover, \( L\left( {E, s}\right) = L\left( {s,\chi }\right) \). | Proof. Assume first that \( p \equiv 2\left( 3\right) \) and \( p \nmid {6D} \). By Theorem \( 4,{N}_{p} = p + 1 \) so that \( {a}_{p} = 0 \). Let \( P = \left( p\right) .P \) is a prime ideal in \( \mathbb{Z}\left\lbrack \omega \right\rbrack \) and \( \chi \left( P\right) = - p \). Thus\n\n\[ 1 - {a}_{p}{p}^{-s} + {p}... | Yes |
Lemma 1. \( {\left\lbrack \left( {\theta }_{1} - {\theta }_{2}\right) \left( {\theta }_{2} - {\theta }_{3}\right) \left( {\theta }_{1} - {\theta }_{3}\right) \right\rbrack }^{2} = - \left( {4{a}^{3} + {27}{b}^{2}}\right) \) . | Proof. By substituting \( x = {\theta }_{i} \) in the formal derivative of \( f\left( x\right) \) one obtains \( 3{\theta }_{i}^{2} + a = \left( {{\theta }_{i} - {\theta }_{j}}\right) \left( {{\theta }_{i} - {\theta }_{k}}\right), i, j, k \) distinct. Multiplication now shows that the negative of the left-hand side of ... | Yes |
Lemma 2. \( \phi \) is a homomorphism. | Proof. If \( P = \left( {\alpha ,\beta }\right) \), since the definition of \( \phi \) is independent of \( \beta \) and \( - P = \left( {\alpha , - \beta }\right) \), then \( \phi \left( P\right) = \phi \left( {-P}\right) \) . Now if \( \rho \) is in \( U/{U}^{2} \), then \( {\rho }^{2} = 1 \) , so that \( \phi \left(... | Yes |
Lemma 3. \( \ker \phi = {2E} \) . | Proof. Since \( \phi \left( {2P}\right) = \phi {\left( P\right) }^{2} = 1 \) we see that \( {2E} \subseteq \ker \phi \) . Thus, consider a point \( P \), which we may assume different from \( \infty \), such that \( \phi \left( P\right) = 1 \) . Write \( P = \left( {\alpha ,\beta }\right) ,\alpha ,\beta \in k \) . Then... | Yes |
Lemma 2. A discrete additive subgroup \( A \) of \( W \) is a lattice. | Proof. Let \( {v}_{1},\ldots ,{v}_{m} \) be a maximal set of \( \mathbb{R} \) -independent elements in \( A \) . Then any element \( a \) of \( A \) may be written in the form \( a = {r}_{1}{v}_{1} + \cdots + \) \( {r}_{m}{v}_{m} \) where \( {r}_{i} \in \mathbb{R} \) . Now \( A \) contains the lattice \( \Gamma = \math... | Yes |
Lemma 3. \( \mu \left( \mathcal{E}\right) \) is a lattice. | Proof. If \( T \) is a compact set in \( {\mathbb{R}}^{s + t} \), then \( S = {\lambda }^{-1}\left( {T \cap \mu \left( \mathcal{E}\right) }\right) \subset \phi \left( \mathbb{D}\right) \), and the comment preceding this lemma together with lemma 1 shows that \( S \) is finite. Hence, \( T \cap \mu \left( \& \right) \) ... | No |
Lemma 1. The set of ideals \( \mathrm{I}\left( P\right) \) is finite. | Proof. \( g\left( x\right) - g\left( \theta \right) = \left( {x - \theta }\right) t\left( x\right) \), where \( t\left( x\right) \) is a linear polynomial with coefficients in \( \mathbb{Z}\left\lbrack \theta \right\rbrack \) . Substituting \( x = \alpha /\beta \) gives \( g\left( \theta \right) {\beta }^{2} = {h}_{\al... | Yes |
Lemma 2. \( \left( {\alpha - {\beta \theta }}\right) = \mathrm{I}\left( P\right) {C}^{2} \) for some ideal \( C \) . | Proof. Since \( \mathbf{I}\left( P\right) \) is the greatest common divisor of \( \alpha - {\beta \theta } \) and \( {h}_{\alpha ,\beta } \), we may write \( \left( {\alpha - {\beta \theta }}\right) = \mathrm{I}\left( P\right) A,\left( {h}_{\alpha ,\beta }\right) = \mathrm{I}\left( P\right) B \), where \( A \) and \( B... | Yes |
Lemma 3. There is a finite set of algebraic integers \( S \) such that for any \( P = \left( {\alpha ,\beta }\right) \in E\left( \mathbb{Q}\right) \) one can write\n\n\[ \alpha - {\beta \theta } = {u\gamma }{\tau }^{2} \]\n\nfor a suitable unit \( u \), an algebraic number \( \tau \), and \( \gamma \in S \) . | Proof. If \( C \) is as in the preceding lemma, then \( C \) is equivalent to \( {C}_{s} \) for some \( s \) . Therefore, \( \mathrm{I}\left( P\right) {C}_{s}^{2} \) is eqivalent to the principal ideal \( \mathrm{I}\left( P\right) {C}^{2} = \) \( \left( {\alpha - {\beta \theta }}\right) \) and is thus a principal ideal... | Yes |
Theorem 19.4.1. E/2E is a finite group. | Proof. It is enough to show that \( \phi \left( E\right) \) is finite. We may assume that \( P \neq \infty \) and that \( P \) does not have order 2 . Then \( \phi \left( P\right) \) is defined as the coset modulo \( {U}^{2} \) of \( \alpha /\beta - x \), where \( P = \left( {\alpha /\beta, w}\right) \), in the group \... | Yes |
Lemma 1. If \( C \) is a fixed constant, then there are at most a finite number of points \( P \) on \( E \), rational over \( \mathbb{Q} \) with \( H\left( P\right) \leqq C \) . | Now fix a point \( Q \neq \infty \) on \( E \), and let \( P \) be any \( \mathbb{Q} \) -rational point on \( E \) such that \( P - Q \) is not of order 2, i.e., \( {2P} \neq {2Q} \) . Write, using homogeneous coordinates as above,\n\n\[ P = \left( {{x}_{1}{z}_{1},{y}_{1},{z}_{1}^{3}}\right) \]\n\n\[ Q = \left( {{ce}, ... | Yes |
Theorem 19.5.1. The group \( E\left( \mathbb{Q}\right) \) of \( \mathbb{Q} \) -rational points on \( E \) is a finitely generated abelian group. | Proof. Let \( P \) be an arbitrary point on \( E \), rational over \( \mathbb{Q} \) . Then \( P + {Q}_{{a}_{1}} = \) \( 2{P}_{1} \) for some \( {a}_{1} \) and \( {P}_{1} \) . We have by the preceding lemma\n\n\[ H\left( {P}_{1}\right) \leqq C{\left( H\left( 2{P}_{1} - {Q}_{{a}_{1}}\right) \right) }^{1/2} = C{\left( H\l... | Yes |
Proposition 20.4.1. For all \( \alpha \in \overline{\mathbb{Q}}, H\left( \alpha \right) \geq 1 \) . Moreover, if \( C \) and \( n \) are given, the set \( \{ \alpha \in \overline{\mathbb{Q}} \mid H\left( \alpha \right) \leq C \) and \( \deg \left( \alpha \right) \leq n\} \) is finite. | Proof. We cannot give the proof of the full result since we have defined \( H\left( \alpha \right) \) only in the case \( \alpha \) is an algebraic integer. We give the proof in this special case and remark that the proof of the general result is quite similar.\n\nThe first assertion is clear from the definition. Now a... | No |
Proposition 20.4.2. Let E/K be an elliptic curve defined over an algebraic number field \( K \) . For all \( C \), the set \( \{ P \in E\left( K\right) \mid h\left( P\right) \leq C\} \) is finite. | Proof. By Proposition 20.4.1, the set \( \left( {\alpha \in K \mid H\left( \alpha \right) \leq {e}^{c}\} }\right. \) is finite. Since for each \( \alpha \in K \) there are at most two values of \( \beta \) such that \( \left( {\alpha ,\beta }\right) \in E\left( K\right) \) , the result follows. | Yes |
Proposition 20.4.3. Let \( E/K \) be an elliptic curve defined over a number field \( K \) . For all \( P, Q \in E\left( \bar{K}\right) \) we have\n\n\[ h\left( {P + Q}\right) + h\left( {P - Q}\right) = {2h}\left( P\right) + {2h}\left( Q\right) + O\left( 1\right) . \] | Proof. We sketch the proof, referring to Silverman [Si1] for details. Assuming that \( K = \mathbb{Q} \), one can use the methods of Chapter 19 to establish that \( h\left( {P + Q}\right) + h\left( {P - Q}\right) \leq {2h}\left( P\right) + {2h}\left( Q\right) + O\left( 1\right) \) . The problem is to show the reverse i... | No |
Theorem 20.4.4. The canonical height \( \widehat{h}\left( P\right) \) satisfies\n\n(i) \( h\left( P\right) = h\left( P\right) + O\left( 1\right) \).\n\n(ii) \( \langle P, Q\rangle = 1/2\left( {\widehat{h}\left( {P + Q}\right) - \widehat{h}\left( P\right) - \widehat{h}\left( Q\right) }\right) \) is bi-additive.\n\n(iii)... | Proof. We sketch the proof, referring the reader to the references for the details (we note parenthetically that the canonical height in [Sil] is half the one defined here).\n\nIn equation (1) set \( m = 0 \) and take the limit of both sides as \( n \) tends to infinity. Since we have shown that the right-hand side is ... | No |
Proposition 20.5.4. Let \( K \) be a quadratic number field with discriminant \( D \), and \( E \) an elliptic curve over \( \mathbb{Q} \) . Then\n\n(a) rank \( E\left( K\right) = \operatorname{rank}E\left( \mathbb{Q}\right) + \operatorname{rank}{E}_{D}\left( \mathbb{Q}\right) \) and\n\n(b) \( L\left( {E/K, s}\right) =... | We are now in a position to sketch the proof of Theorem 20.5.2. Let's consider part (a). The assumption is that \( L\left( {E/\mathbb{Q}, s}\right) \) has a simple zero at \( s = 1 \), i.e., \( L\left( {E/\mathbb{Q},1}\right) = 0 \) and \( {L}^{\prime }\left( {E/\mathbb{Q},1}\right) \neq 0 \) . From Proposition 20.5.4,... | No |
Theorem 20.6.1 (Hecke, Deuring, Mordell, Heilbronn).\n\n\[ h\left( D\right) \rightarrow \infty \text{as}\left| D\right| \rightarrow \infty \text{.} \] | The method of proof here is truly amazing. If the generalized Riemann hypothesis is true, then the theorem is true. If the generalized Riemann hypothesis is false, then the theorem is true. Thus, the theorem is true!! | Yes |
11.3.5. Let \( \mathfrak{A} \) be a closed subalgebra of \( \mathbf{E} \) . Then \( \mathfrak{A} \) contains each \( f \in {\mathbf{L}}_{r\left( \mathfrak{A}\right) }^{1} \) for which \( \mathop{\sum }\limits_{{n \in Z}}\left| {\widehat{f}\left( n\right) }\right| < \infty \) . | Proof. Using the notation introduced at the beginning of the proof of 11.3.2, the absolutely convergent Fourier series of \( f \) may be regrouped to appear as\n\n\[ \mathop{\sum }\limits_{\alpha }\widehat{f}\left( {S}_{\alpha }\right) \mathop{\sum }\limits_{{n \in {S}_{\alpha }}}{e}_{n} = \mathop{\sum }\limits_{\alpha... | Yes |
Lemma 1.1.32\n\n\[ \mathop{\sum }\limits_{{\mathbf{y} \in D}}{\omega }^{\langle \mathbf{x},\mathbf{y}\rangle } = \mathop{\sum }\limits_{{{y}_{{h}_{1}} \in Q\smallsetminus \{ 0\} }}\cdots \mathop{\sum }\limits_{{{y}_{{h}_{k}} \in Q\smallsetminus \{ 0\} }}{\omega }^{{x}_{{h}_{1}}{y}_{{h}_{1}} + \cdots + {x}_{{h}_{k}}{y}_... | \[ = {\left( q - 1\right) }^{k - j}\mathop{\prod }\limits_{{l = 1}}^{j}\mathop{\sum }\limits_{{y \in Q\smallsetminus \{ 0\} }}{\omega }^{{x}_{{h}_{i}}y} = {\left( -1\right) }^{j}{\left( q - 1\right) }^{k - j}. \] | Yes |
Proposition 1.1. Let \( E \) be a finite extension of \( K \) . Let \( \left( {\mathfrak{O},\mathfrak{P}}\right) \) be a discrete valuation ring in \( E \) above \( \left( {0, p}\right) \) in \( K \) . Suppose that \( E = K\left( y\right) \) where \( y \) is the root of a polynomial \( f\left( Y\right) = 0 \) having co... | Proof. There exists a constant \( {y}_{0} \in k \) such that \( y \equiv {y}_{0}{\;\operatorname{mod}\;\mathfrak{P}} \) . By hypothesis, \( {f}^{\prime }\left( {y}_{0}\right) ≢ 0{\;\operatorname{mod}\;\mathfrak{P}} \) . Let \( \left\{ {y}_{n}\right\} \) be the sequence defined recursively by\n\n\[ {y}_{n + 1} = {y}_{n}... | No |
Proposition 1.2. If \( {\mathfrak{o}}_{1} \) and \( {\mathfrak{o}}_{2} \) are two discrete valuation rings with quotient field \( K \), such that \( {\mathfrak{o}}_{1} \subset {\mathfrak{o}}_{2} \), then \( {\mathfrak{o}}_{1} = {\mathfrak{o}}_{2} \) . | Proof. We shall first prove that if \( {\mathfrak{p}}_{1} \) and \( {\mathfrak{p}}_{2} \) are their maximal ideals, then \( {\mathfrak{p}}_{2} \subset {\mathfrak{p}}_{1} \) . Let \( y \in {\mathfrak{p}}_{2} \) . If \( y \notin {\mathfrak{p}}_{1} \), then \( 1/y \in {\mathfrak{o}}_{1} \), whence \( 1/y \in {\mathfrak{p}... | Yes |
Proposition 1.3. There exists an element \( y \) of \( K \) having a zero at \( {\mathfrak{o}}_{1} \) and a pole at \( {\mathfrak{o}}_{j}\left( {j = 2,\ldots, n}\right) \) . | Proof. This will be proved by induction. Suppose \( n = 2 \) . Since there is no inclusion relation between \( {\mathfrak{o}}_{1} \) and \( {\mathfrak{o}}_{2} \), we can find \( y \in {\mathfrak{o}}_{2} \) and \( y \notin {\mathfrak{o}}_{1} \) . Similarly, we can find \( z \in {\mathfrak{o}}_{1} \) and \( z \notin {\ma... | Yes |
Theorem 1.4. Given elements \( {a}_{1},\ldots ,{a}_{n} \) of \( K \), and an integer \( N \), there exists an element \( y \in K \) such that \( {\operatorname{ord}}_{i}\left( {y - {a}_{i}}\right) > N \) . | Proof. For each \( i \), use the corollary to get \( {z}_{i} \) close to 1 at \( {0}_{i} \) and close to 0 at \( {\mathrm{o}}_{j}\left( {j \neq i}\right) \), or rather at the valuations associated with these valuation rings. Then \( {z}_{1}{a}_{1} + \cdots + {z}_{n}{a}_{n} \) has the required property. | No |
Proposition 2.1. Let \( \mathfrak{a} \) and \( \mathfrak{b} \) be two divisors. Then \( \Lambda \left( \mathfrak{a}\right) \supset \Lambda \left( \mathfrak{b}\right) \) if and only if \( \mathfrak{a} \geqq \mathfrak{b} \) . If this is the case, then\n\n1. \( \left( {\Lambda \left( \mathfrak{a}\right) : \Lambda \left( \... | Proof. The first assertion is trivial. Formula 1 is easy to prove as follows. If a point \( P \) appears in \( \alpha \) with multiplicity \( d \) and in \( \mathfrak{b} \) with multiplicity \( e \), then \( d \geqq e \) . If \( t \) is an element of order 1 at \( P \) in \( {K}_{P} \), then the index \( \left( {{t}^{-... | No |
Theorem 2.2. Let \( K \) be the function field of a curve, and \( y \in K \) a nonconstant function. If \( \mathfrak{c} \) is the divisor of poles of \( y \), then \( \deg \left( \mathfrak{c}\right) = \left\lbrack {K : k\left( y\right) }\right\rbrack \) . Hence the degree of a divisor of a function is equal to 0 (a fun... | Proof. If we let \( {\mathfrak{c}}^{\prime } \) be the divisor of zeros of \( y \) then \( {\mathfrak{c}}^{\prime } \) is the divisor of poles of \( 1/y \), and \( \left\lbrack {K : k\left( {1/y}\right) }\right\rbrack = n \) also. | No |
Theorem 2.4. There exists an integer \( g \geqq 0 \) depending only on \( K \) such that for any divisor \( \mathfrak{a} \) we have\n\n\[ l\left( \alpha \right) = \deg \left( \alpha \right) + 1 - g + \delta \left( \alpha \right) \]\n\nwhere \( \delta \left( \mathfrak{a}\right) \geqq 0 \) . | By a differential \( \lambda \) of \( K \) we shall mean a \( k \) -linear functional of \( A \) which vanishes on some \( \Lambda \left( \alpha \right) \), and also vanishes on \( K \) (considered to be embedded in \( A \) ). The first condition means that \( \lambda \) is required to be continuous, when we take the d... | No |
Theorem 2.5. If \( \lambda \) is a differential, there is a maximal parallelotope \( \Lambda \left( \mathfrak{a}\right) \) on which \( \lambda \) vanishes. | Proof. If \( \lambda \) vanishes on \( \Lambda \left( {\alpha }_{1}\right) \) and \( \Lambda \left( {\alpha }_{2}\right) \), and if we put\n\n\[ a = \sup \left( {{a}_{1},{a}_{2}}\right) \]\n\nthen \( \lambda \) vanishes on \( \Lambda \left( \alpha \right) \) . Hence to prove our theorem it will suffice to prove that th... | Yes |
Theorem 2.6. The differentials form a 1-dimensional K-space. | Proof. Suppose we have two differentials \( \lambda \) and \( \mu \) which are linearly independent over \( K \) . Suppose \( {x}_{1},\ldots ,{x}_{n} \) and \( {y}_{1},\ldots ,{y}_{n} \) are two sets of elements of \( K \) which are linearly independent over \( k \) . Then the differentials \( {x}_{1}\lambda ,\ldots ,{... | Yes |
Theorem 2.7. Let \( \mathfrak{a} \) be an arbitrary divisor of \( K \) . Then\n\n\[ l\left( \alpha \right) = \deg \left( \alpha \right) + 1 - g + l\left( {c - \alpha }\right) . \]\n\nwhere \( \mathfrak{c} \) is any divisor of the canonical class. In other words,\n\n\[ \delta \left( \alpha \right) = l\left( {\mathfrak{c... | Proof. Let \( \mathfrak{c} \) be the divisor which is such that \( \Lambda \left( \mathfrak{c}\right) \) is the maximal paral-lelotope on which a non-zero differential \( \lambda \) vanishes. If \( \mathfrak{b} \) is an arbitrary divisor and \( y \in L\left( b\right) \), then we know that \( {y\lambda } \) vanishes on ... | Yes |
Corollary 1. If \( \mathfrak{c} \) is a canonical divisor, then \( l\left( \mathfrak{c}\right) = g \) . | Proof. Put \( \alpha = 0 \) in the Riemann-Roch theorem. Then \( L\left( \alpha \right) \) consists of the constants alone, and so \( l\left( \alpha \right) = 1 \) . Since \( \deg \left( 0\right) = 0 \), we get what we want. | No |
Corollary 2. The degree of the canonical class is \( {2g} - 2 \) . | Proof. Put \( \alpha = c \) in the Riemann-Roch theorem, and use Corollary 1. | No |
Corollary 3. If \( \deg \left( \alpha \right) > {2g} - 2 \), then \( \delta \left( \alpha \right) = 0 \) . | Proof. \( \delta \left( \alpha \right) \) is equal to \( l\left( {c - \alpha }\right) \) . Since a function cannot have more zeros than poles, \( L\left( {\mathfrak{c} - \bar{\alpha }}\right) = 0 \) if \( \deg \left( \alpha \right) > {2g} - 2 \) . | Yes |
Lemma 3.1. Let \( D \) be a derivation of a field \( K \). Let \( x \) be any element in an extension field of \( K \), and let \( f\left( X\right) \) be a generator for the ideal determined by \( x \) in \( K\left\lbrack X\right\rbrack \). Then, if \( u \) is an element of \( K\left( x\right) \) satisfying the equatio... | Proof. The necessity has been shown above. Conversely, if \( g\left( x\right), h\left( x\right) \) are in \( K\left\lbrack x\right\rbrack \), and \( h\left( x\right) \neq 0 \), one verifies immediately that the mapping \( {D}^{ * } \) defined by the formulas\n\n\[ {D}^{ * }g\left( x\right) = {g}^{D}\left( x\right) + g\... | No |
Lemma 3.2. If \( K \) is a function field (in one variable) over the algebraically closed field \( k \), then \( \mathcal{D} \) has dimension 1 over \( K \) . An element \( t \in K \) is such that \( K \) over \( k\left( t\right) \) is separable if and only if \( {dt} \) is a basis of the dual space of \( \mathfrak{D} ... | Proof. If \( K \) is separable over \( k\left( t\right) \), then any derivation on \( K \) is determined by its effect on \( t \) . If \( {Dt} = u \), then \( D = u{D}_{1} \), where \( {D}_{1} \) is the derivation such that \( {D}_{1}t = 1 \) . Thus \( \mathcal{D} \) has dimension 1 over \( K \), and \( {dt} \) is a ba... | Yes |
Proposition 4.1. Let \( x \) and \( y \) be two elements of \( k\left( \left( t\right) \right) \), and let \( u \) be another parameter of \( k\left( \left( t\right) \right) \) . Then\n\n\[ \n{\operatorname{res}}_{u}\left( {y\frac{dx}{du}}\right) = {\operatorname{res}}_{t}\left( {y\frac{dx}{dt}}\right) .\n\] | Proof. It clearly suffices to show that for any element \( y \) of \( k\left( \left( t\right) \right) \) we have \( {\operatorname{res}}_{t}\left( y\right) = {\operatorname{res}}_{u}\left( {{ydt}/{du}}\right) \) . Since the residue is \( k \) -linear as a function of power series, and vanishes on power series which hav... | Yes |
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