Q stringlengths 4 3.96k | A stringlengths 1 3k | Result stringclasses 4
values |
|---|---|---|
Lemma 11.5.20. Set\n\n\[ v\left( {k, p}\right) = {v}_{p}\left( {{B}_{-{2k}, p}{p}^{{2k} + 1}/\left( {{2k}\left( {{2k} + 1}\right) }\right) }\right) .\n\]\n\nThen \( v\left( {k, p}\right) \geq 4 \) except for the following values: \( v\left( {1,2}\right) = 1, v\left( {1,3}\right) = 1 \) , \( v\left( {1, p}\right) = 3 \)... | Proof. Assume first that \( \left( {p - 1}\right) \nmid {2k} \) . By Corollary 11.4.8, \( {B}_{-{2k}, p}/\left( {2k}\right) \) is \( p \) -integral, so that \( v\left( {k, p}\right) \geq {2k} + 1 - {v}_{p}\left( {{2k} + 1}\right) \) . It is immediate to check that for \( k \geq 2 \) we have \( {2k} + 1 - {v}_{p}\left( ... | Yes |
Proposition 11.5.22. Let \( \chi \) be a character modulo \( {p}^{v} \) . If \( x \in {\mathbb{Z}}_{p} \) we have\n\n\[ \n{\int }_{{\mathbb{Z}}_{p}}{\operatorname{log\Gamma }}_{p}\left( {\chi, x + t}\right) {dt} = \left( {x - 1}\right) {\psi }_{p}\left( {\chi, x}\right) + {\zeta }_{p}\left( {{\chi \omega },0, x}\right)... | Proof. As for Proposition 11.5.9, this follows immediately by differentiation with respect to \( s \) of Raabe’s formula for \( {\zeta }_{p}\left( {{\chi \omega }, s, x}\right) \) (Proposition 11.2.22), and is left to the reader. The special case \( \chi = {\chi }_{0} \) then follows from Proposition 11.2.19. | No |
Proposition 11.5.23. Let \( \chi \) be a character modulo \( {q}_{p} \) and let \( x \in {\mathbb{Z}}_{p} \) . We have\n\n\[ \n{\operatorname{Log\Gamma }}_{p}\left( {\chi, x}\right) \equiv \left\{ \begin{matrix} 0 & \left( {{\;\operatorname{mod}\;p}{\mathbb{Z}}_{p}}\right) & \text{ if }\chi \neq {\omega }^{-1}\text{ an... | Proof. Assume first that \( p \geq 3 \) . By Propositions 11.5.12 and 11.5.10 we have\n\n\[ \n\log {\Gamma }_{p}\left( {\chi, x}\right) \equiv c\left( p\right) \mathop{\sum }\limits_{{0 \leq j < p}}{\chi \omega }\left( {x + j}\right) \left( {{\;\operatorname{mod}\;p}{\mathbb{Z}}_{p}}\right) , \n\]\n\nwith \( c\left( p\... | Yes |
Proposition 11.5.24. Let \( \chi \) be a primitive character of conductor \( f \), and denote by \( {f}_{1} \) the conductor of the character \( {\chi }_{1} = \chi {\omega }^{-1} \) .\n\n(1) If \( p \nmid f \) then \( {f}_{1} = {q}_{p}f \) and\n\n\[ {L}_{p}^{\prime }\left( {\chi ,0}\right) = {B}_{1}\left( {\chi }_{1}\r... | Proof. Immediate consequence of Corollary 11.5.5 and Proposition 11.5.12 and left to the reader (Exercise 17). | No |
Corollary 11.5.26. We have\n\n\[ \mathop{\lim }\limits_{{N \rightarrow \infty }}\mathop{\sum }\limits_{{0 \leq a < {p}^{N}}}^{\left( p\right) }\frac{{\left( -1\right) }^{a - 1}}{a} = 2\left( {1 - \frac{1}{p}}\right) {\log }_{p}\left( 2\right) . \] | Proof. Clear from the above theorem for \( p \geq 3 \), and immediate for \( p = 2 \) by grouping terms for \( a \) and \( {p}^{N} - a \) . | No |
Theorem 11.5.27. Let \( r \in {\mathbb{Z}}_{ \geq 0} \) be coprime to \( p \), and let \( z \in {\mathbb{C}}_{p} \) be such that \( {v}_{p}\left( {z - 1}\right) > 0 \) .\n\n(1) We have\n\n\[ \mathop{\lim }\limits_{{N \rightarrow \infty }}\mathop{\sum }\limits_{{1 \leq a \leq {p}^{N}}}\frac{{z}^{ra} - 1}{a}\left\lceil \... | Proof. (1). For any positive real number \( u \) it is clear that \( \lceil u\rceil = \mathop{\sum }\limits_{{0 \leq m < u}}1 \) . It follows that\n\n\[ \mathop{\sum }\limits_{{1 \leq a \leq {p}^{N}}}\frac{{z}^{ra}}{a}\left\lceil \frac{ra}{{p}^{N}}\right\rceil = \mathop{\sum }\limits_{{1 \leq a \leq {p}^{N}}}\frac{{z}^... | Yes |
Lemma 11.5.28. Let \( a \) and \( b \) in \( {\mathbb{Z}}_{p} \), and for \( k \in {\mathbb{Z}}_{ \geq 0} \) set\n\n\[ \n{d}_{k} = \left( \begin{matrix} {p}^{N}a + b \\ k \end{matrix}\right) - \left( \begin{array}{l} b \\ k \end{array}\right) .\n\]\n\nThen \( {v}_{p}\left( {d}_{k}\right) \geq N - \log \left( k\right) /... | Proof. Recall the binomial identity\n\n\[ \n\mathop{\sum }\limits_{{0 \leq j \leq k}}\left( \begin{matrix} {p}^{N}a \\ j \end{matrix}\right) \left( \begin{matrix} b \\ k - j \end{matrix}\right) = \left( \begin{matrix} {p}^{N}a + b \\ k \end{matrix}\right) ,\n\]\n\nwhich is proved by expanding the product of the binomia... | Yes |
Lemma 11.5.29. Let \( z \in {\mathbb{C}}_{p} \) be such that \( {v}_{p}\left( {z - 1}\right) > 0 \) . There exists a constant \( B \) depending on \( z \), but not on \( N \), such that \[ \mathop{\sum }\limits_{{0 \leq a < {p}^{N}}}^{\left( p\right) }\frac{{z}^{a}}{a} = O\left( {p}^{N - B}\right) . \] | Proof. As in the computation of \( S\left( M\right) \) made above, we have \[ \mathop{\sum }\limits_{{1 \leq a \leq {p}^{N}}}\frac{{z}^{a}}{a} = \mathop{\sum }\limits_{{1 \leq a \leq {p}^{N}}}\frac{1}{a} + \mathop{\sum }\limits_{{k \geq 1}}\frac{{\left( z - 1\right) }^{k}}{k}\left( \begin{matrix} {p}^{N} \\ k \end{matr... | Yes |
Theorem 11.5.30. Let \( r \in {\mathbb{Z}}_{ \geq 0} \) be such that \( p \nmid r \), let \( z \in {\mathbb{C}}_{p} \) be such that \( {v}_{p}\left( {z - 1}\right) > 0 \), and set\n\n\[ \n{S}_{N}\left( z\right) = \frac{1}{{p}^{N}}\mathop{\sum }\limits_{{0 \leq a < {p}^{N}}}{z}^{a}{\log }_{p}\left( a\right) .\n\]\n\n(1)... | Proof. (1). By the ultrametric property it is enough to show that \( {S}_{N + 1}\left( z\right) - \) \( {S}_{N}\left( z\right) \) tends to 0 . Writing \( a = {p}^{N}q + b \) with \( 0 \leq q < p \) and \( 0 \leq b < {p}^{N} \) we have\n\n\[ \n{S}_{N + 1}\left( z\right) = \frac{1}{{p}^{N + 1}}\mathop{\sum }\limits_{{0 \... | Yes |
Theorem 11.5.31. Let \( m \geq 1 \), and denote as usual by \( {\zeta }_{m} \) a primitive mth root of unity. For \( r \) and \( u \) in \( \mathbb{Z} \) we set\n\n\[ \n{F}_{N}\left( u\right) = \frac{1}{{p}^{N}m}\mathop{\sum }\limits_{{r \leq a < {p}^{N}m + r}}^{\left( p\right) }{\zeta }_{m}^{ua}{\log }_{p}\left( a\rig... | Proof. Since \( r \geq 0 \), setting \( a = {p}^{N}m + b \) and using the fact that \( {\zeta }_{m}^{m} = 1 \) , we have\n\n\[ \n\mathop{\sum }\limits_{{{p}^{N}m \leq a < {p}^{N}m + r}}^{\left( p\right) }{\zeta }_{m}^{ua}{\log }_{p}\left( a\right) = \mathop{\sum }\limits_{{0 \leq b < r}}^{\left( p\right) }{\zeta }_{m}^... | Yes |
Corollary 11.5.32. Keep the same notation and assumptions, in particular that \( 0 \leq r < m \), and assume in addition that \( p \nmid \gcd \left( {r, m}\right) \) . Then\n\n\[ \mathop{\sum }\limits_{{u{\;\operatorname{mod}\;m}}}{\zeta }_{m}^{-{ur}}F\left( u\right) = - \left( {1 + \frac{{\delta }_{m, p}}{p - 1}}\righ... | Proof. First we have\n\n\[ \mathop{\sum }\limits_{{u{\;\operatorname{mod}\;m}}}{\zeta }_{m}^{-{ur}}L\left( u\right) = \mathop{\sum }\limits_{{0 \leq a < r}}\frac{{}^{\left( p\right) }1}{a}\mathop{\sum }\limits_{{u{\;\operatorname{mod}\;m}}}{\zeta }_{m}^{-u\left( {r - a}\right) }.\]\n\nSince \( 1 \leq r - a \leq r < m \... | Yes |
Theorem 11.5.33. Let \( m \in {\mathbb{Z}}_{ \geq 1} \) be such that \( {q}_{p} \mid m \), denote as usual by \( {\zeta }_{m} \) a primitive mth root of unity in \( \overline{\mathbb{Q}} \subset {\mathbb{C}}_{p} \), and let \( r \in \mathbb{Z} \) be such that \( p \nmid r \) and \( 0 \leq r < m \) . We have the explici... | Proof. By Proposition 11.5.7 we have\n\n\[ \n{\psi }_{p}\left( {r/m}\right) = \mathop{\lim }\limits_{{N \rightarrow \infty }}\frac{1}{{p}^{N}}\mathop{\sum }\limits_{{0 \leq a < {p}^{N}}}{\log }_{p}\left( {a + r/m}\right) \n\]\n\n\[ \n= - {\log }_{p}\left( m\right) + \mathop{\lim }\limits_{{N \rightarrow \infty }}\frac{... | Yes |
Theorem 11.5.34. Let \( m \in {\mathbb{Z}}_{ \geq 1} \) be such that \( p \nmid m \), denote as usual by \( {\zeta }_{m} \) a primitive mth root of unity in \( \overline{\mathbb{Q}} \subset {\mathbb{C}}_{p} \), and let \( r \in \mathbb{Z} \) be such that \( 0 \leq r < m \) . We have the explicit formula\n\n\[ \n{\psi }... | Proof. By Proposition 11.5.18 we have\n\n\[ \n{\psi }_{p}\left( {r/m}\right) = \mathop{\lim }\limits_{{N \rightarrow \infty }}\frac{1}{{p}^{N}}\mathop{\sum }\limits_{\substack{{0 \leq a < {p}^{N}} \\ {{v}_{p}\left( {a + r/m}\right) = 0} }}{\log }_{p}\left( {a + r/m}\right) .\n\]\n\nSince \( p \nmid m \), the condition ... | Yes |
Theorem 11.5.35. Let \( \chi \) be a primitive character modulo \( {p}^{v} \) for some \( v \geq 1 \) , let \( m \in {\mathbb{Z}}_{ \geq 1} \) be such that \( p \nmid m \), and let \( r \in \mathbb{Z} \) be such that \( 0 \leq r < m \) . We have the explicit formula\n\n\[{\psi }_{p}\left( {\chi ,\frac{r}{m}}\right) = \... | Proof. Since the proof is very similar, but using simple properties of Gauss sums, we leave it as an excellent but long exercise for the reader (Exercise 28). | No |
Proposition 11.5.36. Let \( \chi \) be a nontrivial primitive character of conductor \( f \) . We have\n\n\[ \n{L}_{p}\left( {\chi ,1}\right) = - \frac{1}{f}\mathop{\sum }\limits_{{0 \leq r < f}}\chi \left( r\right) {\psi }_{p}\left( \frac{r}{f}\right) .\n\] | Proof. Clear from the definition and Propositions 11.5.6 and 11.5.15. Note that, as in the definition of \( {L}_{p}\left( {\chi, s}\right) \), we use here the function \( {\psi }_{p}\left( x\right) \) for \( x \in {\mathbb{{CZ}}}_{p} \) when \( {q}_{p} \mid f \), and for \( x \in {\mathbb{Z}}_{p} \) when \( p \nmid f \... | No |
Theorem 11.5.37. Let \( \chi \) be a nontrivial even primitive character of conductor \( f \), let \( \zeta = {\zeta }_{f} \) be a primitive \( f \) th root of unity, and as usual let \( \tau \left( \chi \right) = \mathop{\sum }\limits_{{1 \leq a < f}}\chi \left( a\right) {\zeta }^{a} \in {\mathbb{C}}_{p} \) be the Gau... | Proof. We separate the cases \( {q}_{p} \mid f \) and \( p \nmid f \) . Assume first that \( {q}_{p} \mid f \) . By the above proposition and Theorem 11.5.33, and since \( \mathop{\sum }\limits_{{0 \leq r < f}}\chi \left( r\right) = 0 \), we have \( {L}_{p}\left( {\chi ,1}\right) = - \left( {{S}_{1} - {S}_{p}/p}\right)... | Yes |
Corollary 11.5.38. Let \( D > 1 \) be a fundamental discriminant, let \( {\varepsilon }_{D} \) be a fundamental unit of \( \mathbb{Q}\left( \sqrt{D}\right) \), and denote by \( {\chi }_{D} \) the Legendre-Kronecker symbol \( {\chi }_{D}\left( n\right) = \left( \frac{D}{n}\right) \). We have\n\n\[ \n{L}_{p}\left( {{\chi... | Proof. By the above theorem and the basic property of the \( p \) -adic logarithm we have\n\n\[ \n{L}_{p}\left( {{\chi }_{D},1}\right) = - \left( {1 - \frac{{\chi }_{D}\left( p\right) }{p}}\right) \frac{\tau \left( {\chi }_{D}\right) }{D}\mathop{\sum }\limits_{{1 \leq r < D}}\left( \frac{D}{r}\right) {\log }_{p}\left( ... | Yes |
Proposition 11.6.2. Let \( u \) be a p-adic unit.\n\n(1) For all \( s \in {\mathbb{Z}}_{p} \) the quantity \( {u}^{\left\lbrack s - s \smallsetminus p\right\rbrack } \) defined by\n\n\[ {u}^{\left\lbrack s - s \smallsetminus p\right\rbrack } = \mathop{\lim }\limits_{{m \rightarrow s}}\mathop{\prod }\limits_{{1 \leq k \... | Proof. The equality of the last two quantities is clear. Let \( {m}_{i} \) be a sequence of elements of \( {\mathbb{Z}}_{ \geq 0} \) tending to \( s \) . For \( i \) sufficiently large we will have \( {m}_{i} \equiv {a}_{0}\left( s\right) \) \( \left( {\;\operatorname{mod}\;p}\right) \), so that \( {m}_{i} \smallsetmin... | Yes |
Corollary 11.6.3. Assume that \( p > 2 \), let \( s = a/\left( {p - 1}\right) \) with \( a \in \mathbb{Z} \), and let \( u \) be a p-adic unit. Then\n\n\[ \n{u}^{\left\lbrack s - 1 - \left( s - 1\right) \smallsetminus p\right\rbrack } = {u}^{a \smallsetminus p}\omega {\left( u\right) }^{-a}\;\text{ and }\;{u}^{\left\lb... | Proof. Since \( p > 2 \), the proposition shows that \( {u}^{\left\lbrack s - 1 - \left( s - 1\right) \smallsetminus p\right\rbrack } = {u}^{{a}_{0}\left( {s - 1}\right) }E \) with\n\n\[ \nE = {\exp }_{p}\left( {\left( {\left( {s - 1 - {a}_{0}\left( {s - 1}\right) }\right) /p}\right) {\log }_{p}\left( {u}^{p - 1}\right... | No |
Proposition 11.6.4. For any \( a \) and \( N \) in \( {\mathbb{Z}}_{ \geq 1} \) and \( m \in \mathbb{Z} \) we have\n\n\[ \mathop{\prod }\limits_{{m \leq k < m + {p}^{N}a}}^{\left( p\right) }k \equiv {\left( -1\right) }^{{p}^{N}a}\left( {\;\operatorname{mod}\;{p}^{N}}\right) ,\]\n\nexcept for \( \left( {p, N}\right) = \... | Proof. Note that the special case \( a = 1, N = 1 \), and \( m = 0 \) is Wilson’s theorem. We prove this proposition in the same way. Assume first that \( a = 1 \) . Let \( G = {\left( \mathbb{Z}/\left( {p}^{N}\mathbb{Z}\right) \right) }^{ * } \) . The integers \( k \) such that \( m \leq k < m + {p}^{N} \) and \( p \n... | Yes |
Proposition 11.6.6. The above definition makes sense (in other words the limit always exists) for all \( s \in {\mathbb{Z}}_{p} \), and \( {\Gamma }_{p}\left( s\right) \) is a p-adic unit. Furthermore, for all \( s \) and \( t \) in \( {\mathbb{Z}}_{p} \) we have \( {\Gamma }_{p}\left( s\right) \equiv {\Gamma }_{p}\lef... | Proof. Set \( {u}_{m} = {\left( -1\right) }^{m}\mathop{\prod }\limits_{{0 \leq k < m}}^{\left( p\right) }k \), and let \( {m}_{i} \) be any sequence of positive integers tending to \( s \) as \( i \rightarrow \infty \) . This means that \( {m}_{i} \) is a Cauchy sequence,\nin other words because of the ultrametric ineq... | Yes |
Lemma 11.6.7. Let \( {n}_{i} \) and \( {m}_{i} \) be two sequences of elements of \( \mathbb{Z} \) such that \( {n}_{i} \leq {m}_{i} \), converging p-adically respectively to \( s \) and \( t \) . Then\n\n\[ \mathop{\lim }\limits_{{i \rightarrow \infty }}{\left( -1\right) }^{{m}_{i} - {n}_{i}}\mathop{\prod }\limits_{{{... | Proof. If we had \( {n}_{i} \geq 0 \) for all \( i \), the result would be immediate from the definition. Thus, we must handle this difficulty. For this, choose a sequence of exponents \( {N}_{i} \) such that \( {N}_{i} \rightarrow \infty \) and \( {n}_{i} + {p}^{{N}_{i}} \geq 0 \) for all \( i \), and for simplicity o... | Yes |
Corollary 11.6.8. (1) The function \( {\Gamma }_{p}\left( s\right) \) is continuous on \( {\mathbb{Z}}_{p} \) ; more precisely, it satisfies \[ \left| {{\Gamma }_{p}\left( s\right) - {\Gamma }_{p}\left( t\right) }\right| \leq \left| {s - t}\right| \] except for \( p = 2 \), for which this inequality is valid only for \... | Proof. (1) is a restatement of the second part of Proposition 11.6.6. | No |
Corollary 11.6.9. If \( n \) and \( m \) are in \( \mathbb{Z} \) with \( n \leq m \) and \( s \in {\mathbb{Z}}_{p} \), we have\n\n\[ \mathop{\prod }\limits_{\substack{{n \leq k < m} \\ {{v}_{p}\left( {s + k}\right) = 0} }}\left( {s + k}\right) = {\left( -1\right) }^{m - n}\frac{{\Gamma }_{p}\left( {s + m}\right) }{{\Ga... | Proof. Clear from (3) of the above corollary. | No |
Lemma 11.6.10. Let \( {n}_{i} \) and \( {m}_{i} \) be two sequences of elements of \( \mathbb{Z} \) such that \( {n}_{i} \leq {m}_{i} \), converging p-adically respectively to \( s \) and \( t \), and let \( a \in {\mathbb{Z}}_{p} \) . Then\n\n\[ \mathop{\lim }\limits_{{i \rightarrow \infty }}{\left( -1\right) }^{{m}_{... | Proof. By the above corollary we have\n\n\[ \mathop{\prod }\limits_{\substack{{{n}_{i} \leq k < {m}_{i}} \\ {{v}_{p}\left( {k + a}\right) = 0} }}\left( {k + a}\right) = {\left( -1\right) }^{{m}_{i} - {n}_{i}}\frac{{\Gamma }_{p}\left( {a + {m}_{i}}\right) }{{\Gamma }_{p}\left( {a + {n}_{i}}\right) },\]\n\nso the result ... | Yes |
Corollary 11.6.11. Let \( f \geq 1 \), set \( q = {p}^{f} \), and for all \( k \in {\mathbb{Z}}_{ \geq 0} \) write \( k = \) \( {qm} + r \) with \( 0 \leq r < q \) . Then\n\n\[ \mathop{\prod }\limits_{{0 \leq i < f}}{\Gamma }_{p}\left( {-\left\lfloor {k/{p}^{i}}\right\rfloor }\right) = {\left( -p\right) }^{m\left( {q -... | Proof. Set \( {k}_{i} = \left\lfloor {k/{p}^{i}}\right\rfloor \) . By Corollary 11.6.8 (4), we have \( {\Gamma }_{p}\left( {-{k}_{i}}\right) = \) \( {\left( -p\right) }^{{k}_{i + 1}}\frac{{k}_{i + 1}!}{{k}_{i}!} \) . The factorials give a telescoping product, so that\n\n\[ \mathop{\prod }\limits_{{0 \leq i < f}}{\Gamma... | Yes |
Proposition 11.6.12. For all \( s \in {\mathbb{Z}}_{p} \) we have\n\n\[ \n{\Gamma }_{p}\left( s\right) {\Gamma }_{p}\left( {1 - s}\right) = {\left( -1\right) }^{\left\lbrack {s - 1 - \left( {s - 1}\right) \smallsetminus p}\right\rbrack + 1} = \left\{ \begin{array}{ll} {\left( -1\right) }^{{a}_{0}\left( {s - 1}\right) +... | Proof. Let \( {m}_{i} \rightarrow s \) as \( i \rightarrow \infty \) with \( {m}_{i} \geq 0 \) . Then \( 1 - {m}_{i} \rightarrow 1 - s \) . It follows from Lemma 11.6.7 that\n\n\[ \n\frac{{\Gamma }_{p}\left( 0\right) }{{\Gamma }_{p}\left( {1 - s}\right) } = \mathop{\lim }\limits_{{i \rightarrow \infty }}{\left( -1\righ... | Yes |
Corollary 11.6.13. (1) For \( p \neq 2 \) we have \( {\Gamma }_{p}{\left( 1/2\right) }^{2} = {\left( -1\right) }^{\left( {p + 1}\right) /2} \), so that \( {\Gamma }_{p}\left( {1/2}\right) = \pm 1 \) when \( p \equiv 3\left( {\;\operatorname{mod}\;4}\right) \), and \( {\Gamma }_{p}\left( {1/2}\right) = \pm i \) when \( ... | Proof. Note that when \( p \) is odd we have \( - 1/2 = \mathop{\sum }\limits_{{j \geq 0}}\left( {\left( {p - 1}\right) /2}\right) {p}^{j} \) . It follows that \( {a}_{0}\left( {-1/2}\right) = \left( {p - 1}\right) /2 \), so that by the above proposition \( {\Gamma }_{p}{\left( 1/2\right) }^{2} = {\left( -1\right) }^{\... | Yes |
Theorem 11.6.14. Let \( p \) be a prime number and \( n \geq 1 \) such that \( p \nmid n \) . Then for all \( s \in {\mathbb{Z}}_{p} \) we have the distribution formula\n\n\[ \mathop{\prod }\limits_{{0 \leq j < N}}{\Gamma }_{p}\left( {s + \frac{j}{N}}\right) = {c}_{p, N}\frac{{\Gamma }_{p}\left( {Ns}\right) }{{N}^{\lef... | Proof. Let \( {m}_{i} \) be a sequence of positive integers tending to \( s \) . By Lemma 11.6.10 applied to \( {n}_{i} = 0,{m}_{i} \), and \( a = j/N \in {\mathbb{Z}}_{p} \), we have\n\n\[ \mathop{\lim }\limits_{{i \rightarrow \infty }}{\left( -1\right) }^{{m}_{i}}\mathop{\prod }\limits_{\substack{{0 \leq k < {m}_{i}}... | Yes |
Proposition 11.6.15. As in Section 4.2.3, let \( {u}_{k} \) be the sequence of rational numbers defined formally by \( \exp \left( {X + {X}^{p}/p}\right) = \mathop{\sum }\limits_{{k \geq 0}}{u}_{k}{X}^{k} \) . Then we have the following convergent expansions for \( x \in {\mathbb{Z}}_{p} \) :\n\n(1)\n\n\[ \mathop{\sum ... | Proof. (1). Let \( f\left( x\right) \) be the function defined as \( f\left( x\right) = {\Gamma }_{p}\left( x\right) \) for \( \left| x\right| < 1 \) and \( f\left( x\right) = 0 \) for \( \left| x\right| = 1 \) . Since by Corollary 11.6.8 the function \( {\Gamma }_{p}\left( x\right) \) is continuous and since the \( p ... | No |
Corollary 11.6.16. (1) We have\n\n\[ \mathop{\sum }\limits_{{k \geq 0}}{k}^{m}k!{u}_{k} = \left\{ \begin{array}{ll} 0 & \text{ if }0 \leq m \leq p - 2, \\ - 1 & \text{ if }m = p - 1. \end{array}\right. \] | Proof. Left to the reader (Exercise 47). | No |
Proposition 11.6.17. Let \( r \) be fixed such that \( 0 \leq r < q \) . We have the Mahler expansion\n\n\[\n\mathop{\prod }\limits_{{0 \leq i < f}}{\Gamma }_{p}\left( {-\left( {\left\lfloor {r/{p}^{i}}\right\rfloor + {p}^{f - i}x}\right) }\right) = {\pi }^{-{s}_{p}\left( r\right) }\mathop{\sum }\limits_{{k \geq 0}}\fr... | Proof. As usual, if we set\n\n\[\ng\left( x\right) = \mathop{\prod }\limits_{{0 \leq i < f}}{\Gamma }_{p}\left( {-\left( {\left\lfloor {r/{p}^{i}}\right\rfloor + {p}^{f - i}x}\right) }\right) ,\n\]\n\nthen \( g \) is continuous on \( {\mathbb{Z}}_{p} \), so by Mahler’s theorem \( g\left( x\right) = \mathop{\sum }\limit... | Yes |
Theorem 11.6.18. (1) For \( p > 2 \) the function \( {\Gamma }_{p}\left( x\right) \) has a power series expansion around \( x = 0 \) with a radius of convergence at least equal to \( {p}^{-\left( {{2p} - 1}\right) /\left( {p\left( {p - 1}\right) }\right) } \) . | Proof. (1) and (2). In the proof of Proposition 11.6.15 we have seen that\n\n\[ {\Gamma }_{p}\left( {-{px}}\right) = \mathop{\sum }\limits_{{k \geq 0}}{a}_{0, k}\left( \begin{array}{l} x \\ k \end{array}\right) \]\n\nwhere by Corollary 4.2.23 we have \( {v}_{p}\left( {a}_{0, k}\right) \geq \left( {1 - 1/p}\right) \left... | Yes |
Proposition 11.6.19. (1) Let \( x \in 2{\mathbb{Z}}_{2} \) . We have with evident notation\n\n\[ \n{\exp }_{2}\left( {{\log }_{2}\left( {{\Gamma }_{2}\left( x\right) }\right) }\right) = {\left( -1\right) }^{x\left( {x - 2}\right) /8}{\Gamma }_{2}\left( x\right) .\n\] | Proof. (1). By the easy Exercise 41, we know that \( {\Gamma }_{2}\left( {2n}\right) \equiv {\left( -1\right) }^{n\left( {n - 1}\right) /2} \) \( \left( {\;\operatorname{mod}\;4}\right) \) . Taking \( n \rightarrow x/2 \in {\mathbb{Z}}_{2} \) implies that \( {\Gamma }_{2}\left( x\right) \equiv {\left( -1\right) }^{x\le... | No |
For \( f \geq 1 \) set\n\n\[ \n{a}_{f, k} = \mathop{\sum }\limits_{{m = 0}}^{k}{\left( -1\right) }^{k - m}\left( \begin{matrix} k \\ m \end{matrix}\right) 1 \cdot 3\cdots \left( {{2}^{f}m - 1}\right) \;\text{ and } \n\]\n\n\[ \n{b}_{f, k} = \mathop{\sum }\limits_{{m = 0}}^{k}{\left( -1\right) }^{k - m + {2}^{f - 1}m}\l... | Proof. Follows immediately from the above corollary, the explicit formula for the Mahler coefficients (Theorem 4.2.26), and the values on \( \mathbb{Z} \) of the function \( {\Gamma }_{2}\left( s\right) \) (Corollary 11.6.8). | No |
Proposition 11.6.21. Set\n\n\[ \n{D}_{p}\left( {x, y}\right) = \frac{{\log }_{p}\left( {{\Gamma }_{p}\left( {x + y}\right) }\right) - {\log }_{p}\left( {{\Gamma }_{p}\left( x\right) }\right) - {\log }_{p}\left( {{\Gamma }_{p}\left( y\right) }\right) }{{xy}\left( {x + y}\right) }.\n\]\n\nIf \( x \) and \( y \) are in \(... | Proof. By Proposition 11.5.19 we have\n\n\[ \n{D}_{p}\left( {x, y}\right) = - \mathop{\sum }\limits_{{k \geq 1}}\frac{{d}_{k, p}}{{p}^{3}}\frac{{\left( x + y\right) }^{{2k} + 1} - {x}^{{2k} + 1} - {y}^{{2k} + 1}}{{p}^{{2k} - 2}{xy}\left( {x + y}\right) }\n\]\n\nwith \( {d}_{k, p} \) as above. Since the numerator of\n\n... | Yes |
Proposition 11.7.1. For all \( r \) and \( N \) in \( {\mathbb{Z}}_{ \geq 0} \) we have\n\n\[ \left( {1 - q}\right) \mathop{\sum }\limits_{{0 \leq k < N}}{d}_{\left( {q - 1}\right) k + r} = {G}_{r}\left( \frac{r}{1 - q}\right) - {G}_{\left( {q - 1}\right) N + r}\left( {\frac{r}{1 - q} - N}\right) . \] | Proof. We first prove the proposition for \( N = 1 \) . For this, we transform the term \( {G}_{q - 1 + r}\left( {x - 1}\right) \) (where we will set \( x = r/\left( {1 - q}\right) \) later) as follows. First note that by differentiating the defining formula \( \exp \left( {\pi \left( {X - {X}^{q}}\right) }\right) = \m... | Yes |
Corollary 11.7.2. For \( r \in {\mathbb{Z}}_{ \geq 0} \) we have\n\n\[ \left( {1 - q}\right) \mathop{\sum }\limits_{{k \geq 0}}{d}_{\left( {q - 1}\right) k + r} = {G}_{r}\left( \frac{r}{1 - q}\right) . | Proof. Note that since by Proposition 4.4.40 the valuation of \( {d}_{k} \) tends to infinity with \( k \), we already know that the series on the left-hand side converges. In any case, thanks to the proposition, to prove that it converges and that its sum is as given is equivalent to showing that \( {G}_{\left( {q - 1... | Yes |
Proposition 11.7.4. For any \( r \in \mathbb{Z} \) such that \( 0 \leq r < q - 1 \) we have\n\n\[ \n{\tau }_{q}\left( r\right) = \left( {q - 1}\right) \mathop{\sum }\limits_{{m \geq 0}}{d}_{\left( {q - 1}\right) m + r} = - {G}_{r}\left( \frac{r}{1 - q}\right) .\n\] | Proof. Indeed\n\n\[ \n{\tau }_{q}\left( r\right) = \mathop{\sum }\limits_{{{a}^{q - 1} = 1}}{a}^{-r}{D}_{\pi, f}\left( a\right) = \mathop{\sum }\limits_{{{a}^{q - 1} = 1}}{a}^{-r}\mathop{\sum }\limits_{{k \geq 0}}{d}_{k}{a}^{k}\n\]\n\n\[ \n= \mathop{\sum }\limits_{{k \geq 0}}{d}_{k}\mathop{\sum }\limits_{{{a}^{q - 1} =... | Yes |
Theorem 11.7.5 (Gross-Koblitz). If \( 0 \leq r < q - 1 \) then\n\n\[ \n{\tau }_{q}\left( r\right) = - {\pi }^{{s}_{p}\left( r\right) }\mathop{\prod }\limits_{{0 \leq i < f}}{\Gamma }_{p}\left( \frac{{r}^{\left( i\right) }}{q - 1}\right) ,\n\]\n\nwhere \( 0 \leq {r}^{\left( i\right) } < q - 1 \) have base-p expansions o... | Proof. By definition of \( {G}_{r}\left( x\right) \), Proposition 11.6.17 tells us that\n\n\[ \n\mathop{\prod }\limits_{{0 \leq i < f}}{\Gamma }_{p}\left( {-\left( {\left\lfloor {r/{p}^{i}}\right\rfloor + {p}^{f - i}x}\right) }\right) = {\pi }^{-{s}_{p}\left( r\right) }{G}_{r}\left( x\right) .\n\]\n\nSetting \( x = r/\... | Yes |
Corollary 11.7.6. Recall that we define \( s\left( r\right) = {s}_{p}\left( {r{\;\operatorname{mod}\;\left( {q - 1}\right) }}\right) \) and that \( \{ x\} \) denotes the fractional part of \( x \) . For all \( r \in \mathbb{Z} \) we have\n\n\[{\tau }_{q}\left( r\right) = - {\pi }^{s\left( r\right) }\mathop{\prod }\limi... | Proof. Since both sides are now periodic in \( r \) of period \( q - 1 \), it is sufficient to prove the result when \( 0 \leq r < q - 1 \) . In that case we have \( {p}^{f - i}r/\left( {q - 1}\right) = \) \( \left( {r + r/\left( {q - 1}\right) }\right) /{p}^{i} \) and since \( 0 \leq r/\left( {q - 1}\right) < 1 \) we ... | Yes |
Corollary 11.7.7. Let \( m/n \in \mathbb{Q} \) be such that \( n \mid \left( {p - 1}\right) \) .\n\n(1) If \( 0 \leq m/n < 1 \) we have the formula\n\n\[ \n{\Gamma }_{p}\left( {m/n}\right) = - {\pi }^{-r}{\tau }_{p}\left( r\right) \n\] \n\nwhere \( r = m\left( {p - 1}\right) /n \) .\n\n(2) \( {\Gamma }_{p}\left( {m/n}\... | Proof. (1) is evidently the special case \( f = 1 \) of the theorem. Since \( \pi = \) \( {\left( -p\right) }^{1/\left( {p - 1}\right) } \) we have \( {\pi }^{-r} = {\left( -p\right) }^{-m/n} \in \mathbb{Q}\left( {\left( -p\right) }^{1/n}\right) \) . Furthermore, \( {\zeta }_{\pi } \) is a \( p \) th root of unity; in ... | Yes |
Proposition 11.7.9. We have\n\n\[ \n{\Gamma }_{5}\left( {1/4}\right) = \sqrt{-2 + i} \]\n\nwhere \( i \) is the square root of -1 congruent to 3 modulo 5, and the outer square root is chosen congruent to 1 modulo 5 . | Proof. By the theorem and the above proposition, we have\n\n\[ \n{\left( {\Gamma }_{5}\left( 1/4\right) \right) }^{2} = {\left( -{\pi }^{-1}{\tau }_{p}\left( 1\right) \right) }^{2} = {\pi }^{-2}{\tau }_{p}\left( 2\right) \mathop{\sum }\limits_{\substack{{{a}^{p - 1} = 1} \\ {a \neq 1} }}{a}^{-1}\omega {\left( 1 - a\rig... | Yes |
Proposition 11.7.10 (Ferrero-Greenberg). Let \( \chi \) be a primitive even character, let \( {\chi }_{1} = \chi {\omega }^{-1} \), and denote by \( {f}_{1} \) the conductor of \( {\chi }_{1} \) . Assume that \( {\chi }_{1}\left( p\right) = 1 \), so that in particular \( p \nmid {f}_{1} \), let \( u \) be the order of ... | Proof. Since \( {\chi }_{1}\left( p\right) = 1 \), by Propositions 11.3.9 and 11.5.24 we have \( {L}_{p}\left( {\chi ,0}\right) = 0 \) and\n\n\[ \n{L}_{p}^{\prime }\left( {\chi ,0}\right) = \mathop{\sum }\limits_{{0 \leq k < {f}_{1}}}{\chi }_{1}\left( k\right) {\operatorname{log\Gamma }}_{p}\left( {{\chi }_{0},\frac{k}... | Yes |
Proposition 11.7.11. If \( r \in \mathbb{Z} \) is such that \( 0 \leq r < q - 1 \) then\n\n\[{\tau }_{q}\left( r\right) \equiv - {\pi }^{{s}_{p}\left( r\right) }\frac{{\left( -p\right) }^{{v}_{p}\left( {r!}\right) }}{r!} \equiv - \frac{{\pi }^{r}}{r!}\left( {{\;\operatorname{mod}\;{\mathfrak{p}}^{{s}_{p}\left( r\right)... | Proof. We have seen that\n\n\[{r}^{\left( i\right) } = \left( {1 - q}\right) \left\lfloor \frac{r}{{p}^{i}}\right\rfloor + {p}^{f - i}r \equiv \left\lfloor \frac{r}{{p}^{i}}\right\rfloor \left( {\;\operatorname{mod}\;p}\right)\n\nsince \( 0 \leq i < f \) and \( p \mid q \) . Since \( q - 1 \equiv - 1\left( {\;\operator... | Yes |
Lemma 11.7.12. If \( \pi \) is the unique element of \( {\mathbb{Q}}_{p}\left( {\zeta }_{p}\right) \) such that \( {\pi }^{p - 1} = - p \) and \( \pi /\left( {{\zeta }_{p} - 1}\right) \equiv 1\left( {{\;\operatorname{mod}\;\mathfrak{p}}{\mathbb{Z}}_{\mathcal{K}}}\right) \), then with the notation of Section 3.6.2 we ha... | Proof. By definition \( {\omega }_{\mathfrak{P}}\left( x\right) \) is the unique \( \left( {q - 1}\right) \) st root of unity congruent to \( x \) modulo \( \mathfrak{P} \) . Furthermore, \( {\operatorname{Tr}}_{\left( {{\mathbb{Z}}_{L}/\mathfrak{P}}\right) /\left( {{\mathbb{Z}}_{K}/\mathfrak{p}}\right) }\left( x\right... | Yes |
Theorem 11.7.13. With the above notation we have\n\n\[ \frac{\tau \left( {{\omega }_{\mathfrak{P}}^{-r},{\psi }_{1}}\right) }{{\left( {\zeta }_{p} - 1\right) }^{s\left( r\right) }} \equiv - \frac{{\left( -p\right) }^{{v}_{p}\left( {r!}\right) }}{r!}\left( {\;\operatorname{mod}\;\mathfrak{P}}\right) . \] | Proof. As usual by periodicity we may assume that \( 0 \leq r < q - 1 \) . By the above lemma and Proposition 11.7.11 we have (in the \( \mathfrak{p} \) -adic context)\n\n\[ \frac{\tau \left( {{\omega }_{\mathfrak{P}}^{-r},{\psi }_{1}}\right) }{{\left( {\zeta }_{p} - 1\right) }^{{s}_{p}\left( r\right) }} = \frac{{\tau ... | Yes |
Lemma 11.7.14. As usual let \( q = {p}^{f}, m \mid q - 1 \), and \( d = \left( {q - 1}\right) /m \) . For integers \( r \) and \( i \) such that \( 0 \leq r < q - 1 \) and \( 0 \leq i < f \) set\n\n\[ e\left( {r, i}\right) = \left\{ \frac{{p}^{f - i}r}{q - 1}\right\} = \frac{{r}^{\left( i\right) }}{q - 1}, \]\n\nand fo... | Proof. Denote the left-hand side by \( P\left( b\right) \) . It is clear that \( P\left( b\right) \) is periodic in \( b \) of period dividing \( d \), hence we may assume that \( 0 \leq b < d \) . Writing \( {p}^{f - i}\left( {{ad} + b}\right) /\left( {q - 1}\right) = {p}^{f - i}a/m + x \) with \( x = {p}^{f - i}b/\le... | Yes |
Corollary 11.7.15. (1) With the same notation, for all \( b \in \mathbb{Z} \) we have\n\n\[ \mathop{\prod }\limits_{{0 \leq a < m}}{\tau }_{q}\left( {{ad} + b}\right) = - \omega {\left( m\right) }^{s\left( {mb}\right) }{\tau }_{q}\left( {mb}\right) \mathop{\prod }\limits_{{0 \leq a < m}}{\tau }_{q}\left( {ad}\right) ,\... | Proof. (1). Either directly or by Lemma 3.6.7 it is clear that we have \( \mathop{\sum }\limits_{{0 \leq i < f}}e\left( {{mb}, i}\right) = s\left( {mb}\right) /\left( {p - 1}\right) \) . Thus, if we denote by LHS the left hand side of (1), then by the above lemma and the Gross-Koblitz formula (more precisely Corollary ... | Yes |
Theorem 11.7.16. Let \( \rho \) be a multiplicative character of exact order \( m \) dividing \( \left( {q - 1}\right) \), and let \( \psi \) be any nontrivial additive character on \( {\mathbb{F}}_{q} \) . (1) For any multiplicative character \( \chi \) on \( {\mathbb{F}}_{q} \) we have \[ \mathop{\prod }\limits_{{0 \... | Proof. (1) clearly follows from the above corollary, Lemma 11.7.12, and the fact that \( {\omega }_{\mathfrak{P}}\left( m\right) \) can be identified with the Teichmüller character \( \omega \left( m\right) \) in \( {\mathbb{Q}}_{p} \), since they are both \( \left( {p - 1}\right) \) st roots of unity congruent to \( m... | Yes |
Corollary 12.2.2. The only integer solutions to the Diophantine equation\n\n\\[ \n{2}^{m} - {3}^{n} = 5 \n\\]\n\nare \\( \\left( {m, n}\\right) = \\left( {3,1}\\right) \\) and \\( \\left( {m, n}\\right) = \\left( {5,3}\\right) \\), which correspond respectively to \\( 8 - \\) \\( 5 = 3 \\) and to \\( {32} - {27} = 5 \\... | Proof. Applying the above theorem to this equation we get\n\n\\[ \n5 > {2}^{m} \\cdot {\\left( em\\right) }^{-{5.87} \\times {10}^{8}}, \n\\]\n\nwhich implies\n\n\\[ \n\\log 5 > m\\log 2 - {5.87} \\times {10}^{8}\\left( {1 + \\log m}\\right) \n\\]\n\nso that \\( m < {2.1} \\times {10}^{10} \\) and \\( n < m\\log 2/\\lo... | Yes |
Lemma 12.2.4. Let \( B \) be a nonnegative integer such that\n\n\[ \alpha \log B + \beta \geq {\gamma B} \]\n\nIf \( \alpha \geq {e\gamma } \), we have\n\n\[ B \leq \frac{2}{\gamma }\left( {\alpha \log \frac{\alpha }{\gamma } + \beta }\right) \] | Proof. Exercise for the reader. | No |
Lemma 12.8.2. Let \( \varepsilon > 1,\eta > 1,\gamma \), and \( \delta \) be real numbers. Under the assumptions\n\n\[ \left| {\gamma {\varepsilon }^{n} - \delta {\eta }^{m}}\right| \leq C \cdot \max \left( {{\varepsilon }^{-n},{\eta }^{-m}}\right) ,\;n \neq 0,\;\text{ and } \]\n\n\[ m > {M}_{0} \mathrel{\text{:=}} \fr... | Proof. Since \( n, m > 0 \), we have \( \left| {\gamma {\varepsilon }^{n} - \delta {\eta }^{m}}\right| \leq C \), hence \( \left| {\varepsilon }^{n}\right| \leq C + \) \( \left| {\delta {\gamma }^{-1}{\eta }^{m}}\right| \leq 2\left| {\delta {\gamma }^{-1}{\eta }^{m}}\right| \) for \( m > \log \left( {C\left| {\gamma {\... | Yes |
Lemma 12.8.3 (Baker-Davenport). Let \( {x}_{0},{x}_{1},{x}_{2} \) be real numbers, \( {b}_{1} \) an integer with \( \left| {b}_{1}\right| \leq B \), and \( Q \) a nonnegative integer. Then\n\n\[ d\left( {{x}_{0} + {b}_{1}{x}_{1},\mathbb{Z}}\right) \geq \frac{1}{Q}\left\{ {d\left( {Q{x}_{0},\mathbb{Z}}\right) - {Bd}\lef... | Proof. This follows from the chain of inequalities\n\n\[ {Qd}\left( {{x}_{0} + {b}_{1}{x}_{1},\mathbb{Z}}\right) \geq d\left( {Q{x}_{0} + Q{b}_{1}{x}_{1},\mathbb{Z}}\right) \]\n\n\[ \geq d\left( {Q{x}_{0},\mathbb{Z}}\right) - d\left( {Q{b}_{1}{x}_{1},\mathbb{Z}}\right) \]\n\n\[ \geq d\left( {Q{x}_{0},\mathbb{Z}}\right)... | Yes |
Theorem 12.9.1 (Tijdeman). Let \( x, y, m \geq 2 \), and \( n \geq 2 \) be strictly positive integers such that \( {x}^{m} - {y}^{n} = 1 \) . There exists an effectively computable, absolute constant \( C \) such that \( \max \{ x, y, m, n\} < C \) . | Proof. By the results of Lebesgue and Ko Chao (Proposition 6.7.12 and Theorem 6.11.8) we may assume that \( m \) and \( n \) are odd. We consider the equation\n\n\[ \n{x}^{m} - {y}^{n} = \varepsilon \n\]\n\nwhere \( \varepsilon = \pm 1 \) and \( x, y, m, n \) are strictly positive integers with \( n > m > 2 \) .\n\nSin... | Yes |
Theorem 12.10.1. With the above notation, the equation\n\n\\[ \na\\left( {x - {\\alpha }_{1}y}\\right) \\cdots \\left( {x - {\\alpha }_{n}y}\\right) = m\n\\]\n\nhas only finitely many solutions in integers \\( x \\) and \\( y \\), and all of these can be effectively determined. Moreover, the bound for \\( \\max \\{ \\l... | Proof. Let \\( x \\) and \\( y \\) be integers satisfying the above equation. For the sake of simplicity assume that \\( a = 1 \\) . Without loss of generality, we may assume that \\( x \\) and \\( y \\) are very large, and that \\( x/y \\) is very close to \\( {\\alpha }_{1} \\) . More precisely, setting \\( X \\mathr... | No |
Lemma 12.10.2. Let \( \left( {x, y}\right) \) be a solution of the Thue equation of Theorem 12.10.1, and assume that \( \left( {x - {\alpha }_{1}y}\right) /{\gamma }_{1} \) is a unit of \( \mathbb{Q}\left( {\alpha }_{1}\right) \) and \( \mid x - \) \( \left. {{\alpha }_{1}y}\right| \; = \mathop{\min }\limits_{{1 \leq i... | Proof. For \( j \neq 1 \) we have\n\n\[ \n2\left| {x - {\alpha }_{j}y}\right| \geq \left| {x - {\alpha }_{j}y}\right| + \left| {x - {\alpha }_{1}y}\right| \geq \left| y\right| \left| {{\alpha }_{1} - {\alpha }_{j}}\right| .\n\]\n\n(10.5)\n\nThus,\n\n\[ \n\left| {x - {\alpha }_{1}y}\right| = \frac{\left| m\right| }{\mat... | Yes |
Theorem 13.2.13 (Weil). If \( K \) is a number field, \( {J}_{K}\left( \mathcal{C}\right) \) is a finitely generated abelian group. | In the following we will assume that \( K \) is a number field. The structure of \( {J}_{K}\left( \mathcal{C}\right) \) can be computed analogously to the computation of the Mordell-Weil group for elliptic curves that we have studied in Chapter 8, in other words by using descent methods. It is evidently more difficult,... | No |
Theorem 13.2.15 (Mumford’s Representation). Let \( \mathcal{C} \) be a hyperelliptic curve of genus \( g \) defined over \( K \) as in (13.2). Any semireduced \( K \) -rational divisor \( D \) can be represented by a pair of polynomials \( a \) and \( b \) in \( K\left\lbrack x\right\rbrack \) . If \( D = {P}_{1} + \cd... | This means that for each point \( {P}_{i} = \left( {{x}_{i},{y}_{i}}\right) ,{x}_{i} \) is a root of \( a \) with multiplicity \( {\operatorname{ord}}_{{P}_{i}}\left( D\right) \) . The last conditions ensure that \( b\left( {x}_{i}\right) = {y}_{i} \) with the correct multiplicity. In this representation, the neutral e... | Yes |
Corollary 13.3.2. Let \( \mathcal{C} \) be a nonsingular projective curve defined over \( \mathbb{Q} \) . Let \( E \) be an elliptic curve defined over \( \mathbb{Q} \) such that there exist \( l \) independent morphisms from \( \mathcal{C} \) to \( E \) defined over \( \mathbb{Q} \) . If \( l > \operatorname{rk}\left(... | This method is used in [Dem2], [Sil4], and [Kul] to solve some examples or families of examples of curves. We now describe in detail how this corollary can be made explicit in the first nontrivial case, namely when the elliptic curve \( E \) has rank 1 and \( l = 2 \) . Thus, assume that \( E \) has rank 1 over \( \mat... | Yes |
Let \( \mathcal{C} \) be the Fermat quartic curve defined over \( \mathbb{Q} \) by\n\n\[ \mathcal{C} : \;{x}^{4} + {y}^{4} = 2 \]\n\nand let \( E \) be the elliptic curve defined over \( \mathbb{Q} \) by\n\n\[ E : \;{y}^{2} = {x}^{3} - {2x}. \]\n\nThe point \( T = \left( {0,0}\right) \) is the only nontrivial torsion p... | We deduce that if \( {n}_{1} \neq \pm {n}_{2} \) then\n\n\[ \max \left( {\left| {n}_{1}\right| ,\left| {n}_{2}\right| }\right) \leq \frac{1}{2}\left( {\frac{{B}_{1} + {B}_{2} + {B}_{3}}{\widehat{h}\left( G\right) } + 1}\right) \leq {8.42}. \]\n\nIf \( {n}_{1} = \pm {n}_{2} \) we have \( {\phi }_{1}\left( P\right) \pm {... | Yes |
Theorem 13.3.4 (Chabauty). Let \( \mathcal{C} \) be an algebraic curve defined over a number field \( K \), and let \( {J}_{K}\left( \mathcal{C}\right) \) be the group of \( K \) -rational points on the Jacobian of \( \mathcal{C} \) . If the rank of \( {J}_{K}\left( \mathcal{C}\right) \) is strictly less than the genus... | To prove this, Chabauty used \( p \) -adic integration, and Coleman noticed in 1985 [Col] that it is possible to deduce from Chabauty's proof a bound on the number of \( K \) -rational points on the curve. For this, he obtained an upper bound on the number of zeros of an integral of the first kind on \( \mathcal{C}\lef... | Yes |
Let \( \mathcal{C} \) be the curve of genus 2 defined over \( \mathbb{Q} \) by the equation\n\n\[ \n{y}^{2} = x\left( {x - 1}\right) \left( {x - 2}\right) \left( {x - 5}\right) \left( {x - 6}\right) .\n\]\n\nIt can be shown that the Jacobian of \( \mathcal{C} \) has rank 1, so that Chabauty’s condition is satisfied. Th... | We check that \( \left| {\mathcal{C}\left( {\mathbb{F}}_{7}\right) }\right| = 8 \), so that there are at most 10 points on \( \mathcal{C}\left( \mathbb{Q}\right) \) . Since it is easy to find 10 points on \( \mathcal{C}\left( \mathbb{Q}\right) \) this bound is sharp, so we have shown that\n\n\[ \n\mathcal{C}\left( \mat... | Yes |
In his PhD dissertation [Wet], Wetherell solves, thanks to this kind of technique, what seems to be the only curve considered by Diophantus that has genus strictly greater than 1 (Problem 17 of Book VI of the Arabic manuscript of Arithmetica [Ses]). This curve of genus 2 is given by the equation\n\n\\[ \n\\mathcal{C} :... | The Jacobians of these genus-3 curves have rank 1 and 0 , so that the rational points on these curves can be found using Chabauty’s methods. Finally, one finds that the finite rational points on \\( \\mathcal{C} \\) are \\( \\left( {0, \\pm 1}\\right) \\) and \\( \\left( {\\pm 1/2, \\pm 9/8}\\right) \\) . | Yes |
In [Fly-Wet2], Flynn and Wetherell use another method, whose principle has been used for several other Diophantine equations, to solve the Diophantine equation\n\n\\[ \n{x}^{4} + {y}^{4} = {17} \n\\]\n\nproposed by Serre. | They use the map introduced for 2-descent on the Jacobian that generalizes the fundamental 2-descent map for elliptic curves studied in Sections 8.2 and 8.3. Thanks to this map, they obtain a covering collection for the curve defined over \\( \\mathbb{Q} \\) by the equation\n\n\\[ \n{y}^{2} = \\left( {9{x}^{2} - {28x} ... | No |
Proposition 14.2.1. (1) The equation \( {x}^{2} + 3{y}^{2} = {z}^{3} \) in nonzero integers \( x \) , \( y \), and \( z \) with \( x \) and \( y \) coprime can be parametrized by\n\n\[ \left( {x, y, z}\right) = \left( {s\left( {s - {3t}}\right) \left( {s + {3t}}\right) ,{3t}\left( {s - t}\right) \left( {s + t}\right) ,... | Proof. For (1) we set \( {x}_{1} = x + {3y} \) and the equation becomes \( {x}_{1}^{2} - 3{x}_{1}\left( {2y}\right) + \) \( 3{\left( 2y\right) }^{2} = {z}^{3} \) . Thanks to Proposition 6.4.16 of Chapter 6 we know that this equation has three disjoint parametrizations. Among these, only the first gives an even value fo... | Yes |
Theorem 14.5.1. Let \( G \) be the vertices of a regular tetrahedron, octahedron, or icosahedron inscribed in the Riemann sphere, let \( N \) be the north pole of the sphere, and for \( g \in G \) let \( \left( {{\alpha }_{g} : {\beta }_{g}}\right) \in {\mathbb{P}}_{1}\left( \mathbb{C}\right) \) be the point obtained b... | Although for the moment the polynomials are with coefficients in \( \mathbb{C} \), this is exactly what we need for solving the \( \left( {2,3, r}\right) \) equation in the elliptic case. To make this clearer, we look at all three cases. Consider first the regular tetrahedron. Up to rescaling and rotation we can choose... | No |
Proposition 14.6.1. In the parabolic case \( 1/p + 1/q + 1/r = 1 \), the equation \( {x}^{p} + {y}^{q} = {z}^{r} \) has no solutions in nonzero coprime integers, except that the equation \( {x}^{3} + {y}^{6} = {z}^{2} \) has the solutions \( \left( {x, y, z}\right) = \left( {2, \pm 1, \pm 3}\right) \), and the equation... | Proof. The \( \left( {3,3,3}\right) \) case is FLT for exponent 3, which has been proved in Section 6.9, the case of general coefficients having been studied in Sections \( {6.4.2},{6.4.3},{6.4.4} \), and 6.4.5. The \( \left( {2,4,4}\right) \) case corresponds to the equations \( {x}^{4} \pm {y}^{4} = {z}^{2} \) which ... | Yes |
Theorem 14.6.2. For fixed \( p, q \), and \( r \) such that \( 1/p + 1/q + 1/r < 1 \) and fixed nonzero integers \( A, B \), and \( C \), there exist only finitely many solutions to the equation \( A{x}^{p} + B{y}^{q} + C{z}^{r} = 0 \) in integers \( x, y \), and \( z \) with \( x \) and \( y \) coprime. | To prove this theorem, Darmon and Granville succeed in reducing it to Faltings's famous theorem on the finiteness of the number of rational points on a curve of genus greater than or equal to 2 (Mordell's conjecture), which is not a trivial task since \( A{x}^{p} + B{y}^{q} + C{z}^{r} = 0 \) does not a priori represent... | No |
The abc conjecture implies that the total number of nonzero coprime solutions to \( {x}^{p} \pm {y}^{q} \pm {z}^{r} = 0 \) with \( 1/p + 1/q + 1/r < 1 \) is finite, even allowing \( p, q \), and \( r \) to vary. Here, if \( x = \pm 1 \) (respectively \( y = \pm 1 \), respectively \( z = \pm 1 \) ), we identify solution... | Order \( p, q \), and \( r \) such that \( p \leq q \leq r \) . Then the hyperbolic cases correspond to the triples \( \left( {2,3, r}\right) \) for \( r \geq 7,\left( {2,4, r}\right) \) for \( r \geq 5,\left( {2, q, r}\right) \) for \( r \geq q \geq 5,\left( {3,3, r}\right) \) for \( r \geq 4,\left( {3, q, r}\right) \... | Yes |
Proposition 14.6.7. The equation \( {x}^{4} + {y}^{4} = {z}^{5} \) has no solution in nonzero coprime integers \( x, y, z \) . | Proof. Once again we use the solution to the dihedral equation. Our equation is thus equivalent to \( {x}^{2} = s\left( {{s}^{4} - {10}{t}^{2}{s}^{2} + 5{t}^{4}}\right) ,{y}^{2} = t\left( {5{s}^{4} - {10}{t}^{2}{s}^{2} + {t}^{4}}\right) \) , and \( z = {s}^{2} + {t}^{2} \), where \( s \) and \( t \) are coprime integer... | Yes |
Proposition 14.6.8. The equation \( {x}^{6} - {y}^{4} = {z}^{2} \) has no solution in nonzero coprime integers \( x, y, z \) . | Proof. Once again, we use the solution to the dihedral equation \( {X}^{2} + {Y}^{2} = \) \( {Z}^{3} \) . We deduce that \( {x}^{6} - {y}^{4} = {z}^{2} \) is equivalent to \( {x}^{2} = {s}^{2} + {t}^{2},{y}^{2} = s\left( {{s}^{2} - 3{t}^{2}}\right) \) , \( z = t\left( {3{s}^{2} - {t}^{2}}\right) \), where \( s \) and \... | Yes |
Proposition 14.6.9. The equation \( {x}^{4} - {y}^{6} = {z}^{2} \) has no solution in nonzero coprime integers \( x, y, z \) . | Proof. Once again, we use the solution to the dihedral equation. We find that our equation is equivalent to \( {x}^{2} = s\left( {{s}^{2} + 3{t}^{2}}\right), z = t\left( {3{s}^{2} + {t}^{2}}\right) ,{y}^{2} = {s}^{2} - {t}^{2} \) with \( s \) and \( t \) coprime of opposite parity, or to \( {x}^{2} = \pm \left( {2{s}^{... | Yes |
Proposition 14.6.10. The equation \( {x}^{6} + {y}^{4} = {z}^{2} \) has no solution in nonzero coprime integers \( x, y, z \) . | Proof. Once again, we use the solution to the dihedral equation. We find that our equation is equivalent to \( z = s\left( {{s}^{2} + 3{t}^{2}}\right) ,{y}^{2} = t\left( {3{s}^{2} + {t}^{2}}\right) ,{x}^{2} = {s}^{2} - {t}^{2} \) with \( s \) and \( t \) coprime of opposite parity, or to \( z = \pm \left( {2{s}^{3} + {... | Yes |
Corollary 14.6.11. The equations \( \pm {x}^{6} \pm {y}^{4} = {z}^{2} \) have no solutions in nonzero coprime integers \( x, y, z \) . | none | No |
Proposition 14.7.2 (Mason). Let \( A, B, C \) be pairwise coprime polynomials in one variable, not all constant and such that \( A + B + C = 0 \) . Then\n\n\[ \max \left( {\deg \left( A\right) ,\deg \left( B\right) ,\deg \left( C\right) }\right) \leq \deg \left( {\operatorname{rad}\left( {ABC}\right) }\right) - 1. \] | Proof. Let \( f = A/C \) and \( g = B/C \), so that \( f \) and \( g \) are rational functions such that \( f + g + 1 = 0 \) . Note that \( g \) is not constant; otherwise, \( f \) would also be constant and \( A, B \), and \( C \) would be proportional hence constant since they are pairwise coprime. Differentiating, i... | Yes |
Corollary 14.7.3. FLT is true for polynomials in one variable that are not all constant, in other words if \( f, g \), and \( h \) are nonzero polynomials, not all constant and such that \( {f}^{n} + {g}^{n} = {h}^{n} \) then \( n \leq 2 \) . | Proof. Dividing the equation by \( \gcd {\left( f, g\right) }^{n} \) we may assume that \( f, g \) , and \( h \) are pairwise coprime. Setting \( A = {f}^{n}, B = {g}^{n}, C = - {h}^{n} \) we have \( A + B + C = 0 \) and \( \operatorname{rad}\left( {ABC}\right) \mid {fgh} \) . Thus by the above proposition we have\n\n\... | Yes |
Corollary 14.7.4. Let \( p, q, r \) be integers such that \( 2 \leq p \leq q \leq r \), and assume that \( f, g \), and \( h \) are pairwise coprime polynomials, not all constant and satisfying the super-Fermat equation \( {f}^{p} + {g}^{q} = {h}^{r} \). We are then in the elliptic case, in other words \( \left( {p, q,... | Proof. Once again we have \( A + B + C = 0 \) with \( A = {f}^{p}, B = {g}^{q} \), and \( C = - {h}^{r} \), which are pairwise coprime by assumption, and \( \operatorname{rad}\left( {ABC}\right) \mid {fgh} \). If we denote by \( a, b \), and \( c \) respectively the degrees of \( f, g \), and \( h \), the above proposi... | Yes |
Corollary 15.1.2. There are no newforms for levels\n\n\\[ \n1,2,3,4,5,6,7,8,9,{10},{12},{13},{16},{18},{22},{25},{28},{60}.\n\\]\n\nFor all other levels there are newforms. | Proof. Follows from an immediate (computer-aided) computation from the proposition. | No |
Theorem 15.2.8. Assume that \( N \) is squarefree. If \( {\Delta }_{\min } \) is a pth power for some prime \( p \) then \( p \leq 5 \) and \( E \) has a rational point of order \( p \) . | Proof. We prove only the statement \( p \leq 7 \) using the tools that we have introduced. Since \( N \) is squarefree and \( {\Delta }_{\min } \) is a \( p \) th power, the definition shows that \( {N}_{p} = 1 \) . Assume first that \( p \geq {11} \) . Since \( N \) is squarefree, the second condition of Mazur’s theor... | No |
An elliptic curve defined over \( \mathbb{Q} \) can arise from a new-form whose field of definition \( K \) has arbitrarily large degree. | Let \( p \geq 5 \) be a prime, set \( L = {2}^{p + 4} + 1 \), and let \( E \) be the elliptic curve with equation\n\n\[ \n{Y}^{2} = X\left( {X + 1}\right) \left( {X - {2}^{p + 4}}\right) .\n\]\n\nUsing Tate's algorithm we easily compute that the minimal discriminant and conductor are given by\n\n\[ \n{\Delta }_{\min } ... | Yes |
Theorem 15.3.2. Let \( E \) and \( F \) be two elliptic curves defined over \( \mathbb{Q} \). Assume that \( F \) has complex multiplication by some order in an imaginary quadratic field \( L \), and that \( p \) is a prime number such that \( E{ \sim }_{p}F \). (1) If \( p = {11} \) or \( p \geq {17} \) and \( p \) sp... | Remarks. (1) Part (1) was proved by Halberstadt and Kraus in [Hal-Kra1] as a consequence of work of Momose [Mom], and part (2) was proved by Darmon and Merel in [Dar-Mer]. | No |
Proposition 15.3.3. With the assumptions of Theorem 15.3.2 (2), assume in addition that \( {p}^{2} \cdot N \) and that \( p \cdot {N}^{\prime } \), where \( N \) is the conductor of \( E \) and \( {N}^{\prime } \) that of \( F \) . Then in fact \( j\left( E\right) \in \mathbb{Z} \) . | Proof. By (2) we know that \( j\left( E\right) \in \mathbb{Z}\left\lbrack {1/p}\right\rbrack \) . Assume by contradiction that \( j\left( E\right) \notin \mathbb{Z} \), so that the denominator of \( j\left( E\right) \) is divisible by \( p \) . Thus \( p \mid N \) , and since \( {p}^{2} \nmid N \) we have \( p\parallel... | Yes |
Theorem 15.3.5 (Darmon-Merel [Dar-Mer]). If \( p \geq 7 \) is prime there are no nonzero pairwise coprime solutions to \( {x}^{2} = {y}^{p} + {z}^{p} \) . | We will prove this for \( p \neq {13} \) .\n\nProof. First note that if \( z \) (or \( y \) ) is even we can write \( {x}^{2} = {y}^{p} + {2}^{p}{\left( z/2\right) }^{p} \) , which does not have any pairwise coprime solutions for \( p \geq 7 \) by Theorem 15.3.4. We may therefore assume that both \( y \) and \( z \) ar... | Yes |
Proposition 15.4.1. Let \( E/\mathbb{Q} \) be an elliptic curve of conductor \( N \), and let \( t \) be an integer such that \( t\left| \right| {E}_{t}\left( \mathbb{Q}\right) \mid \), where we recall that \( {E}_{t}\left( \mathbb{Q}\right) \) is the torsion subgroup of \( E\left( \mathbb{Q}\right) \) . Let \( f \) be... | Proof. First note that if \( \ell \) is a prime of good reduction, in other words such that \( \ell \nmid N \), then by Proposition 8.1.13, \( \left| {{E}_{t}\left( \mathbb{Q}\right) }\right| \) divides \( E\left( {\mathbb{F}}_{\ell }\right) = \ell + 1 - {a}_{\ell }\left( E\right) \) , so if \( t\left| \right| {E}_{t}\... | Yes |
Proposition 15.4.2. In each of the following cases there are infinitely many \( \ell \) for which \( {B}_{\ell }\left( f\right) \neq 0 \) :\n\n(1) When \( f \) is irrational.\n\n(2) When \( f \) is rational, \( t \) is either a prime number or is equal to 4, and for every elliptic curve \( F \) isogenous to the ellipti... | Note that for (1), if \( f \) is not rational it is easy to show that \( {c}_{\ell } \notin \mathbb{Q} \) for infinitely many \( \ell \), so that \( {B}_{\ell }\left( f\right) \neq 0 \) at least for all such \( \ell \) since \( {\left( \ell + 1\right) }^{2} - {c}_{\ell }^{2} \neq 0 \) (recall that \( \left| {c}_{\ell }... | No |
Lemma 15.5.1. Assume that \( F \) is an elliptic curve defined over \( \mathbb{Q} \) with conductor \( {2L} \), and assume that \( F \) has full 2-torsion. Then \( L \) is either a Mersenne or a Fermat prime (in other words \( L \) is a prime such that \( L = \) \( \left. {{2}^{m} - 1\text{ or }L = {2}^{{2}^{k}} + 1}\r... | Proof. Since \( F \) has full 2-torsion, up to isomorphism it has an equation of the form \( {Y}^{2} = X\left( {X - a}\right) \left( {X + b}\right) \) with \( a \) and \( b \) in \( \mathbb{Z} \) . It is easily shown that we can choose \( a \) and \( b \) such that this equation is minimal at all primes different from ... | Yes |
Theorem 15.5.2 (Mazur). Let \( L \) be an odd prime that is neither a Mersenne nor a Fermat prime. There exists a constant \( {C}_{L} \) such that for any nontrivial solution \( \left( {x, y, z, p}\right) \) to the SMK equation we have \( p \leq {C}_{L} \) . | Proof. By the discussion preceding the lemma we know that \( E{ \sim }_{p}f \) for a newform \( f \) at level \( {N}_{p} = {2L} \) . If \( f \) is irrational then by Proposition 15.4.2 (1) we know that there are infinitely many \( \ell \) for which \( {B}_{\ell }\left( f\right) \neq 0 \) . If \( f \) is rational, then ... | Yes |
Theorem 15.5.3. Suppose that \( 3 \leq L \leq {100} \) is prime. Then the SMK equation does not have any nontrivial solutions, except for \( L = {31} \), in which case \( E{ \sim }_{p}F \), where \( F \) is the curve \( {62A1} \) in \( \left\lbrack {\mathrm{{Cre}}2}\right\rbrack \) . | Proof. This follows from the use of Proposition 15.4.1, the method of Kraus (Proposition 15.6.3 below), and the method of predicting exponents (Section 15.7 below). We will see examples of each of these methods applied to this theorem. | No |
Proposition 15.5.4. The above result is true for \( L = {19} \) . | Proof. From what we have done we already know that \( E{ \sim }_{p}f \) for some newform of level 38. There are two newforms of level 38, and although this cannot be seen purely from the dimension formulas that we have given, both are rational. Their \( q \) -expansions are:\n\n\[ \n{f}_{1} = q - {q}^{2} + {q}^{3} + {q... | Yes |
Lemma 15.6.1. Let \( A, B \), and \( C \) be nonzero integers such that \( A + B + C = \) 0, and let \( E \) be the elliptic curve with equation\n\n\[ \n{Y}^{2} = X\left( {X - A}\right) \left( {X + B}\right) .\n\]\n\nThen any permutation of \( A, B \), and \( C \) gives rise to a curve that is either isomorphic to \( E... | Proof. The exchange of \( A \) and \( B \) clearly changes \( E \) into \( {E}^{\prime } \) . The exchange of \( A \) and \( C \) changes \( E \) into the curve with equation \( {Y}^{2} = X\left( {X + A + B}\right) (X + \) \( B) \), which we see is again isomorphic to \( {E}^{\prime } \) by changing \( X \) into \( X -... | Yes |
Proposition 15.6.3. Let \( p \geq 5 \) be a fixed prime number, and let \( E \) be as above. Assume that for every newform \( f \) of level \( {2L} \) there exists a positive integer \( n \) satisfying the following conditions, where as usual \( {c}_{\ell } \) denotes the lth Fourier coefficient of \( f \) :\n\n(1) \( ... | Proof. If \( \ell \) satisfies (1) and (2) it is a prime different from \( 2, L \), and \( p \), so the above lemma is applicable. Because of (3) and the lemma we have \( \ell \nmid {xyz} \) . But then the reduction modulo \( \ell \) of \( \delta = {\left( x/z\right) }^{p} \) is well defined and is in \( {\mathbb{F}}_{... | Yes |
Theorem 15.6.4. Assume that \( L = {31} \) . Then the SMK equation does not have any nontrivial solution for \( {11} \leq p \leq {10}^{6} \) . | Proof. Thanks to Ribet’s theorem we have seen that \( E{ \sim }_{p}f \) for a newform \( f \) of level \( {2L} = {62} \) . The number of newforms is equal to 3, but one is rational and the other two are conjugate over \( \mathbb{Q}\left( \sqrt{3}\right) \) . If we set \( \theta = \sqrt{3} \), the forms are\n\n\[ \n{f}_... | Yes |
Lemma 15.7.2. If \( y \neq - 1 \) then \( y \geq {\left( \sqrt{p} - 1\right) }^{2} \) . | Proof. Note the trivial fact that if \( y \neq - 1 \) then \( y > 1 \), and furthermore \( y \) is odd, since otherwise \( {x}^{2} \equiv 2\left( {\;\operatorname{mod}\;4}\right) \), which is absurd. Thus there exists an odd prime \( \ell \) such that \( \ell \mid y \) . Since \( N = {2}^{7}\operatorname{rad}\left( y\r... | Yes |
Proposition 15.7.4. Let \( p \geq 5 \) be a fixed prime number, and let \( f \) be the newform of level \( {2L} \) such that \( E{ \sim }_{p}f \) . Assume that there exists a positive integer \( n \) satisfying the following conditions:\n\n(1) \( \ell = {np} + 1 \) is prime.\n\n(2) \( \ell \neq L \) .\n\n(3) \( {L}^{n}... | Proof. By Proposition 15.6.3, if \( \left( {x, y, z}\right) \) is a nontrivial solution then \( \delta = \) \( \overline{z/{x}^{p}} \in {\mathcal{X}}_{\ell } \) as defined above. We now apply the homomorphism \( \Phi \) to the identity \( {L}^{r}{y}^{p} = {x}^{p}\left( {-1 - {\left( z/x\right) }^{p}}\right) \) and note... | Yes |
Corollary 15.7.5. The SMK equation does not have any nontrivial solution for \( L = {31} \) and \( p = 7 \) . | Proof. We use \( n = 6 \) hence \( \ell = {43} \) . An immediate computation shows that the assumptions of the lemma are satisfied, and we find that \( {\mathcal{X}}_{\ell } = \{ 6,{36}\} \) , and then that \( {\mathcal{R}}_{\ell } = \{ 0{\;\operatorname{mod}\;7}\} \) . It follows that \( r \equiv 0\left( {\;\operatorn... | Yes |
Corollary 15.7.6. The SMK equation does not have any nontrivial solution for \( L = {23} \) and \( p \geq 5 \) . | Proof. By Ribet’s theorem we know that if \( E \) is the Frey curve corresponding to a possible solution \( \left( {x, y, z}\right) \) then \( E{ \sim }_{p}f \) for a newform \( f \) of level 46. By Proposition 15.1.1 there is a single newform at that level, necessarily rational, and equal to\n\n\[ f = q - {q}^{2} + {q... | Yes |
Theorem 15.8.1 (Kraus). Under the above assumptions we have \( E{ \sim }_{p}f \) for some newform \( f \) of level \( {N}_{p} = {2}^{\beta }{\operatorname{rad}}_{2}\left( R\right) \), where\n\n\[ \beta = \left\{ \begin{array}{ll} 1 & \text{ if }{v}_{2}\left( R\right) \geq 5\text{ or }{v}_{2}\left( R\right) = 0, \\ 0 & ... | Proof. The proof is left as an exercise to the reader, who is referred to [Kra3]. | No |
Lemma 16.1.1. For all \( i \) such that \( 1 \leq i \leq p - 1 \) set \( {\beta }_{i} = \left( {x - {\zeta }^{i}}\right) /(1 - \) \( \left. {\zeta }^{i}\right) \) . The \( {\beta }_{i} \) are algebraic integers not divisible by \( \mathfrak{p} \) and the ideals that they generate are pairwise coprime and equal to qth p... | Proof. By Cassels’s relations recalled above we have \( p \mid \left( {x - 1}\right) \), hence \( {v}_{\mathfrak{p}}\left( {x - 1}\right) \geq p - 1 \geq 2 \), hence \( {v}_{\mathfrak{p}}\left( {{\beta }_{i} - 1}\right) = {v}_{\mathfrak{p}}\left( {x - 1}\right) - {v}_{\mathfrak{p}}\left( {1 - {\zeta }^{i}}\right) \geq ... | Yes |
Lemma 16.1.2. For any \( \theta \in \left( {1 - \iota }\right) {I}_{s}\left( p\right) \) the element \( {\left( x - \zeta \right) }^{\theta } \) is a qth power in \( K \) . | Proof. Write \( \theta = \left( {1 - \iota }\right) {\theta }_{1} \) with \( {\theta }_{1} \in {I}_{s}\left( p\right) \) . By the preceding lemma we have \( \beta {\mathbb{Z}}_{K} = {\mathfrak{b}}^{q} \) for some ideal \( \mathfrak{b} \), while by Stickelberger’s Theorem 3.6.19 we have \( {\mathfrak{b}}^{{\theta }_{1}}... | Yes |
Subsets and Splits
No community queries yet
The top public SQL queries from the community will appear here once available.