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Lemma 15.16. For each \( n \geq 1 \), let \( 0 \rightarrow {A}_{n}\overset{{f}_{n}}{ \rightarrow }{B}_{n}\overset{{g}_{n}}{ \rightarrow }{C}_{n} \rightarrow 0 \) be an exact sequence of compact groups, and let \( {\phi }_{n + 1, n}^{X} : {X}_{n + 1} \rightarrow {X}_{n} \) for \( X = A, B, C \) be compatible with the ma... | Proof. The only difficulty is the surjectivity of \( g \) . Let \( c = \left( {c}_{n}\right) \in \mathop{\lim }\limits_{ \leftarrow }{C}_{n} \) . For each \( N \geq 1 \), let \( {b}_{N} \in {B}_{N} \) be such that \( {g}_{N}\left( {b}_{N}\right) = {c}_{N} \) . Let \( {b}_{i}^{\left( N\right) } = {\phi }_{N, i}^{B}\left... | Yes |
Lemma 15.17. Let \( X \) be a finitely generated torsion \( \Lambda \) -module.\n\n(1) \( \operatorname{char}\left( X\right) \cdot X \) is finite.\n\n(2) If \( X \) is finite, then \( {\left( p, T\right) }^{n}X = 0 \) for \( n \) sufficiently large; hence, the annihilator of \( X \) is of finite index in \( \Lambda \) ... | Proof. (1) There is an exact sequence \( 0 \rightarrow A \rightarrow X \rightarrow E \) with \( A \) finite and \( E \) elementary. If \( x \in X \), then \( \operatorname{char}\left( X\right) \cdot x \) maps to 0 in \( E \) ; hence lies in \( A \).\n\n(2) If \( f \in \left( {p, T}\right) \) and \( x \in X \), then \( ... | Yes |
Lemma 15.18. Let \( X \sim \oplus \Lambda /\left( {f}_{i}^{{m}_{i}}\right) \) as above. Then\n\n\[ X{ \otimes }_{\Lambda }{\Lambda }_{\mathfrak{p}} = {\bigoplus }_{\left( {f}_{i}\right) = \mathfrak{p}}{\Lambda }_{\mathfrak{p}}/{f}_{i}^{{m}_{i}}{\Lambda }_{\mathfrak{p}} \]\n | Proof. There is an exact sequence\n\n\[ 0 \rightarrow A \rightarrow X \rightarrow E \rightarrow B \rightarrow 0 \]\n\nwith \( A, B \) finite. Since localization preserves exact sequences (see, for example, Atiyah-Macdonald [1, p. 39]),\n\n\[ 0 \rightarrow A \otimes {\Lambda }_{\mathfrak{p}} \rightarrow X \otimes {\Lamb... | Yes |
Corollary 15.19. E is uniquely determined by \( X \) (of course, if we allow reducible \( {f}_{i} \), then we can use Lemma 13.8 to replace \( E \) with other modules; but all yield the same characteristic polynomial). | Proof. \( {\Lambda }_{\mathrm{p}} \) is a principal ideal domain (the ideals are powers of \( \left( f\right) \) ), so the uniqueness of the exponents \( {m}_{i} \) follows from the uniqueness part of the structure theorem for finitely generated modules over a PID. | No |
Corollary 15.20. \( X \) is finite if and only if \( X \otimes {\Lambda }_{\mathfrak{p}} = 0 \) for all height one prime ideals \( \mathfrak{p} \) . | Proof. \( X \) is finite if and only if the corresponding \( E \) is 0 . | No |
Proposition 15.21. A map \( {X}_{1} \rightarrow {X}_{2} \) between finitely generated torsion \( \Lambda \) - modules is a pseudo-isomorphism if and only if the induced map \( {X}_{1} \otimes {\Lambda }_{p} \rightarrow \) \( {X}_{2} \otimes {\Lambda }_{\mathfrak{p}} \) is an isomorphism for all height one prime ideals ... | Proof. This follows immediately from the exactness of localization and Corollary 15.20. | No |
Proposition 15.22. Let \( 0 \rightarrow {X}_{1} \rightarrow {X}_{2} \rightarrow {X}_{3} \rightarrow 0 \) be an exact sequence of finitely generated \( \Lambda \) -modules. Then\n\n\[ \operatorname{char}\left( {X}_{1}\right) \cdot \operatorname{char}\left( {X}_{3}\right) = \operatorname{char}\left( {X}_{2}\right) . \] | Proof. This follows immediately from Lemma 15.18 and the corresponding result for modules over a PID. | No |
Lemma 15.23. Let \( \psi : X \rightarrow { \oplus }_{\mathfrak{p}}\left( {X \otimes {\Lambda }_{\mathfrak{p}}}\right) \) be the natural map. Then \( \operatorname{Ker}\psi \) is finite and is the maximal finite submodule of \( X \) . | Proof. Any finite module is contained in \( \operatorname{Ker}\psi \) by Corollary 15.20. Since \( \Lambda \) is Noetherian and \( X \) is finitely generated, \( \operatorname{Ker}\psi \) is finitely generated. It is therefore finite by Corollary 15.20. | Yes |
Proposition 15.24. The map\n\n\\[ \n\\phi : X{ \\otimes }_{\\Lambda }\\left( {\\bigcup \\frac{1}{{\\sigma }_{n}}\\Lambda }\\right) \\rightarrow {\\bigoplus }_{\\mathfrak{p}}\\left( {X{\\bigotimes }_{\\Lambda }{\\Lambda }_{\\mathfrak{p}}}\\right)\n\\]\n\n\\[ \nx \\otimes \\frac{1}{{\\sigma }_{n}} \\mapsto \\left( {\\ldo... | Proof. Note that every element on the left can be written in the form \\( x \\otimes \\frac{1}{{\\sigma }_{n}} \\) . Suppose \\( \\phi \\left( {x \\otimes \\frac{1}{{\\sigma }_{n}}}\\right) = 0 \\) . Multiplying by \\( {\\sigma }_{n} \\), we find that \\( x \\otimes 1 = 0 \\) in \\( X \\otimes {\\Lambda }_{\\mathfrak{p... | Yes |
Proposition 15.25. Assume \( f \in \Lambda ,\pi \in p{\mathbb{Z}}_{p} \), and \( f\left( \pi \right) \neq 0 \) . Then\n\n\[ \Lambda /\left( f\right) \simeq {\operatorname{Hom}}_{{\mathbb{Z}}_{p}}{\left( \Lambda /\left( f\right) \otimes \Lambda \left\lbrack \frac{1}{T - \pi }\right\rbrack /\Lambda ,{\mathbb{Q}}_{p}/{\ma... | Proof. By the above, the middle term is \( {\operatorname{Hom}}_{{\mathbb{Z}}_{p}}{\left( \operatorname{Coker}\psi ,{\mathbb{Q}}_{p}/{\mathbb{Z}}_{p}\right) }^{ \sim } \), which equals \( \alpha \left( {\Lambda /\left( f\right) }\right) \) by definition. So it remains to prove the first isomorphism.\n\nFor \( g = \math... | Yes |
Lemma 15.26. Suppose \( A \) and \( B \) are \( {\mathbb{Z}}_{p} \) -modules with \( A \simeq {\mathbb{Z}}_{p}^{n} \) . Assume there is a nondegenerate pairing\n\n\[ A \times B \rightarrow {\mathbb{Q}}_{p}/{\mathbb{Z}}_{p} \]\n\nThen \( A \simeq {\operatorname{Hom}}_{{\mathbb{Z}}_{p}}\left( {B,{\mathbb{Q}}_{p}/{\mathbb... | Proof. Let \( \left( {a, b}\right) \) denote the pairing. For \( b \in B \), define \( {\phi }_{b} : A \rightarrow {\mathbb{Q}}_{p}/{\mathbb{Z}}_{p} \) by \( {\phi }_{b}\left( a\right) = \left( {a, b}\right) \) . The nondegeneracy implies that there is an injection \( B \hookrightarrow \) \( \operatorname{Hom}\left( {A... | No |
Proposition 15.28. (i) \( \alpha \left( X\right) \) has no nonzero finite \( \Lambda \) -submodules. | Proof. (i) It suffices to work with \( \widetilde{\alpha }\left( X\right) \) . Choose \( \pi \in {p\Lambda } \) such that \( \left\{ {\left( T - \pi \right) }^{n}\right\} \) forms an admissible sequence for \( X \) . Suppose \( \phi \in \operatorname{Hom}\left( {X \otimes \Lambda \left\lbrack \frac{1}{T - \pi }\right\r... | Yes |
Proposition 15.29. An exact sequence \( 0 \rightarrow X \rightarrow Y \rightarrow Z \rightarrow 0 \) of finitely generated torsion \( \Lambda \) -modules induces an exact sequence\n\n\[ 0 \rightarrow \alpha \left( Z\right) \rightarrow \alpha \left( Y\right) \rightarrow \alpha \left( X\right) \rightarrow \text{ finite. ... | Proof. It suffices to work with \( \widetilde{\alpha }\left( X\right) ,\widetilde{\alpha }\left( Y\right) \), and \( \widetilde{\alpha }\left( Z\right) \) . Consider the diagram \n\nThe bottom row is exact because lo... | Yes |
Proposition 15.30. Let \( X \) and \( Y \) be finitely generated torsion \( \Lambda \) -modules with \( X \sim Y \) . Then \( \alpha \left( Y\right) \sim \alpha \left( X\right) \) . | Proof. There is an exact sequence\n\n\[ \n0 \rightarrow A \rightarrow X \rightarrow Y \rightarrow B \rightarrow 0 \n\]\n\nwith \( A \) and \( B \) finite. From Proposition 15.29,\n\n\[ \n0 \rightarrow \alpha \left( {X/A}\right) \rightarrow \alpha \left( X\right) \rightarrow \alpha \left( A\right) \n\]\n\nis exact, and ... | Yes |
Corollary 15.31. \( X \sim \alpha \left( X\right) \), and \( \alpha \left( X\right) \) is also a finitely generated torsion \( \Lambda \) - module. | Proof. By Theorem 13.12, there is an elementary \( \Lambda \) -module \( E \) with \( X \sim E \) . By Corollary 15.27 and Proposition 15.30, \( X \sim E \simeq \alpha \left( E\right) \sim \alpha \left( X\right) \) . Since \( X \) is finitely generated \( \Lambda \) -torsion, it follows immediately that the same must b... | Yes |
Proposition 15.32. \( \mathop{\lim }\limits_{ \rightarrow }X/{\sigma }_{n}X \simeq X{ \otimes }_{\Lambda }\left( {\bigcup \frac{1}{{\sigma }_{n}}\Lambda /\Lambda }\right) \) | Proof. We have\n\n\[ X/{\sigma }_{n}X \simeq X \otimes \left( {\Lambda /{\sigma }_{n}\Lambda }\right) \simeq X \otimes \left( {\frac{1}{{\sigma }_{n}}\Lambda /\Lambda }\right) .\n\nThe maps \( X/{\sigma }_{n}X \rightarrow X/{\sigma }_{n + 1}X \) correspond to the natural inclusions \( \frac{1}{{\sigma }_{n}}\Lambda /\L... | Yes |
Lemma 15.33. For \( n \geq e \), the natural map \( {A}_{n} \rightarrow {A}_{n + 1} \) corresponds to the map\n\n\[ \nX/{v}_{n, e}{Y}_{e} \rightarrow X/{v}_{n + 1, e}{Y}_{e} \n\]\n\ngiven by \( x \mapsto {v}_{n + 1, n}x \) . | Proof. Let \( x \in X \), so for each \( n \geq e \) there exists \( {I}_{n} \in {A}_{n} \) such that\n\n\[ \nx\left( {{\;\operatorname{mod}\;{v}_{n, e}}{Y}_{e}}\right) = \left\lbrack {{I}_{n},{L}_{n}/{K}_{n}}\right\rbrack \in \operatorname{Gal}\left( {{L}_{n}/{K}_{n}}\right) \n\]\n\nwhere \( \left\lbrack {{I}_{n},{L}_... | Yes |
Proposition 15.34. \( \widetilde{X} \sim {\operatorname{Hom}}_{{\mathbb{Z}}_{p}}\left( {\mathop{\lim }\limits_{ \rightarrow }{A}_{n},{\mathbb{Q}}_{p}/{\mathbb{Z}}_{p}}\right) \) . | Proof. The exact sequence\n\n\[ 0 \rightarrow {Y}_{e}/{v}_{n, e}{Y}_{e} \rightarrow X/{v}_{n, e}{Y}_{e} \rightarrow X/{Y}_{e} \rightarrow 0 \]\n\nyields the exact sequence\n\n\[ 0 \rightarrow \mathop{\lim }\limits_{ \rightarrow }{Y}_{e}/{v}_{n, e}{Y}_{e} \rightarrow \mathop{\lim }\limits_{ \rightarrow }{A}_{n} \rightar... | Yes |
Proposition 15.35. Let the notation be as in Proposition 13.32. Then\n\n\[ \n\widetilde{{\varepsilon }_{i}X} \sim {\operatorname{Hom}}_{{\mathbb{Z}}_{p}}\left( {\mathop{\lim }\limits_{ \rightarrow }{\varepsilon }_{i}{A}_{n},{\mathbb{Q}}_{p}/{\mathbb{Z}}_{p}}\right) .\n\] | The proof of Proposition 15.34 also shows that\n\n\[ \n{\operatorname{Hom}}_{{\mathbb{Z}}_{p}}\left( {\mathop{\lim }\limits_{ \rightarrow }{\varepsilon }_{i}{A}_{n},{\mathbb{Q}}_{p}/{\mathbb{Z}}_{p}}\right) \simeq \widetilde{\alpha }\left( {{\varepsilon }_{i}{Y}_{e}}\right) .\n\] | No |
Proposition 15.37. Suppose \( {\varepsilon }_{i}X \) has characteristic polynomial \( f\left( T\right) \) . Then\n\n\[{\operatorname{Hom}}_{{\mathbb{Z}}_{p}}\left( {\mathop{\lim }\limits_{ \rightarrow }{\varepsilon }_{i}{A}_{n},{\mathbb{Q}}_{p}/{\mathbb{Z}}_{p}}\right)\]\n\nhas characteristic polynomial \( f\left( {{\l... | Proof. The first statement follows from Proposition 15.35 and the definition of the action of \( \Lambda \) on \( \widetilde{{\varepsilon }_{i}X} \) . For the second, note that if \( {\gamma }_{0} \) acts on \( {\varepsilon }_{i}X \) as \( \left( {1 + T}\right) \), then it acts on \( \widetilde{\left( {\varepsilon }_{i... | Yes |
Proposition 15.38. Let \( {\bar{C}}_{1}^{n},{X}_{n},{U}_{1}^{n} \), and \( {\mathcal{X}}_{n} \) be as in Section 15.4. Then\n\n\[ \n{\varepsilon }_{\chi }{\bar{C}}_{1}^{\infty }/{P}_{n} \simeq {\varepsilon }_{\chi }{\bar{C}}_{1}^{n} \]\n\n\[ \n{\varepsilon }_{\chi }X/{P}_{n} \simeq {\varepsilon }_{\chi }{X}_{n} \]\n\n\... | Proof. The result for \( {A}_{n} \simeq {X}_{n} \) follows from Proposition 13.22 and that for \( {\bar{C}}_{1}^{\infty } \) from Proposition 8.11, as in Section 13.8. Section 13.5 treats \( {\mathcal{X}}_{\infty } \) . The result for \( {U}_{1}^{\infty } \) is Proposition 13.54. | No |
Lemma 15.39. Let \( 0 \rightarrow {M}_{1} \rightarrow {M}_{2} \rightarrow {M}_{3} \rightarrow 0 \) be an exact sequence of \( \Lambda \) - modules.\n\n(a) \( \operatorname{Ker}\left( {{M}_{1}/{P}_{n} \rightarrow {M}_{2}/{P}_{n}}\right) \simeq {M}_{3}^{{\Gamma }_{n}}/\operatorname{Im}{M}_{2}^{{\Gamma }_{n}} \) . | Proof. Consider the diagram\n\n\n\nwhere the vertical maps are multiplication by \( {P}_{n} = {\gamma }_{0}^{{p}^{n}} - 1 \) . Note that \( {M}_{i}^{{\Gamma }_{n}} = \) \( \operatorname{Ker}\left( {{M}_{i}\overset{{P... | Yes |
Lemma 15.41. There is an exact sequence\n\n\[ 0 \rightarrow {\varepsilon }_{\chi }{\bar{E}}_{1}^{\infty }\overset{\theta }{ \rightarrow }\Lambda \rightarrow \text{ finite } \rightarrow 0. \] | Proof. We have \( {\varepsilon }_{\chi }{\bar{E}}_{1}^{\infty } \subseteq {\varepsilon }_{\chi }{U}_{1}^{\infty } \simeq \Lambda \) by Theorem 13.54. Since \( \Lambda \) is Noetherian, \( {\varepsilon }_{\gamma }{\bar{E}}_{1}^{\infty } \) is finitely generated and torsion-free. By Theorem 13.12, there is a pseudo-isomo... | Yes |
Proposition 15.42. Let \( \mathfrak{A} \) be as in Proposition 15.40 and let \( \alpha \in \mathfrak{A} \) . Let\n\n\[ \n{h}_{\chi } = \operatorname{char}\left( {{\varepsilon }_{\chi }{\bar{E}}_{1}^{\infty }/{\bar{C}}_{1}^{\infty }}\right) .\n\]\n\nFor each \( n \geq 0 \) there is a map\n\n\[ \n{\theta }_{\alpha }^{n} ... | Proof. The map \( \theta \) in Lemma 15.41 induces an exact sequence\n\n\[ \n0 \rightarrow {\varepsilon }_{\chi }{\bar{E}}_{1}^{\infty }/{\bar{C}}_{1}^{\infty }\overset{\theta }{ \rightarrow }\Lambda /\theta \left( {{\varepsilon }_{\chi }{\bar{C}}_{1}^{\infty }}\right) \rightarrow \text{ finite } \rightarrow 0.\n\]\n\n... | Yes |
Proposition 15.43. Let \( \chi = 1 \) (so \( {\varepsilon }_{\chi } = {\varepsilon }_{0} \) ). Then \( {\varepsilon }_{0}{\bar{E}}_{1}^{n}/{\bar{C}}_{1}^{n} = 1 \) for all \( n \leq \infty \) . Also, \( {\varepsilon }_{0}{X}_{n} = 0 \) for all \( n < \infty \), and \( {\varepsilon }_{0}X = 0 \) . | Proof. Let \( {\mathbb{B}}_{n} \) be the unique subfield of \( \mathbb{Q}\left( {\zeta }_{{p}^{n + 1}}\right) \) of degree \( {p}^{n} \) over \( \mathbb{Q} \) . Corollary 10.7 says that the class number of \( {\mathbb{B}}_{n} \) is not divisible by \( p \) . Since \( \left( {p - 1}\right) {\varepsilon }_{0} \) is the n... | Yes |
Proposition 15.44. Let \( \chi \) be arbitrary (including \( \chi = 1 \) ). There exists a constant \( c > 0 \) such that\n\n\[ \n{c}^{-1}\left\lbrack {{\varepsilon }_{\chi }{\bar{E}}_{1}^{n} : {\varepsilon }_{\chi }{\bar{C}}_{1}^{n}}\right\rbrack \leq \left| {\Lambda /\left( {{P}_{n},{h}_{\chi }}\right) }\right| \leq ... | Proof. The case where \( \chi = 1 \) follows immediately from the previous proposition. Assume now that \( \chi \neq 1 \) . From the proof of Proposition 15.42, there is an exact sequence\n\n\[ \n0 \rightarrow {\varepsilon }_{\chi }{\bar{E}}_{1}^{\infty }/{\bar{C}}_{1}^{\infty } \rightarrow \Lambda /\left( {h}_{\chi }\... | Yes |
Lemma 15.46. \( {\overline{\mathrm{{ind}}}}_{\lambda } \) and \( {\bar{v}}_{\lambda } \) are \( \mathbb{Z}\left\lbrack G\right\rbrack \) -homomorphisms. | Proof. Let \( \tau \in G \) . Then \( \tau \left( \kappa \right) \equiv {s}^{{a}_{\sigma }}\left( {{\;\operatorname{mod}\;\tau }{\sigma \lambda }}\right) \), so \( {\overline{\operatorname{ind}}}_{\lambda }\left( {\tau \kappa }\right) = \sum {a}_{\sigma }{\tau \sigma } = \tau {\overline{\operatorname{ind}}}_{\lambda }\... | Yes |
Proposition 15.47. Let \( p \) be an odd prime. Let \( m \geq 1, F = \mathbb{Q}{\left( {\zeta }_{m}\right) }^{ + } \), and \( G = \) \( \operatorname{Gal}\left( {F/\mathbb{Q}}\right) \) . Let \( \mathfrak{C} \) be an ideal class of \( F \) of order a power of \( p \), let \( M \) be a power of \( p \), and let \( L \ge... | Proof. Let \( H \) be the Hilbert class field of \( F \), so \( \operatorname{Gal}\left( {L/F}\right) \) is isomorphic to the class group of \( F \) . Let \( {F}_{ML} = F\left( {\zeta }_{ML}\right) \) . We first need to identify \( {F}_{ML}\left( {W}^{1/M}\right) \cap H. \)\n\nThere is a natural map \( {F}^{ \times }/{... | Yes |
Lemma 15.48. Let \( m, n \geq 1 \) with \( m \mid n \) . If \( K/\mathbb{Q}\left( {\zeta }_{m}\right) \) is unramified at all primes and \( K \subseteq \mathbb{Q}\left( {\zeta }_{n}\right) \), then \( K = \mathbb{Q}\left( {\zeta }_{m}\right) \) . If \( {K}^{\prime }/\mathbb{Q}{\left( {\zeta }_{m}\right) }^{ + } \) is u... | Proof. Let \( p \) be a prime dividing \( n/m \) . Then \( \mathbb{Q}\left( {\zeta }_{mp}\right) /\mathbb{Q}\left( {\zeta }_{m}\right) \) is totally ramified at the primes above \( p \), so \( K \cap \mathbb{Q}\left( {\zeta }_{mp}\right) = \mathbb{Q}\left( {\zeta }_{m}\right) \) . Therefore \( \left\lbrack {K\left( {\z... | Yes |
Lemma 15.49. Let \( M \geq 1 \) and let \( A \) be a group. Let \( {\phi }_{1},{\phi }_{2} \in \operatorname{Hom}\left( {A,\mathbb{Z}/M\mathbb{Z}}\right) \) . Suppose, for all \( a \in A \), that \( {\phi }_{1}\left( a\right) = 0 \Leftrightarrow {\phi }_{2}\left( a\right) = 0 \) . Then there exists \( u \in \) \( {\lef... | Proof. Let \( {a}_{1} \) be such that \( {\phi }_{1}\left( {a}_{1}\right) \) generates \( \operatorname{Im}{\phi }_{1} \) . Since, for any \( n, n{\phi }_{1}\left( {a}_{1}\right) = \) \( 0 \Leftrightarrow n{\phi }_{2}\left( {a}_{1}\right) = 0 \), it follows that \( {\phi }_{2}\left( {a}_{1}\right) \) has the same order... | No |
Proposition 15.50. Let \( \kappa \left( {\ell L}\right) \) and \( \kappa \left( L\right) \) be as in Section 15.3. Then \[ {\bar{v}}_{\lambda }\left( {{\varepsilon }_{\chi }^{\prime }\kappa \left( {\ell L}\right) }\right) \equiv - {\overline{\operatorname{ind}}}_{\lambda }\left( {{\varepsilon }_{\chi }^{\prime }\kappa ... | Proof. Proposition 15.12 implies that \( {v}_{\sigma \lambda }\left( {\kappa \left( {\ell L}\right) }\right) \equiv {\operatorname{ind}}_{\sigma \lambda }\left( {\kappa \left( L\right) }\right) {\;\operatorname{mod}\;M} \) . The result follows from the definition of \( \bar{v} \) and ind, plus Lemma 15.46. | Yes |
Lemma 16.1. Suppose there is an integer \( c \) with \( {c}^{\left( {n - 1}\right) /2} \equiv - 1\left( {\;\operatorname{mod}\;n}\right) \) . Then\n\n\[ \n{v}_{2}\left( {r - 1}\right) \geq {v}_{2}\left( {n - 1}\right) \n\] \n\nfor all \( r \mid n \) . | Proof. Let \( {x}_{r} \) be the order of \( c{\;\operatorname{mod}\;r} \) . Since \( {c}^{\left( {n - 1}\right) /2} ≢ 1\left( {\;\operatorname{mod}\;r}\right) \) (note that \( n \) is odd, hence \( r \neq 2 \) ), and \( {c}^{n - 1} \equiv 1\left( {\;\operatorname{mod}\;r}\right) \), we have \( {v}_{2}\left( {x}_{r}\rig... | Yes |
Proposition 16.2. Let \( p \) be odd and let \( J \) (and a and \( b \) ) be as above. If \( {J}^{\alpha } \) is not congruent to a \( {p}^{k} \) th root of unity \( {\;\operatorname{mod}\;n}\mathbb{Z}\left\lbrack {\zeta }_{{p}^{k}}\right\rbrack \), then \( n \) is composite. If \( {J}^{\alpha } \equiv \zeta \) \( \lef... | Proof. We need three lemmas. | No |
Lemma 16.3. Assume \( {ab}\left( {a + b}\right) ≢ 0\left( {\;\operatorname{mod}\;p}\right) \) . Let\n\n\[ \beta = \mathop{\sum }\limits_{\substack{{x = 1} \\ {p \nmid x} }}^{{p}^{k}}\left( {\left\lbrack \frac{\left( {a + b}\right) x}{{p}^{k}}\right\rbrack - \left\lbrack \frac{ax}{{p}^{k}}\right\rbrack - \left\lbrack \f... | Proof. Let\n\n\[ \theta = \frac{1}{{p}^{k}}\mathop{\sum }\limits_{\substack{{x = 1} \\ {p \nmid x} }}^{{p}^{k}}x{\sigma }_{x}^{-1} = \mathop{\sum }\limits_{\substack{{x = 1} \\ {p \nmid x} }}^{{p}^{k}}\left\{ \frac{x}{{p}^{k}}\right\} {\sigma }_{x}^{-1} \]\nbe the Stickelberger element, as in Chapter 6, where \( \{ y\}... | Yes |
Lemma 16.4. If \( {\left( a + b\right) }^{p} ≢ {a}^{p} + {b}^{p}\left( {\;\operatorname{mod}\;{p}^{2}}\right) \) and \( {ab}\left( {a + b}\right) ≢ 0\left( {\;\operatorname{mod}\;p}\right) \), then\n\n\[ \mathop{\sum }\limits_{\substack{{x = 1} \\ {p \nmid x} }}^{{p}^{k}}\left( {\left\lbrack \frac{\left( {a + b}\right)... | Proof. If \( x \equiv y\left( {\;\operatorname{mod}\;{p}^{k}}\right) \), then \( {x}^{p} \equiv {y}^{p}\left( {\;\operatorname{mod}\;{p}^{k + 1}}\right) \), so there is a well-defined ring homomorphism\n\n\[ \mathbb{Z}\left\lbrack G\right\rbrack \rightarrow \mathbb{Z}/{p}^{k + 1}\mathbb{Z} \]\n\n\[ \mathop{\sum }\limit... | Yes |
Proposition 16.6. If \( {q}^{\left( {n - 1}\right) /2} ≢ \pm 1\left( {\;\operatorname{mod}\;n}\right) \), then \( n \) is composite. If \( {q}^{\left( {n - 1}\right) /2} \equiv \pm 1 \) \( \left( {\;\operatorname{mod}\;n}\right) \), then \[ {\chi }_{q,2}\left( r\right) = {\chi }_{q,2}{\left( n\right) }^{\ell \left( r\r... | Proof. If \( n \) is prime, \( {q}^{n - 1} \equiv 1\left( {\;\operatorname{mod}\;n}\right) \), so \( {q}^{\left( {n - 1}\right) /2} \equiv \pm 1\left( {\;\operatorname{mod}\;n}\right) \). Assume now that \( n \) is not yet known to be prime. By Lemma 16.5, \( g{\left( {\chi }_{q,2}\right) }^{r - 1} \equiv {\chi }_{q,2}... | Yes |
Theorem 16.7. Let \( n, s \), and \( t \) be as above. Suppose\n\n(1) \( {q}^{\left( {n - 1}\right) /2} \equiv \pm 1\left( {\;\operatorname{mod}\;n}\right) \) for all \( q \mid s \) ;\n\n(2) \( J{\left( {\chi }_{q, p}^{a},{\chi }_{q, p}^{b}\right) }^{\alpha } \equiv \zeta \left( {\;\operatorname{mod}\;n}\right) \) with... | Proof. Let \( r \mid n \) and let \( q \mid s \) . Propositions 16.2 and 16.6 imply that \( {\chi }_{q, p}\left( r\right) = \) \( {\chi }_{q, p}{\left( n\right) }^{\ell \left( r\right) } \) for all \( p \mid q - 1 \) . Since these characters generate the group of Dirichlet characters \( {\;\operatorname{mod}\;q} \) ,\n... | Yes |
Proposition 16.10. Let \( k \) be a field, let \( {X}_{1},\ldots ,{X}_{n}, Z\left( {n \geq 1}\right) \) be indeterminates over \( k \), and let \( {Y}_{1},\ldots ,{Y}_{m}\left( {m \geq 1}\right) \) be nontrivial elements of the group \( \mathop{\prod }\limits_{i}{X}_{i}^{Z} \) generated by \( {X}_{1},\ldots ,{X}_{n} \)... | Proof. Enlarge \( k \) if necessary so that \( {k}^{ \times } \) has an element \( t \) of infinite order. Suppose there is a relation in which not all \( {r}_{j} \) are constant and suppose the \( {r}_{j} \) are chosen so that \( m \) is minimal. Then no \( {r}_{j} \) can be constant, otherwise we could shorten the re... | Yes |
Lemma 16.11. Let \( p \) be prime and let \( \mathbb{F} \) be a field of characteristic \( p \) . Let \( {a}_{1},\ldots ,{a}_{n} \in {\mathbb{Z}}_{p} \) be linearly independent over \( \mathbb{Q} \) . Then \( {\left( 1 + T\right) }^{{a}_{1}},\ldots ,{\left( 1 + T\right) }^{{a}_{n}}, \) regarded as power series in \( \m... | Proof. Suppose we have a relation\n\n\[ \sum {b}_{D}{\left( 1 + T\right) }^{{d}_{1}{a}_{1} + \cdots + {d}_{n}{a}_{n}} = 0,\;{b}_{D} \in {\mathbb{F}}^{n}, \]\n\nwhere the sum is over \( n \) -tuples of non-negative integers and \( {b}_{D} = 0 \) for almost all \( D \) . Changing \( \left( {1 + T}\right) \) to \( {\left(... | Yes |
Lemma 16.13. Let \( {t}_{1},\ldots ,{t}_{s} \in {\mathbb{Z}}_{p} \) be distinct \( {\;\operatorname{mod}\;{p}^{M}} \) for some \( M \geq 1 \) . Suppose there are a primitive \( {p}^{m} \) th root of unity \( {\zeta }_{{p}^{m}} \), with \( m \geq M + c \), and constants \( {c}_{1},\ldots ,{c}_{s} \in \mathcal{O} \) such... | Proof. The hypotheses imply that \( {\zeta }_{{p}^{m}}^{{t}_{i} - {t}_{j}} \notin K\left( {\zeta }_{{p}^{c}}\right) \) for \( i \neq j \) . Therefore\n\n\[ 0 \equiv {\operatorname{Trace}}_{K\left( {{\zeta }_{p}m}\right) /K\left( {{\zeta }_{p}c}\right) }\left( {{\zeta }_{{p}^{m}}^{-{t}_{j}}\mathop{\sum }\limits_{i}{c}_{... | Yes |
The functions \( {U}^{{a}_{1}},\ldots ,{U}^{{a}_{r}} \) are algebraically independent over \( k \) . | Proof. Suppose we have a relation\n\n\[ \mathop{\sum }\limits_{\left( d\right) }{c}_{\left( d\right) }{U}^{\sum {a}_{i}{d}_{i}} = 0,\;{c}_{\left( d\right) } \in k \]\n\nwith \( \left( d\right) = \left( {{d}_{1},\ldots ,{d}_{r}}\right) \) running through finitely many \( r \) -tuples in \( {\mathbb{Z}}^{r} \) . Since \(... | Yes |
Lemma 16.15. Let \( \ell \) be a prime and let \( K/F \) be an extension of number fields of degree prime to \( \ell \) . Let \( {A}_{K} \) and \( {A}_{F} \) be the \( \ell \) -parts of the class groups of \( K \) and \( F \) . Then the natural map \( {A}_{F} \rightarrow {A}_{K} \) is injective and\n\n\[ \n{A}_{K} \sim... | Proof. Let \( n = \left\lbrack {K : F}\right\rbrack \) and let \( N : {A}_{K} \rightarrow {A}_{F} \) be the norm. Since the composition\n\n\[ \n{A}_{F} \rightarrow {A}_{K}\overset{N}{ \rightarrow }{A}_{F}\xrightarrow[]{{n}^{-1}}{A}_{F} \n\]\n\nis the identity, the map \( {A}_{F} \rightarrow {A}_{K} \) is injective and ... | Yes |
Theorem 4. Let \( k \) be a number field and let \( K \) be the maximal unramified (including \( \infty \) ) abelian extension of \( k \) . Then\n\n\[ \operatorname{Gal}\left( {K/k}\right) \simeq \text{ideal class group of}k\text{,}\]\n\nthe isomorphism being induced by the Artin map. (The field \( K \) is called the H... | We now justify a statement made in Section 10.2. Let \( K \) be the Hilbert class field (or \( p \) -class field) of \( k \), let \( F \subseteq k \), and suppose \( k/F \) is Galois. Then \( K/F \) is also Galois, by the maximality of \( K \) . As in Chapter \( {10}, G = \operatorname{Gal}\left( {k/F}\right) \) acts o... | Yes |
Theorem 6 (Kronecker-Weber). Let \( K \) be an abelian extension of \( \mathbb{Q} \). Then \( K \) is contained in a cyclotomic field. | Let \( K/\mathbb{Q} \) be abelian and let \( H \supseteq {P}_{n\infty } \) be the corresponding subgroup. Since\n\n\[ \n{I}_{n\infty }/{P}_{n\infty } \simeq {\left( \mathbb{Z}/n\mathbb{Z}\right) }^{ \times }, \n\]\n\nthe group \( H/{P}_{n\infty } \) corresponds to a subgroup of congruence classes \( {\;\operatorname{mo... | No |
Lemma 2. If \( a, b \in \mathbb{Z} \) and \( b > 0 \), there exist \( q, r \in \mathbb{Z} \) such that \( a = {qb} + r \) with \( 0 \leq r < b \) . | Proof. Consider the set of all integers of the form \( a - {xb} \) with \( x \in \mathbb{Z} \) . This set includes positive elements. Let \( r = a - {qb} \) be the least nonnegative element in this set. We claim that \( 0 \leq r < b \) . If not, \( r = a - {qb} \geq b \) and so \( 0 \leq a - \) \( \left( {q + 1}\right)... | Yes |
Lemma 3. If \( a, b \in \mathbb{Z} \), then there is a \( d \in \mathbb{Z} \) such that \( \left( {a, b}\right) = \left( d\right) \) . | Proof. We may assume that not both \( a \) and \( b \) are zero so that there are positive elements in \( \left( {a, b}\right) \) . Let \( d \) be the smallest positive element in \( \left( {a, b}\right) \) . Clearly \( \left( d\right) \subseteq \left( {a, b}\right) \) . We shall show that the reverse inclusion also ho... | Yes |
Lemma 4. Let \( a, b \in \mathbb{Z} \) . If \( \left( {a, b}\right) = \left( d\right) \) then \( d \) is a greatest common divisor of \( a \) and \( b \) . | Proof. Since \( a \in \left( d\right) \) and \( b \in \left( d\right) \) we see that \( d \) is a common divisor of \( a \) and \( b \) . Suppose that \( c \) is a common divisor. Then \( c \) divides every number of the form \( {ax} + {by} \) . In particular \( c \mid d \) . | Yes |
Proposition 1.1.1. Suppose that \( a \mid {bc} \) and that \( \left( {a, b}\right) = 1 \) . Then \( a \mid c \) . | Proof. Since \( \left( {a, b}\right) = 1 \) there exist integers \( r \) and \( s \) such that \( {ra} + {sb} = 1 \) . Therefore, \( {rac} + {sbc} = c \) . Since \( a \) divides the left-hand side of this equation we have \( a \mid c \) . | Yes |
Corollary 1. If \( p \) is a prime and \( p \mid {bc} \), then either \( p \mid b \) or \( p \mid c \) . | Proof. The only divisors of \( p \) are \( \pm 1 \) and \( \pm p \) . Thus \( \left( {p, b}\right) = 1 \) or \( p \) ; i.e., either \( p \mid b \) or \( p \) and \( b \) are relatively prime. If \( p \mid b \), we are done. If not, \( \left( {p, b}\right) = 1 \) and so, by the proposition, \( p \mid c \) . | Yes |
Corollary 2. Suppose that \( p \) is a prime and that \( a, b \in \mathbb{Z} \) . Then \( {\operatorname{ord}}_{p}{ab} = {\operatorname{ord}}_{p}a \) \( + {\operatorname{ord}}_{p}b \) . | Proof. Let \( \alpha = {\operatorname{ord}}_{p}a \) and \( \beta = {\operatorname{ord}}_{p}b \) . Then \( a = {p}^{\alpha }c \) and \( b = {p}^{\beta }d \), where \( p \nmid c \) and \( p \nmid d \) . Then \( {ab} = {p}^{\alpha + \beta }{cd} \) and by Corollary \( {1p} \nmid {cd} \) . Thus ord \( {}_{p}{ab} = \) \( \al... | Yes |
Lemma 1. Every nonconstant polynomial is the product of irreducible polynomials. | Proof. The proof is by induction on the degree. It is easy to see that polynomials of degree 1 are irreducible. Assume that we have proved the result for all polynomials of degree less than \( n \) and that \( \deg f = n \) . If \( f \) is irreducible, we are done. Otherwise \( f = {gh} \), where \( 1 \leq \deg g \) , ... | Yes |
Theorem 2. Let \( f \in k\left\lbrack x\right\rbrack \) . Then we can write\n\n\[ f = c\mathop{\prod }\limits_{p}{p}^{a\left( p\right) } \]\n\nwhere the product is over all monic irreducible polynomials and \( c \) is a constant. The constant \( c \) and the exponents \( a\left( p\right) \) are uniquely determined by \... | The existence of such a product follows immediately from Lemma 1. As before, the uniqueness is more difficult and the proof will be postponed until we develop a few tools. | No |
Lemma 2. Let \( f, g \in k\left\lbrack x\right\rbrack \) . If \( g \neq 0 \), there exist polynomials \( h, r \in k\left\lbrack x\right\rbrack \) such that \( f = {hg} + r \), where either \( r = 0 \) or \( r \neq 0 \) and \( \deg r < \deg g \) . | Proof. If \( g \mid f \), simply set \( h = f/g \) and \( r = 0 \) . If \( g \nmid f \), let \( r = f - {hg} \) be the polynomial of least degree among all polynomials of the form \( f - {lg} \) with \( l \in k\left\lbrack x\right\rbrack \) . We claim that \( \deg r < \deg g \) . If not, let the leading term of \( r \)... | Yes |
Lemma 3. Given \( f, g \in k\left\lbrack x\right\rbrack \) there is a \( d \in k\left\lbrack x\right\rbrack \) such that \( \left( {f, g}\right) = \left( d\right) \) . | Proof. In the set \( \left( {f, g}\right) \) let \( d \) be an element of least degree. We have \( \left( d\right) \subseteq \left( {f, g}\right) \) and we want to prove the reverse inclusion. Let \( c \in \left( {f, g}\right) \) . If \( d \nmid c \), then there exist polynomials \( h \) and \( r \) such that \( c = {h... | Yes |
Lemma 4. Let \( f, g \in k\left\lbrack x\right\rbrack \) . By Lemma 3 there is a \( d \in k\left\lbrack x\right\rbrack \) such that \( \left( {f, g}\right) = \) (d). \( d \) is a greatest common divisor of \( f \) and \( g \) . | Proof. Since \( f \in \left( d\right) \) and \( g \in \left( d\right) \) we have \( d\left| {f\text{and}d}\right| g \) . Suppose that \( h \mid f \) and that \( h \mid g \) . Then \( h \) divides every polynomial of the form \( {fl} + {gm} \) with \( l, m \in k\left\lbrack x\right\rbrack \) . In particular \( h \mid d ... | Yes |
Proposition 1.2.1. If \( f \) and \( g \) are relatively prime and \( f \mid {gh} \), then \( f \mid h \) . | Proof. If \( f \) and \( g \) are relatively prime, we have \( \left( {f, g}\right) = \left( 1\right) \) so there are polynomials \( l \) and \( m \) such that \( {lf} + {mg} = 1 \) . Thus \( {lfh} + {mgh} = h \) . Since \( f \) divides the left-hand side of this equation \( f \) must divide \( h \) . | Yes |
Corollary 1. If \( p \) is an irreducible polynomial and \( p \mid {fg} \), then \( p \mid f \) or \( p \mid g \) . | Proof. Since \( p \) is irreducible \( \left( {p, f}\right) = \left( p\right) \) or (1). In the first case \( p \mid f \) and we are done. In the second case \( p \) and \( f \) are relatively prime and the result follows from the proposition. | No |
Corollary 2. If \( p \) is a monic irreducible polynomial and \( f, g \in k\left\lbrack x\right\rbrack \), we have \( {\operatorname{ord}}_{p}{fg} = {\operatorname{ord}}_{p}f + {\operatorname{ord}}_{p}g. \) | Proof. The proof is almost word for word the same as the proof to Corollary 2 to Proposition 1.1.1. | No |
Proposition 1.3.1. If \( R \) is a Euclidean domain and \( I \subseteq R \) is an ideal, then there is an element \( a \in R \) such that \( I = {Ra} = \{ {ra} \mid r \in R\} \) . | Proof. Consider the set of nonnegative integers \( \{ \lambda \left( b\right) \mid b \in I, b \neq 0) \) . Since every set of nonnegative integers has a least element there is an \( a \in I, a \neq 0 \) , such that \( \lambda \left( a\right) \leq \lambda \left( b\right) \) for all \( b \in I, b \neq 0 \) . We claim tha... | Yes |
Proposition 1.3.2. Let \( R \) be a PID and \( a, b \in R \) . Then \( a \) and \( b \) have a greatest common divisor \( d \) and \( \left( {a, b}\right) = \left( d\right) \) . | Proof. Form the ideal \( \left( {a, b}\right) \) . Since \( R \) is a PID there is an element \( d \) such that \( \left( {a, b}\right) = \left( d\right) \) . Since \( \left( a\right) \subseteq \left( d\right) \) and \( \left( b\right) \subseteq \left( d\right) \) we have \( d\left| {a\text{and}d}\right| b \) . If \( {... | Yes |
Corollary 2. If \( R \) is a PID and \( p \in R \) is irreducible, then \( p \) is prime. | Proof. Suppose that \( p \mid {ab} \) and that \( p \nmid a \) . Since \( p \nmid a \) it follows that the only common divisors are units. By Corollary \( 1\left( {a, p}\right) = R \) . Thus \( \left( {{ab},{pb}}\right) = \left( b\right) \) . Since \( {ab} \in \left( p\right) \) and \( {pb} \in \left( p\right) \) we ha... | Yes |
Lemma 1. Let \( \left( {a}_{1}\right) \subseteq \left( {a}_{2}\right) \subseteq \left( {a}_{3}\right) \subseteq \cdots \) be an ascending chain of ideals. Then there is an integer \( k \) such that \( \left( {a}_{k}\right) = \left( {a}_{k + l}\right) \) for \( l = 0,1,2,\ldots \) . In other words, the chain breaks off ... | Proof. Let \( I = \mathop{\bigcup }\limits_{{i = 1}}^{\infty }\left( {a}_{i}\right) \) . It is easy to see that \( I \) is an ideal. Thus \( I = \left( a\right) \) for some \( a \in R \) . But \( a \in \mathop{\bigcup }\limits_{{i = 1}}^{\infty }\left( {a}_{i}\right) \) implies that \( a \in \left( {a}_{k}\right) \) fo... | Yes |
Proposition 1.3.3. Every nonzero nonunit of \( R \) is a product of irreducibles. | Proof. Let \( a \in R, a \neq 0, a \) not a unit. We wish to show, to begin with, that \( a \) is divisible by an irreducible element. If \( a \) is irreducible, we are done. Otherwise \( a = {a}_{1}{b}_{1} \), where \( {a}_{1} \) and \( {b}_{1} \) are nonunits. If \( {a}_{1} \) is irreducible, we are done. Otherwise \... | Yes |
Lemma 2. Let \( p \) be a prime and \( a \neq 0 \) . Then there is an integer \( n \) such that \( {p}^{n} \mid a \) but \( {p}^{n + 1} \times a \) . | Proof. If the lemma were false, then for each integer \( m > 0 \) there would be an element \( {b}_{m} \) such that \( a = {p}^{m}{b}_{m} \) . Then \( p{b}_{m + 1} = {b}_{m} \) so that \( \left( {b}_{1}\right) \subset \left( {b}_{2}\right) \subset \) \( \left( {b}_{3}\right) \subset \cdots \) would be an infinite ascen... | Yes |
Lemma 3. If \( a, b \in R \) with \( a, b \neq 0 \), then \( {\operatorname{ord}}_{p}{ab} = {\operatorname{ord}}_{p}a + {\operatorname{ord}}_{p}b \) . | Proof. Let \( \alpha = {\operatorname{ord}}_{p}a \) and \( \beta = {\operatorname{ord}}_{p}b \) . Then \( a = {p}^{\alpha }c \) and \( b = {p}^{\beta }d \) with \( p \nmid c \) and \( p \nmid d \) . Thus \( {ab} = {p}^{\alpha + \beta }{cd} \) . Since \( p \) is prime \( p \nmid {cd} \) . Consequently, \( {\operatorname... | Yes |
Theorem 3. Let R be a PID and S a set of primes with the properties given above. Then if \( a \in R, a \neq 0 \), we can write\n\n\[ a = u\mathop{\prod }\limits_{p}{p}^{e\left( p\right) } \]\n\n(1)\n\nwhere \( u \) is a unit and the product is over all \( p \in S \). The unit \( u \) and the exponents \( e\left( p\righ... | Proof. The existence of such a decomposition follows immediately from Proposition 1.3.3.\n\nTo prove the uniqueness, let \( q \) be a prime in \( S \) and apply ord, to both sides of Equation (1). Using Lemma 3 we get\n\n\[ {\operatorname{ord}}_{q}a = {\operatorname{ord}}_{q}u + \mathop{\sum }\limits_{p}e\left( p\right... | Yes |
Proposition 1.4.1. \( \mathbb{Z}\left\lbrack i\right\rbrack \) is a Euclidean domain. | Proof. For \( a + {bi} \in \mathbb{Q}\left\lbrack i\right\rbrack \) define \( \lambda \left( {a + {bi}}\right) = {a}^{2} + {b}^{2} \) . Let \( \alpha = a + {bi} \) and \( \gamma = c + {di} \) and suppose that \( \gamma \neq 0.\alpha /\gamma = r + {si} \) , where \( r \) and \( s \) are real numbers (they are, in fact, ... | Yes |
Proposition 1.4.2. \( \mathbb{Z}\left\lbrack \omega \right\rbrack \) is a Euclidean domain. | Proof. For \( \alpha = a + {b\omega } \in \mathbb{Z}\left\lbrack \omega \right\rbrack \) define \( \lambda \left( \alpha \right) = {a}^{2} - {ab} + {b}^{2} \) . A simple calculation shows that \( \lambda \left( \alpha \right) = \alpha \bar{\alpha } \) . Now, let \( \alpha ,\beta \in \mathbb{Z}\left\lbrack \omega \right... | Yes |
Theorem 1 (Euclid). In the ring \( \mathbb{Z} \) there are infinitely many prime numbers. | Proof. Let us consider positive primes. Label them in increasing order \( {p}_{1},{p}_{2},{p}_{3},\ldots \) . Thus \( {p}_{1} = 2,{p}_{2} = 3,{p}_{3} = 5 \), etc. Let \( N = \left( {{p}_{1}{p}_{2}\cdots {p}_{n}}\right) + 1 \) . \( N \) is greater than 1 and not divisible by any \( {p}_{i}, i = 1,2,\ldots, n \) . On the... | Yes |
Proposition 2.2.1. If \( n \in \mathbb{Z}, n \) can be written in the form \( n = a{b}^{2} \), where \( a, b \in \mathbb{Z} \) and \( a \) is square-free. | Proof. Let \( n = {p}_{1}^{{a}_{1}}{p}_{2}^{{a}_{2}}\cdots {p}_{l}^{{a}_{l}} \) . One can write \( {a}_{i} = 2{b}_{i} + {r}_{i} \), where \( {r}_{i} = 0 \) or 1 depending on whether \( {a}_{i} \) is even or odd. Set \( a = {p}_{1}^{{r}_{1}}{p}_{2}^{{r}_{2}}\cdots {p}_{l}^{{r}_{l}} \) and \( b = \) \( {p}_{1}^{{b}_{1}}{... | Yes |
Proposition 2.2.2. If \( n \) is a positive integer, let \( n = {p}_{1}^{{a}_{1}}{p}_{2}^{{a}_{2}}\cdots {p}_{l}^{{a}_{l}} \) be its prime decomposition. Then\n\n(a) \( v\left( n\right) = \left( {{a}_{1} + 1}\right) \left( {{a}_{2} + 1}\right) \cdots \left( {{a}_{l} + 1}\right) \).\n\n(b) \( \sigma \left( n\right) = \l... | Proof. To prove part (a) notice that \( m \mid n \) iff \( m = {p}_{1}^{{b}_{1}}{p}_{2}^{{b}_{2}}\cdots {p}_{l}^{{b}_{l}} \) and \( 0 \leq {b}_{i} \leq {a}_{i} \) for \( i = 1,2,\ldots, l \) . Thus the positive divisors of \( n \) are one-to-one correspondence with the \( n \) -tuples \( \left( {{b}_{1},{b}_{2},\ldots ... | Yes |
Proposition 2.2.3. If \( n > 1,\mathop{\sum }\limits_{{d \mid n}}\mu \left( d\right) = 0 \) . | Proof. If \( n = {p}_{1}^{{a}_{1}}{p}_{2}^{{a}_{2}}\cdots {p}_{l}^{{a}_{l}} \), then \( \mathop{\sum }\limits_{{d \mid n}}\mu \left( d\right) = \mathop{\sum }\limits_{\left( {\varepsilon }_{1},\ldots ,{\varepsilon }_{l}\right) }\mu \left( {{p}_{1}^{{\varepsilon }_{1}}\cdots {p}_{l}^{{\varepsilon }_{l}}}\right) \), wher... | Yes |
Theorem 2 (Möbius Inversion Theorem). Let \( F\left( n\right) = \mathop{\sum }\limits_{{d \mid n}}f\left( d\right) \) . Then \( f\left( n\right) = \) \( \mathop{\sum }\limits_{{d \mid n}}\mu \left( d\right) F\left( {n/d}\right) \) . | Proof. \( F = f \circ I \) . Thus \( F \circ \mu = \left( {f \circ I}\right) \circ \mu = f \circ \left( {I \circ \mu }\right) = f \circ \mathbb{I} = f \) . This shows that \( f\left( n\right) = F \circ \mu \left( n\right) = \mathop{\sum }\limits_{{d \mid n}}\mu \left( d\right) F\left( {n/d}\right) \) . | Yes |
Proposition 2.2.4. \( \mathop{\sum }\limits_{{d \mid n}}\phi \left( d\right) = n \) . | Proof. Consider the \( n \) rational numbers \( 1/n,2/n,3/n,\ldots ,\left( {n - 1}\right) /n, n/n \) . Reduce each to lowest terms; i.e., express each number as a quotient of relatively prime integers. The denominators will all be divisors of \( n \) . If \( d \mid n \) , exactly \( \phi \left( d\right) \) of our numbe... | Yes |
Proposition 2.2.5. If \( n = {p}_{1}^{{a}_{1}}{p}_{2}^{{a}_{2}}\cdots {p}_{l}^{{a}_{l}} \), then\n\n\[ \phi \left( n\right) = n\left( {1 - \left( {1/{p}_{1}}\right) }\right) \left( {1 - \left( {1/{p}_{2}}\right) }\right) \cdots \left( {1 - \left( {1/{p}_{l}}\right) }\right) . \] | Proof. Since \( n = \mathop{\sum }\limits_{{d \mid n}}\phi \left( d\right) \) the Möbius inversion theorem implies that \( \phi \left( n\right) = \) \( \mathop{\sum }\limits_{{d \mid n}}\mu \left( d\right) n/d = n - \mathop{\sum }\limits_{i}n/{p}_{i} + \mathop{\sum }\limits_{{i < j}}n/{p}_{i}{p}_{j}\cdots = n\left( {1 ... | Yes |
Theorem 3. \( \sum 1/p \) diverges, where the sum is over all positive primes in \( \mathbb{Z} \) . | Proof. Let \( {p}_{1},{p}_{2},\ldots ,{p}_{l\left( n\right) } \) be all the primes less than \( n \) and define \( \lambda \left( n\right) = \) \( \mathop{\prod }\limits_{{i = 1}}^{{l\left( n\right) }}{\left( 1 - 1/{p}_{i}\right) }^{-1} \) . Since \( {\left( 1 - 1/{p}_{i}\right) }^{-1} = \mathop{\sum }\limits_{{{a}_{i}... | Yes |
Proposition 2.4.1. \( \pi \left( x\right) \geq \log \left( {\log x}\right), x \geq 2 \) . | Proof. Let \( {p}_{n} \) denote the \( n \) th prime. Then since any prime dividing \( {p}_{1}{p}_{2}\cdots {p}_{n} \) +1 is distinct from \( {p}_{1}\ldots ,{p}_{n} \) it follows that \( {p}_{n + 1} \leq {p}_{1}\cdots {p}_{n} + 1 \) . Now \( {p}_{1} < {2}^{\left( {2}^{1}\right) },{p}_{2} < {2}^{\left( {2}^{2}\right) } ... | Yes |
Proposition 2.4.2. \( \pi \left( x\right) \geq \log x/2\log 2 \) . | Proof. For any set of primes \( S \) define \( {f}_{S}\left( x\right) \) to be the number of integers \( n \) , \( 1 \leq n \leq x \), with \( \gamma \left( n\right) \subset S \) . Suppose that \( S \) is a finite set with \( t \) elements. Writing such an \( n \) in the form \( n = {m}^{2}s \) with \( s \) square free... | Yes |
Proposition 2.4.3. \( \theta \left( x\right) < \left( {4\log 2}\right) x \) . | Proof. Consider the binomial coefficient\n\n\[ \left( \begin{matrix} {2n} \\ n \end{matrix}\right) = \frac{\left( {n + 1}\right) \cdots \left( {2n}\right) }{1 \cdot 2\cdots n}. \]\n\nClearly this integer is divisible by all primes \( p, n < p < {2n} \) . Furthermore, since\n\n\[ {\left( 1 + 1\right) }^{2n} = \mathop{\s... | Yes |
Corollary 1. There is a positive constant \( {c}_{1} \) such that \( \pi \left( x\right) < {c}_{1}x/\log x \) for \( x \geq 2 \) . | Proof.\n\[ \theta \left( x\right) \geq \mathop{\sum }\limits_{{p > \sqrt{x}}}^{{p \leq x}}\log p \]\n\[ \geq \left( {\log \sqrt{x}}\right) \left( {\pi \left( x\right) - \pi \left( \sqrt{x}\right) }\right) \]\n\[ \geq \left( {\log \sqrt{x}}\right) \pi \left( x\right) - \sqrt{x}\log \sqrt{x} \]\nThus\n\[ \pi \left( x\rig... | Yes |
Proposition 2.4.4. There is a positive constant \( {c}_{2} \) such that \( \pi \left( x\right) > {c}_{2}\left( {x/\log x}\right) \) . | Proof. By the above we have\n\n\[ \n{2}^{n} \leq \left( \begin{matrix} {2n} \\ n \end{matrix}\right) \leq \mathop{\prod }\limits_{{p < {2n}}}{p}^{{t}_{p}} \n\]\n\nThus\n\n\[ \nn\log 2 \leq \mathop{\sum }\limits_{{p < {2n}}}{t}_{p}\log p = \mathop{\sum }\limits_{{p < {2n}}}\left\lbrack \frac{\log {2n}}{\log p}\right\rbr... | Yes |
Proposition 3.2.1.\n\n(a) \( a \equiv a\left( m\right) \).\n\n(b) \( a \equiv b\left( m\right) \) implies that \( b \equiv a\left( m\right) \).\n\n(c) If \( a \equiv b\left( m\right) \) and \( b \equiv c\left( m\right) \), then \( a \equiv c\left( m\right) \). | Proof.\n\n(a) \( a - a = 0 \) and \( m \mid 0 \).\n\n(b) If \( m\left| {b - a\text{, then}m}\right| a - b \).\n\n(c) If \( m\left| {b - a\text{and}m}\right| c - b \), then \( m \mid c - a = \left( {c - b}\right) + \left( {b - a}\right) \). | Yes |
Proposition 3.2.3. If \( a \equiv c\left( m\right) \) and \( b \equiv d\left( m\right) \), then \( a + b \equiv c + d\left( m\right) \) and \( {ab} \equiv {cd}\left( m\right) \) . | Proof. If \( m\left| {c - a\text{and}m}\right| d - b \), then \( m \mid \left( {c - a}\right) + \left( {d - b}\right) = \left( {c + d}\right) - \left( {a + b}\right) \). Thus \( a + b \equiv c + d\left( m\right) \). Notice that \( {cd} - {ab} = c\left( {d - b}\right) + b\left( {c - a}\right) \). Thus \( m \mid {cd} - {... | Yes |
Proposition 3.3.1. The congruence \( {ax} \equiv b\left( m\right) \) has solutions iff \( d\left| {b\text{. If}d}\right| b \), then there are exactly \( d \) solutions. If \( {x}_{0} \) is a solution, then the other solutions are given by \( {x}_{0} + {m}^{\prime },{x}_{0} + 2{m}^{\prime },\ldots ,{x}_{0} + \left( {d -... | Proof. If \( {x}_{0} \) is a solution, then \( a{x}_{0} - b = m{y}_{0} \) for some integer \( {y}_{0} \) . Thus \( a{x}_{0} - m{y}_{0} = b \) . Since \( d \) divides \( a{x}_{0} - m{y}_{0} \), we must have \( d \mid b \) .\n\nConversely, suppose that \( d \mid b \) . By Lemma 4 on page 4 there exist integers \( {x}_{0}... | Yes |
Corollary 1. If \( a \) and \( m \) are relatively prime, then \( {ax} \equiv b\left( m\right) \) has one and only one solution. | Proof. In this case \( d = 1 \) so clearly \( d \mid b \), and there are \( d = 1 \) solutions. | No |
Corollary 2. If \( p \) is a prime and \( a ≢ 0\left( p\right) \), then \( {ax} \equiv b\left( p\right) \) has one and only one solution. | Proof. Immediate from Corollary 1. | No |
Corollary 1 (Euler’s Theorem). If \( \left( {a, m}\right) = 1 \), then \( {a}^{\phi \left( m\right) } \equiv 1\left( m\right) \) . | Proof. The units in \( \mathbb{Z}/m\mathbb{Z} \) form a group of order \( \phi \left( m\right) \) . If \( \left( {a, m}\right) = 1,\bar{a} \) is a unit. Thus \( {\bar{a}}^{\phi \left( m\right) } = \overline{1} \) or \( {a}^{\phi \left( m\right) } \equiv 1\left( m\right) \) . | Yes |
Corollary 2 (Fermat’s Little Theorem). If \( p \) is a prime and \( p \nmid a \), then \( {a}^{p - 1} \equiv \) \( 1\left( p\right) \) . | Proof. If \( p \nmid a \), then \( \left( {a, p}\right) = 1 \) . Thus \( {a}^{\phi \left( p\right) } \equiv 1\left( p\right) \) . The result follows, since for a prime \( p,\phi \left( p\right) = p - 1 \) . | Yes |
Lemma 1. If \( {a}_{1},\ldots ,{a}_{l} \) are all relatively prime to \( m \), then so is \( {a}_{1}{a}_{2}\cdots {a}_{l} \) . | Proof. \( {\bar{a}}_{i} \in \mathbb{Z}/m\mathbb{Z} \) is a unit. Thus so is \( {\bar{a}}_{1}{\bar{a}}_{2}\cdots {\bar{a}}_{l} = \overline{{a}_{1}{a}_{2}\cdots {a}_{l}} \) . By Proposition 3.3.2, \( {a}_{1}{a}_{2}\cdots {a}_{l} \) is relatively prime to \( m \) . | Yes |
Lemma 2. Suppose that \( {a}_{1},\ldots ,{a}_{t} \) all divide \( n \) and that \( \left( {{a}_{i},{a}_{j}}\right) = 1 \) for \( i \neq j \) . Then \( {a}_{1}{a}_{2}\cdots {a}_{t} \) divides \( n \) . | Proof. The proof is by induction on \( t \) . If \( t = 1 \), there is nothing to do. Suppose that \( t > 1 \) and that the lemma is true for \( t - 1 \) . Then \( {a}_{1}{a}_{2}\cdots {a}_{t - 1} \) divides \( n \) . By Lemma \( 1,{a}_{t} \) is prime to \( {a}_{1}{a}_{2}\cdots {a}_{t - 1} \) . Thus there are integers ... | Yes |
Theorem 1 (Chinese Remainder Theorem). Suppose that \( m = {m}_{1}{m}_{2}\cdots {m}_{t} \) and that \( \left( {{m}_{i},{m}_{j}}\right) = 1 \) for \( i \neq j \) . Let \( {b}_{1},{b}_{2},\ldots ,{b}_{t} \) be integers and consider the system of congruences:\n\n\[ x \equiv {b}_{1}\left( {m}_{1}\right), x \equiv {b}_{2}\l... | Proof. Let \( {n}_{i} = m/{m}_{i} \) . By Lemma \( 1,\left( {{m}_{i},{n}_{i}}\right) = 1 \) . Thus there are integers \( {r}_{i} \) and \( {s}_{i} \) such that \( {r}_{i}{m}_{i} + {s}_{i}{n}_{i} = 1 \) . Let \( {e}_{i} = {s}_{i}{n}_{i} \) . Then \( {e}_{i} \equiv 1\left( {m}_{i}\right) \) and \( {e}_{i} \equiv 0\left( ... | Yes |
Proposition 3.4.1. If \( S = {R}_{1} \oplus {R}_{2} \oplus \cdots \oplus {R}_{n} \), then \( U\left( S\right) = U\left( {R}_{1}\right) \times U\left( {R}_{2}\right) \times U\left( {R}_{3}\right) \times \cdots \times U\left( {R}_{n}\right) \). | Proof. Immediate from Theorem 1' and Proposition 3.4.1. | No |
Lemma 1. Let \( f\left( x\right) \in k\left\lbrack x\right\rbrack, k \) a field. Suppose that \( \deg f\left( x\right) = n \) . Then \( f \) has at most \( n \) distinct roots. | Proof. The proof goes by induction on \( n \) . For \( n = 1 \) the assertion is trivial. Assume that the lemma is true for polynomials of degree \( n - 1 \) . If \( f\left( x\right) \) has no roots in \( k \), we are done. If \( \alpha \) is a root, \( f\left( x\right) = q\left( x\right) \left( {x - \alpha }\right) + ... | Yes |
Proposition 4.1.1. \( {x}^{p - 1} - 1 \equiv \left( {x - 1}\right) \left( {x - 2}\right) \cdots \left( {x - p + 1}\right) \left( p\right) \) . | Proof. If \( \bar{a} \) denotes the residue class of an integer \( a \) in \( \mathbb{Z}/p\mathbb{Z} \), an equivalent way of stating the proposition is \( {x}^{p - 1} - \overline{1} = \left( {x - \overline{1}}\right) \left( {x - \overline{2}}\right) \cdots \left( {x - \left( {p - 1}\right) }\right) \) in \( \mathbb{Z}... | Yes |
Proposition 4.1.2. If \( d \mid p - 1 \), then \( {x}^{d} \equiv 1\\left( p\\right) \) has exactly \( d \) solutions. | Proof. Let \( d{d}^{\\prime } = p - 1 \) . Then\n\n\\[ \n\\frac{{x}^{p - 1} - 1}{{x}^{d} - 1} = \\frac{{\\left( {x}^{d}\\right) }^{{d}^{\\prime }} - 1}{{x}^{d} - 1} = {\\left( {x}^{d}\\right) }^{{d}^{\\prime } - 1} + {\\left( {x}^{d}\\right) }^{{d}^{\\prime } - 2} + \\cdots + {x}^{d} + 1 = g\\left( x\\right) .\n\\]\n\n... | Yes |
Theorem 1. \( U\left( {\mathbb{Z}/p\mathbb{Z}}\right) \) is a cyclic group. | Proof. For \( d \mid p - 1 \) let \( \psi \left( d\right) \) be the number of elements in \( U\left( {\mathbb{Z}/p\mathbb{Z}}\right) \) of order \( d \) . By Proposition 4.1.2 we see that the elements of \( U\left( {\mathbb{Z}/p\mathbb{Z}}\right) \) satisfying \( {x}^{d} \equiv \overline{1} \) form a group of order \( ... | Yes |
Lemma 2. If \( p \) is a prime and \( 1 \leq k < p \), then the binomial coefficient \( \left( \begin{array}{l} p \\ k \end{array}\right) \) is divisible by \( p \) . | Proof. We give two proofs.\n\n(a) By definition\n\n\[ \left( \begin{array}{l} p \\ k \end{array}\right) = \frac{p!}{k!\left( {p - k}\right) !}\text{ so that }p! = k!\left( {p - k}\right) !\left( \begin{array}{l} p \\ k \end{array}\right) .\n\]\n\nNow, \( p \) divides \( p \) !, but \( p \) does not divide \( k!\left( {... | Yes |
Lemma 3. If \( l \geq 1 \) and \( a \equiv b\left( {p}^{l}\right) \), then \( {a}^{p} \equiv {b}^{p}\left( {p}^{l + 1}\right) \) . | Proof. We may write \( a = b + c{p}^{l}, c \in \mathbb{Z} \) . Thus \( {a}^{p} = {b}^{p} + \left( \begin{array}{l} p \\ 1 \end{array}\right) {b}^{p - 1}c{p}^{l} + A \) , where \( A \) is an integer divisible by \( {p}^{l + 2} \) . The second term is clearly divisible by \( {p}^{l + 1} \) . Thus \( {a}^{p} \equiv {b}^{p... | Yes |
Corollary 1. If \( l \geq 2 \) and \( p \neq 2 \), then \( {\left( 1 + ap\right) }^{{p}^{l - 2}} \equiv 1 + a{p}^{l - 1}\left( {p}^{l}\right) \) for all \( a \in \mathbb{Z} \) . | Proof. The proof is by induction on \( l \) . For \( l = 2 \) the assertion is trivial. Suppose that it is true for some \( l \geq 2 \) . We show that it is then true for \( l + 1 \) . Applying Lemma 3 we obtain\n\n\[ \n{\left( 1 + ap\right) }^{{p}^{l - 1}} \equiv {\left( 1 + a{p}^{l - 1}\right) }^{p}\left( {p}^{l + 1}... | Yes |
Corollary 2. If \( p \neq 2 \) and \( p \nmid a \), then \( {p}^{l - 1} \) is the order of 1 + ap \( {\;\operatorname{mod}\;{p}^{l}} \) . | Proof. By Corollary \( 1,{\left( 1 + ap\right) }^{{p}^{l - 1}} \equiv 1 + a{p}^{l}\left( {p}^{l + 1}\right) \), implying that \( (1 + \) \( {ap}{)}^{{p}^{l - 1}} \equiv 1\left( {p}^{l}\right) \) and thus that \( 1 + {ap} \) has order dividing \( {p}^{l - 1} \cdot {\left( 1 + ap\right) }^{{p}^{l - 2}} \equiv \) \( 1 + a... | Yes |
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