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Proposition 1.10.6. We have \( e\left( {\mathfrak{P}/{\mathfrak{P}}_{I}}\right) = e\left( {\mathfrak{P}/{\mathfrak{P}}_{D}}\right) = e\left( {\mathfrak{P}/\mathfrak{p}}\right), e\left( {{\mathfrak{P}}_{I}/{\mathfrak{P}}_{D}}\right) = 1, f\left( {\mathfrak{P}/f{p}_{I}}\right) = \) 1 and \( f\left( {{\mathfrak{P}}_{I}/{\... | Proof. Since \( \left\{ {\sigma \in \operatorname{Gal}\left( {L/{U}_{\mathfrak{P}}}\right) \mid \sigma \left( x\right) \equiv x\left( {\;\operatorname{mod}\;\mathfrak{P}}\right) }\right\} = \operatorname{Gal}\left( {L/{U}_{\mathfrak{P}}}\right) \), applying Proposition 1.10.5 to the case \( K = {U}_{\mathfrak{P}} \), w... | Yes |
Lemma 1.10.10. Let \( {L}_{1},{L}_{2} \) be finite Galois extensions of \( K \). (1) Let \( \mathfrak{p} \) be a prime ideal of \( {\mathcal{O}}_{K} \), suppose \( \mathfrak{p} \) is unramified in \( {L}_{1} \) and \( {L}_{2} \), show \( \mathfrak{p} \) is unramified in \( {L}_{1}{L}_{2} \). | Proof. Exercise. | No |
Theorem 1.10.12. Let \( H \) be the Hilbert class field of \( K \), then the morphism \( {J}_{K} \rightarrow \) \( \operatorname{Gal}\left( {L/K}\right) \) factors through an isomorphism | \[ {C}_{K} \cong {J}_{K}/{P}_{K}\overset{ \sim }{ \rightarrow }\operatorname{Gal}\left( {L/K}\right) \] | Yes |
Proposition 2.1.2. \( {\mathbb{Z}}_{p} \) is a domain, and the unique maximal ideal of \( {\mathbb{Z}}_{p} \) is \( p{\mathbb{Z}}_{p} \) . | Proof. Suppose \( \left( {x}_{n}\right) \left( {y}_{n}\right) = 0 \), and \( \left( {x}_{n}\right) \neq 0,\left( {y}_{n}\right) \neq 0 \) . There exists \( k \) such that \( {x}_{k} \neq 0 \) , \( {y}_{k} \neq 0 \) . So there exist \( i, j < k \) such that \( {x}_{k} \in {p}^{i}\mathbb{Z}/{p}^{k} \smallsetminus {p}^{i ... | Yes |
Lemma 2.1.5. \( {\mathbb{Z}}_{p} \) is a topological ring, i.e. the operations\n\n\[ \left\{ \begin{array}{l} {\mathbb{Z}}_{p} \times {\mathbb{Z}}_{p} \rightarrow {\mathbb{Z}}_{p},\left( {x, y}\right) \mapsto x + y, \\ {\mathbb{Z}}_{p} \times {\mathbb{Z}}_{p} \rightarrow {\mathbb{Z}}_{p},\left( {x, y}\right) \mapsto {x... | Proof. Exercises. | No |
Proposition 2.1.6. \( {\mathbb{Z}}_{p} \) is complete, i.e. any Cauchy sequence has a limit in \( {\mathbb{Z}}_{p} \) . And \( \mathbb{Z} \) is dense in \( {\mathbb{Z}}_{p} \) . | Proof. Let \( \left\{ {a}_{n}\right\} \) be a Cauchy sequence in \( {\mathbb{Z}}_{p} \) . Thus for any \( m \in {\mathbb{Z}}_{ \geq 1} \), there exists \( N \) such that for any \( {n}_{1},{n}_{2} > N\left( m\right) ,{a}_{{n}_{1}} - {a}_{{n}_{2}} \in {p}^{m}{\mathbb{Z}}_{p} \) . This implies that for any \( n > N\left(... | No |
Proposition 2.1.7. We have:\n\n(1) \( {\left| x\right| }_{p} = 0 \) if and only \( x = 0 \) .\n\n(2) \( {\left| xy\right| }_{p} = {\left| x\right| }_{p}{\left| y\right| }_{p} \) .\n\n(3) (Non-archimedean) \( {\left| x + y\right| }_{p} \leq \max \left\{ {{\left| x\right| }_{p},{\left| y\right| }_{p}}\right\} \) .\n\nThu... | Proof. (1) and (2) are clear. For (3), we have (suppose \( r \leq s \) )\n\n\[ \n{p}^{r}\frac{a}{b} + {p}^{s}\frac{c}{d} = {p}^{r}\left( {\frac{a}{b} + {p}^{s - r}\frac{c}{d}}\right) = {p}^{r}\left( \frac{{ad} + {p}^{s - r}{bc}}{bd}\right) , \n\]\n\nhence (the equality follows from (2) and the inequality follows from \... | Yes |
Proposition 2.1.9. \( {\mathbb{Q}}_{p} \) is the completion of \( \mathbb{Q} \) with respect to \( {\left| \cdot \right| }_{p} \) | Proof. Let \( \left\{ {a}_{n}\right\} \) be a Cauchy sequence in \( \mathbb{Q} \) . Then the set \( \left\{ {a}_{n}\right\} \) is bounded. There exists thus \( M \in {\mathbb{Z}}_{ \geq 0} \) such that \( {\left| {p}^{M}{a}_{n}\right| }_{p} \leq 1 \) for all \( n \) .\n\nClaim: If \( a \in \mathbb{Q},{\left| a\right| }... | Yes |
Lemma 2.1.10. \( {\mathbb{Z}}_{p} = \left\{ {x \in {\mathbb{Q}}_{p}\left| \right| x{ \mid }_{p} \leq 1}\right\} \) . | Proof. Let \( x \in {\mathbb{Q}}_{p} \) such that \( {\left| x\right| }_{p} \leq 1 \) . By the claim in the proof of Proposition 2.1.9, we can find a Cauchy sequence in \( \mathbb{Z} \) that converges to \( x \), hence \( x \in {\mathbb{Z}}_{p} \) . | Yes |
Lemma 2.1.12. Every element \( x \) in \( {\mathbb{Q}}_{p} \) can be written uniquely of the form \( x = \mathop{\sum }\limits_{{n \gg - \infty }}{a}_{n}{p}^{n} \) , with \( {a}_{n} \in \{ 0,\cdots, p - 1\} \) . | Proof. For the uniqueness, if \( \mathop{\sum }\limits_{{n \gg - \infty }}{a}_{n}{p}^{n} = \mathop{\sum }\limits_{{n \gg - \infty }}{b}_{n}{p}^{n} \), letting \( m \) be the minimal integer such that \( {b}_{m} \neq {a}_{m} \), we see \( 0 = \mathop{\sum }\limits_{{i = m}}^{{+\infty }}\left( {{b}_{i} - {a}_{i}}\right) ... | Yes |
Proposition 2.2.3. Let \( \left( {K,\left| \cdot \right| }\right) \) be a valuation field. Then there exists a unique valuation field \( \left( {\widehat{K},{\left| \cdot \right| }_{\widehat{K}}}\right) \) such that\n\n1. \( K \) is a subfield of \( \widehat{K} \), and the restriction of \( {\left. \left| \cdot \right|... | Proof. Define \( \widehat{K} \) to be the completion of \( k \) via \( \left| \cdot \right| \) . Let \( \left( {a}_{n}\right) ,\left( {b}_{n}\right) \) be two Cauchy sequences in \( K \), one can check\n\n- \( \left( {{a}_{n}{b}_{n}}\right) ,\left( {{a}_{n} + {b}_{n}}\right) \) are Cauchy sequences in \( K \),\n\n- if ... | Yes |
Lemma 2.3.1. \( \left| {x + y}\right| = \max \{ \left| x\right| ,\left| y\right| \} \) if \( \left| x\right| \neq \left| y\right| \) | Proof. Suppose \( \left| x\right| > \left| y\right| \), then \( \left| x\right| \leq \max \{ \left| {x + y}\right| ,\left| {-y}\right| \} \) . Note \( \left| {-y}\right| = \left| {-1}\right| \left| y\right| = \left| y\right| \) \( \left( {\left| {1 \cdot x}\right| = \left| x\right| = \left| 1\right| \left| x\right| \Ri... | No |
Lemma 2.3.2. Let \( r > 0 \), then \( \{ x \in K\left| \right| x \mid \leq r\} \) is an open subset of \( K \) . | Proof. For any \( a \in \{ x \in K\left| \right| x \mid \leq r\} ,0 < s < r \), we have \( \{ x \in K\left| \right| x - a \mid < s\} \subset \{ x \in \) \( K \mid \left| x\right| \leq r\} \) . | Yes |
Lemma 2.3.3. \( {\mathcal{O}}_{K} \) is a subring of \( K \) . | Proof. Let \( x, y \in {\mathcal{O}}_{K} \), then \( \left| {xy}\right| \leq 1 \), and \( \left| {x \pm y}\right| \leq 1 \) . | No |
Proposition 2.3.4. Two non-trivial non-archimedean norms \( {\left| \cdot \right| }_{1} \) and \( {\left| \cdot \right| }_{2} \) on a field \( K \) are equivalent if and only if their valuation rings are the same. | Proof. \ | No |
Proposition 2.3.8. Let \( \left( {K,\left| \cdot \right| }\right) \) be a non-archimedean valuation field.\n\n(1) \( {\mathcal{O}}_{K} \) is integrally closed and is a local ring with maximal ideal \( {\mathfrak{m}}_{K} \mathrel{\text{:=}} \{ x \in K\left| \right| x \mid < \) \( 1\} \) (a ring is called local if it has... | Proof. (1) Let \( x \in K \) and suppose \( {x}^{n} + {a}_{n - 1}{x}^{n - 1} + \cdots + {a}_{0} = 0,{a}_{i} \in {\mathcal{O}}_{K} \) . If \( \left| x\right| > 1 \), then \( \left| {x}^{n}\right| > \left| {{a}_{i}{x}^{i}}\right| \) for all \( i = 0,\cdots, n - 1 \) . So \( 0 = \left| {{x}^{n} + \cdots + {a}_{0}}\right| ... | Yes |
Proposition 2.3.10. For a domain \( R, R \) is a discrete valuation ring if and only if \( R \) is a local Dedekind domain. | Proof. The \ | No |
Lemma 2.3.11. For all \( x \in K, x \) can be uniquely written as \( x = \mathop{\sum }\limits_{{n \gg - \infty }}{a}_{n}{\pi }^{n},{a}_{n} \in S \) . | Proof. It suffices to show for \( x \in {\mathcal{O}}_{K}, x \) can be uniquely written as \( \mathop{\sum }\limits_{{n = 0}}^{{+\infty }}{a}_{n}{\pi }^{n} \) . Let \( {a}_{0} \in S \) be the unique element such that \( x - {a}_{0} \in \pi {\mathcal{O}}_{K} \) . We use induction to construct \( \left\{ {a}_{i}\right\} ... | Yes |
Lemma 2.3.12. For any \( x \in {\mathcal{O}}_{K},{\left( 1 + \pi x\right) }^{{p}^{n}} \in 1 + {\pi }^{n + 1}{\mathcal{O}}_{K} \) . | Proof. We have \( {\left( 1 + {\pi }^{n}x\right) }^{p} = \mathop{\sum }\limits_{{i = 0}}^{p}\left( \begin{matrix} p \\ i \end{matrix}\right) {\left( p{i}^{n}x\right) }^{i} \) . Since \( p \in \pi {\mathcal{O}}_{K} \), we see \( {\left( 1 + {\pi }^{n}x\right) }^{p} \in \) \( 1 + {\pi }^{n + 1}{\mathcal{O}}_{K} \) . The ... | No |
Proposition 2.3.13. The sequence \( {\widetilde{a}}^{{q}^{n}} \) converges. Let \( \left\lbrack a\right\rbrack \mathrel{\text{:=}} \mathop{\lim }\limits_{n}{\widetilde{a}}^{{q}^{n}} \), then \( \left\lbrack a\right\rbrack \equiv a \) \( \left( {\;\operatorname{mod}\;{\mathfrak{m}}_{K}}\right) \) . The map \( k \rightar... | Proof. We have \( {\widetilde{a}}^{{q}^{m}} - {\widetilde{a}}^{{q}^{n}} = {\widetilde{a}}^{{q}^{n}}\left( {{\widetilde{a}}^{{q}^{m} - {q}^{n}} - 1}\right) = {\widetilde{a}}^{{q}^{n}}\left( {{\left( {\widetilde{a}}^{{q}^{m - n} - 1}\right) }^{{q}^{m}} - 1}\right) \) . We have \( {\widetilde{a}}^{{q}^{m - n} - 1} \equiv ... | Yes |
Proposition 2.3.15. (1) \( { \cap }_{n}{U}_{K}^{n} = 1 \) . | Proof. (1) (2) are clear. | No |
Theorem 2.4.1 (Hensel’s Lemma). Let \( \left( {K,\left| \cdot \right| }\right) \) be a complete non-archimedean field, \( f\left( x\right) \in {\mathcal{O}}_{K}\left\lbrack x\right\rbrack \) such that \( 0 \neq \overline{f\left( x\right) } \in k\left\lbrack x\right\rbrack \) (such polynomial \( f\left( x\right) \) is c... | Proof. Let \( r \mathrel{\text{:=}} \deg \bar{u}\left( x\right), s \mathrel{\text{:=}} \deg f - r \) (thus \( s \geq \deg \bar{v}\left( x\right) \) ). Let \( {u}_{0}\left( x\right) ,{v}_{0}\left( x\right) \in {\mathcal{O}}_{K}\left\lbrack x\right\rbrack \) such that \( \overline{{u}_{0}\left( x\right) } \equiv \bar{u}\... | Yes |
Corollary 2.4.2. Let \( f\left( x\right) \in {\mathcal{O}}_{K}\left\lbrack x\right\rbrack ,{\alpha }_{0} \in {\mathcal{O}}_{K} \) such that \( f\left( {\alpha }_{0}\right) \equiv 0\left( {\;\operatorname{mod}\;{\mathfrak{m}}_{K}}\right) \) and \( {f}^{\prime }\left( {\alpha }_{0}\right) \neq 0\left( {\;\operatorname{mo... | Proof. Exercise. | No |
Corollary 2.4.3. Let \( f\left( x\right) = \mathop{\sum }\limits_{{i = 0}}^{n}{a}_{i}{x}^{i} \in K\left\lbrack x\right\rbrack \) and suppose \( f\left( x\right) \) is irreducible. Then \( \max \left\{ \left| {a}_{i}\right| \right\} = \max \left\{ {\left| {a}_{0}\right| ,\left| {a}_{k}\right| }\right\} \) | Proof. Exercise. | No |
Lemma 2.4.6. Let \( V \) be a finite dimensional vector space over \( \left( {K,\left| \cdot \right| }\right) \) . Then any two norms on \( V \) are equivalent and \( V \) is complete. | Proof. Let \( {e}_{1},\cdots ,{e}_{n} \) be a basis of \( V \) over \( K \) . For \( x = \mathop{\sum }\limits_{{i = 1}}^{n}{a}_{i}{e}_{i} \), put \( \parallel x\parallel \mathrel{\text{:=}} \mathop{\max }\limits_{{1 \leq i \leq n}}\left\{ \left| {a}_{i}\right| \right\} \) . This defines a norm on \( V \), and it is ea... | Yes |
Lemma 2.4.8 (Krasner’s lemma). Let \( \left( {K,\left| \cdot \right| }\right) \) be a complete non-archimedean valuation field. Let \( \alpha ,\beta \in \bar{K} \) such that \( \left| {\alpha - \beta }\right| < \left| {\alpha - {\alpha }^{\prime }}\right| \) for all Galois conjugate \( {\alpha }^{\prime } \) of \( \alp... | Proof. It suffices to show that for any \( \sigma : K\left( {\alpha ,\beta }\right) \hookrightarrow \bar{K} \), if \( \sigma \left( \beta \right) = \beta \) then \( \sigma \left( \alpha \right) = \alpha \) . However, we have \( \left| {\sigma \left( \alpha \right) - \alpha }\right| = \left| {\sigma \left( \alpha \right... | Yes |
Corollary 2.4.9. Let \( f\left( x\right) = {x}^{n} + \cdots + {a}_{0} \in {\mathcal{O}}_{K}\left\lbrack x\right\rbrack \) be an irreducible monic polynomial of degree \( n \) . Put \( {d}_{0} \mathrel{\text{:=}} \mathop{\min }\limits_{{\alpha \neq {\alpha }^{\prime }}}\left\{ \left| {\alpha - {\alpha }^{\prime }}\right... | Proof. Let \( \beta \) be a root of \( g\left( x\right) \) . Then \( \left| {\mathop{\prod }\limits_{{\alpha }^{\prime }}\left( {\beta - {\alpha }^{\prime }}\right) }\right| = \left| {f\left( \beta \right) }\right| = \left| {f\left( \beta \right) - g\left( \beta \right) }\right| \leq \mathop{\max }\limits_{{0 \leq i \l... | Yes |
Corollary 2.4.10. The algebraic closure \( \left( {\overline{{\mathbb{Q}}_{p}},\left| \cdot \right| }\right) \) is not complete. | Proof. Let \( {\left\{ {\alpha }_{i}\right\} }_{i = 1}^{\infty } \) be a set of elements in \( \overline{{\mathbb{Q}}_{p}} \) that are linearly independent over \( {\mathbb{Q}}_{p} \) . It is easy to inductively find \( {c}_{i} \in {\mathbb{Q}}_{p}^{ \times } \) such that \( \left| {{c}_{n}{\alpha }_{n}}\right| \righta... | Yes |
Corollary 2.4.11. \( {\mathbb{C}}_{p} \) is algebraic closed. | Proof. Let \( p\left( x\right) = {x}^{d} + \cdots + {a}_{0} \in {\mathbb{C}}_{p}\left\lbrack x\right\rbrack, d > 1 \) . It suffices to show \( p\left( x\right) \) has a root in \( {\mathbb{C}}_{p} \) . We can and do assume \( {a}_{i} \in {\mathcal{O}}_{{\mathbb{C}}_{p}} \) (e.g. by replacing \( x \) by \( x/{p}^{k} \) ... | Yes |
Corollary 2.4.12. Let \( K \) be a finite extension of \( {\mathbb{Q}}_{p} \) . Then there exists a (irreducible) polynomial \( f\left( x\right) \in \mathbb{Q}\left\lbrack x\right\rbrack \) such that \( K \cong {\mathbb{Q}}_{p}\left\lbrack x\right\rbrack /f\left( x\right) \) . | Proof. Exercise. | No |
Lemma 2.5.1. \( {\mathcal{O}}_{L} \) is a finite free \( {\mathcal{O}}_{K} \) -module of rank \( \left\lbrack {L : K}\right\rbrack \) . | Proof. Let \( {b}_{i} \) in \( {\mathcal{O}}_{L} \) such that \( \left\{ \overline{{b}_{i}}\right\} \) form a basis of \( {\mathcal{O}}_{L}/{\varpi }_{K} \) over \( {\mathcal{O}}_{K}/{\varpi }_{K} = k \) .\n\nWe show first \( \left\{ {b}_{i}\right\} \) are linearly independent of \( K \) . Indeed, if not, suppose \( \m... | Yes |
Proposition 2.5.3. \( {fe} = \left\lbrack {L : K}\right\rbrack = \left\lbrack {{\mathcal{O}}_{L} : {\mathcal{O}}_{K}}\right\rbrack \) . | Proof. We know \( {\mathcal{O}}_{L}/{\pi }_{K} = {\mathcal{O}}_{L}/{\pi }_{L}^{e} \) is a \( k \) -vector space of dimension \( \left\lbrack {{\mathcal{O}}_{L} : {\mathcal{O}}_{K}}\right\rbrack \) . And \( {\mathcal{O}}_{L}/{\pi }_{L}^{e} \) is isomorphic to a successive extension of \( {\pi }_{L}^{i}{\mathcal{O}}_{L}/... | Yes |
Lemma 2.5.4. Let \( {\alpha }_{1},\cdots ,{\alpha }_{f} \in {\mathcal{O}}_{L} \) such that \( \overline{{\alpha }_{i}} \in {k}_{L} \) form a basis of \( {k}_{L} \) over \( k \) . Then \( \left\{ {{\alpha }_{i}{\pi }_{L}^{j} \mid 1 \leq i \leq f,0 \leq j \leq e - 1}\right\} \) form a basis of \( {\mathcal{O}}_{L} \) ove... | Proof. Using dévissage \( \left( {0 \rightarrow {\mathcal{O}}_{L}/{\pi }_{L}^{j - 1}\xrightarrow[]{{\pi }_{L}}{\mathcal{O}}_{L}/{\pi }_{L}^{j} \rightarrow {k}_{L} \rightarrow 0}\right) \) and an induction argument, it is easy to see \( \left\{ {{\alpha }_{i}{\pi }_{L}^{j} \mid 1 \leq i \leq f,0 \leq j \leq e - 1}\right... | No |
Proposition 2.5.5. Suppose \( {k}_{L}/k \) is separable, then there exists \( \alpha \in {\mathcal{O}}_{L} \) such that \( {\mathcal{O}}_{L} = \) \( {\mathcal{O}}_{K}\left\lbrack \alpha \right\rbrack \) | Proof. Let \( \beta \in {\mathcal{O}}_{L} \) such that \( {k}_{L} = k\left( \bar{\beta }\right) \), then \( {\mathcal{O}}_{L} \) is generated by \( {\left\{ {\beta }^{i}{\pi }_{L}^{j}\right\} }_{\begin{matrix} {0 \leq i \leq f - 1} \\ {0 \leq j \leq e - 1} \end{matrix}} \). Claim: There exists \( \beta \in {\mathcal{O}... | Yes |
Lemma 2.6.1. We have \( {N}_{L/K}\left( {\mathfrak{m}}_{L}\right) = {\pi }_{K}^{f}{\mathcal{O}}_{K} \) . | Proof. Let \( {L}_{0} \) be the maximal unramified subextension of \( L \) over \( K \) . Then \( {N}_{L/K}\left( {\pi }_{L}\right) = \) \( {N}_{{L}_{0}/K} \circ {N}_{L/{L}_{0}}\left( {\pi }_{L}\right) \) . Since \( L \) is totally ramified over \( {L}_{0} \), we deduce by Exercise 2.5.6 that \( {N}_{L/{L}_{0}}\left( {... | No |
Lemma 2.6.2. Let \( {e}_{1},\cdots ,{e}_{d} \) be a basis of \( {\mathcal{O}}_{L} \) over \( {\mathcal{O}}_{K},\mathfrak{a} \subset {\mathcal{O}}_{L},{\alpha }_{1},\cdots ,{\alpha }_{d} \) be a basis of \( \mathfrak{a} \) over \( {\mathcal{O}}_{K} \) . Let \( A \in {M}_{d}\left( K\right) \) such that \( \left( {{\alpha... | Proof. The both sides of the equation is multiplicative, hence we reduce to the case where \( \mathfrak{a} = {\pi }_{L}{\mathcal{O}}_{L} \) . However, if \( {\pi }_{L}\left( {{e}_{1},\cdots ,{e}_{d}}\right) = \left( {{e}_{1},\cdots ,{e}_{d}}\right) A \), then by definition \( {N}_{L/K}\left( {\pi }_{L}\right) = \) \( \... | Yes |
Proposition 2.6.3. Let \( \mathfrak{a} \) (resp. \( \mathfrak{b} \) ) be a fractional ideal of \( K \) (resp. \( L \) ), then \( {\operatorname{Tr}}_{L/K}\left( \mathfrak{b}\right) \subset \mathfrak{a} \) if and only if \( \mathfrak{b} \subset \mathfrak{a}{\mathcal{D}}_{L/K}^{-1} \) . | Proof. We have\n\n\[ \n{\operatorname{Tr}}_{L/K}\left( \mathfrak{b}\right) \subset \mathfrak{a} \Leftrightarrow {\operatorname{Tr}}_{L/K}\left( {{\mathfrak{a}}^{-1}\mathfrak{b}}\right) \subset {\mathcal{O}}_{K} \Leftrightarrow {\mathfrak{a}}^{-1}\mathfrak{b} \subset {\mathcal{D}}_{L/K}^{-1} \Leftrightarrow \mathfrak{b}... | Yes |
Corollary 2.6.4. Let \( M/L/K \) be separable extensions of finite degrees. Then \( {\mathcal{D}}_{M/K} = \) \( {\mathcal{D}}_{M/L}{\mathcal{D}}_{L/K} \) | Proof. We have\n\n\[ \mathfrak{a} \subset {\mathcal{D}}_{M/K}^{-1} \Leftrightarrow {\operatorname{Tr}}_{L/K} \circ {\operatorname{Tr}}_{M/L}\left( \mathfrak{a}\right) = {\operatorname{Tr}}_{M/K}\left( \mathfrak{a}\right) \subset {\mathcal{O}}_{K} \Leftrightarrow {\operatorname{Tr}}_{M/L}\left( \mathfrak{a}\right) \subs... | Yes |
Proposition 2.6.5. Suppose there exists \( \alpha \in {\mathcal{O}}_{L} \) such that \( {\mathcal{O}}_{L} = {\mathcal{O}}_{K}\left\lbrack \alpha \right\rbrack \) . Let \( f\left( x\right) \in {\mathcal{O}}_{K}\left\lbrack x\right\rbrack \) be the minimal polynomial of \( \alpha \) over \( K \), then \( {\mathcal{D}}_{L... | Proof. Let \( \left\{ {e}_{i}\right\} \) be the dual basis of \( \left\{ {1,\cdots ,{\alpha }^{n - 1}}\right\} \) with respect to \( \langle \) , \( \rangle ,{thenwehave} \n\n\[ \n\left( {1,\cdots ,{\alpha }^{n - 1}}\right) = \left( {{e}_{1},\cdots ,{e}_{n - 1}}\right) {\left( {\operatorname{Tr}}_{L/K}\left( {\alpha }^... | Yes |
Corollary 2.6.6. (1) Assume \( L/K \) is totally ramified of degree \( e \), then \( {\mathcal{D}}_{L/K} \subset {\mathfrak{m}}_{L}^{e - 1} \) . Moreover the equality holds if and only if \( e \) is prime to \( \operatorname{char}k \) . | Proof. (1) Since \( L/K \) is totally ramified, \( {\mathcal{O}}_{L} = {\mathcal{O}}_{K}\left\lbrack {\pi }_{L}\right\rbrack \) . Recall the minimal polynomial \( f\left( x\right) = {x}^{e} + \cdots + {a}_{0} \) of \( {\pi }_{L} \) over \( K \) is Eisenstein. We have \( {f}^{\prime }\left( {\pi }_{L}\right) = e{\pi }_{... | Yes |
Lemma 2.7.1. Let \( {\pi }_{L} \) be a uniformizer of \( {\mathcal{O}}_{L} \), and \( \sigma \in \operatorname{Gal}\left( {L/K}\right) \) . Then the followings are equivalent:\n\n(1) \( {v}_{L}\left( {\sigma \left( x\right) - x}\right) \geq n + 1,\forall x \in {\mathcal{O}}_{L} \),\n\n(2) \( {v}_{L}\left( {\sigma \left... | Proof. \( \left( 1\right) \Rightarrow \left( 2\right) \) is clear. \( \left( 2\right) \Rightarrow \left( 1\right) \) follows from the fact \( {\mathcal{O}}_{L} = {\mathcal{O}}_{K}\left\lbrack {\pi }_{L}\right\rbrack \) . | No |
Corollary 2.7.2. We have\n\n\[ \n{G}_{n} = \left\{ {g \in \operatorname{Gal}\left( {L/K}\right) \mid \frac{g\left( {\pi }_{L}\right) }{{\pi }_{L}} \in {U}_{L}^{n}}\right\} .\n\] | We define a map\n\n\[ \n{G}_{n} \rightarrow {U}_{L}^{n}/{U}_{L}^{n + 1}, g \mapsto \frac{g\left( {\pi }_{L}\right) }{{\pi }_{L}}.\n\] | No |
Lemma 2.7.3. The map (2.2) is a group homomorphism and factors through an injection \( {G}_{n}/{G}_{n + 1} \hookrightarrow {U}_{L}^{n}/{U}_{L}^{n + 1} \) . Moreover, the map is independent of the choice of \( {\pi }_{L} \) . | Proof. Let \( g, h \in {G}_{n} \), we have\n\n\[ \frac{{gh}\left( {\pi }_{L}\right) }{{\pi }_{L}} = \frac{g\left( {h\left( {\pi }_{L}\right) }\right) }{h\left( {\pi }_{L}\right) }\frac{h\left( {\pi }_{L}\right) }{{\pi }_{L}}. \]\n\n(2.3)\n\nFor another uniformizer \( {\pi }_{L}^{\prime } \) of \( L \), let \( s \in {\m... | Yes |
Proposition 2.7.8. A profinite group is totally disconnected, compact and Hausdorff. | Proof. Let \( G \cong \mathop{\lim }\limits_{{i \in I}}{G}_{i} \) . Since \( {G}_{i} \) is compact, so is \( \mathop{\prod }\limits_{i}{G}_{i} \) . We show \( G \) is closed in \( \mathop{\prod }\limits_{i}{G}_{i} \) . Let \( x = \left( {x}_{i}\right) \in \mathop{\prod }\limits_{i}{G}_{i} \smallsetminus \mathop{\lim }\... | Yes |
Lemma 2.7.11. Let \( L/K \) be a Galois extension (not necessarily finite), \( G \mathrel{\text{:=}} \operatorname{Gal}\left( {L/K}\right) \) . If \( F/K \) is a normal subextension of \( L/K \), then \( H = \operatorname{Gal}\left( {L/F}\right) \) is a nomral subgroup of \( G \) and \( F = {L}^{H} \) . And we have an ... | Proof. Since \( F/K \) is normal, we have a natural morphism\n\n\[ \operatorname{Gal}\left( {L/K}\right) \rightarrow \operatorname{Gal}\left( {F/K}\right) \]\n\n(2.4)\n\nand let \( H \) be its kernel.\n\nClaim: The map (2.4) is surjective.\n\nLet \( {F}^{\prime } \) be a normal extension over \( K \) such that \( F \su... | Yes |
Theorem 2.7.12. (1) We have \( \operatorname{Gal}\left( {L/K}\right) \cong {\underline{\lim }}_{F}\operatorname{Gal}\left( {F/K}\right) \) where \( F \) runs though finite normal extensions of \( K \) in \( L \) . | Proof. By the above lemma, we have a natural map \( \operatorname{Gal}\left( {L/K}\right) \rightarrow {\underline{\lim }}_{F}\operatorname{Gal}\left( {F/K}\right) \) such that for each \( F \), the map \( \operatorname{Gal}\left( {L/K}\right) \rightarrow \operatorname{Gal}\left( {F/K}\right) \) is surjective. We deduce... | Yes |
Lemma 2.7.15. The map \( R \rightarrow \mathop{\lim }\limits_{{x \mapsto {x}^{p}}}{\mathcal{O}}_{{\mathbb{C}}_{p}}/p,\left( {x}^{\left( n\right) }\right) \mapsto \left( \overline{{x}^{\left( n\right) }}\right) \) is bijective. | Proof. We construct an inverse of the map. Let \( \left( {x}_{n}\right) \in \mathop{\lim }\limits_{{x \rightarrow {x}^{p}}}{\mathcal{O}}_{{\mathbb{C}}_{p}}/p \), and for each \( n \) , let \( {\widetilde{x}}_{n} \) be a lifting of \( {x}_{n} \) in \( {\mathcal{O}}_{{\mathbb{C}}_{p}} \) . Put \( {x}^{\left( n\right) } \... | No |
Theorem 2.8.1. (1) Any two of the absolute valuations \( {\left| \cdot \right| }_{v} \) for places \( v \) of \( K \) are not equivalent to each other. | Proof. (1) If \( {v}_{1} \) is an archimedean norm, and \( {v}_{2} \) is a non-archimedean norm. Then there exist non-zero \( x, y \in K \) such that \( {\left| x + y\right| }_{{v}_{1}} > \max \left\{ {{\left| x\right| }_{{v}_{1}},{\left| y\right| }_{{v}_{1}}}\right\} \), hence \( {\left| \left( x + y\right) /x\right| ... | Yes |
Proposition 2.8.2. Let \( S = \left\{ {{v}_{1},\cdots ,{v}_{r}}\right\} \) be a finite set of places of \( K \) . Then the image of the diagonal map \( K \hookrightarrow \mathop{\prod }\limits_{{{v}_{i} \in S}}{K}_{{v}_{i}} \) is dense. | Proof. We need to show for any \( \left( {z}_{i}\right) \in \prod {K}_{{v}_{i}} \), and any \( \epsilon > 0 \), there exists \( z \in K \) such that \( {\left| z - {z}_{i}\right| }_{{v}_{i}} \leq \epsilon \) . Since \( K \) is dense in \( {K}_{{v}_{i}} \) for all \( {v}_{i} \), we can and do assume \( {z}_{i} \in K \) ... | Yes |
Lemma 2.8.4. We have \( e\left( {{\mathfrak{P}}_{w}/{\mathfrak{p}}_{v}}\right) = e\left( {{L}_{w}/{K}_{v}}\right) = : e\left( {w/v}\right) \) and \( f\left( {{\mathfrak{P}}_{w}/{\mathfrak{p}}_{v}}\right) = f\left( {{L}_{w}/{K}_{v}}\right) = : \) \( f\left( {w/v}\right) \) . | Proof. Let \( {\pi }_{v} \in {\mathfrak{p}}_{v} \smallsetminus {\mathfrak{p}}_{v}^{2},{\pi }_{w} \in {\mathfrak{P}}_{w} \smallsetminus {\mathfrak{P}}_{w}^{2} \) . Then \( {\pi }_{v} \) (resp. \( {\pi }_{w} \) ) is a uniformizer of \( {\mathcal{O}}_{{K}_{v}} \) (resp. \( {\mathcal{O}}_{{L}_{w}} \) ). By prime factorizat... | Yes |
Proposition 2.8.5. The morphism \( \iota : {K}_{v}{ \otimes }_{K}L \rightarrow \mathop{\prod }\limits_{{w \mid v}}{L}_{w}, x \otimes y \mapsto {\left( xy\right) }_{w} \) is an isomorphism of \( {K}_{v} \) -algebras. | Proof. It is clear \( \operatorname{Im}\iota \) is a finite dimensional \( {K}_{v} \) -vector subspace, hence is complete and closed in \( \mathop{\prod }\limits_{{w \mid v}}{L}_{w} \) . Consider the map \( L \rightarrow \mathop{\prod }\limits_{{w \mid v}}{L}_{w} \), we know this map has dense image. Hence \( \operator... | Yes |
Proposition 2.8.8. Let \( {e}_{1},\cdots ,{e}_{d} \) be a basis of \( L/K \) . For a non-archimedean place \( v \) of \( K \), let \( {M}_{v} \) be the free \( {\mathcal{O}}_{{K}_{v}} \) -submodule of \( {K}_{v}{ \otimes }_{K}L \) generated by \( {e}_{1},\cdots ,{e}_{d} \) . Then for all but finitely many places \( v \... | Proof. Let \( \alpha \in {K}^{ \times } \) such that \( \alpha {e}_{1},\cdots ,\alpha {e}_{d} \in {\mathcal{O}}_{L} \) . If \( {\left| \alpha \right| }_{v} = 1 \) (note this hods for all but finitely many \( v \) ), then \( {M}_{v} \cong {\mathcal{O}}_{{K}_{v}}\left( {\alpha {e}_{1}}\right) \oplus \cdots \oplus {\mathc... | Yes |
Proposition 2.8.11. Suppose \( L \) is Galois over \( K \) . Let \( w \) be a finite place of \( L \) and \( {\mathfrak{P}}_{w} \) be the associated prime ideal of \( {\mathcal{O}}_{L} \) . Then the natural restriction map \( j : \operatorname{Gal}\left( {{L}_{w}/{K}_{v}}\right) \rightarrow \) \( \operatorname{Gal}\lef... | Proof. Since \( L \) is dense in \( {L}_{w} \), we see \( j \) is injective (noting \( \sigma \in \operatorname{Gal}\left( {{L}_{w}/{K}_{v}}\right) \) acts continuously on \( {L}_{w} \) ). Let \( \sigma \in \operatorname{Gal}\left( {{L}_{w}/{K}_{v}}\right) \), then \( \sigma \left( {\mathfrak{m}}_{w}\right) = {\mathfra... | No |
Proposition 3.1.6. \( \mathop{\prod }\limits_{{j \in J}}^{\prime }{G}_{j} \) is a locally compact group. | Proof. Exercise. | No |
Lemma 3.1.8. The diagonal map \( K \hookrightarrow \mathop{\prod }\limits_{{v \in J}}{K}_{v} \) factors though an injection \( K \hookrightarrow {\mathbb{A}}_{K} \) . | Proof. For any \( x \in K \), there are only finitely many \( v \in J \smallsetminus {J}_{\infty } \) such that \( x \notin {\mathcal{O}}_{{K}_{v}} \) . The lemma follows. | No |
Proposition 3.1.9. We have \( K + {\mathbb{A}}_{{J}_{\infty }} = {\mathbb{A}}_{K} \) . | Proof. We need to show that for all \( {\left( {x}_{v}\right) }_{v \in J} \in {\mathbb{A}}_{K} \), there exists \( x \in K \) such that \( \left( {x - {x}_{v}}\right) \in \) \( {\mathbb{A}}_{{J}_{\infty }} \) . Let \( m \in {\mathcal{O}}_{K} \smallsetminus \{ 0\} \) such that \( m{x}_{v} \in {\mathcal{O}}_{{K}_{v}} \) ... | Yes |
Proposition 3.1.10. Let \( {e}_{1},\cdots ,{e}_{d} \) be a basis of \( L \) over \( K \) . The morphism\n\n\[ f : {\mathbb{A}}_{K}{e}_{1} \oplus \cdots \oplus {\mathbb{A}}_{k}{e}_{d} \rightarrow {\mathbb{A}}_{L},\sum {a}_{i}{e}_{i} \mapsto \sum \iota \left( {a}_{i}\right) {e}_{i} \]\n\n is well defined and is an isomor... | Proof. Recall for all places \( v \) of \( K \), the morphism \( {\iota }_{v} : {K}_{v}{e}_{1} \oplus \cdots {K}_{v}{e}_{d} \rightarrow \mathop{\prod }\limits_{{w \mid v}}{L}_{w} \) is an isomorphism. Moreover, for all but finitely many finite places \( v,{\iota }_{v} \) induces an isomorphism \( {\mathcal{O}}_{{K}_{v}... | Yes |
Theorem 3.1.11. \( K \) is a discrete, cocompact subgroup of \( {\mathbb{A}}_{K} \) (i.e. \( {\mathbb{A}}_{K}/K \) is compact). | Proof. Let \( {e}_{1},\cdots ,{e}_{d} \) be a basis of \( K \) over \( \mathbb{Q} \) . We have by the above proposition\n\n\n\nIt thus suffices to show the statement for \( K = \mathbb{Q} \) .\n\nLet \( U \mathrel{\tex... | Yes |
Lemma 3.2.1. As a set, we have \( {I}_{K} = {\mathbb{A}}_{K}^{ \times } \) . | Proof. For \( \left( {x}_{v}\right) \in {I}_{K} \subset {\mathbb{A}}_{K} \), we have \( {x}_{v}^{-1} \in {\mathcal{O}}_{{K}_{v}} \) for all but finitely many places \( v \) . Hence \( \left( {x}_{v}^{-1}\right) \in {\mathbb{A}}_{K} \), so \( {I}_{K} \subset {\mathbb{A}}_{K}^{ \times } \) . Conversely, if \( \left( {x}_... | Yes |
Lemma 3.2.2. If we equip \( {\mathbb{A}}_{K}^{ \times } \) with the topology induced from \( {\mathbb{A}}_{K} \), then the morphism \( \iota : {\mathbb{A}}_{K}^{ \times } \rightarrow {\mathbb{A}}_{K}^{ \times }, x \mapsto {x}^{-1} \) is not continuous. | Proof. If \( \iota \) is continuous, then it is a homemorphism. For the induced topology, we see \( U \mathrel{\text{:=}} {\mathbb{A}}_{K}^{ \times } \cap \left( {{U}_{S} \times \mathop{\prod }\limits_{{v \notin S}}\left( {{\mathcal{O}}_{{K}_{v}}\smallsetminus \{ 0\} }\right) }\right) \), where \( S \) runs through fin... | Yes |
Theorem 3.2.5 (Product formula). The natural embedding \( {K}^{ \times } \hookrightarrow {I}_{K} \) factors through \( {I}_{K}^{1} \) . | Proof. By the discussion above the theorem, for \( \left( {x}_{v}\right) \in {I}_{K},{\left| {\left( {x}_{v}\right) }_{v}\right| }_{K} = {\left| {\left( \mathop{\prod }\limits_{{v \mid p}}{N}_{{K}_{v}/{\mathbb{Q}}_{p}}\left( {x}_{v}\right) \right) }_{p}\right| }_{\mathbb{Q}} \) . Thus for \( x \in K \), we have \( {\le... | Yes |
Example 3.3.4. (1) Let \( K \) be a number field, \( v \) be a finite place of \( K \), and \( G = \left( {{K}_{v}, + }\right) \) . Let \( {\mu }_{v} \) be a Haar measure on \( {K}_{v} \) (the existence following from the above theorem). We have \( {\mu }_{v}\left( {\mathcal{O}}_{{K}_{v}}\right) \neq 0 \), since otherw... | Indeed, let \( {\pi }_{v} \) be a uniformizer and suppose \( x \in {\pi }_{v}^{n}{\mathcal{O}}_{{K}_{v}}^{ \times } \) and \( n \geq 0 \) (the case \( n < 0 \) being similar). Let \( \left\{ {{x}_{1},\cdots ,{x}_{{q}_{v}^{n}}}\right\} \) be a set of representatives of \( {\mathcal{O}}_{{K}_{v}}/{\pi }_{v}^{n} \) in \( ... | Yes |
Lemma 3.3.5. Let \( x = \left( {x}_{v}\right) \in {I}_{K},\mu \) be a Haar measure on \( {\mathbb{A}}_{K} \), and let \( E \) be a compact subset in \( {\mathbb{A}}_{K} \) such that \( \mu \left( E\right) \neq 0 \) . Then \( \mu \left( {xE}\right) = {\left| x\right| }_{K}\mu \left( E\right) \) (noting by Exercise 3.1.7... | Proof. Since \( \mu \) is a Haar measure, we deduce \( U \mapsto \mu \left( {xU}\right) \) is also a Haar measure. Actually, by Exercise 3.1.7, the element \( x \) induces a homemorphism \( {\mathbb{A}}_{K} \rightarrow {\mathbb{A}}_{K}, a \mapsto {xa} \) . By Theorem 3.3.3, it suffices to show the statement for a singl... | Yes |
Corollary 3.3.7. The group \( {C}_{K} \) is finite. | Let \( C \mathrel{\text{:=}} \left\{ {\left( {x}_{v}\right) \in {I}_{K} \mid {\left| {x}_{v}\right| }_{v} = 1}\right\} = \mathop{\prod }\limits_{{v \in {J}_{\infty }}}\left\{ {{\left| x\right| }_{v} = 1}\right\} \times \mathop{\prod }\limits_{{v \notin {J}_{\infty }}}{\mathcal{O}}_{{K}_{v}}^{ \times } \) . We see \( C ... | No |
Proposition 3.3.8. We have \( C \cap {K}^{ \times } = \left\{ {x \in {K}^{ \times } \mid {x}^{n} = 1}\right. \) for some \( \left. n\right\} \) . | Proof. Since \( {K}^{ \times } \) is a discrete subgroup of \( {I}_{K} \), we see \( C \cap {K}^{ \times } \) (with the induced topology) is both compact and discrete, and hence is a finite set. So \( C \cap {K}^{ \times } \) is a finite (hence cyclic) subgroup of \( {K}^{ \times } \) . Any element in \( C \cap {K}^{ \... | Yes |
Lemma 3.3.12. Let \( S \) be a finite set of places of \( K \) containing all archimedean places such that \( {\left\{ {\mathfrak{p}}_{v}\right\} }_{v \in S \smallsetminus {J}_{\infty }} \) can generate \( {C}_{K} \) . Then \( {K}^{ \times }{I}_{K, S} = {I}_{K} \) . | Proof. Let \( x = \left( {x}_{v}\right) \in {I}_{K} \), and consider the fractional ideal \( {\mathfrak{a}}_{x} \mathrel{\text{:=}} \mathop{\prod }\limits_{{v \nmid \infty }}{\mathfrak{p}}_{v}^{{\operatorname{ord}}_{v}\left( {x}_{v}\right) } \) . By the assumption on \( S \), there exists \( \alpha \in {K}^{ \times } \... | Yes |
Proposition 3.3.13. We have an isomorphism \( {\mathcal{C}}_{K} \cong {I}_{K}/\overline{{K}^{ \times }{\left( {K}_{\infty }^{ \times }\right) }^{o}} \) (where \( \overline{\left( \cdot \right) } \) denotes the closure). | Proof. It suffices to show \( {I}_{K}/\overline{{K}^{ \times }{\left( {K}_{\infty }^{ \times }\right) }^{o}} \) is profinite. Let \( S \) be as in the above lemma, we have \( {I}_{K, S}/\overline{{\mathcal{O}}_{K, S}^{ \times }{\left( {K}_{\infty }^{ \times }\right) }^{o}}\overset{ \sim }{ \rightarrow }{I}_{K}/\overlin... | No |
Lemma 3.3.16. The natural map \( {I}_{K}\left( \mathfrak{m}\right) \rightarrow {I}_{K} \) induces an isomorphism\n\n\[ {I}_{K}\left( \mathfrak{m}\right) /\left( {{I}_{K}\left( \mathfrak{m}\right) \cap {K}^{ \times }}\right) \overset{ \sim }{ \rightarrow }{I}_{K}/{K}^{ \times } \] | Proof. The injectivity is trivial. For \( \left( {x}_{v}\right) \in {I}_{K} \), since \( K \) is dense in \( \mathop{\prod }\limits_{{v \mid \mathfrak{m}}}{K}_{v} \), there exists \( \alpha \in {K}^{ \times } \) such that \( \alpha \) is arbitrarily close to \( {x}_{v} \) for all \( v \mid \mathfrak{m} \) . In particul... | Yes |
Proposition 3.3.17. The ray class group \( {J}_{K}\left( \mathfrak{m}\right) /{P}_{K}\left( \mathfrak{m}\right) \) is finite. | Proof. Exercise. | No |
Proposition 4.1.1. There are infinitely many primes. | Proof. Suppose there are only finitely many prime numbers \( {p}_{1},\cdots ,{p}_{r} \) . Then we have\n\n\[ \mathop{\sum }\limits_{{n = 1}}^{{{p}_{1}^{N}\cdots {p}_{r}^{N}}}\frac{1}{n} = \mathop{\prod }\limits_{{i = 1}}^{r}\left( {1 + \frac{1}{{p}_{i}} + \cdots + \frac{1}{{p}_{i}^{N}}}\right) . \]\n\nWe then deduce \(... | No |
Proposition 4.1.2. \( \mathop{\sum }\limits_{{n = 1}}^{\infty }\frac{1}{{n}^{s}} \) is absolutely convergent if \( \operatorname{Re}s > 1 \) and \( \zeta \left( s\right) \mathrel{\text{:=}} \mathop{\sum }\limits_{{n = 1}}^{\infty }\frac{1}{{n}^{s}} \) is holomorphic for \( \operatorname{Re}s > 1 \) . | Proof. We have \( \left| \frac{1}{{n}^{s}}\right| = \left| \frac{1}{{n}^{\operatorname{Re}s}}\right| \) and \( \mathop{\sum }\limits_{{n = 1}}^{\infty }\frac{1}{{n}^{a}} \) is absolutely convergent for \( a \in {\mathbb{R}}_{ > 1} \) : \( \mathop{\sum }\limits_{{n = 1}}^{\infty }\frac{1}{{n}^{a}} \sim \) \( {\int }_{1}... | Yes |
Proposition 4.1.4. The product \( \mathop{\prod }\limits_{p}{\left( 1 - \frac{1}{{p}^{s}}\right) }^{-1} \) is absolutely convergent for \( \operatorname{Re}s > 1 \) and \( \mathop{\prod }\limits_{p}{\left( 1 - \frac{1}{{p}^{s}}\right) }^{-1} = \zeta \left( s\right) \) | Proof. We have \( \left| {\zeta \left( s\right) - \mathop{\prod }\limits_{{\mathfrak{p} \leq {\mathfrak{p}}_{N}}}\frac{1}{\left( 1 - \frac{1}{{p}^{s}}\right) }}\right| \leq \mathop{\sum }\limits_{{n > {\mathfrak{p}}_{N}}}\left| \frac{1}{{n}^{s}}\right| \rightarrow 0 \) as \( {\mathfrak{p}}_{N} \rightarrow \infty \) . T... | No |
Theorem 4.1.6. The Riemann zeta function \( \zeta \left( s\right) \) extends (uniquely) to a meromorphic function on \( \mathbb{C} \), holomorphic everywhere except for a simple pole at \( s = 1 \) . We have the so-called functional equation:\n\n\[ \n{\pi }^{-\frac{s}{2}}\Gamma \left( \frac{s}{2}\right) \zeta \left( s\... | Proof of Theorem 4.1.6. We have \( {\pi }^{\frac{s}{2}}\Gamma \left( \frac{s}{2}\right) \frac{1}{{n}^{s}} = {\int }_{0}^{\infty }{t}^{\frac{s}{2} - 1}{e}^{-{n}^{2}{\pi t}}{dt} \) (for Re \( s > 0 \) ). Thus\n\n\[ \n{\pi }^{-\frac{s}{2}}\Gamma \left( \frac{s}{2}\right) \zeta \left( s\right) = {\int }_{0}^{\infty }{t}^{\... | Yes |
Lemma 4.1.7. We have \( w\left( {1/t}\right) = - \frac{1}{2} + \frac{1}{2}{t}^{\frac{1}{2}} + {t}^{\frac{1}{2}}w\left( t\right) \) . | Let \( \theta \left( x\right) \mathrel{\text{:=}} \mathop{\sum }\limits_{{n = - \infty }}^{\infty }{e}^{-{n}^{2}{\pi x}} \), then \( {2w}\left( x\right) = \theta \left( x\right) - 1 \) . It suffices to show \( \theta \left( {1/x}\right) = {x}^{\frac{1}{2}}\theta \left( x\right) \) . | No |
Theorem 4.1.9 (Poisson formula). Let \( f : \mathbb{R} \rightarrow \mathbb{C} \) be a Schwartz function, then\n\n\[ \mathop{\sum }\limits_{{n \in \mathbb{Z}}}f\left( n\right) = \mathop{\sum }\limits_{{n \in \mathbb{Z}}}\widehat{f}\left( n\right) \] | Proof. Let \( F\left( x\right) \mathrel{\text{:=}} \mathop{\sum }\limits_{{n \in \mathbb{Z}}}f\left( {n + x}\right) \) . Note since \( f \) is Schwartz, the series is absolutely convergent and uniformly continuous for compact sets in \( \mathbb{R} \) . Hence \( F\left( x\right) \) is continuous, and \( F\left( {x + 1}\... | Yes |
Lemma 4.2.2. If \( \vartheta \left( x\right) \sim x \), then \( \pi \left( x\right) \sim \frac{x}{\log \left( x\right) } \) . | Proof. It is clear that \( \vartheta \left( x\right) \leq \pi \left( x\right) \log \left( x\right) \) . On the other hand, for any \( \epsilon > 0,\vartheta \left( x\right) \geq \) \( \mathop{\sum }\limits_{{{x}^{1 - \epsilon } < p \leq x}}\log p \geq \left( {1 - \epsilon }\right) \mathop{\sum }\limits_{{{x}^{1 - \epsi... | Yes |
Lemma 4.2.3. We have \( \vartheta \left( x\right) = O\left( x\right) \) . | Proof. We have\n\n\[ \n{2}^{2n} \geq \left( \begin{matrix} {2n} \\ n \end{matrix}\right) \geq \mathop{\prod }\limits_{{n < p \leq {2n}}}p = {e}^{\vartheta \left( {2n}\right) - \vartheta \left( n\right) }.\n\]\n\nHence \( \vartheta \left( {2n}\right) - \vartheta \left( n\right) \leq {2n}\log 2 \), thus \( \vartheta \lef... | Yes |
Lemma 4.2.4. If \( {\int }_{1}^{\infty }\frac{\vartheta \left( x\right) - x}{{x}^{2}}{dx} \) exists, then \( \vartheta \left( x\right) \sim x \) . | Proof. Let \( \lambda > 1 \), and \( x > 0 \) such that \( \vartheta \left( x\right) \geq {\lambda x} \) then\n\n\[ \mathop{\sum }\limits_{x}^{{\lambda x}}\frac{\vartheta \left( t\right) - t}{{t}^{2}}{dt} \geq {\int }_{x}^{\lambda x}\frac{{\lambda x} - t}{{t}^{2}}{dt} = {\int }_{1}^{\lambda }\frac{\lambda - t}{{t}^{2}}... | Yes |
Lemma 4.2.6. We have \( {\int }_{0}^{\infty }{e}^{-\left( {s + 1}\right) t}\vartheta \left( {e}^{t}\right) {dt} = \frac{\Phi \left( {s + 1}\right) }{s + 1} \) (for \( \operatorname{Re}s > 0 \) ). | Proof. We have (for \( \operatorname{Re}s \gg 0 \) )\n\n\[ \Phi \left( s\right) = \mathop{\sum }\limits_{{i = 1}}^{\infty }\left( {\left( {\mathop{\sum }\limits_{{j \leq i}}\log {p}_{j}}\right) \left( {\frac{1}{{p}_{i}^{s}} - \frac{1}{{p}_{i + 1}^{s}}}\right) }\right) = - \mathop{\sum }\limits_{{i = 1}}^{\infty }{\int ... | Yes |
Lemma 4.2.7. We have (for \( \operatorname{Re}s > 1 \) ) \[ - \frac{{\zeta }^{\prime }\left( s\right) }{\zeta \left( s\right) } = \Phi \left( s\right) + \mathop{\sum }\limits_{p}\frac{\log p}{{p}^{s}\left( {{p}^{s} - 1}\right) }.\] | Proof. We have \[ {\left( \mathop{\prod }\limits_{{p \leq {p}_{N}}}{\left( 1 - {p}^{-s}\right) }^{-1}\right) }^{\prime } = \mathop{\sum }\limits_{{p \leq {p}_{N}}}\left( {-\mathop{\prod }\limits_{\substack{{{p}^{\prime } \leq {p}_{N}} \\ {{p}^{\prime } \neq p} }}{\left( 1 - {\left( {p}^{\prime }\right) }^{-s}\right) }^... | Yes |
Proposition 4.2.8. We have \( \zeta \left( s\right) \neq 0 \) for any \( \operatorname{Re}s = 1 \) and \( s \neq 1 \) . | Proof. First by (4.2) (or (4.1)), if \( \zeta \left( s\right) = 0 \), then \( \zeta \left( \bar{s}\right) = 0 \) for \( \operatorname{Re}s > 0 \) . Suppose \( 1 + {\alpha i} \) is a zero of \( \zeta \left( s\right) \) of order \( \lambda \), and \( 1 + {2\alpha i} \) is a zero of \( \zeta \left( s\right) \) of order \(... | Yes |
Lemma 4.3.1. The series \( \mathop{\sum }\limits_{{\mathfrak{a} \subset {\mathcal{O}}_{K}}}\frac{1}{N{\left( \mathfrak{a}\right) }^{s}} \) is absolutely and uniformly convergent on any compact subset of \( \operatorname{Re}s > 1 \), and we have\n\n\[{\zeta }_{K}\left( s\right) = \mathop{\prod }\limits_{\mathfrak{p}}{\l... | Proof. We first show \( \mathop{\prod }\limits_{\mathfrak{p}}{\left( 1 - N{\left( \mathfrak{p}\right) }^{-s}\right) }^{-1} \) is absolutely and uniformly convergent on any compact subset of \( \operatorname{Re}s > 1 \) . It is sufficient show the same convergence for \( \mathop{\sum }\limits_{\mathfrak{p}}\frac{1}{N{\l... | Yes |
Proposition 4.3.2. Let \( f\left( s\right) = \mathop{\sum }\limits_{{n = 1}}^{\infty }\frac{{a}_{n}}{{n}^{s}} \) that absolutely converges for \( \operatorname{Re}s \) sufficiently large. Let \( {S}_{t} = \mathop{\sum }\limits_{{n \leq t}}{a}_{n} \) for \( t > 0 \) . Suppose there exists \( \kappa \in \mathbb{C} \) and... | Proof. Consider \( f\left( s\right) - {\kappa \zeta }\left( s\right) = : \mathop{\sum }\limits_{n}\frac{{b}_{n}}{{n}^{s}} \), and put \( {S}_{t}^{\prime } \mathrel{\text{:=}} \mathop{\sum }\limits_{{n \leq t}}{b}_{n} \) . So \( {S}_{t}^{\prime } = O\left( {t}^{1 - \delta }\right) \) . It is sufficient to show \( \matho... | Yes |
Lemma 4.3.5. Let \( J \) be a fractional ideal such that \( \left\lbrack {J}^{-1}\right\rbrack = \mathcal{C} \) . Then the following map is a bijection\n\n\[ \left\{ {\alpha \in J \mid N\left( \alpha \right) \leq {tN}\left( J\right) }\right\} /{\mathcal{O}}_{K}^{ \times } \rightarrow \left\{ {\mathfrak{a} \subset {\mat... | Proof. The map is clearly well-defined and injective. For \( \mathfrak{a} \subset {\mathcal{O}}_{K} \) such that \( \left\lbrack \mathfrak{a}\right\rbrack = \mathcal{C} \) and \( N\left( \mathfrak{a}\right) \leq t \), there exists \( \alpha \in {K}^{ \times } \) such that \( \mathfrak{a} = \alpha {J}^{-1} \) . Since \(... | Yes |
Suppose \( K/\mathbb{Q} \) is an imaginary quadratic field. Let \( \sigma : K \hookrightarrow \mathbb{C} \) be an embedding. Recall \( \sigma \left( J\right) \) is a lattice in \( \mathbb{C} \cong {\mathbb{R}}^{2} \), and we have \( \operatorname{Vol}\left( {{\mathbb{R}}^{2}/\sigma \left( J\right) }\right) = \frac{1}{2... | Let \( {e}_{1},{e}_{2} \) be a basis of \( \lambda \left( J\right) \), and \( \Omega = \left\{ {{x}_{1}{e}_{1} + {x}_{2}{e}_{2} \mid {x}_{i} \in ( - 1/2,1/2\rbrack }\right\} \) . For \( s > 0 \), denote by \( {m}_{s}^{ + } \) (resp. \( {m}_{s}^{ - } \) ) be the cardinality of the set consisting of \( \alpha \in J \) su... | Yes |
Lemma 4.3.8. Let \( B \) be a bounded region in \( {\mathbb{R}}^{n} \) such that \( \partial B \) is \( \left( {n - 1}\right) \) -Lipschitz parametrizable, and \( \Lambda \subset {\mathbb{R}}^{n} \) be a complete lattice. Then for \( a > 1 \), we have\n\n\[ \n\# \left( {\Lambda \cap {aB}}\right) = \frac{\mu \left( B\ri... | Proof. We give a sketch of the proof. First the map \( x \mapsto {ax} \) induces a bijection between \( \left( {\frac{1}{a}\Lambda }\right) \cap B \rightarrow \Lambda \cap {aB} \) . Let \( {e}_{1},\cdots ,{e}_{n} \) be a basis of \( \Lambda \), and \( \Omega \mathrel{\text{:=}} \left\{ {\mathop{\sum }\limits_{i}{x}_{i}... | Yes |
Lemma 4.3.9. For any \( x \in {X}_{t}^{\text{Log }} \), there exists a unique \( y \in {D}_{t}^{\text{Log }} \) such that \( x - y \in \ell \left( {\mathcal{O}}_{K}^{ \times }\right) \) . | Proof. The lemma follows easily from the fact \( \left\{ {\mathop{\sum }\limits_{{i = 1}}^{{r + s - 1}}{t}_{i}\ell \left( {\varepsilon }_{i}\right) \mid {t}_{i} \in \lbrack 0,1)}\right\} \) is a fundamental mesh of the lattice \( \oplus \mathbb{Z}\ell \left( {\varepsilon }_{i}\right) \) in \( H \) . | No |
Lemma 4.3.10. For any \( x \in {X}_{t}^{ * } \), there exists a unique \( y \in {D}_{t} \) such that \( x{y}^{-1} \in \mathop{\prod }\limits_{{i = 1}}^{{r + s - 1}}{\varepsilon }_{i}^{\mathbb{Z}} \hookrightarrow \) \( {\mathcal{O}}_{K}^{ \times } \) . | Consequently, we have a bijection\n\n\[ \left( {J \cap \left\{ {\alpha \in K \mid N\left( \alpha \right) \leq {tN}\left( J\right) }\right\} }\right) /{\mathcal{O}}_{K}^{ \times } \leftrightarrow \left( {\lambda \left( J\right) \cap {D}_{t}}\right) /{\mu }_{K} \]\n\nWe have \( {D}_{t} = {t}^{\frac{1}{n}}{D}_{1} \) and \... | Yes |
Lemma 4.4.1. (1) We have a (non-canonical) isomoprhism \( \widehat{G} \cong G \) . | Proof. (1) By the structure theorem of finite abelian groups, it suffices to show \( \widehat{\mathbb{Z}/n\mathbb{Z}} \cong \) \( \mathbb{Z}/n\mathbb{Z} \) . However, it is clear that \( \chi \in \overline{\mathbb{Z}/n\mathbb{Z}} \) is determined by \( \chi \left( 1\right) \in {\mu }_{n}\left( \mathbb{C}\right) = \left... | Yes |
Proposition 4.4.2. Let \( \chi \in \widehat{G},\chi \neq 1 \), then \( \mathop{\sum }\limits_{{g \in G}}\chi \left( g\right) = 0 \) | Proof. Let \( h \in G \) such that \( \chi \left( h\right) \neq 1 \), then \( \chi \left( h\right) \mathop{\sum }\limits_{{g \in G}}\chi \left( g\right) = \mathop{\sum }\limits_{{g \in G}}\chi \left( {hg}\right) = \mathop{\sum }\limits_{{g \in G}}\chi \left( g\right) \) . The proposition follows. | Yes |
Proposition 4.4.4. If \( \chi \neq 1 \), then \( L\left( {\chi, s}\right) \) can extend to a holomorphic function on \( \operatorname{Re}s > \) 0 . | Proof. By Proposition 4.4.2, the absolute value of \( {S}_{t} \mathrel{\text{:=}} \mathop{\sum }\limits_{{n < t}}\chi \left( n\right) \) is bounded by \( {f}_{\chi } \) . The proposition then follows from Proposition 4.3.2 (with \( \kappa = 0,\delta = 1 \) ). | No |
Theorem 4.4.5. We have\n\n\\[ \n\\mathop{\\prod }\\limits_{{\\chi \\in \\widehat{H}}}L\\left( {\\chi, s}\\right) = {\\zeta }_{K}\\left( s\\right) \n\\]\n\n(4.5) | Proof. It suffices to prove the equality for \\( \\operatorname{Re}s \\gg 0 \\) . Using the formulas of Euler product for both \\( L\\left( {\\chi, s}\\right) \\) and \\( {\\zeta }_{K}\\left( s\\right) \\), we reduce to show\n\n\\[ \n\\mathop{\\prod }\\limits_{{\\chi \\in \\widehat{H}}}\\left( {1 - \\chi \\left( p\\rig... | Yes |
Lemma 4.4.10. (1) \( {\tau }_{a}\left( \chi \right) = \bar{\chi }\left( a\right) \tau \left( \chi \right) \) . | Proof. (1) Assume first \( \left( {a, f}\right) = 1 \) . Then we have\n\n\[ \n{\tau }_{a}\left( \chi \right) = \mathop{\sum }\limits_{{z \in \mathbb{Z}/f\mathbb{Z}}}\chi \left( z\right) {\zeta }_{f}^{az} = \bar{\chi }\left( a\right) \mathop{\sum }\limits_{{z \in \mathbb{Z}/f\mathbb{Z}}}\chi \left( {az}\right) {\zeta }_... | Yes |
Suppose \( k = \mathbb{R}, f = {e}^{-\pi {x}^{2}} \) and \( \chi = {\left| \cdot \right| }^{s} \). Then | \[ \zeta \left( {f,\chi }\right) = {\int }_{{\mathbb{R}}^{ \times }}{e}^{-\pi {x}^{2}}{\left| x\right| }^{s - 1}{dx} = 2{\int }_{0}^{+\infty }{e}^{-\pi {x}^{2}}{x}^{s - 1}{dx} = {\pi }^{-\frac{s}{2}}\Gamma \left( \frac{s}{2}\right) .\] | Yes |
Lemma 1.2.6. Let \( F, G, H \) be formal groups over \( A \), and let \( f : F \rightarrow G, g : G \rightarrow H \) be morphisms. Then \( g \circ f \mathrel{\text{:=}} g\left( {f\left( T\right) }\right) \in {TA}\left\lbrack \left\lbrack T\right\rbrack \right\rbrack \) is a morphism from \( F \) to \( H \) . | Proof. \( g\left( {f\left( {F\left( {X, Y}\right) }\right) }\right) = g\left( {G\left( {f\left( X\right), f\left( Y\right) }\right) }\right) = H\left( {g\left( {f\left( X\right) }\right), g\left( {f\left( Y\right) }\right) }\right) \) . | Yes |
Proposition 1.3.3. For any \( f \in {\mathcal{F}}_{\varpi } \), there exists a unique formal group law \( {F}_{f} \in \) \( {\mathcal{O}}_{K}\left\lbrack \left\lbrack {X, Y}\right\rbrack \right\rbrack \) such that \( f \in \operatorname{End}\left( {F}_{f}\right) \) . | Proof. Applying the previous lemma to \( {\Phi }_{1} = X + Y \) and \( g = f \), we obtain \( {F}_{f} \mathrel{\text{:=}} \Phi \left( {X, Y}\right) \in \) \( {\mathcal{O}}_{K}\left\lbrack \left\lbrack {X, Y}\right\rbrack \right\rbrack \) . We need to show \( {F}_{f} \) is a formal group law. By definition \( {F}_{f}\le... | No |
Proposition 1.3.5. \( {\left\lbrack a\right\rbrack }_{g, f}\left( T\right) \) is a morphism from \( {F}_{f} \) to \( {F}_{g} \) . | Proof. We need to show \( {\Phi }^{1}\left( {X, Y}\right) \mathrel{\text{:=}} {\left\lbrack a\right\rbrack }_{g, f}\left( {{F}_{f}\left( {X, Y}\right) }\right) = {F}_{g}\left( {{\left\lbrack a\right\rbrack }_{g, f}\left( X\right) ,{\left\lbrack a\right\rbrack }_{g, f}\left( Y\right) }\right) = : {\Phi }^{2}\left( {X, Y... | Yes |
Proposition 1.3.6. (1) \( {\left\lbrack a + b\right\rbrack }_{g, f} = {\left\lbrack a\right\rbrack }_{g, f} + {}_{{F}_{g}}{\left\lbrack b\right\rbrack }_{g, f} \) . | Proof. (1) \( {\left\lbrack a + b\right\rbrack }_{g, f}\left( T\right) \equiv \left( {a + b}\right) T\left( {{\;\operatorname{mod}\;\deg } \geq 2}\right) \equiv {\left\lbrack a\right\rbrack }_{g, f} + {}_{{F}_{g}}{\left\lbrack b\right\rbrack }_{g, f} \) . We have \( \lbrack a + \) \( b{\rbrack }_{g, f}\left( {f\left( T... | Yes |
Corollary 1.3.7. For \( f, g \in {\mathcal{F}}_{\varpi } \), we have \( {F}_{f} \cong {F}_{g} \) . | Proof. Let \( u \in {\mathcal{O}}_{K}^{ \times } \), then \( {\left\lbrack u\right\rbrack }_{g, f} \) defines a morphism from \( {F}_{f} \) to \( {F}_{g} \) . By (2) of the above proposition, we have \( {\left\lbrack {u}^{-1}\right\rbrack }_{f, g} \circ \left\lbrack {u}_{g, f}\right\rbrack = {\left\lbrack 1\right\rbrac... | Yes |
Lemma 1.3.11. \( {\left\lbrack \varpi \right\rbrack }_{f}\left( T\right) = f\left( T\right) \) . | Proof. We have \( f \equiv \varpi T \equiv {\left\lbrack \varpi \right\rbrack }_{f}\left( T\right) \left( {{\;\operatorname{mod}\;\deg } \geq 2}\right) \), and \( f\left( {f\left( T\right) }\right) = f\left( {f\left( T\right) }\right) \) . By Lemma 1.3.2, the lemma follows. | No |
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