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Lemma 3.6.21. Denote by \( j \) the canonical isomorphism from \( {\left( \mathbb{Z}/m\mathbb{Z}\right) }^{ * } \) to \( G \) such that \( j\left( t\right) = {\sigma }_{t} \), and complex conjugation \( {\sigma }_{-1} \) by \( \iota \) . We have \( \Theta {e}_{\chi } = {\lambda }_{\chi }{e}_{\chi } \) with\n\n\[{\lambd... | Proof. By definition of \( {e}_{\chi } \) we have\n\n\[{\sigma }_{t}{e}_{\chi } = \frac{1}{\phi \left( m\right) }\mathop{\sum }\limits_{{\sigma \in G}}\chi \left( \sigma \right) {\sigma }_{t}{\sigma }^{-1} = \frac{1}{\phi \left( m\right) }\mathop{\sum }\limits_{{\sigma \in G}}\chi \left( {\sigma {\sigma }_{t}}\right) {... | Yes |
Lemma 3.6.22. (1) \( A\mathbb{C} \) -basis of \( \mathbb{C}{\left\lbrack G\right\rbrack }^{ - } \) is given by the \( {e}_{\chi } \) for \( \chi \left( \iota \right) = - 1 \) , in other words such that \( \chi \circ j \) is an odd character. | Proof. (1). Since \( {\sigma }_{t}{e}_{\chi } = \chi \left( {\sigma }_{t}\right) {e}_{\chi } \) we have \( \iota {e}_{\chi } = \chi \left( \iota \right) {e}_{\chi } = - {e}_{\chi } \) if \( \chi \left( \iota \right) = \) -1, so such \( {e}_{\chi } \) are in \( \mathbb{C}{\left\lbrack G\right\rbrack }^{ - } \) . When \(... | Yes |
Lemma 3.6.24. (1) \( {\left( \frac{x}{\mathrm{p}}\right) }_{m} \) is characterized by the fact that it is a character of order \( m \) such that\n\n\[{\left( \frac{x}{\mathfrak{p}}\right) }_{m} \equiv {x}^{\left( {q - 1}\right) /m}\left( {\;\operatorname{mod}\;\mathfrak{p}}\right) .\] | Proof. (1) is the translation of the corresponding properties of the character \( {\omega }_{\mathfrak{P}} \), and (2) from the fact that the natural inclusion map from \( {K}_{m} \) to \( L \) induces a canonical isomorphism between \( {\mathbb{Z}}_{{K}_{m}}/\mathfrak{p} \) and \( {\mathbb{Z}}_{L}/\mathfrak{P} \) . | No |
Proposition 3.6.26. We have:\n\n(1) \( {\left| G\left( \mathfrak{a}\right) \right| }^{2} = \mathcal{N}{\left( \mathfrak{a}\right) }^{m} \) .\n\n(2) \( G\left( \mathfrak{a}\right) {\mathbb{Z}}_{{K}_{m}} = {\mathfrak{a}}^{\gamma } \), where \( \gamma = {m\Theta } \) is as above.\n\n(3) If \( \alpha \in {\mathbb{Z}}_{{K}_... | Proof. If \( \mathfrak{p} \) is a prime, by Proposition 2.5.9 we have \( {\left| G\left( \mathfrak{p}\right) \right| }^{2} = {\left| \tau \left( {\omega }_{\mathfrak{P}}^{-d}\right) \right| }^{2m} = \) \( {q}^{m} = \mathcal{N}{\left( \mathfrak{p}\right) }^{m} \), so (1) follows by multiplicativity. Statement (2) for a ... | Yes |
Lemma 3.6.27. For any integral ideal a prime to \( m \) and any \( \sigma \in \operatorname{Gal}\left( {{K}_{m}/\mathbb{Q}}\right) \) we have \( G{\left( \mathfrak{a}\right) }^{\sigma } = G\left( {\mathfrak{a}}^{\sigma }\right) \) . | Proof. By definition, if \( \mathfrak{p} \) is a prime ideal coprime to \( m \) we have \( G\left( \mathfrak{p}\right) = \) \( \tau {\left( {\omega }_{\mathfrak{P}}^{-d}\right) }^{m} \) . Thus by Lemma 3.6.3 we have \( G{\left( \mathfrak{p}\right) }^{\sigma } = {\tau }_{(}{\omega }_{\sigma \left( \mathfrak{P}\right) }^... | Yes |
Lemma 3.6.28. If \( \alpha \in {\mathbb{Z}}_{{K}_{m}} \) then \( {\left| {\alpha }^{\gamma }\right| }^{2} = \mathcal{N}{\left( \alpha \right) }^{m} \) . | Proof. Since \( {\sigma }_{-1} \) sends \( {\zeta }_{m} \) to \( {\zeta }_{m}^{-1} \) it is complex conjugation. Thus\n\n\[ \n{\left| {\alpha }^{\gamma }\right| }^{2} = {\alpha }^{\gamma }{\sigma }_{-1}\left( {\alpha }^{\gamma }\right) = {\alpha }^{\left( {1 + {\sigma }_{-1}}\right) \gamma }.\n\]\n\nDenoting by \( \mat... | Yes |
Proposition 3.6.29. The element \( \varepsilon \left( \alpha \right) \) is a root of unity. In other words, there exist \( i \in \mathbb{Z} \) and a sign \( \pm \) such that \( G\left( \alpha \right) = \pm {\zeta }_{m}^{i}{\alpha }^{\gamma } \) . | Proof. By Proposition 3.6.26 (1) and (3) we have\n\n\[ \n{\left| G\left( \alpha \right) \right| }^{2} = {\left| \mathcal{N}\left( \alpha \right) \right| }^{m} = {\left| \varepsilon \left( \alpha \right) \right| }^{2}{\left| {\alpha }^{\gamma }\right| }^{2} = {\left| \varepsilon \left( \alpha \right) \right| }^{2}\mathc... | Yes |
Proposition 3.6.30. Let \( {p}_{1} \) and \( {p}_{2} \) be two distinct prime numbers not dividing \( m \) and let \( {\mathfrak{p}}_{1} \) and \( {\mathfrak{p}}_{2} \) be prime ideals of \( {K}_{m} \) above \( {p}_{1} \) and \( {p}_{2} \) respectively.\n\nThen\n\[ \n{\left( \frac{G\left( {\mathfrak{p}}_{1}\right) }{{\... | Proof. Let \( {f}_{1} \) and \( {f}_{2} \) respectively be the orders of \( {p}_{1} \) and \( {p}_{2} \) modulo \( m \) , so that by Proposition 3.5.18 we have \( f\left( {{\mathfrak{p}}_{i}/p}\right) = {f}_{i} \) . Set \( {q}_{i} = \mathcal{N}\left( {\mathfrak{p}}_{i}\right) = {p}_{i}^{{f}_{i}} \equiv 1 \) \( \left( {... | Yes |
Corollary 3.6.31. For all ideals \( \mathfrak{a} \) and \( \mathfrak{b} \) coprime to \( m \) such that \( \mathcal{N}\left( \mathfrak{a}\right) \) and \( \mathcal{N}\left( \mathfrak{b}\right) \) are coprime we have \( {\left( \frac{G\left( \mathfrak{a}\right) }{\mathfrak{b}}\right) }_{m} = {\left( \frac{\mathcal{N}\le... | Proof. Clear since by definition \( {\left( \frac{ \cdot }{\mathfrak{a}}\right) }_{m} \) is a character and is multiplicative in \( \mathfrak{a} \), and \( G\left( \mathfrak{a}\right) \) and \( \mathcal{N}\left( \mathfrak{a}\right) \) are also multiplicative. | Yes |
Corollary 3.6.32. In addition to the assumptions of the preceding corollary, assume that \( \mathfrak{a} = \alpha {\mathbb{Z}}_{{K}_{m}} \) is a principal ideal. Then\n\n\[ \n{\left( \frac{\mathcal{N}\left( \mathfrak{b}\right) }{\alpha }\right) }_{m} = {\left( \frac{\varepsilon \left( \alpha \right) }{\mathfrak{b}}\rig... | Proof. By multiplicativity we have\n\n\[ \n{\left( \frac{G\left( \alpha \right) }{\mathfrak{b}}\right) }_{m} = {\left( \frac{\varepsilon \left( \alpha \right) }{\mathfrak{b}}\right) }_{m}{\left( \frac{{\alpha }^{\gamma }}{\mathfrak{b}}\right) }_{m} = {\left( \frac{\varepsilon \left( \alpha \right) }{\mathfrak{b}}\right... | Yes |
Lemma 3.6.33. If \( \mathfrak{a} \) is an ideal coprime to \( \ell \) then \( G\left( \mathfrak{a}\right) \equiv \pm 1\left( {\;\operatorname{mod}\;\ell R}\right) \) . | Proof. If \( \mathfrak{p} \) is a prime ideal coprime to \( \ell \) we have as above\n\n\[ G\left( \mathfrak{p}\right) = \tau {\left( {\chi }_{\mathfrak{p}}\right) }^{\ell } \equiv \tau \left( {{\chi }_{\mathfrak{p}}^{\ell },{\psi }_{\ell }}\right) \equiv \mathop{\sum }\limits_{{t \in {\left( \mathbb{Z}/\ell \mathbb{Z}... | Yes |
Lemma 3.6.35. If \( \alpha \in R \) is coprime to \( \ell \) there exists \( i \in \mathbb{Z} \), unique modulo \( \ell \), such that \( {\zeta }_{\ell }^{i}\alpha \) is primary. | Proof. Since \( f\left( {\mathcal{L}/\ell }\right) = 1 \) we have \( R/\mathcal{L} \simeq \mathbb{Z}/\ell \mathbb{Z} \), so there exists \( a \in \mathbb{Z} \) , evidently unique modulo \( \ell \), such that \( \alpha \equiv a\left( {\;\operatorname{mod}\;\mathcal{L}}\right) \) . Thus \( \beta = \left( {\alpha - a}\rig... | Yes |
Lemma 3.6.36. If \( \alpha \) is primary we have \( \varepsilon \left( \alpha \right) = \pm 1 \) . | Proof. Since \( \mathcal{L} \) is the unique prime ideal above \( \ell \), for all \( \sigma \in \operatorname{Gal}\left( {\mathbb{Q}\left( {\zeta }_{\ell }\right) /\mathbb{Q}}\right) \) we have \( \sigma \left( \mathcal{L}\right) = \mathcal{L} \) . By definition of \( \gamma \) it follows that \( {\mathcal{L}}^{\gamma... | Yes |
Proposition 3.6.37. If \( \alpha \) is primary and \( \mathfrak{b} \) is an ideal coprime to \( \ell \) such that \( \mathcal{N}\left( \mathfrak{b}\right) \) is coprime to \( {\alpha R} \) then\n\n\[{\left( \frac{\alpha }{\mathcal{N}\left( \mathfrak{b}\right) }\right) }_{\ell } = {\left( \frac{\mathcal{N}\left( \mathfr... | Proof. Since \( \alpha \) is primary the above lemma tells us that \( \varepsilon \left( \alpha \right) = \pm 1 \) . On the other hand, for all \( \mathfrak{p} \) coprime to \( \ell ,{\chi }_{\mathfrak{p}} \) is a character of order \( \ell \), we have \( {\chi }_{\mathfrak{p}}{\left( \pm 1\right) }^{\ell } = 1 \) and ... | Yes |
Theorem 3.6.38 (Eisenstein). Let \( \ell \) be an odd prime, \( a \in \mathbb{Z} \) an integer not divisible by \( \ell ,\alpha \in R \) a primary element, and assume that \( {\alpha R} \) and \( {aR} \) are coprime. Then\n\n\[{\left( \frac{\alpha }{a}\right) }_{\ell } = {\left( \frac{a}{\alpha }\right) }_{\ell }\] | Proof. As usual, by multiplicativity it is enough to prove this when \( a = p \) is a prime number different from \( \ell \) and prime to \( {\alpha R} \) . Let \( \mathfrak{p} \) be a prime ideal of \( R \) above \( p, f = f\left( {\mathfrak{p}/p}\right) \), so that \( \mathcal{N}\left( \mathfrak{p}\right) = {p}^{f} \... | Yes |
Proposition 3.7.1. Recall that \( \{ z\} \) denotes the fractional part of \( z \in \mathbb{R} \) .\n\n(1) If \( k \) and \( m \) are coprime integers and \( z \in \mathbb{R} \) we have\n\n\[ \mathop{\sum }\limits_{{0 \leq a < m}}\left\{ \frac{{ka} + z}{m}\right\} = \frac{m - 1}{2} + \{ z\} .\n\] | Proof. (1). The expression \( \{ \left( {{ka} + z}\right) /m\} \) depends only on \( a \) modulo \( m \) ; hence we may replace the summation by \( \mathop{\sum }\limits_{{a \in \mathbb{Z}/m\mathbb{Z}}} \) . Since \( k \) is coprime to \( m \) , multiplication by \( k \) is an automorphism of \( \mathbb{Z}/m\mathbb{Z} ... | Yes |
Proposition 3.7.2. If \( m \mid \left( {q - 1}\right) \) and \( d = \left( {q - 1}\right) /m \), then for all \( b \in \mathbb{Z} \) we have \[ \frac{{m}^{mb}\mathop{\prod }\limits_{{0 \leq a < m}}t\left( {{da} + b}\right) }{t\left( {mb}\right) \mathop{\prod }\limits_{{0 \leq a < m}}t\left( {da}\right) } \equiv 1\left(... | Proof. Consider the left-hand side as a function \( F\left( b\right) \) of \( b \) . We have evidently \( F\left( 0\right) = 1 \) . Furthermore, \( t\left( {m\left( {b + d}\right) }\right) = t\left( {{mb} + q - 1}\right) = t\left( {mb}\right) \) , \( \mathop{\prod }\limits_{{0 \leq a < m}}t\left( {{da} + b + d}\right) ... | Yes |
Corollary 3.7.5. If \( {\chi }_{1},\ldots ,{\chi }_{k} \) are multiplicative characters on \( {\mathbb{F}}_{q} \) that are not all trivial, and \( \psi \) is a nontrivial additive character, then\n\n\[ \n{J}_{k}\left( {{\chi }_{1}^{\left( n\right) },\ldots ,{\chi }_{k}^{\left( n\right) }}\right) = {\left( -1\right) }^{... | Proof. Immediate by inspection of the cases of Proposition 2.5.14. | No |
Corollary 3.7.6. Assume that \( p \) is odd, let \( q = {p}^{f} \), and let \( {\chi }_{q} \) be the unique multiplicative character of order 2 of \( {\mathbb{F}}_{q} \) . We have \[ \tau \left( {{\chi }_{q},{\psi }_{1}}\right) = \left\{ \begin{array}{ll} {\left( -1\right) }^{f - 1}{q}^{1/2} & \text{ if }p \equiv 1\lef... | Proof. Denote by \( {\chi }_{p} \) the Legendre symbol modulo \( p \) . We have \( {\mathcal{N}}_{{\mathbb{F}}_{q}/{\mathbb{F}}_{p}}\left( x\right) = \) \( \mathop{\prod }\limits_{{0 \leq i < f}}{x}^{{p}^{i}} = {x}^{\left( {q - 1}\right) /\left( {p - 1}\right) } \), hence \[ {\chi }_{p} \circ {\mathcal{N}}_{{\mathbb{F}... | Yes |
Theorem 3.7.7. Assume that \( d \mid \left( {q - 1}\right) \), and for \( 1 \leq i \leq k \) let \( {a}_{i} \in {\mathbb{F}}_{q}^{ * } \) . For \( n \geq 1 \) denote by \( {M}_{n}\left( q\right) \) the number of projective solutions in \( {\mathbb{F}}_{{q}^{n}} \) of \( \mathop{\sum }\limits_{{1 \leq i \leq k}}{a}_{i}{... | Proof. We use the notation of Section 2.5.5 and also keep the notation used in the proof of the HD lifting relation. By Corollary 2.5.24 we have\n\n\[ \n{M}_{n}\left( q\right) = \frac{{q}^{n\left( {k - 1}\right) } - 1}{{q}^{n} - 1} + {R}_{n}\left( q\right) \;\text{ with } \n\]\n\n\[ \n{R}_{n}\left( q\right) = \mathop{\... | Yes |
Corollary 3.7.8. Define the zeta function of our hypersurface by \( {Z}_{q}\left( T\right) = \) \( \exp \left( {\mathop{\sum }\limits_{{n \geq 1}}{M}_{n}\left( q\right) {T}^{n}/n}\right) \) . Then \( {Z}_{q}\left( T\right) \) is a rational function of \( T \) of the form\n\n\[ \n{Z}_{q}\left( T\right) = \frac{P{\left( ... | Proof. Using the formal expansion of the logarithm and the explicit formula for \( {M}_{n}\left( q\right) \) (where we replace \( \left( {{q}^{n\left( {k - 1}\right) } - 1}\right) /\left( {{q}^{n} - 1}\right) \) by \( \mathop{\sum }\limits_{{0 \leq i \leq k - 2}}{q}^{in} \) ) we find that\n\n\[ \n\mathop{\sum }\limits_... | Yes |
Lemma 4.1.3. Two absolute values \( \parallel {\parallel }_{1} \) and \( \parallel {\parallel }_{2} \) are equivalent if and only if there exists \( c > 0 \) such that \( \parallel x{\parallel }_{2} = \parallel x{\parallel }_{1}^{c} \) for all \( x \in K \) . | Proof. If \( \parallel x{\parallel }_{2} = \parallel x{\parallel }_{1}^{c} \), for each \( R > 0 \) we have \( {B}_{1}\left( {0, R}\right) = {B}_{2}\left( {0,{R}^{c}}\right) \) , where \( {B}_{i}\left( {0, R}\right) \) denotes the open ball of radius \( R \) for the metric induced by \( \parallel {\parallel }_{i} \) ; ... | Yes |
Corollary 4.1.6. Let \( K \) be a number field and \( \left( {{r}_{1},{r}_{2}}\right) \) its signature. There exist exactly \( {r}_{1} + {r}_{2} \) inequivalent Archimedean absolute values on \( K \), corresponding to the \( {r}_{1} \) real embeddings and to the \( {r}_{2} \) pairs of complex conjugate embeddings. | Proof. By the above lemma, an Archimedean absolute value is equivalent to one defined by \( \parallel \alpha \parallel = \left| {\sigma \left( \alpha \right) }\right| \) for some fixed embedding \( \sigma \) of \( K \) into \( \mathbb{C} \) , and conversely, any such embedding gives rise to an Archimedean absolute valu... | Yes |
Lemma 4.1.7. An absolute value \( \parallel \parallel \) on a field \( K \) is non-Archimedean if and only if it satisfies the so-called ultrametric inequality\n\n\[ \parallel x + y\parallel \leq \max \left( {\parallel x\parallel ,\parallel y\parallel }\right) . \] | Proof. Assume first that \( \parallel \parallel \) satisfies the ordinary triangle inequality, i.e., that we can choose \( a = 1 \) in Definition 4.1.1. Then by the binomial theorem, for any positive integer \( N \) we have\n\n\[ \parallel x + y{\parallel }^{N} = \begin{Vmatrix}{\mathop{\sum }\limits_{{0 \leq n \leq N}... | Yes |
Corollary 4.1.8. Let \( \parallel \parallel \) be an ultrametric absolute value.\n\n(1) When \( \parallel x\parallel \neq \parallel y\parallel \) we have equality in the ultrametric inequality:\n\n\[ \parallel x + y\parallel = \max \left( {\parallel x\parallel ,\parallel y\parallel }\right) . \] | Proof. (1) Assume for instance that \( \parallel y\parallel < \parallel x\parallel \) . Writing \( x = \left( {x + y}\right) + \left( {-y}\right) \) , we see that \( \parallel x\parallel \leq \max \left( {\parallel x + y\parallel ,\parallel y\parallel }\right) \) . The maximum cannot be equal to \( \parallel y\parallel... | Yes |
Proposition 4.1.9. Let \( \parallel \parallel \) be an ultrametric absolute value on a field \( K \) . Let \( A = \{ x \in K/\parallel x\parallel \leq 1\} \) and \( \mathfrak{p} = \{ x \in K/\parallel x\parallel < 1\} \) . Then \( A \) is a local ring with maximal ideal \( \mathfrak{p} \) and field of fractions \( K \)... | Proof. The fact that \( A \) is a ring immediately follows from the ultrametric inequality. It is clear that if \( x \in {K}^{ * } \) then either \( x \) or \( 1/x \) belongs to \( A \), so \( K \) is the field of fractions of \( A \) . Finally, \( x \in A \) is invertible in \( A \) if and only if \( \parallel x\paral... | Yes |
Theorem 4.1.13 (Ostrowski). The places of a number field \( K \) are in one-to-one correspondence with on the one hand the \( {r}_{1} \) real embeddings and \( {r}_{2} \) pairs of nonreal complex conjugate embeddings of \( K \) into \( \mathbb{C} \), corresponding to the Archimedean absolute values (called the infinite... | Although places are determined up to equivalence, there is a canonical way to define them uniquely. This comes from the action of multiplication by an element on a Haar measure, but we simply give the formulas.\n\nIf \( \sigma \) is one of the \( {r}_{1} \) real embeddings of \( K \), we choose\n\n\[ \parallel \alpha {... | Yes |
Proposition 4.1.14 (Product formula). With the above normalizations, for any \( \alpha \in {K}^{ * } \) the set of places \( v \) of \( K \) for which \( \parallel \alpha \parallel \neq 1 \) is finite, and we have\n\n\[ \mathop{\prod }\limits_{v}\parallel \alpha {\parallel }_{v} = 1 \]\n\nwhere \( v \) runs over all th... | Proof. Let \( \alpha {\mathbb{Z}}_{K} = \mathop{\prod }\limits_{\mathfrak{p}}{\mathfrak{p}}^{{v}_{\mathfrak{p}}\left( \alpha \right) } \) be the decomposition of the ideal \( \alpha {\mathbb{Z}}_{K} \) into a product of powers of prime ideals. Thus, \( \parallel \alpha {\parallel }_{v} = 1 \) if \( v \) is neither an i... | Yes |
Lemma 4.1.18. Let \( K \) be a field with an ultrametric absolute value. A sequence \( \left( {a}_{n}\right) \) in \( K \) is a Cauchy sequence if and only if \( \begin{Vmatrix}{{a}_{n + 1} - {a}_{n}}\end{Vmatrix} \) tends to 0 as \( n \) tends to infinity. In particular, if \( K \) is complete, a series \( \mathop{\su... | Proof. The condition is of course necessary. Conversely, by the ultrametric inequality, for \( m \geq n \) we have\n\n\[ \begin{Vmatrix}{{a}_{m} - {a}_{n}}\end{Vmatrix} = \begin{Vmatrix}{\mathop{\sum }\limits_{{n \leq j < m}}\left( {{a}_{j + 1} - {a}_{j}}\right) }\end{Vmatrix} \leq \mathop{\max }\limits_{{n \leq j < m}... | Yes |
Lemma 4.1.19. Let \( K \) be a number field and \( \mathfrak{p} \) a (nonzero) prime ideal of \( {\mathbb{Z}}_{K} \) . If we denote by \( {K}_{\mathfrak{p}} \) the completion of \( K \) for the \( \mathfrak{p} \) -adic absolute value, then for all \( x \in {K}_{\mathfrak{p}}^{ * } \) there exists \( k \in \mathbb{Z} \)... | Proof. Let \( \left( {x}_{n}\right) \) be a sequence of elements of \( K \) converging to \( x \) . Since \( x \neq 0 \), without loss of generality we may assume that \( {x}_{n} \neq 0 \) for all \( n \) . Set \( {k}_{n} = \log \left( {\left| {x}_{n}\right| }_{\mathfrak{p}}\right) /\log \left( {\mathcal{N}\mathfrak{p}... | Yes |
Proposition 4.1.21. The set \( {\mathbb{Z}}_{\mathfrak{p}} \) is a discrete valuation ring: more precisely, the only nonzero ideals of \( {\mathbb{Z}}_{\mathfrak{p}} \) are the \( {\mathfrak{p}}^{k}{\mathbb{Z}}_{\mathfrak{p}} \) for \( k \geq 0 \), the only maximal ideal is \( \mathfrak{p}{\mathbb{Z}}_{\mathfrak{p}} \)... | Proof. Since \( {\left| \right| }_{\mathfrak{p}} \) is an ultrametric absolute value, it is clear that \( {\mathbb{Z}}_{\mathfrak{p}} \) is a subring of \( {K}_{\mathfrak{p}} \) . Moreover, \( x \in {\mathbb{Z}}_{\mathfrak{p}} \) is invertible in \( {\mathbb{Z}}_{\mathfrak{p}} \) if and only if \( {\left| x\right| }_{\... | Yes |
Proposition 4.1.22. Let \( A \) and \( B \) be two monic polynomials of \( {K}_{\mathfrak{p}}\left\lbrack X\right\rbrack \) such that \( B \mid A \) . If \( A \in {\mathbb{Z}}_{\mathfrak{p}}\left\lbrack X\right\rbrack \), then also \( B \in {\mathbb{Z}}_{\mathfrak{p}}\left\lbrack X\right\rbrack \) . | Proof. Write \( A = {BC} \), and let \( \pi \) be a uniformizer of \( \mathfrak{p} \) . Let \( {v}_{B} \) and \( {v}_{C} \) be the smallest exponents such that \( {\pi }^{{v}_{B}}B\left( X\right) \in {\mathbb{Z}}_{\mathfrak{p}}\left\lbrack X\right\rbrack \) and \( {\pi }^{{v}_{C}}C\left( X\right) \in {\mathbb{Z}}_{\mat... | Yes |
Lemma 4.1.23. Under the above identification \( {\mathbb{Z}}_{K} \) is a subring of \( {\mathbb{Z}}_{\mathfrak{p}} \) and is dense in \( {\mathbb{Z}}_{\mathfrak{p}} \) . | Proof. If \( x \in {\mathbb{Z}}_{K} \) then \( x \) is a root of \( {x}^{n} + \mathop{\sum }\limits_{{0 \leq i \leq n - 1}}{a}_{i}{x}^{i} = 0 \) with \( {a}_{i} \in \mathbb{Z} \) . Applying the ultrametric inequality we see that\n\n\[{\left| x\right| }_{\mathfrak{p}}^{n} = {\left| {x}^{n}\right| }_{\mathfrak{p}} = {\le... | Yes |
Proposition 4.1.24. For any \( k \geq 1 \), the natural map from \( {\mathbb{Z}}_{K} \) to \( {\mathbb{Z}}_{\mathfrak{p}} \) induces \( a \) ring isomorphism from \( {\mathbb{Z}}_{K}/{\mathfrak{p}}^{k} \) to \( {\mathbb{Z}}_{\mathfrak{p}}/{\mathfrak{p}}^{k}{\mathbb{Z}}_{\mathfrak{p}} \) . | Proof. It is clear that the map induces a ring homomorphism \( \phi \) from \( {\mathbb{Z}}_{K}/{\mathfrak{p}}^{k} \) to \( {\mathbb{Z}}_{\mathfrak{p}}/{\mathfrak{p}}^{k}{\mathbb{Z}}_{\mathfrak{p}} \) . Furthermore, \( \bar{x} \in {\mathbb{Z}}_{K}/{\mathfrak{p}}^{k} \) is such that \( \phi \left( x\right) = \overline{0... | Yes |
Proposition 4.1.25. Let \( R \) be a set of representatives in \( {\mathbb{Z}}_{K} \) of the elements of \( {\mathbb{Z}}_{K}/\mathfrak{p} \), and for each \( m \in \mathbb{Z} \) let \( {\pi }_{m} \in {\mathfrak{p}}^{m} \smallsetminus {\mathfrak{p}}^{m + 1} \) be an element of exact \( \mathfrak{p} \) -adic valuation \(... | Proof. Since \( {a}_{m}{\pi }_{m} \) tends to 0 for the \( \mathfrak{p} \) -adic topology, any series of this form converges. Dividing by \( {\pi }^{{v}_{\mathfrak{p}}\left( x\right) } \), we may assume that \( {v}_{\mathfrak{p}}\left( x\right) = 0 \), i.e., that \( {x}_{0} = x \) is an invertible element of \( {\mathb... | Yes |
Corollary 4.1.26. The field \( {K}_{\mathfrak{p}} \) is locally compact and totally discontinuous. The ring of \( \mathfrak{p} \) -adic integers \( {\mathbb{Z}}_{\mathfrak{p}} \) is compact. | Proof. We first check that the open unit ball \( B = \left\{ {x \in {K}_{\mathfrak{p}}/{\left| x\right| }_{\mathfrak{p}} < 1}\right\} \) is compact. By the proposition, \( x \in B \) if and only if \( x = \mathop{\sum }\limits_{{m \geq 1}}{a}_{m}{\pi }_{m} \) for \( {a}_{m} \in R \) . Thus, if \( \left( {x}_{n}\right) ... | Yes |
Corollary 4.1.27. Let \( K \) be a number field, let \( L \) be a finite extension of \( K \), let \( \mathfrak{p} \) be a (nonzero) prime ideal of \( {\mathbb{Z}}_{K} \), and finally let \( \mathfrak{P} \) be a prime ideal of \( {\mathbb{Z}}_{L} \) above \( \mathfrak{p} \). Denote as usual by \( e\left( {\mathfrak{P}/... | Proof. Let \( \Pi \in \mathfrak{P} \smallsetminus {\mathfrak{P}}^{2} \) be a uniformizer of \( \mathfrak{P} \) in \( L \), let \( \pi \in \mathfrak{p} \smallsetminus {\mathfrak{p}}^{2} \) be a uniformizer of \( \mathfrak{p} \) in \( K \), and for simplicity write \( e = e\left( {\mathfrak{P}/\mathfrak{p}}\right) \) and... | Yes |
Corollary 4.1.28. With the same notation, if in addition \( L/K \) is a Galois extension then \( {L}_{\mathfrak{P}}/{K}_{\mathfrak{p}} \) is also a Galois extension with Galois group isomorphic to the decomposition group \( D\left( {\mathfrak{P}/\mathfrak{p}}\right) \) of \( \mathfrak{P} \) over \( \mathfrak{p} \) . | Proof. Immediate and left to the reader. | No |
Proposition 4.1.29 (Hensel). Let \( f \in {\mathbb{Z}}_{\mathfrak{p}}\left\lbrack X\right\rbrack \), and assume that \( \bar{f}\left( X\right) = \) \( {\phi }_{1}\left( X\right) {\phi }_{2}\left( X\right) \) with \( {\phi }_{i} \in \left( {{\mathbb{Z}}_{K}/\mathfrak{p}}\right) \left\lbrack X\right\rbrack \) coprime. Th... | Proof. We prove by induction on \( k \) that there exist polynomials \( {A}_{k},{B}_{k} \) , \( {U}_{k},{V}_{k} \) in \( {\mathbb{Z}}_{\mathfrak{p}}\left\lbrack X\right\rbrack \) such that \( f\left( X\right) \equiv {A}_{k}\left( X\right) {B}_{k}\left( X\right) \left( {\;\operatorname{mod}\;{\mathfrak{p}}^{k}),{U}_{k}\... | Yes |
Lemma 4.1.30. Let \( L/K \) be an extension of number fields, let \( \mathfrak{p} \) be a prime ideal of \( {\mathbb{Z}}_{K} \), and let \( \mathfrak{P} \) be a prime ideal of \( {\mathbb{Z}}_{L} \) above \( \mathfrak{p} \). Finally, let \( \alpha \in {L}_{\mathfrak{P}} \) and let \( f\left( X\right) = {X}^{m} + {f}_{m... | Proof. The proof is identical to that of the first part of Lemma 4.1.23 and is left to the reader. | No |
Proposition 4.1.31. Let \( L/K \) be an extension of number fields, let \( \mathfrak{p} \) be a prime ideal of \( {\mathbb{Z}}_{K} \), and let \( \mathfrak{P} \) be a prime ideal of \( {\mathbb{Z}}_{L} \) above \( \mathfrak{p} \) . Finally, let \( \alpha \in {L}_{\mathfrak{P}} \) and let \( f\left( X\right) = {f}_{m}{X... | Proof. Note first that the fact that \( {L}_{\mathfrak{P}} \) is a finite extension of \( {K}_{\mathfrak{p}} \) follows from Corollary 4.1.27. Assume first that \( \alpha \in {\mathbb{Z}}_{\mathfrak{P}} \), and by contradiction that not all \( {f}_{k} \) belong to \( {\mathbb{Z}}_{\mathfrak{p}} \) . Let \( - v \) with ... | Yes |
Corollary 4.1.32. Let \( L/K \) be an extension of number fields, let \( \mathfrak{p} \) be a prime ideal of \( {\mathbb{Z}}_{K} \), and let \( \mathfrak{P} \) be a prime ideal of \( {\mathbb{Z}}_{L} \) above \( \mathfrak{p} \) . Then any element \( \alpha \in {\mathbb{Z}}_{\mathfrak{P}} \) is integral over \( {\mathbb... | Proof. Clear from the above proposition. | No |
Proposition 4.1.33. Let \( f\left( X\right) = {f}_{m}{X}^{m} + {f}_{m - 1}{X}^{m - 1} + \cdots + {f}_{0} \in {K}_{\mathfrak{p}}\left\lbrack X\right\rbrack \) be a monic polynomial (so that \( {f}_{m} = 1 \) ). If \( f \) is irreducible in \( {K}_{\mathfrak{p}}\left\lbrack X\right\rbrack \), then either \( {v}_{\mathfra... | Proof. Let \( v = \mathop{\min }\limits_{k}{v}_{\mathfrak{p}}\left( {f}_{k}\right) \leq 0 \) since \( {f}_{m} = 1 \), and let \( {j}_{0} \) be the smallest index \( j \) such that \( {v}_{\mathfrak{p}}\left( {f}_{{j}_{0}}\right) = v \), so that in particular \( {v}_{\mathfrak{p}}\left( {f}_{j}\right) > {v}_{\mathfrak{p... | Yes |
Corollary 4.1.34. With the same assumptions as in the proposition, if \( {f}_{0} \in \) \( {\mathbb{Z}}_{\mathfrak{p}} \) then \( f\left( X\right) \in {\mathbb{Z}}_{\mathfrak{p}}\left\lbrack X\right\rbrack \), and if \( {f}_{0} \in \mathfrak{p}{\mathbb{Z}}_{\mathfrak{p}} \) then \( {f}_{j} \in \mathfrak{p}{\mathbb{Z}}_... | Proof. Clear. | No |
Proposition 4.1.35 (Eisenstein). Let \( f\left( X\right) = {X}^{n} + {f}_{n - 1}{X}^{n - 1} + \cdots + {f}_{0} \in {\mathbb{Z}}_{\mathfrak{p}}\left\lbrack X\right\rbrack \) be a monic polynomial with \( \mathfrak{p} \) -integral coefficients. Assume that \( {v}_{\mathfrak{p}}\left( {f}_{j}\right) \geq 1 \) for all \( j... | Proof. Note first that if \( \alpha \) is a root of \( f\left( X\right) = 0 \) in some finite extension of \( {\mathbb{Z}}_{\mathfrak{p}} \), then by the same reasoning as that used in the \ | No |
Corollary 4.1.36. The polynomial\n\n\[ \n{\Phi }_{{p}^{n}}\left( X\right) = \frac{{X}^{{p}^{n}} - 1}{{X}^{{p}^{n - 1}} - 1} = {X}^{{p}^{n - 1}\left( {p - 1}\right) } + {X}^{{p}^{n - 1}\left( {p - 2}\right) } + \cdots + 1 \]\n\nis irreducible in \( {\mathbb{Q}}_{p}\left\lbrack X\right\rbrack \) . | Proof. It suffices to show that \( {\Phi }_{{p}^{n}}\left( {X + 1}\right) \) is an Eisenstein polynomial. For \( n = 1 \), by the binomial theorem we have directly\n\n\[ \n{\Phi }_{p}\left( {X + 1}\right) = {X}^{p - 1} + \mathop{\sum }\limits_{{2 \leq j \leq p - 2}}\left( \begin{array}{l} p \\ j \end{array}\right) {X}^... | Yes |
Corollary 4.1.39. Let \( P\left( \underline{X}\right) \in {\mathbb{Z}}_{p}\left\lbrack {{X}_{1},\ldots ,{X}_{n}}\right\rbrack \) be a homogeneous polynomial in \( n \) variables, and let \( \left( {{x}_{1},\ldots ,{x}_{n}}\right) \in {\mathbb{F}}_{p}^{n} \) be a nontrivial nonsingular solution of \( \bar{P}\left( \unde... | Proof. By definition, there exists \( j \) such that \( \frac{\partial P}{\partial {X}_{j}}\left( {{x}_{1},\ldots ,{x}_{n}}\right) ≢ 0\left( {{\;\operatorname{mod}\;p}{\mathbb{Z}}_{p}}\right) \) ; hence it is a \( p \) -adic unit, while \( P\left( {{x}_{1},\ldots ,{x}_{n}}\right) \in p{\mathbb{Z}}_{p} \) . We can thus ... | Yes |
Corollary 4.1.40. (1) Let \( P\left( \underline{X}\right) \in {\mathbb{Z}}_{p}\left\lbrack {{X}_{1},\ldots ,{X}_{n}}\right\rbrack \) be a homogeneous polynomial in \( n \) variables of degree \( d \), and assume that the reduction \( \bar{P} \) modulo \( p{\mathbb{Z}}_{p} \) is nonsingular over \( {\mathbb{F}}_{p} \) a... | Proof. (1). When we reduce the polynomial \( P \) modulo \( p{\mathbb{Z}}_{p} \), the degree may decrease; hence the condition \( d < n \) is still satisfied for the reduced polynomial. Thus we may apply Chevalley-Warning's Theorem 2.5.2 to deduce the existence of a nontrivial solution in \( {\mathbb{F}}_{p} \) . Since... | Yes |
Lemma 4.1.41. Let \( p \) be the prime number below \( \mathfrak{p} \), denote by \( e = e\left( {\mathfrak{p}/p}\right) \) the absolute ramification index, and assume given \( \mathfrak{p} \) -adic units \( {\alpha }_{0} \) and \( \beta \) such that \( \beta \equiv {\alpha }_{0}^{p}\left( {\;\operatorname{mod}\;{\math... | Proof. We again construct a sequence \( {\alpha }_{n} \) of \( \mathfrak{p} \) -adic units such that \( \beta \equiv {\alpha }_{n}^{p} \) \( \left( {\;\operatorname{mod}\;{\mathfrak{p}}^{e + r + n}}\right) \) and \( {\alpha }_{n + 1} \equiv {\alpha }_{n}\left( {\;\operatorname{mod}\;{\mathfrak{p}}^{n + r}}\right) \) . ... | Yes |
Lemma 4.2.1. Let \( \left( {b}_{i, j}\right) \) be a double sequence of elements of \( \mathcal{K} \) . Assume that for all \( \varepsilon > 0 \) there exists \( N\left( \varepsilon \right) \) such that \( \left| {b}_{i, j}\right| < \varepsilon \) when \( \max \left( {i, j}\right) \geq N\left( \varepsilon \right) \) . ... | Proof. Immediate from the ultrametric inequality and left to the reader. | No |
Lemma 4.2.3. Keep the notation of the above lemma, and let \( y \in \mathcal{D} \) . For any \( m \geq 0 \), set\n\n\[ \n{g}_{m} = \mathop{\sum }\limits_{{n \geq m}}\left( \begin{array}{l} n \\ m \end{array}\right) {f}_{n}{y}^{n - m}.\n\]\n\nThen the series \( g\left( X\right) = \mathop{\sum }\limits_{{m \geq 0}}{g}_{m... | Proof. Since \( {f}_{n}{y}^{n} \) tends to 0, we note that the series defining \( {g}_{m} \) converges. The above lemma shows that the series \( g\left( x\right) \) converges and the interchange of summation justified by the same lemma shows that \( g\left( x\right) = f\left( {x + y}\right) \) by the binomial theorem. ... | Yes |
Corollary 4.2.4. A function defined by a power series is infinitely differentiable in its domain of convergence. In particular, if two power series with strictly positive radius of convergence coincide on some open ball with nonzero radius, their coefficients coincide. | Proof. Indeed, the \( k \) th derivative of \( g\left( x\right) \) is equal to \( \mathop{\sum }\limits_{{m \geq k}}m(m - \) 1) \( \cdots \left( {m - k + 1}\right) {g}_{m}{x}^{m - k} \), and since \( \left| {m\left( {m - 1}\right) \cdots \left( {m - k + 1}\right) }\right| \leq 1 \), the radius of convergence of this se... | No |
Lemma 4.2.5. For all \( n \in {\mathbb{Z}}_{ \geq 0} \) denote by \( {s}_{p}\left( n\right) \) the sum of the digits of \( n \) in base \( p \), and let \( k \in {\mathbb{Z}}_{ \geq 0} \) . (1) We have \( {s}_{p}\left( k\right) \leq k \) . | Proof. (1) follows from the formula \( 0 \leq {v}_{p}\left( {k!}\right) = \left( {k - {s}_{p}\left( k\right) }\right) /\left( {p - 1}\right) \) that we shall prove below (Lemma 4.2.8), and (2) from \[ 0 \leq {v}_{p}\left( \left( \begin{matrix} k \\ i \end{matrix}\right) \right) = \frac{{s}_{p}\left( i\right) + {s}_{p}\... | No |
Proposition 4.2.6. Let \( A\left( X\right) = \mathop{\sum }\limits_{{k \geq 0}}{a}_{k}{X}^{k} \) and \( B\left( X\right) = \mathop{\sum }\limits_{{k \geq 0}}{b}_{k}{X}^{k} \) be two power series with coefficients in \( \mathcal{K} \), and set \( C\left( X\right) = A\left( X\right) B\left( X\right) = \) \( \mathop{\sum ... | Proof. We have \( {c}_{k} = \mathop{\sum }\limits_{{0 \leq i \leq k}}{a}_{i}{b}_{k - i} \), hence\n\n\[ \n{v}_{p}\left( {c}_{k}\right) \geq \mathop{\min }\limits_{{0 \leq i \leq k}}\left( {{v}_{p}\left( {a}_{i}\right) + {v}_{p}\left( {b}_{k - i}\right) }\right) \geq \mathop{\min }\limits_{{0 \leq i \leq k}}\left( {-{\a... | Yes |
Proposition 4.2.7. Let \( f\left( X\right) = \mathop{\sum }\limits_{{n \geq 0}}{a}_{n}{X}^{n} \) and \( g\left( X\right) = \mathop{\sum }\limits_{{m \geq 1}}{b}_{m}{X}^{m} \) be two formal power series, let \( F = f \circ g \) be defined as above, and let \( R \) be the radius of convergence of \( f \) . If \( x \in \m... | Proof. We use the above notation. Consider the double series \( \mathop{\sum }\limits_{{i, j}}{c}_{i, j}{x}^{j} \) . By definition of the \( {c}_{i, j} \) we have\n\n\[ \n{c}_{n, m}{x}^{m} = \mathop{\sum }\limits_{\substack{{{k}_{1},\ldots ,{k}_{n} \geq 1} \\ {{k}_{1} + \cdots + {k}_{n} = m} }}{a}_{n}{b}_{{k}_{1}}{x}^{... | Yes |
Lemma 4.2.8. (1) We have\n\n\[ \n{v}_{p}\left( {n!}\right) = \mathop{\sum }\limits_{{k = 1}}^{{\lfloor \log \left( n\right) /\log \left( p\right) \rfloor }}\left\lfloor \frac{n}{{p}^{k}}\right\rfloor = \frac{n - {s}_{p}\left( n\right) }{p - 1}.\n\]\n\nIn particular, for \( n \geq 1 \) we have \( {v}_{p}\left( {n!}\righ... | Proof. (1). We have\n\n\[ \n{v}_{p}\left( {n!}\right) = \mathop{\sum }\limits_{{1 \leq i \leq n}}{v}_{p}\left( i\right) = \mathop{\sum }\limits_{{k \geq 0}}k\mathop{\sum }\limits_{{1 \leq i \leq n,{v}_{p}\left( i\right) = k}}1\n\]\n\n\[ \n= \mathop{\sum }\limits_{{k \geq 0}}k\left( {\left\lfloor {n/{p}^{k}}\right\rfloo... | No |
Lemma 4.2.9. As in the preceding lemma, when \( {v}_{p}\left( a\right) \geq 0 \), write \( {v}_{p}\left( a\right) = \) \( q - \theta \) with \( 0 \leq \theta < 1 \) . Then\n\n\[ \mathop{\max }\limits_{{m \geq 1}}\left( \frac{1/\left( {p - 1}\right) - {v}_{p}\left( a\right) + {v}_{p}\left( m\right) }{m}\right) = \left\{... | Proof. For simplicity, set \( A = 1/\left( {p - 1}\right) - {v}_{p}\left( a\right) \) and \( m = {p}^{u}n \) with \( p \nmid n \) . We must compute \( \mathop{\max }\limits_{{u \geq 0, n \geq 1}}\left( {A + u}\right) /\left( {{p}^{u}n}\right) \) . Consider for the moment \( n \) fixed, and set \( f\left( u\right) = \le... | Yes |
(1) The series obtained by replacing the formal variable \( X \) by \( x \in \mathcal{K} \) in \( \exp \left( X\right) \) converges if and only if \( \left| x\right| < {r}_{p} = {p}^{-1/\left( {p - 1}\right) } \) (or equivalently, \( \left. {{v}_{p}\left( x\right) > 1/\left( {p - 1}\right) }\right) \), and its sum is d... | Proof. (1) and (2). By the above lemma, we have \( \lim \sup \left( {{v}_{p}\left( {n!}\right) /n}\right) = \) \( 1/\left( {p - 1}\right) \), so the radius of convergence of \( \exp \left( X\right) \) is \( 1/\limsup \left( {\left| 1/n!\right| }^{1/n}\right) = \) \( {p}^{-1/\left( {p - 1}\right) } = {r}_{p} \) . Since ... | Yes |
Proposition 4.2.11. Assume that \( x \in 2{\mathbb{Z}}_{2} \) . Then \( \left| {{\log }_{2}\left( {1 + x}\right) }\right| < {r}_{2} \), but \( {\exp }_{2}\left( {{\log }_{2}\left( {1 + x}\right) }\right) = \varepsilon \left( {1 + x}\right) \) with \( \varepsilon = 1 \) if \( x \in 4{\mathbb{Z}}_{2} \) and \( \varepsilo... | Proof. Left to the reader (Exercise 11). | No |
Lemma 4.2.12. Let \( f\left( x\right) = \mathop{\sum }\limits_{{n \geq 0}}{a}_{n}{x}^{n}/n! \), and let \( k \geq 1 \) . Assume that there exists \( z \in \mathbb{R} \) such that \( {v}_{p}\left( {a}_{n}\right) \geq z \) for all \( n \geq k \), and that \( {v}_{p}\left( x\right) > 1/\left( {p - 1}\right) \). Then the s... | Proof. By Lemma 4.2.8 for \( n \geq k \) we have by assumption \[ {v}_{p}\left( {{a}_{n}{x}^{n}/n!}\right) \geq k{v}_{p}\left( x\right) + z + \left( {n - k}\right) {v}_{p}\left( x\right) - \left( {n - {s}_{p}\left( n\right) }\right) /\left( {p - 1}\right) . \] Furthermore, if \( {s}_{p}\left( n\right) \geq k \) we have... | Yes |
Corollary 4.2.13. If \( {v}_{p}\left( x\right) > 1/\left( {p - 1}\right) \) then\n\n\[ \n{v}_{p}\left( {{\log }_{p}\left( {1 + x}\right) - x}\right) > 2{v}_{p}\left( x\right) - 1/\left( {p - 1}\right) \;\text{ and }\n\]\n\n\[ \n{v}_{p}\left( {{\exp }_{p}\left( x\right) - 1 - x}\right) > 2{v}_{p}\left( x\right) - 1/\lef... | Proof. Clear. | No |
Proposition 4.2.14. Let \( x \) be such that \( v = {v}_{p}\left( x\right) > 0 \), assume that \( v \leq \) \( 1/\left( {p - 1}\right) \), and set\n\n\[ \n{k}_{0} = \left\lceil \frac{-\log \left( {\left( {p - 1}\right) v}\right) }{\log p}\right\rceil .\n\]\n\n(1) When \( v \leq 1/\left( {p - 1}\right) \) then\n\n\[ \n{... | Proof. Let \( {u}_{n} = {\left( -1\right) }^{n - 1}{x}^{n}/n \) be the \( n \) th summand of the series for \( {\log }_{p}\left( {1 + x}\right) \) . If we write \( n = {p}^{k}m \) with \( p \nmid m \), then \( {v}_{p}\left( {u}_{n}\right) = {p}^{k}{mv} - k \) . This is a function of the two integer variables \( m \) an... | Yes |
Corollary 4.2.15. Let \( a \in \mathcal{K} \) be fixed and consider the power series\n\n\[{\left( 1 + X\right) }^{a} = \mathop{\sum }\limits_{{n \geq 0}}\left( \begin{array}{l} a \\ n \end{array}\right) {X}^{n}\]\n\nDefine\n\n\[V\left( a\right) = \left\{ \begin{array}{ll} 1/\left( {p - 1}\right) - {v}_{p}\left( a\right... | Proof. (3) and the first and last statement of (1) immediately follow from Lemma 4.2.8. In addition, for \( {v}_{p}\left( a\right) \geq 0 \) we have \( V\left( a\right) \leq 1/\left( {p - 1}\right) \), while \( V\left( a\right) = 1/\left( {p - 1}\right) - {v}_{p}\left( a\right) \) for \( {v}_{p}\left( a\right) < 0 \) .... | Yes |
Corollary 4.2.16. Assume that \( {v}_{p}\left( x\right) = v > 1/\left( {p - 1}\right) \) . If \( {v}_{p}\left( a\right) > 1/(p - \) 1) \( - v \) the series \( {\left( 1 + x\right) }^{a} \) converges, and \( {v}_{p}\left( {\left( 1 + x\right) }^{a}\right) = 0 \) . | Proof. By the above corollary, if \( {v}_{p}\left( a\right) \geq 0 \) the series converges when \( {v}_{p}\left( x\right) > 1/\left( {p - 1}\right) \), and if \( {v}_{p}\left( a\right) < 0 \) the series converges when \( {v}_{p}\left( x\right) > \) \( 1/\left( {p - 1}\right) - {v}_{p}\left( a\right) \), in other words ... | Yes |
Corollary 4.2.17. Let \( a \in {\mathbb{Z}}_{p} \) .\n\n(1) If \( p \geq 3 \) then for all \( x \in \mathcal{K} \) such that \( {v}_{p}\left( x\right) \geq 1/\left( {p - 2}\right) \) we have\n\n\[{v}_{p}\left( {{\left( 1 + x\right) }^{a} - 1 - {ax}}\right) \geq 2{v}_{p}\left( x\right) + {v}_{p}\left( a\right) ,\]\nin o... | Proof. By the above corollary (or by Lemma 4.2.12) we have the convergent series \( {\left( 1 + x\right) }^{a} = \mathop{\sum }\limits_{{n \geq 0}}\left( \begin{array}{l} a \\ n \end{array}\right) {x}^{n} \), and since \( {v}_{p}\left( {n!\left( \begin{array}{l} a \\ n \end{array}\right) }\right) = {v}_{p}(a\left( {a -... | Yes |
Corollary 4.2.18. Let \( a \in \mathcal{K} \) be fixed.\n\n(1) When \( {v}_{p}\left( a\right) > 1/\left( {p - 1}\right) \) the power series in \( x \)\n\n\[ {\phi }_{a}\left( x\right) = {\left( 1 + a\right) }^{x} = {\exp }_{p}\left( {x{\log }_{p}\left( {1 + a}\right) }\right) \]\n\nconverges if and only if \( {v}_{p}\l... | Proof. This immediately follows from Proposition 4.2.14 since \( {\exp }_{p}\left( x\right) \) converges when \( {v}_{p}\left( x\right) > 1/\left( {p - 1}\right) \), and since \( {v}_{2}\left( {{\log }_{2}\left( {1 + a}\right) }\right) \geq 2 \) when \( a \in 2{\mathbb{Z}}_{2} \) . | No |
Lemma 4.2.20. Let \( S\left( X\right) = \mathop{\sum }\limits_{{k \geq 0}}{s}_{k}{X}^{k} \in 1 + X{\mathbb{Q}}_{p}\left\lbrack \left\lbrack X\right\rbrack \right\rbrack \) be a power series with coefficients in \( {\mathbb{Q}}_{p} \) such that \( {s}_{0} = 1 \) . Then \( {s}_{k} \in {\mathbb{Z}}_{p} \) for all \( k \) ... | Proof. If \( S\left( X\right) = \mathop{\sum }\limits_{{k \geq 0}}{s}_{k}{X}^{k} \in 1 + X{\mathbb{Z}}_{p}\left\lbrack \left\lbrack X\right\rbrack \right\rbrack \) then clearly\n\n\[ S{\left( X\right) }^{p} \equiv \mathop{\sum }\limits_{{k \geq 0}}{s}_{k}^{p}{X}^{pk} \equiv \mathop{\sum }\limits_{{k \geq 0}}{s}_{k}{X}^... | Yes |
Corollary 4.2.21. The Artin-Hasse power series \( {E}_{p}\left( X\right) \) defined above has p-integral coefficients (in other words belongs to \( 1 + X{\mathbb{Z}}_{p}\left\lbrack \left\lbrack X\right\rbrack \right\rbrack \) ), and in particular has radius of convergence greater than or equal to 1 . | Proof. Indeed, we have \( {E}_{p}{\left( X\right) }^{p} = \exp \left( {{pX} + \mathop{\sum }\limits_{{n \geq 1}}{X}^{{p}^{n}}/{p}^{n - 1}}\right) \) and \( {E}_{p}\left( {X}^{p}\right) = \exp \left( {\mathop{\sum }\limits_{{n \geq 0}}{X}^{{p}^{n + 1}}/{p}^{n}}\right) \), hence\n\n\[ \frac{{E}_{p}{\left( X\right) }^{p}}... | Yes |
Theorem 4.2.22. Set \( \exp \left( {X + {X}^{p}/p}\right) = \mathop{\sum }\limits_{{k \geq 0}}{u}_{k}{X}^{k} \) . Then\n\n\[ \n{v}_{p}\left( {u}_{k}\right) \geq - \frac{{2p} - 1}{\left( {p - 1}\right) {p}^{2}}\left( {k - {s}_{p}\left( k\right) }\right) .\n\] | Proof. We trivially have\n\n\[ \nexp \left( {X + {X}^{p}/p}\right) = {E}_{p}\left( X\right) \mathop{\prod }\limits_{{n \geq 2}}\exp \left( {-{X}^{{p}^{n}}/{p}^{n}}\right) .\n\]\n\nWrite \( \exp \left( {-{X}^{{p}^{n}}/{p}^{n}}\right) = \mathop{\sum }\limits_{{k \geq 0}}{c}_{n, k}{X}^{k} \) . We have \( {c}_{n, k} = 0 \)... | Yes |
Corollary 4.2.23. (1) Set\n\n\[ \n{u}_{k} = \mathop{\sum }\limits_{{j = 0}}^{{\lfloor k/p\rfloor }}\frac{1}{{p}^{j}j!\left( {k - {pj}}\right) !}. \]\n\nThen\n\[ \n{v}_{p}\left( {u}_{k}\right) \geq - \frac{{2p} - 1}{\left( {p - 1}\right) {p}^{2}}\left( {k - {s}_{p}\left( k\right) }\right) . \]\n\n(2) For \( 0 \leq r < p... | Proof. The result for \( {u}_{k} \) is simply a reformulation of the theorem since the formula for \( {u}_{k} \) is obtained by computing the product of the series \( \exp \left( X\right) \) and \( \exp \left( {{X}^{p}/p}\right) \) . For \( {a}_{r, k} \) it is immediate that \( {a}_{r, k} = {\left( -p\right) }^{k}k!{u}... | Yes |
Lemma 4.2.24. For any sequence \( {a}_{k} \) such that \( {a}_{k} \in {\mathbb{Z}}_{p} \) and tending to 0 p-adically as \( k \rightarrow \infty \) the function\n\n\[ f\left( x\right) = \mathop{\sum }\limits_{{k \geq 0}}{a}_{k}\left( \begin{array}{l} x \\ k \end{array}\right) \]\n\nis well defined and is continuous on ... | Proof. Since by Lemma 4.2.8 we have \( \left( \begin{array}{l} x \\ k \end{array}\right) \in {\mathbb{Z}}_{p} \), the term \( {a}_{k}\left( \begin{array}{l} x \\ k \end{array}\right) \) tends to 0 as \( k \rightarrow \infty \) so that \( f\left( x\right) \) is well defined. Furthermore, for the same reason (i.e., \( {\... | Yes |
Proposition 4.2.25. If\n\n\[ \n{b}_{n} = \mathop{\sum }\limits_{{0 \leq k \leq n}}{a}_{k}\left( \begin{array}{l} n \\ k \end{array}\right) \n\]\n\nthen\n\[ \n{a}_{k} = \mathop{\sum }\limits_{{0 \leq m \leq k}}{\left( -1\right) }^{k - m}{b}_{m}\left( \begin{matrix} k \\ m \end{matrix}\right) .\n\] | Proof. One can of course prove this directly. However, the simplest proof is by using generating functions: if \( A\left( T\right) = \mathop{\sum }\limits_{{k \geq 0}}{a}_{k}{T}^{k}/k \) ! and \( B\left( T\right) = \) \( \mathop{\sum }\limits_{{k \geq 0}}{b}_{k}{T}^{k}/k! \), then by definition \( B\left( T\right) = \e... | Yes |
Corollary 4.2.27. Let \( {b}_{n} \) be a sequence of elements of \( {\mathbb{Z}}_{p} \) . The following conditions are equivalent.\n\n(1) This sequence is p-adically continuous; in other words, for all \( k \geq 0 \) there exists \( j \geq 0 \) such that \( {v}_{p}\left( {n - m}\right) \geq j \) implies that \( {v}_{p}... | Proof. Clear. | No |
Proposition 4.2.28. Let \( f\left( x\right) = \mathop{\sum }\limits_{{k \geq 0}}{a}_{k}\left( \begin{array}{l} x \\ k \end{array}\right) \), and assume that for all \( k \) we have \( {v}_{p}\left( {a}_{k}\right) \geq {\alpha k} + {\alpha }^{\prime }{s}_{p}\left( k\right) + {\alpha }^{\prime \prime } \) for some consta... | Proof. Define integers \( s\left( {k, j}\right) \) (called Stirling numbers of the first kind) by the formula\n\n\[ X\left( {X - 1}\right) \cdots \left( {X - k + 1}\right) = \mathop{\sum }\limits_{{j \geq 0}}{\left( -1\right) }^{k - j}s\left( {k, j}\right) {X}^{j}, \]\n\nwith \( s\left( {k, j}\right) = 0 \) for \( j > ... | Yes |
Corollary 4.2.29. Let \( {\left( {c}_{k}\right) }_{k \geq 0} \) be a sequence of elements of \( \mathbb{Z} \), set\n\n\[ \n{a}_{k} = \mathop{\sum }\limits_{{0 \leq m \leq k}}{\left( -1\right) }^{k - m}\left( \begin{matrix} k \\ m \end{matrix}\right) {c}_{m},\n\]\n\nand assume that as \( k \rightarrow \infty \) we have ... | Proof. Clear from the above results. | No |
Proposition 4.3.1. An element \( x \in {\mathbb{Z}}_{\mathfrak{p}} \) is invertible in \( {\mathbb{Z}}_{\mathfrak{p}} \) if and only if \( {v}_{p}\left( x\right) = 0 \), i.e., \( {\left| x\right| }_{\mathfrak{p}} = 1 \) . | Proof. Clear since \( x \in {\mathbb{Z}}_{\mathfrak{p}} \) if and only if \( {\left| x\right| }_{\mathfrak{p}} \leq 1 \) . | No |
Proposition 4.3.2. Let \( {\zeta }_{1} \) and \( {\zeta }_{2} \) in \( \overline{\mathbb{Q}} \) be distinct roots of unity of order not divisible by \( p \) . Then \( {\zeta }_{1} - {\zeta }_{2} \) is invertible modulo \( p \) ; in other words, there exists an algebraic integer \( \alpha \) such that \( \alpha \left( {... | Proof. Let \( m \) be the LCM of the orders of \( {\zeta }_{1} \) and \( {\zeta }_{2} \), so that \( {\zeta }_{1}^{m} = {\zeta }_{2}^{m} = 1 \) and \( p \nmid m \) . From the identity\n\n\[ \frac{{X}^{m} - 1}{X - {\zeta }_{1}} = \mathop{\prod }\limits_{{{\zeta }^{m} = 1,\zeta \neq {\zeta }_{1}}}\left( {X - \zeta }\righ... | Yes |
Lemma 4.3.3. Set\n\n\[ \n{k}_{0} = \left\lfloor \frac{\log \left( {e/\left( {p - 1}\right) }\right) }{\log \left( p\right) }\right\rfloor + 1 \]\n\nThen if \( x \equiv 1\left( {{\;\operatorname{mod}\;\mathfrak{p}}{\mathbb{Z}}_{\mathfrak{p}}}\right) \), for \( k \geq {k}_{0} \) we have \( {v}_{p}\left( {{x}^{{p}^{k}} - ... | Proof. We write \( x = 1 + y \) with \( {v}_{\mathfrak{p}}\left( y\right) \geq 1 \), in other words \( {v}_{p}\left( y\right) \geq 1/e \), so\n\nthat\n\[ \n{x}^{{p}^{k}} = 1 + {y}^{{p}^{k}} + \mathop{\sum }\limits_{{1 \leq m \leq {p}^{k} - 1}}\left( \begin{matrix} {p}^{k} \\ m \end{matrix}\right) {y}^{m}. \]\n\nFor \( ... | Yes |
Proposition 4.3.4. Let \( x \in {U}_{0} \) be a \( \mathfrak{p} \) -adic unit.\n\n(1) The sequence \( {x}^{\mathcal{N}{\mathfrak{p}}^{n}} \) converges to some \( \omega \left( x\right) \in {U}_{0} \) characterized by the two properties \( \omega \left( x\right) \equiv x\left( {{\;\operatorname{mod}\;\mathfrak{p}}{\math... | Proof. (1) and (2). Since \( \left| {\left( {\mathbb{Z}}_{\mathfrak{p}}/{\mathfrak{p}}^{a}{\mathbb{Z}}_{\mathfrak{p}}\right) }^{ * }\right| = \mathcal{N}{\mathfrak{p}}^{a - 1}\left( {\mathcal{N}\mathfrak{p} - 1}\right) \), when \( x \in {U}_{0} \) we have \( {x}^{\mathcal{N}{\mathfrak{p}}^{a - 1}\left( {\mathcal{N}\mat... | Yes |
Corollary 4.3.5. There exists a canonical group isomorphism \( \bar{\omega } \) between \( {\left( {\mathbb{Z}}_{K}/\mathfrak{p}\right) }^{ * } \) and the subgroup \( {\mu }_{\mathfrak{p}} \) of \( z \in {U}_{0} \) such that \( {z}^{\mathcal{N}\mathfrak{p} - 1} = 1 \) . In particular, \( {K}_{\mathfrak{p}} \) contains ... | Proof. Let \( \phi \) be the canonical isomorphism from \( {\mathbb{Z}}_{K}/\mathfrak{p} \) to \( {\mathbb{Z}}_{\mathfrak{p}}/\mathfrak{p}{\mathbb{Z}}_{\mathfrak{p}} \) given by Proposition 4.1.24. The map \( \bar{\omega } = \omega \circ \phi \) is then a canonical group homomorphism from \( {\left( {\mathbb{Z}}_{K}/\m... | Yes |
Corollary 4.3.6. The Teichmüller character induces a canonical isomorphism from \( {\left( {\mathbb{Z}}_{K}/\mathfrak{p}\right) }^{ * } \times {U}_{1} \) to \( {U}_{0} \) . | Proof. For \( \left( {a, x}\right) \in {\left( {\mathbb{Z}}_{K}/\mathfrak{p}\right) }^{ * } \times {U}_{1} \) define \( {\psi }_{1}\left( \left( {a, x}\right) \right) = \bar{\omega }\left( a\right) \cdot x \in {U}_{0} \), and for \( x \in {U}_{0} \) define \( {\psi }_{2}\left( x\right) = \left( {{\bar{\omega }}^{-1}\le... | Yes |
Corollary 4.3.7. We have\n\n\[ \n{K}_{\mathfrak{p}}^{ * } = {\pi }^{\mathbb{Z}} \times {\mu }_{\mathfrak{p}} \times {U}_{1} \simeq \mathbb{Z} \times {\left( {\mathbb{Z}}_{K}/\mathfrak{p}\right) }^{ * } \times {U}_{1}, \]\n\nwhere \( {\mu }_{\mathfrak{p}} \) is the group of \( \left( {\mathcal{N}\mathfrak{p} - 1}\right)... | Proof. Clear. | No |
Lemma 4.3.9. For every \( i \geq 1 \), the multiplicative group \( {U}_{i}/{U}_{i + 1} \) is non-canonically isomorphic to the additive group \( {\mathbb{Z}}_{K}/\mathfrak{p} \), hence to \( {\left( \mathbb{Z}/p\mathbb{Z}\right) }^{f\left( {\mathfrak{p}/p}\right) } \) . In particular, for every \( i \geq 1 \) we have \... | Proof. Let \( \pi \) be a uniformizer of \( \mathfrak{p} \) . If \( x \) and \( y \) are in \( {U}_{i} \) we have\n\n\[ \frac{{xy} - 1}{{\pi }^{i}} - \frac{x - 1}{{\pi }^{i}} - \frac{y - 1}{{\pi }^{i}} = \frac{\left( {x - 1}\right) \left( {y - 1}\right) }{{\pi }^{i}} \equiv 0\left( {\;\operatorname{mod}\;\mathfrak{p}}\... | Yes |
Proposition 4.3.11. If we are not in the special case \( p = 2 \) and \( K = {\mathbb{Q}}_{2} \) then \( \langle x\rangle \) is the unique element of \( {U}_{1} \) such that \( x/\langle x\rangle \) is an \( \left( {\mathcal{N}\mathfrak{p} - 1}\right) \) st root of unity. On the other hand, if \( p = 2 \) and \( K = {\... | Proof. Immediate and left to the reader. | No |
Proposition 4.3.12. Set \( z\left( \mathfrak{p}\right) = \left\lfloor \frac{e\left( {\mathfrak{p}/p}\right) }{p - 1}\right\rfloor + 1 \) . Then for all \( i \geq z\left( \mathfrak{p}\right) \) the \( \mathfrak{p} \) - adic logarithm and exponential give inverse isomorphisms between the multiplicative group \( {U}_{i} \... | Proof. If \( x \in {U}_{z\left( \mathfrak{p}\right) } \) then \( \left| {x - 1}\right| < 1 \), so that the logarithm converges. Furthermore, writing as usual \( e = e\left( {\mathfrak{p}/p}\right) \), we have\n\n\[ \n{v}_{p}\left( {{\left( x - 1\right) }^{k - 1}/k}\right) = \left( {k - 1}\right) {v}_{p}\left( {x - 1}\r... | Yes |
Corollary 4.3.13. (1) For every \( i \geq 1,{U}_{i} \) has a natural \( {\mathbb{Z}}_{p} \) -module structure. | Proof. (1). For \( x = 1 + y \in {U}_{i} \) with \( i \geq 1 \) and \( \alpha \in {\mathbb{Z}}_{p} \), we set directly\n\n\[{\left( 1 + y\right) }^{\alpha } = \mathop{\sum }\limits_{{n \geq 0}}\left( \begin{array}{l} \alpha \\ n \end{array}\right) {y}^{n}\]\n\nBy Corollary 4.2.15 this series converges and we have \( {x... | Yes |
Corollary 4.3.14. As usual set \( e = e\left( {\mathfrak{p}/p}\right) \) and \( f = f\left( {\mathfrak{p}/p}\right) \) . As abelian groups we have the isomorphism\n\n\[ \n{K}_{\mathfrak{p}}^{ * } \simeq {\mu }_{\mathfrak{p}}^{\prime } \times \mathbb{Z} \times {\mathbb{Z}}_{p}^{ef} \n\] \n\nwhere \( {\mu }_{\mathfrak{p}... | Proof. By Corollary 4.3.7 and the above corollary we have \( {K}_{\mathfrak{p}}^{ * } \simeq {\mu }_{\mathfrak{p}} \times {\mu }_{\mathfrak{p},1} \times \) \( \mathbb{Z} \times {\mathbb{Z}}_{p}^{ef} \) and \( \left| {\mu }_{\mathfrak{p}}\right| = \mathcal{N}\mathfrak{p} - 1 \), while \( \left| {\mu }_{\mathfrak{p},1}\r... | Yes |
Lemma 4.3.15 (Nakayama). Let \( M \) be a finitely generated \( {\mathbb{Z}}_{\mathfrak{p}} \) -module. The equality \( M = \mathfrak{p}M \) implies that \( M = 0 \) . In particular, a set \( \left( {x}_{i}\right) \) of elements of \( M \) is a generating set for \( M \) if and only if the classes modulo \( \mathfrak{p... | Proof. Assume that \( M = \mathfrak{p}M \neq 0 \), and let \( {\left( {m}_{i}\right) }_{1 \leq i \leq k} \) be a system of generators of \( M \) with \( k \) minimal. If \( \pi \) is a uniformizer of \( \mathfrak{p} \), we have \( {\pi M} = \mathfrak{p}M \) since any element of \( \mathfrak{p} \) can be written as \( {... | Yes |
Proposition 4.3.16. Let \( \pi \) be a uniformizer of \( {\mathbb{Z}}_{\mathfrak{p}} \), and let \( x = 1 + u{\pi }^{i} \) with \( {ua}\mathfrak{p} \) -adic unit be an element of \( {U}_{i} \smallsetminus {U}_{i + 1} \) . Denote by \( e = e\left( {\mathfrak{p}/p}\right) \) the absolute ramification index of \( \mathfra... | Proof. Simply look at the valuations in the binomial expansion. | No |
Theorem 4.3.18. Let \( {\zeta }_{p} \) be a primitive pth root of unity.\n\n(1) We have equality of \( \mathfrak{p} \)-adic fields\n\n\[ \n{\mathbb{Q}}_{p}\left( {\left( -p\right) }^{1/\left( {p - 1}\right) }\right) = {\mathbb{Q}}_{p}\left( {\zeta }_{p}\right) \n\]\n\nin other words, the field extensions of \( {\mathbb... | Proof. (1). The polynomial \( {X}^{p - 1} + p \) is an Eisenstein polynomial, and the polynomial \( {X}^{p - 1} + \cdots + 1 \) is a shifted Eisenstein polynomial (see Corollary 4.1.36), so both are irreducible in \( {\mathbb{Q}}_{p}\left\lbrack X\right\rbrack \) hence define extensions of the same degree \( p - 1 \) .... | Yes |
Corollary 4.3.19. Let \( {K}_{\mathfrak{p}} \) be a regular field, and denote by \( e \) and \( f \) its ramification and residual indices. Let \( \pi \) be a uniformizer, and let \( {\zeta }_{1},\ldots ,{\zeta }_{f} \) be lifts to \( {\mu }_{\mathfrak{p}} \) of an \( {\mathbb{F}}_{p} \) -basis of the residue field \( ... | Proof. By Lemma 4.3.9, the \( {\eta }_{i, j} \) for fixed \( i \) and \( 1 \leq j \leq f \) form a generating set of \( {U}_{i} \) modulo \( {U}_{i + 1} \) . Let us call this (for this proof only) a generating set of level \( i \) . With the notation of Proposition 4.3.12, when \( i = 1,2,\ldots ,\lfloor e/\left( {p - ... | Yes |
Proposition 4.3.20. Let \( \mathfrak{p} \) be a prime ideal, let \( p \) be the prime number below \( \mathfrak{p} \), let \( \pi \) be a uniformizer of \( \mathfrak{p} \), and set \( e = e\left( {\mathfrak{p}/p}\right) \) and \( f = f\left( {\mathfrak{p}/p}\right) \) . (1) If \( \mathfrak{p} \) is not above 2 then \( ... | Proof. (1). By Proposition 4.3.7 we have \( {K}_{\mathfrak{p}}^{ * } = {\pi }^{\mathbb{Z}} \times {\mu }_{\mathfrak{p}} \times {U}_{1} \) . By Corollary 4.2.15 for \( \mathfrak{p} \nmid 2 \) the series \( {\left( 1 + x\right) }^{1/2} \) converges for \( \left| x\right| < 1 \) ; hence every element of \( {U}_{1} \) is a... | Yes |
Corollary 4.3.21. Up to isomorphism there are exactly three quadratic extensions of \( {K}_{\mathfrak{p}} \) when \( \mathfrak{p} \nmid 2 \), and \( {2}^{{ef} + 2} - 1 \) when \( \mathfrak{p} \mid 2 \) . | Proof. Clear since quadratic extensions of a field \( L \) of characteristic different from 2 are in one-to-one correspondence with the classes in \( {L}^{ * }/{L}^{*2} \) other than the unit class. | No |
Proposition 4.3.22. Assume that \( p \geq 3 \), and let \( a \in {\mathbb{Q}}_{p}^{ * } \) . A necessary and sufficient condition for a to be a square in \( {\mathbb{Q}}_{p}^{ * } \) is that \( v\left( a\right) \) be even and \( \left( \frac{a/{p}^{v\left( a\right) }}{p}\right) = 1 \) . Furthermore, a system of represe... | Proof. Clear. | No |
Proposition 4.3.23. A necessary and sufficient condition for some \( a \in {\mathbb{Q}}_{2}^{ * } \) to be a square in \( {\mathbb{Q}}_{2}^{ * } \) is that \( v\left( a\right) \) be even and \( a/{p}^{v\left( a\right) } \equiv 1\left( {\;\operatorname{mod}\;8}\right) \) . Furthermore, a system of representatives of \( ... | Proof. Since the square of an odd number is congruent to 1 modulo 8, as above the condition is necessary. For the converse, note that by Lemma 4.2.8\n\nwe have\n\[ \n{v}_{2}\left( \left( \begin{matrix} 1/2 \\ k \end{matrix}\right) \right) = - \left( {k + {v}_{2}\left( {k!}\right) }\right) \geq - {2k} \n\]\n\nhence the ... | Yes |
Proposition 4.4.1. Let \( \mathcal{K} \) be a complete field with respect to some absolute value \( \left| \cdot \right| \), and let \( V \) be a finite-dimensional \( \mathcal{K} \) -vector space. Any two norms on \( V \) are equivalent, and the corresponding topology makes \( V \) into a complete metric space. | Proof. Let \( {e}_{1},\ldots ,{e}_{n} \) be a fixed \( \mathcal{K} \) -basis of \( V \) . We consider the sup norm\n\n\[{\begin{Vmatrix}\mathop{\sum }\limits_{{1 \leq i \leq n}}{x}_{i}{e}_{i}\end{Vmatrix}}_{\infty } = \mathop{\max }\limits_{{1 \leq i \leq n}}\left| {x}_{i}\right|\]\n\nIt is indeed clearly a norm, and \... | Yes |
Corollary 4.4.2. Let \( \mathcal{K} \) be a complete field with respect to some nontrivial absolute value ||, and let \( \mathcal{L} \) be a finite extension of \( \mathcal{K} \). There exists at most one extension \( \parallel \parallel \) of \( \parallel \) to \( \mathcal{L} \). In addition, \( \mathcal{L} \) is comp... | Proof. We can consider \( \mathcal{L} \) as a finite-dimensional \( \mathcal{K} \) -vector space, and an absolute value \( \parallel \parallel \) on \( \mathcal{L} \) clearly satisfies the conditions of a norm. By the above proposition, any two such absolute values \( \parallel {\parallel }_{1} \) and \( \parallel {\pa... | Yes |
Corollary 4.4.4. Let \( \mathcal{K} \) be a completion of a number field \( K \) with respect to some nontrivial absolute value \( \left| \right| \) . There exists an extension of \( \left| \right| \) to the algebraic closure \( \overline{\mathcal{K}} \) of \( \mathcal{K} \), and this extension is unique. | Proof. Since the algebraic closure is the union of all finite extensions included in it, the result is clear. | No |
Proposition 4.4.5. Let \( \mathcal{K} = {K}_{\mathfrak{p}} \), let \( L/K \) be an extension, let \( \mathfrak{P} \) be a prime ideal of \( L \) above \( \mathfrak{p} \), let \( \mathcal{L} = {L}_{\mathfrak{P}} \), and set \( n = \left\lbrack {\mathcal{L} : \mathcal{K}}\right\rbrack = e\left( {\mathfrak{P}/\mathfrak{p}... | Proof. Since both norms define the same topology on \( \mathcal{L} \) they are equivalent, so that one is a power of the other. To determine which power, we simply choose \( x = \pi \), where \( \pi \) is a uniformizer of \( \mathfrak{p} \), and the result immediately follows. | No |
Proposition 4.4.6 (Krasner’s lemma). Let \( \mathcal{K} \) be a completion of a number field \( K \) with respect to some nontrivial absolute value \( \mid \mid \) and let \( \mathcal{L} \) be a finite extension of \( \mathcal{K} \) . (1) If \( x \) and \( y \) in \( \mathcal{L} \) are conjugate over \( \mathcal{K} \) ... | Proof. (1) is clear since \( x \) and \( y \) have the same norm. If (2) were not true, in other words if for some \( a \in \mathcal{K} \) we had \( \left| {x - y}\right| > \left| {a - x}\right| \), we would have \[ \left| {a - y}\right| = \max \left( {\left| {a - x}\right| ,\left| {x - y}\right| }\right) = \left| {x -... | Yes |
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