Q stringlengths 4 3.96k | A stringlengths 1 3k | Result stringclasses 4
values |
|---|---|---|
Proposition 1. In order that \( \mathrm{K} \) be locally compact, it is necessary and sufficient that its residue field \( \overline{\mathrm{K}} = \mathrm{A}/\pi \mathrm{A} \) be a finite field and \( \mathrm{K} \) be complete. | If \( \mathrm{K} \) is locally compact, it is complete. And as the \( {\pi }^{n}\mathrm{\;A} \) form a fundamental system of closed neighborhoods of 0 , at least one of them is compact, so multiplying by \( {\pi }^{-n} \) shows that \( \mathrm{A} \) is compact. The quotient \( \overline{\mathrm{K}} = \mathrm{A}/\pi \ma... | Yes |
Proposition 2. Let \( \mathrm{K} \) be a field satisfying the conditions of prop. 1, and let \( \mu \) be a Haar measure on the locally compact additive group underlying \( \mathrm{K} \) . Then for every measurable subset \( \mathrm{E} \) of \( \mathrm{K} \) and every \( x \in \mathrm{K} \) one has \[ \mu \left( {x\mat... | One may assume \( x \neq 0 \) ; the homothety \( y \mapsto {xy} \) is then an automorphism of the additive group of \( \mathrm{K} \), hence transforms the Haar measure \( \mu \) into one of its multiples \( \chi \left( x\right) .\mu \), and one must verify that the multiplier \( \chi \left( x\right) \) is equal to \( \... | Yes |
Proposition 3. Let \( \mathrm{K} \) be a field on which a discrete valuation \( v \) is defined, having valuation ring A. Assume \( \mathrm{K} \) to be complete in the topology defined by \( v \) . Let \( \mathrm{L}/\mathrm{K} \) be a finite extension of \( \mathrm{K} \), and let \( \mathrm{B} \) be the integral closur... | We begin with the case L/K separable. Condition (F) of Chap. I, §4 is then automatically satisfied; as A is principal, it follows that B is a free A-module of rank \( n \) . Let \( {\mathfrak{P}}_{i} \) be the prime ideals of \( \mathrm{B} \), with \( {w}_{i} \) the corresponding valuations. Each \( {w}_{i} \) defines ... | No |
Corollary 1. If \( e \) (resp. \( f \) ) denotes the ramification index (resp. the residue degree) of \( \mathrm{L} \) over \( \mathrm{K} \), then ef \( = n \) . | That follows from prop. 10 of Chapter I, which is applicable because we have shown that \( \mathrm{B} \) is an \( \mathrm{A} \) -module of finite type. | No |
Corollary 2. There is a unique valuation \( w \) of \( \mathrm{L} \) that prolongs \( v \) . | This is just a reformulation of part of the proposition. | No |
Corollary 3. Two elements of \( \mathrm{L} \) that are conjugate over \( \mathrm{K} \) have the same valuation. | Enlarging \( \mathrm{L} \) if necessary, we can assume \( \mathrm{L}/\mathrm{K} \) to be normal. If \( s \in \mathrm{G}\left( {\mathrm{L}/\mathrm{K}}\right) \) , \( w \circ s \) prolongs \( v \), hence coincides with \( w \) (cor. 2); the corollary then results from the fact that the conjugates of \( x \in \mathrm{L} \... | Yes |
Corollary 4. For every \( x \in \mathrm{L}, w\left( x\right) = \left( {1/f}\right) v\left( {{\mathrm{\;N}}_{\mathrm{L}/\mathrm{K}}\left( x\right) }\right) \) . | Here again one reduces to the case \( \mathrm{L}/\mathrm{K} \) normal, where the assertion results from cor. 3. [One could just as well directly apply prop. 14 of Chap. I.] | No |
Theorem 1. Let \( \mathrm{L}/\mathrm{K} \) be an extension of finite degree \( n, v \) a discrete valuation of \( \mathrm{K} \) with ring \( \mathrm{A} \), and \( \mathrm{B} \) the integral closure of \( \mathrm{A} \) in \( \mathrm{L} \) . Suppose that the \( \mathrm{A} \) - module \( \mathrm{B} \) is finitely generate... | Statement (ii) is evident, taking \( §2 \) into account, and it implies statement (i). On the other hand, the product topology makes \( \prod {\widehat{\mathrm{L}}}_{i} \) into a Hausdorff topological vector space of dimension \( n \) over \( \widehat{\mathrm{K}} \) ; by the approximation lemma (Chap. I,§3), \( \varphi... | Yes |
Corollary 1. The fields \( {\widehat{\mathrm{L}}}_{i} \) are the composites of the extensions \( \widehat{\mathrm{K}} \) and \( \mathrm{L} \) of \( \mathrm{K} \) . | One knows that those composites are the quotient fields of the tensor product \( \mathrm{L}{ \otimes }_{\mathrm{K}}\widehat{\mathrm{K}} \) (cf. Bourbaki, Alg., Chap. VIII,§8). | Yes |
Corollary 2. If \( x \in \mathrm{L} \), the characteristic polynomial \( \mathrm{F} \) of \( x \) in \( \mathrm{L}/\mathrm{K} \) is equal to the product of the characteristic polynomials \( {\mathrm{F}}_{i} \) of \( x \) in the \( {\widehat{\mathbf{L}}}_{i}/\widehat{\mathbf{K}} \). In particular, if \( \mathrm{{Tr}} \)... | The polynomial \( \mathrm{F} \) is also the characteristic polynomial of \( x \) in the \( \mathrm{K} \) - algebra \( \mathrm{L}{ \otimes }_{\mathrm{K}}\widehat{\mathrm{K}} \). The formula \( \mathrm{F} = \prod {\mathrm{F}}_{i} \) follows from the isomorphism (iii), and the trace and norm formulas are an immediate cons... | Yes |
Corollary 3. If \( \mathrm{L}/\mathrm{K} \) is separable (in which case the finiteness hypothesis made on \( \mathrm{B} \) is automatically satisfied), the \( {\widehat{\mathrm{L}}}_{i}/\widehat{\mathrm{K}} \) are also. | For we have \( {\widehat{\mathrm{L}}}_{i} = \mathrm{L}\widehat{\mathrm{K}} \) . | No |
Corollary 4. If \( \mathrm{L}/\mathrm{K} \) is Galois with group \( \mathrm{G} \), and if \( {\mathrm{D}}_{i} \) denotes the decomposition group of \( {w}_{i} \) in \( \mathrm{G} \) (cf. Chap. I,§7), the extension \( {\widehat{\mathrm{L}}}_{i}/\widehat{\mathrm{K}} \) is Galois with Galois group \( {\mathrm{D}}_{i} \) . | Every element of \( {\mathrm{D}}_{i} \) extends by continuity to a \( \widehat{\mathrm{K}} \) -automorphism of \( {\widehat{\mathrm{L}}}_{i} \) , and the corollary results from the fact that \( {\mathrm{D}}_{i} \) has order \( \left\lbrack {{\widehat{\mathrm{L}}}_{i} : \widehat{\mathrm{K}}}\right\rbrack \) . | Yes |
Proposition 4. With the hypotheses and notation of theorem 1, let \( {\mathbf{B}}_{i} \) be the ring of the valuation \( {w}_{i} \) . The canonical homomorphism\n\n\[ \varphi : \mathrm{B}{ \otimes }_{\mathrm{A}}\widehat{\mathrm{A}} \rightarrow \mathop{\prod }\limits_{i}{\widehat{\mathrm{B}}}_{i} \]\n\nis then an isomor... | Both sides are free Â-modules of rank \( n \) . To show that \( \varphi \) is bijective, it suffices therefore to see that it is when one reduces modulo the maximal ideal \( \widehat{m} \) of \( \widehat{\mathbf{A}} \) . One gets \( \mathbf{B}/m\mathbf{B} \) for the left side, and \( \prod \mathbf{B}/{m}_{i}^{{e}_{i}}\... | Yes |
Proposition 5. Every element \( a \in \mathrm{A} \) can be written uniquely as a convergent series\n\n\[ a = \mathop{\sum }\limits_{{n = 0}}^{\infty }{s}_{n}{\pi }^{n},\;\text{ with }{s}_{n} \in \mathrm{S}. \] | The second assertion results from the first by multiplying by a suitable negative power of \( \pi \) . Thus let \( a \in \mathrm{A} \) ; by definition of \( \mathrm{S} \), there is an \( {s}_{0} \in \mathrm{S} \) such that \( a - {s}_{0} \equiv 0{\;\operatorname{mod}\;.}\pi \) ; if one writes \( a = {s}_{0} + \pi {a}_{... | Yes |
Theorem 2. Let \( \mathrm{A} \) be a complete discrete valuation ring with residue field \( \bar{\mathrm{K}} \) . Suppose that \( \mathrm{A} \) and \( \overline{\mathrm{K}} \) have the same characteristic and that \( \overline{\mathrm{K}} \) is perfect. Then \( \mathrm{A} \) is isomorphic to \( \overline{\mathrm{K}}\le... | It all comes down to showing that A contains a system of representatives which is a field. We will distinguish two cases, depending on the characteristic:\n\n## (i) The characteristic of \( \overline{\mathrm{K}} \) is 0 .\n\nThe existence of a field of representatives is then true for local rings that are far more gene... | No |
Proposition 6. Let \( \mathrm{A} \) be a local ring that is Hausdorff and complete for the topology defined by a decreasing sequence \( {\mathfrak{a}}_{1} \supset {\mathfrak{a}}_{2} \supset \cdots \) of ideals such that \( {\mathfrak{a}}_{n}.{\mathfrak{a}}_{m} \subset {\mathfrak{a}}_{n + m} \) . Suppose that \( \overli... | As \( \mathbf{Z} \rightarrow \mathrm{A} \rightarrow \overline{\mathrm{K}} \) is injective, the homomorphism \( \mathbf{Z} \rightarrow \mathrm{A} \) extends to \( \mathbf{Q} \), and we see that A contains Q. By Zorn's lemma, there exists a maximal subfield \( \mathrm{S} \) of \( \mathrm{A} \) ; if \( \overline{\mathrm{S... | Yes |
Proposition 7. Let \( \mathrm{A} \) be a local ring that is Hausdorff and complete for the topology defined by a decreasing sequence \( {\mathfrak{a}}_{1} \supset {\mathfrak{a}}_{2} \supset \cdots \) of ideals such that \( {\mathfrak{a}}_{n} \cdot {\mathfrak{a}}_{m} \subset {\mathfrak{a}}_{n + m} \) . Suppose that \( {... | If \( x \) is such a root, one has \( f\left( \mathrm{X}\right) = \left( {\mathrm{X} - x}\right) g\left( \mathrm{X}\right) \), with \( \bar{g}\left( \lambda \right) \neq 0 \) ; if \( {x}^{\prime } \) is also such a root, substituting \( {x}^{\prime } \) for \( \mathrm{X} \) yields \( 0 = \left( {{x}^{\prime } - x}\righ... | Yes |
Proposition 8. Let \( \mathrm{A} \) be a ring that is Hausdorff and complete for the topology defined by a decreasing sequence \( {\mathfrak{a}}_{1} \supset {\mathfrak{a}}_{2} \supset \cdots \) of ideals such that \( {\mathfrak{a}}_{n} \cdot {\mathfrak{a}}_{m} \subset \) \( {\mathfrak{a}}_{n + m} \) . Assume that the r... | Let \( \lambda \in \overline{\mathrm{K}} \) ; for all \( n \geq 0 \), denote by \( {\mathrm{L}}_{n} \) the inverse image of \( {\lambda }^{{p}^{-n}} \) in \( \mathrm{A} \), and by \( {\mathrm{U}}_{n} \) the set of all \( {x}^{{p}^{n}}, x \in {\mathrm{L}}_{n} \) ; the \( {\mathrm{U}}_{n} \) are contained in the residue ... | Yes |
Lemma 1. If \( a \equiv b{\;\operatorname{mod}\;.}{\mathfrak{a}}_{n} \), then \( {a}^{p} \equiv {b}^{p}{\;\operatorname{mod}\;.}{\mathfrak{a}}_{n + 1} \) . | This lemma results from the binomial formula, taking into account that \( p \in {a}_{1} \), whence \( p{\mathfrak{a}}_{n} \subset {\mathfrak{a}}_{n + 1} \) . | Yes |
Theorem 3. For every perfect field \( k \) of characteristic \( p \), there exists a complete discrete valuation ring and only one (up to unique isomorphism) which is absolutely unramified and has \( k \) as its residue field. | In what follows, this ring will be denoted \( \mathbf{W}\left( k\right) \) . It is \ | No |
Theorem 4. Let \( \\mathrm{A} \) be a complete discrete valuation ring of characteristic unequal to that of its residue field \( k \) . Let \( e \) be its absolute ramification index. Then there exists a unique homomorphism of \( \\mathrm{W}\\left( k\\right) \) into \( \\mathrm{A} \) which makes commutative the diagram... | [By applying prop. 18 of Chap. I, one sees that A is obtained by adjoining to \( \\mathrm{W}\\left( k\\right) \) an element \( \\pi \) satisfying an \ | No |
Proposition 9. Let \( \\mathrm{A} \) be a p-ring with residue ring \( k \) and let \( f : k \\rightarrow \\mathrm{A} \) be the system of multiplicative representatives in A. Let \( \\left\\{ {\\alpha }_{i}\\right\\} \) and \( \\left\\{ {\\beta }_{i}\\right\\} \) be two sequences of elements of \( k \) . Then\n\n\[ \n\\... | One sees immediately that there is a homomorphism \( \\theta \) of \( \\mathbf{Z}\\left\\lbrack {{\\mathrm{X}}_{i}^{{p}^{-\\infty }},{\\mathrm{Y}}_{i}^{{p}^{-\\infty }}}\\right\\rbrack \) into A which maps \( {\\mathrm{X}}_{i} \) to \( f\\left( {\\alpha }_{i}\\right) \) and \( {\\mathrm{Y}}_{i} \) to \( f\\left( {\\bet... | Yes |
Proposition 10. Let \( \mathrm{A} \) and \( {\mathrm{A}}^{\prime } \) be two p-rings with residue rings \( k \) and \( {k}^{\prime } \), and suppose that \( \mathrm{A} \) is strict. For every homomorphism \( \phi : k \rightarrow {k}^{\prime } \), there exists a unique homomorphism \( g : \mathrm{A} \rightarrow {\mathrm... | We have already remarked that every homomorphism of \( A \) into \( {A}^{\prime } \) commutes with multiplicative representatives. If \( a \in \mathrm{A} \) is an element with coordinates \( \left\{ {\alpha }_{i}\right\} \), we must have:\n\n\[ g\left( a\right) = \mathop{\sum }\limits_{{i = 0}}^{\infty }g\left( {{f}_{\... | Yes |
Lemma 2. Let \( \varphi : k \rightarrow {k}^{\prime } \) be a surjective homomorphism, the rings \( k \) and \( {k}^{\prime } \) being perfect of characteristic \( p \) . If there exists a strict \( p \) -ring \( \mathrm{A} \) with residue ring \( k \), then there also exists a strict p-ring \( {\mathrm{A}}^{\prime } \... | We define \( {\mathrm{A}}^{\prime } \) as a quotient of \( \mathrm{A} \) . If \( a \) and \( b \) are two elements of \( \mathrm{A} \) with coordinates \( {\alpha }_{i},{\beta }_{i} \) in \( k \), we write \( a \equiv b \) if \( \varphi \left( {\alpha }_{i}\right) = \varphi \left( {\beta }_{i}\right) \) for all \( i \)... | Yes |
Theorem 5. For every perfect ring \( k \) of characteristic \( p \), there exists a unique strict \( p \) -ring \( \mathrm{W}\left( k\right) \) with residue ring \( k \) . | Uniqueness has already been proved. As for existence: if \( k \) has the form \( {\mathbf{F}}_{p}\left\lbrack {\mathrm{X}}_{\alpha }^{{p}^{-\infty }}\right\rbrack \), for an arbitrary family of indeterminates \( {\mathrm{X}}_{\alpha } \), one takes \( \mathrm{W}\left( k\right) = \) \( \widehat{\mathbf{Z}}\left\lbrack {... | No |
Theorem 7. The laws of composition defined above make \( {\mathrm{A}}^{\mathbf{N}} \) into a commutative unitary ring (called the ring of Witt vectors with coefficients in A and denoted \( \mathrm{W}\left( \mathrm{A}\right) \) ). | Note first that if one assigns to a Witt vector \( \mathfrak{a} = \left( {{a}_{0},\ldots ,{a}_{n},\ldots }\right) \) the element of the product ring \( {\mathrm{A}}^{\mathbf{N}} \) having the \( {\mathrm{W}}_{n}\left( \mathrm{a}\right) \) as coordinates, one gets a homomorphism\n\n\[ \n{\mathrm{W}}_{ * } : \mathrm{W}\l... | No |
Theorem 8. If \( k \) is a perfect ring of characteristic \( p,\mathrm{\;W}\left( k\right) \) is a strict p-ring with residue ring \( k \) . | Let \( \mathrm{H} \) be the strict \( p \) -ring with residue ring \( k \), and let \( f : k \rightarrow \mathrm{H} \) be the multiplicative system of representatives of \( \mathrm{H} \) . Associate to a Witt vector \( \mathfrak{a} = \left( {{a}_{0},\ldots ,{a}_{n},\ldots }\right) \) the element \( \theta \left( \mathf... | Yes |
Lemma 1. If \( {\mathrm{X}}_{1} \) and \( {\mathrm{X}}_{2} \) are lattices of \( \mathrm{V} \), then the fractional ideal \( \chi \left( {{\mathrm{X}}_{1}/{\mathrm{X}}_{3}}\right) .\chi {\left( {\mathrm{X}}_{2}/{\mathrm{X}}_{3}\right) }^{-1} \), defined for every lattice \( {\mathrm{X}}_{3} \subset {\mathrm{X}}_{1} \ca... | Indeed, if one sets \( {\mathrm{X}}_{4} = {\mathrm{X}}_{1} \cap {\mathrm{X}}_{2} \), the exact sequence\n\n\[ 0 \rightarrow {\mathrm{X}}_{4}/{\mathrm{X}}_{3} \rightarrow {\mathrm{X}}_{1}/{\mathrm{X}}_{3} \rightarrow {\mathrm{X}}_{1}/{\mathrm{X}}_{4} \rightarrow 0 \]\n\nshows that \( \chi \left( {{\mathrm{X}}_{1}/{\math... | Yes |
Proposition 1. The following formulas are valid:\n\n(a) \( \chi \left( {{\mathrm{X}}_{1},{\mathrm{X}}_{2}}\right) \cdot \chi \left( {{\mathrm{X}}_{2},{\mathrm{X}}_{3}}\right) \cdot \chi \left( {{\mathrm{X}}_{3},{\mathrm{X}}_{1}}\right) = 1 \) .\n\n(b) \( \chi \left( {{\mathrm{X}}_{1},{\mathrm{X}}_{2}}\right) \cdot \chi... | Formula (a) is proved by choosing a lattice \( \mathrm{X} \) contained in \( {\mathrm{X}}_{1} \cap {\mathrm{X}}_{2} \cap {\mathrm{X}}_{3} \) , and by writing \( \chi \left( {{\mathrm{X}}_{i},{\mathrm{X}}_{j}}\right) \) in the form \( \chi \left( {{\mathrm{X}}_{i}/\mathrm{X}}\right) \cdot \chi {\left( {\mathrm{X}}_{j}/\... | No |
Proposition 2. If \( u \) is a \( \mathrm{K} \)-automorphism of \( \mathrm{V} \) and \( \mathrm{X} \) a lattice of \( \mathrm{V} \), then \( \chi \left( {\mathrm{X}, u\mathrm{X}}\right) = \left( {\det \left( u\right) }\right) \) (principal ideal generated by \( \det \left( u\right) \) ). | By localising and multiplying \( u \) by a constant, we are reduced to the case where \( \mathrm{X} = {\mathrm{A}}^{n} \) and \( u\mathrm{X} \subset \mathrm{X} \) ; the proposition follows in that case from lemma 3 of Chap. I, §5. | No |
Proposition 3. If \( \mathrm{X} \) is a free A-module with basis \( \mathrm{S} = \left\{ {{e}_{1},\ldots ,{e}_{n}}\right\} \), then \( {\mathfrak{d}}_{\mathrm{X},\mathrm{T}} \) is the principal ideal generated by the discriminant \( {\mathbf{D}}_{\mathrm{T}}\left( \mathrm{S}\right) \) (in the sense of Bourbaki, Alg., C... | [Recall that \( {\mathrm{D}}_{\mathrm{T}}\left( \mathrm{S}\right) = \det \left( {\mathrm{T}\left( {{e}_{i},{e}_{j}}\right) }\right) \) .] Indeed, it is known that in this case \( {\mathrm{X}}_{\mathrm{w}} \) is a free A-module with basis \( \{ e\} \) , where \( e = {e}_{1} \land \cdots \land {e}_{n} \), and that \( \ma... | Yes |
Proposition 4. Let \( \\mathrm{X} \) be a lattice of \( \\mathrm{V} \), and let \( {\\mathrm{X}}_{\\mathrm{T}}^{ * } \) be the set of all \( y \\in \\mathrm{V} \) such that \( \\mathrm{T}\\left( {x, y}\\right) \\in \\mathrm{A} \) for all \( x \\in \\mathrm{V} \). Then \( {\\mathrm{X}}_{\\mathrm{T}}^{ * } \) is a lattic... | Localising reduces us to the case where \( \\mathrm{X} \) is free with basis \( \\left\\{ {e}_{i}\\right\\} \); then \( {\\mathrm{X}}_{\\mathrm{T}}^{ * } \) is free with the basis \( \\left\\{ {e}_{i}^{ * }\\right\\} \) defined by the relations\n\n\[ \n\\mathrm{T}\\left( {{e}_{i},{e}_{j}^{ * }}\\right) = {\\delta }_{ij... | Yes |
Proposition 5. If \( \mathrm{X} \) and \( {\mathrm{X}}^{\prime } \) are lattices of \( \mathrm{V} \), then \[ {\mathfrak{d}}_{{\mathrm{X}}^{\prime },\mathrm{T}} = {\mathfrak{d}}_{\mathrm{X},\mathrm{T}} \cdot \chi {\left( \mathrm{X},{\mathrm{X}}^{\prime }\right) }^{2} \] | Let \( \mathfrak{a} = \chi \left( {\mathrm{X},{\mathrm{X}}^{\prime }}\right) \) . We saw in \( §1 \) that \( {\mathrm{X}}_{\mathrm{W}}^{\prime } = \mathfrak{a}.{\mathrm{X}}_{\mathrm{W}} \) in \( \mathrm{W} \) ; the image of \( {\mathrm{X}}_{\mathrm{W}}^{\prime } \otimes {\mathrm{X}}_{\mathrm{W}}^{\prime } \) under the ... | Yes |
Proposition 6. \( {\mathfrak{d}}_{\mathrm{B}/\mathrm{A}} = {\chi }_{\mathrm{A}}\left( {{\mathrm{B}}^{ * }/\mathrm{B}}\right) = {\mathrm{N}}_{\mathrm{L}/\mathrm{K}}\left( {\mathfrak{D}}_{\mathrm{B}/\mathrm{A}}\right) \) . | The equality \( {b}_{\mathrm{B}/\mathrm{A}} = {\chi }_{\mathrm{A}}\left( {{\mathrm{B}}^{ * }/\mathrm{B}}\right) \) follows from prop. 4. On the other hand, \( {\chi }_{\mathrm{B}}\left( {{\mathrm{B}}^{ * }/\mathrm{B}}\right) = {\mathfrak{D}}_{\mathrm{B}/\mathrm{A}} \), and we know that \( {\chi }_{\mathrm{A}} = {\mathr... | Yes |
Proposition 7. Let \( \\mathfrak{a} \) (resp. b) be a fractional ideal of \( \\mathrm{K} \) (resp. L) relative to \( \\mathrm{A} \) (resp. B). The following two properties are equivalent:\n\n(i) \( \\operatorname{Tr}\\left( b\\right) \\subset a \) .\n\n(ii) \( \\mathrm{b} \\subset \\mathfrak{a}.\\mathfrak{D}_{\\mathrm{... | The case \( \\mathfrak{a} = 0 \) is trivial. When \( \\mathfrak{a} \\neq 0 \), the proposition follows from the equivalences:\n\n\[ \n\\operatorname{Tr}\\left( b\\right) \\subset \\mathfrak{a} \\Leftrightarrow {\\mathfrak{a}}^{-1}\\operatorname{Tr}\\left( b\\right) \\subset \\mathrm{A} \\Leftrightarrow \\operatorname{T... | Yes |
Proposition 8. Let \( \\mathrm{M}/\\mathrm{L} \) be a separable extension of finite degree \( n,\\mathrm{C} \) the integral closure of \( \\mathrm{A} \) in \( \\mathrm{M} \) . Then\n\n\[ \n{\\mathfrak{D}}_{\\mathrm{C}/\\mathrm{A}} = {\\mathfrak{D}}_{\\mathrm{C}/\\mathrm{B}} \\cdot {\\mathfrak{D}}_{\\mathrm{B}/\\mathrm{... | Put \( \\theta = {\\operatorname{Tr}}_{\\mathbf{M}/\\mathbf{K}},{\\theta }^{\\prime } = {\\operatorname{Tr}}_{\\mathbf{L}/\\mathbf{K}},{\\theta }^{\\prime \\prime } = {\\operatorname{Tr}}_{\\mathbf{M}/\\mathbf{L}} \) ; then \( \\theta = {\\theta }^{\\prime } \\circ {\\theta }^{\\prime \\prime } \) . Let \( \\mathfrak{c... | Yes |
Proposition 9. If \( \mathrm{S} \) is a multiplicative subset of \( \mathrm{A} \), then\n\n\[{\mathrm{S}}^{-1}{\mathfrak{D}}_{\mathrm{B}/\mathrm{A}} = {\mathfrak{D}}_{{\mathrm{S}}^{-1}\mathrm{\;B}/{\mathrm{S}}^{-1}\mathrm{\;A}}\;\text{ and }\;{\mathrm{S}}^{-1}{\mathfrak{d}}_{\mathrm{B}/\mathrm{A}} = {\mathfrak{d}}_{{\m... | In view of the formula \( {\left( {\mathrm{S}}^{-1}b\right) }^{-1} = {\mathrm{S}}^{-1}{b}^{-1} \) (used several times in Chap. I), it suffices to show that \( {\mathrm{S}}^{-1}{\mathrm{\;B}}^{ * } = {\left( {\mathrm{S}}^{-1}\mathrm{\;B}\right) }^{ * } \) . If \( x = {s}^{-1}y \), with \( s \in \mathrm{S}, y \in {\mathr... | Yes |
Proposition 10. Let \( \mathfrak{P} \) be a prime ideal of \( \mathrm{B} \), and let \( \mathfrak{p} = \mathfrak{P} \cap \mathrm{A} \) . Let \( {\widehat{\mathfrak{D}}}_{\mathfrak{B}} \) be the ideal generated by the different \( {\mathfrak{D}}_{\mathrm{B}/\mathrm{A}} \) in the completion \( {\widehat{\mathrm{B}}}_{\ma... | By applying prop. 9 with \( \mathrm{S} = \mathrm{A} - \mathfrak{p} \), we are reduced to the case where \( \mathrm{A} \) is a discrete valuation ring; we denote by \( \widehat{\mathrm{A}} \) (resp. \( \widehat{\mathrm{K}} \) ) its completion (resp. that of the field \( \mathrm{K} \) ). Similarly, if \( {\left\{ {\mathf... | Yes |
Theorem 1. Let \( \mathfrak{P} \) be a prime ideal of \( \mathrm{B} \), and let \( \mathfrak{p} = \mathfrak{P} \cap \mathrm{A} \). In order that the extension \( \mathrm{L}/\mathrm{K} \) be unramified at \( \mathfrak{P} \) (cf. Chap. I,§4), it is necessary and sufficient that \( \mathfrak{P} \) does not divide the diff... | Propositions 9 and 10 permit us to reduce to the case where \( \mathrm{A} \) is a complete discrete valuation ring, with residue field \( k \); in this case, \( \mathrm{B} \) is also a discrete valuation ring. To say that \( \mathfrak{P} \) is unramified is then equivalent to saying that \( \mathrm{B}/\mathrm{{pB}} \) ... | Yes |
Corollary 1. Let \( \mathfrak{p} \) be a prime ideal of \( \mathrm{A} \). In order that \( \mathrm{L}/\mathrm{K} \) be unramified at \( \mathfrak{p} \), it is necessary and sufficient that \( \mathfrak{p} \) does not divide the discriminant \( {\mathfrak{d}}_{\mathrm{B}/\mathrm{A}} \). | This follows from the fact that \( {\mathfrak{d}}_{\mathrm{B}/\mathrm{A}} = \mathrm{N}\left( {\mathfrak{D}}_{\mathrm{B}/\mathrm{A}}\right) \). | Yes |
Corollary 2. Almost all the prime ideals of \( \mathrm{B}\left( {\text{or of}\mathrm{A}}\right) \) are unramified in the extension \( \mathrm{L}/\mathrm{K} \) . | Obvious. | No |
Theorem 2. Let \( {k}^{\prime }/k \) be a finite separable extension. Then there exists a finite unramified extension \( {\mathrm{K}}^{\prime }/\mathrm{K} \) whose corresponding residue extension is isomorphic to \( {k}^{\prime }/k \) ; this extension is unique, up to unique isomorphism. It is Galois if and only if \( ... | Since \( {k}^{\prime }/k \) is finite and separable, it is generated by a primitive element \( \xi \) ; let \( \varphi \) be its minimal polynomial over \( k \), which has degree \( n = \left\lbrack {{k}^{\prime } : k}\right\rbrack \) . Let \( \bar{f} \in \mathrm{A}\left\lbrack \mathrm{X}\right\rbrack \) be a monic pol... | Yes |
Theorem 3. Let \( {\mathrm{K}}^{\prime }/\mathrm{K} \) be a finite unramified extension, with residue extension \( {k}^{\prime }/k \), and let \( {\mathrm{K}}^{\prime \prime }/\mathrm{K} \) be an arbitrary finite extension, with residue extension \( {k}^{\prime \prime }/k \) . The set of \( \mathrm{K} \) -isomorphisms ... | Let us denote by \( {\operatorname{Hom}}_{\mathrm{A}}^{al}\left( {\mathrm{\;B},\mathrm{C}}\right) \) the set of \( \mathrm{A} \) -algebra homomorphisms of \( \mathrm{B} \) into \( \mathrm{C} \) (for any ring \( \mathrm{A} \) ). If \( {\mathrm{A}}^{\prime } \) and \( {\mathrm{A}}^{\prime \prime } \) are the integral clo... | Yes |
Corollary 1. Let \( {k}_{\mathrm{s}} \) be the separable closure of \( k \) (i.e., the largest separable extension of \( k \) within a given algebraic closure of \( k \) ), and let \( {\mathrm{K}}_{nr} \) be the inductive (direct) limit of the unramified extensions of \( \mathrm{K} \) that correspond to the finite sube... | This is clear. | No |
Corollary 2. Let \( {\mathrm{K}}^{\prime \prime }/\mathrm{K} \) be a finite extension, with residue extension \( {k}^{\prime \prime }/k \) . The subextensions \( {\mathrm{K}}^{\prime }/\mathrm{K} \) of \( {\mathrm{K}}^{\prime \prime }/\mathrm{K} \) which are unramified over \( \mathrm{K} \) are in one-to-one correspond... | This is clear. | No |
Corollary 3. With the hypotheses of cor. 2, there exists a maximal unramified subextension \( {\mathrm{K}}^{\prime }/\mathrm{K} \) of \( {\mathrm{K}}^{\prime \prime }/\mathrm{K} \). Its residue extension \( {k}^{\prime }/k \) is the largest separable subextension of \( {k}^{\prime \prime }/k \). We have \( e\left( {{\m... | This follows from cor. 2. | No |
Proposition 11. Let \( n = \left\lbrack {\mathrm{\;L} : \mathrm{K}}\right\rbrack \) and let \( \mathrm{C} \) be a subring of \( \mathrm{B} \) containing \( \mathrm{A} \) and having an A-basis consisting of the powers \( {x}^{i},0 \leq i \leq n - 1 \), of a single element \( x \) . Let \( f \) be the characteristic poly... | The coefficients of \( f \) are integral over \( \mathrm{A} \) and belong to \( \mathrm{K} \) ; as \( \mathrm{A} \) is integrally closed, they belong to \( A \), which proves (i). (Note that the ring \( C \) is isomorphic to the ring \( {\mathrm{B}}_{f} = \mathrm{A}\left\lbrack \mathrm{X}\right\rbrack /\left( f\right) ... | No |
Lemma 2 (Euler). \( \operatorname{Tr}\left( {{x}^{i}/{f}^{\prime }\left( x\right) }\right) = 0 \) for \( i = 0,\ldots, n - 2 \), and \( \operatorname{Tr}\left( {{x}^{n - 1}/{f}^{\prime }\left( x\right) }\right) = 1 \) . | Let \( {x}_{k}, k = 1,\ldots, n \), be the conjugates of \( x \) in a suitable extension of \( \mathrm{L} \) . We must compute the sums \( \mathop{\sum }\limits_{k}{\left( {x}_{k}\right) }^{i}/{f}^{\prime }\left( {x}_{k}\right) \) . Now we have the identity\n\n(*) \n\n\[ \frac{1}{f\left( \mathrm{\;T}\right) } = \mathop... | No |
Corollary 1. With the preceding notation and hypotheses,\n\n\[ \mathfrak{r} = {f}^{\prime }\left( x\right) \cdot {\mathfrak{D}}_{\mathbf{B}/\mathbf{A}}^{-1} \] | To simplify the writing, put \( b = {f}^{\prime }\left( x\right) \) . We have the following equivalences for \( t \in \mathrm{L} \) :\n\n\[ t \in \mathrm{r} \Leftrightarrow t\mathrm{\;B} \subset \mathrm{C} \Leftrightarrow {b}^{-1}t\mathrm{\;B} \subset {\mathrm{C}}^{ * } \Leftrightarrow \operatorname{Tr}\left( {{b}^{-1}... | Yes |
Corollary 2. The different \( {\mathfrak{D}}_{\mathrm{B}/\mathrm{A}} \) divides the principal ideal \( \left( {{f}^{\prime }\left( x\right) }\right) \) . For these two ideals to be equal, it is necessary and sufficient that \( \mathbf{B} = \mathbf{C} \) (i.e., that \( \mathrm{B} = \mathrm{A}\left\lbrack x\right\rbrack ... | The first assertion results from the formula \( \left( {{f}^{\prime }\left( x\right) }\right) = \mathrm{r} \) . \( {\mathfrak{D}}_{\mathrm{B}/\mathrm{A}} \) . That same formula shows that \( {\mathfrak{D}}_{\mathrm{B}/\mathrm{A}} = \left( {{f}^{\prime }\left( x\right) }\right) \) if and only if \( \mathrm{r} = 1 \), i.... | Yes |
Proposition 12. Suppose that B (hence also A) is a discrete valuation ring; if \( \overline{\mathrm{L}} \) and \( \overline{\mathrm{K}} \) denote the residue fields of these two rings, suppose also that the extension \( \overline{\mathrm{L}}/\overline{\mathrm{K}} \) is separable. Then \( \mathrm{B} \) has a basis over ... | Let \( e \) be the ramification index of \( \mathrm{L}/\mathrm{K} \), and let \( f = \left\lbrack {\overline{\mathrm{L}} : \overline{\mathrm{K}}}\right\rbrack \), so that \( n = {ef} \) . Let \( \pi \) be a uniformizer of \( \mathrm{B} \), and let \( x \in \mathrm{B} \) represent a primitive element \( \bar{x} \) for t... | No |
Lemma 3. The products \( {x}^{i}{\pi }^{j},0 \leq i < f,0 \leq j < e \), form a basis for the A-module B. | The number of these products being \( {ef} \), it suffices to show that they span \( \mathbf{B} \) , and even that their classes span \( \mathrm{B}/\mathfrak{p}\mathrm{B} \) (where \( \mathfrak{p} \) is the maximal ideal of \( \mathrm{A} \) ). We have \( \mathfrak{p}\mathrm{B} = {\pi }^{e}\mathrm{B} \) . Hence it is en... | No |
Lemma 4. The element \( x \) may be chosen so that there exists a monic polynomial \( \mathrm{R}\left( \mathrm{X}\right) \), of degree \( f \), with coefficients in \( \mathrm{A} \), such that \( \mathrm{R}\left( x\right) \) is a uniformizer of \( \mathrm{B} \) . | Let us first choose \( x \) such that \( \overline{\mathrm{L}} = \overline{\mathrm{K}}\left( \bar{x}\right) \) . Let \( \mathrm{R} \) be a monic polynomial with coefficients in A whose reduction \( \overline{\mathbf{R}} \) is the minimal polynomial of \( \bar{x} \) . If \( w \) denotes the valuation of \( \mathbf{B} \)... | Yes |
Proposition 13. Let \( \mathfrak{P} \) be a non-zero prime ideal of \( \mathrm{B} \), and let \( \mathfrak{p} = \mathfrak{P} \cap \mathrm{A} \) . Let \( {\mathrm{L}}_{\mathfrak{P}} \) and \( {\overline{\mathrm{K}}}_{\mathfrak{p}} \) be the corresponding residue fields. Suppose that the residue extension is separable. T... | By localising and completing, we reduce to the case where A and B are complete discrete valuation rings; using cor. 3 to th. 3 , we may further suppose that \( \mathrm{L}/\mathrm{K} \) is totally ramified. If \( \pi \) is a uniformizer of \( \mathrm{B} \), we know (cf. Chap. I, §6) that \( \pi \) satisfies an Eisenstei... | Yes |
Proposition 14. Let A be a Dedekind domain, with field of fractions K; let L be a finite separable extension of \( \mathrm{K} \), and let \( \mathrm{B} \) be the integral closure of \( \mathrm{A} \) in L. Suppose that for every prime ideal \( \mathfrak{P} \) of \( \mathrm{B} \), the corresponding residue extension is s... | It is easy to show that if \( {\mathrm{A}}^{\prime } \) is an arbitrary commutative A-algebra, and if \( {\mathrm{B}}^{\prime } = \mathrm{B}{ \otimes }_{\mathrm{A}}{\mathrm{A}}^{\prime } \), then \( {\Omega }_{{\mathrm{A}}^{\prime }}\left( {\mathrm{B}}^{\prime }\right) = {\Omega }_{\mathrm{A}}\left( \mathrm{B}\right) {... | No |
Lemma 1. Let \( s \in \mathbf{G} \), and let \( i \) be an integer \( \geq - 1 \) . Then the three following conditions are equivalent:\n\na) s operates trivially on the quotient ring \( {\mathrm{A}}_{\mathrm{L}}/{\mathfrak{p}}_{\mathrm{L}}^{i + 1} \) .\n\nb) \( {v}_{\mathrm{L}}\left( {s\left( a\right) - a}\right) \geq... | The equivalence of a) and b) is trivial. On the other hand, the image \( {x}_{i} \) of \( x \) in \( {\mathrm{A}}_{\mathrm{L}}/{\mathfrak{p}}_{\mathrm{L}}^{i + 1} = {\mathrm{A}}_{i} \) generates \( {\mathrm{A}}_{i} \) as an \( {\mathrm{A}}_{\mathrm{K}} \) -algebra. Hence \( s\left( {x}_{i}\right) = {x}_{i} \) is the ne... | Yes |
Proposition 1. For each integer \( i \geq - 1 \), let \( {\mathrm{G}}_{i} \) be the set of \( s \in \mathrm{G} \) satisfying conditions a), b), c) of lemma 1. Then the \( {\mathrm{G}}_{i} \) form a decreasing sequence of normal subgroups of \( \mathrm{G};{\mathrm{G}}_{-1} = \mathrm{G},{\mathrm{G}}_{0} \) is the inertia... | Condition a) shows that the \( {\mathrm{G}}_{i} \) are normal subgroups of \( \mathrm{G} \), and they obviously decrease with \( i \) . Condition c) shows that if\n\n\[ i \geq \sup \left\{ {{v}_{\mathrm{L}}\left( {s\left( x\right) - x}\right) }\right\} \;\text{ for }s \neq 1, \]\n\nthen \( {\mathrm{G}}_{i} \) is trivia... | No |
Proposition 2. For every \( s \in \mathrm{H},{i}_{\mathrm{H}}\left( s\right) = {i}_{\mathrm{G}}\left( s\right) \), and \( {\mathrm{H}}_{i} = {G}_{i} \cap \mathrm{H} \). | This follows from condition a) of lemma 1. | No |
For every \( \sigma \in \mathrm{G}/\mathrm{H} \), \[ {i}_{\mathrm{G}/\mathrm{H}}\left( \sigma \right) = \frac{1}{{e}^{\prime }}\mathop{\sum }\limits_{{s \rightarrow \sigma }}{i}_{\mathrm{G}}\left( s\right) \] where \( {e}^{\prime } = {e}_{\mathrm{L}/\mathrm{K}} \). | (Proof After J. Tate). For \( \sigma = 1 \), both sides are equal to \( + \infty \), so the equation holds. Suppose \( \sigma \neq 1 \). Let \( x \) (resp. \( y \) ) be an \( {\mathrm{A}}_{\mathrm{K}} \) -generator of \( {\mathrm{A}}_{\mathrm{L}} \) (resp. \( {\mathrm{A}}_{{\mathrm{K}}^{\prime }} \) ). By definition, \... | Yes |
If \( {\mathfrak{D}}_{\mathrm{L}/\mathrm{K}} \) denotes the different of \( \mathrm{L}/\mathrm{K} \), then\n\n\[ \n{v}_{\mathrm{L}}\left( {\mathfrak{D}}_{\mathrm{L}/\mathrm{K}}\right) = \mathop{\sum }\limits_{{s \neq 1}}{i}_{\mathrm{G}}\left( s\right) = \mathop{\sum }\limits_{{i = 0}}^{{i = \infty }}\left( {\operatorna... | Again let \( x \) be an \( {\mathrm{A}}_{\mathrm{K}} \) -generator of \( {\mathrm{A}}_{\mathrm{L}} \), and let \( f \) be its minimal polynomial over \( \mathrm{K} \). According to cor. 2 to prop. 11 of Chap. III, \( {\mathfrak{D}}_{\mathrm{L}/\mathrm{K}} \) is generated by \( {f}^{\prime }\left( x\right) \). But \( f\... | Yes |
Proposition 5. Let \( i \) be a non-negative integer. In order that an element \( s \) of the inertia group \( {\mathrm{G}}_{0} \) belong to \( {\mathrm{G}}_{i} \), it is necessary and sufficient that\n\n\[ s\left( \pi \right) /\pi \equiv 1{\;\operatorname{mod}\;.}{\mathfrak{p}}_{\mathbf{L}}^{i} \] | Replacing \( \mathrm{G} \) by \( {\mathrm{G}}_{0} \) and \( \mathrm{K} \) by \( {\mathrm{K}}_{r} \) reduces us to the case of a totally ramified extension. By prop. 18 of Chapter I, \( §6 \), the element \( \pi \) is then an \( {\mathrm{A}}_{\mathrm{K}} \) -generator of \( {\mathrm{A}}_{\mathrm{L}} \) . Therefore\n\n\[... | Yes |
Proposition 6. (b) For \( i \geq 1 \), the group \( {\mathrm{U}}_{\mathrm{L}}^{i}/{\mathrm{U}}_{\mathrm{L}}^{i + 1} \) is canonically isomorphic to the group \( {\mathfrak{p}}_{\mathrm{L}}^{i}/{\mathfrak{p}}_{\mathrm{L}}^{i + 1} \), which is itself isomorphic (non-canonically) to the additive group of the residue field... | To prove (b), let correspond to each \( x \in {\mathfrak{P}}_{\mathrm{L}}^{i} \) the element \( 1 + x \) of \( {\mathrm{U}}_{\mathrm{L}}^{i} \) ; this gives, by passage to the quotient, the canonical isomorphism. Moreover, as \( {\mathfrak{p}}_{\mathrm{L}}^{i}/{\mathfrak{p}}_{\mathrm{L}}^{i + 1} \) is a one-dimensional... | Yes |
Proposition 7. The map which, to \( s \in {\mathrm{G}}_{i} \), assigns \( s\left( \pi \right) /\pi \), induces by passage to the quotient an isomorphism \( {\theta }_{i} \) of the quotient group \( {\mathrm{G}}_{i}/{\mathrm{G}}_{i + 1} \) onto a subgroup of the group \( {\mathrm{U}}_{\mathrm{L}}^{\mathrm{i}}/{\mathrm{U... | If \( {\pi }^{\prime } \) is another uniformizer, then \( {\pi }^{\prime } = {\pi u} \), with \( u \in {\mathrm{U}}_{\mathrm{L}} \), whence\n\n\[ s\left( {\pi }^{\prime }\right) /{\pi }^{\prime } = s\left( \pi \right) /\pi \cdot s\left( u\right) /u. \]\n\nIf \( s \in {\mathrm{G}}_{i} \), then \( s\left( u\right) \equiv... | Yes |
Corollary 1. The group \( {\mathrm{G}}_{0}/{\mathrm{G}}_{1} \) is cyclic, and is mapped isomorphically by \( {\theta }_{0} \) onto a subgroup of the group of roots of unity contained in \( \overline{\mathbf{L}} \) . Its order is prime to the characteristic of the residue field \( \overline{\mathbf{L}} \) . | Indeed, \( {\mathrm{U}}_{\mathrm{L}}/{\mathrm{U}}_{\mathrm{L}}^{1} = {\mathrm{L}}^{ * } \), which shows that \( {\theta }_{0}\left( {{\mathrm{G}}_{0}/{\mathrm{G}}_{1}}\right) \) is a finite subgroup of the group of roots of unity in \( \mathbf{L} \) ; the fact that it is cyclic and of order prime to the characteristic ... | Yes |
Corollary 2. If the characteristic of \( \mathrm{L} \) is zero, then \( {\mathrm{G}}_{1} = \{ 1\} \), and the group \( {\mathrm{G}}_{0} \) is cyclic. | Indeed, if \( i \geq 1,{\mathrm{U}}_{\mathrm{L}}^{i}/{\mathrm{U}}_{\mathrm{L}}^{i + 1} \) is isomorphic to \( \mathrm{L} \), which has no non-trivial finite subgroups. Hence \( {\mathrm{G}}_{i} = {\mathrm{G}}_{i + 1} \), and since \( {\mathrm{G}}_{i} = \{ 1\} \) for large \( i \), we do have \( {\mathrm{G}}_{1} = \{ 1\... | Yes |
Corollary 3. If the characteristic of \( \overline{\mathrm{L}} \) is \( p \neq 0 \), the quotients \( {\mathrm{G}}_{i}/{\mathrm{G}}_{i + 1}, i \geq 1 \) , are abelian groups, and are direct products of cyclic groups of order p. The group \( {\mathrm{G}}_{1} \) is a p-group. | Indeed, for \( i \geq 1,{\mathrm{U}}_{\mathrm{L}}^{i}/{\mathrm{U}}_{\mathrm{L}}^{i + 1} \) is isomorphic to the additive group of \( \mathrm{L} \) , and every subgroup of \( \overline{\mathbf{L}} \) is a vector space over the prime field of \( p \) elements, hence is a direct sum of cyclic groups of order \( p \) . As ... | Yes |
If the characteristic of \( \mathrm{L} \) is \( p \neq 0 \), the inertia group \( {\mathrm{G}}_{0} \) has the following property:\n\n\( \left( {\mathbf{R}}_{p}\right) \) It is the semi-direct product of a cyclic group of order prime to \( p \) with a normal subgroup whose order is a power of \( p \) . | By corollaries 1 and \( 3,{\mathrm{G}}_{1} \) is a \( p \) -group and \( {\mathrm{G}}_{0}/{\mathrm{G}}_{1} \) is cyclic of order prime to \( p \) ; thus it all comes down to showing the existence of a subgroup \( \mathrm{H} \) of \( {\mathrm{G}}_{0} \) that projects isomorphically onto \( {\mathrm{G}}_{0}/{\mathrm{G}}_... | Yes |
Corollary 5. The group \( {\mathrm{G}}_{0} \) is solvable. If \( \overline{\mathrm{K}} \) is a finite field, then \( \mathrm{G} \) too is solvable. | The first assertion follows from the solvability of \( p \) -groups, cyclic groups and extensions of solvable groups. The second follows from this and the fact that \( \mathrm{G}/{\mathrm{G}}_{0} = \mathrm{G}\left( {\mathrm{L}/\overline{\mathrm{K}}}\right) \) is cyclic when \( \overline{\mathrm{K}} \) is finite. | Yes |
Proposition 8. Let \( k \) be an algebraically closed field of characteristic zero, and let \( \mathrm{K} = k\left( \left( \mathrm{\;T}\right) \right) \) . Then the algebraic closure \( {\mathrm{K}}_{a} \) of the field \( \mathrm{K} \) is the union of the fields \( {\mathrm{K}}_{n} = k\left( \left( {\mathrm{T}}^{1/n}\r... | Let \( \mathrm{L} \subset {\mathrm{K}}_{a} \) be a Galois extension of \( \mathrm{K} \) of finite degree, \( \mathrm{G} \) its Galois group. Since \( \overline{\mathrm{K}} = k \) is algebraically closed, \( \mathrm{G} = {\mathrm{G}}_{0} \) (Chap. I,§7, prop. 20); cor. 2 then shows that \( \mathrm{G} \) is cyclic. Let \... | Yes |
Proposition 9. If \( s \in {\mathrm{G}}_{0} \) and \( \tau \in {\mathrm{G}}_{i}/{\mathrm{G}}_{i + 1}, i \geq 1 \), then\n\n\[{\theta }_{i}\left( {{s\tau }{s}^{-1}}\right) = {\theta }_{0}{\left( s\right) }^{i}{\theta }_{i}\left( \tau \right)\] | [The formula makes sense because \( {\theta }_{i}\left( \tau \right) \) belongs to \( {\mathfrak{p}}_{\mathrm{L}}^{i}/{\mathfrak{p}}_{\mathrm{L}}^{i + 1} \) which is a one-dimensional vector space over \( L \), and \( {\theta }_{0}\left( s\right) \) is an element of the multiplicative group of L.]\n\nLet \( t \in {\mat... | Yes |
Corollary 1. If \( s \in {\mathrm{G}}_{0} \) and \( t \in {\mathrm{G}}_{i}, i \geq 1 \), then \( {st}{s}^{-1}{t}^{-1} \in {\mathrm{G}}_{i + 1} \) if and only if \( {s}^{i} \in {\mathrm{G}}_{1} \) or \( t \in {\mathrm{G}}_{i + 1}. \) | Indeed, \( {st}{s}^{-1}{t}^{-1} \in {\mathrm{G}}_{i + 1} \) is equivalent to \( {st}{s}^{-1} \equiv t{\;\operatorname{mod}\;.}{G}_{i + 1} \), which in turn is equivalent to\n\n\[ \n{\theta }_{i}\left( {{st}{s}^{-1}}\right) = {\theta }_{i}\left( t\right) \n\]\n\nwhence the result follows from the proposition. | No |
Corollary 2. Suppose \( \mathrm{G} \) is abelian, and let \( {e}_{0} \) be the order of \( {\mathrm{G}}_{0}/{\mathrm{G}}_{1} \) . Then for any integer \( i \) not divisible by \( {e}_{0},{\mathrm{G}}_{i} = {\mathrm{G}}_{i + 1} \) . | Indeed, if \( t \in {\mathrm{G}}_{i} \), and if \( s \) is an element of \( {\mathrm{G}}_{0} \) which generates \( {\mathrm{G}}_{0}/{\mathrm{G}}_{1} \) , then by hypothesis \( {st}{s}^{-1}{t}^{-1} = 1 \), and since \( {s}^{i} \notin {\mathrm{G}}_{1} \), cor. \( 1\mathrm{\;{im}} \equiv s \in {\mathrm{G}}_{i + 1} \) . | No |
Lemma 2. With the hypotheses of prop. 10, we have st \( {s}^{-1}{t}^{-1} \in {G}_{i + j} \) and\n\n\[ \n{\theta }_{i + j}\left( {{st}{s}^{-1}{t}^{-1}}\right) = \left( {j - i}\right) {\theta }_{i}\left( s\right) {\theta }_{j}\left( t\right) .\n\] | [This formula is meaningful because, as was already noted above, the direct sum of the \( {\mathfrak{p}}_{\mathrm{L}}^{i}/{\mathfrak{p}}_{\mathrm{L}}^{i + 1} \) has a natural structure of graded L-algebra.]\n\n\[ \n\text{We put}s\left( \pi \right) = \pi \left( {1 + a}\right), t\left( \pi \right) = \pi \left( {1 + b}\ri... | Yes |
Lemma 3. \( {\varphi }_{\mathrm{L}/\mathrm{K}}\left( u\right) = \frac{1}{{g}_{0}}\mathop{\sum }\limits_{{s \in \mathrm{G}}}\operatorname{Inf}\left( {{i}_{\mathrm{G}}\left( s\right), u + 1}\right) - 1 \) . | Let \( \theta \left( u\right) \) be the function defined by the right side of this equation; it is a continuous piecewise-linear function that vanishes at \( u = 0 \) . If \( m < u < m + 1 \) , where \( m \) is an integer, the derivative \( {\theta }^{\prime }\left( u\right) \) is equal to the number of \( s \in \mathr... | Yes |
Lemma 4. Let \( \sigma \in \mathrm{G}/\mathrm{H} \), and let \( j\left( \sigma \right) \) be the upper bound of the integers \( {i}_{\mathrm{G}}\left( s\right) \) as s runs through the pre-image of \( \sigma \) in \( \mathrm{G} \). Then \[ {i}_{\mathrm{G}/\mathrm{H}}\left( \sigma \right) - 1 = {\varphi }_{\mathrm{L}/{\... | Let \( s \in \mathrm{G} \) have image \( \sigma \) and \( {i}_{\mathrm{G}}\left( s\right) = j\left( \sigma \right) \). Put \( m = {i}_{\mathrm{G}}\left( s\right) \). If \( t \in \mathrm{H} \) belongs to \( {\mathrm{H}}_{m - 1} \), then \( {i}_{\mathrm{G}}\left( t\right) \geq m \) whence \( {i}_{\mathrm{G}}\left( {st}\r... | Yes |
Lemma 5. If \( v = {\phi }_{\mathrm{L}/{\mathrm{K}}^{\prime }}\left( u\right) \) then \( {\mathrm{G}}_{u}\mathrm{H}/\mathrm{H} = {\left( \mathrm{G}/\mathrm{H}\right) }_{v} \) . | Keep the notation of the preceding lemma. Then\n\n\[ \sigma \in {\mathrm{G}}_{u}\mathrm{H}/\mathrm{H} \Leftrightarrow j\left( \sigma \right) - 1 \geq u \Leftrightarrow {\varphi }_{\mathrm{L}/{\mathrm{K}}^{\prime }}\left( {j\left( \sigma \right) - 1}\right) \geq {\varphi }_{\mathrm{L}/{\mathrm{K}}^{\prime }}\left( u\rig... | Yes |
Proposition 16. Let \( \mathrm{K} \) be a field complete under a discrete valuation, and having finite residue field \( k = {\mathrm{F}}_{q} \), with \( q = {p}^{f} \) . Given an integer \( n \) prime to \( p \) . Let \( {\mathrm{K}}_{n} \) (resp. \( {k}_{n} \) ) be the field obtained by adjoining to \( \mathrm{K} \) (... | Let \( \mathrm{L} \) be the unramified extension of \( \mathrm{K} \) which has \( {k}_{n} \) as residue field (cf. Chap. III, \( §5 \) ), and let \( \mathrm{S} \) be the set of multiplicative representatives in \( \mathrm{L} \) (cf. Chap. II,§4). Let \( \bar{\zeta } \) be a primitive \( n \) th root of unity in \( {k}_... | Yes |
Corollary 1. The degree \( \left\lbrack {{\mathrm{K}}_{n} : \mathrm{K}}\right\rbrack \) is equal to the smallest integer \( r \geq 1 \) such that \( {q}^{r} \equiv 1{\;\operatorname{mod}\;.}n \) . | Indeed, \( r \) is the smallest integer such that \( {s}^{r} = 1 \) . | No |
Corollary 2. The maximal unramified extension \( {\mathrm{K}}_{nr} \) of \( \mathrm{K} \) is obtained by adjoining to \( \mathrm{K} \) all the roots of unity of order prime to \( p \) . Its Galois group can be identified with \( \widehat{\mathbf{Z}} \) ; it admits a generator \( s \) such that \( s\left( z\right) = {z}... | This results from prop. 16 by passage to the limit on \( n \), once we note that the union of the \( {k}_{n} \) is the algebraic closure of \( k \) . | No |
Proposition 17. Let \( {\mathrm{K}}_{n} \) be the field obtained from \( \mathrm{K} = {\mathbf{Q}}_{p} \) by adjoining a primitive \( n \) th root of unity \( \zeta \), with \( n = {p}^{m} \) . Then\ni) \( \left\lbrack {{\mathrm{K}}_{n} : \mathrm{K}}\right\rbrack = \varphi \left( n\right) = \left( {p - 1}\right) {p}^{m... | We know a priori that \( \mathrm{G}\left( {{\mathrm{K}}_{n}/\mathrm{K}}\right) \) can be identified with a subgroup of \( \mathrm{G}\left( n\right) \) (cf. Bourbaki, Alg., Chap. V,§11); as the order of \( \mathrm{G}\left( n\right) \) is \( \varphi \left( n\right) \), we see that assertions i) and ii) are equivalent.\n\... | Yes |
Proposition 18. The ramification groups \( {\mathrm{G}}_{u} \) of \( \mathrm{G}\left( {{\mathrm{K}}_{n}/\mathrm{K}}\right) \) are:\n\n\[ \n{\mathrm{G}}_{0} = \mathrm{G} \]\n\n\[ \n\text{if}1 \leq u \leq p - 1,\;{\mathrm{G}}_{u} = \mathrm{G}{\left( n\right) }^{1} \]\n\n\[ \n\text{if}p \leq u \leq {p}^{2} - 1,\;{\mathrm{... | Let \( a \neq 1 \) be an element of \( \mathrm{G}\left( n\right) \), and let \( {s}_{a} \) be the corresponding element of \( \mathrm{G} \) . Let \( v \) be the largest integer such that \( a \equiv 1{\;\operatorname{mod}\;{p}^{v}} \) ; then \( a \in \mathrm{G}{\left( n\right) }^{v} \) and \( a \notin \mathrm{G}{\left(... | Yes |
Lemma 1. Let\n\n\[ \n0 \rightarrow {\mathrm{A}}^{\prime } \rightarrow \mathrm{A} \rightarrow {\mathrm{A}}^{\prime \prime } \rightarrow 0 \]\n\n\[ \n\begin{matrix} {f}^{\prime } \downarrow \;f \downarrow \;{f}^{\prime \prime } \downarrow \end{matrix} \]\n\n\[ \n0 \rightarrow {\mathrm{B}}^{\prime } \rightarrow \mathrm{B}... | [Recall that \( \operatorname{Ker}f = {f}^{-1}\left( 0\right) \) is the kernel of \( f \), and \( \operatorname{Coker}f = \mathrm{B}/f\left( \mathrm{\;A}\right) \) is its cokernel.]\n\nIf \( {a}^{\prime \prime } \in \operatorname{Ker}{f}^{\prime \prime } \), choose an element \( a \in \mathrm{A} \) projecting onto \( {... | No |
Lemma 2. Let A (resp. A') be an abelian group provided with a decreasing sequence of subgroups \( {\mathrm{A}}_{n} \) (resp. \( {\mathrm{A}}_{n}^{\prime } \) ). Suppose that \( {\mathrm{A}}_{0} = \mathrm{A},{\mathrm{A}}_{0}^{\prime } = {\mathrm{A}}^{\prime } \), and that A and \( {\mathrm{A}}^{\prime } \) are complete ... | We quickly recall the proof of this result (cf. Bourbaki, Alg. comm., Chap. III, §2):\n\nIf the \( {u}_{n} \) are injective, then \( \operatorname{Ker}\left( u\right) \cap {\mathrm{A}}_{n} = \operatorname{Ker}\left( u\right) \cap {\mathrm{A}}_{n + 1} \), whence by induction on \( n,\operatorname{Ker}\left( u\right) \su... | Yes |
Proposition 1. If \( \mathrm{L}/\mathrm{K} \) is unramified, then \( \mathrm{N} \) maps \( {\mathrm{U}}_{\mathrm{L}}^{n} \) into \( {\mathrm{U}}_{\mathrm{K}}^{n} \) for all \( n \) . | Let \( x = 1 + y \), with \( y \in {\mathfrak{p}}_{\mathrm{L}}^{n} \) . Then \( s\left( x\right) = 1 + s\left( y\right) \) for all \( s \in \mathrm{G} \), and \( s\left( y\right) \in {\mathfrak{p}}_{\mathrm{L}}^{n} \) . Hence\n\n\( \left( *\right) \)\n\n\[ \mathrm{N}x = \mathop{\prod }\limits_{{s \in \mathrm{G}}}\left(... | Yes |
Proposition 2. Suppose \( \mathrm{L}/\mathrm{K} \) is unramified. Then\ni) The map \( {\mathrm{N}}_{0} : {\overline{\mathrm{L}}}^{ * } \rightarrow {\overline{\mathrm{K}}}^{ * } \) is just the norm in the residue extension \( \overline{\mathrm{L}}/\overline{\mathrm{K}} \).\nii) For \( n \geq 1 \), the map \( {\mathrm{N}... | Assertion i) is trivial. Assertion ii) follows from formula (*) proved above. | No |
Lemma 3. The different \( \mathfrak{D} \) of the extension \( \mathrm{L}/\mathrm{K} \) is \( {\mathfrak{p}}_{\mathrm{L}}^{m} \), where \( m = \left( {t + 1}\right) \left( {l - 1}\right) \) . | This is a consequence of prop. 4 of Chap. IV. | No |
Lemma 4. For every integer \( n \geq 0,\operatorname{Tr}\left( {\mathfrak{p}}_{\mathrm{L}}^{n}\right) = {\mathfrak{p}}_{\mathrm{K}}^{r} \), where \( r = \left\lbrack {\left( {m + n}\right) /l}\right\rbrack \) and \( m = \left( {t + 1}\right) \left( {l - 1}\right) \). | [Recall that the symbol \( \left\lbrack x\right\rbrack \) denotes the largest integer \( \leq x \).]\n\nSince the trace is \( {\mathrm{A}}_{\mathrm{K}} \) -linear, \( \operatorname{Tr}\left( {\mathfrak{p}}_{\mathrm{L}}^{n}\right) \) is an ideal in \( {\mathrm{A}}_{\mathrm{K}} \). If \( r \) is any integer, prop. 7 of C... | Yes |
Lemma 5. If \( x \in {\mathfrak{p}}_{\mathrm{L}}^{n} \), then\n\n\[ \mathrm{N}\left( {1 + x}\right) \equiv 1 + \operatorname{Tr}\left( x\right) + \mathrm{N}\left( x\right) {\;\operatorname{mod}\;.}\operatorname{Tr}\left( {\mathfrak{p}}_{\mathrm{L}}^{2n}\right) . \] | For the computation to follow, it is convenient to use the exponential notation for the group \( \mathrm{G} \), i.e., to denote by \( {x}^{s} \) the transform of \( x \) by \( s \in \mathrm{G} \) . Then\n\n\[ \mathrm{N}\left( {1 + x}\right) = \mathop{\prod }\limits_{{s \in \mathrm{G}}}\left( {1 + {x}^{s}}\right) \]\n\n... | Yes |
Proposition 5. i) For \( n = 0 \), the map \( {\mathrm{N}}_{0} : {\overline{\mathrm{K}}}^{ * } \rightarrow {\overline{\mathrm{K}}}^{ * } \) is given by \( {\mathrm{N}}_{0}\left( \xi \right) = {\xi }^{l} \) . If \( t \neq 0 \) (where \( t \) is the last index for which the corresponding ramification group is non-trivial... | It is easy to see that \( \mathrm{N}\left( {\mathrm{U}}_{\mathrm{L}}\right) \subset {\mathrm{U}}_{\mathrm{K}},\mathrm{N}\left( {\mathrm{U}}_{\mathrm{L}}^{1}\right) \subset {\mathrm{U}}_{\mathrm{K}}^{1} \), and that \( {\mathrm{N}}_{0}\left( \xi \right) = {\xi }^{l} \) . If \( t \neq 0 \), then \( l = p \), and \( {\mat... | Yes |
Corollary 1. The homomorphism \( {\mathrm{N}}_{n} \) is injective for all \( n \) except for \( n = t \) , in which case there is an exact sequence:\n\n\[ 0 \rightarrow \mathrm{G}\overset{{\theta }_{\iota }}{ \rightarrow }{\mathrm{U}}_{\mathrm{L}}^{t}/{\mathrm{U}}_{\mathrm{L}}^{t + 1}\overset{{\mathrm{N}}_{\iota }}{ \r... | Obvious. | No |
Corollary 2. The homomorphism \( {\mathrm{N}}_{n} \) is surjective for \( n > t \), and, if \( \overline{\mathrm{K}} \) is perfect, for \( n < t \) . If \( \overline{\mathrm{K}} \) is algebraically closed, it is surjective for all \( n \) . | Obvious. | No |
Corollary 3. We have \( \mathrm{N}\left( {\mathrm{U}}_{\mathrm{L}}^{\psi \left( n\right) }\right) = {\mathrm{U}}_{\mathrm{K}}^{n} \) for \( n > t \), and \( \mathrm{N}\left( {\mathrm{U}}_{\mathrm{L}}^{\psi \left( n\right) + 1}\right) = {\mathrm{U}}_{\mathrm{K}}^{n + 1} \) for \( n \geq t \) . When \( \overline{\mathrm{... | Filter \( {\mathrm{U}}_{\mathrm{L}}^{\psi \left( n\right) } \) by the \( {\mathrm{U}}_{\mathrm{L}}^{\psi \left( m\right) } \), and \( {\mathrm{U}}_{\mathrm{K}}^{n} \) by the \( {\mathrm{U}}_{\mathrm{K}}^{m} \) ; by passage to the quotient, we get homomorphisms\n\n\[ \n{\mathrm{U}}_{\mathrm{L}}^{\psi \left( m\right) }/{... | Yes |
Corollary 4. We have \( \mathrm{N}\left( {\mathrm{U}}_{\mathrm{L}}^{\psi \left( v\right) }\right) = {\mathrm{U}}_{\mathrm{K}}^{v} \) if \( v \) is a real number \( > t \) or if \( \overline{\mathrm{K}} \) is algebraically closed. | Indeed, suppose that \( n < v \leq n + 1 \), where \( n \) is an integer. Then\n\n\[ \psi \left( n\right) < \psi \left( v\right) \leq \psi \left( {n + 1}\right) . \]\n\nIf \( m \) is the smallest integer \( \geq \psi \left( v\right) \), then\n\n\[ \psi \left( n\right) + 1 \leq m \leq \psi \left( {n + 1}\right) \]\n\nWe... | Yes |
Corollary 5. If \( t = 0,\operatorname{Coker}\left( {\mathrm{N}}_{t}\right) = {\overline{\mathrm{K}}}^{ * }/{\overline{\mathrm{K}}}^{*l} \) . If \( t \neq 0,\operatorname{Coker}\left( {\mathrm{N}}_{t}\right) \) is isomorphic to \( \overline{\mathrm{K}}/\wp \left( \overline{\mathrm{K}}\right) \), with \( \wp \left( \xi ... | If \( t = 0 \), we have seen that \( {\mathrm{N}}_{t}\left( \xi \right) = {\xi }^{l} \), whence the first assertion. If \( t \neq 0 \) , we must show that \( \operatorname{Coker}\left( {\mathrm{N}}_{t}\right) \) is isomorphic to \( \operatorname{Coker}\left( \wp \right) \) . Now we have\n\n\[{\mathrm{N}}_{t}\left( \xi ... | Yes |
Corollary 6. If \( \overline{\mathrm{K}} \) is perfect, \( \mathrm{N} : {\mathrm{U}}_{\mathrm{L}}/{\mathrm{U}}_{\mathrm{L}}^{n} \rightarrow {\mathrm{U}}_{\mathrm{K}}/{\mathrm{U}}_{\mathrm{K}}^{n} \) is an isomorphism for all \( n \leq t \) . | By cors. 1 and \( 2,{\mathrm{\;N}}_{n} \) is surjective for \( n < t \) . The result follows by induction on \( n \), using lemma 1. | No |
Corollary 7. If \( \overline{\mathrm{K}} \) is perfect, the following three canonical homomorphism are isomorphisms: | a) Apply lemma 1 to the diagram\n\n\[ 0 \rightarrow {\mathrm{U}}_{\mathrm{L}}^{t + 1} \rightarrow {\mathrm{U}}_{\mathrm{L}}^{t} \rightarrow {\mathrm{U}}_{\mathrm{L}}^{t}/{\mathrm{U}}_{\mathrm{L}}^{t + 1} \rightarrow 0 \]\n\n\[ 0 \rightarrow {\mathrm{U}}_{\mathrm{K}}^{t + 1} \rightarrow {\mathrm{U}}_{\mathrm{K}}^{t} \ri... | Yes |
Proposition 6. With the hypotheses above, let \( x \in {\mathrm{K}}^{ * } \) . Then there exists an extension \( {\overline{\mathrm{K}}}^{\prime }/\overline{\mathrm{K}} \), of degree \( \leq l \), such that \( x \) is a norm in the corresponding extended extension \( {\mathrm{L}}^{\prime }/{\mathrm{K}}^{\prime } \) . | By cor. 7 to prop. 5, we may assume that \( x \in {\mathrm{U}}_{\mathrm{K}}^{t} \) . For \( x \) to be a norm in this case, it is necessary and sufficient that its class \( \xi \) in \( {\mathrm{U}}_{\mathrm{K}}^{t}/{\mathrm{U}}_{\mathrm{K}}^{t + 1} \) have the form \( {\mathrm{N}}_{t}\left( \eta \right) \), where \( \... | Yes |
Lemma 6. Suppose that \( \mathrm{L} \) is a field that is the union of an increasing directed family of subfields \( {\left\{ {\mathbf{L}}_{i}\right\} }_{i \in 1} \), and let \( \mathbf{M} \) be a finite extension of \( \mathbf{L} \) of degree \( n \) . Then there exists an index \( i \in \mathbf{I} \) and an extension... | Let \( \left\{ {m}_{\alpha }\right\} ,\alpha = 1,\ldots, n \), be a basis of \( \mathrm{M} \) over \( \mathrm{L} \) ; then\n\n\[ \n{m}_{\alpha } \cdot {m}_{\beta } = \sum {c}_{\alpha \beta }^{\gamma }{m}_{\gamma },\;\text{ with }{c}_{\alpha \beta }^{\gamma } \in \mathrm{L}.\n\]\n\nChoose \( i \) so large that the \( {c... | Yes |
Lemma 7. Let \( \mathrm{K} \) be a field complete under a discrete valuation, having perfect residue field \( \overline{\mathrm{K}} \). Let \( {\mathrm{K}}_{nr} \) be the maximal unramified extension of \( \mathrm{K} \), and let \( \mathrm{E} \) be a finite extension of \( {\mathrm{K}}_{nr} \) of degree \( n \). Then t... | The residue field \( {\overline{\mathrm{K}}}_{nr} \) of \( {\mathrm{K}}_{nr} \) is an algebraic closure of \( \overline{\mathrm{K}} \). Let \( {\left\{ {\overline{\mathrm{K}}}_{i}\right\} }_{i \in 1} \) be the family of finite subextensions of \( {\overline{\mathbf{K}}}_{nr} \), and let \( {\left\{ {\mathbf{K}}_{i}\rig... | Yes |
Proposition 7. Let \( \mathrm{K} \) be a field complete under a discrete valuation, with perfect residue field \( \overline{\mathbf{K}} \) . Let \( {\mathbf{K}}_{nr} \) be the maximal unramified extension of \( \mathbf{K} \) , and let \( \mathrm{F} \supset \mathrm{E} \supset {\mathrm{K}}_{nr} \) be two finite extension... | We may assume \( \mathrm{F}/\mathrm{E} \) to be Galois (enlarging \( \mathrm{F} \) if necessary). By lemma 7, there is a finite subextension \( {\mathrm{K}}^{\prime } \) of \( {\mathrm{K}}_{nr} \) such that \( \mathrm{F} = {\mathrm{F}}_{nr}^{\prime },\mathrm{E} = {\mathrm{E}}_{nr}^{\prime } \), where \( {\mathrm{F}}^{\... | No |
Subsets and Splits
No community queries yet
The top public SQL queries from the community will appear here once available.