Q stringlengths 4 3.96k | A stringlengths 1 3k | Result stringclasses 4
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(i) \( \mathcal{Z}\left( {\mathfrak{a}\mathfrak{b}}\right) = \mathcal{Z}\left( \mathfrak{a}\right) \cup \mathcal{Z}\left( \mathfrak{b}\right) \) . | Proof. Exercise. See Corollary 2.3 of Chapter X. | No |
Theorem 5.3. Let \( A \) be a Noetherian commutative ring. Then every closed set \( C \) can be expressed as a finite union of irreducible closed sets, and this expression is unique if in the union\n\n\[ C = {C}_{1} \cup \cdots \cup {C}_{r} \]\n\nof irreducible closed sets, we have \( {C}_{i} ⊄ {C}_{j} \) if \( i \neq ... | Proof. We give the proof as an example to show how the version of Theorem 2.2 has an immediate translation in the more general context of \( \operatorname{spec}\left( A\right) \) . Suppose the family of closed sets which cannot be represented as a finite union of irreducible ones is not empty. Translating the Noetheria... | Yes |
Proposition 5.4. Let \( C \) be a closed subset of \( \operatorname{spec}\left( A\right) \) . Then \( C \) is irreducible if and only if \( C = \mathcal{L}\left( \mathfrak{p}\right) \) for some prime ideal \( \mathfrak{p} \) . | Proof. Exercise. | No |
Proposition 1.1. Let \( M \) be a Noetherian \( A \) -module. Then every submodule and every factor module of \( M \) is Noetherian. | Proof. Our assertion is clear for submodules (say from the first condition). For the factor module, let \( N \) be a submodule and \( f : M \rightarrow M/N \) the canonical homomorphism. Let \( {\bar{M}}_{1} \subset {\bar{M}}_{2} \subset \cdots \) be an ascending chain of submodules of \( M/N \) and let \( {M}_{i} = {f... | Yes |
Proposition 1.2. Let \( M \) be a module, \( N \) a submodule. Assume that \( N \) and \( M/N \) are Noetherian. Then \( M \) is Noetherian. | Proof. With every submodule \( L \) of \( M \) we associate the pair of modules\n\n\[ L \mapsto \left( {L \cap N,\left( {L + N}\right) /N}\right) .\n\]\n\nWe contend: If \( E \subset F \) are two submodules of \( M \) such that their associated pairs are equal, then \( E = F \) . To see this, let \( x \in F \) . By the... | Yes |
Corollary 1.3. Let \( M \) be a module, and let \( N,{N}^{\prime } \) be submodules. If \( M = N + {N}^{\prime } \) and if both \( N,{N}^{\prime } \) are Noetherian, then \( M \) is Noetherian. \( A \) finite direct sum of Noetherian modules is Noetherian. | Proof. We first observe that the direct product \( N \times {N}^{\prime } \) is Noetherian since it contains \( N \) as a submodule whose factor module is isomorphic to \( {N}^{\prime } \) , and Proposition 1.2 applies. We have a surjective homomorphism \[ N \times {N}^{\prime } \rightarrow M \] such that the pair \( \... | Yes |
Proposition 1.4. Let \( A \) be a Noetherian ring and let \( M \) be a finitely generated module. Then \( M \) is Noetherian. | Proof. Let \( {x}_{1},\ldots ,{x}_{n} \) be generators of \( M \) . There exists a homomorphism\n\n\[ f : A \times A \times \cdots \times A \rightarrow M \]\n\nof the product of \( A \) with itself \( n \) times such that\n\n\[ f\left( {{a}_{1},\ldots ,{a}_{n}}\right) = {a}_{1}{x}_{1} + \cdots + {a}_{n}{x}_{n}. \]\n\nT... | No |
Proposition 1.5. Let \( A \) be a ring which is Noetherian, and let \( \varphi : A \rightarrow B \) be a surjective ring-homomorphism. Then \( B \) is Noetherian. | Proof. Let \( {\mathfrak{b}}_{1} \subset \cdots \subset {\mathfrak{b}}_{n} \subset \cdots \) be an ascending chain of left ideals of \( B \) and let \( {\mathfrak{a}}_{i} = {\varphi }^{-1}\left( {\mathfrak{b}}_{i}\right) \) . Then the \( {\mathfrak{a}}_{i} \) form an ascending chain of left ideals of \( A \) which must... | No |
Proposition 1.6. Let \( A \) be a commutative Noetherian ring, and let \( S \) be a multiplicative subset of \( A \) . Then \( {S}^{-1}A \) is Noetherian. | Proof. We leave the proof as an exercise. | No |
Proposition 2.1. Let \( S \) be a multiplicative subset of \( A \), and assume that \( S \) does not contain 0 . Then there exists an ideal of \( A \) which is maximal in the set of ideals not intersecting \( S \), and any such ideal is prime. | Proof. The existence of such an ideal \( \mathfrak{p} \) follows from Zorn’s lemma (the set of ideals not meeting \( S \) is not empty, because it contains the zero ideal, and is clearly inductively ordered). Let \( \mathfrak{p} \) be maximal in the set. Let \( a, b \in A,{ab} \in \mathfrak{p} \) , but \( a \notin \mat... | Yes |
Corollary 2.2. An element \( a \) of \( A \) is nilpotent if and only if it lies in every prime ideal of \( A \) . | Proof. If \( {a}^{n} = 0 \), then \( {a}^{n} \in \mathfrak{p} \) for every prime \( \mathfrak{p} \), and hence \( a \in \mathfrak{p} \) . If \( {a}^{n} \neq 0 \) for any positive integer \( n \), we let \( S \) be the multiplicative subset of powers of \( a \) , namely \( \left\{ {1, a,{a}^{2},\ldots }\right\} \), and ... | Yes |
Corollary 2.3. An element a of \( A \) lies in the radical of an ideal \( \mathfrak{a} \) if and only if it lies in every prime ideal containing \( \mathfrak{a} \) . | Proof. Corollary 2.3 is equivalent to Corollary 2.2 applied to the ring \( A/a \) . | No |
Lemma 2.4. Let \( x \) be an element of a module \( M \), and let \( \mathfrak{a} \) be its annihilator.\n\nLet \( \mathfrak{p} \) be a prime ideal of \( A \) . Then \( {\left( Ax\right) }_{\mathfrak{p}} \neq 0 \) if and only if \( \mathfrak{p} \) contains \( \mathfrak{a} \) . | Proof. The lemma is an immediate consequence of the definitions, and will be left to the reader. | No |
Proposition 2.5. Let \( M \) be a module, \( a \in A \) . Then \( {a}_{M} \) is locally nilpotent if and only if \( a \) lies in every prime ideal \( \mathfrak{p} \) such that \( {M}_{\mathfrak{p}} \neq 0 \) . | Proof. Assume that \( {a}_{M} \) is locally nilpotent. Let \( \mathfrak{p} \) be a prime of \( A \) such that \( {M}_{\mathfrak{p}} \neq 0 \) . Then there exists \( x \in M \) such that \( {\left( Ax\right) }_{\mathfrak{p}} \neq 0 \) . Let \( n \) be a positive integer such that \( {a}^{n}x = 0 \) . Let \( \mathfrak{a}... | Yes |
Proposition 2.6. Let \( M \) be a module \( \neq 0 \) . Let \( \mathfrak{p} \) be a maximal element in the set of ideals which are annihilators of elements \( x \in M, x \neq 0 \) . Then \( \mathfrak{p} \) is prime. | Proof. Let \( \mathfrak{p} \) be the annihilator of the element \( x \neq 0 \) . Then \( \mathfrak{p} \neq A \) . Let \( a, b \in A,{ab} \in \mathfrak{p}, a \notin \mathfrak{p} \) . Then \( {ax} \neq 0 \) . But the ideal \( \left( {b,\mathfrak{p}}\right) \) annihilates \( {ax} \), and contains \( \mathfrak{p} \) . Sinc... | Yes |
Corollary 2.7. If \( A \) is Noetherian and \( M \) is a module \( \neq 0 \), then there exists a prime associated with \( M \) . | Proof. The set of ideals as in Proposition 2.6 is not empty since \( M \neq 0 \) , and has a maximal element because \( A \) is Noetherian. | Yes |
Corollary 2.8. Assume that both \( A \) and \( M \) are Noetherian, \( M \neq 0 \) . Then there exists a sequence of submodules\n\n\[ \nM = {M}_{1} \supset {M}_{2} \supset \cdots \supset {M}_{r} = 0 \n\] \n\nsuch that each factor module \( {M}_{i}/{M}_{i + 1} \) is isomorphic to \( A/{\mathfrak{p}}_{i} \) for some prim... | Proof. Consider the set of submodules having the property described in the corollary. It is not empty, since there exists an associated prime \( \mathfrak{p} \) of \( M \) , and if \( \mathfrak{p} \) is the annihilator of \( x \), then \( {Ax} \approx A/\mathfrak{p} \) . Let \( N \) be a maximal element in the set. If ... | Yes |
Proposition 2.9. Let \( A \) be Noetherian, and \( a \in A \) . Let \( M \) be a module. Then \( {a}_{M} \) is injective if and only if a does not lie in any associated prime of \( M \) . | Proof. Assume that \( {a}_{M} \) is not injective, so that \( {ax} = 0 \) for some \( x \in M \) , \( x \neq 0 \) . By Corollary 2.7, there exists an associated prime \( \mathfrak{p} \) of \( {Ax} \), and \( a \) is an element of \( \mathfrak{p} \) . Conversely, if \( {a}_{M} \) is injective, then \( a \) cannot lie in... | Yes |
Proposition 2.10. Let \( A \) be Noetherian, and let \( M \) be a module. Let \( a \in A \) . The following conditions are equivalent:\n\n(i) \( {a}_{M} \) is locally nilpotent.\n\n(ii) a lies in every associated prime of \( M \) .\n\n(iii) a lies in every prime \( \mathfrak{p} \) such that \( {M}_{\mathfrak{p}} \neq 0... | Proof. The fact that (i) implies (ii) is obvious from the definitions, and does not need the hypothesis that \( A \) is Noetherian. Neither does the fact that (iii) implies (i), which has been proved in Proposition 2.5. We must therefore prove that (ii) implies (iii) which is actually implied by the last statement. The... | Yes |
Corollary 2.11. Let \( A \) be Noetherian, and let \( M \) be a module. The following conditions are equivalent:\n\n(i) There exists only one associated prime of \( M \) .\n\n(ii) We have \( M \neq 0 \), and for every \( a \in A \), the homomorphism \( {a}_{M} \) is injective, or locally nilpotent.\n\nIf these conditio... | Proof. Immediate consequence of Propositions 2.9 and 2.10. | No |
Proposition 2.12. Let \( N \) be a submodule of \( M \) . Every associated prime of \( N \) is associated with \( M \) also. An associated prime of \( M \) is associated with \( N \) or with \( M/N \) . | Proof. The first assertion is obvious. Let \( \mathfrak{p} \) be an associated prime of \( M \) , and say \( \mathfrak{p} \) is the annihilator of the element \( x \neq 0 \) . If \( {Ax} \cap N = 0 \), then \( {Ax} \) is isomorphic to a submodule of \( M/N \), and hence \( \mathfrak{p} \) is associated with \( M/N \) .... | Yes |
Proposition 3.1. Let \( M \) be a module, and \( {Q}_{1},\ldots ,{Q}_{r} \) submodules which are \( \mathfrak{p} \) -primary for the same prime \( \mathfrak{p} \) . Then \( {Q}_{1} \cap \cdots \cap {Q}_{r} \) is also \( \mathfrak{p} \) -primary. | Proof. Let \( Q = {Q}_{1} \cap \cdots \cap {Q}_{r} \) . Let \( a \in \mathfrak{p} \) . Let \( {n}_{i} \) be such that \( {\left( {a}_{M/{Q}_{i}}\right) }^{{n}_{i}} = 0 \) for each \( i = 1,\ldots, r \) and let \( n \) be the maximum of \( {n}_{1},\ldots ,{n}_{r} \) . Then \( {a}_{M/Q}^{{n}^{ * }} = 0 \) , so that \( {a... | Yes |
Theorem 3.2. Let \( N \) be a submodule of \( M \), and let\n\n\[ N = {Q}_{1} \cap \cdots \cap {Q}_{r} = {Q}_{1}^{\prime } \cap \cdots \cap {Q}_{s}^{\prime } \]\n\nbe a reduced primary decomposition of \( N \) . Then \( r = s \) . The set of primes belonging to \( {Q}_{1},\ldots ,{Q}_{r} \) and \( {Q}_{1}^{\prime },\ld... | Proof. The uniqueness of the number of terms in a reduced decomposition and the uniqueness of the family of primes belonging to the primary components will be a consequence of Theorem 3.5 below.\n\nThere remains to prove the uniqueness of the primary module belonging to an isolated prime, say \( {\mathfrak{p}}_{1} \) .... | Yes |
Theorem 3.3. Let \( M \) be a Noetherian module. Let \( N \) be a submodule of M. Then \( N \) admits a primary decomposition. | Proof. We consider the set of submodules of \( M \) which do not admit a primary decomposition. If this set is not empty, then it has a maximal element because \( M \) is Noetherian. Let \( N \) be this maximal element. Then \( N \) is not primary, and there exists \( a \in A \) such that \( {a}_{M/N} \) is neither inj... | Yes |
Proposition 3.4. Let \( A \) and \( M \) be Noetherian. A submodule \( Q \) of \( M \) is primary if and only if \( M/Q \) has exactly one associated prime \( \mathfrak{p} \), and in that case, \( \mathfrak{p} \) belongs to \( Q \), i.e. \( Q \) is \( \mathfrak{p} \)-primary. | Proof. Immediate consequence of the definitions, and Corollary 2.11. | No |
Theorem 3.5. Let \( A \) and \( M \) be Noetherian. The associated primes of \( M \) are precisely the primes which belong to the primary modules in a reduced primary decomposition of 0 in \( M \) . In particular, the set of associated primes of \( M \) is finite. | Proof. Let\n\n\[ 0 = {Q}_{1} \cap \cdots \cap {Q}_{r} \]\n\nbe a reduced primary decomposition of 0 in \( M \) . We have an injective homomorphism\n\n\[ M \rightarrow {\bigoplus }_{i = 1}^{r}M/{Q}_{i} \]\n\nBy Proposition 2.12 and Proposition 3.4, we conclude that every associated prime of \( M \) belongs to some \( {Q... | Yes |
Theorem 3.6. Let \( A \) be a Noetherian ring. Then the set of divisors of zero in \( A \) is the set-theoretic union of all primes belonging to primary ideals in a reduced primary decomposition of 0. | Proof. An element of \( a \in A \) is a divisor of 0 if and only if \( {a}_{A} \) is not injective. According to Proposition 2.9, this is equivalent to \( a \) lying in some associated prime of \( A \) (viewed as module over itself). Applying Theorem 3.5 concludes the proof. | Yes |
Lemma 4.1. Let \( \mathfrak{a} \) be an ideal of \( A \) which is contained in every maximal ideal of \( A \) . Let \( E \) be a finitely generated \( A \) -module. Suppose that \( \mathfrak{a}E = E \) . Then \( E = \{ 0\} \) . | Proof. Induction on the number of generators of \( E \) . Let \( {x}_{1},\ldots ,{x}_{s} \) be generators of \( E \) . By hypothesis, there exist elements \( {a}_{1},\ldots ,{a}_{s} \in \mathfrak{a} \) such that\n\n\[ \n{x}_{s} = {a}_{1}{x}_{1} + \cdots + {a}_{s}{x}_{s} \n\]\n\nso there is an element \( a \) (namely \(... | Yes |
Lemma 4.2. Let \( A \) be a local ring, let \( E \) be a finitely generated \( A \) -module, and \( F \) a submodule. If \( E = F + {mE} \), then \( E = F \) . | Proof. Apply Lemma 4.1 to \( E/F \) . | No |
Lemma 4.3. Let \( A \) be a local ring. Let \( E \) be a finitely generated \( A \) -module. If \( {x}_{1},\ldots ,{x}_{n} \) are generators for \( E{\;\operatorname{mod}\;\mathfrak{m}}E \), then they are generators for \( E \) . | Proof. Take \( F \) to be the submodule generated by \( {x}_{1},\ldots ,{x}_{n} \) . | No |
Theorem 4.4. Let \( A \) be a local ring and \( E \) a finite projective \( A \) -module. Then \( E \) is free. In fact, if \( {x}_{1},\ldots ,{x}_{n} \) are elements of \( E \) whose residue classes \( {\bar{x}}_{1},\ldots ,{\bar{x}}_{n} \) are a basis of \( E/\mathfrak{m}E \) over \( A/\mathfrak{m} \), then \( {x}_{1... | Proof. I am indebted to George Bergman for the following proof of the first statement. Let \( F \) be a free module with basis \( {e}_{1},\ldots ,{e}_{n} \), and let \( f : F \rightarrow E \) be the homomorphism mapping \( {e}_{i} \) to \( {x}_{i} \) . We want to prove that \( f \) is an isomorphism. By Lemma 4.3, \( f... | Yes |
Proposition 4.5. Let \( f : E \rightarrow F \) be a homomorphism of modules, finite over a local ring \( A \) . Then:\n\n(i) If \( {f}_{\left( \mathrm{m}\right) } \) is surjective, so is \( f \) .\n\n(ii) Assume \( f \) is injective. If \( {f}_{\left( m\right) } \) is surjective, then \( f \) is an isomorphism.\n\n(iii... | Proof. The proofs are immediate consequences of Nakayama's lemma and will be left to the reader. For instance, in the first statement, consider the exact sequence\n\n\[ E \rightarrow F \rightarrow F/\operatorname{Im}f \rightarrow 0 \]\n\nand apply Nakayama to the term on the right. In (iii), use the lifting of bases as... | No |
Proposition 5.1. Let \( \left\{ {E}_{n}\right\} \) and \( \left\{ {E}_{n}^{\prime }\right\} \) be stable a-filtrations of \( E \) . Then there exists a positive integer \( d \) such that\n\n\[ \n{E}_{n + d} \subset {E}_{n}^{\prime }\;\text{ and }\;{E}_{n + d}^{\prime } \subset {E}_{n} \n\] \n\nfor all \( n \geqq 0 \) . | Proof. It suffices to prove the proposition when \( {E}_{n}^{\prime } = {\mathfrak{a}}^{n}E \) . Since \( \mathfrak{a}{E}_{n} \subset {E}_{n + 1} \) for all \( n \), we have \( {\mathfrak{a}}^{n}E \subset {E}_{n} \) . By the stability hypothesis, there exists \( d \) such that \n\n\[ \n{E}_{n + d} = {\mathfrak{a}}^{n}{... | Yes |
Proposition 5.2. Let \( A \) be a graded ring. Then \( A \) is Noetherian if and only if \( {A}_{0} \) is Noetherian, and \( A \) is finitely generated as \( {A}_{0} \) -algebra. | Proof. A finitely generated algebra over a Noetherian ring is Noetherian, because it is a homomorphic image of the polynomial ring in finitely many variables, and we can apply Hilbert's theorem.\n\nConversely, suppose that \( A \) is Noetherian. The sum\n\n\[ \n{A}^{ + } = {\bigoplus }_{n = 1}^{\infty }{A}_{n} \n\]\n\n... | Yes |
Lemma 5.3. Let \( A \) be a Noetherian ring, and \( E \) a finitely generated module, with an \( \mathfrak{a} \) -filtration. Then \( {E}_{S} \) is finite over \( S \) if and only if the filtration of \( E \) is a-stable. | Proof. Let\n\n\[ \n{F}_{n} = {\bigoplus }_{i = 0}^{n}{E}_{i} \n\]\n\nand let\n\n\[ \n{G}_{n} = {E}_{0} \oplus \cdots \oplus {E}_{n} \oplus \mathfrak{a}{E}_{n} \oplus {\mathfrak{a}}^{2}{E}_{n} \oplus {\mathfrak{a}}^{3}{E}_{n} \oplus \cdots \n\]\n\nThen \( {G}_{n} \) is an \( S \) -submodule of \( {E}_{S} \), and is fini... | Yes |
Theorem 5.4. (Artin-Rees). Let \( A \) be a Noetherian ring, a an ideal, \( E \) a finite A-module with a stable a-filtration. Let \( F \) be a submodule, and let \( {F}_{n} = F \cap {E}_{n} \) . Then \( \left\{ {F}_{n}\right\} \) is a stable \( \mathfrak{a} \) -filtration of \( F \) . | Proof. We have\n\n\[ \mathfrak{a}\left( {F \cap {E}_{n}}\right) \subset \mathfrak{a}F \cap \mathfrak{a}{E}_{n} \subset F \cap {E}_{n + 1}, \]\n\nso \( \left\{ {F}_{n}\right\} \) is an a-filtration of \( F \) . We can then form the associated graded \( S \) -module \( {F}_{S} \), which is a submodule of \( {E}_{S} \), a... | No |
Corollary 5.5. Let \( A \) be a Noetherian ring, \( E \) a finite \( A \) -module, and \( F \) a submodule. Let \( \mathfrak{a} \) be an ideal. There exists an integer \( s \) such that for all integers \( n \geqq s \) we have\n\n\[{\mathfrak{a}}^{n}E \cap F = {\mathfrak{a}}^{n - s}\left( {{\mathfrak{a}}^{s}E \cap F}\r... | Proof. Special case of Theorem 5.4 and the definitions. | No |
Theorem 5.6. (Krull). Let \( A \) be a Noetherian ring, and let \( \mathfrak{a} \) be an ideal contained in every maximal ideal of \( A \) . Let \( E \) be a finite \( A \) -module. Then\n\n\[ \mathop{\bigcap }\limits_{{n = 1}}^{\infty }{\mathfrak{a}}^{n}E = 0 \] | Proof. Let \( F = \bigcap {\mathfrak{a}}^{n}E \) and apply Nakayama’s lemma to conclude the proof. | No |
Corollary 5.7. Let \( \mathfrak{o} \) be a local Noetherian ring with maximal ideal \( \mathfrak{m} \) . Then\n\n\[\n\mathop{\bigcap }\limits_{{n = 1}}^{\infty }{m}^{n} = 0\n\] | Proof. Special case of Theorem 5.6 when \( E = A \) . | Yes |
Proposition 5.8. Assume that \( A \) is Noetherian, and let \( \mathfrak{a} \) be an ideal of \( A \) . Then \( {\operatorname{gr}}_{\mathfrak{a}}\left( A\right) \) is Noetherian. If \( E \) is a finite \( A \) -module with a stable \( \mathfrak{a} \) -filtration, then \( \operatorname{gr}\left( E\right) \) is a finite... | Proof. Let \( {x}_{1},\ldots ,{x}_{s} \) be generators of a. Let \( {\bar{x}}_{i} \) be the residue class of \( {x}_{i} \) in \( \mathfrak{a}/{\mathfrak{a}}^{2} \) . Then \[ {\operatorname{gr}}_{\mathfrak{a}}\left( A\right) = \left( {A/\mathfrak{a}}\right) \left\lbrack {{\bar{x}}_{1},\ldots ,{\bar{x}}_{s}}\right\rbrack... | Yes |
Theorem 6.1. (Hilbert-Serre). Let \( s \) be the number of generators of \( A \) as \( {A}_{0} \) -algebra. Then \( P\left( {E, t}\right) \) is a rational function of type\n\n\[ P\left( {E, t}\right) = \frac{f\left( t\right) }{\mathop{\prod }\limits_{{i = 1}}^{s}\left( {1 - {t}^{{d}_{i}}}\right) } \]\n\nwith suitable p... | Proof. Induction on \( s \) . For \( s = 0 \) the assertion is trivially true. Let \( s \geqq 1 \) . Let \( A = {A}_{0}\left\lbrack {{x}_{1},\ldots ,{x}_{s}}\right\rbrack \), deg. \( {\mathrm{x}}_{i} = {d}_{i} \geqq 1 \) . Multiplication by \( {x}_{s} \) on \( E \) gives rise to an exact sequence\n\n\[ 0 \rightarrow {K... | Yes |
Theorem 6.2. Assume that \( A \) is generated as an \( {A}_{0} \) -algebra by homogeneous elements of degree 1 . Let \( d \) be the order of the pole of \( P\left( {E, t}\right) \) at \( t = 1 \) . Then for all sufficiently large \( n,\varphi \left( {E}_{n}\right) \) is a polynomial in \( n \) of degree \( d - 1 \) . (... | Proof. By Theorem 6.1, \( \varphi \left( {E}_{n}\right) \) is the coefficient of \( {t}^{n} \) in the rational function\n\n\[ P\left( {E, t}\right) = f\left( t\right) /{\left( 1 - t\right) }^{s}. \]\n\nCancelling powers of \( 1 - t \), we write \( P\left( {E, t}\right) = h\left( t\right) /{\left( 1 - t\right) }^{d} \),... | Yes |
Theorem 6.3. Let \( A \) be a Noetherian local ring with maximal ideal \( \mathfrak{m} \) . Let \( \mathfrak{q} \) be an \( \mathfrak{m} \) -primary ideal, and let \( E \) be a finitely generated \( A \) -module, with a stable \( \mathfrak{q} \) -filtration. Then:\n\n(i) \( E/{E}_{n} \) has finite length for \( n \geqq... | Proof. Let\n\n\[ G = {\operatorname{gr}}_{\mathfrak{q}}\left( A\right) = \bigoplus {\mathfrak{q}}^{n}/{\mathfrak{q}}^{n + 1}. \]\n\nThen \( \operatorname{gr}\left( E\right) = \bigoplus {E}_{n}/{E}_{n + 1} \) is a graded \( G \) -module, and \( {G}_{0} = A/\mathfrak{q} \) . By Proposition 5.8, \( G \) is Noetherian and ... | No |
Lemma 6.4. Let \( P \in \mathbf{Q}\left\lbrack T\right\rbrack \) be a polynomial of degree \( d \) with rational coefficients.\n\n(a) If \( P\left( n\right) \in \mathbf{Z} \) for all sufficiently large integers \( n \), then there exist integers \( {c}_{0},\ldots ,{c}_{d} \) such that\n\n\[ P\left( T\right) = {c}_{0}\l... | Proof. We prove (a) by induction. If the degree of \( P \) is 0, then the assertion is obvious. Suppose \( \deg P \geqq 1 \) . By (1) there exist rational numbers \( {c}_{0},\ldots ,{c}_{d} \) such that \( P\left( T\right) \) has the expression given in (a). But \( {\Delta P} \) has degree strictly smaller than deg \( ... | Yes |
Proposition 6.5. Let \( \mathfrak{a},\mathfrak{b} \) be homogeneous ideals in \( A \) . Then\n\n\[ \varphi \left( {n,\mathfrak{a} + \mathfrak{b}}\right) = \varphi \left( {n,\mathfrak{a}}\right) + \varphi \left( {n,\mathfrak{b}}\right) - \varphi \left( {n,\mathfrak{a} \cap \mathfrak{b}}\right) \]\n\n\[ \chi \left( {n,\m... | Proof. The first is immediate, and the second follows from the definition of \( \chi \) . | No |
Theorem 6.6. Let \( F \) be a homogeneous polynomial of degree \( d \) . Assume that \( F \) is not a divisor of zero mod \( \mathfrak{a} \), that is: if \( G \in A,{FG} \in \mathfrak{a} \), then \( G \in \mathfrak{a} \) . Then\n\n\[ \chi \left( {n,\mathfrak{a} + \left( F\right) }\right) = \chi \left( {n,\mathfrak{a}}\... | Proof. First observe that trivially\n\n\[ \varphi \left( {n,\left( F\right) }\right) = \varphi \left( {n - d}\right) ,\]\n\nbecause the degree of a product is the sum of the degrees. Next, using the hypothesis that \( F \) is not divisor of 0 mod \( \mathfrak{a} \), we conclude immediately\n\n\[ \varphi \left( {n,\math... | Yes |
Lemma 6.8. Let \( V, W \) be varieties over a field \( k \) . \[ \\text{If}V \\supset W\\text{and}\\dim V = \\dim W\\text{, then}V = W\\text{.} \] | Proof. Say \( V, W \) are in affine space \( {\\mathbf{A}}^{N} \) . Let \( {\\mathfrak{p}}_{V} \) and \( {\\mathfrak{p}}_{W} \) be the respective prime ideals of \( V \) and \( W \) in \( k\\left\\lbrack X\\right\\rbrack \) . Then we have a canonical homomorphism \[ k\\left\\lbrack X\\right\\rbrack /{\\mathfrak{p}}_{V}... | Yes |
Theorem 6.9. Let \( \\mathfrak{a} \) be a homogeneous ideal in \( A \) . Let \( r \) be the maximum dimension of the irreducible components of the algebraic space in projective space defined by \( \\mathfrak{a} \) . Then there exists a polynomial \( P \\in \\mathbf{Q}\\left\\lbrack T\\right\\rbrack \) of degree \( \\le... | Proof. By Proposition 6.7(c), we may assume that no primary component in the primary decomposition of \( \\mathfrak{a} \) is irrelevant. Let \( Z \) be the algebraic space of zeros of \( \\mathfrak{a} \) in projective space. We may assume \( k \) algebraically closed as noted previously. Then there exists a homogeneous... | Yes |
Example 2. Let \( A \) be a commutative Noetherian local ring with maximal ideal \( m \), and let \( q \) be an \( m \) -primary ideal. Then for every positive integer \( n \) , \( A/{\mathfrak{q}}^{n} \) is Artinian. | Indeed, \( A/{\mathfrak{q}}^{n} \) has a Jordan-Hölder filtration in which each factor is a finite dimensional vector space over the field \( A/m \), and is a module of finite length. See Proposition 7.2. | No |
Proposition 7.1. Let \( A \) be a ring, and let\n\n\[ 0 \rightarrow {E}^{\prime } \rightarrow E \rightarrow {E}^{\prime \prime } \rightarrow 0 \]\n\nbe an exact sequence of \( A \) -modules. Then \( E \) is Artinian if and only if \( {E}^{\prime } \) and \( {E}^{\prime \prime } \) are Artinian. | We leave the proof to the reader. The proof is the same as in the Noetherian case, reversing the inclusion relations between modules. | No |
Proposition 7.2. A module \( E \) has a finite simple filtration if and only if \( E \) is both Noetherian and Artinian. | Proof. A simple module is generated by one element, and so is Noetherian. Since it contains no proper submodule \( \neq 0 \), it is also Artinian. Proposition 7.2 is then immediate from Proposition 7.1. | No |
Proposition 7.3. (Fitting's Lemma). Assume that \( E \) is Noetherian and Artinian. Let \( u \in \operatorname{End}\left( E\right) \) . Then \( E \) has a direct sum decomposition\n\n\[ E = \operatorname{Im}{u}^{\infty } \oplus \operatorname{Ker}{u}^{\infty }.\]\n\nFurthermore, the restriction of \( u \) to \( \operato... | Proof. Choose \( n \) such that \( \operatorname{Im}{u}^{\infty } = \operatorname{Im}{u}^{n} \) and \( \operatorname{Ker}{u}^{\infty } = \operatorname{Ker}{u}^{n} \) . We have\n\n\[ \operatorname{Im}{u}^{\infty } \cap \operatorname{Ker}{u}^{\infty } = \{ 0\} \]\n\nfor if \( x \) lies in the intersection, then \( x = {u... | Yes |
Proposition 7.4. Let \( E \) be an indecomposable module over the ring \( A \). Assume E Noetherian and Artinian. Any endomorphism of \( E \) is either nilpotent or an automorphism. Furthermore \( \operatorname{End}\left( E\right) \) is local. | Proof. By Fitting’s lemma, we know that for any endomorphism \( u \), we have \( E = \operatorname{Im}{u}^{\infty } \) or \( E = \operatorname{Ker}{u}^{\infty } \). So we have to prove that \( \operatorname{End}\left( E\right) \) is local. Let \( u \) be an endomorphism which is not a unit, so \( u \) is nilpotent. For... | Yes |
Lemma 7.6. Let \( M, N \) be modules, and assume \( N \) indecomposable. Let \( u : M \rightarrow N \) and \( v : N \rightarrow M \) be such that \( {vu} \) is an automorphism. Then \( u, v \) are isomorphisms. | Proof. Let \( e = u{\left( vu\right) }^{-1}v \) . Then \( {e}^{2} = e \) is an idempotent, lying in \( \operatorname{End}\left( N\right) \) , and therefore equal to 0 or 1 since \( N \) is assumed indecomposable. But \( e \neq 0 \) because \( {\mathrm{{id}}}_{M} \neq 0 \) and\n\n\[ 0 \neq {\mathrm{{id}}}_{M} = {\mathrm... | Yes |
Proposition 1.1. Let \( K \) be an ordered field and \( F \) a subfield. Let \( \mathfrak{o} \) be the valuation ring determined by the ordering of \( K/F \), and let \( \mathfrak{m} \) be its maximal ideal. Then \( \mathfrak{o}/\mathfrak{m} \) is a real field. | Proof. Otherwise, we could write\n\n\[ \n- 1 = \sum {\alpha }_{i}^{2} + a \n\] \n\nwith \( {\alpha }_{i} \in \mathfrak{v} \) and \( a \in \mathfrak{m} \) . Since \( \sum {\alpha }_{i}^{2} \) is positive and \( a \) is infinitely small, such a relation is clearly impossible. | No |
Proposition 2.1. Let \( K \) be a real field.\n\n(i) If \( a \in K \), then \( K\left( \sqrt{a}\right) \) or \( K\left( \sqrt{-a}\right) \) is real. If \( a \) is a sum of squares in \( K \) , then \( K\left( \sqrt{a}\right) \) is real. If \( K\left( \sqrt{a}\right) \) is not real, then \( - a \) is a sum of squares in... | Proof. Let \( a \in K \) . If \( a \) is a square in \( K \), then \( K\left( \sqrt{a}\right) = K \) and hence is real by assumption. Assume that \( a \) is not a square in \( K \) . If \( K\left( \sqrt{a}\right) \) is not real, then there exist \( {b}_{i},{c}_{i} \in K \) such that\n\n\[ - 1 = \sum {\left( {b}_{i} + {... | Yes |
Theorem 2.2. Let \( K \) be a real field. Then there exists a real closure of \( K \) . If \( R \) is real closed, then \( R \) has a unique ordering. The positive elements are the squares of \( R \) . Every positive element is a square, and every polynomial of odd degree in \( R\left\lbrack X\right\rbrack \) has a roo... | Proof. By Zorn’s lemma, our field \( K \) is contained in some real closed field algebraic over \( K \) . Now let \( R \) be a real closed field. Let \( P \) be the set of non-zero elements of \( R \) which are sums of squares. Then \( P \) is closed under addition and multiplication. By Proposition 2.1, every element ... | Yes |
Corollary 2.3. Let \( K \) be a real field and a an element of \( K \) which is not a sum of squares. Then there exists an ordering of \( K \) in which a is negative. | Proof. The field \( K\left( \sqrt{-a}\right) \) is real by Proposition 1.1 and hence has an ordering as a subfield of a real closure. In this ordering, \( - a > 0 \) and hence \( a \) is negative. | Yes |
Proposition 2.4. Let \( R \) be a field such that \( R \neq {R}^{\mathrm{a}} \) but \( {R}^{\mathrm{a}} = R\left( \sqrt{-1}\right) \) . Then \( R \) is real and hence real closed. | Proof. Let \( P \) be the set of elements of \( R \) which are squares and \( \neq 0 \) . We contend that \( P \) is an ordering of \( R \) . Let \( a \in R, a \neq 0 \) . Suppose that \( a \) is not a square in \( R \) . Let \( \alpha \) be a root of \( {X}^{2} - a = 0 \) . Then \( R\left( \alpha \right) = R\left( \sq... | Yes |
Theorem 2.5. Let \( R \) be a real closed field, and \( f\left( X\right) \) a polynomial in \( R\left\lbrack X\right\rbrack \) . Let \( a, b \in R \) and assume that \( f\left( a\right) < 0 \) and \( f\left( b\right) > 0 \) . Then there exists \( c \) between \( a \) and \( b \) such that \( f\left( c\right) = 0 \) . | Proof. Since \( R\left( \sqrt{-1}\right) \) is algebraically closed, it follows that \( f \) splits into a product of irreducible factors of degree 1 or 2 . If \( {X}^{2} + {\alpha X} + \beta \) is irreducible \( \left( {\alpha ,\beta \in R}\right) \) then it is a sum of squares, namely \[ {\left( X + \frac{\alpha }{2}... | Yes |
Lemma 2.6. Let \( K \) be a subfield of an ordered field \( E \) . Let \( \alpha \in E \) be algebraic over \( K \), and a root of the polynomial \[ f\left( X\right) = {X}^{n} + {a}_{n - 1}{X}^{n - 1} + \cdots + {a}_{0} \] with coefficients in \( K \) . Then \( \left| \alpha \right| \leqq 1 + \left| {a}_{n - 1}\right| ... | Proof. If \( \left| \alpha \right| \leqq 1 \), the assertion is obvious. If \( \left| \alpha \right| > 1 \), we express \( {\left| \alpha \right| }^{n} \) in terms of the terms of lower degree, divide by \( {\left| \alpha \right| }^{n - 1} \), and get a proof for our lemma. | No |
Theorem 2.7. (Sturm’s Theorem). The number of roots of \( f \) between \( u \) and \( v \) is equal to \( {W}_{S}\left( u\right) - {W}_{S}\left( v\right) \) for any Sturm sequence \( S \) . | Proof. We observe that if \( {\alpha }_{1} < {\alpha }_{2} < \cdots < {\alpha }_{r} \) is the ordered sequence of roots of the polynomials \( {f}_{j} \) in \( \left\lbrack {u, v}\right\rbrack \left( {j = 0,\ldots, m - 1}\right) \), then \( {W}_{S}\left( x\right) \) is constant on the open intervals between these roots,... | Yes |
Corollary 2.8. Let \( K \) be an ordered field, \( f \) an irreducible polynomial of degree \( \geqq 1 \) over \( K \) . The number of roots of \( f \) in two real closures of \( K \) inducing the given ordering on \( K \) is the same. | Proof. We can take \( v \) sufficiently large positive and \( u \) sufficiently large negative in \( K \) so that all roots of \( f \) and all roots of the polynomials in the Sturm sequence lie between \( u \) and \( v \), using Lemma 2.6. Then \( {W}_{S}\left( u\right) - {W}_{S}\left( v\right) \) is the total number o... | Yes |
Theorem 2.9. Let \( K \) be an ordered field, and let \( R,{R}^{\prime } \) be real closures of \( K \) , whose orderings induce the given ordering on \( K \) . Then there exists a unique isomorphism \( \sigma : R \rightarrow {R}^{\prime } \) over \( K \), and this isomorphism is order-preserving. | Proof. We first show that given a finite subextension \( E \) of \( R \) over \( K \), there exists an embedding of \( E \) into \( {R}^{\prime } \) over \( K \) . Let \( E = K\left( \alpha \right) \), and let\n\n\[ f\left( X\right) = \operatorname{Irr}\left( {\alpha, K, X}\right) .\n\]\n\nThen \( f\left( \alpha \right... | Yes |
Proposition 2.10. Let \( K \) be an ordered field, \( {K}^{\prime } \) an extension such that there is no relation\n\n\[ - 1 = \mathop{\sum }\limits_{{i = 1}}^{n}{a}_{i}{\alpha }_{i}^{2} \]\n\nwith \( {a}_{i} \in K,{a}_{i} > 0 \), and \( {\alpha }_{i} \in {K}^{\prime } \) . Let \( L \) be the field obtained from \( {K}... | Proof. If not, there exists a relation of type\n\n\[ - 1 = \mathop{\sum }\limits_{{i = 1}}^{n}{a}_{i}{\alpha }_{i}^{2} \]\n\nwith \( {a}_{i} \in K,{a}_{i} > 0 \), and \( {\alpha }_{i} \in L \) . (We can take \( {a}_{i} = 1 \) .) Let \( r \) be the smallest integer such that we can write such a relation with \( {\alpha ... | Yes |
Theorem 2.11. Let \( K \) be an ordered field. There exists a real closure \( R \) of \( K \) inducing the given ordering on \( K \) . | Proof. Take \( {K}^{\prime } = K \) in Proposition 2.10. Then \( L \) is real, and is contained in a real closure. Our assertion is clear. | No |
Corollary 2.12. Let \( K \) be an ordered field, and \( {K}^{\prime } \) an extension field. In order that there exist an ordering on \( {K}^{\prime } \) inducing the given ordering of \( K \), it is necessary and sufficient that there is no relation of type\n\n\[ - 1 = \mathop{\sum }\limits_{{i = 1}}^{n}{a}_{i}{\alpha... | Proof. If there is no such relation, then Proposition 2.10 states that \( L \) is contained in a real closure, whose ordering induces an ordering on \( {K}^{\prime } \), and the given ordering on \( K \), as desired. The converse is clear. | No |
Corollary 3.2. Notation being as in the theorem, let \( {y}_{1},\ldots ,{y}_{m} \in k\left\lbrack x\right\rbrack \) and assume\n\n\[ \n{y}_{1} < {y}_{2} < \cdots < {y}_{m} \n\]\n\nis the given ordering of \( K \) . Then one can choose \( \varphi \) such that\n\n\[ \n\varphi {y}_{1} < \cdots < \varphi {y}_{m} \n\] | Proof. Let \( {\gamma }_{i} \in {K}^{\mathrm{a}} \) be such that \( {\gamma }_{i}^{2} = {y}_{i + 1} - {y}_{i} \) . Then \( K\left( {{\gamma }_{1},\ldots ,{\gamma }_{n - 1}}\right) \) has an ordering inducing the given ordering on \( K \) . We apply the theorem to the ring\n\n\[ \nk\left\lbrack {{x}_{1},\ldots ,{x}_{n},... | Yes |
Corollary 3.3. (Artin). Let \( k \) be a real field admitting only one ordering. Let \( f\left( {{X}_{1},\ldots ,{X}_{n}}\right) \in k\left( X\right) \) be a rational function having the property that for all \( \left( a\right) = \left( {{a}_{1},\ldots ,{a}_{n}}\right) \in {R}_{k}^{\left( n\right) } \) such that \( f\l... | Proof. Assume that our conclusion is false. By Corollary 2.3, there exists an ordering of \( k\left( X\right) \) in which \( f \) is negative. Apply Corollary 3.2 to the ring \[ k\left\lbrack {{X}_{1},\ldots ,{X}_{n}, h{\left( X\right) }^{-1}}\right\rbrack \] where \( h\left( X\right) \) is a polynomial denominator for... | Yes |
Lemma 3.4. Let \( R \) be a real closed field and let \( {R}_{0} \) be a subfield which is algebraically closed in \( R \) (i.e. such that every element of \( R \) not in \( {R}_{0} \) is transcendental over \( {R}_{0} \) ). Then \( {R}_{0} \) is real closed. | Proof. Let \( f\left( X\right) \) be an irreducible polynomial over \( {R}_{0} \) . It splits in \( R \) into linear and quadratic factors. Its coefficients in \( R \) are algebraic over \( {R}_{0} \), and hence must lie in \( {R}_{0} \) . Hence \( f\left( X\right) \) is linear itself, or quadratic irreducible already ... | Yes |
Lemma 3.5. Let \( R \) be a real closed field, and \( \left\{ {{h}_{i}\left( x\right) }\right\} \) a finite set of rational functions in one variable with coefficients in \( R \) . Suppose the rational field \( R\left( x\right) \) ordered in some way, so that each \( {h}_{i}\left( x\right) \) has a sign attached to it.... | Proof. Considering the numerators and denominators of the rational functions, we may assume without loss of generality that the \( {h}_{i} \) are polynomials. We then write\n\n\[ \n{h}_{i}\left( x\right) = \alpha \prod \left( {x - \lambda }\right) \prod p\left( x\right) \n\]\n\nwhere the first product is extended over ... | Yes |
Theorem 3.6. Let \( k \) be a real field, \( K = k\left( {{x}_{1},\ldots ,{x}_{n}, y}\right) = k\left( {x, y}\right) \) a finitely generated extension such that \( {x}_{1},\ldots ,{x}_{n} \) are algebraically independent over \( k \), and \( y \) is algebraic over \( k\left( x\right) \) . Let \( f\left( {X, Y}\right) \... | Proof. Let \( {t}_{1},\ldots ,{t}_{n} \) be algebraically independent over \( R \) . Inductively, we can put an ordering on \( R\left( {{t}_{1},\ldots ,{t}_{n}}\right) \) such that each \( {t}_{i} \) is infinitely small with respect to \( R \) ,(cf. the example in \( §1 \) ). Let \( {R}^{\prime } \) be a real closure o... | Yes |
Theorem 1.2. (Approximation Theorem). (Artin-Whaples). Let \( K \) be a field and \( {\left. \left| {}_{1},\ldots ,\right| {}_{s}\right| }_{s} \) non-trivial pairwise independent absolute values on \( K \) . Let \( {x}_{1},\ldots ,{x}_{s} \) be elements of \( K \), and \( \epsilon > 0 \) . Then there exists \( x \in K ... | Proof. Consider first two of our absolute values, say \( {v}_{1} \) and \( {v}_{2} \) . By hypothesis we can find \( \alpha \in K \) such that \( {\left| \alpha \right| }_{1} < 1 \) and \( {\left| \alpha \right| }_{s} \geqq 1 \) . Similarly, we can find \( \beta \in K \) such that \( {\left| \beta \right| }_{1} \geqq 1... | Yes |
Proposition 2.2. Let \( K \) be a complete field under a non-trivial absolute value, and let \( E \) be a finite-dimensional space over \( K \). Then any two norms on \( E \) (compatible with the given absolute value on \( K \)) are equivalent. | Proof. We shall first prove that the topology on \( E \) is that of a product space, i.e. if \( {\omega }_{1},\ldots ,{\omega }_{n} \) is a basis of \( E \) over \( K \), then a sequence\n\n\[ \n{\xi }^{\left( v\right) } = {x}_{1}^{\left( v\right) }{\omega }_{1} + \cdots + {x}_{n}^{\left( v\right) }{\omega }_{n},\;{x}_... | Yes |
Corollary 2.4. Let \( K \) be a field, which is an extension of \( \mathbf{R} \), and has an absolute value extending the ordinary absolute value on \( \mathbf{R} \) . Then \( K = \mathbf{R} \) or \( K = \mathbf{C} \) . | Proof. Assume first that \( K \) contains \( \mathbf{C} \) . Then the assumption that \( K \) is a field and Theorem 2.3 imply that \( K = \mathbf{C} \) .\n\nIf \( K \) does not contain \( \mathbf{C} \), in other words, does not contain a square root of -1, we let \( L = K\left( j\right) \) where \( {j}^{2} = - 1 \) . ... | Yes |
Proposition 2.5. Let \( K \) be complete with respect to a nontrivial absolute value \( v \) . If \( E \) is any algebraic extension of \( K \), then \( v \) has a unique extension to E. If \( E \) is finite over \( K \), then \( E \) is complete. | Proof. In the archimedean case, the existence is obvious since we deal with the real and complex numbers. In the non-archimedean case, we postpone the existence proof to a later section. It uses entirely different ideas from the present ones. As to uniqueness, we may assume that \( E \) is finite over \( K \) . By Prop... | No |
Proposition 3.1. Let \( E \) be a finite extension of \( K \) . Let \( w \) be an absolute value on \( E \) extending \( v \), and let \( {E}_{w} \) be the completion. Let \( {K}_{w} \) be the closure of \( K \) in \( {E}_{w} \) and identify \( E \) in \( {E}_{w} \) . Then \( {E}_{w} = E{K}_{w} \) (the composite field)... | Proof. We observe that \( {K}_{w} \) is a completion of \( K \), and that the composite field \( E{K}_{w} \) is algebraic over \( {K}_{w} \) and therefore complete by Proposition 2.5. Since it contains \( E \), it follows that \( E \) is dense in it, and hence that \( {E}_{w} = E{K}_{w} \) . | Yes |
Proposition 3.2. Let \( E \) be an algebraic extension of \( K \) . Two embeddings \( \sigma ,\tau : E \rightarrow {K}_{v}^{\mathrm{a}} \) give rise to the same absolute value on \( E \) if and only if they are conjugate over \( {K}_{v} \) . | Proof. Suppose they are conjugate over \( {K}_{v} \) . Then the uniqueness of the extension of the absolute value from \( {K}_{v} \) to \( {K}_{v}^{\mathrm{a}} \) guarantees that the induced absolute values on \( E \) are equal. Conversely, suppose this is the case. Let \( \lambda : {\tau E} \rightarrow {\sigma E} \) b... | Yes |
Proposition 3.3. Let \( E \) be a finite separable extension of \( K \), of degree \( N \) . Then\n\n\[ N = \mathop{\sum }\limits_{{w \mid v}}{N}_{w} \] | Proof. We can write \( E = K\left( \alpha \right) \) for a single element \( \alpha \) . Let \( f\left( X\right) \) be its irreducible polynomial over \( K \) . Then over \( {K}_{v} \), we have a decomposition\n\n\[ f\left( X\right) = {f}_{1}\left( X\right) \cdots {f}_{r}\left( X\right) \]\n\ninto irreducible factors \... | Yes |
Proposition 3.4. Let \( E \) be a finite extension of \( K \). Then\n\n\[ \mathop{\sum }\limits_{{w \mid v}}\left\lbrack {{E}_{w} : {K}_{v}}\right\rbrack \leqq \left\lbrack {E : K}\right\rbrack \]\n\nIf \( E \) is purely inseparable over \( K \), then there exists only one absolute value won\n\nE extending \( v \). | Proof. Let us first prove the second statement. If \( E \) is purely inseparable over \( K \), and \( {p}^{r} \) is its inseparable degree, then \( {\alpha }^{{p}^{r}} \in K \) for every \( \alpha \) in \( E \). Hence \( v \) has a unique extension to \( E \). Consider now the general case of a finite extension, and le... | Yes |
Proposition 3.8. Let \( E \) be a finite extension of \( K \), and assume that \( v \) is well behaved. Let \( \alpha \in E \). Then:\n\n\[ \n{N}_{K}^{E}\left( \alpha \right) = \mathop{\prod }\limits_{{w \mid v}}{N}_{{K}_{v}}^{{E}_{w}}\left( \alpha \right)\n\]\n\n\[ \n{\operatorname{Tr}}_{K}^{E}\left( \alpha \right) = ... | Proof. Suppose first that \( E = K\left( \alpha \right) \), and let \( f\left( X\right) \) be the irreducible polynomial of \( \alpha \) over \( K \). If we factor \( f\left( X\right) \) into irreducible terms over \( {K}_{v} \), then\n\n\[ \nf\left( X\right) = {f}_{1}\left( X\right) \cdots {f}_{r}\left( X\right)\n\]\n... | No |
Theorem 4.1. Let \( K \) be a subfield of a field \( L \) . Then a valuation on \( K \) has an extension to a valuation on \( L \) . | Proof. Let \( \mathfrak{o} \) be the valuation ring on \( K \) corresponding to the given valuation. Let \( \varphi : \mathfrak{o} \rightarrow \mathfrak{o}/\mathfrak{m} \) be the canonical homomorphism on the residue class field, and extend \( \varphi \) to a homomorphism of a valuation ring \( \mathfrak{O} \) of \( L ... | Yes |
Proposition 4.2. Let \( L \) be a finite extension of \( K \), of degree \( n \) . Let \( w \) be a valuation of \( L \) with value group \( {\Gamma }^{\prime } \) . Let \( \Gamma \) be the value group of \( K \) . Then \( \left( {{\Gamma }^{\prime } : \Gamma }\right) \leqq n \) . | Proof. Let \( {y}_{1},\ldots ,{y}_{r} \) be elements of \( L \) whose values represent distinct cosets of \( \Gamma \) in \( {\Gamma }^{\prime } \) . We shall prove that the \( {y}_{j} \) are linearly independent over \( K \) . In a relation \( {a}_{1}{y}_{1} + \cdots + {a}_{r}{y}_{r} = 0 \) with \( {a}_{j} \in K,{a}_{... | Yes |
Corollary 4.3. There exists an integer \( e \geqq 1 \) such that the map \( \gamma \mapsto {\gamma }^{e} \) induces an injective homomorphism of \( {\Gamma }^{\prime } \) into \( \Gamma \) . | Proof. Take \( e \) to be the index \( \left( {{\Gamma }^{\prime } : \Gamma }\right) \) . | No |
Corollary 4.4. If \( K \) is a field with a valuation \( v \) whose value group is an ordered subgroup of the ordered group of positive real numbers, and if \( L \) is an algebraic extension of \( K \), then there exists an extension of \( v \) to \( L \) whose value group is also an ordered subgroup of the positive re... | Proof. We know that we can extend \( v \) to a valuation \( w \) of \( L \) with some value group \( {\Gamma }^{\prime } \), and the value group \( \Gamma \) of \( v \) can be identified with a subgroup of \( {\mathbf{R}}^{ + } \) . By Corollary 4.3, every element of \( {\Gamma }^{\prime } \) has finite period modulo \... | Yes |
Corollary 4.5. If \( L \) is finite over \( K \), and if \( \Gamma \) is infinite cyclic, then \( {\Gamma }^{\prime } \) is also infinite cyclic. | Proof. Use Corollary 4.3 and the fact that a subgroup of a cyclic group is cyclic. | No |
Proposition 4.6. Let \( L \) be a finite extension of degree \( n \) of a field \( K \), and let \( \mathfrak{O} \) be a valuation ring of \( L \) . Let \( \mathfrak{M} \) be its maximal ideal, let \( \mathfrak{o} = \mathfrak{O} \cap K \), and let \( \mathfrak{m} \) be the maximal ideal of \( \mathfrak{o} \), i.e. \( \... | Proof. Let \( {y}_{1},\ldots ,{y}_{e} \) be representatives in \( {L}^{ * } \) of distinct cosets of \( {\Gamma }^{\prime }/\Gamma \) and let \( {z}_{1},\ldots ,{z}_{s} \) be elements of \( \mathfrak{O} \) whose residue classes \( {\;\operatorname{mod}\;\mathfrak{M}} \) are linearly independent over \( \mathfrak{o}/\ma... | Yes |
Proposition 4.7. Let \( K \) be a field with a valuation \( v \), and let \( K \subset E \subset L \) be finite extensions of \( K \) . Let \( w \) be an extension of \( v \) to \( E \) and let \( u \) be an extension of \( w \) to \( L \) . Then\n\n\[ e\left( {u \mid w}\right) e\left( {w \mid v}\right) = e\left( {u \m... | Proof. Obvious. | No |
Proposition 4.8. Let \( \mathfrak{o} \) be a valuation ring in a field \( K \) . Let \( L \) be a finite extension of \( K \) . Let \( \mathfrak{O} \) be a valuation ring of \( L \) lying above \( \mathfrak{o} \), and \( \mathfrak{M} \) its maximal ideal. Let \( B \) be the integral closure of \( \mathfrak{o} \) in \( ... | Proof. It is clear that \( {B}_{\mathfrak{P}} \) is contained in \( \mathfrak{O} \) . Conversely, let \( x \) be an element of \( \mathfrak{O} \) . Then \( x \) satisfies an equation with coefficients in \( K \), not all 0, say\n\n\[ \n{a}_{n}{x}^{n} + \cdots + {a}_{0} = 0,\;{a}_{i} \in K.\n\] \n\nSuppose that \( {a}_{... | Yes |
Corollary 4.9. Let the notation be as in the proposition. Then there is only a finite number of valuation rings of \( L \) lying above \( \mathfrak{P} \) . | Proof. This comes from the fact that there is only a finite number of maximal ideals \( \mathfrak{P} \) of \( B \) lying above the maximal ideal of \( \mathfrak{o} \) (Corollary of Proposition 2.1, Chapter VII). | Yes |
Corollary 4.10. Let the notation be as in the proposition. Assume in addition that \( L \) is Galois over \( K \) . If \( \mathfrak{O} \) and \( {\mathfrak{O}}^{\prime } \) are two valuation rings of \( L \) lying above \( \mathfrak{o} \) , with maximal ideals \( \mathfrak{M},{\mathfrak{M}}^{\prime } \) respectively, t... | Proof. Let \( \mathfrak{P} = \mathfrak{O} \cap B \) and \( {\mathfrak{P}}^{\prime } = {\mathfrak{O}}^{\prime } \cap B \) . By Proposition 2.1 of Chapter VII, we know that there exists an automorphism \( \sigma \) of \( L \) over \( K \) such that \( \sigma \mathfrak{P} = {\mathfrak{P}}^{\prime } \) . From this our asse... | Yes |
Corollary 6.2. Let \( \alpha \in E,\alpha \neq 0 \) . Let \( v \) be the valuation on \( K \) and \( w \) its extension to \( E \) . Then\n\n\[{\operatorname{ord}}_{v}{N}_{K}^{E}\left( \alpha \right) = f\left( {w \mid v}\right) {\operatorname{ord}}_{w}\alpha .\] | Proof. This is immediate from the formula\n\n\[ \left| {{N}_{K}^{E}\left( \alpha \right) }\right| = {\left| \alpha \right| }^{ef} \]\n\nand the definitions. | Yes |
Corollary 6.3. Let \( K \) be any field and \( v \) a discrete valuation on \( K \) . Let \( E \) be a finite extension of \( K \) . If \( v \) is well behaved in \( E \) (for instance if \( E \) is separable over \( K \) ), then\n\n\[ \mathop{\sum }\limits_{{w \mid v}}e\left( {w \mid v}\right) f\left( {w \mid v}\right... | Proof. Our first assertion comes from our assumption, and Proposition 3.3. If \( E \) is Galois over \( K \), we know from Corollary 4.10 that any two valuations of \( E \) lying above \( v \) are conjugate. Hence all ramification indices are equal, and similarly for the residue class degrees. Our relation efr \( = \le... | Yes |
Proposition 7.1. If \( g \) is sufficiently close to \( f \), and \( {\beta }_{1},\ldots ,{\beta }_{s} \) are the roots of \( g \) belonging to \( \alpha \) (counting multiplicities), then \( s = {r}_{1} \) is the multiplicity of \( \alpha \) in \( f \) . | Proof. Assume the contrary. Then we can find a sequence \( {g}_{v} \) of polynomials approaching \( f \) with precisely \( s \) roots \( {\beta }_{1}^{\left( v\right) },\ldots ,{\beta }_{s}^{\left( v\right) } \) belonging to \( \alpha \), but with \( s \neq r \) . (We can take the same multiplicity \( s \) since there ... | Yes |
Proposition 7.2. Let \( f\left( X\right) \in \mathfrak{o}\left\lbrack X\right\rbrack \) . Let \( r \) be an integer \( \geqq 1 \) and let \( A \in {\mathfrak{o}}^{\left( n\right) } \) be such that\n\n\[ f\left( A\right) \equiv 0\;\left( {\;\operatorname{mod}\;{\mathfrak{p}}^{{2r} - 1}}\right) \]\n\n\[ {D}_{i}f\left( A\... | Proof. The proof is shorter than the statement of the proposition. Write \( Y = B + {\pi }^{r + v}C \) . By Taylor’s expansion,\n\n\[ f\left( {B + {\pi }^{r + v}C}\right) = f\left( B\right) + {\pi }^{r + v}\operatorname{grad}f\left( B\right) \cdot C\;\left( {\;\operatorname{mod}\;{\mathfrak{p}}^{{2r} + {2v}}}\right) . ... | Yes |
Corollary 7.3. Assumptions being as in Proposition 7.2, there exists a zero of \( f \) in \( {\mathfrak{o}}^{\left( n\right) } \) which is congruent to \( A{\;\operatorname{mod}\;{\mathfrak{p}}^{r}} \) . | Proof. We can write this zero as a convergent sum\n\n\[ A + {\pi }^{r + 1}{C}_{1} + {\pi }^{r + 2}{C}_{2} + \cdots \]\n\nsolving for \( {C}_{1},{C}_{2},\ldots \) inductively as in the proposition. | Yes |
Corollary 7.4. Let \( f \) be a polynomial in one variable in \( \mathfrak{o}\left\lbrack X\right\rbrack \), and let \( a \in \mathfrak{o} \) be such that \( f\left( a\right) \equiv 0\left( {\;\operatorname{mod}\;\mathfrak{p}}\right) \) but \( {f}^{\prime }\left( a\right) ≢ 0\left( {\;\operatorname{mod}\;\mathfrak{p}}\... | Proof. Take \( n = 1 \) and \( r = 1 \) in the proposition, and apply Corollary 7.3. | No |
Corollary 7.5. Let \( m \) be a positive integer not divisible by the characteristic of \( K \) . There exists an integer \( r \) such that for any \( a \in \mathfrak{o}, a \equiv 1\\left( {\\;\\operatorname{mod}\\;{\\mathfrak{p}}^{r}}\\right) \), the equation \( {X}^{m} - a = 0 \) has a root in \( K \) . | Proof. Apply the proposition. | No |
Proposition 7.6. Let \( K \) be a complete under a non-archimedean absolute value (nontrivial). Let \( \mathfrak{o} \) be the valuation ring and let \( f\left( X\right) \in \mathfrak{o}\left\lbrack X\right\rbrack \) be a polynomial in one variable. Let \( {\alpha }_{0} \in \mathfrak{o} \) be such that \[ \left| {f\left... | Proof. Let \( c = \left| {f\left( {\alpha }_{0}\right) /{f}^{\prime }{\left( {\alpha }_{0}\right) }^{2}}\right| < 1 \) . We show inductively that: 1. \( \left| {\alpha }_{i}\right| \leqq 1 \) , 2. \( \left| {{\alpha }_{i} - {\alpha }_{0}}\right| \leqq c \) 3. \( \left| \frac{f\left( {\alpha }_{i}\right) }{{f}^{\prime }... | Yes |
Proposition 3.1. Let \( E \) be a free module over \( R \), and let \( \left\{ {{x}_{1},\ldots ,{x}_{n}}\right\} \) be a basis. Let \( {y}_{1},\ldots ,{y}_{n} \) be elements of \( E \) . Let \( A \) be the matrix in \( R \) such that\n\n\[ A\left( \begin{matrix} {x}_{1} \\ \vdots \\ {x}_{n} \end{matrix}\right) = \left(... | Proof. Let \( X, Y \) be the column vectors of our elements. Then \( {AX} = Y \) . Suppose \( Y \) is a basis. Then there exists a matrix \( C \) in \( R \) such that \( {CY} = X \) .\n\nThen \( {CAX} = X \), whence \( {CA} = I \) and \( A \) is invertible. Conversely, assume that \( A \) is invertible. Then \( X = {A}... | Yes |
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