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(i) \( \mathcal{Z}\left( {\mathfrak{a}\mathfrak{b}}\right) = \mathcal{Z}\left( \mathfrak{a}\right) \cup \mathcal{Z}\left( \mathfrak{b}\right) \) .
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Proof. Exercise. See Corollary 2.3 of Chapter X.
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No
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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 \( i \neq j \) .
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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 Noetherian hypothesis in this case shows that there exists a minimal such set \( C \) . Then \( C \) cannot be irreducible, and we can write \( C \) as a union of closed sets\n\n\[ C = {C}^{\prime } \cup {C}^{\prime \prime } \]\n\nwith \( {C}^{\prime } \neq C \) and \( {C}^{\prime \prime } \neq C \) . Since \( {C}^{\prime } \) and \( {C}^{\prime \prime } \) are strictly smaller than \( C \), then we can express \( {C}^{\prime } \) and \( {C}^{\prime \prime } \) as finite unions of irreducible closed sets, thus getting a similar expression for \( C \), and a contradiction which proves existence.\n\nAs to uniqueness, let\n\n\[ C = {C}_{1} \cup \cdots \cup {C}_{r} = {Z}_{1} \cup \cdots \cup {Z}_{s} \]\n\nbe an expression of \( C \) as union of irreducible closed sets, without inclusion relations. For each \( {Z}_{j} \) we can write\n\n\[ {Z}_{j} = \left( {{Z}_{j} \cap {C}_{1}}\right) \cup \cdots \cup \left( {{Z}_{j} \cap {C}_{r}}\right) \]\n\nSince each \( {Z}_{j} \cap {C}_{i} \) is a closed set, we must have \( {Z}_{j} = {Z}_{j} \cap {C}_{i} \) for some \( i \) . Hence \( {Z}_{j} = {C}_{i} \) for some \( i \) . Similarly, \( {C}_{i} \) is contained in some \( {Z}_{k} \) . Since there is no inclusion relation among the \( {Z}_{j} \) ’s, we must have \( {Z}_{j} = {C}_{i} = {Z}_{k} \) . This argument can be carried out for each \( {Z}_{j} \) and each \( {C}_{i} \) . This proves that each \( {Z}_{j} \) appears among the \( {C}_{i} \) ’s and each \( {C}_{i} \) appears among the \( {Z}_{j} \) ’s, and proves the uniqueness of our representation. This proves the theorem.
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Yes
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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 \( \mathfrak{p} \) .
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Proof. Exercise.
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No
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Proposition 1.1. Let \( M \) be a Noetherian \( A \) -module. Then every submodule and every factor module of \( M \) is Noetherian.
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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 \( {M}_{i} = {f}^{-1}\left( {\bar{M}}_{i}\right) \) . Then \( {M}_{1} \subset {M}_{2} \subset \cdots \) is an ascending chain of submodules of \( M \), which must have a maximal element, say \( {M}_{r} \), so that \( {M}_{i} = {M}_{r} \) for \( r \geqq i \) . Then \( f\left( {M}_{i}\right) = {\bar{M}}_{i} \) and our assertion follows.
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Yes
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Proposition 1.2. Let \( M \) be a module, \( N \) a submodule. Assume that \( N \) and \( M/N \) are Noetherian. Then \( M \) is Noetherian.
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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 hypothesis that \( \left( {E + N}\right) /N = \left( {F + N}\right) /N \) there exist elements \( u, v \in N \) and \( y \in E \) such that \( y + u = x + v \) . Then\n\n\[ x - y = u - v \in F \cap N = E \cap N.\n\]\n\nSince \( y \in E \), it follows the \( x \in E \) and our contention is proved. If we have an ascending sequence\n\n\[ {E}_{1} \subset {E}_{2} \subset \cdots\n\]\n\nthen the associated pairs form an ascending sequence of submodules of \( N \) and \( M/N \) respectively, and these sequences must stop. Hence our sequence \( {E}_{1} \subset {E}_{2}\cdots \) also stops, by our preceding contention.
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Yes
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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.
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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 \( \left( {x,{x}^{\prime }}\right) \) with \( x \in N \) and \( {x}^{\prime } \in {N}^{\prime } \) maps on \( x + {x}^{\prime } \) . By Proposition 1.1, it follows that \( M \) is Noetherian. Finite products (or sums) follow by induction.
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Yes
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Proposition 1.4. Let \( A \) be a Noetherian ring and let \( M \) be a finitely generated module. Then \( M \) is Noetherian.
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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\nThis homomorphism is surjective. By the corollary of the preceding proposition, the product is Noetherian, and hence \( M \) is Noetherian by Proposition 1.1.
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No
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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.
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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 stop, say at \( {\mathfrak{a}}_{r} \) . Since \( \varphi \left( {\mathfrak{a}}_{i}\right) = {\mathfrak{b}}_{i} \) for all \( i \), our proposition is proved.
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No
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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.
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Proof. We leave the proof as an exercise.
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No
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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.
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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 \( a \notin \mathfrak{p} \) and \( b \notin \mathfrak{p} \) . By hypothesis, the ideals \( \left( {a,\mathfrak{p}}\right) \) and \( \left( {b,\mathfrak{p}}\right) \) generated by \( a \) and \( \mathfrak{p} \) (or \( b \) and \( \mathfrak{p} \) respectively) meet \( S \), and there exist therefore elements \( s,{s}^{\prime } \in S, c,{c}^{\prime }, x,{x}^{\prime } \in A, p,{p}^{\prime } \in \mathfrak{p} \) such that\n\n\[ s = {ca} + {xp}\text{ and }{s}^{\prime } = {c}^{\prime }b + {x}^{\prime }{p}^{\prime }.\]\n\nMultiplying these two expressions, we obtain\n\n\[ s{s}^{\prime } = c{c}^{\prime }{ab} + {p}^{\prime \prime }\]\n\nwith some \( {p}^{\prime \prime } \in \mathfrak{p} \), whence we see that \( s{s}^{\prime } \) lies in \( \mathfrak{p} \) . This contradicts the fact that \( \mathfrak{p} \) does not intersect \( S \), and proves that \( \mathfrak{p} \) is prime.
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Yes
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Corollary 2.2. An element \( a \) of \( A \) is nilpotent if and only if it lies in every prime ideal of \( A \) .
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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 find a prime ideal as in the proposition to prove the converse.
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Yes
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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 \( \mathfrak{a} \) .
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Proof. Corollary 2.3 is equivalent to Corollary 2.2 applied to the ring \( A/a \) .
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No
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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 \( \mathfrak{a} \) .
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Proof. The lemma is an immediate consequence of the definitions, and will be left to the reader.
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No
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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 \( {M}_{\mathfrak{p}} \neq 0 \) .
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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 \( \mathfrak{a} \) be the annihilator of \( x \) . Then \( {a}^{n} \in \mathfrak{a} \), and hence we can apply the lemma, and Corollary 4.3 to conclude that \( a \) lies in every prime \( \mathfrak{p} \) such that \( {M}_{\mathfrak{p}} \neq 0 \) . Conversely, suppose \( {a}_{M} \) is not locally nilpotent, so there exists \( x \in M \) such that \( {a}^{n}x = 0 \) for all \( n \geqq 0 \) . Let \( S = \left\{ {1, a,{a}^{2},\ldots }\right\} \), and using Proposition 2.1 let \( \mathfrak{p} \) be a prime not intersecting \( S \) . Then \( {\left( Ax\right) }_{\mathfrak{p}} \neq 0 \), so \( {M}_{\mathfrak{p}} \neq 0 \) and \( a \notin \mathfrak{p} \), as desired.
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Yes
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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.
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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 \( \mathfrak{p} \) . Since \( \mathfrak{p} \) is maximal, it follows that \( b \in \mathfrak{p} \), and hence \( \mathfrak{p} \) is prime.
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Yes
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Corollary 2.7. If \( A \) is Noetherian and \( M \) is a module \( \neq 0 \), then there exists a prime associated with \( M \) .
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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.
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Yes
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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 prime \( {\mathfrak{p}}_{i} \) .
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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 \( N \neq M \), then by the preceding argument applied to \( M/N \), there exists a submodule \( {N}^{\prime } \) of \( M \) containing \( N \) such that \( {N}^{\prime }/N \) is isomorphic to \( A/\mathfrak{p} \) for some \( \mathfrak{p} \), and this contradicts the maximality of \( N \) .
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Yes
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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 \( M \) .
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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 any associated prime because \( a \) does not annihilate any non-zero element of \( M \) .
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Yes
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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 \( {M}_{\mathfrak{p}} \neq 0 \) .
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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 latter is proved as follows. Let \( \mathfrak{p} \) be a prime such that \( {M}_{\mathfrak{p}} \neq 0 \) . Then there exists \( x \in M \) such that \( {\left( Ax\right) }_{\mathrm{p}} \neq 0 \) . By Corollary 2.7, there exists an associated prime \( \mathfrak{q} \) of \( {\left( Ax\right) }_{\mathfrak{p}} \) in \( A \) . Hence there exists an element \( y/s \) of \( {\left( Ax\right) }_{\mathfrak{p}} \), with \( y \in {Ax} \) , \( s \notin \mathfrak{p} \), and \( y/s \neq 0 \), such that \( \mathfrak{q} \) is the annihilator of \( y/s \) . It follows that \( \mathfrak{q} \subset \mathfrak{p} \) , for otherwise, there exists \( b \in \mathfrak{q}, b \notin \mathfrak{p} \), and \( 0 = {by}/s \), whence \( y/s = 0 \), contradiction. Let \( {b}_{1},\ldots ,{b}_{n} \) be generators for \( \mathfrak{q} \) . For each \( i \), there exists \( {s}_{i} \in A \) , \( {s}_{i} \notin \mathfrak{p} \), such that \( {s}_{i}{b}_{i}y = 0 \) because \( {b}_{i}y/s = 0 \) . Let \( t = {s}_{1}\cdots {s}_{n} \) . Then it is trivially verified that \( \mathfrak{q} \) is the annihilator of \( {ty} \) in \( A \) . Hence \( \mathfrak{q} \subset \mathfrak{p} \), as desired.
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Yes
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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 conditions are satisfied, then the set of elements \( a \in A \) such that \( {a}_{M} \) is locally nilpotent is equal to the associated prime of \( M \) .
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Proof. Immediate consequence of Propositions 2.9 and 2.10.
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No
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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 \( M/N \) .
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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 \( M/N \) . Suppose \( {Ax} \cap N \neq 0 \) . Let \( y = {ax} \in N \) with \( a \in A \) and \( y \neq 0 \) . Then \( \mathfrak{p} \) annihilates \( y \) . We claim \( \mathfrak{p} = \operatorname{ann}\left( y\right) \) . Let \( b \in A \) and \( {by} = 0 \) . Then \( {ba} \in \mathfrak{p} \) but \( a \notin \mathfrak{p} \), so \( b \in \mathfrak{p} \) . Hence \( \mathfrak{p} \) is the annihilator of \( y \) in \( A \), and therefore is associated with \( N \), as was to be shown.
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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.
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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 \( {a}_{M/Q} \) is nilpotent. Conversely, suppose \( a \notin \mathfrak{p} \) . Let \( x \in M, x \notin {Q}_{j} \) for some \( j \) . Then \( {a}^{n}x \notin {Q}_{j} \) for all positive integers \( n \), and consequently \( {a}_{M/Q} \) is injective. This proves our proposition.
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Yes
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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 \( {Q}_{1}^{\prime },\ldots ,{Q}_{r}^{\prime } \) is the same. If \( \left\{ {{\mathfrak{p}}_{1},\ldots ,{\mathfrak{p}}_{m}}\right\} \) is the set of isolated primes belonging to these decompositions, then \( {Q}_{i} = {Q}_{i}^{\prime } \) for \( i = 1,\ldots, m \), in other words, the primary modules corresponding to isolated primes are uniquely determined.
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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 \( {\mathfrak{p}}_{1} \) . By definition, for each \( j = 2,\ldots, r \) there exists \( {a}_{j} \in {\mathfrak{p}}_{j} \) and \( {a}_{j} \notin {\mathfrak{p}}_{1} \) . Let \( a = {a}_{2}\cdots {a}_{r} \) be the product. Then \( a \in {\mathfrak{p}}_{j} \) for all \( j > 1 \) , but \( a \notin {\mathfrak{p}}_{1} \) . We can find an integer \( n \geqq 1 \) such that \( {a}_{M/{Q}_{j}}^{n} = 0 \) for \( j = 2,\ldots, r \) . Let\n\n\[ {N}_{1} = \text{set of}x \in M\text{such that}{a}^{n}x \in N\text{.} \]\n\nWe contend that \( {Q}_{1} = {N}_{1} \) . This will prove the desired uniqueness. Let \( x \in {Q}_{1} \) . Then \( {a}^{n}x \in {Q}_{1} \cap \cdots \cap {Q}_{r} = N \), so \( x \in {N}_{1} \) . Conversely, let \( x \in {N}_{1} \), so that \( {a}^{n}x \in N \), and in particular \( {a}^{n}x \in {Q}_{1} \) . Since \( a \notin {\mathfrak{p}}_{1} \), we know by definition that \( {a}_{M/{Q}_{1}} \) is injective. Hence \( x \in {Q}_{1} \), thereby proving our theorem.
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Yes
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Theorem 3.3. Let \( M \) be a Noetherian module. Let \( N \) be a submodule of M. Then \( N \) admits a primary decomposition.
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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 injective nor nilpotent. The increasing sequence of modules\n\n\[ \operatorname{Ker}{a}_{M/N} \subset \operatorname{Ker}{a}_{M/N}^{2} \subset \operatorname{Ker}{a}_{M/N}^{3} \subset \cdots \]\n\nstops, say at \( {a}_{M/N}^{r} \) . Let \( \varphi : M/N \rightarrow M/N \) be the endomorphism \( \varphi = {a}_{M/N}^{r} \) . Then \( \operatorname{Ker}{\varphi }^{2} = \operatorname{Ker}\varphi \) . Hence \( 0 = \operatorname{Ker}\varphi \cap \operatorname{Im}\varphi \) in \( M/N \), and neither the kernel nor the image of \( \varphi \) is 0 . Taking the inverse image in \( M \), we see that \( N \) is the intersection of two submodules of \( M \), unequal to \( N \) . We conclude from the maximality of \( N \) that each one of these submodules admits a primary decomposition, and therefore that \( N \) admits one also, contradiction.
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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.
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Proof. Immediate consequence of the definitions, and Corollary 2.11.
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No
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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.
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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 \( {Q}_{i} \) . Conversely, let \( N = {Q}_{2} \cap \cdots \cap {Q}_{r} \) . Then \( N \neq 0 \) because our decomposition is reduced. We have\n\n\[ N = N/\left( {N \cap {Q}_{1}}\right) \approx \left( {N + {Q}_{1}}\right) /{Q}_{1} \subset M/{Q}_{1}. \]\n\nHence \( N \) is isomorphic to a submodule of \( M/{Q}_{1} \), and consequently has an associated prime which can be none other than the prime \( {\mathfrak{p}}_{1} \) belonging to \( {Q}_{1} \) . This proves our theorem.
|
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.
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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 \( E = \{ 0\} \) .
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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 \( {a}_{s} \) ) in a such that \( \left( {1 + a}\right) {x}_{s} \) lies in the module generated by the first \( s - 1 \) generators. Furthermore \( 1 + a \) is a unit in \( A \) , otherwise \( 1 + a \) is contained in some maximal ideal, and since \( a \) lies in all maximal ideals, we conclude that 1 lies in a maximal ideal, which is not possible. Hence \( {x}_{s} \) itself lies in the module generated by \( s - 1 \) generators, and the proof is complete by induction.
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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 \( E = F \) .
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Proof. Apply Lemma 4.1 to \( E/F \) .
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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 \( E \) .
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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 \( {x}_{1},\ldots ,{x}_{n} \) are a basis of \( E \) over \( A \) . If \( {x}_{1},\ldots ,{x}_{r} \) are such that \( {\bar{x}}_{1},\ldots ,{\bar{x}}_{r} \) are linearly independent over \( A/\mathfrak{m} \), then they can be completed to a basis of \( E \) over \( A \) .
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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, \( f \) is surjective. Since \( E \) is projective, it follows that \( f \) splits, i.e. we can write \( F = {P}_{0} \oplus {P}_{1} \), where \( {P}_{0} = \operatorname{Ker}f \) and \( {P}_{1} \) is mapped isomorphically onto \( E \) by \( f \) . Now the linear independence of \( {x}_{1},\ldots ,{x}_{n} \) mod \( \mathrm{m}E \) shows that\n\n\[ \n{P}_{0} \subset \mathfrak{m}F = \mathfrak{m}{P}_{0} \oplus \mathfrak{m}{P}_{1} \n\]\n\nHence \( {P}_{0} \subset m{P}_{0} \) . Also, as a direct summand in a finitely generated module, \( {P}_{0} \) is finitely generated. So by Lemma 4.3, \( {P}_{0} = \left( 0\right) \) and \( f \) is an isomorphism, as was to be proved.\n\nAs to the second statement, it is immediate since we can complete a given sequence \( {x}_{1},\ldots ,{x}_{r} \) with \( {\bar{x}}_{1},\ldots ,{\bar{x}}_{r} \) linearly independent over \( A/\mathfrak{m} \), to a sequence \( {x}_{1},\ldots ,{x}_{n} \) with \( {\bar{x}}_{1},\ldots ,{x}_{n} \) linearly independent over \( A/\mathfrak{m} \), and then we can apply the first part of the proof. This concludes the proof of the theorem.
|
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) Assume that \( E, F \) are free. If \( {f}_{\left( m\right) } \) is injective (resp. an isomorphism) then \( f \) is injective (resp. an isomorphism).
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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 in Theorem 4.4.
|
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 \( n \geqq 0 \) .
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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\[ \n{E}_{n + d} = {\mathfrak{a}}^{n}{E}_{d} \subset {\mathfrak{a}}^{n}E \n\] \n\nwhich proves the proposition.
|
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.
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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 is an ideal of \( A \), whose residue class ring is \( {A}_{0} \), which is thus a homomorphic image of \( A \), and is therefore Noetherian. Furthermore, \( {A}^{ + } \) has a finite number of generators \( {x}_{1},\ldots ,{x}_{s} \) by hypothesis. Expressing each generator as a sum of homogeneous elements, we may assume without loss of generality that these generators are homogeneous, say of degrees \( {d}_{1},\ldots ,{d}_{s} \) respectively, with all \( {d}_{i} > 0 \) . Let \( B \) be the subring of \( A \) generated over \( {A}_{0} \) by \( {x}_{1},\ldots ,{x}_{s} \) . We claim that \( {A}_{n} \subset B \) for all \( n \) . This is certainly true for \( n = 0 \) . Let \( n > 0 \) . Let \( x \) be homogeneous of degree \( n \) . Then there exist elements \( {a}_{i} \in {A}_{n - {d}_{i}} \) such that\n\n\[ \nx = \mathop{\sum }\limits_{{i = 1}}^{s}{a}_{i}{x}_{i} \n\]\n\nSince \( {d}_{i} > 0 \) by induction, each \( {a}_{i} \) is in \( {A}_{0}\left\lbrack {{x}_{1},\ldots ,{x}_{s}}\right\rbrack = B \), so this shows \( x \in B \) also, and concludes the proof.
|
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.
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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 finite over \( S \) since \( {F}_{n} \) is finite over \( A \) . We have\n\n\[ \n{G}_{n} \subset {G}_{n + 1}\;\text{ and }\;\bigcup {G}_{n} = {E}_{S} \n\]\n\nSince \( S \) is Noetherian, we get:\n\n\[ \n{E}_{S}\text{is finite over}S \Leftrightarrow {E}_{S} = {G}_{N}\text{for some}N \n\]\n\n\[ \n\Leftrightarrow {E}_{N + m} = {\mathfrak{a}}^{m}{E}_{N}\text{for all}m \geqq 0 \n\]\n\n\[ \n\Leftrightarrow \text{the filtration of}E\text{is a-stable.} \n\]\n\nThis proves the lemma.
|
Yes
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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 \( F \) .
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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 \( {E}_{S} \), and is finite over \( S \) since \( S \) is Noetherian. We apply Lemma 5.3 to conclude the proof.
|
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\[{\mathfrak{a}}^{n}E \cap F = {\mathfrak{a}}^{n - s}\left( {{\mathfrak{a}}^{s}E \cap F}\right) .\]
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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\[ \mathop{\bigcap }\limits_{{n = 1}}^{\infty }{\mathfrak{a}}^{n}E = 0 \]
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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 \( {\operatorname{gr}}_{a}\left( A\right) \) -module.
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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 \[ {\operatorname{gr}}_{\mathfrak{a}}\left( A\right) = \left( {A/\mathfrak{a}}\right) \left\lbrack {{\bar{x}}_{1},\ldots ,{\bar{x}}_{s}}\right\rbrack \] is Noetherian, thus proving the first assertion. For the second assertion, we have for some \( d \) , \[ {E}_{d + m} = {\mathfrak{a}}^{m}{E}_{d}\;\text{ for all }m \geqq 0. \] Hence \( \operatorname{gr}\left( E\right) \) is generated by the finite direct sum \[ \operatorname{gr}{\left( E\right) }_{0} \oplus \cdots \oplus \operatorname{gr}{\left( E\right) }_{d} \] But each \( \operatorname{gr}{\left( E\right) }_{n} = {E}_{n}/{E}_{n + 1} \) is finitely generated over \( A \), and annihilated by \( \mathfrak{a} \) , so is a finite \( A/\mathfrak{a} \) -module. Hence the above finite direct sum is a finite \( A/\mathfrak{a} \) - module, so \( \operatorname{gr}\left( E\right) \) is a finite \( {\operatorname{gr}}_{a}\left( A\right) \) -module, thus concluding the proof of the proposition.
|
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 positive integers \( {d}_{i} \), and \( f\left( t\right) \in \mathbf{Z}\left\lbrack t\right\rbrack \) .
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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\[ 0 \rightarrow {K}_{n} \rightarrow {E}_{n}\xrightarrow[]{{x}_{s}}{E}_{n + {d}_{s}} \rightarrow {L}_{n + {d}_{s}} \rightarrow 0. \]\n\nLet\n\n\[ K = \bigoplus {K}_{n}\text{ and }L = \bigoplus {L}_{n}. \]\n\nThen \( K, L \) are finite \( A \) -modules (being submodules and factor modules of \( E \) ), and are annihilated by \( {x}_{s} \), so are in fact graded \( {A}_{0}\left\lbrack {{x}_{1},\ldots ,{x}_{s - 1}}\right\rbrack \) -modules. By definition of an Euler-Poincaré function, we get\n\n\[ \varphi \left( {K}_{n}\right) - \varphi \left( {E}_{n}\right) + \varphi \left( {E}_{n + {d}_{s}}\right) - \varphi \left( {L}_{n + {d}_{s}}\right) = 0. \]\n\nMultiplying by \( {t}^{n + {d}_{s}} \) and summing over \( n \), we get\n\n\[ \left( {1 - {t}^{{d}_{s}}}\right) P\left( {E, t}\right) = P\left( {L, t}\right) - {t}^{{d}_{s}}P\left( {K, t}\right) + g\left( t\right) , \]\n\nwhere \( g\left( t\right) \) is a polynomial in \( \mathbf{Z}\left\lbrack t\right\rbrack \) . The theorem follows by induction.
|
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 \( d - 1 \) . (For this statement, the zero polynomial is assumed to have degree -1 .)
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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 \( P\left( {E, t}\right) = h\left( t\right) /{\left( 1 - t\right) }^{d} \), and \( h\left( 1\right) \neq 0 \), with \( h\left( t\right) \in \mathbf{Z}\left\lbrack t\right\rbrack \) . Let\n\n\[ h\left( t\right) = \mathop{\sum }\limits_{{k = 0}}^{m}{a}_{k}{t}^{k} \]\n\nWe have the binomial expansion\n\n\[ {\left( 1 - t\right) }^{-d} = \mathop{\sum }\limits_{{k = 0}}^{\infty }\left( \begin{matrix} d + k - 1 \\ d - 1 \end{matrix}\right) {t}^{k} \]\n\nFor convenience we let \( \left( \begin{array}{r} n \\ - 1 \end{array}\right) = 0 \) for \( n \geqq 0 \) and \( \left( \begin{array}{r} n \\ - 1 \end{array}\right) = 1 \) for \( n = - 1 \) . We then get\n\n\[ \varphi \left( {E}_{n}\right) = \mathop{\sum }\limits_{{k = 0}}^{m}{a}_{k}\left( \begin{matrix} d + n - k - 1 \\ d - 1 \end{matrix}\right) \;\text{ for all }n \geqq m. \]\n\nThe sum on the right-hand side is a polynomial in \( n \) with leading term\n\n\[ \left( {\sum {a}_{k}}\right) \frac{{n}^{d - 1}}{\left( {d - 1}\right) !} \neq 0 \]\n\nThis proves the theorem.
|
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 \( n \geqq 0 \) .\n\n(ii) For all sufficiently large \( n \), this length is a polynomial \( g\left( n\right) \) of degree \( \leqq s \) , where \( s \) is the least number of generators of \( \mathfrak{q} \) .\n\n(iii) The degree and leading coefficient of \( g\left( n\right) \) depend only on \( E \) and \( \mathfrak{q} \), but not on the chosen filtration.
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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 \( \operatorname{gr}\left( E\right) \) is a finite \( G \) -module. By the remarks preceding the theorem, \( E/{E}_{n} \) has finite length, and if \( \varphi \) denotes the length, then\n\n\[ \varphi \left( {E/{E}_{n}}\right) = \mathop{\sum }\limits_{{j = 1}}^{n}\varphi \left( {{E}_{j - 1}/{E}_{j}}\right) \]\n\nIf \( {x}_{1},\ldots ,{x}_{s} \) generate \( \mathfrak{q} \), then the images \( {\bar{x}}_{1},\ldots ,{\bar{x}}_{s} \) in \( \mathfrak{q}/{\mathfrak{q}}^{2} \) generate \( G \) as \( A/\mathfrak{q} \) - algebra, and each \( {\bar{x}}_{i} \) has degree 1 . By Theorem 6.2 we see that\n\n\[ \varphi \left( {{E}_{n}/{E}_{n + 1}}\right) = h\left( n\right) \]\n\nis a polynomial in \( n \) of degree \( \leqq s - 1 \) for sufficiently large \( n \) . Since\n\n\[ \varphi \left( {E/{E}_{n + 1}}\right) - \varphi \left( {E/{E}_{n}}\right) = h\left( n\right) ,\]\n\nit follows by Lemma 6.4 below that \( \varphi \left( {E/{E}_{n}}\right) \) is a polynomial \( g\left( n\right) \) of degree \( \leqq s \) for all large \( n \) . The last statement concerning the independence of the degree of \( g \) and its leading coefficient from the chosen filtration follows immediately from Proposition 5.1, and will be left to the reader. This concludes the proof.
|
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\[ P\left( T\right) = {c}_{0}\left( \begin{array}{l} T \\ d \end{array}\right) + {c}_{1}\left( \begin{matrix} T \\ d - 1 \end{matrix}\right) + \ldots + {c}_{d}. \]
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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 \( P \) . Using (2) and induction, we conclude that \( {c}_{0},\ldots ,{c}_{d - 1} \) must be integers. Finally \( {c}_{d} \) is an integer because \( P\left( n\right) \in \mathbf{Z} \) for \( n \) sufficiently large. This proves (a).
|
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\[ \chi \left( {n,\mathfrak{a} + \mathfrak{b}}\right) = \chi \left( {n,\mathfrak{a}}\right) + \chi \left( {n,\mathfrak{b}}\right) - \chi \left( {n,\mathfrak{a} \cap \mathfrak{b}}\right) . \]
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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\[ \chi \left( {n,\mathfrak{a} + \left( F\right) }\right) = \chi \left( {n,\mathfrak{a}}\right) - \chi \left( {n - d,\mathfrak{a}}\right) . \]
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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\[ \varphi \left( {n,\mathfrak{a} \cap \left( F\right) }\right) = \varphi \left( {n - d,\mathfrak{a}}\right) . \]\n\nFinally, by Proposition 6.5 (the formula for \( \chi \) ), we obtain:\n\n\[ \chi \left( {n,\mathfrak{a} + \left( F\right) }\right) = \chi \left( {n,\mathfrak{a}}\right) + \chi \left( {n,\left( F\right) }\right) - \chi \left( {n,\mathfrak{a} \cap \left( F\right) }\right) \]\n\n\[ = \chi \left( {n,\mathfrak{a}}\right) + \varphi \left( n\right) - \varphi \left( {n,\left( F\right) }\right) - \varphi \left( n\right) + \varphi \left( {n,\mathfrak{a} \cap \left( F\right) }\right) \]\n\n\[ = \chi \left( {n,\mathfrak{a}}\right) - \varphi \left( {n - d}\right) + \varphi \left( {n - d,\mathfrak{a}}\right) \]\n\n\[ = \chi \left( {n,\mathfrak{a}}\right) - \chi \left( {n - d,\mathfrak{a}}\right) \]\n\nthus proving the theorem.
|
Yes
|
Lemma 6.8. Let \( V, W \) be varieties over a field \( k \) . \[ \\text{If}V \\supset W\\text{and}\\dim V = \\dim W\\text{, then}V = W\\text{.} \]
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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 \[ k\\left\\lbrack X\\right\\rbrack /{\\mathfrak{p}}_{V} \\approx k\\left\\lbrack x\\right\\rbrack \\rightarrow k\\left\\lbrack y\\right\\rbrack \\approx k\\left\\lbrack X\\right\\rbrack /{\\mathfrak{p}}_{W} \] from the affine coordinate ring of \( V \) onto the affine coordinate ring of \( W \) . If the transcendence degree of \( k\\left( x\\right) \) is the same as that of \( k\\left( y\\right) \), and say \( {y}_{1},\\ldots ,{y}_{r} \) form a transcendence basis of \( k\\left( y\\right) \) over \( k \), then \( {x}_{1},\\ldots ,{x}_{r} \) is a transcendence basis of \( k\\left( x\\right) \) over \( k \), the homomorphism \( k\\left\\lbrack x\\right\\rbrack \\rightarrow k\\left\\lbrack y\\right\\rbrack \) induces an isomorphism \[ k\\left\\lbrack {{x}_{1},\\ldots ,{x}_{r}}\\right\\rbrack \\overset{ \\approx }{ \\rightarrow }k\\left\\lbrack {{y}_{1},\\ldots ,{y}_{r}}\\right\\rbrack \] and hence an isomorphism on the finite extension \( k\\left\\lbrack x\\right\\rbrack \) to \( k\\left\\lbrack y\\right\\rbrack \), as desired.
|
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 \( \\leqq r \) , such that \( P\\left( \\mathbf{Z}\\right) \\subset \\mathbf{Z} \), and such that\n\n\[ P\\left( n\\right) = \\chi \\left( {n,\\mathfrak{a}}\\right) \]\n\nfor all \( n \) sufficiently large.
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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 polynomial \( L \\in k\\left\\lbrack X\\right\\rbrack \) of degree 1 (a linear form) which does not lie in any of the prime ideals belonging to the primary ideals in the given decomposition. In particular, \( L \) is not a divisor of zero mod a. Then the components of the algebraic space of zeros of \( \\mathfrak{a} + \\left( L\\right) \) must have dimension \( \\leqq r - 1 \) . By induction and Theorem 6.6, we conclude that the difference\n\n\[ \\chi \\left( {n,\\mathfrak{a}}\\right) - \\chi \\left( {n - 1,\\mathfrak{a}}\\right) \]\n\nsatisfies the conditions of Lemma 6.4(b), which concludes the proof.
|
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.
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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.
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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 \( \operatorname{Im}{u}^{\infty } \) is an automorphism, and the restriction of \( u \) to \( \operatorname{Ker}{u}^{\infty } \) is nilpotent.
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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 \( x = {u}^{n}\left( y\right) \) for some \( y \in E \), and then \( 0 = {u}^{n}\left( x\right) = {u}^{2n}\left( y\right) \) . So \( y \in \operatorname{Ker}{u}^{2n} = \operatorname{Ker}{u}^{n} \), whence \( x = {u}^{n}\left( y\right) = 0 \) .\n\nSecondly, let \( x \in E \) . Then for some \( y \in {u}^{n}\left( E\right) \) we have\n\n\[ {u}^{n}\left( x\right) = {u}^{n}\left( y\right) \]\n\nThen we can write\n\n\[ x = x - {u}^{n}\left( y\right) + {u}^{n}\left( y\right) \]\n\nwhich shows that \( E = \operatorname{Im}{u}^{\infty } + \operatorname{Ker}{u}^{\infty } \) . Combined with the first step of the proof, this shows that \( E \) is a direct sum as stated.\n\nThe final assertion is immediate, since the restriction of \( u \) to \( \operatorname{Im}{u}^{\infty } \) is surjective, and its kernel is 0 by the first part of the proof. The restriction of \( u \) to Ker \( {u}^{\infty } \) is nilpotent because Ker \( {u}^{\infty } = \operatorname{Ker}{u}^{n} \) . This concludes the proof of the proposition.
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Yes
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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.
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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 any endomorphism \( v \) it follows that \( {uv} \) and \( {vu} \) are not surjective or injective respectively, so are not automorphisms. Let \( {u}_{1},{u}_{2} \) be endomorphisms which are not units. We have to show \( {u}_{1} + {u}_{2} \) is not a unit. If it is a unit in \( \operatorname{End}\left( E\right) \), let \( {v}_{i} = {u}_{i}{\left( {u}_{1} + {u}_{2}\right) }^{-1} \). Then \( {v}_{1} + {v}_{2} = 1 \). Furthermore, \( {v}_{1} = 1 - {v}_{2} \) is invertible by the geometric series since \( {v}_{2} \) is nilpotent. But \( {v}_{1} \) is not a unit by the first part of the proof, contradiction. This concludes the proof.
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Yes
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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.
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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\[ 0 \neq {\mathrm{{id}}}_{M} = {\mathrm{{id}}}_{M}^{2} = {\left( vu\right) }^{-1}{vu}{\left( vu\right) }^{-1}{vu}. \]\n\nSo \( e = {\mathrm{{id}}}_{N} \) . Then \( u \) is injective because \( {vu} \) is an automorphism; \( v \) is injective because \( e = {\mathrm{{id}}}_{N} \) is injective; \( u \) is surjective because \( e = {\mathrm{{id}}}_{N} \) ; and \( v \) is surjective because \( {vu} \) is an automorphism. This concludes the proof of the lemma.
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Yes
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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.
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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.
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No
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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 \( K \) .\n\n(ii) If \( f \) is an irreducible polynomial of odd degree \( n \) in \( K\left\lbrack X\right\rbrack \) and if \( \alpha \) is a root of \( f \), then \( K\left( \alpha \right) \) is real.
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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\[ - 1 = \sum {\left( {b}_{i} + {c}_{i}\sqrt{a}\right) }^{2} \]\n\n\[ = \sum \left( {{b}_{i}^{2} + 2{c}_{i}{b}_{i}\sqrt{a} + {c}_{i}^{2}a}\right) \text{.} \]\n\nSince \( \sqrt{a} \) is of degree 2 over \( K \), it follows that\n\n\[ - 1 = \sum {b}_{i}^{2} + a\sum {c}_{i}^{2} \]\n\nIf \( a \) is a sum of squares in \( K \), this yields a contradiction. In any case, we conclude that\n\n\[ - a = \frac{1+\sum {b}_{i}^{2}}{\sum {c}_{i}^{2}} \]\n\nis a quotient of sums of squares, and by a previous remark, that \( - a \) is a sum of squares. Hence \( K\left( \sqrt{a}\right) \) is real, thereby proving our first assertion.\n\nAs to the second, suppose \( K\left( \alpha \right) \) is not real. Then we can write\n\n\[ - 1 = \sum {g}_{i}{\left( \alpha \right) }^{2} \]\n\nwith polynomials \( {g}_{i} \) in \( K\left\lbrack X\right\rbrack \) of degree \( \leqq n - 1 \) . There exists a polynomial \( h \) in \( K\left\lbrack X\right\rbrack \) such that\n\n\[ - 1 = \sum {g}_{i}{\left( X\right) }^{2} + h\left( X\right) f\left( X\right) . \]\n\nThe sum of \( {g}_{i}{\left( X\right) }^{2} \) has even degree, and this degree must be \( > 0 \), otherwise -1 is a sum of squares in \( K \) . This degree is \( \leqq {2n} - 2 \) . Since \( f \) has odd degree \( n \), it follows that \( h \) has odd degree \( \leqq n - 2 \) . If \( \beta \) is a root of \( h \) then we see that -1 is a sum of squares in \( K\left( \beta \right) \) . Since \( \deg h < \deg f \), our proof is finished by induction.
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Yes
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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 root in \( R \) . We have \( {R}^{\mathrm{a}} = R\left( \sqrt{-1}\right) \) .
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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 of \( P \) is a square in \( R \), and given \( a \in R, a \neq 0 \), we must have \( a \in P \) or \( - a \in P \) . Thus \( P \) defines an ordering. Again by Proposition 2.1, every polynomial of odd degree over \( R \) has a root in \( R \) . Our assertion follows by Example 5 of Chapter VI, §2.
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Yes
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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.
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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.
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Yes
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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.
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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 \( R\left( \alpha \right) = R\left( \sqrt{-1}\right) \), and hence there exist \( c, d \in R \) such that \( \alpha = c + d\sqrt{-1} \) . Then\n\n\[ \n{\alpha }^{2} = {c}^{2} + {2cd}\sqrt{-1} - {d}^{2} \n\]\n\nSince \( 1,\sqrt{-1} \) are linearly independent over \( R \), it follows that \( c = 0 \) (because \( a \notin {R}^{2} \) ), and hence \( - a \) is a square.\n\nWe shall now prove that a sum of squares is a square. For simplicity, write \( i = \sqrt{-1} \) . Since \( R\left( i\right) \) is algebraically closed, given \( a, b \in R \) we can find \( c, d \in R \) such that \( {\left( c + di\right) }^{2} = a + {bi} \) . Then \( a = {c}^{2} - {d}^{2} \) and \( b = {2cd} \) . Hence\n\n\[ \n{a}^{2} + {b}^{2} = {\left( {c}^{2} + {d}^{2}\right) }^{2} \n\]\n\nas was to be shown.\n\nIf \( a \in R, a \neq 0 \), then not both \( a \) and \( - a \) can be squares in \( R \) . Hence \( P \) is an ordering and our proposition is proved.
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Yes
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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 \( f\left( c\right) = 0 \) .
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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 \[ {\left( X + \frac{\alpha }{2}\right) }^{2} + \left( {\beta - \frac{{\alpha }^{2}}{4}}\right) \] and we must have \( {4\beta } > {\alpha }^{2} \) since our factor is assumed irreducible. Hence the change of sign of \( f \) must be due to the change of sign of a linear factor, which is trivially verified to be a root lying between \( a \) and \( b \) .
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Yes
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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 \( \left| \alpha \right| \leqq 1 + \left| {a}_{n - 1}\right| + \cdots + \left| {a}_{0}\right| \) .
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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.
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No
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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 \( S \) .
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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, by Theorem 2.5. Hence it will suffice to prove that if there is precisely one element \( \alpha \) such that \( u < \alpha < v \) and \( \alpha \) is a root of some \( {f}_{j} \), then \( {W}_{S}\left( u\right) - {W}_{S}\left( v\right) = 1 \) if \( \alpha \) is a root of \( f \), and 0 otherwise. Suppose that \( \alpha \) is a root of some \( {f}_{j} \), for \( 1 \leqq j \leqq m - 1 \) . Then \( {f}_{j - 1}\left( \alpha \right) ,{f}_{j + 1}\left( \alpha \right) \) have opposite signs by ST 3, and these signs do not change when we replace \( \alpha \) by \( u \) or \( v \) . Hence the variation of signs in\n\n\[ \left\{ {{f}_{j - 1}\left( u\right) ,{f}_{j}\left( u\right) ,{f}_{j + 1}\left( u\right) }\right\} \text{ and }\left\{ {{f}_{j - 1}\left( v\right) ,{f}_{j}\left( v\right) ,{f}_{j + 1}\left( v\right) }\right\} \]\n\nis the same, namely equal to 2 . If \( \alpha \) is not a root of \( f \), we conclude that\n\n\[ {W}_{S}\left( u\right) = {W}_{S}\left( v\right) \]\n\nIf \( \alpha \) is a root of \( f \), then \( f\left( u\right) \) and \( f\left( v\right) \) have opposite signs, but \( {f}^{\prime }\left( u\right) \) and \( {f}^{\prime }\left( v\right) \) have the same sign, namely, the sign of \( {f}^{\prime }\left( \alpha \right) \) . Hence in this case,\n\n\[ {W}_{S}\left( u\right) = {W}_{S}\left( v\right) + 1 \]\n\nThis proves our theorem.
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Yes
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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.
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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 of roots of \( f \) in any real closure of \( K \) inducing the given ordering.
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Yes
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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.
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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 \( f\left( \alpha \right) = 0 \) and the corollary of Sturm’s Theorem (Corollary 2.8) shows that \( f \) has a root \( \beta \) in \( {R}^{\prime } \) . Thus there exists an isomorphism of \( K\left( \alpha \right) \) on \( K\left( \beta \right) \) over \( K \) , mapping \( \alpha \) on \( \beta \) .\n\nLet \( {\alpha }_{1},\ldots ,{\alpha }_{n} \) be the distinct roots of \( f \) in \( R \), and let \( {\beta }_{1},\ldots ,{\beta }_{m} \) be the distinct roots of \( f \) in \( {R}^{\prime } \) . Say\n\n\[ {\alpha }_{1} < \cdots < {\alpha }_{n}\;\text{in the ordering of}R\text{,}\n\]\n\n\[ {\beta }_{1} < \cdots < {\beta }_{m}\;\text{in the ordering of}{R}^{\prime }\text{.}\n\]\n\nWe contend that \( m = n \) and that we can select an embedding \( \sigma \) of \( K\left( {{\alpha }_{1},\ldots ,{\alpha }_{n}}\right) \) into \( {R}^{\prime } \) such that \( \sigma {\alpha }_{i} = {\beta }_{i} \) for \( i = 1,\ldots, n \) . Indeed, let \( {\gamma }_{i} \) be an element of \( R \) such that\n\n\[ {\gamma }_{i}^{2} = {\alpha }_{i + 1} - {\alpha }_{i}\;\text{ for }\;i = 1,\ldots, n - 1\n\]\n\nand let \( {E}_{1} = K\left( {{\alpha }_{1},\ldots ,{\alpha }_{n},{\gamma }_{1},\ldots ,{\gamma }_{n - 1}}\right) \) . By what we have seen, there exists an embedding \( \sigma \) of \( {E}_{1} \) into \( {R}^{\prime } \), and then \( \sigma {\alpha }_{i + 1} - \sigma {\alpha }_{i} \) is a square in \( {R}^{\prime } \) . Hence\n\n\[ \sigma {\alpha }_{1} < \cdots < \sigma {\alpha }_{n}\n\]\n\nThis proves that \( m \geqq n \) . By symmetry, it follows that \( m = n \) . Furthermore, the condition that \( \sigma {\alpha }_{i} = {\beta }_{i} \) for \( i = 1,\ldots, n \) determines the effect of \( \sigma \) on\n\n\( K\left( {{\alpha }_{1},\ldots ,{\alpha }_{n}}\right) \) . We contend that \( \sigma \) is order-preserving. Let \( y \in K\left( {{\alpha }_{1},\ldots ,{\alpha }_{n}}\right) \) and \( 0 < y \) . Let \( \gamma \in R \) be such that \( {\gamma }^{2} = y \) . There exists an embedding of\n\n\[ K\left( {{\alpha }_{1},\ldots ,{\alpha }_{n},{\gamma }_{1},\ldots ,{\gamma }_{n - 1},\gamma }\right)\n\]\n\ninto \( {R}^{\prime } \) over \( K \) which must induce \( \sigma \) on \( K\left( {{\alpha }_{1},\ldots ,{\alpha }_{n}}\right) \) and is such that \( {\sigma y} \) is a square, hence \( > 0 \), as contended.\n\nUsing Zorn’s lemma, it is now clear that we get an isomorphism of \( R \) onto \( {R}^{\prime } \) over \( K \) . This isomorphism is order-preserving because it maps squares on squares, thereby proving our theorem.
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Yes
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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 \( {K}^{\prime } \) by adjoining the square roots of all positive elements of \( K \) . Then \( L \) is real.
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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 \( {\alpha }_{i} \) in a subfield of \( L \), of type\n\n\[ {K}^{\prime }\left( {\sqrt{{b}_{1}},\ldots ,\sqrt{{b}_{r}}}\right) \]\n\nwith \( {b}_{j} \in K,{b}_{j} > 0 \) . Write\n\n\[ {\alpha }_{i} = {x}_{i} + {y}_{i}\sqrt{{b}_{r}} \]\n\nwith \( {x}_{i},{y}_{i} \in {K}^{\prime }\left( {\sqrt{{b}_{1}},\ldots ,\sqrt{{b}_{r - 1}}}\right) \) . Then\n\n\[ - 1 = \sum {a}_{i}{\left( {x}_{i} + {y}_{i}\sqrt{{b}_{r}}\right) }^{2} \]\n\n\[ = \sum {a}_{i}\left( {{x}_{i}^{2} + 2{x}_{i}{y}_{i}\sqrt{{b}_{r}} + {y}_{i}^{2}{b}_{r}}\right) . \]\n\nBy hypothesis, \( \sqrt{{b}_{r}} \) is not in \( {K}^{\prime }\left( {{b}_{1},\ldots ,\sqrt{{b}_{r - 1}}}\right) \) . Hence\n\n\[ - 1 = \sum {a}_{i}{x}_{i}^{2} + \sum {a}_{i}{b}_{r}{y}_{i}^{2} \]\n\ncontradicting the minimality of \( r \) .
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Yes
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Theorem 2.11. Let \( K \) be an ordered field. There exists a real closure \( R \) of \( K \) inducing the given ordering on \( K \) .
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Proof. Take \( {K}^{\prime } = K \) in Proposition 2.10. Then \( L \) is real, and is contained in a real closure. Our assertion is clear.
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No
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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\[ - 1 = \mathop{\sum }\limits_{{i = 1}}^{n}{a}_{i}{\alpha }_{i}^{2} \]\n\nwith \( {a}_{i} \in K,{a}_{i} > 0 \), and \( {\alpha }_{i} \in {K}^{\prime } \) .
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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.
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No
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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\[ \n\varphi {y}_{1} < \cdots < \varphi {y}_{m} \n\]
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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\[ \nk\left\lbrack {{x}_{1},\ldots ,{x}_{n},{\gamma }_{1}^{-1},\ldots ,{\gamma }_{m - 1}^{-1},{\gamma }_{1},\ldots ,{\gamma }_{m - 1}}\right\rbrack \text{.} \n\]
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Yes
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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 \( f\left( a\right) \) is defined, we have \( f\left( a\right) \geqq 0 \) . Then \( f\left( X\right) \) is a sum of squares in \( k\left( X\right) \) .
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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 \( f\left( X\right) \) . We can find a homomorphism \( \varphi \) of this ring into \( {R}_{k} \) (inducing the identity on \( k \) ) such that \( \varphi \left( f\right) < 0 \) . But \[ \varphi \left( f\right) = f\left( {\varphi {X}_{1},\ldots ,\varphi {X}_{n}}\right) . \] contradiction. We let \( {a}_{i} = \varphi \left( {X}_{i}\right) \) to conclude the proof.
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Yes
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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.
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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 over \( {R}_{0} \) . By the intermediate value theorem, we may assume that \( f \) is positive definite, i.e. \( f\left( a\right) > 0 \) for all \( a \in {R}_{0} \) . Without loss of generality, we may assume that \( f\left( X\right) = {X}^{2} + {b}^{2} \) for some \( b \in {R}_{0} \) . Any root of this polynomial will bring \( \sqrt{-1} \) with it and therefore the only algebraic extension of \( {R}_{0} \) is \( {R}_{0}\left( \sqrt{-1}\right) \) . This proves that \( {R}_{0} \) is real closed.
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Yes
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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. Then there exist infinitely many special values \( c \) of \( x \) in \( R \) such that \( {h}_{i}\left( c\right) \) is defined and has the same sign as \( {h}_{i}\left( x\right) \), for all \( i \) .
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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 all roots \( \lambda \) of \( {h}_{i} \) in \( R \), and the second product is over positive definite quadratic factors over \( R \) . For any \( \xi \in R, p\left( \xi \right) \) is positive. It suffices therefore to show that the signs of \( \left( {x - \lambda }\right) \) can be preserved for all \( \lambda \) by substituting infinitely many values \( \alpha \) for \( x \) . We order all values of \( \lambda \) and of \( x \) and obtain\n\n\[ \n\cdots < {\lambda }_{1} < x < {\lambda }_{2} < \cdots \n\]\n\nwhere possibly \( {\lambda }_{1} \) or \( {\lambda }_{2} \) is omitted if \( x \) is larger or smaller than any \( \lambda \) . Any value \( \alpha \) of \( x \) in \( R \) selected between \( {\lambda }_{1} \) and \( {\lambda }_{2} \) will then satisfy the requirements of our lemma.
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Yes
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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 \( f\left( {X, Y}\right) \) be the irreducible polynomial in \( k\left\lbrack {X, Y}\right\rbrack \) such that \( f\left( {x, y}\right) = 0 \) . Let \( R \) be a real closed field containing \( k \) , and assume that there exists \( \left( {a, b}\right) \in {R}^{\left( n + 1\right) } \) such that \( f\left( {a, b}\right) = 0 \) but\n\n\[{D}_{n + 1}f\left( {a, b}\right) \neq 0.\]\n\nThen \( K \) is real.
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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 of \( R\left( {{t}_{1},\ldots ,{t}_{n}}\right) \) preserving the ordering. Let \( {u}_{i} = {a}_{i} + {t}_{i} \) for each \( i = 1,\ldots, n \) . Then \( f\left( {u, b + h}\right) \) changes sign for small \( h \) positive and negative in \( R \), and hence \( f\left( {u, Y}\right) \) has a root in \( {R}^{\prime } \), say \( v \) . Since \( f \) is irreducible, the isomorphism of \( k\left( x\right) \) on \( k\left( u\right) \) sending \( {x}_{i} \) on \( {u}_{i} \) extends to an embedding of \( k\left( {x, y}\right) \) into \( {R}^{\prime } \), and hence \( K \) is real, as was to be shown.
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Yes
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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 \( x \in K \) such that\n\n\[ \n{\left| x - {x}_{i}\right| }_{i} < \epsilon \n\]\n\nfor all \( i \) .
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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 \( {\left| \beta \right| }_{1} \geqq 1 \) and \( {\left| \beta \right| }_{s} < 1 \) . Put \( y = \beta /\alpha \) . Then \( {\left| y\right| }_{1} > 1 \) and \( {\left| y\right| }_{s} < 1 \) .\n\nWe shall now prove that there exists \( z \in K \) such that \( {\left| z\right| }_{1} > 1 \) and \( {\left| z\right| }_{j} < 1 \) for \( j = 2,\ldots, s \) . We prove this by induction, the case \( s = 2 \) having just been proved. Suppose we have found \( z \in K \) satisfying\n\n\[ \n{\left| z\right| }_{1} > 1\text{ and }{\left| z\right| }_{j} < 1\;\text{ for }\;j = 2,\ldots, s - 1.\n\]\n\nIf \( {\left| z\right| }_{s} \leqq 1 \) then the element \( {z}^{n}y \) for large \( n \) will satisfy our requirements.\n\nIf \( {\left| z\right| }_{s} > 1 \), then the sequence\n\n\[ \n{t}_{n} = \frac{{z}^{n}}{1 + {z}^{n}}\n\]\n\ntends to 1 at \( {v}_{1} \) and \( {v}_{s} \), and tends to 0 at \( {v}_{j}\left( {j = 2,\ldots, s - 1}\right) \) . For large \( n \), it is then clear that \( {t}_{n}y \) satisfies our requirements.\n\nUsing the element \( z \) that we have just constructed, we see that the sequence \( {z}^{n}/\left( {1 + {z}^{n}}\right) \) tends to 1 at \( {v}_{1} \) and to 0 at \( {v}_{j} \) for \( j = 2,\ldots, s \) . For each \( i = 1,\ldots, s \) we can therefore construct an element \( {z}_{i} \) which is very close to 1 at \( {v}_{i} \) and very close to 0 at \( {v}_{j}\left( {j \neq i}\right) \) . The element\n\n\[ \nx = {z}_{1}{x}_{1} + \cdots + {z}_{s}{x}_{s}\n\]\n\nthen satisfies the requirement of the theorem.
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Yes
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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.
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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\[ \n{\xi }^{\left( v\right) } = {x}_{1}^{\left( v\right) }{\omega }_{1} + \cdots + {x}_{n}^{\left( v\right) }{\omega }_{n},\;{x}_{i}^{\left( v\right) } \in K,\n\]\n\nis a Cauchy sequence in \( E \) only if each one of the \( n \) sequences \( {x}_{i}^{\left( v\right) } \) is a Cauchy sequence in \( K \). We do this by induction on \( n \). It is obvious for \( n = 1 \). Assume \( n \geqq 2 \). We consider a sequence as above, and without loss of generality, we may assume that it converges to 0 . (If necessary, consider \( {\xi }^{\left( v\right) } - {\xi }^{\left( \mu \right) } \) for \( v,\mu \rightarrow \infty \).) We must then show that the sequences of the coefficients converge to 0 also. If this is not the case, then there exists a number \( a > 0 \) such that we have for some \( j \), say \( j = 1 \),\n\n\[ \n\left| {x}_{1}^{\left( v\right) }\right| > a\n\]\n\nfor arbitrarily large \( v \). Thus for a subsequence of \( \left( v\right) ,{\xi }^{\left( v\right) }/{x}_{1}^{\left( v\right) } \) converges to 0, and we can write\n\n\[ \n\frac{{\xi }^{\left( v\right) }}{{x}_{1}^{\left( v\right) }} - {\omega }_{1} = \frac{{x}_{2}^{\left( v\right) }}{{x}_{1}^{\left( v\right) }}{\omega }_{2} + \cdots + \frac{{x}_{n}^{\left( v\right) }}{{x}_{1}^{\left( v\right) }}{\omega }_{n}\n\]\n\nWe let \( {\eta }^{\left( v\right) } \) be the right-hand side of this equation. Then the subsequence \( {\eta }^{\left( v\right) } \) converges (according to the left-hand side of our equation). By induction, we\n\nconclude that its coefficients in terms of \( {\omega }_{2},\ldots ,{\omega }_{n} \) also converge in \( K \), say to \( {y}_{2},\ldots ,{y}_{n} \). Taking the limit, we get\n\n\[ \n{\omega }_{1} = {y}_{2}{\omega }_{2} + \cdots + {y}_{n}{\omega }_{n}\n\]\n\ncontradicting the linear independence of the \( {\omega }_{i} \).\n\nWe must finally see that two norms inducing the same topology are equivalent. Let \( {\left. \mid \right| }_{1} \) and \( {\left. \mid \right| }_{2} \) be these norms. There exists a number \( C > 0 \) such that for any \( \xi \in E \) we have\n\n\[ \n{\left| \xi \right| }_{1} \leqq C\text{ implies }{\left| \xi \right| }_{2} \leqq 1\n\]\n\nLet \( a \in K \) be such that \( 0 < \left| a\right| < 1 \). For every \( \xi \in E \) there exists a unique integer \( s \) such that\n\n\[ \nC\left| a\right| < {\left| {a}^{s}\xi \right| }_{1} \leqq C.\n\]\n\nHence \( {\left| {a}^{s}\xi \right| }_{2} \leqq 1 \) whence we get at once\n\n\[ \n{\left| \xi \right| }_{2} \leqq {C}^{-1}{\left| a\right| }^{-1}{\left| \xi \right| }_{1}\n\]\n\nThe other inequality follows by symmetry, with a similar constant.
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Yes
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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 \( K = \mathbf{C} \) .
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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 \( {j}^{2} = - 1 \) . We define a norm on \( L \) (as an R-space) by putting\n\n\[ \left| {x + {yj}}\right| = \left| x\right| + \left| y\right| \]\n\nfor \( x, y \in K \) . This clearly makes \( L \) into a normed \( \mathbf{R} \) -space. Furthermore, if \( z = x + {yj} \) and \( {z}^{\prime } = {x}^{\prime } + {y}^{\prime }j \) are in \( L \), then\n\n\[ \left| {z{z}^{\prime }}\right| = \left| {x{x}^{\prime } - y{y}^{\prime }}\right| + \left| {x{y}^{\prime } + {x}^{\prime }y}\right| \]\n\n\[ \leqq \left| {x{x}^{\prime }}\right| + \left| {y{y}^{\prime }}\right| + \left| {x{y}^{\prime }}\right| + \left| {{x}^{\prime }y}\right| \]\n\n\[ \leqq \left| x\right| \left| {x}^{\prime }\right| + \left| y\right| \left| {y}^{\prime }\right| + \left| x\right| \left| {y}^{\prime }\right| + \left| {x}^{\prime }\right| \left| y\right| \]\n\n\[ \leqq \left( {\left| x\right| + \left| y\right| }\right) \left( {\left| {x}^{\prime }\right| + \left| {y}^{\prime }\right| }\right) \]\n\n\[ \leqq \left| z\right| \left| {z}^{\prime }\right| \]\n\nand we can therefore apply Theorem 2.3 again to conclude the proof.
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Yes
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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.
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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 Proposition 2.2, an extension of \( v \) to \( E \) defines the same topology as the max norm obtained in terms of a basis as above. Given a Cauchy sequence \( {\xi }^{\left( v\right) } \) in \( E \) ,\n\n\[{\xi }^{\left( v\right) } = {x}_{v1}{\omega }_{1} + \cdots + {x}_{vn}{\omega }_{n}\]\n\nthe \( n \) sequences \( \left\{ {x}_{vi}\right\} \left( {i = 1,\ldots, n}\right) \) must be Cauchy sequences in \( K \) by the definition of the max norm. If \( \left\{ {x}_{vi}\right\} \) converges to an element \( {z}_{i} \) in \( K \), then it is clear that the sequence \( {\xi }^{\left( v\right) } \) converges to \( {z}_{1}{\omega }_{1} + \cdots + {z}_{n}{\omega }_{n} \) . Hence \( E \) is complete. Furthermore, since any two extensions of \( v \) to \( E \) are equivalent, we can apply Proposition 1.1, and we see that we must have \( \lambda = 1 \), since the extensions induce the same absolute value \( v \) on \( K \) . This proves what we want.
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No
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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).
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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 \( {E}_{w} = E{K}_{w} \) .
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Yes
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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 \( {K}_{v} \) .
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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 \( \lambda : {\tau E} \rightarrow {\sigma E} \) be an isomorphism over \( K \) . We shall prove that \( \lambda \) extends to an isomorphism of \( {\tau E} \cdot {K}_{v} \) onto \( {\sigma E} \cdot {K}_{v} \) over \( {K}_{v} \) . Since \( {\tau E} \) is dense in \( {\tau E} \cdot {K}_{v} \) , an element \( x \in {\tau E} \cdot {K}_{v} \) can be written\n\n\[ x = \lim \tau {x}_{n} \]\n\nwith \( {x}_{n} \in E \) . Since the absolute values induced by \( \sigma \) and \( \tau \) on \( E \) coincide, it follows that the sequence \( {\lambda \tau }{x}_{n} = \sigma {x}_{n} \) converges to an element of \( {\sigma E} \cdot {K}_{v} \) which we denote by \( {\lambda x} \) . One then verifies immediately that \( {\lambda x} \) is independent of the particular sequence \( \tau {x}_{n} \) used, and that the map \( \lambda : {\tau E} \cdot {K}_{v} \rightarrow {\sigma E} \cdot {K}_{v} \) is an isomorphism, which clearly leaves \( {K}_{v} \) fixed. This proves our proposition.
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Yes
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Proposition 3.3. Let \( E \) be a finite separable extension of \( K \), of degree \( N \) . Then\n\n\[ N = \mathop{\sum }\limits_{{w \mid v}}{N}_{w} \]
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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 \( {f}_{i}\left( X\right) \) . They all appear with multiplicity 1 according to our hypothesis of separability. The embeddings of \( E \) into \( {K}_{v}^{\mathrm{a}} \) correspond to the maps of \( \alpha \) onto the roots of the \( {f}_{i} \) . Two embeddings are conjugate if and only if they map \( \alpha \) onto roots of the same polynomial \( {f}_{i} \) . On the other hand, it is clear that the local degree in each case is precisely the degree of \( {f}_{i} \) . This proves our proposition.
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Yes
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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 \( v \).
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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 let \( F = {E}^{{p}^{r}}K \). Then \( F \) is separable over \( K \) and \( E \) is purely inseparable over \( F \). By the preceding proposition,\n\n\[ \mathop{\sum }\limits_{{w \mid v}}\left\lbrack {{F}_{w} : {K}_{v}}\right\rbrack = \left\lbrack {F : K}\right\rbrack \]\n\nand for each \( w \), we have \( \left\lbrack {{E}_{w} : {F}_{w}}\right\rbrack \leqq \left\lbrack {E : F}\right\rbrack \). From this our inequality in the statement of the proposition is obvious.
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Yes
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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\[ \n{\operatorname{Tr}}_{K}^{E}\left( \alpha \right) = \mathop{\sum }\limits_{{w \mid v}}{\operatorname{Tr}}_{{K}_{v}}^{{E}_{w}}\left( \alpha \right)\n\]
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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\nwhere each \( {f}_{i}\left( X\right) \) is irreducible, and the \( {f}_{i} \) are distinct because of our hypothesis that \( v \) is well behaved. The norm \( {N}_{K}^{E}\left( \alpha \right) \) is equal to \( {\left( -1\right) }^{\deg f} \) times the constant term of \( f \), and similarly for each \( {f}_{i} \). Since the constant term of \( f \) is equal to the product of the constant terms of the \( {f}_{i} \), we get the first part of the proposition. The statement for the trace follows by looking at the penultimate coefficient of \( f \) and each \( {f}_{i} \).\n\nIf \( E \) is not equal to \( K\left( \alpha \right) \), then we simply use the transitivity of the norm and trace. We leave the details to the reader.\n\nOne can also argue directly on the embeddings. Let \( {\sigma }_{1},\ldots ,{\sigma }_{m} \) be the distinct embeddings of \( E \) into \( {K}_{v}^{\mathrm{a}} \) over \( K \), and let \( {p}^{r} \) be the inseparable degree of \( E \) over \( K \). The inseparable degree of \( {\sigma E} \cdot {K}_{v} \) over \( {K}_{v} \) for any \( \sigma \) is at most equal to \( {p}^{r} \). If we separate \( {\sigma }_{1},\ldots ,{\sigma }_{m} \) into distinct conjugacy classes over \( {K}_{v} \), then from our hypothesis that \( v \) is well behaved, we conclude at once that the inseparable degree of \( {\sigma }_{i}E \cdot {K}_{v} \) over \( {K}_{v} \) must be equal to \( {p}^{r} \) also, for each \( i \). Thus the formula giving the norm as a product over conjugates with multiplicity \( {p}^{r} \) breaks up into a product of factors corresponding to the conjugacy classes over \( {K}_{v} \).
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No
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Theorem 4.1. Let \( K \) be a subfield of a field \( L \) . Then a valuation on \( K \) has an extension to a valuation on \( L \) .
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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 \( L \) as in \( §3 \) of Chapter VII. Let \( \mathfrak{M} \) be the maximal ideal of \( \mathfrak{O} \) . Since \( \mathfrak{M} \cap \mathfrak{o} \) contains \( m \) but does not contain 1, it follows that \( \mathfrak{M} \cap \mathfrak{o} = \mathfrak{m} \) . Let \( {U}^{\prime } \) be the group of units of \( \mathfrak{O} \) . Then \( {U}^{\prime } \cap K = U \) is the group of units of \( \mathfrak{o} \) . Hence we have a canonical injection\n\n\[ \n{K}^{ * }/U \rightarrow {L}^{ * }/{U}^{\prime }\n\]\n\nwhich is immediately verified to be order-preserving. Identifying \( {K}^{ * }/U \) in \( {L}^{ * }/{U}^{\prime } \) we have obtained an extension of our valuation of \( K \) to a valuation of \( L \) .
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Yes
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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 \( \left( {{\Gamma }^{\prime } : \Gamma }\right) \leqq n \) .
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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 \( {a}_{j} \in K,{a}_{j} \neq 0 \) two terms must have the same value, say \( \left| {{a}_{i}{y}_{i}}\right| = \left| {{a}_{j}{y}_{j}}\right| \) with \( i \neq j \), and hence\n\n\[ \left| {y}_{i}\right| = \left| {{a}_{i}^{-1}{a}_{j}}\right| \left| {y}_{j}\right| \]\n\nThis contradicts the assumption that the values of \( {y}_{i},{y}_{j}\left( {i \neq j}\right) \) represent distinct cosets of \( \Gamma \) in \( {\Gamma }^{\prime } \), and proves our proposition.
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Yes
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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 \( \Gamma \) .
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Proof. Take \( e \) to be the index \( \left( {{\Gamma }^{\prime } : \Gamma }\right) \) .
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No
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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 reals.
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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 \( \Gamma \) . Since every element of \( {\mathbf{R}}^{ + } \) has a unique \( e \) -th root for every integer \( e \geqq 1 \), we can find in an obvious way an order-preserving embedding of \( {\Gamma }^{\prime } \) into \( {\mathbf{R}}^{ + } \) which induces the identity on \( \Gamma \) . In this way we get our extension of \( v \) to an absolute value on \( L \) .
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Yes
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Corollary 4.5. If \( L \) is finite over \( K \), and if \( \Gamma \) is infinite cyclic, then \( {\Gamma }^{\prime } \) is also infinite cyclic.
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Proof. Use Corollary 4.3 and the fact that a subgroup of a cyclic group is cyclic.
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No
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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. \( \mathfrak{m} = \mathfrak{M} \cap \mathfrak{o} \) . Then the residue class degree \( \left\lbrack {\mathfrak{O}/\mathfrak{M} : \mathfrak{o}/\mathfrak{m}}\right\rbrack \) is finite. If we denote it by \( f \), and if \( e \) is the ramification index, then ef \( \leqq n \) .
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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 \( \mathfrak{o}/\mathfrak{m} \) . Consider a relation\n\n\[ \mathop{\sum }\limits_{{i, j}}{a}_{ij}{z}_{j}{y}_{i} = 0 \]\n\nwith \( {a}_{ij} \in K \), not all \( {a}_{ij} = 0 \) . In an inner sum\n\n\[ \mathop{\sum }\limits_{{j = 1}}^{s}{a}_{ij}{z}_{j} \]\n\ndivide by the coefficient \( {a}_{iv} \) having the biggest valuation. We obtain a linear combination of \( {z}_{1},\ldots ,{z}_{s} \) with coefficients in \( \mathfrak{o} \), and at least one coefficient equal to a unit. Since \( {z}_{1},\ldots ,{z}_{s} \) are linearly independent \( {\;\operatorname{mod}\;\mathfrak{M}} \) over \( \mathfrak{o}/\mathfrak{m} \), it follows that our linear combination is a unit. Hence\n\n\[ \left| {\mathop{\sum }\limits_{{j = 1}}^{s}{a}_{ij}{z}_{j}}\right| = \left| {a}_{iv}\right| \]\n\nfor some index \( v \) . In the sum\n\n\[ \mathop{\sum }\limits_{{i = 1}}^{e}\left( {\mathop{\sum }\limits_{{j = 1}}^{s}{a}_{ij}{z}_{j}}\right) {y}_{i} = 0 \]\n\nviewed as a sum on \( i \), at least two terms have the same value. This contradicts the independence of \( \left| {y}_{1}\right| ,\ldots ,\left| {y}_{e}\right| {\;\operatorname{mod}\;\Gamma } \) just as in the proof of Proposition 4.2.
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Yes
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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\[ e\left( {u \mid w}\right) e\left( {w \mid v}\right) = e\left( {u \mid v}\right) \]\n\n\[ f\left( {u \mid w}\right) f\left( {w \mid v}\right) = f\left( {u \mid v}\right) .
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Proof. Obvious.
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No
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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 \( L \), and let \( \mathfrak{P} = \mathfrak{M} \cap B \) . Then \( \mathfrak{O} \) is equal to the local ring \( {B}_{\mathfrak{P}} \) .
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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 \( {a}_{s} \) is the coefficient having the biggest value among the \( {a}_{i} \) for the valuation associated with the valuation ring \( o \), and that it is the coefficient farthest to the left having this value. Let \( {b}_{i} = {a}_{i}/{a}_{s} \) . Then all \( {b}_{i} \in \mathfrak{o} \) and\n\n\[ \n{b}_{n},\ldots ,{b}_{s + 1} \in \mathfrak{M}.\n\] \n\nDivide the equation by \( {x}^{s} \) . We get\n\n\[ \n\left( {{b}_{n}{x}^{n - s} + \cdots + {b}_{s + 1}x + 1}\right) + \frac{1}{x}\left( {{b}_{s - 1} + \cdots + {b}_{0}\frac{1}{{x}^{s - 1}}}\right) = 0.\n\] \n\nLet \( y \) and \( z \) be the two quantities in parentheses in the preceding equation, so that we can write\n\n\[ \n- y = z/x\text{ and } - {xy} = z.\n\] \n\nTo prove our proposition it will suffice to show that \( y \) and \( z \) lie in \( B \) and that \( y \) is not in \( \mathfrak{P} \) .\n\nWe use Proposition 3.5 of Chapter VII. If a valuation ring of \( L \) above\n\ncontains \( x \), then it contains \( y \) because \( y \) is a polynomial in \( x \) with coefficients in\n\nHence such a valuation ring also contains \( z = - {xy} \) . If on the other hand the valuation ring of \( L \) above contains \( 1/x \), then it contains \( z \) because \( z \) is a polynomial in \( 1/x \) with coefficients in . Hence this valuation ring also contains \( y \) . From this we conclude by Chapter VII, Proposition 3.5, that \( y, z \) lie in \( B \) .\n\nFurthermore, since \( x \in \mathfrak{O} \), and \( {b}_{n},\ldots ,{b}_{s + 1} \) are in \( \mathfrak{M} \) by construction, it follows that \( y \) cannot be in \( \mathfrak{M} \), and hence cannot be in \( \mathfrak{P} \) . This concludes the proof.
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Yes
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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 \( \mathfrak{P} \) .
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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).
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Yes
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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, then there exists an automorphism \( \sigma \) of \( L \) over \( K \) such that \( \sigma \mathfrak{O} = {\mathfrak{O}}^{\prime } \) and \( \sigma \mathfrak{M} = {\mathfrak{M}}^{\prime } \) .
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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 assertion is obvious.
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Yes
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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\[{\operatorname{ord}}_{v}{N}_{K}^{E}\left( \alpha \right) = f\left( {w \mid v}\right) {\operatorname{ord}}_{w}\alpha .\]
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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.
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Yes
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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\[ \mathop{\sum }\limits_{{w \mid v}}e\left( {w \mid v}\right) f\left( {w \mid v}\right) = \left\lbrack {E : K}\right\rbrack . \]\n\nIf \( E \) is Galois over \( K \), then all \( {e}_{w} \) are equal to the same number \( e \), all \( {f}_{w} \) are\n\nequal to the same number \( f \), and so\n\n\[ \text{ efr } = \left\lbrack {E : K}\right\rbrack \text{,} \]\n\nwhere \( r \) is the number of extensions of \( v \) to \( E \) .
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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 \( = \left\lbrack {E : K}\right\rbrack \) is then obvious.
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Yes
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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 \( f \) .
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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 is only a finite number of choices for such multiplicities.) Furthermore, the other roots of \( g \) also belong to roots of \( f \), and we may suppose that these roots are bunched together, according to which root of \( f \) they belong to. Since \( \lim {g}_{v} = f \), we conclude that \( \alpha \) must have multiplicity \( s \) in \( f \), contradiction.
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Yes
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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\[ {D}_{i}f\left( A\right) \equiv 0\;\left( {\;\operatorname{mod}\;{\mathfrak{p}}^{r - 1}}\right) ,\;\text{ for all }\;i = 1,\ldots, n, \]\n\n\[ {D}_{i}f\left( A\right) ≢ 0\;\left( {\;\operatorname{mod}\;{\mathfrak{p}}^{r}}\right) ,\;\text{ for some }i = 1,\ldots, n. \]\n\nLet \( v \) be an integer \( \geqq 0 \) and let \( B \in {\mathfrak{o}}^{\left( n\right) } \) be such that\n\n\[ B \equiv A\;\left( {\;\operatorname{mod}\;{\mathfrak{p}}^{r}}\right) \;\text{ and }\;f\left( B\right) \equiv 0\;\left( {{\;\operatorname{mod}\;{\mathfrak{p}}^{{2r} - 1 + v}}\text{ 为任意常数 }}\right) . \]\n\nA vector \( Y \in {\mathfrak{o}}^{\left( n\right) } \) satisfies\n\n\[ Y \equiv B\;\left( {\;\operatorname{mod}\;{\mathfrak{p}}^{r + v}}\right) \;\text{ and }\;f\left( Y\right) \equiv 0\;\left( {\;\operatorname{mod}\;{\mathfrak{p}}^{{2r} + v}}\right) \]\n\nif and only if \( Y \) can be written in the form \( Y = B + {\pi }^{r + v}C \), with some \( C \in {\mathfrak{o}}^{\left( n\right) } \) satisfying the proper congruence\n\n\[ f\left( B\right) + {\pi }^{r + v}\operatorname{grad}f\left( B\right) \cdot C \equiv 0\;\left( {\;\operatorname{mod}\;{\mathfrak{p}}^{{2r} + v}}\right) . \]
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Proof. The proof is shorter than the statement of the proposition. Write \( Y = B + {\pi }^{r + v}C \) . By Taylor’s expansion,\n\n\[ f\left( {B + {\pi }^{r + v}C}\right) = f\left( B\right) + {\pi }^{r + v}\operatorname{grad}f\left( B\right) \cdot C\;\left( {\;\operatorname{mod}\;{\mathfrak{p}}^{{2r} + {2v}}}\right) . \]\n\nTo solve this last congruence \( {\;\operatorname{mod}\;{\mathfrak{p}}^{{2r} + v}} \), we obtain a proper congruence by hypothesis, because grad \( f\left( B\right) \equiv \operatorname{grad}f\left( A\right) \equiv 0\left( {\;\operatorname{mod}\;{\mathfrak{p}}^{r - 1}}\right. \) ).
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Yes
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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 \( A{\;\operatorname{mod}\;{\mathfrak{p}}^{r}} \) .
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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.
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Yes
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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 \( {f}^{\prime }\left( a\right) ≢ 0\left( {\;\operatorname{mod}\;\mathfrak{p}}\right) \) . Then there exists \( b \in \mathfrak{o}, b \equiv a\left( {\;\operatorname{mod}\;\mathfrak{p}}\right) \) such that \( f\left( b\right) = 0 \) .
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Proof. Take \( n = 1 \) and \( r = 1 \) in the proposition, and apply Corollary 7.3.
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No
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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 \( K \) .
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Proof. Apply the proposition.
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No
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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 \[ \left| {f\left( {\alpha }_{0}\right) }\right| < \left| {{f}^{\prime }{\left( {\alpha }_{0}\right) }^{2}}\right| \] (here \( {f}^{\prime } \) denotes the formal derivative of \( f \) ). Then the sequence \[ {\alpha }_{i + 1} = {\alpha }_{i} - \frac{f\left( {\alpha }_{i}\right) }{{f}^{\prime }\left( {\alpha }_{i}\right) } \] converges to a root \( \alpha \) of \( f \) in \( \mathfrak{o} \), and we have \[ \left| {\alpha - {\alpha }_{0}}\right| \leqq \left| \frac{f\left( {\alpha }_{0}\right) }{{f}^{\prime }{\left( {\alpha }_{0}\right) }^{2}}\right| < 1 \]
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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. \( \left| \frac{f\left( {\alpha }_{i}\right) }{{f}^{\prime }{\left( {\alpha }_{i}\right) }^{2}}\right| \leqq {c}^{{2}^{i}} \) . These three conditions obviously imply our proposition. If \( i = 0 \), they are hypotheses. By induction, assume them for \( i \) . Then: 1. \( \left| {f\left( {\alpha }_{i}\right) /{f}^{\prime }{\left( {\alpha }_{i}\right) }^{2}}\right| \leqq {c}^{{2}^{i}} \) gives \( \left| {{\alpha }_{i + 1} - {\alpha }_{i}}\right| \leqq {c}^{{2}^{i}} < 1 \), whence \( \left| {\alpha }_{i + 1}\right| \leqq 1 \) . 2. \( \left| {{\alpha }_{i + 1} - {\alpha }_{0}}\right| \leqq \max \left\{ {\left| {{\alpha }_{i + 1} - {\alpha }_{i}}\right| ,\left| {{\alpha }_{i} - {\alpha }_{0}}\right| }\right\} = c \) . 3. By Taylor's expansion, we have \[ f\left( {\alpha }_{i + 1}\right) = f\left( {\alpha }_{i}\right) - {f}^{\prime }\left( {\alpha }_{i}\right) \frac{f\left( {\alpha }_{i}\right) }{{f}^{\prime }\left( {\alpha }_{i}\right) } + \beta {\left( \frac{f\left( {\alpha }_{i}\right) }{{f}^{\prime }\left( {\alpha }_{i}\right) }\right) }^{2} \] for some \( \beta \in \mathfrak{o} \), and this is less than or equal to \[ {\left| \frac{f\left( {\alpha }_{i}\right) }{{f}^{\prime }\left( {\alpha }_{i}\right) }\right| }^{2} \] in absolute value. Using Taylor’s expansion on \( {f}^{\prime }\left( {\alpha }_{i + 1}\right) \) we conclude that \[ \left| {{f}^{\prime }\left( {\alpha }_{i + 1}\right) }\right| = \left| {{f}^{\prime }\left( {\alpha }_{i}\right) }\right| . \] From this we get \[ \left| \frac{f\left( {\alpha }_{i + 1}\right) }{{f}^{\prime }{\left( {\alpha }_{i + 1}\right) }^{2}}\right| \leqq {c}^{{2}^{i + 1}} \] as desired.
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Yes
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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\[ A\left( \begin{matrix} {x}_{1} \\ \vdots \\ {x}_{n} \end{matrix}\right) = \left( \begin{matrix} {y}_{1} \\ \vdots \\ {y}_{n} \end{matrix}\right) \]\n\nThen \( \left\{ {{y}_{1},\ldots ,{y}_{n}}\right\} \) is a basis of \( E \) if and only if \( A \) is invertible.
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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 \( X = {A}^{-1}Y \) and hence \( {x}_{1},\ldots ,{x}_{n} \) are in the module generated by \( {y}_{1},\ldots ,{y}_{n} \) . Suppose that we have a relation\n\n\[ {b}_{1}{y}_{1} + \cdots + {b}_{n}{y}_{n} = 0 \]\n\nwith \( {b}_{i} \in R \) . Let \( B \) be the row vector \( \left( {{b}_{1},\ldots ,{b}_{n}}\right) \) . Then\n\n\[ {BY} = 0 \]\n\nand hence \( {BAX} = 0 \) . But \( \left\{ {{x}_{1},\ldots ,{x}_{n}}\right\} \) is a basis. Hence \( {BA} = 0 \), and hence \( {BA}{A}^{-1} = B = 0 \) . This proves that the components of \( Y \) are linearly independent over \( R \), and proves our proposition.
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Yes
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