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Let \( S \) be a multiplicative subset of \( A \), then \( A\left\lbrack {S}^{-1}\right\rbrack \) is flat over \( A \). | It is not faithfully flat in general, however; see Theorem 3.5.6. | No |
Example 3.3.4. A routine fact is that for any family \( {\left( {M}_{ \bullet }^{\left( i\right) }\right) }_{i \in I} \) of complexes of \( A \) -modules, we have\n\n\[ \forall i \in I,{M}_{ \bullet }^{\left( i\right) }\text{ is exact } \Leftrightarrow {\bigoplus }_{i \in I}{M}_{ \bullet }^{\left( i\right) }\text{ is e... | Recall that \( \otimes \) preserves direct sums. It follows that a direct sum of modules is flat if and only if each summand is flat. From this we deduce the flatness of free modules since \( A\underset{\lambda }{ \otimes }M \simeq M \) functorially for each \( M \) . Furthermore, projective modules are flat as they ar... | No |
Proposition 3.3.7. The following are equivalent for an R-module \( N \) . (i) \( N \) is flat over \( R \) , (ii) \( {N}_{\mathfrak{p}} \) is flat over \( {R}_{\mathfrak{p}} \) for all prime ideal \( \mathfrak{p} \) ,(iii) \( {N}_{\mathfrak{m}} \) is flat over \( {R}_{\mathfrak{m}} \) for all maximal ideal \( \mathfrak... | Proof. Direct consequence of the exactness of localization and Remark 3.3.6. | No |
Lemma 3.3.8. Let \( \varphi : R \rightarrow {R}^{\prime } \) be a ring homomorphism, \( {\mathfrak{p}}^{\prime } \in \operatorname{Spec}\left( {R}^{\prime }\right) \) maps to \( \mathfrak{p} \in \operatorname{Spec}\left( R\right) \) under \( {\varphi }^{\sharp } \) . If \( \varphi \) is flat, so is the induced homomorp... | Proof. Set \( S = R \smallsetminus \mathfrak{p} \) so that \( \varphi \left( S\right) \subset {R}^{\prime } \smallsetminus {\mathfrak{p}}^{\prime } \) . Factorize \( {R}_{\mathfrak{p}} \rightarrow {R}_{{\mathfrak{p}}^{\prime }} \) as\n\n\[ \n{R}_{\mathfrak{p}} \rightarrow \underset{\text{as a ring }}{\underbrace{{R}^{\... | Yes |
Proposition 3.3.9. Let \( \varphi : R \rightarrow {R}^{\prime } \) be a ring homomorphism. The following are equivalent:\n\n(i) \( {R}^{\prime } \) is flat over \( R \) ,\n\n(ii) \( {R}_{{\mathfrak{p}}^{\prime }}^{\prime } \) is flat over \( {R}_{\mathfrak{p}} \) for all \( {\mathfrak{p}}^{\prime } \in \operatorname{Sp... | Proof. (i) \( \Rightarrow \) (ii): Set \( S \mathrel{\text{:=}} R \smallsetminus \mathfrak{p} \) . Base change implies \( {R}^{\prime }\left\lbrack {S}^{-1}\right\rbrack \) is flat over \( R\left\lbrack {S}^{-1}\right\rbrack = {R}_{\mathfrak{p}} \) . Since \( {R}_{{\mathfrak{p}}^{\prime }}^{\prime } \) is a localizatio... | No |
Lemma 3.3.10 (Equational criterion of flatness). An R-module \( N \) is flat if and only if for all \( r \geq 1,{a}_{1},\ldots ,{a}_{r} \in R,{x}_{1},\ldots ,{x}_{r} \in N \) verifying \( \mathop{\sum }\limits_{{i = 1}}^{r}{a}_{i}{x}_{i} = 0 \), there exist \( s \in {\mathbb{Z}}_{ \geq 1} \), an \( R \) -valued matrix ... | Proof. Suppose \( M \) is flat and consider the exact sequence \( 0 \rightarrow \ker \left( f\right) \rightarrow {R}^{\oplus r}\overset{f}{ \rightarrow }R \) where \( f\left( {{t}_{1},\ldots ,{t}_{r}}\right) = \mathop{\sum }\limits_{i}{a}_{i}{t}_{i} \) . We obtain an exact\n\n\[ 0 \rightarrow \ker \left( f\right) \unde... | Yes |
Proposition 3.4.1. The following are equivalent for an R-module \( N \) :\n\n(i) \( N \) is flat,\n\n(ii) \( {\operatorname{Tor}}_{i}^{R}\left( {N, - }\right) = 0 \) for all \( i > 0 \),\n\n(iii) \( {\operatorname{Tor}}_{1}^{R}\left( {N, - }\right) = 0 \),\n\n(iv) \( {\operatorname{Tor}}_{1}^{R}\left( {N, R/\mathfrak{a... | Proof. First, the equivalence mentioned in (iv) is a consequence of the exact sequence\n\n\[ \underset{ = 0}{\underline{{\operatorname{Tor}}_{1}^{R}\left( {N, R}\right) }} \rightarrow {\operatorname{Tor}}_{1}^{R}\left( {N, R/\mathfrak{a}}\right) \rightarrow N\underset{R}{ \otimes }\mathfrak{a} \rightarrow N \rightarrow... | Yes |
Corollary 3.4.2. If \( r \in R \) is not a zero divisor, then \( r \) is not a zero divisor for any flat \( R \) -module \( N \) . | Proof. Take \( \mathfrak{a} \mathrel{\text{:=}} {Rr} \), which is \( \simeq R \), and contemplate on \( N \simeq \mathfrak{a}\underset{R}{ \otimes }N \hookrightarrow N \) . | No |
Lemma 3.4.5. Suppose \( A \rightarrow B \) is flat. Write \( {M}_{B} \mathrel{\text{:=}} B \otimes M \) for any \( A \) -module \( B \) . Then there are natural isomorphisms \( {\operatorname{Tor}}_{i}^{B}\left( {{M}_{B},{N}_{B}}\right) \simeq B \otimes {\operatorname{Tor}}_{i}^{A}\left( {M, N}\right) \) for all \( i \... | Proof. Take a projective resolution \( 0 \leftarrow M \leftarrow {P}_{ \bullet } \) of \( A \) -modules. Since \( B \) is flat over \( A \) , its base-change \( 0 \leftarrow {M}_{B} \leftarrow {P}_{\bullet, B} \) to \( B \) is still a projective resolution; we are using the fact that base-change preserves projectivity.... | No |
Theorem 3.4.6. Let \( R \) be a local ring with maximal ideal \( \mathfrak{m} \) . Let \( M \) be a finitely generated R-module. The following are equivalent:\n\n\( \diamond M \) is free,\n\n\( \diamond M \) is projective. | Proof. Free modules are known to be projective. Now let \( M \) be finitely generated projective, and take a basis \( {\bar{x}}_{1},\ldots {\bar{x}}_{n} \) of the \( R/\mathfrak{m} \) -vector space \( M/\mathfrak{m}M \), together with liftings \( M \ni {x}_{i} \mapsto {\bar{x}}_{i} \) . Nakayama’s Lemma then implies th... | Yes |
Corollary 3.4.7. Let \( M \) be a finitely presented \( R \) -module. Then \( M \) is projective if and only \( {M}_{\mathfrak{m}} \) is free for every maximal ideal \( \mathfrak{m} \) . | Proof. In view of Theorem 3.4.6, it suffices to show \( M \) is projective if and only if \( {M}_{\mathfrak{m}} \) is for all maximal ideal \( \mathfrak{m} \) . One direction is easy: if \( M \) is a direct summand of a free module, then so is \( {M}_{\mathfrak{m}} \) .\n\nConversely, the assumption on finite presentat... | Yes |
Lemma 3.5.1. Let \( F : C \rightarrow {C}^{\prime } \) be an additive functor between abelian categories. The following are equivalent.\n\n(i) \( F \) is exact and faithful;\n\n(ii) \( F \) is exact and \( \left( {{FM} = 0 \Leftrightarrow M = 0}\right) \) for every object \( M \) of \( C \) ;\n\n(iii) a sequence \( {M}... | Proof. (i) \( \Rightarrow \) (ii): If \( {FM} = 0 \) then \( F\left( {\mathrm{{id}}}_{M}\right) = {\mathrm{{id}}}_{FM} = 0 \), and the faithfulness implies \( {\operatorname{id}}_{M} = 0 \) in \( {\operatorname{End}}_{\mathcal{C}}\left( M\right) \) ; this is possible only when \( M = 0 \) .\n\n(ii) \( \Rightarrow \) (i... | Yes |
Proposition 3.5.2. The following are equivalent for an R-module \( N \) .\n\n(i) \( N \) is faithfully flat.\n\n(ii) The functor \( - \underset{R}{ \otimes }N \) is exact and faithful.\n\n(iii) \( N \) is flat and for every maximal ideal \( \mathfrak{m} \) of \( R \), we have \( N/\mathfrak{m}N \simeq N\underset{R}{ \o... | Proof. The equivalence (i) \( \Leftrightarrow \) (ii) has just been established. An \( R \) -module \( M \) is nonzero if and only if there exist exact sequences\n\n\[ 0 \rightarrow R/\mathfrak{a} \rightarrow M,\;R/\mathfrak{a} \rightarrow R/\mathfrak{m} \rightarrow 0 \]\n\nwhere \( \mathfrak{a} \) is a proper ideal an... | No |
Corollary 3.5.3. Let \( R \rightarrow {R}^{\prime } \) be a local homomorphism \( {}^{1} \) between local rings and let \( M \) be a finitely generated \( {R}^{\prime } \) -module, \( M \neq 0 \) . Then as an \( R \) -module, \( M \) is faithfully flat if and only if it is flat. | Proof. Let \( \mathfrak{m},{\mathfrak{m}}^{\prime } \) be the maximal ideals of \( R,{R}^{\prime } \), respectively. Since \( \mathfrak{m} \) is mapped into \( {\mathfrak{m}}^{\prime } = \operatorname{rad}\left( {R}^{\prime }\right) \), the assertion follows from Proposition 3.5.2 together with Nakayama’s Lemma. | No |
Proposition 3.5.5. Suppose \( \varphi : A \rightarrow B \) is a faithfully flat ring homomorphism and regard \( B \) as an A-algebra.\n\n- For any A-module \( M \), the natural map \( M \rightarrow M \otimes B \) is injective; in particular, \( \varphi \) is seen to be injective by taking \( M = A \) .\n\n\( \diamond \... | Proof. Let \( N \) be the kernel of \( M \rightarrow M \otimes B \) . The \( B \) -module homomorphism \( N \otimes B \rightarrow M \otimes B \) is injective by flatness; it is zero on \( N \otimes 1 \), hence identically zero and we obtain \( N = \{ 0\} \) by faithful flatness (Lemma 3.5.1).\n\nLet \( \mathfrak{a} \su... | Yes |
Theorem 3.5.6. The following are equivalent for a ring homomorphism \( \varphi : A \rightarrow B \) .\n\n(i) \( \varphi \) is faithfully flat;\n\n(ii) \( \varphi \) is flat and \( {\varphi }^{\sharp } : \operatorname{Spec}\left( B\right) \rightarrow \operatorname{Spec}\left( A\right) \) is surjective;\n\n(iii) \( \varp... | Proof. (i) \( \Rightarrow \) (ii) is contained in Proposition 3.5.5. As for (ii) \( \Rightarrow \) (iii), take any \( \mathfrak{q} \in \operatorname{Spec}\left( B\right) \) that pulls back to \( \mathfrak{p} \in \operatorname{Spec}\left( A\right) \), then any maximal over-ideal \( \mathfrak{m} \) of \( \mathfrak{q} \) ... | Yes |
Lemma 4.1.1 (Existence of minimal over-primes). Let \( \mathfrak{a} \) be a proper ideal in a ring \( R \) . There exists a prime ideal \( \mathfrak{p} \) which is minimal among all primes containing \( \mathfrak{a} \) . If \( \mathfrak{P} \) is an ideal containing \( \mathfrak{a} \), one can choose \( \mathfrak{p} \su... | Proof. One easily reduces to the case \( \mathfrak{a} = \{ 0\} \) . We want to use Zorn’s Lemma to find a minimal prime. It boils down to show that any chain \( {\left( {\mathfrak{p}}_{i}\right) }_{i \in I} \) of prime ideals ( \( I \) : totally ordered set with \( j > i \Rightarrow {\mathfrak{p}}_{i} \supset {\mathfra... | Yes |
Lemma 4.1.2. The going-down property for \( \varphi \) is equivalent to the following: for every \( \mathfrak{p} \in \) \( \operatorname{Spec}\left( A\right) \) with \( \varphi \left( \mathfrak{p}\right) B \neq B \) and any minimal over-prime \( \mathfrak{q} \) of \( \varphi \left( \mathfrak{p}\right) B \), we have \( ... | Proof. Assuming going-down for \( \varphi \), let \( \mathfrak{p},\mathfrak{q} \) be as above. Evidently \( {\varphi }^{\sharp }\left( \mathfrak{q}\right) \supset {\varphi }^{-1}\left( {\varphi \left( \mathfrak{p}\right) B}\right) \supset \) \( \mathfrak{p} \) . If we have \( \supsetneq \), then going-down guarantees t... | Yes |
Theorem 4.1.3. Going-down holds for flat \( \varphi : A \rightarrow B \) . | Proof. Consider \( \mathfrak{p} \subset {\mathfrak{p}}^{\prime } \) and \( {\mathfrak{q}}^{\prime } \) lying over \( {\mathfrak{p}}^{\prime } \) in the setting of going-down. First, \( {B}_{{\mathfrak{q}}^{\prime }} \) is flat over \( {A}_{{\mathfrak{p}}^{\prime }} \) by Proposition 3.3.9. Secondly, \( {A}_{{\mathfrak{... | Yes |
Theorem 4.1.4 (Krull-Cohen-Seidenberg). Suppose the ring B is integral over its subring A. The following holds.\n\n(i) The map \( \operatorname{Spec}\left( B\right) \rightarrow \operatorname{Spec}\left( A\right) \) given by \( \mathfrak{q} \mapsto \mathfrak{q} \cap A \) is surjective.\n\n(ii) There are no inclusion rel... | Proof. (iv): Let \( \mathfrak{q} \in \operatorname{MaxSpec}\left( B\right) \) and \( {\mathfrak{p}}_{0} \mathrel{\text{:=}} \mathfrak{q} \cap A \) . We know \( B/\mathfrak{q} \) is a field, integral over its subring \( A/{\mathfrak{p}}_{0} \) . We claim that \( A/{\mathfrak{p}}_{0} \) is also a field, therefore \( {\ma... | Yes |
Proposition 4.2.1. Let \( \varphi : A \rightarrow B \) be a ring homomorphism with going-up property and suppose \( B \) is Noetherian. Then \( {\varphi }^{\sharp } : \operatorname{Spec}\left( B\right) \rightarrow \operatorname{Spec}\left( A\right) \) is a closed map: it maps closed subsets to closed subsets. | Proof. Consider a closed subset \( V\left( \mathfrak{b}\right) \) of \( \operatorname{Spec}\left( B\right) \) . First, every \( \mathfrak{q} \in \operatorname{Spec}\left( B\right) \) with \( \mathfrak{q} \supset \mathfrak{b} \) lies over a minimal over-prime of \( \mathfrak{b} \), by Lemma 4.1.1. Secondly, \( B \) is N... | Yes |
Corollary 4.2.2. Suppose \( B \) is Noetherian and integral over a subring \( A \) . Then \( \operatorname{Spec}\left( B\right) \rightarrow \operatorname{Spec}\left( A\right) \) is a closed surjection with finite fibers. | Proof. Apply Theorem 4.1.4 with Proposition 4.2.1 to show that \( \operatorname{Spec}\left( B\right) \rightarrow \operatorname{Spec}\left( A\right) \) is closed and surjective.\n\nTo show the finiteness of the fiber over \( \mathfrak{p} \in \operatorname{Spec}\left( A\right) \), note that the preimage of \( V\left( \ma... | Yes |
Lemma 4.2.4. With respect to the Zariski topology, \( \mathfrak{p} \) is a generalization of \( {\mathfrak{p}}^{\prime } \) if and only if \( {\mathfrak{p}}^{\prime } \in \overline{\{ \mathfrak{p}\} } \) . | Proof. The condition \( {\mathfrak{p}}^{\prime } \in \overline{\{ \mathfrak{p}\} } \) means that for every ideal \( \mathfrak{a} \), if \( \mathfrak{p} \supset \mathfrak{a} \) then \( {\mathfrak{p}}^{\prime } \supset \mathfrak{a} \) . Taking \( \mathfrak{a} = \mathfrak{p} \) yields \( {\mathfrak{p}}^{\prime } \supset \... | Yes |
Lemma 4.2.9. Let \( E \) be a subset of \( \operatorname{Spec}\left( R\right) \) where \( R \) is a Noetherian ring. The following are equivalent:\n\n(i) \( E \) is constructible;\n\n(ii) for every irreducible subset \( Z \) of \( \operatorname{Spec}\left( R\right) \), either \( Z \cap E \) is not dense in \( Z \) or \... | Proof. Omitted. See [11, (6.C)]. | No |
Theorem 4.2.10 (C. Chevalley). Let \( \varphi : A \rightarrow B \) be a ring homomorphism such that \( A \) is Noetherian and B is a finitely generated A-algebra. Then \( {\varphi }^{\sharp } \) maps constructible subsets to constructible subsets. | Proof. Omitted. See [11, (6.E)]. | No |
Proposition 4.2.11. Suppose \( R \) is Noetherian and \( E \) is a constructible subset of \( \operatorname{Spec}\left( R\right) \) . If \( E \) is stable under specialization (resp. generalization), then \( E \) is closed (resp. open) in \( \operatorname{Spec}\left( R\right) \) . | Proof. It suffices to treat the specialization-stable case by taking complements. Write the Zariski-closure \( \bar{E} \) as a finite union of irreducibles components \( Z \), without inclusion relations. For each irreducible component \( Z \), notice that \( Z \cap E \) is dense in \( Z \) for all \( Z \) , since othe... | Yes |
Proposition 4.2.12. Consider a ring homomorphism \( \varphi : A \rightarrow B \) satisfying going-down. Suppose \( A \) is Noetherian and \( B \) is a finitely generated \( A \)-algebra, then \( {\varphi }^{\sharp } : \operatorname{Spec}\left( B\right) \rightarrow \operatorname{Spec}\left( A\right) \) is an open map. | Proof. Let \( U = \operatorname{Spec}\left( B\right) \smallsetminus V\left( \mathfrak{a}\right) \) be an open subset. Going-down implies that \( {\varphi }^{\sharp }\left( U\right) \) is stable under generalization. It suffices to show \( {\varphi }^{\sharp }\left( U\right) \) is constructible, and this is the content ... | Yes |
Lemma 4.3.5. Let \( M \) be a \( \mathbb{Z} \) -graded module over a \( \mathbb{Z} \) -graded ring \( R \) . If \( x \in M \) and \( \mathfrak{p} \mathrel{\text{:=}} \operatorname{ann}\left( x\right) \) is a prime ideal, then\n\n(i) \( \mathfrak{p} \) is a homogeneous ideal, and\n\n(ii) \( \mathfrak{p} = \operatorname{... | Proof. We begin with (i). Let \( t \in \mathfrak{p} \) and \( x \in M \) be such that \( \mathfrak{p} = \operatorname{ann}\left( M\right) \) . Write\n\n\[ t = \mathop{\sum }\limits_{{\gamma \in \mathcal{A}}}{t}_{\gamma },\;x = \mathop{\sum }\limits_{{\eta \in \mathcal{B}}}{x}_{\eta } \]\n\nwhere \( {t}_{\gamma } \in {R... | Yes |
Proposition 4.4.4. Suppose \( \mathfrak{a} \) is a proper ideal of \( R \) . If \( R \) is Noetherian, so is \( \operatorname{gr}\left( R\right) \) with respect to the \( \mathfrak{a} \) -adic filtration. In fact \( \operatorname{gr}\left( R\right) \) is finitely generated over \( {\operatorname{gr}}^{0}\left( R\right)... | Proof. Let \( {x}_{1},\ldots ,{x}_{n} \) be generators of \( \mathfrak{a} \), with images \( {\bar{x}}_{i} \in \mathfrak{a}/{\mathfrak{a}}^{2} \) . Then \( \operatorname{gr}\left( R\right) \) is generated by \( {\bar{x}}_{1},\ldots {\bar{x}}_{n} \) over \( R/\mathfrak{a} = {\operatorname{gr}}^{0}R \), which is also Noe... | Yes |
Proposition 4.4.5. Let \( \mathfrak{a} \) be a proper ideal of \( R \) and \( M \) a finitely generated \( R \) -module. Suppose \( M \) is endowed with an \( \mathfrak{a} \) -stable filtration such that \( {F}^{i}M \) is finitely generated for each \( i \), and \( {F}^{ \leq 0}M = M \) . Then \( \operatorname{gr}\left... | Proof. Take \( n \) such that \( \mathfrak{a} \cdot {F}^{i}M = {F}^{i + 1}M \) for all \( i \geq n \) . Then in \( \operatorname{gr}\left( M\right) = {\bigoplus }_{i}{\operatorname{gr}}^{i}M \) we have\n\n\[ \left( {\mathfrak{a}/{\mathfrak{a}}^{2}}\right) \cdot {\operatorname{gr}}^{i}M = {\operatorname{gr}}^{i + 1}M,\;... | Yes |
Proposition 4.5.3. Suppose that \( \varphi : M \rightarrow N \) is a morphism between filtered R-modules. If \( M \) is exhaustive and separating, and \( \operatorname{gr}\varphi \) is injective, then \( \varphi \) is also injective. | Proof. Let \( x \in \ker \left( \varphi \right) \) . There exists \( n \) such that \( x \in {F}^{n}M \) . Regard \( x + {F}^{n + 1}M \) as an element of \( {\operatorname{gr}}^{n}M \) . Then \( \operatorname{gr}\left( \varphi \right) \left( {x + {F}^{n + 1}M}\right) = \varphi \left( x\right) + {F}^{n + 1}N = 0 \), so ... | Yes |
Lemma 4.5.6. Consider a ring \( R \) with proper ideal \( \mathfrak{a} \), together with a filtered \( R \) -module \( M \) , assume furthermore that each \( {F}^{i}M \) is finitely generated over \( R \) . The following are equivalent:\n\n(i) \( \mathrm{{Bl}}\left( M\right) \) is finitely generated over \( {\mathrm{{B... | Proof. (i) \( \Rightarrow \) (ii): Choose homogeneous generators \( {x}_{1},\ldots ,{x}_{n} \in \mathrm{{Bl}}\left( M\right) \) with degrees \( {d}_{1},\ldots ,{d}_{n} \) respectively. It is then routine to see that\n\n\[ i \geq \max \left\{ {{d}_{1},\ldots ,{d}_{n}}\right\} \Longrightarrow \left( {{F}^{i + 1}M}\right)... | Yes |
Theorem 4.5.7 (Artin-Rees). Let \( R \) be a Noetherian ring endowed with a-adic filtration. Let \( M \) be a finitely generated \( R \) -module and \( N \subset M \) an \( R \) -submodule. Then the filtration on \( N \) induced by the \( \mathfrak{a} \) -adic filtration of \( M \), namely \( {F}^{i}N \mathrel{\text{:=... | Proof. Define \( \mathrm{{Bl}}\left( N\right) \) using the induced filtration \( {F}^{i}N \mathrel{\text{:=}} {\mathfrak{a}}^{i}M \cap N \), which is a submodule of the finitely generated \( {\mathrm{{Bl}}}_{\mathfrak{a}}R \) -module \( \mathrm{{Bl}}\left( M\right) \) (Lemma 4.5.6). If \( \mathfrak{a} = \left( {{a}_{1}... | Yes |
Theorem 4.5.8. For \( R,\mathfrak{a}, M \) as in the previous theorem, we set \( N \mathrel{\text{:=}} \mathop{\bigcap }\limits_{{n \geq 0}}{\mathfrak{a}}^{n}M \) . Then \( \mathfrak{a}N = \) \( N \) . | Proof. Since the induced filtration on \( N \) is \( \mathfrak{a} \) -stable by Theorem 4.5.7, for \( n \gg 0 \) we have\n\n\[ N = {\mathfrak{a}}^{n}M \cap N = \mathfrak{a} \cdot \left( {{\mathfrak{a}}^{n - 1}M \cap N}\right) = \mathfrak{a}N. \]\n\nThe assertion follows. | Yes |
Corollary 4.5.9 (Krull). If \( \mathfrak{a} \subset \operatorname{rad}\left( R\right) \), then \( \mathop{\bigcap }\limits_{{n \geq 0}}{\mathfrak{a}}^{n}M = \{ 0\} \) for any finitely generated \( R \) -module \( M \) . In particular \( \mathop{\bigcap }\limits_{{n \geq 0}}{\mathfrak{a}}^{n} = \{ 0\} \) whenever \( \ma... | Proof. Theorem 4.5.8 together with Nakayama’s lemma imply \( N = \{ 0\} \) . | No |
Corollary 4.5.10 (Krull's Intersection Theorem). Let \( R \) be a Noetherian domain and a a proper ideal. Then \( \mathop{\bigcap }\limits_{{n \geq 0}}{\mathfrak{a}}^{n} = \{ 0\} \) . | Proof. Define \( N \mathrel{\text{:=}} \mathop{\bigcap }\limits_{{n > 0}}{\mathfrak{a}}^{n} \subset R \) . By Theorem 4.5.8 we have \( \mathfrak{a}N = N \), thus there exists \( r \in \mathfrak{a} \) with \( 1 + r \in \operatorname{ann}\left( N\right) \) by Nakayama’s Lemma (Theorem 1.3.5). As \( \mathfrak{a} \) is pro... | Yes |
Lemma 5.1.5. Let \( M \) be a complete \( R \) -module with respect to some filtration \( {F}^{ \bullet }M \) . For any submodule \( N \), the quotient \( M/N \) is also complete with respect to the quotient topology, or equivalently with respect to the quotient filtration \( \left( {{F}^{ \bullet }M + N}\right) /N \) ... | Proof. Let \( {\bar{x}}_{n} \) be a Cauchy sequence in \( M/N \) . Choose preimages \( M \ni {x}_{n} \mapsto {\bar{x}}_{n} \) for all \( n \) . We have \( {\bar{x}}_{n + 1} - {\bar{x}}_{n} \in {F}^{i\left( n\right) }M + N \) where \( \mathop{\lim }\limits_{{n \rightarrow \infty }}i\left( n\right) = \infty \), therefore... | Yes |
Proposition 5.1.6. Let \( R \) be a Noetherian ring and \( I \subsetneq R \) an ideal. Suppose \( 0 \rightarrow {M}^{\prime } \rightarrow M \rightarrow \) \( {M}^{\prime \prime } \rightarrow 0 \) is an exact sequence of finitely generated \( R \) -modules, each term equipped with the I-adic topology. The completed sequ... | Proof. We shall show\n\n(i) the topology on \( {M}^{\prime } \) induced from \( M \) is the same as the \( I \) -adic topology;\n\n(ii) the quotient topology on \( {M}^{\prime \prime } = M/{M}^{\prime } \) is \( I \) -adic;\n\n(iii) the completion \( {\widehat{M}}^{\prime } \) is naturally identified with the closure o... | Yes |
Theorem 5.1.7. Suppose \( R \) is Noetherian. Let \( M \) be a finitely generated \( R \) -module endowed with the I-adic topology. The homomorphism \( M\underset{R}{ \otimes }\widehat{R} \rightarrow \widehat{M} \) is then an isomorphism. | Proof. Write down a finite presentation \( {R}^{\oplus a} \rightarrow {R}^{\oplus b} \rightarrow M \rightarrow 0 \) . By the naturality of the homomorphism above, the right-exactness of \( \otimes \) and Proposition 5.1.6, we have a commutative diagram\n\n is flat. | Proof. Flatness can be tested on short exact sequences of the form \( 0 \rightarrow \mathfrak{a} \rightarrow R \rightarrow R/\mathfrak{a} \rightarrow \) 0 where \( \mathfrak{a} \) is a finitely generated ideal of \( R \) . Its base-change to \( \widehat{R} \) is the same as completion, and completion is an exact functo... | Yes |
Corollary 5.1.10. Assume \( R \) is I-adically complete Hausdorff. Then every finitely generated \( R \) -module \( M \) is I-adically complete Hausdorff, and any submodule \( N \subset M \) is closed. | Proof. For the first assertion: the completion of \( M \) can be identified with the composite \( M = M\underset{R}{ \otimes }R \rightarrow M\underset{R}{ \otimes }\widehat{R} \rightarrow \widehat{M} \), which is bijective. Therefore every Cauchy sequence in \( M \) has a limit in \( M \) .\n\nAs to the second assertio... | Yes |
Proposition 5.2.1. Let \( \mathfrak{a} \) be an ideal, then \( \mathfrak{a}\widehat{M} = \widehat{\mathfrak{a}M} = \widehat{\mathfrak{a}}\widehat{M} \) as submodules of \( \widehat{M} \) . Consequently \( \widehat{M}/\widehat{\mathfrak{a}}\widehat{M} \simeq {\left( M/\mathfrak{a}M\right) }^{ \land } \) canonically. | Proof. By the exactness of completion (Proposition 5.1.6), we may realize \( \widehat{\mathfrak{a}} \) as an ideal of \( \widehat{R} \) ; in fact it is the image \( \mathfrak{a}\widehat{R} \) of \( \mathfrak{a}\underset{R}{ \otimes }\widehat{R} \rightarrow \widehat{R}\underset{R}{ \otimes }\widehat{R} = \widehat{R} \) ... | Yes |
Proposition 5.2.2. For any finitely generated R-module \( M \), the topology on \( \widehat{M} \) coincides with the \( \widehat{\widehat{I}} \) -adic one. | Proof. Consider the closure of the image of \( {I}^{n}M \) in \( \widehat{M} \) . It is readily seen to be \( \left\{ {{\left( {x}_{k}\right) }_{k} : i \leq }\right. \) \( n \Rightarrow {x}_{i} = 0\} = {F}^{n}\widehat{M} \) . On the other hand, we have seen that this closure is \( \widehat{{I}^{n}M} \subset \widehat{M}... | Yes |
Lemma 5.2.3. Consider a homomorphism \( \varphi : L \rightarrow N \) between filtered modules over some ring, such that \( L \) is complete, \( N \) is Hausdorff and exhaustive (see §4.5) with respect to their filtrations, and \( \operatorname{gr}\left( \varphi \right) : \operatorname{gr}\left( L\right) \rightarrow \op... | Proof. Let \( y \in {F}^{d}N \), we may take \( x \in {F}^{d}L \) such that \( {y}^{\prime } \mathrel{\text{:=}} y - \varphi \left( x\right) \in {F}^{d + 1}N \) . Next, take \( {x}^{\prime } \in {F}^{d + 1}L \) with \( {y}^{\prime \prime } \mathrel{\text{:=}} {y}^{\prime } - \varphi \left( {x}^{\prime }\right) \in {F}^... | Yes |
Proposition 5.2.4. The ring \( \widehat{R} \) is also Noetherian, and \( \widehat{I} \subset \operatorname{rad}\left( \widehat{R}\right) \) . | Proof. Let \( \mathfrak{A} \) be any ideal of \( \widehat{R} \), equipped with the filtration \( {F}^{n}\mathfrak{A} \mathrel{\text{:=}} {\widehat{I}}^{n} \cap \mathfrak{A} \) . We have to show \( \mathfrak{A} \) is finitely generated. Since \( {\operatorname{gr}}_{I}\left( R\right) = {\operatorname{gr}}_{\widehat{I}}\... | Yes |
Proposition 5.2.5. The map \( \mathfrak{p} \mapsto \widehat{\mathfrak{p}} \) furnishes an injection from \( V\left( I\right) \) to \( \operatorname{Spec}\left( \widehat{R}\right) \) satisfying \( R/\mathfrak{p} \simeq \widehat{R}/\widehat{\mathfrak{p}} \) (as rings). It restricts to a bijection \( \operatorname{MaxSpec... | Proof. Since \( \mathfrak{p} \supset I \), the \( I \) -adic topology on \( R/\mathfrak{p} \) is discrete. By Proposition 5.2.1, \( \widehat{R}/\widehat{p} \simeq \) \( {\left( R/\mathfrak{p}\right) }^{ \land } = R/\mathfrak{p} \), and here the isomorphism even respects ring structures. Therefore \( \widehat{\mathfrak{... | Yes |
Lemma 5.3.1. For \( R \) as above and \( M \) a finitely generated graded \( R \) -module, each graded piece \( {M}_{\gamma } \) is an \( {R}_{0} \) -module of finite length. | Proof. Using (5-2) this is readily reduced to the case \( M = R\left( {-\eta }\right) \), and then to \( M = R \) . Write \( R = {R}_{0}\left\lbrack {{x}_{1},\ldots ,{x}_{n}}\right\rbrack \) where each \( {x}_{i} \) is homogeneous of degree \( {d}_{i} \) . Given \( \gamma \), the \( {R}_{0} \) - module \( {M}_{\gamma }... | Yes |
Theorem 5.3.3. Assume \( \Gamma = {\mathbb{Z}}_{ \geq 0} \) . Suppose \( R \) is generated by homogeneous elements \( {x}_{1},\ldots ,{x}_{n} \) over \( {R}_{0} \) . There exists a unique quasi-polynomial \( {H}_{M} \) of degree \( \leq n - 1 \), with coefficients in \( \mathbb{Q} \) and period \( e \mathrel{\text{:=}}... | Proof. Uniqueness is clear. We construct \( {H}_{M} \) by induction on the minimal number of generators \( n \) . If \( n = 0 \) then \( R = {R}_{0} \) and \( {M}_{\gamma } = 0 \) for \( \left| \gamma \right| \gg 0 \), in which case \( {H}_{M} = 0 \) .\n\nFor \( n \geq 1 \), write \( R = {R}_{0}\left\lbrack {{x}_{1},\l... | Yes |
For \( R = M = \mathbb{k}\left\lbrack {{X}_{1},\ldots ,{X}_{n}}\right\rbrack \) graded by total degree, where \( \mathbb{k} \) is a field, our assumptions are readily verified. We see \( \chi \left( {M,\gamma }\right) = {\dim }_{\mathbb{k}}\mathbb{k}{\left\lbrack {X}_{1},\ldots ,{X}_{n}\right\rbrack }_{\deg = \gamma } ... | Hence the Hilbert-Samuel polynomial is \( {H}_{M}\left( X\right) = \left( \begin{matrix} X + n - 1 \\ n - 1 \end{matrix}\right) \in \mathbb{Q}\left\lbrack X\right\rbrack \) . | Yes |
Lemma 5.4.4. Suppose \( R \) is Noetherian. The following are equivalent for a finitely generated \( R \) -module \( M \neq \{ 0\} \).\n\n(i) \( \dim M = 0 \).\n\n(ii) \( R/\operatorname{ann}\left( M\right) \) is Artinian.\n\n(iii) M has finite length. | Proof. (i) \( \Leftrightarrow \) (ii) is already known: recall that a Noetherian ring is Artinian if and only if its prime ideals are all maximal (Corollary 1.4.3). Let us show (i) or (ii) \( \Rightarrow \) (iii). By writing \( M = {M}_{1} + \cdots + {M}_{n} \) where each \( {M}_{i} \) is generated by one element, we m... | Yes |
Lemma 5.4.7. An ideal \( I \subset \operatorname{rad}\left( R\right) \) is a parameter ideal for \( M \) if and only if there exists \( k \) with \( \operatorname{rad}{\left( R\right) }^{k} \subset I + \operatorname{ann}\left( M\right) \) . In this case \( R/\left( {I + \operatorname{ann}\left( M\right) }\right) \) is ... | Proof. First we claim that \( V\left( {\operatorname{ann}\left( {M/{IM}}\right) }\right) = \operatorname{Supp}\left( {M/{IM}}\right) \) equals \( V\left( {I + \operatorname{ann}\left( M\right) }\right) \) . By the exactness of localizations together with Nakayama's Lemma, we have\n\n\[ \operatorname{Supp}\left( {M/{IM}... | Yes |
Lemma 5.4.8. Let \( M \neq \{ 0\} \) be a finitely generated \( R \) -module with parameter ideal \( I \) .\n\n(i) The degree \( d\left( M\right) \) of \( {H}_{I}\left( {M, \cdot }\right) \) is independent of the choice of the parameter ideal \( I \) .\n\n(ii) In a short exact sequence \( 0 \rightarrow {M}^{\prime } \r... | Proof. (i): To compare the graded objects associated to two parameter ideals \( I, J \), we apply the characterization in Lemma 5.4.7: it suffices to take \( J = \operatorname{rad}\left( R\right) \), so that\n\n\[ {J}^{m} + \operatorname{ann}\left( M\right) \subset I + \operatorname{ann}\left( M\right) \subset J + \ope... | Yes |
For any finitely generated nonzero \( R \) -module \( M \), we have \( \dim M = s\left( M\right) = \) \( d\left( M\right) \) . In particular \( \dim M \) is finite. | Proof. We argue inductively on \( d\left( M\right) \) to show \( \dim M \leq d\left( M\right) \) . If \( d\left( M\right) = 0 \) then \( {I}^{n}M = \) \( {I}^{n + 1}M = \cdots \) for \( n \gg 0 \) . Corollary 4.5.9 implies \( {I}^{n}M = \{ 0\} \), hence \( M = M/{I}^{n}M \) has finite length and \( \dim M = 0 \) by Lem... | No |
Lemma 5.4.8 (ii) implies that \( d\left( {R/{Rt} + \mathfrak{p}}\right) < d\left( {R/\mathfrak{p}}\right) \) . | By induction we deduce \( d\left( {R/\mathfrak{p}}\right) > \) \( d\left( {R/{Rt} + \mathfrak{p}}\right) \geq \dim \left( {R/{Rt} + \mathfrak{p}}\right) \geq m - 1 \), hence \( d\left( {R/\mathfrak{p}}\right) \geq m \) as required. | No |
Corollary 5.4.10. Under the same assumptions, we have \( {\dim }_{R}M = {\dim }_{\widehat{R}}\widehat{M} \), where we take I-adic completions. | Proof. Proposition 5.2.1 gives identifications \( {\operatorname{gr}}_{I}\left( R\right) = {\operatorname{gr}}_{\widehat{I}}\left( \widehat{R}\right) \) and \( {\operatorname{gr}}_{I}\left( M\right) = {\operatorname{gr}}_{\widehat{I}}\left( \widehat{M}\right) \) ; moreover \( \widehat{R} \) is still semi-local and \( \... | Yes |
Theorem 5.5.1 (Krull). Suppose \( \mathfrak{a} = \left( {{t}_{1},\ldots ,{t}_{r}}\right) \) is a proper ideal of \( R \), then for every minimal over-prime ideal \( \mathfrak{p} \) of \( \mathfrak{a} \), we have \( \operatorname{ht}\left( \mathfrak{p}\right) \leq r \) . | Proof. We work in \( {R}_{\mathfrak{p}} \) and \( I \mathrel{\text{:=}} \mathfrak{a}{R}_{\mathfrak{p}} = \left( {{t}_{1},\ldots ,{t}_{r}}\right) \) . Since \( I \subset \operatorname{rad}\left( {R}_{\mathfrak{p}}\right) \) and \( {R}_{\mathfrak{p}}/I \) has dimension zero, thus Artinian, \( I \) is a parameter ideal. W... | No |
Proposition 5.5.2. Suppose \( R \) is a Noetherian local ring with a system of parameters \( {t}_{1},\ldots ,{t}_{d} \) , which generate a parameter ideal \( I \) . For any \( 0 \leq i \leq d \) we have \( \dim \left( {R/\left( {{t}_{1},\ldots ,{t}_{i}}\right) }\right) = d - i \) , and \( {t}_{i + 1},\ldots ,{t}_{d} \)... | Proof. Consider the sequence \( R, R/\left( {t}_{1}\right), R/\left( {{t}_{1},{t}_{2}}\right) ,\ldots, R/\left( {{t}_{1},\ldots ,{t}_{d}}\right) \) . Recall the formalism in Theorem 5.4.9: at each stage \( d\left( {R/\cdots }\right) \) drops at most by one, by \( \left( {5 - 3}\right) \) . After \( d \) steps we arrive... | Yes |
For any Noetherian local ring \( R \) with maximal ideal \( \mathfrak{m} \) and residue field \( \mathbb{k} \), we have \( \dim R \leq {\dim }_{\mathbb{k}}\mathfrak{m}/{\mathfrak{m}}^{2} \) . Equality holds if and only if \( R \) is a regular local ring. | Proof. By Nakayama’s Lemma (more precisely, Corollary 1.3.6), \( \mathfrak{m}/{\mathfrak{m}}^{2} \) can be generated over \( \mathbb{k} \) by \( s \) elements if and only if \( \mathfrak{m} \) can be generated over \( R \) by \( s \) elements, for any \( s \in {\mathbb{Z}}_{ \geq 0} \) . Hence Theorem 5.4.9 imposes the... | Yes |
Theorem 5.5.6. Regular local rings are integral domains. | Proof. Induction on \( d \mathrel{\text{:=}} \dim R \) . If \( \dim R = 0 \) then \( \mathfrak{m} = \{ 0\} \), hence \( R \) is a field. Assume hereafter that \( d > 0 \) . We know there are only finitely many minimal prime ideals and \( {\dim }_{\mathbb{k}}\mathfrak{m}/{\mathfrak{m}}^{2} \geq 1 \) . By prime avoidance... | Yes |
Proposition 6.1.1. Assume \( A, B \) to be Noetherian. Let \( \mathfrak{q} \in \operatorname{Spec}\left( B\right) \) and \( \mathfrak{p} \mathrel{\text{:=}} {\varphi }^{\sharp }\left( \mathfrak{q}\right) \) . We have\n\n(i) \( \dim \left( {B}_{\mathfrak{q}}\right) \leq \dim \left( {A}_{\mathfrak{p}}\right) + \dim {B}_{... | Proof. We have an induced local homomorphism \( {A}_{\mathfrak{p}} \rightarrow {B}_{\mathfrak{q}} \) since \( \mathfrak{q} \mapsto \mathfrak{p} \) . Since (i) and (ii) depend only on this induced homomorphism, we may assume from the outset that \( A, B \) are local with maximal ideals \( \mathfrak{p},\mathfrak{q} \), a... | Yes |
Proposition 6.1.3. Suppose \( B \) is integral over a subring \( A \) . (i) We have \( \dim A = \dim B \) . | Proof. Going-up holds and \( \operatorname{Spec}\left( B\right) \rightarrow \operatorname{Spec}\left( A\right) \) in the situation of (i) by Theorem 4.1.4, hence \( \dim B \geq \dim A \) by lifting prime chains. To prove \( \leq \), observe that \( \mathfrak{q} \subsetneq {\mathfrak{q}}^{\prime } \) implies \( \mathfra... | Yes |
Theorem 6.2.1. Let \( A \) be a Noetherian ring, we have \( \dim A\left\lbrack {{X}_{1},\ldots ,{X}_{n}}\right\rbrack = \dim A + n \) for any \( n \geq 0 \). In particular, \( \dim A\left\lbrack {{X}_{1},\ldots ,{X}_{n}}\right\rbrack = n \) if \( A \) is Artinian (eg. a field). | Proof. Evidently we may assume \( n = 1 \). We shall apply Proposition 6.1.1 to \( A \hookrightarrow B = \) \( A\left\lbrack X\right\rbrack \). Take any \( \mathfrak{p} \in \operatorname{Spec}\left( A\right) \) and let \( \mathfrak{q} \) be a maximal element in \( \left\{ {{\mathfrak{q}}^{\prime } \in \operatorname{Spe... | Yes |
Corollary 6.2.2. Let \( \mathbb{k} \) be a field, then for every \( 0 \leq i \leq n \) we have \( \operatorname{ht}\left( {{X}_{1},\ldots ,{X}_{i}}\right) = i \) in \( \mathbb{k}\left\lbrack {{X}_{1},\ldots ,{X}_{n}}\right\rbrack \) | Proof. The prime chain \( \{ 0\} \subset \left( {X}_{1}\right) \subsetneq \cdots \subsetneq \left( {{X}_{1},\ldots ,{X}_{n}}\right) \) has length \( n = \dim \mathbb{k}\left\lbrack {{X}_{1},\ldots ,{X}_{n}}\right\rbrack \) . Thus for each \( 0 \leq i \leq n \), the chain \( \{ 0\} \subset \left( {X}_{1}\right) \subsetn... | Yes |
Corollary 6.2.3. Let \( \mathbb{k} \) be a field. Any \( \mathbb{k} \) -algebra A with \( n \) generators has finite dimension \( \leq n \) . | Proof. Writing \( A = \mathbb{k}\left\lbrack {{X}_{1},\ldots ,{X}_{n}}\right\rbrack /I \) for some ideal \( I \), we have \( \dim A \leq \mathbb{k}\left\lbrack {{X}_{1},\ldots ,{X}_{n}}\right\rbrack \) since every prime chain in \( A \) lifts to \( \mathbb{k}\left\lbrack {{X}_{1},\ldots ,{X}_{n}}\right\rbrack \) . Now ... | Yes |
Lemma 6.3.1. Suppose \( \mathbb{k} \) is a field and \( t \in \mathbb{k}\left\lbrack {{X}_{1},\ldots {X}_{e}}\right\rbrack \smallsetminus \mathbb{k} \). There exist \( {t}_{1},\ldots ,{t}_{e - 1} \in \) \( \mathbb{k}\left\lbrack {{X}_{1},\ldots ,{X}_{e}}\right\rbrack \) such that \( \mathbb{k}\left\lbrack {{X}_{1},\ldo... | Proof. We seek \( {t}_{i} \) of the form \( {X}_{i} - {X}_{e}^{{k}^{i}} \) where \( k \) is a large integer. Then \( t \) can be uniquely expressed as a polynomial of \( {t}_{1},\ldots ,{t}_{e - 1},{X}_{e} \). We claim that upon modifying \( t \) by \( {\mathbb{k}}^{ \times } \), which is clearly harmless, one can choo... | Yes |
Lemma 6.3.2. Let \( \mathfrak{a} \) be a nonzero ideal of a domain \( R \), then \( \dim \left( {R/\mathfrak{a}}\right) + 1 \leq \dim R \) . | Proof. Any prime chain \( {\mathfrak{p}}_{0} \supsetneq \cdots \supsetneq {\mathfrak{p}}_{n} \) in \( R \) with \( {\mathfrak{p}}_{n} \supset \mathfrak{a} \) can be extended to a prime chain length \( n + 1 \), namely by adjoining \( {\mathfrak{p}}_{n + 1} \mathrel{\text{:=}} \{ 0\} \) . | Yes |
Corollary 6.3.4 (Dimension formula). Let B be a finitely generated algebra over a field \( \mathbb{k} \) . Suppose \( B \) is a domain, then for all \( \mathfrak{q} \in \operatorname{Spec}\left( B\right) \) we have\n\n\[ \dim \left( {B/\mathfrak{q}}\right) + \operatorname{ht}\left( \mathfrak{q}\right) = \dim B \] | Proof. Choose a subalgebra \( A \) of \( B \) as in Theorem 6.3.3 and put \( \mathfrak{p} = \mathfrak{q} \cap A \) . Since \( B \) is a domain and \( A \) is normal, the Cohen-Seidenberg Theorem 4.1.4 asserts the going-down property for \( A \hookrightarrow B \) . Proposition 6.1.3 implies that \( \dim A = \dim B,\dim ... | Yes |
Corollary 6.3.5. Suppose \( B \) is a domain finitely generated over a field \( \mathbb{k} \) . Set \( L \mathrel{\text{:=}} \operatorname{Frac}\left( B\right) \) . Then \( \dim B = {\operatorname{tr.deg}}_{\mathbb{k}}\left( L\right) \) and it is the common length of maximal prime chains. | Proof. Choose a subalgebra \( A \) of \( B \) as in Theorem 6.3.3. Since the field \( \operatorname{Frac}\left( B\right) \) is a finite extension of \( \operatorname{Frac}\left( A\right) \) and \( \dim A = \dim B \) by Proposition 6.1.3, the first assertion reduces immediately to the case \( B = \mathbb{k}\left\lbrack ... | Yes |
Lemma 7.1.3. Let \( R \) be a discrete valuation ring with valuation \( v \) and uniformizer \( t \) . Every ideal \( \mathfrak{a} \neq \{ 0\} \) of \( R \) has the form \( \left( {t}^{r}\right) \) for a unique \( r \geq 0 \) . In particular, \( R \) is a local principal ideal domain which is not a field, hence is of d... | Proof. Take \( r \mathrel{\text{:=}} \min \{ v\left( x\right) : x \in \mathfrak{a}\} \) . | No |
Proposition 7.1.6. Suppose \( t \) is a regular parameter in a regular local ring \( R \) of dimension one, then \( R \) is a domain, and every element \( x \in R \smallsetminus \{ 0\} \) can be uniquely written as \( x = {t}^{r}u \) with \( r \geq 0 \) and \( u \in {R}^{ \times } \). This makes \( R \) into a discrete... | Proof. From Theorem 5.5.6 we know regular local rings are Noetherian domains. By applying Krull’s Intersection Theorem (Corollary 4.5.10) to the powers of \( \left( t\right) \), we see that \( r \mathrel{\text{:=}} \sup \left\{ {k \geq 0 : x \in {\left( t\right) }^{k}}\right\} \) is finite. Write \( x = {t}^{r}u \). Si... | Yes |
Lemma 7.2.1. Let \( M \) be a finitely generated \( R \) -module. An element \( x \in M \) is zero if and only if its image in \( {M}_{\mathfrak{p}} \) is zero for every maximal element \( \mathfrak{p} \) in \( \operatorname{Ass}\left( M\right) \) . | Proof. Suppose \( x \neq 0 \) . Since \( M \) is Noetherian, among ideals of the form ann \( \left( y\right) \) there is a maximal one containing ann \( \left( x\right) \), and we have seen in Lemma 2.2.3 that such an ideal \( \mathfrak{p} \) belongs to \( \operatorname{Ass}\left( M\right) \) . Since \( \operatorname{a... | Yes |
Lemma 7.2.2. Suppose \( R \) is reduced, then \( \operatorname{Ass}\left( R\right) \) consists of minimal primes. | Proof. As \( R \) is reduced, \( \{ 0\} = \sqrt{{0}_{R}} \) is the intersection of minimal prime ideals \( {\mathfrak{p}}_{1},{\mathfrak{p}}_{2},\ldots \) (all lying in Ass \( \left( R\right) \), hence finite in number). By the theory of primary decompositions, one infers that \( \operatorname{Ass}\left( R\right) = \le... | No |
Lemma 7.2.3. Let \( R \) be reduced. Then \( K\left( R\right) \overset{ \sim }{ \rightarrow }\mathop{\prod }\limits_{\mathfrak{p}}K\left( {R/\mathfrak{p}}\right) \) as \( R \) -algebras, where \( \mathfrak{p} \) ranges over the minimal prime ideals of \( R \) . For any multiplicative subset \( S \subset R \) there is a... | Proof. Each element in \( K\left( R\right) = R\left\lbrack {T}^{-1}\right\rbrack \) is either a zero-divisor or invertible. The set of zero-divisors of \( K\left( R\right) \) is the union of minimal prime ideals \( {\mathfrak{p}}_{i}K\left( R\right) \) of \( K\left( R\right) \) (where \( \operatorname{Ass}\left( R\righ... | Yes |
Lemma 7.2.4. Suppose \( R \) is reduced. Then \( x \in K\left( R\right) \) belongs to \( R \) if and only if its image in \( K{\left( R\right) }_{\mathfrak{p}} = K\left( {R}_{\mathfrak{p}}\right) \) belongs to \( {R}_{\mathfrak{p}} \) for every prime \( \mathfrak{p} \) associated to a non zero-divisor. | Proof. Only the \ | No |
Proposition 7.3.2. Let \( R \) be a Noetherian domain. Then \( R \) is normal if and only if for every principal ideal \( \left( t\right) \subset R \) and every \( \mathfrak{p} \in \operatorname{Ass}\left( {R/\left( t\right) }\right) \), the ideal \( \mathfrak{p}{R}_{\mathfrak{p}} \) is principal. | Proof. Assume the conditions above. To prove the normality of \( R \), it suffices to use \( R = \bigcap {R}_{\mathfrak{p}} \) where \( \mathfrak{p} \) ranges over the primes associated to nonzero principal ideals (consequence of Lemma 7.2.4). Indeed, each \( {R}_{\mathfrak{p}} \) is regular, hence normal by Propositio... | Yes |
Corollary 7.3.3. The following are equivalent for a local domain \( R \) :\n\n(i) \( R \) is normal of dimension 1;\n\n(ii) \( R \) is a regular local ring of dimension 1 ;\n\n(iii) \( R \) is a discrete valuation ring. | Proof. (iii) \( \Rightarrow \) (i). We have seen that discrete valuation rings are principal ideal rings of dimension 1, therefore also normal by unique factorization property.\n\n(i) \( \Rightarrow \) (ii). Under the normality assumption, choose any \( t \in R \smallsetminus \{ 0\} \) . Since \( \dim R = 1 \) and \( \... | Yes |
Corollary 7.3.4. Let \( R \) be a Noetherian normal domain. Then\n\n\[ R = \mathop{\bigcap }\limits_{{\mathrm{{ht}}\left( \mathfrak{p}\right) = 1}}{R}_{\mathfrak{p}} \]\n\ninside \( \operatorname{Frac}\left( R\right) \) . | Proof. Evidently \( \subset \) holds. By Lemma 7.2.4 together with Proposition 7.3.2, \( R \) can be written as an intersection of \( {R}_{\mathfrak{p}} \) where \( \mathfrak{p} \) is associated to some non zero-divisor, such that \( \mathfrak{p}{R}_{\mathfrak{p}} \) is principal; it suffices to show \( \operatorname{h... | Yes |
Lemma 7.4.1. Let \( R \) be a Noetherian ring. Suppose that\n\n- the primes in \( \operatorname{Ass}\left( R\right) \) are all minimal, and\n\n- \( {R}_{\mathfrak{p}} \) is a field for every minimal prime ideal \( \mathfrak{p} \) ,\nthen \( R \) is reduced. | Proof. Take a minimal primary decomposition \( \{ 0\} = {I}_{1} \cap \cdots \cap {I}_{m} \) with \( \operatorname{Ass}\left( {R/{I}_{j}}\right) = \) \( \left\{ {{\mathfrak{p}}_{j} = \sqrt{{I}_{j}}}\right\} \) and \( \operatorname{Ass}\left( R\right) = \left\{ {{\mathfrak{p}}_{1},\ldots ,{\mathfrak{p}}_{m}}\right\} \) .... | Yes |
Theorem 7.4.2 (J.-P. Serre). A Noetherian ring \( R \) is a finite direct product of normal domains if and only if the following two conditions hold.\n\nD R1 The localization of \( R \) at every prime ideal of height 1 (resp. 0) is a discrete valuation ring (resp. a field).\n\nD S2 For every non zero-divisor \( t \) of... | Proof. We begin with the \( \Rightarrow \) direction. Suppose \( R = {R}_{1} \times \cdots \times {R}_{n} \) where each \( {R}_{i} \) is a normal domain. As is well-known, \( \operatorname{Spec}\left( R\right) = \mathop{\bigsqcup }\limits_{{i = 1}}^{n}\operatorname{Spec}\left( {R}_{i}\right) \) as topological spaces: t... | Yes |
Corollary 7.4.5. A Noetherian domain \( R \) is normal if and only if \( \mathbf{R1} \) and S2 hold for \( R \) . | Proof. Immediate from the previous exercise and Theorem 7.4.2. | No |
Proposition 7.5.2. For \( I, M \) as above, the following are equivalent:\n\n(i) \( {\operatorname{depth}}_{I}\left( M\right) = 0 \) ;\n\n(ii) for all \( x \in I \), the homomorphism \( M\overset{x}{ \rightarrow }M \) is not injective;\n\n(iii) \( \operatorname{Ass}\left( M\right) \cap V\left( I\right) \neq \varnothing... | Proof. In each case we have \( M \neq \{ 0\} \) . If (i) holds, then \( M\overset{x}{ \rightarrow }M \) vanishes on the image of some nonzero \( R/I \rightarrow M \), hence (ii). If (ii) holds, the union of \( \operatorname{Ass}\left( M\right) \) will cover \( I \) , and (iii) follows by prime avoidance. Finally, suppo... | Yes |
Lemma 7.5.4. Let \( M \) be an R-module, \( {x}_{1},\ldots ,{x}_{r} \) be an \( M \) -regular sequence lying in an ideal \( I \subsetneq R \) . We have \( {\operatorname{depth}}_{I}\left( M\right) = r + {\operatorname{depth}}_{I}\left( {M/\left( {{x}_{1},\ldots ,{x}_{r}}\right) M}\right) \) . | Proof. The case \( r = 1 \) follows by staring at the long exact sequence attached to \( 0 \rightarrow \) \( M\overset{{x}_{1}}{ \rightarrow }M \rightarrow M/{x}_{1}M \rightarrow 0 \) . The general case follows by induction on \( r \) . | No |
Theorem 7.5.5. Assume \( R \) Noetherian, \( M \) finitely generated and \( I \subsetneq R \) .\n\n(i) \( {\operatorname{depth}}_{I}\left( M\right) \) is the supremum of the lengths of \( M \) -regular sequences with elements in \( I \) .\n\n(ii) Suppose \( {\operatorname{depth}}_{I}\left( M\right) < + \infty \) . Ever... | Proof. To prove (i) and (ii), by the previous Lemma we are reduced to show that \( {\operatorname{depth}}_{I}\left( M\right) > 0 \) implies the existence of \( x \in I \) which is not a zero-divisor of \( M \) ; this follows from Proposition 7.5.2.\n\nNow pass to the word \ | No |
Corollary 7.5.6. With the same assumptions, let \( \left( {{x}_{1},\ldots ,{x}_{r}}\right) \) be an \( M \) -regular sequence with \( {x}_{i} \in I \) . It is of length \( {\operatorname{depth}}_{I}\left( M\right) \) if and only if \( \operatorname{Ass}\left( {M/\left( {{x}_{1},\ldots ,{x}_{r}}\right) M}\right) \cap V\... | Proof. The sequence has length \( {\operatorname{depth}}_{I}\left( M\right) \) if and only if \( R \) -module \( M/\left( {{x}_{1},\ldots ,{x}_{r}}\right) M \) has depth zero, so it remains to apply Proposition 7.5.2. | Yes |
Corollary 7.5.7. Let \( R \) be a Noetherian local ring with maximal ideal \( \mathfrak{m} \), and \( M \neq \{ 0\} \) a finitely generated \( R \) -module, then \( {\operatorname{depth}}_{\mathfrak{m}}\left( M\right) \leq \dim M \) . | Proof. Consider the following situation: \( x \in \mathfrak{m} \) is not a zero-divisor for \( M \neq \{ 0\} \) . In the discussion of dimensions, we have seen that \( d\left( M\right) \geq d\left( {M/{xM}}\right) \geq d\left( {M/{xM}}\right) - 1 \) , where \( d\left( \cdot \right) \) is the degree of Hilbert-Samuel po... | Yes |
Proposition 8.1.2 (Homotopy formula). For any \( x \in L \) and \( \omega \in \mathop{\bigwedge }\limits^{n}L \), we have\n\n\[ \n{i}_{u}\left( {x \land \omega }\right) + x \land \left( {{i}_{u}\left( \omega \right) }\right) = u\left( x\right) \omega .\n\] | Proof. Consider \( \omega = {x}_{1} \land \cdots \land {x}_{n} \) . Put \( {x}_{0} \mathrel{\text{:=}} x \) . The left-hand side equals\n\n\[ \n\mathop{\sum }\limits_{{i = 0}}^{n}{\left( -1\right) }^{i} \cdot u\left( {x}_{i}\right) \cdots \land \widehat{{x}_{i}} \land \cdots \n\]\n\nwhereas the right-hand side equals\n... | Yes |
Proposition 8.1.3. Set \( \mathfrak{q} \mathrel{\text{:=}} u\left( L\right) \), which is an ideal of \( R \) . Then \( \mathfrak{q} \) annihilates each homology (resp. cohomology) of \( {K}_{ \bullet }\left( {u;M}\right) \) (resp. \( {K}^{ \bullet }\left( {u;M}\right) \) ). Again, this generalizes to general complexes ... | Proof. Given \( t \in \mathfrak{q} \), the homotopy formula implies that the endomorphism \( \omega \mapsto {t\omega } \) of \( {K}_{ \bullet }\left( u\right) \) is homotopic to zero, hence so are the induced endomorphisms of \( {K}_{ \bullet }\left( {u;M}\right) \) and \( {K}^{ \bullet }\left( {u;M}\right) \) by stand... | Yes |
Proposition 8.1.4. Suppose \( L \) is projective over \( R \) and \( 0 \rightarrow {M}^{\prime } \rightarrow M \rightarrow {M}^{\prime \prime } \rightarrow 0 \) is exact. Then there is a natural short exact sequence of complexes\n\n\[ 0 \rightarrow {K}^{ \bullet }\left( {u;{M}^{\prime }}\right) \rightarrow {K}^{ \bulle... | Proof. Standard. It suffices to note that \( L \) is projective implies each graded piece \( \mathop{\bigwedge }\limits^{n}L \) of \( \bigwedge L \) is projective as well. | No |
Proposition 8.2.1. For every family \( {\left\{ {M}_{\beta }\right\} }_{\beta \in \mathcal{B}} \) of R-modules, we have\n\n\[ \n{\operatorname{depth}}_{I}\left( {\mathop{\prod }\limits_{\beta }{M}_{\beta }}\right) = \mathop{\inf }\limits_{\beta }{\operatorname{depth}}_{I}\left( {M}_{\beta }\right) \n\] | Proof. This follows from \( {\operatorname{Ext}}_{R}^{n}\left( {R/I,\mathop{\prod }\limits_{i}{M}_{i}}\right) = \mathop{\prod }\limits_{{i \in I}}{\operatorname{Ext}}_{R}^{n}\left( {R/I,{M}_{i}}\right) \), as is easily seen by taking a projective resolution of \( R/I \) and using the fact the \( {\mathrm{{Hom}}}_{R} \)... | Yes |
Proposition 8.2.3. Suppose \( N \) is an R-module annihilated by some \( {I}^{m} \), where \( m \geq 1 \) . Then \( {\operatorname{Ext}}_{R}^{i}\left( {N, M}\right) = 0 \) whenever \( i < {\operatorname{depth}}_{I}\left( M\right) . | Proof. To show \( {\operatorname{Ext}}_{R}^{i}\left( {N, M}\right) = 0 \) for \( i < {\operatorname{depth}}_{I}\left( M\right) \), we begin with the case \( m = 1 \) . This case follows by a dimension-shifting argument based on the short exact sequence\n\n\[ 0 \rightarrow K \rightarrow {\left( R/I\right) }^{\oplus I} \... | Yes |
Lemma 8.2.4. Suppose \( J \) is an ideal satisfying \( J \supset {I}^{m} \) for some \( m \geq 1 \) . Then \( {\operatorname{depth}}_{I}\left( M\right) \leq \) \( {\operatorname{depth}}_{J}\left( M\right) \) . | Proof. From the previous Proposition, we have \( {\operatorname{Ext}}_{R}^{i}\left( {R/J, M}\right) = 0 \) for \( i < {\operatorname{depth}}_{I}\left( M\right) \) since \( {I}^{m} \) annihilates \( R/J \) . The assertion follows upon recalling the definition of depth. | No |
Proposition 8.2.5. Let \( {C}^{ \bullet } \) be a cochain complex of \( R \) -modules such that \( n \ll 0 \Rightarrow {C}^{n} = \) 0 . Let \( h \in \mathbb{Z} \) such that for all integers \( n \leq k \leq h \), the depth of \( {C}^{n} \) with respect to \( {J}_{k} \mathrel{\text{:=}} \) \( \operatorname{ann}\left( {{... | Proof. Assume on the contrary that there exists \( k \leq h \) with \( {H}^{ < k} = 0 \) whereas \( {H}^{k} \neq 0 \) . Write \( J = {J}_{k} \) . As \( J \) annihilates the nonzero \( R \) -module \( {H}^{k} \), the criterion of depth-zero modules (Proposition 7.5.2) implies that \( {\operatorname{depth}}_{I}\left( {H}... | Yes |
Corollary 8.2.6. Let \( I \subset R \) be an ideal, \( {C}^{ \bullet } \) be a cochain complex with \( n \ll 0 \Rightarrow {C}_{n} = 0 \) , and \( h \in \mathbb{Z} \) . Suppose that \( n \leq h \) implies \( I \cdot {H}^{n}\left( {C}^{ \bullet }\right) = 0 \) and \( {\operatorname{depth}}_{I}\left( {C}^{n}\right) > h -... | Proof. For \( k \leq h \) we have \( {J}_{k} \mathrel{\text{:=}} \operatorname{ann}\left( {{H}^{k}\left( {C}^{ \bullet }\right) }\right) \supset I \) . Hence Lemma 8.2.4 entails that\n\n\[ \n n \leq k \leq h \Rightarrow {\operatorname{depth}}_{{J}_{k}}\left( {C}^{n}\right) \geq {\operatorname{depth}}_{I}\left( {C}^{n}\... | Yes |
Lemma 8.3.1. Each cohomology of \( {K}^{ \bullet }\left( {\mathbf{x};M}\right) \) is annihilated by \( I \) . | Proof. Apply Proposition 8.1.3 by observing that \( \mathfrak{q} = I \) in our setting. | No |
For each \( n \geq 1 \), the following estimates hold for \( {\theta }_{n} \), \[ {\theta }_{n} < \frac{1}{{\alpha }_{n}} < \frac{1}{n\pi } \] \[ {\theta }_{n} < \frac{1}{{n\pi } + \frac{\pi }{2}}\left( {1 + {\theta }_{n}^{2}}\right) \] \[ {\theta }_{n} < \frac{{2n} + 1}{4}\pi - \sqrt{{\left( \frac{{2n} + 1}{4}\pi \rig... | Proof. Since \( {\alpha }_{n} \) is a solution for the equation \( \varphi \left( t\right) = 0 \), we have \[ \sin {\alpha }_{n} = {\alpha }_{n} \cdot \cos {\alpha }_{n} \] Using \( {\alpha }_{n} = {n\pi } + \frac{\pi }{2} - {\theta }_{n} \), we have \[ 1 = {\alpha }_{n} \cdot \tan {\theta }_{n} > {\alpha }_{n} \cdot {... | No |
For each \( n \geq 1 \), we have the following estimate\n\n\[ \n{\theta }_{n} > \sin {\theta }_{n} > \frac{1}{{n\pi } + \frac{\pi }{2}}. \n\] | Proof. Let \( {\beta }_{n} = {n\pi } + \frac{\pi }{2} - {\eta }_{n} \), where \( {\eta }_{n} \in \left( {0,\frac{\pi }{2}}\right) \) satisfies\n\n\[ \n\sin {\eta }_{n} = \frac{1}{{n\pi } + \frac{\pi }{2}} \n\]\n\nWe have\n\n\[ \n{\left( -1\right) }^{n} \cdot \varphi \left( {\beta }_{n}\right) = \cos {\eta }_{n} - \left... | Yes |
For each \( n \geq 1 \), we have the following estimate\n\n\[ 0 < {\theta }_{n} - {\theta }_{n + 1} < \frac{\pi }{{\alpha }_{n} \cdot {\alpha }_{n + 1}}. \] | Proof. By (1.4) we have\n\n\[ \tan {\theta }_{n} = \frac{1}{{\alpha }_{n}} > \frac{1}{{\alpha }_{n + 1}} = \tan {\theta }_{n + 1}, \]\n\nwhich yields \( {\theta }_{n} > {\theta }_{n + 1} \) .\n\nOn the other hand,\n\n\[ \tan \left( {{\theta }_{n} - {\theta }_{n + 1}}\right) = \frac{\tan {\theta }_{n} - \tan {\theta }_{... | Yes |
For \( \theta \in \left( {0,\frac{\pi }{2}}\right) \), we have\n\n\[ \sin \theta - \theta \cos \theta < \frac{1}{3}{\theta }^{3} \] | Proof. Let \( p\left( \theta \right) = \sin \theta - \theta \cos \theta - \frac{1}{3}{\theta }^{3},\theta \in \left( {0,\frac{\pi }{2}}\right) \) . We have\n\n\[ {p}^{\prime }\left( \theta \right) = \theta \sin \theta - {\theta }^{2} < 0, \]\n\nthus \( p\left( \theta \right) < p\left( 0\right) = 0 \) . | Yes |
For \( x, y \in \left\lbrack {\frac{1}{{\alpha }_{n + 1}},\frac{1}{{\alpha }_{n}}}\right\rbrack ,\left( {n > 1}\right) \), we have\n\n\[ \left| {f\left( y\right) - f\left( x\right) }\right| \leq \sqrt{2\left| {y - x}\right| }.\] | Proof. By Cauchy’s inequality, for \( x, y \in \left\lbrack {\frac{1}{{\alpha }_{n + 1}},\frac{1}{{\alpha }_{n}}}\right\rbrack \) we have\n\n\[ {\left| f\left( y\right) - f\left( x\right) \right| }^{2} = {\left| {\int }_{x}^{y}{f}^{\prime }\left( t\right) dt\right| }^{2} \]\n\n\[ \leq \left| {y - x}\right| \cdot \left|... | Yes |
Proposition 2.3. For \( x, y \in \left\lbrack {\frac{1}{\pi },\infty }\right) \), we have\n\n\[ \left| {f\left( y\right) - f\left( x\right) }\right| \leq \sqrt{2\left| {y - x}\right| }.\] | Proof. From (2.1) we see that \( f \) is concave in \( \left\lbrack {\frac{1}{\pi },\infty }\right) \), and \( \mathop{\max }\limits_{\left\lbrack \frac{1}{\pi },\infty \right) }{f}^{\prime } = \pi \) . First, we prove the following inequality\n\n\[ f\left( x\right) \leq \sqrt{2\left( {x - \frac{1}{\pi }}\right) },\;\f... | Yes |
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