The Stacks project

\begin{equation*} \DeclareMathOperator\Coim{Coim} \DeclareMathOperator\Coker{Coker} \DeclareMathOperator\Ext{Ext} \DeclareMathOperator\Hom{Hom} \DeclareMathOperator\Im{Im} \DeclareMathOperator\Ker{Ker} \DeclareMathOperator\Mor{Mor} \DeclareMathOperator\Ob{Ob} \DeclareMathOperator\Sh{Sh} \DeclareMathOperator\SheafExt{\mathcal{E}\mathit{xt}} \DeclareMathOperator\SheafHom{\mathcal{H}\mathit{om}} \DeclareMathOperator\Spec{Spec} \newcommand\colim{\mathop{\mathrm{colim}}\nolimits} \newcommand\lim{\mathop{\mathrm{lim}}\nolimits} \newcommand\Qcoh{\mathit{Qcoh}} \newcommand\Sch{\mathit{Sch}} \newcommand\QCohstack{\mathcal{QC}\!\mathit{oh}} \newcommand\Cohstack{\mathcal{C}\!\mathit{oh}} \newcommand\Spacesstack{\mathcal{S}\!\mathit{paces}} \newcommand\Quotfunctor{\mathrm{Quot}} \newcommand\Hilbfunctor{\mathrm{Hilb}} \newcommand\Curvesstack{\mathcal{C}\!\mathit{urves}} \newcommand\Polarizedstack{\mathcal{P}\!\mathit{olarized}} \newcommand\Complexesstack{\mathcal{C}\!\mathit{omplexes}} \newcommand\Pic{\mathop{\mathrm{Pic}}\nolimits} \newcommand\Picardstack{\mathcal{P}\!\mathit{ic}} \newcommand\Picardfunctor{\mathrm{Pic}} \newcommand\Deformationcategory{\mathcal{D}\!\mathit{ef}} \end{equation*}

15.89 Miscellany

Some results which do not fit anywhere else.

Lemma 15.89.1. Let $A$ be a Noetherian ring. Let $I \subset A$ be an ideal. Let $M$, $N$ be finite $A$-modules with $N$ annihilated by $I$. For each $p > 0$ there exists an $n$ such that the map $\mathop{\mathrm{Ext}}\nolimits _ A^ p(M, N) \to \text{Ext}_ A^ p(I^ nM, N)$ is zero.

Proof. The result is clear for $p = 0$ (with $n = 1$). Choose a short exact sequence $0 \to K \to A^{\oplus t} \to M \to 0$. For $n$ pick a short exact sequence $0 \to L \to A^{\oplus s} \to I^ nM \to 0$. It is clear that we can construct a map of short exact sequences

\[ \xymatrix{ 0 \ar[r] & L \ar[r] \ar[d] & A^{\oplus s} \ar[r] \ar[d] & I^ nM \ar[r] \ar[d] & 0 \\ 0 \ar[r] & K \ar[r] & A^{\oplus s} \ar[r] & M \ar[r] & 0 } \]

such that $A^{\oplus s} \to A^{\oplus t}$ has image in $(I^ n)^{\oplus t}$. By Artin-Rees (Algebra, Lemma 10.50.2) we see that $L \to K$ has image contained in $I^{n - c}K$ if $n \geq c$. At this point the exact sequence

\[ \mathop{\mathrm{Hom}}\nolimits _ A(A^{\oplus t}, N) \to \mathop{\mathrm{Hom}}\nolimits _ A(K, N) \to \mathop{\mathrm{Ext}}\nolimits ^1_ A(M, N) \to 0 \]

and the corresponding sequence for $\mathop{\mathrm{Ext}}\nolimits ^1_ A(I^ nM, N)$ show that the lemma holds for $p = 1$ with $n = c + 1$. Moreover, we see that the result for $p - 1$ and the module $K$ implies the result for $p$ and the module $M$ by the commutativity of the diagram

\[ \xymatrix{ & \mathop{\mathrm{Ext}}\nolimits ^{p - 1}_ A(L, N) \ar[r]_{\cong } \ar[ld] & \mathop{\mathrm{Ext}}\nolimits _ A^ p(I^ nM, N) \ar[d] \\ \mathop{\mathrm{Ext}}\nolimits ^{p - 1}_ A(I^{n - c}K, N) \ar[r] & \mathop{\mathrm{Ext}}\nolimits ^{p - 1}_ A(K, N) \ar[r] & \text{Ext}_ A^ p(M, N) } \]

for $p > 1$. Some details omitted. $\square$

Lemma 15.89.2. Let $A$ be a Noetherian ring. Let $I \subset A$ be an ideal. Let $K \in D(A)$ be pseudo-coherent and let $M$ be a finite $A$-module. For each $p \in \mathbf{Z}$ there exists an $c$ such that the image of $\mathop{\mathrm{Ext}}\nolimits _ A^ p(K, I^ nM) \to \text{Ext}_ A^ p(K, M)$ is contained in $I^{n - c}\text{Ext}_ A^ p(K, M)$ for $n \geq c$.

Proof. Choose a bounded above complex $P^\bullet $ of finite free $A$-modules representing $K$. Then $\text{Ext}_ A^ p(K, M)$ is the cohomology of

\[ \mathop{\mathrm{Hom}}\nolimits _ A(F^{-p + 1}, M) \xrightarrow {a} \mathop{\mathrm{Hom}}\nolimits _ A(F^{-p}, M) \xrightarrow {b} \mathop{\mathrm{Hom}}\nolimits _ A(F^{-p - 1}, M) \]

and $\mathop{\mathrm{Ext}}\nolimits _ A^ p(K, I^ nM)$ is computed by replacing these finite $A$-modules by $I^ n$ times themselves. Thus we want to prove $\mathop{\mathrm{Ker}}(b) \cap I^ n\mathop{\mathrm{Hom}}\nolimits _ A(F^{-p}, M) \subset I^{n - c}\mathop{\mathrm{Ker}}(b)$. This follows from Artin-Rees (Algebra, Lemma 10.50.2). $\square$

Lemma 15.89.3. Let $A$ be a Noetherian ring. Let $I \subset A$ be an ideal. Let $K \in D(A)$ be pseudo-coherent. Let $a \in \mathbf{Z}$. Assume that for every finite $A$-module $M$ the modules $\text{Ext}^ i_ A(K, M)$ are $I$-power torsion for $i \geq a$. Then for $i \geq a$ and $M$ finite the system $\text{Ext}^ i_ A(K, M/I^ nM)$ is essentially constant with value

\[ \text{Ext}^ i_ A(K, M) = \mathop{\mathrm{lim}}\nolimits \text{Ext}^ i_ A(K, M/I^ nM) \]

Proof. Let $M$ be a finite $A$-module. Since $K$ is pseudo-coherent we see that $\text{Ext}^ i_ A(K, M)$ is a finite $A$-module. Thus for $i \geq a$ it is annihilated by $I^ t$ for some $t \geq 0$. By Lemma 15.89.2 we see that the image of $\text{Ext}^ i_ A(K, I^ nM) \to \text{Ext}^ i_ A(K, M)$ is zero for some $n > 0$. The short exact sequence $0 \to I^ nM \to M \to M/I^ n M \to 0$ gives a long exact sequence

\[ \text{Ext}^ i_ A(K, I^ nM) \to \text{Ext}^ i_ A(K, M) \to \text{Ext}^ i_ A(K, M/I^ nM) \to \text{Ext}^{i + 1}_ A(K, I^ nM) \]

The systems $\text{Ext}^ i_ A(K, I^ nM)$ and $\text{Ext}^{i + 1}_ A(K, M/I^ nM)$ are essentially constant with value $0$ by what we just said (applied to the finite $A$-modules $I^ mM$). A diagram chase shows $\text{Ext}^ i_ A(K, M/I^ nM)$ is essentially constant with value $\text{Ext}^ i_ A(K, M)$. $\square$

Lemma 15.89.4. Let $A$ be a Noetherian ring. Let $I \subset A$ be an ideal. Let $M$ be a finite $A$-module. There exists an integer $n > 0$ such that $I^ nM \to M$ factors through the map $I \otimes _ A^\mathbf {L} M \to M$ in $D(A)$.

Proof. Consider the distinguished triangle

\[ I \otimes _ A^\mathbf {L} M \to M \to A/I \otimes _ A^\mathbf {L} M \to I \otimes _ A^\mathbf {L} M[1] \]

By the axioms of a triangulated category it suffices to prove that $I^ nM \to A/I \otimes _ A^\mathbf {L} M$ is zero in $D(A)$ for some $n$. Choose generators $f_1, \ldots , f_ r$ of $I$ and let $K = K_\bullet (A, f_1, \ldots , f_ r)$ be the Koszul complex and consider the factorization $A \to K \to A/I$ of the quotient map. Then we see that it suffices to show that $I^ nM \to K \otimes _ A M$ is zero in $D(A)$ for some $n > 0$. Suppose that we have found an $n > 0$ such that $I^ nM \to K \otimes _ A M$ factors through $\tau _{\geq t}(K \otimes _ A M)$ in $D(A)$. Then the obstruction to factoring through $\tau _{\geq t + 1}(K \otimes _ A M)$ is an element in $\mathop{\mathrm{Ext}}\nolimits ^ t(I^ nM, H_ t(K \otimes _ A M))$. The finite $A$-module $H_ t(K \otimes _ A M)$ is annihilated by $I$. Then by Lemma 15.89.1 we can after increasing $n$ assume this obstruction element is zero. Repeating this a finite number of times we find $n$ such that $I^ nM \to K \otimes _ A M$ factors through $0 = \tau _{\geq r + 1}(K \otimes _ A M)$ in $D(A)$ and we win. $\square$

Lemma 15.89.5. Let $R$ be a ring. Let $K \in D(R)$ be pseudo-coherent. Let $(M_ n)$ be an inverse system of $R$-modules. Then $R\mathop{\mathrm{lim}}\nolimits K \otimes _ R^\mathbf {L} M_ n = K \otimes _ R^\mathbf {L} R\mathop{\mathrm{lim}}\nolimits M_ n$.

Proof. Consider the defining distinguished triangle

\[ R\mathop{\mathrm{lim}}\nolimits M_ n \to \prod M_ n \to \prod M_ n \to R\mathop{\mathrm{lim}}\nolimits M_ n[1] \]

and apply Lemma 15.62.6. $\square$

Lemma 15.89.6. Let $R$ be a Noetherian local ring. Let $I \subset R$ be an ideal and let $E$ be a nonzero module over $R/I$. If $R/I$ has finite projective dimension and $E$ has finite projective dimension over $R/I$, then $E$ has finite projective dimension over $R$ and

\[ \text{pd}_ R(E) = \text{pd}_ R(R/I) + \text{pd}_{R/I}(E) \]

Proof. We will use that, for a finite module, having finite projective dimension over $R$, resp. $R/I$ is the same as being a perfect module, see discussion following Definition 15.69.1. We see that $E$ has finite projective dimension over $R$ by Lemma 15.69.7. Thus we can apply Auslander-Buchsbaum (Algebra, Proposition 10.110.1) to see that

\[ \text{pd}_ R(E) + \text{depth}(E) = \text{depth}(R),\quad \text{pd}_{R/I}(E) + \text{depth}(E) = \text{depth}(R/I), \]

and

\[ \text{pd}_ R(R/I) + \text{depth}(R/I) = \text{depth}(R) \]

Note that in the first equation we take the depth of $E$ as an $R$-module and in the second as an $R/I$-module. However these depths are the same (this is trivial but also follows from Algebra, Lemma 10.71.10). This concludes the proof. $\square$


Comments (0)


Post a comment

Your email address will not be published. Required fields are marked.

In your comment you can use Markdown and LaTeX style mathematics (enclose it like $\pi$). A preview option is available if you wish to see how it works out (just click on the eye in the toolbar).

Unfortunately JavaScript is disabled in your browser, so the comment preview function will not work.

All contributions are licensed under the GNU Free Documentation License.




In order to prevent bots from posting comments, we would like you to prove that you are human. You can do this by filling in the name of the current tag in the following input field. As a reminder, this is tag 0926. Beware of the difference between the letter 'O' and the digit '0'.