
## 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. $\square$

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