The Stacks project

15.101 Miscellany

Some results which do not fit anywhere else.

Lemma 15.101.1. 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 $\mathop{\mathrm{Ext}}\nolimits ^ i_ A(K, M)$ are $I$-power torsion for $i \geq a$. Then for $i \geq a$ and $M$ finite the system $\mathop{\mathrm{Ext}}\nolimits ^ i_ A(K, M/I^ nM)$ is essentially constant with value

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

Proof. Let $M$ be a finite $A$-module. Since $K$ is pseudo-coherent we see that $\mathop{\mathrm{Ext}}\nolimits ^ i_ A(K, M)$ is a finite $A$-module. Thus for $i \geq a$ it is annihilated by $I^ t$ for some $t \geq 0$. By Lemma 15.100.4 we see that the image of $\mathop{\mathrm{Ext}}\nolimits ^ i_ A(K, I^ nM) \to \mathop{\mathrm{Ext}}\nolimits ^ i_ A(K, M)$ 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

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

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

Lemma 15.101.2. Let $A$ be a Noetherian ring. Let $I \subset A$ be an ideal. Let $M$ be a finite $A$-module. Let $N$ be an $A$-module annihilated by $I$. There exists an integer $n > 0$ such that $\text{Tor}^ A_ p(I^ nM, N) \to \text{Tor}^ A_ p(M, N)$ is zero for all $p \geq 0$.

Proof. By Lemma 15.100.7 we can factor $I^ nM \to M$ as $I^ nM \to M \otimes _ A^\mathbf {L} I \to M$. We claim the composition

\[ I^ nM \otimes _ A^\mathbf {L} N \to (M \otimes _ A^\mathbf {L} I) \otimes _ A^\mathbf {L} N \to M \otimes _ A^\mathbf {L} N \]

is zero. Namely, the diagram

\[ \xymatrix{ (M \otimes _ A^\mathbf {L} I) \otimes _ A^\mathbf {L} N \ar[rr] \ar[rd] & & M \otimes _ A^\mathbf {L} (I \otimes _ A^\mathbf {L} N) \ar[ld] \\ & M \otimes _ A^\mathbf {L} N } \]

commutes (details omitted) and the map $I \otimes _ A^\mathbf {L} N \to N$ is zero as $N$ is annihilated by $I$. $\square$

Lemma 15.101.3. 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 15.64.5. $\square$

Lemma 15.101.4. 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 15.73.1. We see that $E$ has finite projective dimension over $R$ by Lemma 15.73.7. Thus we can apply Auslander-Buchsbaum (Algebra, Proposition 10.111.1) to see that

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


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

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 10.72.11). This concludes the proof. $\square$

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