The Stacks project

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 15.102.4 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

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 $\mathop{\mathrm{Ext}}\nolimits ^ i_ A(K, M)$. $\square$

Proof. By Lemma 15.102.7 we can factor $I^ nM \to M$ as $I^ nM \to M \otimes _ A^\mathbf {L} I \to M$. We claim the composition

is zero. Namely, the diagram

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

Proof. Consider the defining distinguished triangle

and apply Lemma 15.66.5. $\square$

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 15.75.1. We see that $E$ has finite projective dimension over $R$ by Lemma 15.75.7. Thus we can apply Auslander-Buchsbaum (Algebra, Proposition 10.111.1) to see that

and

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. $\square$

Proof. Let $\kappa = \max (|A|, |B|, \aleph _0)$. Set $|M^\bullet | = |\bigcup M^ i|$ and similarly for other complexes. With this notation we have

for the quantity used in the statement of the lemma. We are going to use this and other observations coming from arithmetic of cardinals without further mention.

First, let us show that there are plenty of “small” subcomplexes. For every pair of collections $E = \{ E^ i\} $ and $F = \{ F^ i\} $ of finite subsets $E^ i \subset M^ i$, $i \in \mathbf{Z}$ and $F^ i \subset N^ i$, $i \in \mathbf{Z}$ we can let

be the smallest subcomplexes of $A$ and $B$-modules such that $a(M_1(E, F)^\bullet ) \subset N_1(E, F)^\bullet $ and such that $E^ i \subset M_1(E, F)^ i$ and $F^ i \subset M_2(E, F)^ i$. Then it is easy to see that

Details omitted. It is clear that we have

and the colimits are (termwise) filtered colimits.

There exists a resolution $\ldots \to F^{-1} \to F^0 \to B$ by free $A$-modules $F_ i$ with $|F_ i| \leq \kappa $ (details omitted). The cohomology modules of $M_1^\bullet \otimes _ A^\mathbf {L} B$ are computed by $\text{Tot}(M_1^\bullet \otimes _ A F^\bullet )$. It follows that $|H^ i(M_1^\bullet \otimes _ A^\mathbf {L} B)| \leq \max (\kappa , |M_1^\bullet |)$.

Let $i \in \mathbf{Z}$ and let $\xi \in H^ i(M_1^\bullet \otimes _ A^\mathbf {L} B)$ be an element which maps to zero in $H^ i(N_1^\bullet )$. Then $\xi $ maps to zero in $H^ i(N^\bullet )$ and hence $\xi $ maps to zero in $H^ i(M^\bullet \otimes _ A^\mathbf {L} B)$. Since derived tensor product commutes with filtered colimits, we can find finite collections $E_\xi $ and $F_\xi $ as above such that $\xi $ maps to zero in $H^ i(M_1(E_\xi , F_\xi )^\bullet \otimes _ A^\mathbf {L} B)$.

Let $i \in \mathbf{Z}$ and let $\eta \in H^ i(N_1^\bullet )$. Then the image of $\eta $ in $H^ i(N^\bullet )$ is in the image of $H^ i(M^\bullet \otimes _ A^\mathbf {L} B) \to H^ i(N^\bullet )$. Hence as before, we can find finite collections $E_\eta $ and $F_\eta $ as above such that $\eta $ maps to an element of $H^ i(N_1(E_\eta , F_\eta )$ which is in the image of the map $H^ i(M_1(E_\eta , F_\eta )^\bullet \otimes _ A^\mathbf {L} B) \to H^ i(N_1(E_\eta , F_\eta )$.

Now we simply define

where the sum is over $\xi $ and $\eta $ as in the previous two paragraphs and the sum is taken inside $M^\bullet $. Similarly we set

where the sum is taken inside $N^\bullet $. By construction we will have properties (1) and (2) with these choices. The bound (3) also follows as the set of $\xi $ and $\eta $ has cardinality at most $\max (\kappa , |M_1^\bullet |, |N_1^\bullet |)$. $\square$

Proof. Denote $M_ f = M \otimes _ R^\mathbf {L} R_ f$ and choose a distinguished triangle $F \to M \to M_ f$. By assumption $H^ i(M_ f) = H^ i(M)_ f = 0$ for $i < 0$. Thus it suffices to show that $H^ i(F) = 0$ for $i < 0$. Note that $H^ i(F)$ is $f$-power torsion for all $i \in \mathbf{Z}$. On the other hand, since $f : M_ f \to M_ f$ is an isomorphism, we see that $C$ is isomorphic to the cone of $f : F \to F$ (use Derived Categories, Proposition 13.4.23). Now, if $H^ i(F) \not= 0$, then the kernel of $f : H^ i(F) \to H^ i(F)$ is nonzero, which implies that $H^{i - 1}(C)$ is nonzero. Our assumption implies this cannot happen if $i - 1 < -1$ which finishes the proof. $\square$

Proof. Let $K$ be an $R$-module. It suffices to show that $N = M \otimes _ R^\mathbf {L} K$ is isomorphic to an $R$-module placed in degree $0$, see Section 15.67. By assumption (1) we see that $N$ only has nonvanishing cohomology in degrees $\leq 0$. Thus, by Lemma 15.103.6, it suffices to show that $H^ i(N)_ f = 0$ for $i < 0$ and that $H^ i(C) = 0$ for $i < -1$ where $C$ is the cone of $f : N \to N$. For the first, we note that

By our assumption (2) this is zero, except if $i = 0$ and then one gets $H^0(M)_ f \otimes _ R K$. For the second, let $C'$ be the cone of $f : K \to K$. Then

Now $C'$ only has nonzero cohomology in degrees $0$ and $-1$ equal to $K/fK$ and $K[f]$; in other words, there is a distinguished triangle

On the other hand, for any $R$-module $K'$ annihilated by $f$ we have

By our assumption (3) this is equal to the module $H^0(M \otimes _ R^\mathbf {L} R/fR) \otimes _{R/fR} K'$ placed in degree $0$. Combining the above we conclude that $C$ only has nonzero cohomology in degrees $0$ and $-1$ and the proof is complete. $\square$