Lemma 15.65.1. Let $R$ be a ring. Let $M = \mathop{\mathrm{colim}}\nolimits M_ i$ be a filtered colimit of $R$-modules. Let $K \in D(R)$ be $m$-pseudo-coherent. Then $\mathop{\mathrm{colim}}\nolimits \mathop{\mathrm{Ext}}\nolimits ^ n_ R(K, M_ i) = \mathop{\mathrm{Ext}}\nolimits ^ n_ R(K, M)$ for $n < -m$ and $\mathop{\mathrm{colim}}\nolimits \mathop{\mathrm{Ext}}\nolimits ^{-m}_ R(K, M_ i) \to \mathop{\mathrm{Ext}}\nolimits ^{-m}_ R(K, M)$ is injective.

## 15.65 Pseudo-coherent modules, II

We continue the discussion started in Section 15.64.

**Proof.**
By definition we can find a distinguished triangle

in $D(R)$ such that $E$ is represented by a bounded complex of finite free $R$-modules and such that $H^ i(L) = 0$ for $i \geq m$. Then $\mathop{\mathrm{Ext}}\nolimits ^ n_ R(L, N) = 0$ for any $R$-module $N$ and $n \leq -m$, see Derived Categories, Lemma 13.27.3. By the long exact sequence of $\mathop{\mathrm{Ext}}\nolimits $ associated to the distinguished triangle we see that $\mathop{\mathrm{Ext}}\nolimits ^ n_ R(K, N) \to \mathop{\mathrm{Ext}}\nolimits ^ n_ R(E, N)$ is an isomorphism for $n < -m$ and injective for $n = -m$. Thus it suffices to prove that $M \mapsto \mathop{\mathrm{Ext}}\nolimits _ R^ n(E, M)$ commutes with filtered colimits when $E$ can be represented by a bounded complex of finite free $R$-modules $E^\bullet $. The modules $\mathop{\mathrm{Ext}}\nolimits ^ n_ R(E, M)$ are computed by the complex $\mathop{\mathrm{Hom}}\nolimits _ R(E^\bullet , M)$, see Derived Categories, Lemma 13.19.8. The functor $M \mapsto \mathop{\mathrm{Hom}}\nolimits _ R(E^ p, M)$ commutes with filtered colimits as $E^ p$ is finite free. Thus $\mathop{\mathrm{Hom}}\nolimits _ R(E^\bullet , M) = \mathop{\mathrm{colim}}\nolimits \mathop{\mathrm{Hom}}\nolimits _ R(E^\bullet , M_ i)$ as complexes. Since filtered colimits are exact (Algebra, Lemma 10.8.8) we conclude. $\square$

Lemma 15.65.2. Let $R$ be a ring. Let $K \in D^-(R)$. Let $m \in \mathbf{Z}$. Then $K$ is $m$-pseudo-coherent if and only if for any filtered colimit $M = \mathop{\mathrm{colim}}\nolimits M_ i$ of $R$-modules we have $\mathop{\mathrm{colim}}\nolimits \mathop{\mathrm{Ext}}\nolimits ^ n_ R(K, M_ i) = \mathop{\mathrm{Ext}}\nolimits ^ n_ R(K, M)$ for $n < -m$ and $\mathop{\mathrm{colim}}\nolimits \mathop{\mathrm{Ext}}\nolimits ^{-m}_ R(K, M_ i) \to \mathop{\mathrm{Ext}}\nolimits ^{-m}_ R(K, M)$ is injective.

**Proof.**
One implication was shown in Lemma 15.65.1. Assume for any filtered colimit $M = \mathop{\mathrm{colim}}\nolimits M_ i$ of $R$-modules we have $\mathop{\mathrm{colim}}\nolimits \mathop{\mathrm{Ext}}\nolimits ^ n_ R(K, M_ i) = \mathop{\mathrm{Ext}}\nolimits ^ n_ R(K, M)$ for $n < -m$ and $\mathop{\mathrm{colim}}\nolimits \mathop{\mathrm{Ext}}\nolimits ^{-m}_ R(K, M_ i) \to \mathop{\mathrm{Ext}}\nolimits ^{-m}_ R(K, M)$ is injective. We will show $K$ is $m$-pseudo-coherent.

Let $t$ be the maximal integer such that $H^ t(K)$ is nonzero. We will use induction on $t$. If $t < m$, then $K$ is $m$-pseudo-coherent by Lemma 15.64.7. If $t \geq m$, then since $\mathop{\mathrm{Hom}}\nolimits _ R(H^ t(K), M) = \mathop{\mathrm{Ext}}\nolimits ^{-t}_ R(K, M)$ we conclude that $\mathop{\mathrm{colim}}\nolimits \mathop{\mathrm{Hom}}\nolimits _ R(H^ t(K), M_ i) \to \mathop{\mathrm{Hom}}\nolimits _ R(H^ t(K), M)$ is injective for any filtered colimit $M = \mathop{\mathrm{colim}}\nolimits M_ i$. This implies that $H^ t(K)$ is a finite $R$-module by Algebra, Lemma 10.11.1. Choose a finite free $R$-module $F$ and a surjection $F \to H^ t(K)$. We can lift this to a morphism $F[-t] \to K$ in $D(R)$ and choose a distinguished triangle

in $D(R)$. Then $H^ i(L) = 0$ for $i \geq t$. Moreover, the long exact sequence of $\mathop{\mathrm{Ext}}\nolimits $ associated to this distinguished triangle shows that $L$ inherts the assumption we made on $K$ by a small argument we omit. By induction on $t$ we conclude that $L$ is $m$-pseudo-coherent. Hence $K$ is $m$-pseudo-coherent by Lemma 15.64.2. $\square$

Lemma 15.65.3. Let $R$ be a ring. Let $L$, $M$, $N$ be $R$-modules.

If $M$ is finitely presented and $L$ is flat, then the canonical map $\mathop{\mathrm{Hom}}\nolimits _ R(M, N) \otimes _ R L \to \mathop{\mathrm{Hom}}\nolimits _ R(M, N \otimes _ R L)$ is an isomorphism.

If $M$ is $(-m)$-pseudo-coherent and $L$ is flat, then the canonical map $\mathop{\mathrm{Ext}}\nolimits ^ i_ R(M, N) \otimes _ R L \to \mathop{\mathrm{Ext}}\nolimits ^ i_ R(M, N \otimes _ R L)$ is an isomorphism for $i < m$.

**Proof.**
Choose a resolution $F_\bullet \to M$ whose terms are free $R$-modules, see Algebra, Lemma 10.71.1. The complex $\mathop{\mathrm{Hom}}\nolimits _ R(F_\bullet , N)$ computes $\mathop{\mathrm{Ext}}\nolimits ^ i_ R(M, N)$ and the complex $\mathop{\mathrm{Hom}}\nolimits _ R(F_\bullet , N \otimes _ R L)$ computes $\mathop{\mathrm{Ext}}\nolimits ^ i_ R(M, N \otimes _ R L)$. There always is a map of cochain complexes

which induces canonical maps $\mathop{\mathrm{Ext}}\nolimits ^ i_ R(M, N) \otimes _ R L \to \mathop{\mathrm{Ext}}\nolimits ^ i_ R(M, N \otimes _ R L)$ for all $i \geq 0$ (canonical for example in the sense that these maps do not depend on the choice of the resolution $F_\bullet $). If $L$ is flat, then the complex $\mathop{\mathrm{Hom}}\nolimits _ R(F_\bullet , N) \otimes _ R L$ computes $\mathop{\mathrm{Ext}}\nolimits ^ i_ R(M, N) \otimes _ R L$ since taking cohomology commutes with tensoring by $L$.

Having said all of the above, if $M$ is $(-m)$-pseudo-coherent, then we may choose $F_\bullet $ such that $F_ i$ is finite free for $i = 0, \ldots , m$. Then the map of cochain complexes displayed above is an isomorphism in degrees $\leq m$ and hence an isomorphism on cohomology groups in degrees $< m$. This proves (2). If $M$ is finitely presented, then $M$ is $(-1)$-pseudo-coherent by Lemma 15.64.4 and we get the result because $\mathop{\mathrm{Hom}}\nolimits = \mathop{\mathrm{Ext}}\nolimits ^0$. $\square$

Lemma 15.65.4. Let $R \to R'$ be a flat ring map. Let $M$, $N$ be $R$-modules.

If $M$ is a finitely presented $R$-module, then $\mathop{\mathrm{Hom}}\nolimits _ R(M, N) \otimes _ R R' = \mathop{\mathrm{Hom}}\nolimits _{R'}(M \otimes _ R R', N \otimes _ R R')$.

If $M$ is $(-m)$-pseudo-coherent, then $\mathop{\mathrm{Ext}}\nolimits ^ i_ R(M, N) \otimes _ R R' = \mathop{\mathrm{Ext}}\nolimits ^ i_{R'}(M \otimes _ R R', N \otimes _ R R')$ for $i < m$.

In particular if $R$ is Noetherian and $M$ is a finite module this holds for all $i$.

**Proof.**
By Algebra, Lemma 10.73.1 we have $\mathop{\mathrm{Ext}}\nolimits ^ i_{R'}(M \otimes _ R R', N \otimes _ R R') = \mathop{\mathrm{Ext}}\nolimits ^ i_ R(M, N \otimes _ R R')$. Combined with Lemma 15.65.3 we conclude (1) and (2) holds. The final statement follows from this and Lemma 15.64.17.
$\square$

Lemma 15.65.5. Let $R$ be a ring. Let $K \in D^-(R)$. The following are equivalent:

$K$ is pseudo-coherent,

for every family $(Q_{\alpha })_{\alpha \in A}$ of $R$-modules, the canonical map

\[ \alpha : K \otimes _ R^\mathbf {L} \left( \prod \nolimits _\alpha Q_{\alpha } \right) \longrightarrow \prod \nolimits _\alpha (K \otimes _ R^\mathbf {L} Q_{\alpha }) \]is an isomorphism in $D(A)$,

for every $R$-module $Q$ and every set $A$, the canonical map

\[ \beta : K \otimes _ R^\mathbf {L} Q^ A \longrightarrow (K \otimes _ R^\mathbf {L} Q)^ A \]is an isomorphism in $D(A)$, and

for every set $A$, the canonical map

\[ \gamma : K \otimes _ R^\mathbf {L} R^ A \longrightarrow K^ A \]is an isomorphism in $D(A)$.

Given $m \in \mathbf{Z}$ the following are equivalent

$K$ is $m$-pseudo-coherent,

for every family $(Q_{\alpha })_{\alpha \in A}$ of $R$-modules, with $\alpha $ as above $H^ i(\alpha )$ is an isomorphism for $i > m$ and surjective for $i = m$,

for every $R$-module $Q$ and every set $A$, with $\beta $ as above $H^ i(\beta )$ is an isomorphism for $i > m$ and surjective for $i = m$,

for every set $A$, with $\gamma $ as above $H^ i(\gamma )$ is an isomorphism for $i > m$ and surjective for $i = m$.

**Proof.**
If $K$ is pseudo-coherent, then $K$ can be represented by a bounded above complex of finite free $R$-modules. Then the derived tensor products are computed by tensoring with this complex. Also, products in $D(A)$ are given by taking products of any choices of representative complexes. Hence (1) implies (2), (3), (4) by the corresponding fact for modules, see Algebra, Proposition 10.89.3.

In the same way (using the tensor product is right exact) the reader shows that (a) implies (b), (c), and (d).

Assume (4) holds. To show that $K$ is pseudo-coherent it suffices to show that $K$ is $m$-pseudo-coherent for all $m$ (Lemma 15.64.5). Hence to finish then proof it suffices to prove that (d) implies (a).

Assume (d). Let $i$ be the largest integer such that $H^ i(K)$ is nonzero. If $i < m$, then we are done. If not, then from (d) and the description of products in $D(A)$ given above we find that $H^ i(K) \otimes _ R R^ A \to H^ i(K)^ A$ is surjective. Hence $H^ i(K)$ is a finitely generated $R$-module by Algebra, Proposition 10.89.2. Thus we may choose a complex $L$ consisting of a single finite free module sitting in degree $i$ and a map of complexes $L \to K$ such that $H^ i(L) \to H^ i(K)$ is surjective. In particular $L$ satisfies (1), (2), (3), and (4). Choose a distinguished triangle

Then we see that $H^ j(M) = 0$ for $j \geq i$. On the other hand, $M$ still has property (d) by a small argument which we omit. By induction on $i$ we find that $M$ is $m$-pseudo-coherent. Hence $K$ is $m$-pseudo-coherent by Lemma 15.64.2. $\square$

Lemma 15.65.6. Let $R$ be a ring. Let $K \in D(R)$ be pseudo-coherent. Let $i \in \mathbf{Z}$. There exists a finitely presented $R$-module $M$ and a map $K \to M[-i]$ in $D(R)$ which induces an injection $H^ i(K) \to M$.

**Proof.**
By Definition 15.64.1 we may represent $K$ by a complex $P^\bullet $ of finite free $R$-modules. Set $M = \mathop{\mathrm{Coker}}(P^{i - 1} \to P^ i)$.
$\square$

Lemma 15.65.7. Let $A$ be a Noetherian ring. Let $K \in D(A)$ be pseudo-coherent, i.e., $K \in D^-(A)$ with finite cohomology modules. Let $\mathfrak m$ be a maximal ideal of $A$. If $H^ i(K)/\mathfrak m H^ i(K) \not= 0$, then there exists a finite $A$-module $E$ annihilated by a power of $\mathfrak m$ and a map $K \to E[-i]$ which is nonzero on $H^ i(K)$.

**Proof.**
(The equivalent formulation of pseudo-coherence in the statement of the lemma is Lemma 15.64.17.) Choose $K \to M[-i]$ as in Lemma 15.65.6. By Artin-Rees (Algebra, Lemma 10.51.2) we can find an $n$ such that $H^ i(K) \cap \mathfrak m^ n M \subset \mathfrak m H^ i(K)$. Take $E = M/\mathfrak m^ n M$.
$\square$

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