## 15.62 Pseudo-coherent modules

Suppose that $R$ is a ring. Recall that an $R$-module $M$ is of finite type if there exists a surjection $R^{\oplus a} \to M$ and of finite presentation if there exists a presentation $R^{\oplus a_1} \to R^{\oplus a_0} \to M \to 0$. Similarly, we can consider those $R$-modules for which there exists a length $n$ resolution

15.62.0.1
$$\label{more-algebra-equation-pseudo-coherent} R^{\oplus a_ n} \to R^{\oplus a_{n - 1}} \to \ldots \to R^{\oplus a_0} \to M \to 0$$

by finite free $R$-modules. A module is called pseudo-coherent of we can find such a resolution for every $n$. Here is the formal definition.

Definition 15.62.1. Let $R$ be a ring. Denote $D(R)$ its derived category. Let $m \in \mathbf{Z}$.

1. An object $K^\bullet$ of $D(R)$ is $m$-pseudo-coherent if there exists a bounded complex $E^\bullet$ of finite free $R$-modules and a morphism $\alpha : E^\bullet \to K^\bullet$ such that $H^ i(\alpha )$ is an isomorphism for $i > m$ and $H^ m(\alpha )$ is surjective.

2. An object $K^\bullet$ of $D(R)$ is pseudo-coherent if it is quasi-isomorphic to a bounded above complex of finite free $R$-modules.

3. An $R$-module $M$ is called $m$-pseudo-coherent if $M[0]$ is an $m$-pseudo-coherent object of $D(R)$.

4. An $R$-module $M$ is called pseudo-coherent1 if $M[0]$ is a pseudo-coherent object of $D(R)$.

As usual we apply this terminology also to complexes of $R$-modules. Since any morphism $E^\bullet \to K^\bullet$ in $D(R)$ is represented by an actual map of complexes, see Derived Categories, Lemma 13.19.8, there is no ambiguity. It turns out that $K^\bullet$ is pseudo-coherent if and only if $K^\bullet$ is $m$-pseudo-coherent for all $m \in \mathbf{Z}$, see Lemma 15.62.5. Also, if the ring is Noetherian the condition can be understood as a finite generation condition on the cohomology, see Lemma 15.62.18. Let us first relate this to the informal discussion above.

Lemma 15.62.2. Let $R$ be a ring and $m \in \mathbf{Z}$. Let $(K^\bullet , L^\bullet , M^\bullet , f, g, h)$ be a distinguished triangle in $D(R)$.

1. If $K^\bullet$ is $(m + 1)$-pseudo-coherent and $L^\bullet$ is $m$-pseudo-coherent then $M^\bullet$ is $m$-pseudo-coherent.

2. If $K^\bullet , M^\bullet$ are $m$-pseudo-coherent, then $L^\bullet$ is $m$-pseudo-coherent.

3. If $L^\bullet$ is $(m + 1)$-pseudo-coherent and $M^\bullet$ is $m$-pseudo-coherent, then $K^\bullet$ is $(m + 1)$-pseudo-coherent.

Proof. Proof of (1). Choose $\alpha : P^\bullet \to K^\bullet$ with $P^\bullet$ a bounded complex of finite free modules such that $H^ i(\alpha )$ is an isomorphism for $i > m + 1$ and surjective for $i = m + 1$. We may replace $P^\bullet$ by $\sigma _{\geq m + 1}P^\bullet$ and hence we may assume that $P^ i = 0$ for $i < m + 1$. Choose $\beta : E^\bullet \to L^\bullet$ with $E^\bullet$ a bounded complex of finite free modules such that $H^ i(\beta )$ is an isomorphism for $i > m$ and surjective for $i = m$. By Derived Categories, Lemma 13.19.11 we can find a map $\alpha : P^\bullet \to E^\bullet$ such that the diagram

$\xymatrix{ K^\bullet \ar[r] & L^\bullet \\ P^\bullet \ar[u] \ar[r]^\alpha & E^\bullet \ar[u] }$

is commutative in $D(R)$. The cone $C(\alpha )^\bullet$ is a bounded complex of finite free $R$-modules, and the commutativity of the diagram implies that there exists a morphism of distinguished triangles

$(P^\bullet , E^\bullet , C(\alpha )^\bullet ) \longrightarrow (K^\bullet , L^\bullet , M^\bullet ).$

It follows from the induced map on long exact cohomology sequences and Homology, Lemmas 12.5.19 and 12.5.20 that $C(\alpha )^\bullet \to M^\bullet$ induces an isomorphism on cohomology in degrees $> m$ and a surjection in degree $m$. Hence $M^\bullet$ is $m$-pseudo-coherent.

Assertions (2) and (3) follow from (1) by rotating the distinguished triangle. $\square$

Lemma 15.62.3. Let $R$ be a ring. Let $K^\bullet$ be a complex of $R$-modules. Let $m \in \mathbf{Z}$.

1. If $K^\bullet$ is $m$-pseudo-coherent and $H^ i(K^\bullet ) = 0$ for $i > m$, then $H^ m(K^\bullet )$ is a finite type $R$-module.

2. If $K^\bullet$ is $m$-pseudo-coherent and $H^ i(K^\bullet ) = 0$ for $i > m + 1$, then $H^{m + 1}(K^\bullet )$ is a finitely presented $R$-module.

Proof. Proof of (1). Choose a bounded complex $E^\bullet$ of finite projective $R$-modules and a map $\alpha : E^\bullet \to K^\bullet$ which induces an isomorphism on cohomology in degrees $> m$ and a surjection in degree $m$. It is clear that it suffices to prove the result for $E^\bullet$. Let $n$ be the largest integer such that $E^ n \not= 0$. If $n = m$, then the result is clear. If $n > m$, then $E^{n - 1} \to E^ n$ is surjective as $H^ n(E^\bullet ) = 0$. As $E^ n$ is finite projective we see that $E^{n - 1} = E' \oplus E^ n$. Hence it suffices to prove the result for the complex $(E')^\bullet$ which is the same as $E^\bullet$ except has $E'$ in degree $n - 1$ and $0$ in degree $n$. We win by induction on $n$.

Proof of (2). Choose a bounded complex $E^\bullet$ of finite projective $R$-modules and a map $\alpha : E^\bullet \to K^\bullet$ which induces an isomorphism on cohomology in degrees $> m$ and a surjection in degree $m$. As in the proof of (1) we can reduce to the case that $E^ i = 0$ for $i > m + 1$. Then we see that $H^{m + 1}(K^\bullet ) \cong H^{m + 1}(E^\bullet ) = \mathop{\mathrm{Coker}}(E^ m \to E^{m + 1})$ which is of finite presentation. $\square$

Lemma 15.62.4. Let $R$ be a ring. Let $M$ be an $R$-module. Then

1. $M$ is $0$-pseudo-coherent if and only if $M$ is a finite $R$-module,

2. $M$ is $(-1)$-pseudo-coherent if and only if $M$ is a finitely presented $R$-module,

3. $M$ is $(-d)$-pseudo-coherent if and only if there exists a resolution

$R^{\oplus a_ d} \to R^{\oplus a_{d - 1}} \to \ldots \to R^{\oplus a_0} \to M \to 0$

of length $d$, and

4. $M$ is pseudo-coherent if and only if there exists an infinite resolution

$\ldots \to R^{\oplus a_1} \to R^{\oplus a_0} \to M \to 0$

by finite free $R$-modules.

Proof. If $M$ is of finite type (resp. of finite presentation), then $M$ is $0$-pseudo-coherent (resp. $(-1)$-pseudo-coherent) as follows from the discussion preceding Definition 15.62.1. Conversely, if $M$ is $0$-pseudo-coherent, then $M = H^0(M[0])$ is of finite type by Lemma 15.62.3. If $M$ is $(-1)$-pseudo-coherent, then it is $0$-pseudo-coherent hence of finite type. Choose a surjection $R^{\oplus a} \to M$ and denote $K = \mathop{\mathrm{Ker}}(R^{\oplus a} \to M)$. By Lemma 15.62.2 we see that $K$ is $0$-pseudo-coherent, hence of finite type, whence $M$ is of finite presentation.

To prove the third and fourth statement use induction and an argument similar to the above (details omitted). $\square$

Lemma 15.62.5. Let $R$ be a ring. Let $K^\bullet$ be a complex of $R$-modules. The following are equivalent

1. $K^\bullet$ is pseudo-coherent,

2. $K^\bullet$ is $m$-pseudo-coherent for every $m \in \mathbf{Z}$, and

3. $K^\bullet$ is quasi-isomorphic to a bounded above complex of finite projective $R$-modules.

If (1), (2), and (3) hold and $H^ i(K^\bullet ) = 0$ for $i > b$, then we can find a quasi-isomorphism $F^\bullet \to K^\bullet$ with $F^ i$ finite free $R$-modules and $F^ i = 0$ for $i > b$.

Proof. We see that (1) $\Rightarrow$ (3) as a finite free module is a finite projective $R$-module. Conversely, suppose $P^\bullet$ is a bounded above complex of finite projective $R$-modules. Say $P^ i = 0$ for $i > n_0$. We choose a direct sum decompositions $F^{n_0} = P^{n_0} \oplus C^{n_0}$ with $F^{n_0}$ a finite free $R$-module, and inductively

$F^{n - 1} = P^{n - 1} \oplus C^ n \oplus C^{n - 1}$

for $n \leq n_0$ with $F^{n_0}$ a finite free $R$-module. As a complex $F^\bullet$ has maps $F^{n - 1} \to F^ n$ which agree with $P^{n - 1} \to P^ n$, induce the identity $C^ n \to C^ n$, and are zero on $C^{n - 1}$. The map $F^\bullet \to P^\bullet$ is a quasi-isomorphism (even a homotopy equivalence) and hence (3) implies (1).

Assume (1). Let $E^\bullet$ be a bounded above complex of finite free $R$-modules and let $E^\bullet \to K^\bullet$ be a quasi-isomorphism. Then the induced maps $\sigma _{\geq m}E^\bullet \to K^\bullet$ from the stupid truncation of $E^\bullet$ to $K^\bullet$ show that $K^\bullet$ is $m$-pseudo-coherent. Hence (1) implies (2).

Assume (2). Since $K^\bullet$ is $0$-pseudo-coherent we see in particular that $K^\bullet$ is bounded above. Let $b$ be an integer such that $H^ i(K^\bullet ) = 0$ for $i > b$. By descending induction on $n \in \mathbf{Z}$ we are going to construct finite free $R$-modules $F^ i$ for $i \geq n$, differentials $d^ i : F^ i \to F^{i + 1}$ for $i \geq n$, maps $\alpha : F^ i \to K^ i$ compatible with differentials, such that (1) $H^ i(\alpha )$ is an isomorphism for $i > n$ and surjective for $i = n$, and (2) $F^ i = 0$ for $i > b$. Picture

$\xymatrix{ & F^ n \ar[r] \ar[d]^\alpha & F^{n + 1} \ar[d]^\alpha \ar[r] & \ldots \\ K^{n - 1} \ar[r] & K^ n \ar[r] & K^{n + 1} \ar[r] & \ldots }$

The base case is $n = b + 1$ where we can take $F^ i = 0$ for all $i$. Induction step. Let $C^\bullet$ be the cone on $\alpha$ (Derived Categories, Definition 13.9.1). The long exact sequence of cohomology shows that $H^ i(C^\bullet ) = 0$ for $i \geq n$. By Lemma 15.62.2 we see that $C^\bullet$ is $(n - 1)$-pseudo-coherent. By Lemma 15.62.3 we see that $H^{n - 1}(C^\bullet )$ is a finite $R$-module. Choose a finite free $R$-module $F^{n - 1}$ and a map $\beta : F^{n - 1} \to C^{n - 1}$ such that the composition $F^{n - 1} \to C^{n - 1} \to C^ n$ is zero and such that $F^{n - 1}$ surjects onto $H^{n - 1}(C^\bullet )$. Since $C^{n - 1} = K^{n - 1} \oplus F^ n$ we can write $\beta = (\alpha ^{n - 1}, -d^{n - 1})$. The vanishing of the composition $F^{n - 1} \to C^{n - 1} \to C^ n$ implies these maps fit into a morphism of complexes

$\xymatrix{ & F^{n - 1} \ar[d]^{\alpha ^{n - 1}} \ar[r]_{d^{n - 1}} & F^ n \ar[r] \ar[d]^\alpha & F^{n + 1} \ar[d]^\alpha \ar[r] & \ldots \\ \ldots \ar[r] & K^{n - 1} \ar[r] & K^ n \ar[r] & K^{n + 1} \ar[r] & \ldots }$

Moreover, these maps define a morphism of distinguished triangles

$\xymatrix{ (F^ n \to \ldots ) \ar[r] \ar[d] & (F^{n - 1} \to \ldots ) \ar[r] \ar[d] & F^{n - 1} \ar[r] \ar[d]_\beta & (F^ n \to \ldots )[1] \ar[d] \\ (F^ n \to \ldots ) \ar[r] & K^\bullet \ar[r] & C^\bullet \ar[r] & (F^ n \to \ldots )[1] }$

Hence our choice of $\beta$ implies that the map of complexes $(F^{n - 1} \to \ldots ) \to K^\bullet$ induces an isomorphism on cohomology in degrees $\geq n$ and a surjection in degree $n - 1$. This finishes the proof of the lemma. $\square$

Lemma 15.62.6. Let $R$ be a ring. Let $K \in D^-(R)$. The following are equivalent:

1. $K$ is pseudo-coherent,

2. 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 $D(A)$,

3. 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

4. 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

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

2. 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 $i = m$,

3. 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 $i = m$,

4. 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 10.88.3.

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 15.62.5). 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 10.88.2. 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

$L \to K \to M \to L[1]$

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 15.62.2. $\square$

Lemma 15.62.7. Let $R$ be a ring. Let $(K^\bullet , L^\bullet , M^\bullet , f, g, h)$ be a distinguished triangle in $D(R)$. If two out of three of $K^\bullet , L^\bullet , M^\bullet$ are pseudo-coherent then the third is also pseudo-coherent.

Lemma 15.62.8. Let $R$ be a ring. Let $K^\bullet$ be a complex of $R$-modules. Let $m \in \mathbf{Z}$.

1. If $H^ i(K^\bullet ) = 0$ for all $i \geq m$, then $K^\bullet$ is $m$-pseudo-coherent.

2. If $H^ i(K^\bullet ) = 0$ for $i > m$ and $H^ m(K^\bullet )$ is a finite $R$-module, then $K^\bullet$ is $m$-pseudo-coherent.

3. If $H^ i(K^\bullet ) = 0$ for $i > m + 1$, the module $H^{m + 1}(K^\bullet )$ is of finite presentation, and $H^ m(K^\bullet )$ is of finite type, then $K^\bullet$ is $m$-pseudo-coherent.

Proof. It suffices to prove (3). Set $M = H^{m + 1}(K^\bullet )$. Note that $\tau _{\geq m + 1}K^\bullet$ is quasi-isomorphic to $M[- m - 1]$. By Lemma 15.62.4 we see that $M[- m - 1]$ is $m$-pseudo-coherent. Since we have the distinguished triangle

$(\tau _{\leq m}K^\bullet , K^\bullet , \tau _{\geq m + 1}K^\bullet )$

(Derived Categories, Remark 13.12.4) by Lemma 15.62.2 it suffices to prove that $\tau _{\leq m}K^\bullet$ is pseudo-coherent. By assumption $H^ m(\tau _{\leq m}K^\bullet )$ is a finite type $R$-module. Hence we can find a finite free $R$-module $E$ and a map $E \to \mathop{\mathrm{Ker}}(d_ K^ m)$ such that the composition $E \to \mathop{\mathrm{Ker}}(d_ K^ m) \to H^ m(\tau _{\leq m}K^\bullet )$ is surjective. Then $E[-m] \to \tau _{\leq m}K^\bullet$ witnesses the fact that $\tau _{\leq m}K^\bullet$ is $m$-pseudo-coherent. $\square$

Lemma 15.62.9. Let $R$ be a ring. Let $m \in \mathbf{Z}$. If $K^\bullet \oplus L^\bullet$ is $m$-pseudo-coherent (resp. pseudo-coherent) so are $K^\bullet$ and $L^\bullet$.

Proof. In this proof we drop the superscript ${}^\bullet$. Assume that $K \oplus L$ is $m$-pseudo-coherent. It is clear that $K, L \in D^{-}(R)$. Note that there is a distinguished triangle

$(K \oplus L, K \oplus L, L \oplus L[1]) = (K, K, 0) \oplus (L, L, L \oplus L[1])$

see Derived Categories, Lemma 13.4.10. By Lemma 15.62.2 we see that $L \oplus L[1]$ is $m$-pseudo-coherent. Hence also $L[1] \oplus L[2]$ is $m$-pseudo-coherent. By induction $L[n] \oplus L[n + 1]$ is $m$-pseudo-coherent. By Lemma 15.62.8 we see that $L[n]$ is $m$-pseudo-coherent for large $n$. Hence working backwards, using the distinguished triangles

$(L[n], L[n] \oplus L[n - 1], L[n - 1])$

we conclude that $L[n], L[n - 1], \ldots , L$ are $m$-pseudo-coherent as desired. The pseudo-coherent case follows from this and Lemma 15.62.5. $\square$

Lemma 15.62.10. Let $R$ be a ring. Let $m \in \mathbf{Z}$. Let $K^\bullet$ be a bounded above complex of $R$-modules such that $K^ i$ is $(m - i)$-pseudo-coherent for all $i$. Then $K^\bullet$ is $m$-pseudo-coherent. In particular, if $K^\bullet$ is a bounded above complex of pseudo-coherent $R$-modules, then $K^\bullet$ is pseudo-coherent.

Proof. We may replace $K^\bullet$ by $\sigma _{\geq m - 1}K^\bullet$ (for example) and hence assume that $K^\bullet$ is bounded. Then the complex $K^\bullet$ is $m$-pseudo-coherent as each $K^ i[-i]$ is $m$-pseudo-coherent by induction on the length of the complex: use Lemma 15.62.2 and the stupid truncations. For the final statement, it suffices to prove that $K^\bullet$ is $m$-pseudo-coherent for all $m \in \mathbf{Z}$, see Lemma 15.62.5. This follows from the first part. $\square$

Lemma 15.62.11. Let $R$ be a ring. Let $m \in \mathbf{Z}$. Let $K^\bullet \in D^{-}(R)$ such that $H^ i(K^\bullet )$ is $(m - i)$-pseudo-coherent (resp. pseudo-coherent) for all $i$. Then $K^\bullet$ is $m$-pseudo-coherent (resp. pseudo-coherent).

Proof. Assume $K^\bullet$ is an object of $D^{-}(R)$ such that each $H^ i(K^\bullet )$ is $(m - i)$-pseudo-coherent. Let $n$ be the largest integer such that $H^ n(K^\bullet )$ is nonzero. We will prove the lemma by induction on $n$. If $n < m$, then $K^\bullet$ is $m$-pseudo-coherent by Lemma 15.62.8. If $n \geq m$, then we have the distinguished triangle

$(\tau _{\leq n - 1}K^\bullet , K^\bullet , H^ n(K^\bullet )[-n])$

(Derived Categories, Remark 13.12.4) Since $H^ n(K^\bullet )[-n]$ is $m$-pseudo-coherent by assumption, we can use Lemma 15.62.2 to see that it suffices to prove that $\tau _{\leq n - 1}K^\bullet$ is $m$-pseudo-coherent. By induction on $n$ we win. (The pseudo-coherent case follows from this and Lemma 15.62.5.) $\square$

Lemma 15.62.12. Let $A \to B$ be a ring map. Assume that $B$ is pseudo-coherent as an $A$-module. Let $K^\bullet$ be a complex of $B$-modules. The following are equivalent

1. $K^\bullet$ is $m$-pseudo-coherent as a complex of $B$-modules, and

2. $K^\bullet$ is $m$-pseudo-coherent as a complex of $A$-modules.

The same equivalence holds for pseudo-coherence.

Proof. Assume (1). Choose a bounded complex of finite free $B$-modules $E^\bullet$ and a map $\alpha : E^\bullet \to K^\bullet$ which is an isomorphism on cohomology in degrees $> m$ and a surjection in degree $m$. Consider the distinguished triangle $(E^\bullet , K^\bullet , C(\alpha )^\bullet )$. By Lemma 15.62.8 $C(\alpha )^\bullet$ is $m$-pseudo-coherent as a complex of $A$-modules. Hence it suffices to prove that $E^\bullet$ is pseudo-coherent as a complex of $A$-modules, which follows from Lemma 15.62.10. The pseudo-coherent case of (1) $\Rightarrow$ (2) follows from this and Lemma 15.62.5.

Assume (2). Let $n$ be the largest integer such that $H^ n(K^\bullet ) \not= 0$. We will prove that $K^\bullet$ is $m$-pseudo-coherent as a complex of $B$-modules by induction on $n - m$. The case $n < m$ follows from Lemma 15.62.8. Choose a bounded complex of finite free $A$-modules $E^\bullet$ and a map $\alpha : E^\bullet \to K^\bullet$ which is an isomorphism on cohomology in degrees $> m$ and a surjection in degree $m$. Consider the induced map of complexes

$\alpha \otimes 1 : E^\bullet \otimes _ A B \to K^\bullet .$

Note that $C(\alpha \otimes 1)^\bullet$ is acyclic in degrees $\geq n$ as $H^ n(E) \to H^ n(E^\bullet \otimes _ A B) \to H^ n(K^\bullet )$ is surjective by construction and since $H^ i(E^\bullet \otimes _ A B) = 0$ for $i > n$ by the spectral sequence of Example 15.60.4. On the other hand, $C(\alpha \otimes 1)^\bullet$ is $m$-pseudo-coherent as a complex of $A$-modules because both $K^\bullet$ and $E^\bullet \otimes _ A B$ (see Lemma 15.62.10) are so, see Lemma 15.62.2. Hence by induction we see that $C(\alpha \otimes 1)^\bullet$ is $m$-pseudo-coherent as a complex of $B$-modules. Finally another application of Lemma 15.62.2 shows that $K^\bullet$ is $m$-pseudo-coherent as a complex of $B$-modules (as clearly $E^\bullet \otimes _ A B$ is pseudo-coherent as a complex of $B$-modules). The pseudo-coherent case of (2) $\Rightarrow$ (1) follows from this and Lemma 15.62.5. $\square$

Lemma 15.62.13. Let $A \to B$ be a ring map. Let $K^\bullet$ be an $m$-pseudo-coherent (resp. pseudo-coherent) complex of $A$-modules. Then $K^\bullet \otimes _ A^{\mathbf{L}} B$ is an $m$-pseudo-coherent (resp. pseudo-coherent) complex of $B$-modules.

Proof. First we note that the statement of the lemma makes sense as $K^\bullet$ is bounded above and hence $K^\bullet \otimes _ A^{\mathbf{L}} B$ is defined by Equation (15.56.0.2). Having said this, choose a bounded complex $E^\bullet$ of finite free $A$-modules and $\alpha : E^\bullet \to K^\bullet$ with $H^ i(\alpha )$ an isomorphism for $i > m$ and surjective for $i = m$. Then the cone $C(\alpha )^\bullet$ is acyclic in degrees $\geq m$. Since $-\otimes _ A^{\mathbf{L}} B$ is an exact functor we get a distinguished triangle

$(E^\bullet \otimes _ A^{\mathbf{L}} B, K^\bullet \otimes _ A^{\mathbf{L}} B, C(\alpha )^\bullet \otimes _ A^{\mathbf{L}} B)$

of complexes of $B$-modules. By the dual to Derived Categories, Lemma 13.16.1 we see that $H^ i(C(\alpha )^\bullet \otimes _ A^{\mathbf{L}} B) = 0$ for $i \geq m$. Since $E^\bullet$ is a complex of projective $R$-modules we see that $E^\bullet \otimes _ A^{\mathbf{L}} B = E^\bullet \otimes _ A B$ and hence

$E^\bullet \otimes _ A B \longrightarrow K^\bullet \otimes _ A^{\mathbf{L}} B$

is a morphism of complexes of $B$-modules that witnesses the fact that $K^\bullet \otimes _ A^{\mathbf{L}} B$ is $m$-pseudo-coherent. The case of pseudo-coherent complexes follows from the case of $m$-pseudo-coherent complexes via Lemma 15.62.5. $\square$

Lemma 15.62.14. Let $A \to B$ be a flat ring map. Let $M$ be an $m$-pseudo-coherent (resp. pseudo-coherent) $A$-module. Then $M \otimes _ A B$ is an $m$-pseudo-coherent (resp. pseudo-coherent) $B$-module.

Proof. Immediate consequence of Lemma 15.62.13 and the fact that $M \otimes _ A^{\mathbf{L}} B = M \otimes _ A B$ because $B$ is flat over $A$. $\square$

The following lemma also follows from the stronger Lemma 15.62.15.

Lemma 15.62.15. Let $R$ be a ring. Let $f_1, \ldots , f_ r \in R$ be elements which generate the unit ideal. Let $m \in \mathbf{Z}$. Let $K^\bullet$ be a complex of $R$-modules. If for each $i$ the complex $K^\bullet \otimes _ R R_{f_ i}$ is $m$-pseudo-coherent (resp. pseudo-coherent), then $K^\bullet$ is $m$-pseudo-coherent (resp. pseudo-coherent).

Proof. We will use without further mention that $- \otimes _ R R_{f_ i}$ is an exact functor and that therefore

$H^ i(K^\bullet )_{f_ i} = H^ i(K^\bullet ) \otimes _ R R_{f_ i} = H^ i(K^\bullet \otimes _ R R_{f_ i}).$

Assume $K^\bullet \otimes _ R R_{f_ i}$ is $m$-pseudo-coherent for $i = 1, \ldots , r$. Let $n \in \mathbf{Z}$ be the largest integer such that $H^ n(K^\bullet \otimes _ R R_{f_ i})$ is nonzero for some $i$. This implies in particular that $H^ i(K^\bullet ) = 0$ for $i > n$ (and that $H^ n(K^\bullet ) \not= 0$) see Algebra, Lemma 10.22.2. We will prove the lemma by induction on $n - m$. If $n < m$, then the lemma is true by Lemma 15.62.8. If $n \geq m$, then $H^ n(K^\bullet )_{f_ i}$ is a finite $R_{f_ i}$-module for each $i$, see Lemma 15.62.3. Hence $H^ n(K^\bullet )$ is a finite $R$-module, see Algebra, Lemma 10.22.2. Choose a finite free $R$-module $E$ and a surjection $E \to H^ n(K^\bullet )$. As $E$ is projective we can lift this to a map of complexes $\alpha : E[-n] \to K^\bullet$. Then the cone $C(\alpha )^\bullet$ has vanishing cohomology in degrees $\geq n$. On the other hand, the complexes $C(\alpha )^\bullet \otimes _ R R_{f_ i}$ are $m$-pseudo-coherent for each $i$, see Lemma 15.62.2. Hence by induction we see that $C(\alpha )^\bullet$ is $m$-pseudo-coherent as a complex of $R$-modules. Applying Lemma 15.62.2 once more we conclude. $\square$

Lemma 15.62.16. Let $R$ be a ring. Let $m \in \mathbf{Z}$. Let $K^\bullet$ be a complex of $R$-modules. Let $R \to R'$ be a faithfully flat ring map. If the complex $K^\bullet \otimes _ R R'$ is $m$-pseudo-coherent (resp. pseudo-coherent), then $K^\bullet$ is $m$-pseudo-coherent (resp. pseudo-coherent).

Proof. We will use without further mention that $- \otimes _ R R'$ is an exact functor and that therefore

$H^ i(K^\bullet ) \otimes _ R R' = H^ i(K^\bullet \otimes _ R R').$

Assume $K^\bullet \otimes _ R R'$ is $m$-pseudo-coherent. Let $n \in \mathbf{Z}$ be the largest integer such that $H^ n(K^\bullet )$ is nonzero; then $n$ is also the largest integer such that $H^ n(K^\bullet \otimes _ R R')$ is nonzero. We will prove the lemma by induction on $n - m$. If $n < m$, then the lemma is true by Lemma 15.62.8. If $n \geq m$, then $H^ n(K^\bullet ) \otimes _ R R'$ is a finite $R'$-module, see Lemma 15.62.3. Hence $H^ n(K^\bullet )$ is a finite $R$-module, see Algebra, Lemma 10.82.2. Choose a finite free $R$-module $E$ and a surjection $E \to H^ n(K^\bullet )$. As $E$ is projective we can lift this to a map of complexes $\alpha : E[-n] \to K^\bullet$. Then the cone $C(\alpha )^\bullet$ has vanishing cohomology in degrees $\geq n$. On the other hand, the complex $C(\alpha )^\bullet \otimes _ R R'$ is $m$-pseudo-coherent, see Lemma 15.62.2. Hence by induction we see that $C(\alpha )^\bullet$ is $m$-pseudo-coherent as a complex of $R$-modules. Applying Lemma 15.62.2 once more we conclude. $\square$

Lemma 15.62.17. Let $R$ be a ring. Let $K, L$ be objects of $D(R)$.

1. If $K$ is $n$-pseudo-coherent and $H^ i(K) = 0$ for $i > a$ and $L$ is $m$-pseudo-coherent and $H^ j(L) = 0$ for $j > b$, then $K \otimes _ R^\mathbf {L} L$ is $t$-pseudo-coherent with $t = \max (m + a, n + b)$.

2. If $K$ and $L$ are pseudo-coherent, then $K \otimes _ R^\mathbf {L} L$ is pseudo-coherent.

Proof. Proof of (1). We may assume there exist bounded complexes $K^\bullet$ and $L^\bullet$ of finite free $R$-modules and maps $\alpha : K^\bullet \to K$ and $\beta : L^\bullet \to L$ with $H^ i(\alpha )$ and isomorphism for $i > n$ and surjective for $i = n$ and with $H^ i(\beta )$ and isomorphism for $i > m$ and surjective for $i = m$. Then the map

$\alpha \otimes ^\mathbf {L} \beta : \text{Tot}(K^\bullet \otimes _ R L^\bullet ) \to K \otimes _ R^\mathbf {L} L$

induces isomorphisms on cohomology in degree $i$ for $i > t$ and a surjection for $i = t$. This follows from the spectral sequence of tors (details omitted). Part (2) follows from part (1) and Lemma 15.62.5. $\square$

Lemma 15.62.18. Let $R$ be a Noetherian ring. Then

1. A complex of $R$-modules $K^\bullet$ is $m$-pseudo-coherent if and only if $K^\bullet \in D^{-}(R)$ and $H^ i(K^\bullet )$ is a finite $R$-module for $i \geq m$.

2. A complex of $R$-modules $K^\bullet$ is pseudo-coherent if and only if $K^\bullet \in D^{-}(R)$ and $H^ i(K^\bullet )$ is a finite $R$-module for all $i$.

3. An $R$-module is pseudo-coherent if and only if it is finite.

Proof. In Algebra, Lemma 10.70.1 we have seen that any finite $R$-module is pseudo-coherent. On the other hand, a pseudo-coherent module is finite, see Lemma 15.62.4. Hence (3) holds. Suppose that $K^\bullet$ is an $m$-pseudo-coherent complex. Then there exists a bounded complex of finite free $R$-modules $E^\bullet$ such that $H^ i(K^\bullet )$ is isomorphic to $H^ i(E^\bullet )$ for $i > m$ and such that $H^ m(K^\bullet )$ is a quotient of $H^ m(E^\bullet )$. Thus it is clear that each $H^ i(K^\bullet )$, $i \geq m$ is a finite module. The converse implication in (1) follows from Lemma 15.62.11 and part (3). Part (2) follows from (1) and Lemma 15.62.5. $\square$

Lemma 15.62.19. Let $R$ be a coherent ring (Algebra, Definition 10.89.1). Let $K \in D^-(R)$. The following are equivalent

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

2. $H^ m(K)$ is a finite $R$-module and $H^ i(K)$ is coherent for $i > m$, and

3. $H^ m(K)$ is a finite $R$-module and $H^ i(K)$ is finitely presented for $i > m$.

Thus $K$ is pseudo-coherent if and only if $H^ i(K)$ is a coherent module for all $i$.

Proof. Recall that an $R$-module $M$ is coherent if and only if it is of finite presentation (Algebra, Lemma 10.89.4). This explains the equivalence of (2) and (3). If so and if we choose an exact sequence $0 \to N \to R^{\oplus m} \to M \to 0$, then $N$ is coherent by Algebra, Lemma 10.89.3. Thus in this case, repeating this procedure with $N$ we find a resolution

$\ldots \to R^{\oplus n} \to R^{\oplus m} \to M \to 0$

by finite free $R$-modules. In other words, $M$ is pseudo-coherent. The equivalence of (1) and (2) follows from this and Lemmas 15.62.11 and 15.62.4. The final assertion follows from the equivalence of (1) and (2) combined with Lemma 15.62.5. $\square$

Remark 15.62.20. Let $R$ be ring map. Let $L$, $M$, $N$ be $R$-modules. Consider the canonical map

$\mathop{\mathrm{Hom}}\nolimits _ R(M, N) \otimes _ R L \to \mathop{\mathrm{Hom}}\nolimits _ R(M, N \otimes _ R L)$

Choose a two term free resolution $F_1 \to F_0 \to M \to 0$. Assuming $L$ flat over $R$ we obtain a commutative diagram

$\xymatrix{ 0 \ar[r] & \mathop{\mathrm{Hom}}\nolimits _ R(M, N) \otimes _ R L \ar[r] \ar[d] & \mathop{\mathrm{Hom}}\nolimits _ R(F_0, N) \otimes _ R L \ar[r] \ar[d] & \mathop{\mathrm{Hom}}\nolimits _ R(F_1, N) \otimes _ R L \ar[d] \\ 0 \ar[r] & \mathop{\mathrm{Hom}}\nolimits _ R(M, N \otimes _ R L) \ar[r] & \mathop{\mathrm{Hom}}\nolimits _ R(F_0, N \otimes _ R L) \ar[r] & \mathop{\mathrm{Hom}}\nolimits _ R(F_1, N \otimes _ R L) }$

with exact rows. We conclude that if $F_0$ and $F_1$ are finite free, i.e., if $M$ is finitely presented, then the first displayed map is an isomorphism. Similarly, if $M$ is $(-m)$-pseudo-coherent and still assuming $L$ is flat over $R$, then the 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 $i < m$.

Remark 15.62.21. Let $R$ be ring map. Let $M$, $N$ be $R$-modules. Let $R \to R'$ be a flat ring map. By Algebra, Lemma 10.72.1 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 Remark 15.62.20 we conclude that

$\mathop{\mathrm{Hom}}\nolimits _ R(M, N) \otimes _ R R' = \mathop{\mathrm{Hom}}\nolimits _{R'}(M \otimes _ R R', N \otimes _ R R')$

if $M$ is a finitely presented $R$-module and that

$\mathop{\mathrm{Ext}}\nolimits ^ i_ R(M, N) \otimes _ R R' = \mathop{\mathrm{Ext}}\nolimits ^ i_{R'}(M \otimes _ R R', N \otimes _ R R')$

is an isomorphism for $i < m$ if $M$ is $(-m)$-pseudo-coherent. In particular if $R$ is Noetherian and $M$ is a finite module this holds for all $i$.

[1] This clashes with what is meant by a pseudo-coherent module in .

Comment #4661 by Hailong Dao on

Can you give a bit more details on the last sentence "Similarly,..." of Remark 087Q? I tried to take $M_1$ to be a syzygy of $M$ to prove the case $m=2$, but then get stuck as we only know that $Hom_R(M_1,N)$ surjects onto $Ext^1_R(M,N)$.

Comment #4662 by on

If $M$ is $(-m)$-pseudo-coherent then we have a resolution $F_\bullet \to M$ with $F_i$ finite free for $i = 0, 1, 2, \ldots, m$ (I may be off by $1$ here). Then we get a map of complexes as depicted in Remark 15.62.20, like so $Hom(F_\bullet, N) \otimes L \to Hom(F_\bullet, N \otimes L)$. Since $F_i$ is finite free in those degrees it is an isomorphism in those degrees. Since $L$ is flat we see that we get what was stated in the remark. OK?

Comment #4663 by Hailong Dao on

I think I see it now. The map of complexes is always injective, thus we can use Lemma 0111. Thanks.

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