## 15.63 Pseudo-coherent modules, I

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

15.63.0.1
\begin{equation} \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 \end{equation}

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 15.63.1. 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 $M$ is an $m$-pseudo-coherent object of $D(R)$.

4. An $R$-module $M$ is called pseudo-coherent1 if $M$ 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 15.63.5. Also, if the ring is Noetherian the condition can be understood as a finite generation condition on the cohomology, see Lemma 15.63.17. Let us first relate this to the informal discussion above.

Lemma 15.63.2. 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 15.63.3. 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 15.63.4. 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 15.63.1. Conversely, if $M$ is $0$-pseudo-coherent, then $M = H^0(M)$ is of finite type by Lemma 15.63.3. 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 15.63.2 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 15.63.5. 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 15.63.2 we see that $C^\bullet$ is $(n - 1)$-pseudo-coherent. By Lemma 15.63.3 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

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

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 15.63.6. 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 15.63.7. 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 15.63.4 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 15.63.2 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 15.63.8. 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

$(K \oplus L, K \oplus L, L \oplus L) = (K, K, 0) \oplus (L, L, L \oplus L)$

see Derived Categories, Lemma 13.4.10. By Lemma 15.63.2 we see that $L \oplus L$ is $m$-pseudo-coherent. Hence also $L \oplus L$ is $m$-pseudo-coherent. By induction $L[n] \oplus L[n + 1]$ is $m$-pseudo-coherent. By Lemma 15.63.7 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 15.63.5. $\square$

Lemma 15.63.9. 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 15.63.2 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 15.63.5. This follows from the first part. $\square$

Lemma 15.63.10. 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 15.63.7. 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 15.63.2 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 15.63.5.) $\square$

Lemma 15.63.11. 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 15.63.7 $C(\alpha )^\bullet$ 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 15.63.9. The pseudo-coherent case of (1) $\Rightarrow$ (2) follows from this and Lemma 15.63.5.

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 15.63.7. 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 15.61.4. 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 15.63.9) are so, see Lemma 15.63.2. 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 15.63.2 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 15.63.5. $\square$

Lemma 15.63.12. 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 (15.57.0.2). 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 15.63.5. $\square$

Lemma 15.63.13. 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 15.63.12 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 15.63.15.

Lemma 15.63.14. 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 10.23.2. We will prove the lemma by induction on $n - m$. If $n < m$, then the lemma is true by Lemma 15.63.7. If $n \geq m$, then $H^ n(K^\bullet )_{f_ i}$ is a finite $R_{f_ i}$-module for each $i$, see Lemma 15.63.3. Hence $H^ n(K^\bullet )$ is a finite $R$-module, see Algebra, Lemma 10.23.2. 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 15.63.2. Hence by induction we see that $C(\alpha )^\bullet$ is $m$-pseudo-coherent as a complex of $R$-modules. Applying Lemma 15.63.2 once more we conclude. $\square$

Lemma 15.63.15. 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 15.63.7. If $n \geq m$, then $H^ n(K^\bullet ) \otimes _ R R'$ is a finite $R'$-module, see Lemma 15.63.3. Hence $H^ n(K^\bullet )$ is a finite $R$-module, see Algebra, Lemma 10.83.2. 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 15.63.2. Hence by induction we see that $C(\alpha )^\bullet$ is $m$-pseudo-coherent as a complex of $R$-modules. Applying Lemma 15.63.2 once more we conclude. $\square$

Lemma 15.63.16. 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 15.63.5. $\square$

Lemma 15.63.17. 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 10.71.1 we have seen that any finite $R$-module is pseudo-coherent. On the other hand, a pseudo-coherent module is finite, see Lemma 15.63.4. 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 15.63.10 and part (3). Part (2) follows from (1) and Lemma 15.63.5. $\square$

Lemma 15.63.18. Let $R$ be a coherent ring (Algebra, Definition 10.90.1). 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 10.90.4). 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 10.90.3. 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 15.63.10 and 15.63.4. The final assertion follows from the equivalence of (1) and (2) combined with Lemma 15.63.5. $\square$

 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 15.64.3, 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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