The Stacks project

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 $\gamma : P^\bullet \to E^\bullet$ such that the diagram

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

It follows from the induced map on long exact cohomology sequences and Homology, Lemmas 12.5.19 and 12.5.20 that $C(\gamma )^\bullet \to M^\bullet$ 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. $\square$

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. $\square$

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 15.65.1. Conversely, if $M$ is $0$-pseudo-coherent, then $M = H^0(M[0])$ is of finite type by Lemma 15.65.3. 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 15.65.2 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). $\square$

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

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

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 15.65.2 we see that $C^\bullet$ is $(n - 1)$-pseudo-coherent. By Lemma 15.65.3 we see that $H^{n - 1}(C^\bullet )$ is a finite $R$-module. In particular, we see that the kernel of $H^ n(F^\bullet ) \to H^ n(K^\bullet )$ is a finite $R$-module. Choose a finite free $R$-module $F^{n - 1}$ and a map $F^{n - 1} \to \mathop{\mathrm{Ker}}(F^ n \to F^{n - 1})$ such that $F^{n - 1}$ surjects onto the kernel of $H^ n(F^\bullet ) \to H^ n(K^\bullet )$. We extend our of complexes to

Note that $\alpha ^{n - 1}$ exists because the image of $F^{n - 1} \to F^ n \to K^ n$ is in the image of $K^{n - 1} \to K^ n$ by construction. Dnote again $C^\bullet$ the cone of this extended map of complexes. At this point we see that we get an exact sequence

In other words, we see that the cokernel of $H^{n - 1}(F^\bullet ) \to H^{n - 1}(K^\bullet )$ is a finite $R$-module, say generated by the classes of $\xi _1, \ldots , \xi _ r \in \mathop{\mathrm{Ker}}(K^{n - 1} \to K^ n)$. Then we replace $F^{n - 1}$ by $F^{n - 1} \oplus R^{\oplus r}$ where the basis elements in the free summand map to zero in $F^ n$ and the to $\xi _ i$ in $K^{n - 1}$. This finishes the proof of the induction step. $\square$

Proof. Combine Lemmas 15.65.2 and 15.65.5. $\square$

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 15.65.4 we see that $M[- m - 1]$ is $m$-pseudo-coherent. Since we have the distinguished triangle

(Derived Categories, Remark 13.12.4) by Lemma 15.65.2 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. $\square$

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

see Derived Categories, Lemma 13.4.10. By Lemma 15.65.2 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 15.65.7 we see that $L[n]$ is $m$-pseudo-coherent for large $n$. Hence working backwards, using the distinguished triangles

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 15.65.5. $\square$

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 15.65.2 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 15.65.5. This follows from the first part. $\square$

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 15.65.7. If $n \geq m$, then we have the distinguished triangle

(Derived Categories, Remark 13.12.4) Since $H^ n(K^\bullet )[-n]$ is $m$-pseudo-coherent by assumption, we can use Lemma 15.65.2 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 15.65.5.) $\square$

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 15.65.7 $C(\alpha )^\bullet$ 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 15.65.9. The pseudo-coherent case of (1) $\Rightarrow$ (2) follows from this and Lemma 15.65.5.

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 15.65.7. 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

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 15.62.4. 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 15.65.9) are so, see Lemma 15.65.2. 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 15.65.2 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 15.65.5. $\square$

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

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

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 15.65.5. $\square$

Proof. Immediate consequence of Lemma 15.65.12 and the fact that $M \otimes _ A^{\mathbf{L}} B = M \otimes _ A B$ because $B$ is flat over $A$. $\square$

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

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 15.65.7. If $n \geq m$, then $H^ n(K^\bullet )_{f_ i}$ is a finite $R_{f_ i}$-module for each $i$, see Lemma 15.65.3. 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 15.65.2. Hence by induction we see that $C(\alpha )^\bullet$ is $m$-pseudo-coherent as a complex of $R$-modules. Applying Lemma 15.65.2 once more we conclude. $\square$

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

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 15.65.7. If $n \geq m$, then $H^ n(K^\bullet ) \otimes _ R R'$ is a finite $R'$-module, see Lemma 15.65.3. 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 15.65.2. Hence by induction we see that $C(\alpha )^\bullet$ is $m$-pseudo-coherent as a complex of $R$-modules. Applying Lemma 15.65.2 once more we conclude. $\square$

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

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 15.65.5. $\square$

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 15.65.4. 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 15.65.10 and part (3). Part (2) follows from (1) and Lemma 15.65.5. $\square$

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

by finite free $R$-modules. In other words, $M$ is pseudo-coherent. The equivalence of (1) and (2) follows from this and Lemmas 15.65.10 and 15.65.4. The final assertion follows from the equivalence of (1) and (2) combined with Lemma 15.65.5. $\square$