The Stacks project

Proof. Choose a K-flat resolution $P^\bullet \to K^\bullet $ over $R$ such that $P^ n$ is a flat $R$-module for all $n$, see Lemma 15.59.10. Then $P^\bullet \otimes _ R R[x]$ is a K-flat complex of $R[x]$-modules whose terms are flat $R[x]$-modules, see Lemma 15.59.3 and Algebra, Lemma 10.39.7. In particular $x : P^ n \otimes _ R R[x] \to P^ n \otimes _ R R[x]$ is injective with cokernel isomorphic to $P^ n$. Thus

is a double complex of $R[x]$-modules whose associated total complex is quasi-isomorphic to $P^\bullet $ and hence $K^\bullet $. Moreover, this associated total complex is a K-flat complex of $R[x]$-modules for example by Lemma 15.59.4 or by Lemma 15.59.5. Hence

as desired. $\square$

Proof. This is a special case of Lemma 15.64.11. We also prove it in another way as follows.

Note that $0 \to R[x] \to R[x] \to R \to 0$ is exact. Hence $R$ is pseudo-coherent as an $R[x]$-module. Thus one implication of the lemma follows from Lemma 15.64.11. To prove the other implication, assume that $K^\bullet $ is $m$-pseudo-coherent as a complex of $R[x]$-modules. By Lemma 15.64.12 we see that $K^\bullet \otimes ^{\mathbf{L}}_{R[x]} R$ is $m$-pseudo-coherent as a complex of $R$-modules. By Lemma 15.81.1 we see that $K^\bullet \oplus K^\bullet [1]$ is $m$-pseudo-coherent as a complex of $R$-modules. Finally, we conclude that $K^\bullet $ is $m$-pseudo-coherent as a complex of $R$-modules from Lemma 15.64.8. $\square$

Proof. If $\alpha : R[x_1, \ldots , x_ n] \to A$ and $\beta : R[y_1, \ldots , y_ m] \to A$ are presentations. Choose $f_ j \in R[x_1, \ldots , x_ n]$ with $\alpha (f_ j) = \beta (y_ j)$ and $g_ i \in R[y_1, \ldots , y_ m]$ with $\beta (g_ i) = \alpha (x_ i)$. Then we get a commutative diagram

After a change of coordinates the ring homomorphism $R[x_1, \ldots , x_ n, y_1, \ldots , y_ m] \to R[x_1, \ldots , x_ n]$ is isomorphic to the ring homomorphism which maps each $y_ i$ to zero. Similarly for the left vertical map in the diagram. Hence, by induction on the number of variables this lemma follows from Lemma 15.81.2. The pseudo-coherent case follows from this and Lemma 15.64.5. $\square$

Proof. Choose a surjection $R[x_1, \ldots , x_ n] \to A$. Choose $y_1, \ldots , y_ m \in B$ which generate $B$ over $A$. As $A \to B$ is finite each $y_ i$ satisfies a monic equation with coefficients in $A$. Hence we can find monic polynomials $P_ j(T) \in R[x_1, \ldots , x_ n][T]$ such that $P_ j(y_ j) = 0$ in $B$. Then we get a commutative diagram

The top horizontal arrow and the top right vertical arrow satisfy the assumptions of Lemma 15.64.11. Hence $K^\bullet $ is $m$-pseudo-coherent (resp. pseudo-coherent) as a complex of $R[x_1, \ldots , x_ n]$-modules if and only if $K^\bullet $ is $m$-pseudo-coherent (resp. pseudo-coherent) as a complex of $R[x_1, \ldots , x_ n, y_1, \ldots , y_ m]$-modules. $\square$

Proof. Follows immediately from Lemma 15.64.2 and the definitions. $\square$

Proof. Follows immediately from Lemma 15.64.4 and the definitions. $\square$

Proof. Immediate from Lemma 15.64.8 and the definitions. $\square$

Proof. Immediate from Lemma 15.64.9 and the definitions. $\square$

Proof. Immediate from Lemma 15.64.10 and the definitions. $\square$

Proof. First we show that $K^\bullet $ is $m$-pseudo-coherent relative to $R$. Namely, suppose $R_ f[x_1, \ldots , x_ n] \to A$ is surjective. Write $R_ f = R[x_0]/(fx_0 - 1)$. Then $R[x_0, x_1, \ldots , x_ n] \to A$ is surjective, and $R_ f[x_1, \ldots , x_ n]$ is pseudo-coherent as an $R[x_0, \ldots , x_ n]$-module. Hence by Lemma 15.64.11 we see that $K^\bullet $ is $m$-pseudo-coherent as a complex of $R[x_0, x_1, \ldots , x_ n]$-modules.

Choose an element $g' \in R[x_0, x_1, \ldots , x_ n]$ which maps to $g \in A$. By Lemma 15.64.12 we see that

is $m$-pseudo-coherent as a complex of $R[x_0, x_1, \ldots , x_ n, \frac{1}{g'}]$-modules. write

As $R[x_0, x_1, \ldots , x_ n, \frac{1}{g'}]$ is pseudo-coherent as a $R[x_0, \ldots , x_ n, x_{n + 1}]$-module we conclude (see Lemma 15.64.11) that $K^\bullet \otimes _ A A_ g$ is $m$-pseudo-coherent as a complex of $R[x_0, \ldots , x_ n, x_{n + 1}]$-modules as desired. $\square$

Proof. Choose a surjection $R[x_1, \ldots , x_ n] \to A$. Note that

by Lemma 15.61.2 applied twice. Hence we win by Lemma 15.64.12. $\square$

Proof. Choose a surjection $A[y_1, \ldots , y_ m] \to B$. Choose a surjection $R[x_1, \ldots , x_ n] \to A$. Combined we get a surjection $R[x_1, \ldots , x_ n, y_1, \ldots y_ m] \to B$. Choose a resolution $E^\bullet \to B$ of $B$ by a complex of finite free $A[y_1, \ldots , y_ n]$-modules (which is possible by our assumption on the ring map $A \to B$). We may assume that $K^\bullet $ is a bounded above complex of flat $A$-modules. Then

in $D(A[y_1, \ldots , y_ m])$. The quasi-isomorphism $\cong $ comes from an application of Lemma 15.59.7. Thus we have to show that $\text{Tot}(K^\bullet \otimes _ A E^\bullet )$ is $m$-pseudo-coherent as a complex of $R[x_1, \ldots , x_ n, y_1, \ldots y_ m]$-modules. Note that $\text{Tot}(K^\bullet \otimes _ A E^\bullet )$ has a filtration by subcomplexes with successive quotients the complexes $K^\bullet \otimes _ A E^ i[-i]$. Note that for $i \ll 0$ the complexes $K^\bullet \otimes _ A E^ i[-i]$ have zero cohomology in degrees $\leq m$ and hence are $m$-pseudo-coherent (over any ring). Hence, applying Lemma 15.81.6 and induction, it suffices to show that $K^\bullet \otimes _ A E^ i[-i]$ is pseudo-coherent relative to $R$ for all $i$. Note that $E^ i = 0$ for $i > 0$. Since also $E^ i$ is finite free this reduces to proving that $K^\bullet \otimes _ A A[y_1, \ldots , y_ m]$ is $m$-pseudo-coherent relative to $R$ which follows from Lemma 15.81.12 for instance. $\square$

Proof. Immediate from Lemma 15.81.13. $\square$

Proof. Choose a surjection $R[x_1, \ldots , x_ n] \to A$. Choose a surjection $A[y_1, \ldots , y_ m] \to B$. Then we get a surjection

which is a flat base change of $R[x_1, \ldots , x_ n] \to A$. By assumption $A$ is a pseudo-coherent module over $R[x_1, \ldots , x_ n]$ hence by Lemma 15.64.13 we see that $A[y_1, \ldots , y_ m]$ is pseudo-coherent over $R[x_1, \ldots , x_ n, y_1, \ldots , y_ m]$. Thus the lemma follows from Lemma 15.64.11 and the definitions. $\square$

Proof. The implication (2) $\Rightarrow $ (1) is in Lemma 15.81.11. Assume (1). Write $1 = \sum f_ ig_ i$ in $A$. Choose a surjection $R[x_1, \ldots , x_ n, y_1, \ldots , y_ r, z_1, \ldots , z_ r] \to A$. such that $y_ i$ maps to $f_ i$ and $z_ i$ maps to $g_ i$. Then we see that there exists a surjection

Note that $P$ is pseudo-coherent as an $R[x_1, \ldots , x_ n, y_1, \ldots , y_ r, z_1, \ldots , z_ r]$-module and that $P[1/y_ i]$ is pseudo-coherent as an $R[x_1, \ldots , x_ n, y_1, \ldots , y_ r, z_1, \ldots , z_ r, 1/y_ i]$-module. Hence by Lemma 15.64.11 we see that $K^\bullet \otimes _ A A_{f_ i}$ is an $m$-pseudo-coherent complex of $P[1/y_ i]$-modules for each $i$. Thus by Lemma 15.64.14 we see that $K^\bullet $ is pseudo-coherent as a complex of $P$-modules, and Lemma 15.64.11 shows that $K^\bullet $ is pseudo-coherent as a complex of $R[x_1, \ldots , x_ n, y_1, \ldots , y_ r, z_1, \ldots , z_ r]$-modules. $\square$

Proof. Immediate consequence of Lemma 15.64.17 and the definitions. $\square$