The Stacks project

Proof. This follows from Lemmas 15.64.2, 15.64.8, 15.66.5, and 15.66.7. $\square$

Proof. The first statement follows from the second by taking $M = A$. The second statement follows from Lemmas 15.74.2, 15.66.10, and 15.64.16. $\square$

Proof. To prove (2) implies (1) it suffices by Lemma 15.83.2 to show that an $R$-flat, finitely presented $A$-module $M$ defines an $R$-perfect object of $D(A)$. Since $M$ has finite tor dimension over $R$, it suffices to show that $M$ is pseudo-coherent. By Algebra, Lemma 10.168.1 there exists a finite type $\mathbf{Z}$-algebra $R_0 \subset R$ and a flat finite type ring map $R_0 \to A_0$ and a finite $A_0$-module $M_0$ flat over $R_0$ such that $A = A_0 \otimes _{R_0} R$ and $M = M_0 \otimes _{R_0} R$. By Lemma 15.64.17 we see that $M_0$ is pseudo-coherent $A_0$-module. Choose a resolution $P_0^\bullet \to M_0$ by finite free $A_0$-modules $P_0^ n$. Since $A_0$ is flat over $R_0$, this is a flat resolution. Since $M_0$ is flat over $R_0$ we find that $P^\bullet = P_0^\bullet \otimes _{R_0} R$ still resolves $M = M_0 \otimes _{R_0} R$. (You can use Lemma 15.61.2 to see this.) Hence $P^\bullet$ is a finite free resolution of $M$ over $A$ and we conclude that $M$ is pseudo-coherent.

Assume (1). We can represent $K$ by a bounded above complex $P^\bullet$ of finite free $A$-modules. Assume that $K$ viewed as an object of $D(R)$ has tor amplitude in $[a, b]$. By Lemma 15.66.2 we see that $\tau _{\geq a}P^\bullet$ is a complex of $R$-flat, finitely presented $A$-modules representing $K$. $\square$

Proof. By Lemma 15.64.12 we see that $K \otimes _ A^\mathbf {L} A'$ is pseudo-coherent. By Lemma 15.61.2 we see that $K \otimes _ A^\mathbf {L} A'$ is equal to $K \otimes _ R^\mathbf {L} R'$ in $D(R')$. Then we can apply Lemma 15.66.13 to see that $K \otimes _ R^\mathbf {L} R'$ in $D(R')$ has finite tor dimension. $\square$

Proof. The existence of $P^\bullet$ is the definition of a pseudo-coherent complex. We first represent $L$ by a bounded above complex $F^\bullet$ of free $A$-modules (this is possible because bounded tor dimension in particular implies bounded). Next, say $L$ viewed as an object of $D(R)$ has tor amplitude in $[a, b]$. Then, after replacing $F^\bullet$ by $\tau _{\geq a}F^\bullet$, we get a complex as in (2). This follows from Lemma 15.66.2.

Proof of (a). Since $F^\bullet$ is bounded an since $P^\bullet$ is bounded above, we see that $E^ n = 0$ for $n \ll 0$ and that $E^ n$ is a finite (!) direct sum

and since $P^{-q}$ is finite free, this is indeed an $R$-flat $A$-module. The fact that $E^\bullet$ represents $R\mathop{\mathrm{Hom}}\nolimits _ A(K, L)$ follows from Lemma 15.73.2.

Proof of (b). Let $R \to R'$ be a ring map and $A' = A \otimes _ R R'$. By Lemma 15.61.2 the object $L \otimes _ A^\mathbf {L} A'$ is represented by $F^\bullet \otimes _ R R'$ viewed as a complex of $A'$-modules (by flatness of $F^ p$ over $R$). Similarly for $P^\bullet \otimes _ R R'$. As above $R\mathop{\mathrm{Hom}}\nolimits _{A'}(K \otimes _ A^\mathbf {L} A', L \otimes _ A^\mathbf {L} A')$ is represented by

The equality holds by looking at the terms of the complex individually and using that $\mathop{\mathrm{Hom}}\nolimits _{A'}(P^{-q} \otimes _ R R', F^ p \otimes _ R R') = \mathop{\mathrm{Hom}}\nolimits _ A(P^{-q}, F^ p) \otimes _ R R'$. $\square$

Proof. By Algebra, Lemma 10.127.6 the category of finitely presented $A$-modules is the colimit of the categories of finitely presented $A_ i$-modules. Given this, Algebra, Lemma 10.168.1 tells us that category of $R$-flat, finitely presented $A$-modules is the colimit of the categories of $R_ i$-flat, finitely presented $A_ i$-modules. Thus the characterization in Lemma 15.83.4 proves that (1) is true.

To prove (2) we choose $P_0^\bullet$ representing $K_0$ and $F_0^\bullet$ representing $L_0$ as in Lemma 15.83.6. Then $E_0^\bullet = \mathop{\mathrm{Hom}}\nolimits ^\bullet (P_0^\bullet , F_0^\bullet )$ satisfies

and

by the lemma. Thus the result because tensor product commutes with colimits and filtered colimits are exact (Algebra, Lemma 10.8.8). $\square$

Proof. Observe that $A' \to A$ has nilpotent kernel and that by flatness of $R' \to A'$ we have $K = K' \otimes _{R'}^\mathbf {L} R$ (see Section 15.61). Hence the lemma follows by combining Lemmas 15.75.4 and 15.66.20. $\square$

Proof. By Lemma 15.82.8 $K$ is pseudo-coherent as a complex of $R[x_1, \ldots , x_ d]$-modules. Therefore we may assume $A = R[x_1, \ldots , x_ d]$. Applying Lemma 15.77.6 to $R_\mathfrak p \to A_\mathfrak q$ and the complex $K_\mathfrak q$ using our assumption, we find that $K_\mathfrak q$ is perfect in $D(A_\mathfrak q)$ with tor amplitude in $[a - d, b]$. Since $R_\mathfrak p \to A_\mathfrak q$ is flat, we conclude by Lemma 15.66.11. $\square$

Proof. Observe that (1) implies (2) as an $R$-perfect complex has bounded tor dimension as a complex of $R$-modules by definition. Let us prove the other implication.

Write $A = R[x_1, \ldots , x_ d]/I$. Denote $L$ in $D(R[x_1, \ldots , x_ d])$ the restriction of $K$. By Lemma 15.82.8 we see that $L$ is pseudo-coherent. Since $L$ and $K$ have the same image in $D(R)$ we see that $L$ is $R$-perfect if and only if $K$ is $R$-perfect. Also $L \otimes _ R^\mathbf {L} \kappa (\mathfrak p)$ and $K \otimes _ R^\mathbf {L} \kappa (\mathfrak p)$ are the same objects of $D(\kappa (\mathfrak p))$. This reduces us to the case $A = R[x_1, \ldots , x_ d]$.

Say $A = R[x_1, \ldots , x_ d]$ and $K$ satisfies (2). Let $\mathfrak q \subset A$ be a prime lying over a prime $\mathfrak p \subset R$. By Lemma 15.77.6 applied to $R_\mathfrak p \to A_\mathfrak q$ and the complex $K_\mathfrak q$ using our assumption, we find that $K_\mathfrak q$ is perfect in $D(A_\mathfrak q)$. Since $K$ is bounded below, we see that $K$ is perfect in $D(A)$ by Lemma 15.77.3. This implies that $K$ is $R$-perfect by Lemma 15.83.3 and the proof is complete. $\square$