**Proof.**
The problem is local on $X$ and $S$. Hence we may assume that $X = \mathop{\mathrm{Spec}}(B)$, $S = \mathop{\mathrm{Spec}}(A)$ and $f$ corresponds to a pseudo-coherent ring map $A \to B$.

If (1) holds, then $B$ has finite tor dimension $d$ as $A$-module. Then $B_\mathfrak q$ has tor dimension $d$ as an $A_\mathfrak p$-module for all primes $\mathfrak q \subset B$ with $\mathfrak p = A \cap \mathfrak q$, see More on Algebra, Lemma 15.66.15. Then $\mathcal{O}_ X$ has tor dimension $d$ as a sheaf of $f^{-1}\mathcal{O}_ S$-modules by Cohomology, Lemma 20.45.5. Thus (1) implies (2).

By Cohomology, Lemma 20.45.5 (2) implies (3).

Assume (3). We cannot use More on Algebra, Lemma 15.66.15 to conclude as we are not given that the tor dimension of $B_\mathfrak q$ over $A_\mathfrak p$ is bounded independent of $\mathfrak q$. Choose a presentation $A[x_1, \ldots , x_ n] \to B$. Then $B$ is pseudo-coherent as a $A[x_1, \ldots , x_ n]$-module. Let $\mathfrak q \subset A[x_1, \ldots , x_ n]$ be a prime ideal lying over $\mathfrak p \subset A$. Then either $B_\mathfrak q$ is zero or by assumption it has finite tor dimension as an $A_\mathfrak p$-module. Since the fibres of $A \to A[x_1, \ldots , x_ n]$ have finite global dimension, we can apply More on Algebra, Lemma 15.77.5 to $A_\mathfrak p \to A[x_1, \ldots , x_ n]_\mathfrak q$ to see that $B_\mathfrak q$ is a perfect $A[x_1, \ldots , x_ n]_\mathfrak q$-module. Hence $B$ is a perfect $A[x_1, \ldots , x_ n]$-module by More on Algebra, Lemma 15.77.3. Thus $A \to B$ is a perfect ring map by definition.
$\square$

## Comments (0)