Lemma 15.82.2. A ring map $A \to B$ is perfect if and only if $B = A[x_1, \ldots , x_ n]/I$ and $B$ as an $A[x_1, \ldots , x_ n]$-module has a finite resolution by finite projective $A[x_1, \ldots , x_ n]$-modules.

Proof. If $A \to B$ is perfect, then $B = A[x_1, \ldots , x_ n]/I$ and $B$ is pseudo-coherent as an $A[x_1, \ldots , x_ n]$-module and has finite tor dimension as an $A$-module. Hence Lemma 15.77.5 implies that $B$ is perfect as a $A[x_1, \ldots , x_ n]$-module, i.e., it has a finite resolution by finite projective $A[x_1, \ldots , x_ n]$-modules (Lemma 15.74.3). Conversely, if $B = A[x_1, \ldots , x_ n]/I$ and $B$ as an $A[x_1, \ldots , x_ n]$-module has a finite resolution by finite projective $A[x_1, \ldots , x_ n]$-modules then $B$ is pseudo-coherent as an $A[x_1, \ldots , x_ n]$-module, hence $A \to B$ is pseudo-coherent. Moreover, the given resolution over $A[x_1, \ldots , x_ n]$ is a finite resolution by flat $A$-modules and hence $B$ has finite tor dimension as an $A$-module. $\square$

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