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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