Lemma 16.4.4. In Situation 16.4.1 assume that R \to A is smooth at \mathfrak q and that we have a surjection of R-algebras B \to A with kernel I. Assume R \to B smooth at \mathfrak p_ B = (B \to A)^{-1}\mathfrak p. If the cokernel of
I/I^2 \otimes _ A \Lambda \to \Omega _{B/R} \otimes _ B \Lambda
is a free \Lambda -module, then R \to A is smooth at \mathfrak p.
Proof.
The cokernel of the map I/I^2 \to \Omega _{B/R} \otimes _ B A is \Omega _{A/R}, see Algebra, Lemma 10.131.9. Let d = \dim _\mathfrak q(A/R) be the relative dimension of R \to A at \mathfrak q, i.e., the dimension of \mathop{\mathrm{Spec}}(A[1/\pi ]) at \mathfrak q. See Algebra, Definition 10.125.1. Then \Omega _{A/R, \mathfrak q} is free over A_\mathfrak q of rank d (Algebra, Lemma 10.140.3). Thus if the hypothesis of the lemma holds, then \Omega _{A/R} \otimes _ A \Lambda is free of rank d. It follows that \Omega _{A/R} \otimes _ A \kappa (\mathfrak p) has dimension d (as it is true upon tensoring with \Lambda /\pi \Lambda ). Since R \to A is flat and since \mathfrak p is a specialization of \mathfrak q, we see that \dim _\mathfrak p(A/R) \geq d by Algebra, Lemma 10.125.6. Then it follows that R \to A is smooth at \mathfrak p by Algebra, Lemmas 10.137.17 and 10.140.3.
\square
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