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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