Lemma 10.136.16. Let $R \to S$ be a ring map. Let $\mathfrak q \subset S$ be a prime lying over the prime $\mathfrak p$ of $R$. Assume

there exists a $g \in S$, $g \not\in \mathfrak q$ such that $R \to S_ g$ is of finite presentation,

the local ring homomorphism $R_{\mathfrak p} \to S_{\mathfrak q}$ is flat,

the fibre $S \otimes _ R \kappa (\mathfrak p)$ is smooth over $\kappa (\mathfrak p)$ at the prime corresponding to $\mathfrak q$.

Then $R \to S$ is smooth at $\mathfrak q$.

**Proof.**
By Lemmas 10.135.15 and 10.136.5 we see that there exists a $g \in S$ such that $S_ g$ is a relative global complete intersection. Replacing $S$ by $S_ g$ we may assume $S = R[x_1, \ldots , x_ n]/(f_1, \ldots , f_ c)$ is a relative global complete intersection. For any subset $I \subset \{ 1, \ldots , n\} $ of cardinality $c$ consider the polynomial $g_ I = \det (\partial f_ j/\partial x_ i)_{j = 1, \ldots , c, i \in I}$ of Lemma 10.136.15. Note that the image $\overline{g}_ I$ of $g_ I$ in the polynomial ring $\kappa (\mathfrak p)[x_1, \ldots , x_ n]$ is the determinant of the partial derivatives of the images $\overline{f}_ j$ of the $f_ j$ in the ring $\kappa (\mathfrak p)[x_1, \ldots , x_ n]$. Thus the lemma follows by applying Lemma 10.136.15 both to $R \to S$ and to $\kappa (\mathfrak p) \to S \otimes _ R \kappa (\mathfrak p)$.
$\square$

## Comments (0)