The Stacks project

Proof. By Morphisms, Lemma 29.32.12 the first dimension is finite. The first equality follows as the restriction of $\Omega _{X/S}$ to the fibre is the module of differentials from Morphisms, Lemma 29.32.10. The last equality follows from Lemma 33.16.4. Thus we see that the statement makes sense.

To prove the lemma write $S = \mathop{\mathrm{Spec}}(A)$ and $X = \mathop{\mathrm{Spec}}(B)$ and let $A \to B$ be the ring map corresponding to $f$. Let $\mathfrak q \subset B$ be the prime ideal corresponding to $x$. Choose a surjection of $A$-algebras $A[x_1, \ldots , x_ t] \to B$. Since $\Omega _{B/A}$ is generated by $\text{d}x_1, \ldots , \text{d}x_ t$ we see that their images in $\Omega _{X/S, x} \otimes _{\mathcal{O}_{X, x}} \kappa (x)$ generate this as a $\kappa (x)$-vector space. After renumbering we may assume that $\text{d}x_1, \ldots , \text{d}x_ r$ map to a basis of $\Omega _{X/S, x} \otimes _{\mathcal{O}_{X, x}} \kappa (x)$. We claim that $P = A[x_1, \ldots , x_ r] \to B$ is unramified at $\mathfrak q$. To see this it suffices to show that $\Omega _{B/P, \mathfrak q} = 0$ (Algebra, Lemma 10.151.3). Note that $\Omega _{B/P}$ is the quotient of $\Omega _{B/A}$ by the submodule generated by $\text{d}x_1, \ldots , \text{d}x_ r$. Hence $\Omega _{B/P, \mathfrak q} \otimes _{B_\mathfrak q} \kappa (\mathfrak q) = 0$ by our choice of $x_1, \ldots , x_ r$. By Nakayama's lemma, more precisely Algebra, Lemma 10.20.1 part (2) which applies as $\Omega _{B/P}$ is finite (see reference above), we conclude that $\Omega _{B/P, \mathfrak q} = 0$. $\square$