The Stacks project

Proof. We may remove the (finitely many) irreducible components of $S$ which do not contain $s$. We may replace $S$ by an affine open neighbourhood of $s$. We may replace $X$ by an affine open neighbourhood of $x$. Say $S = \mathop{\mathrm{Spec}}(A)$ and $X = \mathop{\mathrm{Spec}}(B)$. Let $\mathfrak q \subset B$, resp. $\mathfrak p \subset A$ be the prime ideal corresponding to $x$, resp. $s$. As $A$ is a reduced and all of the minimal primes of $A$ are contained in $\mathfrak p$ we see that $A \subset A_\mathfrak p$. As $X \to S$ is of finite type, $B$ is of finite type over $A$. Let $b_1, \ldots , b_ n \in B$ be elements which generate $B$ over $A$ Since $A_\mathfrak p = B_\mathfrak q$ we can find $f \in A$, $f \not\in \mathfrak p$ and $a_ i \in A$ such that $b_ i$ and $a_ i/f$ have the same image in $B_\mathfrak q$. Thus we can find $g \in B$, $g \not\in \mathfrak q$ such that $g(fb_ i - a_ i) = 0$ in $B$. It follows that the image of $A_ f \to B_{fg}$ contains the images of $b_1, \ldots , b_ n$, in particular also the image of $g$. Choose $n \geq 0$ and $f' \in A$ such that $f'/f^ n$ maps to the image of $g$ in $B_{fg}$. Since $A_\mathfrak p = B_\mathfrak q$ we see that $f' \not\in \mathfrak p$. We conclude that $A_{ff'} \to B_{fg}$ is surjective. Finally, as $A_{ff'} \subset A_\mathfrak p = B_\mathfrak q$ (see above) the map $A_{ff'} \to B_{fg}$ is injective, hence an isomorphism. $\square$