The Stacks project

Proof. There is an immediate reduction to the case where $X$ and $Y$ are irreducible which we omit. Moreover, after shrinking further and we may assume $X$ and $Y$ are affine, say $X = \mathop{\mathrm{Spec}}(A)$ and $Y = \mathop{\mathrm{Spec}}(B)$. By assumption $A$, resp. $B$ has a unique minimal prime $\mathfrak p$, resp. $\mathfrak q$, the prime $\mathfrak p$ lies over $\mathfrak q$, and $B_\mathfrak q = A_\mathfrak p$. By Lemma 29.50.3 we have $B_\mathfrak q = A_\mathfrak q = A_\mathfrak p$.

Suppose $B \to A$ is of finite type, say $A = B[x_1, \ldots , x_ n]$. There exist a $b_ i \in B$ and $g_ i \in B \setminus \mathfrak q$ such that $b_ i/g_ i$ maps to the image of $x_ i$ in $A_\mathfrak q$. Hence $b_ i - g_ ix_ i$ maps to zero in $A_{g_ i'}$ for some $g_ i' \in B \setminus \mathfrak q$. Setting $g = \prod g_ i g'_ i$ we see that $B_ g \to A_ g$ is surjective. If moreover $Y$ is reduced, then the map $B_ g \to B_\mathfrak q$ is injective and hence $B_ g \to A_ g$ is injective as well. This proves case (1).

Proof of (2). By the argument given in the previous paragraph we may assume that $B \to A$ is surjective. As $f$ is locally of finite presentation the kernel $J \subset B$ is a finitely generated ideal. Say $J = (b_1, \ldots , b_ r)$. Since $B_\mathfrak q = A_\mathfrak q$ there exist $g_ i \in B \setminus \mathfrak q$ such that $g_ i b_ i = 0$. Setting $g = \prod g_ i$ we see that $B_ g \to A_ g$ is an isomorphism. $\square$