The Stacks project

Proof. Let $Y' \subset Y$ be the reduction of $Y$. Let $X' \to Y'$ be the base change of $f$. Note that $Y' \to Y$ induces a bijection on points and that $X' \to X$ identifies fibres. Hence we may assume that $Y'$ is reduced, i.e., integral, see Properties, Lemma 28.3.4. We may also replace $Y$ by an affine open. Hence we may assume that $Y = \mathop{\mathrm{Spec}}(A)$ with $A$ a domain. Denote $K$ the fraction field of $A$. Pick an affine open $\mathop{\mathrm{Spec}}(B) = U \subset X$ and a section $h_\eta \in \Gamma (U_\eta , \mathcal{O}_{U_\eta }) = B_ K$ which is nonzero and nilpotent. After shrinking $Y$ we may assume that $h$ comes from $h \in \Gamma (U, \mathcal{O}_ U) = B$. After shrinking $Y$ a bit more we may assume that $h$ is nilpotent. Let $I = \{ b \in B \mid hb = 0\} $ be the annihilator of $h$. Then $C = B/I$ is a finite type $A$-algebra whose generic fiber $(B/I)_ K$ is nonzero (as $h_\eta \not= 0$). We apply generic flatness to $A \to C$ and $A \to B/hB$, see Algebra, Lemma 10.118.3, and we obtain a $g \in A$, $g \not= 0$ such that $C_ g$ is free as an $A_ g$-module and $(B/hB)_ g$ is flat as an $A_ g$-module. Replace $Y$ by $D(g) \subset Y$. Now we have the short exact sequence

with $B/hB$ flat over $A$ and with $C$ nonzero free as an $A$-module. It follows that for any homomorphism $A \to \kappa $ to a field the ring $C \otimes _ A \kappa $ is nonzero and the sequence

is exact, see Algebra, Lemma 10.39.12. Note that $B/hB \otimes _ A \kappa = (B \otimes _ A \kappa ) / h(B \otimes _ A \kappa )$ by right exactness of tensor product. Thus we conclude that multiplication by $h$ is not zero on $B \otimes _ A \kappa $. This clearly means that for any point $y \in Y$ the element $h$ restricts to a nonzero element of $U_ y$, whence $X_ y$ is nonreduced. $\square$