Lemma 37.26.1. Let $f : X \to Y$ be a morphism of schemes. Assume $Y$ irreducible with generic point $\eta$ and $f$ of finite type. If $X_\eta$ is nonreduced, then there exists a nonempty open $V \subset Y$ such that for all $y \in V$ the fibre $X_ y$ is nonreduced.

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

$0 \to C \to B \to B/hB \to 0.$

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

$0 \to C \otimes _ A \kappa \to B \otimes _ A \kappa \to B/hB \otimes _ A \kappa \to 0$

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. $\square$

In your comment you can use Markdown and LaTeX style mathematics (enclose it like $\pi$). A preview option is available if you wish to see how it works out (just click on the eye in the toolbar).