Let $f:X \to Y$ be a projective morphism between irreducible Noetherian schemes. If a fiber over a closed point of $Y$ is reduced is the generic fiber reduced?
[Math] Reduced special fiber implies reduced generic fiber for a projective morphism
ag.algebraic-geometry
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If $Y$ is non-reduced, then $X = Y_{red} \times \mathbf{P}^{n}$ is a counterexample.
Let me repeat the assumptions to make sure we agree. Say $f : X \to Y$ is a morphism of varieties over an algebraically closed field $k$ such that (a) $Y$ is affine and smooth of dimension $m$, (b) $(X_y)_{red}$ is smooth of fixed dimension $n$ for all $y \in Y(k)$, (c) the maximal open $V_y \subset X_y$ which is a reduced scheme is dense in $X_y$ for all $y \in Y(k)$.
Claim: $f$ is smooth.
Step 1. Let $V \subset X$ be the open locus where the morphism $f$ is smooth. By assumptions (a), (b), (c) we see that $V_y$ is the fibre of $V \to Y$ over $y$. Namely, $f$ is smooth in points of $V_y$ by 2.8 of the paper by de Jong on alterations. The converse inclusion is obvious.
Step 2. Let $\nu : X' \to X$ be the normalization morphism. Then $V$ is an open subscheme of $X'$. For $y \in Y(k)$ we consider $(X'_y)_{red} \to (X_y)_{red}$. This is a finite morphism which is an isomorphism over the dense open $V_y$. Also every irreducible component of $(X'_y)_{red}$ has dimension $n$ by Krull's height theorem. Hence $(X'_y)_{red} \to (X_y)_{red}$ is birational and hence an isomorphism as the target is normal.
Step 3. In particular the assumptions are true for $f' : X' \to Y$. If $Z \subset Y$ is a smooth effective Cartier divisor, then we can consider the morphism $(f')^{-1}(Z) \to Z$. Using that $X'$ is normal, it is straightforward to show that $(f')^{-1}(Z)$ is reduced, using the criterion $(R_0) + (S_1)$ for reducedness. If $m > 1$, then for any $y \in Y(k)$ we can pick $Z$ such that $(f')^{-1}(Z)$ is irreducible by a Bertini theorem (a la Jouanolou).
Step 4. By induction on $m$ we see that $(f')^{-1}(Z) \to Z$ is smooth. Hence all the fibres of $f'$ are smooth. Hence $X'$ is smooth. Since we have seen above that $X' \to X$ is a bijection on closed points, it suffices to show that no tangent vectors get collapsed. Such a tangent vector would have to be vertical. But this would mean that $(X_y)_{red}$ cannnot be smooth.
Answer to first comment: the fibres are nonempty by assumption (b) or they are all empty if $n < 0$ and then the result is true also. Answer to second comment: forgot to say $y \in Z$. The induction works because we've checked (f')^{-1}(Z) is a variety (except in the case $n = 1$ you get that it might be a disjoint union of varieties). Anyway, others can add more details to this answer if they so desire.
Best Answer
Let's impose a flatness hypothesis (whose necessity is explained in Jason Starr's answer). The answer is still negative, but to explain the context for the counterexample it is instructive to first record some necessary features of any counterexample, so we know where to look.
In view of my above comment about geometric fibers, any search for a counterexample will necessarily have to involve a special fiber that is reduced yet not geometrically reduced. In other words, any counterexample will have to involve an imperfect residue field at the special point in the base.
Moreover, the answer is affirmative if the base $Y$ is Dedekind. Indeed, we can assume in such a case that the base $Y$ is Spec($R$) for a dvr $R$ with uniformizer $t$, so $X$ is a proper flat $R$-scheme having reduced special fiber $X_0$. If $N$ is the coherent sheaf of nilpotents inside $O_X$ then for any section $f \in N(U)$ over an open $U \subset X$ we know that $f_0 = f|_{U_0}$ vanishes since $X_0$ is reduced, so $f$ is a section of $tO_X$. In other words, $N \subset tO_X$. By $R$-flatness of $X$, it is clear (check!) that $N \cap tO_X = tN$, so $N \subset tN$. Hence, by Nakayama's Lemma along points of the special fiber, it follows that $N$ has vanishing stalks along the special fiber $X_0 \subset X$. The closed support of the coherent $N$ inside the $R$-proper $X$ is therefore disjoint from the special fiber. But every non-empty closed set in $X$ must meet $X_0$ due to $R$-properness of $X$ and locality of $R$. Thus, the support is empty, which is to say $N = 0$. Then $X$ is reduced, so its localization given by the generic fiber over the integral base $Y$ is also reduced. That settles the question affirmatively when $Y$ is Dedekind (and $X$ is any proper flat $Y$-scheme).
In view of the preceding observations, an "optimal" counterexample should involve a local base $Y = {\rm{Spec}}(R)$ where $R$ is a non-Dedekind 1-dimensional local noetherian domain having imperfect residue field with characteristic $p > 0$. And in fact there are counterexamples over such $R$, as we now construct. This shows that the EGA result with reducedness for geometric fibers is essentially "best possible". We will give counterexamples in equicharacteristic $p$. Maybe someone else can address the story with generic characteristic 0. (SEE THE END FOR SUCH AN EXAMPLE)
The idea is to make such an $R$ that is an order in a dvr of characteristic $p$ whose fraction field contains a certain $p$th root but for which $R$ does not contain that $p$th root. Let $k$ be a field of characteristic $p > 0$ and let $A = k(t)[x]_{(x)}$, a dvr with uniformizer $x$ and residue field $k(t)$ that is obviously not perfect. Let $F = {\rm{Frac}}(A) = k(t,x)$ and let $A' = A[T]/(T^p - t) = k(T)[x]_{(x)}$, so the dvr $A'$ with uniformizer $x$ is the integral closure of $A$ in $F' = F[T]/(T^p - t) = F(t^{1/p})$. Note that $A \rightarrow A'$ is an example of a finite extension of discrete valuation ring whose ramification degree is 1 but residual extension is not separable (so it is not "unramified").
The residue field of $A'$ is $k(T)$ in which the element $t$ from the residue field of $A$ has image $T^p$ that is a $p$th power, but we can "fix" that (or rather, "ruin" it) by considering the order $R = A + xA'$. Concretely, $R$ is the preimage of $k(t)$ under the reduction map $A' \twoheadrightarrow k(T)$.
Clearly $R$ is a 1-dimensional local noetherian domain whose fraction field is $F'$, residue field is $k(t)$, and normalization is $A'$. Note that the element $T = t^{1/p}$ in the fraction field $F'$ of $R$ does not lie in $R$.
Inside $\mathbf{P}^2_A$ with homogeneous coordinates $[U,V,W]$, consider the closed subscheme $Z$ defined by $U^p + t V^p$. This is $A$-flat since $A$ is Dedekind. Its special and generic fibers are the "same" projective scheme over the residue field $k(t)$ and fraction field $k(t,x)$ respectively, and as such these fibers are both reduced (as $t$ is not a $p$th power in either the residue field or fraction field of $A$, each of characteristic $p$) and are also both geometrically irreducible. The base change $X = Z_R$ is certainly $R$-flat (since $Z$ is $A$-flat) and projective with special fiber that is the same as that of $Z$ since $R$ and $A$ have the same residue field by design. In particular, $X_0$ is reduced (though not geometrically so!). The generic fiber $X_{F'} = Z_{F'}$ is non-reduced since $t$ is a $p$th power in $F'$. Note however that the $R$-flat $X$ has irreducible generic fiber, so $X$ is also irreducible. Voila.
EDIT: Here is an even better example in the same spirit, but with a twist on the idea so that it works in generic characteristic 0. It will use an order in $\mathbf{Z}[T]_{(p)}$ as I had been hoping. This example was pointed out to me by somebody "offline".
Let $A = \mathbf{Z}[t]_{(p)}$, a dvr with uniformizer $p$, residue field $\mathbf{F}_p(t)$, and fraction field $F = \mathbf{Q}(t)$. Let $A'= \mathbf{Z}[T]_{(p)}$ made into an $A$-algebra via $t \mapsto T^p$, so its fraction field $F'$ is $F(t^{1/p})$. Let $R = A + pA'$, a 1-dimensional local noetherian domain with residue field $\mathbf{F}_p(t)$ and fraction field $F'$.
Consider the polynomial $f(U,V) = (U + T V)^p \in U^p + t V^p + pA'[U,V] \subset R[U,V]$. Obviously $f$ is a $p$th power over $F' = \mathbf{Q}(T)$, but it is not a (unit multiple of a) $p$th power over $R$ since the reduction of $f$ over the residue field of $R$ is $U^p + t V^p \in \mathbf{F}_p(t)[U,V]$, which is even irreducible.
Let $X \subset \mathbf{P}^1_R$ be defined by the vanishing of the homogeneous $f$. (The equicharacteristic-$p$ example above ought to have been considered inside $\mathbf{P}^1$ just as well; I mistakenly thought being in $\mathbf{P}^2$ would provide better irreducibility properties, but those hold already with $\mathbf{P}^1$.) The $R$-scheme is $R$-flat because each of $U^p$ and $V^p$ have coefficients that are units in $R$ (ensuring that after dehomogenization, the coordinate ring of the affine open chart of $X$ is a finite free $R$-module). In fact, we see that $X$ is finite flat over $R$, with degree $p$.
The special fiber of $X$ is the zero scheme of $U^p + t V^p$ in $\mathbf{P}^1$ over $\mathbf{F}_p(t)$, and this is visibly reduced. On the other hand, the generic fiber is $(U+TV)^p=0$ in $\mathbf{P}^1_{F'}$, and this is visibly not reduced (but is irreducible, and likewise $X$ is irreducible; in fact, $X_{\rm{red}} = {\rm{Spec}}(R)$!).