Let $X, Y$ be quasi-projective Noetherian schemes and $f:X \to Y$ be a projective surjective morphism. Assume that every fiber of $f$ is isomorphic to a projective space $\mathbb{P}^n$ for a fixed $n$. Is it then true that $f$ is flat?
[Math] Is projective morphism with projective fiber flat
ag.algebraic-geometrydeformation-theoryhilbert-schemes
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Work locally, assume that $f: R\to S$ is a local homomorphism. Let $\operatorname{cmd}(R) =\operatorname{dim} R-\operatorname{depth} R$ (this is the so-called Cohen-Macaulay defect of $R$). Claim: $\operatorname{cmd}$ is preserved by l.c.i maps (easy, essentially because both depth and dimension drop by one when you kill a regular element).
Now since the map $\phi: \operatorname{Spec}(S) \to \operatorname{Spec} (R)$ is finite and surjective, $\operatorname{dim} R= \operatorname{dim} S$, which combines with the last claim to show that $\operatorname{depth} R = \operatorname{depth} S$. But since l.c.i also implies finite flat dimension, we have $\operatorname{depth} R -\operatorname{depth} S = pd_RS$, so $S$ is flat over $R$.
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, $X_{\rm{red}} = {\rm{Spec}}(R)$!).
Best Answer
If $Y$ is non-reduced, then $X = Y_{red} \times \mathbf{P}^{n}$ is a counterexample.