When are all the fibers of a morphism reduced?

Question

This is a sort-of follow up to this question, which I asked before I became confused about if things were reduced.

More specifically, suppose $\varphi:X\to Y$ is a surjective morphism of finite presentation between algebraic varieties (reduced, irreducible, separated schemes, finite type over an algebraically closed field), and furthermore assume that $Y\cong \mathbb{A}^r$ is an affine space.

I am able to show that the underlying reduced schemes of all the fibers $X_y$ for $y\in Y$ are smooth and all equidimensional of dimension $k>0$ (In my case $k=r$, so $\dim(X)=2r$), and also that each fiber is generically reduced. I even know that the generic fiber is reduced, so I know $X_y$ is reduced for $y\in U\subset Y$ in some nonempty open.

Can I conclude that the fibers are reduced?

More generally, under relatively simple circumstances, one can get that fibers over a Zariski-open are reduced. Is there some simple criteria as to when this may be strengthened to all fibers? (short of assuming $\varphi$ is smooth, for example.)

EDIT: As pointed out by Snowball, $X$ being Cohen-Macaulay would suffice. Perhaps a better formulation of the question is

Is $X$ necessarily Cohen-Macaulay, and if not are their known counterexamples?

Answer 1

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.

Answer 2

If you know that $X$ is Cohen-Macaulay (e.g. if $X$ is smooth or $X$ is a local complete intersection) then you know that each fiber is Cohen-Macaulay (because the fibers are complete intersections in a Cohen-Macaulay scheme). You can check that a Cohen-Macaulay scheme is reduced by checking at the generic point of each component, so you would be done by your assumptions.

This is the best criterion I can imagine, but perhaps there are more creative people out there.