This is a sort-of follow up to [this question](https://mathoverflow.net/questions/312047/smooth-equidimensional-fibers-over-a-smooth-base), 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](https://stacks.math.columbia.edu/tag/0574), 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?
> 
> 
> 



When are all the fibers of a morphism reduced?
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.


