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?

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    $\begingroup$ There are very nice (smooth, projective, rational...) surfaces admitting a fibration over $\Bbb{P}^1$ with some multiple fibers. $\endgroup$
    – abx
    Oct 8, 2018 at 18:24
  • $\begingroup$ This seems to contradict Snowball's answer below, since such a surface is clearly Cohen-Macaulay. Wouldn't such a multiple fiber have to be everywhere non-reduced? $\endgroup$ Oct 8, 2018 at 18:32
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    $\begingroup$ Yes, of course. So what? $\endgroup$
    – abx
    Oct 8, 2018 at 18:49
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    $\begingroup$ When you say "I am able to show that the underlying reduced schemes...", I assume you are using additional assumptions you haven't listed? $\endgroup$
    – Will Chen
    Oct 8, 2018 at 18:52
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    $\begingroup$ @abx, I have assumed that "each fiber is generically reduced." $\endgroup$ Oct 8, 2018 at 19:01

2 Answers 2


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.

  • $\begingroup$ You need to assume $f$ surjective (or dominating). $\endgroup$ Oct 9, 2018 at 7:00
  • $\begingroup$ Thank you very much for the answer! I have a few questions as I try to understand: In step 3, you say ``for any $y\in Y(k)$ we can pick $Z$...'' what is the relation between $y$ and $Z$? Also, in step 4, can you elaborate on the induction argument? $\endgroup$ Oct 9, 2018 at 11:20
  • $\begingroup$ I want to accept this answer. I have been able to understand the entire argument, except for the reference to ``a Bertini theorem (a la Jouanolou).'' I don't have access to his book, and am not sure which result you're referring to. In particular, are you assuming $char(k)=0$? $\endgroup$ Oct 16, 2018 at 18:45
  • $\begingroup$ Nevermind: Got a copy. The claim follows from Theorem 6.3(4) of Jouanolou's book. $\endgroup$ Oct 16, 2018 at 19:44

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.

  • $\begingroup$ Thank you for the answer. I have not been able to see that $X$ is Cohen-Macaulay, which is why I did not include that assumption. The question I linked to assumed that the fibers were smooth, and the conclusion was that $X$ was CM. $\endgroup$ Oct 8, 2018 at 17:33

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