# When are all the fibers of a morphism reduced?

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?

• There are very nice (smooth, projective, rational...) surfaces admitting a fibration over $\Bbb{P}^1$ with some multiple fibers.
– abx
Oct 8 '18 at 18:24
• 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? Oct 8 '18 at 18:32
• Yes, of course. So what?
– abx
Oct 8 '18 at 18:49
• When you say "I am able to show that the underlying reduced schemes...", I assume you are using additional assumptions you haven't listed? Oct 8 '18 at 18:52
• @abx, I have assumed that "each fiber is generically reduced." Oct 8 '18 at 19:01

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.

• You need to assume $f$ surjective (or dominating). Oct 9 '18 at 7:00
• 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? Oct 9 '18 at 11:20
• 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 $char(k)=0$? Oct 16 '18 at 18:45
• Nevermind: Got a copy. The claim follows from Theorem 6.3(4) of Jouanolou's book. Oct 16 '18 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.

• 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. Oct 8 '18 at 17:33