MathOverflow is a question and answer site for professional mathematicians. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Given a flat map $f: X \rightarrow Y$ such that $X$ is a projective variety and $Y$ is a smooth curve. Each generic fiber is isomorphic to an irreducible projective variety $A$ of dimension $d$.

The special fiber of $f$ must also be of dimension $d$ because of flatness, but does the special fiber have to be equi-dimensional? If yes, any proofs or references? If no, any counter-examples? Thanks!

share|cite|improve this question
up vote 6 down vote accepted

It must be set-theoretically equidimensional, but not scheme-theoretically. (Consider two lines in space colliding, developing an embedded point at the intersection.)

For the positive statement, let $d\leq e$ be the smallest and largest dimensions occurring among the components. Slice all fibers with the same general codimension $d$ plane, to get $X'$. Now the special fiber has some isolated points, that were in its $d$-component. If $e>d$, then the general fiber is still irreducible by Bertini, so connected. But Zariski's Main Theorem prevents you from degenerating connected to disconnected.

There should be a classical reference, I presume, but I don't know one, and just put this argument in proposition 2 of this paper on degenerations.

(Also, you probably don't mean "each generic fiber" -- the generic fiber is the fiber over $Y$'s (only) generic point. I'm guessing you mean "each general fiber". To see the confusion available here, consider the squaring map $\mathbb A^1 \to \mathbb A^1$, whose general fibers have two points but whose generic fiber is irreducible.)

share|cite|improve this answer
1  
I think Zariski's Main Theorem may be an overkill here, because the statement holds without the projectivity assumption. In fact, can't you just say that for every point $x\in X$, dimension of the fiber at x and dimension of X at x differ by 1 (Hartshorne, Proposition III.9.5). From this one easily sees that every component of X must dominate Y, therefore X has pure dimension d+1, and each fiber has pure dimension d. – t3suji Mar 2 at 20:24
    
As t3suji says, it's easier without projectivity: can localize on $X$! For $f:X \rightarrow Y$ flat surjective of finite type with $Y$ noetherian and connected, if one fiber is equidimensional of dimension $d$, we claim all are. Link irreducible components of $Y$ and specialize from its generic points to reduce to base a dvr. Closures of generic fiber irred. components of $X$ are $Y$-flat, and they cover $X$ since $X$ is $Y$-flat, so enough to show for $X$ irred. with generic fiber of dimension $d$ that dim$(X_0)=d$ (so $X_0$ equidim'l by Zariski-localizing on $X$). Now EGA IV$_3$, 14.3.10. – nfdc23 Mar 2 at 22:00
    
I suppose I remembered the argument I gave because it also gets you that the special fiber is connected in codimension $1$ (which wouldn't be true without the properness assumption). – Allen Knutson Mar 3 at 12:58

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.