# Equi-dimensionality of special fibers in a flat family

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!

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.
(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.)
• 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 '16 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 '16 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 '16 at 12:58