Suppose that $X$ and $Y$ are smooth varieties over a field $k$ (not necessarily algebraically closed), of dimension $m$ and $n$. Suppose we are given a morphism $\pi:X\rightarrow Y$.
We know that if $\pi$ is smooth, it is flat (by definition) and all the fibers are smooth of dimension $m-n$.
Under these assumptions:
- if all the (geometric) fibers are smooth of dimension $m-n$, does it follow that $\pi$ is flat (and hence smooth, e.g. by Vakil, Theorem 25.2.2)
- more generally, can we say something about flatness of $\pi$ even if the fibers are not smooth, but still of constant dimension?
The first part is exercise 25.2.F from Vakil.
We can readily perform a number of reductions to simplify the question: the question is local on the target and on the source, so locally we can replace $Y$ by $\mathbb{A}^n_k$, by choosing an étale surjective map to $\mathbb{A}^n_k$. We can also replace $X$ by a smooth affine variety. It should also be sufficient to check this at closed points. Hence we have a morphism of rings $B \colon= k[y_1,\dots,y_n]\rightarrow A$, where A is a finitely generated $k$-algebra of dimension $m$ with $\Omega_{\mathrm{Spec }A/k}$ locally free of rank $m$, such that all the tensor products with maximal ideals of $B$ are (possibly smooth) of dimension $m-n$.
Does it follow that $A$ is a flat $B$-algebra?
I must say that I don't really know how to approach this. I can't seem to relate the flatness and smoothness conditions, and the answer to this question is clearly false if $Y$ is not smooth, think of the normalization of the node, where all fibers are smooth of dimension $0$ (and probably also if $X$ isn't, although I don't have a counterexample in my head).
Best Answer
If you know that both $X$ and $Y$ are regular (even Cohen-Macauly), than constant fiber dimension implies flatness. See this for example. I believe this answers your question.