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).

  • $\begingroup$ I misread your question, answered. Reread your question incorrectly again, answered again. Finally I think I've fixed it. I believe below is what you want. If you know that $X$ and $Y$ are both smooth over $k$, then the constant fibral dimension implies flatness, and so implies smoothness. Let me know if I have (for a third time!) misinterpreted you. :) $\endgroup$ – Alex Youcis Nov 22 '14 at 12:07
  • $\begingroup$ This time it's correct. Thank you! $\endgroup$ – Jakob Oesinghaus Nov 22 '14 at 12:08
  • $\begingroup$ No problem. Sorry for the (initial) confusion. :) $\endgroup$ – Alex Youcis Nov 22 '14 at 12:09

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.


I guess that one should add some condition - that $X$ should be connected. Consider $(\mathbb{A}^1 - 0) \coprod pt \to \mathbb{A}^1$; Source and target are smooth of dimension 1, fibers are smooth of dimension 0, target and fibers are all connected, but the morphism is not flat.

  • $\begingroup$ Sorry, this is not correct. Scheme-theoretic fibers are not smooth. $\endgroup$ – Sasha Nov 24 '19 at 1:56

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