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This is related to this old question, which doesn't seem to address my exact situation.

Suppose I have a smooth morphism $\varphi: X\to Y$ of smooth varieties over an algebraically closed field, and let $E\subset X$ and $F\subset Y$ be such that

1) $F\subset Y$ is an irreducible, smooth closed subscheme,

2) $E\subset X$ is irreducible, reduced closed subscheme, and

3) the restriction $\varphi|_{E}:E\to F$ is surjective with smooth equidimensional (though not necessarily connected) fibers.

My question is: is this map flat? smooth? or are there simply counterexamples to either? The above cited question did not assume the equidimensionality of the fibers (hence, the counterexample of a blow up), nor the smoothness of the base.

The answers to this other related question do not help me either, since here I assume that $F$ and the fibers of the morphism are smooth.

If I knew that $E$ were Cohen-Macaulay, then the restriction would be flat by miracle flatness, so I might hope that this is forced by the restrictions, but am not sure how to prove it.

Spencer Leslie
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  • If $\varphi|_E$ is flat and of finite presentation, then it is smooth - since finitely presented should be either automatic or easy to check in your case, the whole thing boils down to flatness. If $E\subset X\times_Y F$ is a flat subscheme, everything is fine because $\varphi|_E$ is equal to the composite $E\to X\times_Y F\to F$ which is a composition of flat morphisms. It's hard to see how to use this perspective when $E\subset X\times_Y F$ isn't flat, though, which would be the interesting case. – KReiser Oct 02 '18 at 20:25
  • Thank you for your comment. This point is why I hoped to see that the map was flat. In general, the map $E\subset X\times_Y F$ will not be flat, but I was hopeful that the situation still allowed me to see that $E$ was Cohen-Macaulay. Are there counter examples where $E\to F$ is not flat? – Spencer Leslie Oct 03 '18 at 15:03

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