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Let $f:X\to Y$ be a universal homeomorphism of regular (excellent finite-dimensional) schemes, $Z\subset Y$ be a regular subscheme. Is $f^{-1}(Z)$ necessarily regular?

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2 Answers 2

up vote 7 down vote accepted

Well, $f^{-1}Z$ could easily be non-reduced (for example, take the relative Frobenius morphism $\mathbb A^1_k \to \mathbb A^1_k$, defined by the embedding $k[y] = k[x^p] \subseteq k[x]$, where $k$ is a field of characteristic $p > 0$, and let $Z \subseteq \mathbb A^1$ be defined by $y = 0$), so I would guess that the question should be interpreted as asking whether $f^{-1}Z$ with its reduced structure is regular. But the answer is negative even in this case. For example, take the morphism $\mathbb A^2_k \to \mathbb A^2_k$ defined by the embedding $k[y,t] = k[x^p, t] \subseteq k[x, t]$, and let $Z \subseteq \mathbb A^2_k $ be defined by $y + t^{p+1} = 0$.

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Can't we base-change $f : X \to Y$ with $Z$ and obtain: $g : f^{-1}(Z) \to Z$? This also is a universal homeomorphism by construction, right? So now we have a universal homeomorphism to a regular scheme, but a regular scheme is weakly normal, see A. Andreotti and E. Bombieri, ``Sugli omeomorfismi delle varietà algebriche''.

Therefore $g$ is an isomorphism at least as long as $f^{-1}(Z)$ is reduced and the map $g$ is birational.

EDIT: My argument that $f^{-1}(Z)$ was reduced was junk. I shouldn't have tried to do math while on the run. But as long as $f^{-1}(Z)$ is reduced and $g$ is birational, then I think things are ok.

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Sorry, why $f$ being a universal homeomorphism contradicts its generic inseparablity? –  Mikhail Bondarko Mar 20 '11 at 13:09
I was being dumb. Nevermind. –  Karl Schwede Mar 20 '11 at 18:57

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