MathOverflow is a question and answer site for professional mathematicians. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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?

share|cite|improve this question
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 $y + t^{p+1} = 0$.

share|cite|improve this answer

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.

share|cite|improve this answer
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

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.