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

Hi, everybody.

Consider an ${\rm S}_{1}$- morphism $f:X\rightarrow S$ of reduced complex spaces. Assume that $f$ is open (universally open in Alg.geom), equidimensional with $n$-pure dimensional fiber, surjectiv. Let $U$ be the flat locus of $f$ (which is a dense open set).

Question: It is true that the codimension of $(X-U)\cap X_{s}$ is of codimension 2 in the fiber $X_{s}$ ?

Remark: We can refer to the Thm 15.2.2, p.226 and Prop 4.7.10 of [EGA].

Thank you very much...

share|cite|improve this question
[EGA]? $\mbox{}$ – Joseph O'Rourke Sep 2 '10 at 0:04
"Éléments de géométrie algébrique" is the bible of this sect. – Donu Arapura Sep 3 '10 at 13:08

Take $X=\mathbb{C}$, $S$= the cuspidal plane cubic $y^2=x^3$, and $f$= the normalization map $t\mapsto (t^2,t^3)$. This is a universal homeomorphism. The flat locus is $U=\mathbb{C}^*$, so $X\setminus U$ is the whole fiber at $(0,0)$.

share|cite|improve this answer

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.