3
$\begingroup$

Let $X,Y$ be complex projective varieties with $X$ irreducible, and let $f:X\dashrightarrow Y$ be a rational map. If $U\subseteq X$ is the largest open set where $f$ can be defined, is it true that $\mathrm{codim}_{X}(X\setminus U)\geq 2$. I know this is true if $X$ is smooth.

EDIT: In view of the inkspot's answer, I add: if $D\subset X$ is an irreducible divisor on $X$, with $D\cap\left(X\setminus\mathrm{sing}(X)\right)\neq\emptyset$, can it happen that $D\cap U=\emptyset$?

Thanks.

Improve this question
$\endgroup$

2 Answers 2

Reset to default
1
$\begingroup$

If $D'=D\cap$X\sing(X) is not empty then it is an irreducible divisor on X\sing(X) (which is smooth). So you can apply the very same theorem to X\sing(X) and D' to get that the thing you describe in the edit can't happen and f can be defined almost everywhere on D'.

Improve this answer
$\endgroup$
0
5
$\begingroup$

More generally, the answer is yes if $X$ is normal. But if $X$ is a curve with non-empty singular locus $S$ and normalization $Y$, then the natural (bi)rational map $X- \to Y$ is undefined on $S$, which has codimension $1$.

Improve this answer
$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .