1
$\begingroup$

I am trying to understand a paper, and I have a finite dominant rational map between two varieties $X$ and $Y$, where $Y$ is irreducible, but there is no such condition on $X$. The author goes on to talk about the degree of this finite dominant map. What is the definition of this, when $X$ is not irreducible?

I think that the definition may be: for all the irreducible components $X_{i}$ of $X$ such that $X_{i}$ maps dominantly to $Y$, $\sum \deg(X_{i}/Y) := \deg(X/Y)$. Does that seem reasonable?

Thanks!

EDIT: Here is the reference, http://msp.org/pjm/2001/200-1/pjm-v200-n1-p10-p.pdf on page 10, Theorem 2.7 (Going up), item (v).

$\endgroup$
0

1 Answer 1

0
$\begingroup$

I think I might know what's going on now.. We are in characteristic $0$: If we consider the ring of rational functions $k(X)$, where $X$ is our variety, then we can replace $X$ with the union of all the irreducible components $X_{i}$ such that $X_{i}$ maps dominantly to $Y$. Then, $k(X) \cong \oplus k(X_{i})$ where $k(X_{i})$ is a finite separable extension of the field $k(Y)$. That is, $k(X)$ is an etale algebra over $k(Y)$, and the degree of the map should be the degree of this algebra (notice that the degree of this map will also correctly count the cardinality of the preimages in the union of the components where $X$ maps dominantly to $Y$).

EDIT: I confirmed this with a faculty member at my university: it is the number of preimages in general position which may be calculated in the way I presented above.

$\endgroup$

You must log in to answer this question.

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