Hi,

I have a general question concerning the closed points $X^{0}$ of a variety $X$ over a field $k$: one always hears that the properties of $X^{0}$ are "equivalent" to those of $X$, because it is dense in $X$.

Can you give me some hint what is meant with that rough statement?

For example, I have the following problem:

if I have two morphisms $f,g$ from $X$ to $Y$ ($Y$ another variety) which coincide on $X^0$ (as maps of sets or as morphisms?), then they are equal. How would you formally prove this? The problem for me is that $X^0$ is not open in $X$, so a number of theorems about equality of morphisms dont work here.

Thanks!

morphisms, not just as maps of sets. – A. Pascal Aug 22 '11 at 9:05reduced, isn't it the same even if they coincide just as maps of sets? – Qfwfq Aug 22 '11 at 11:12