# Closed points on a variety

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!

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The formal statement is that the category of varieties in the scheme-sense is equivalent to the category of varieties in the classical sense (where each point is closed). You can find this everywhere, e.g. in Hartshorne chapter II. – Martin Brandenburg Aug 22 '11 at 8:01
Thank you a lot, Martin, I think that's what I wanted! – Descartes Aug 22 '11 at 8:22
It's important that the maps $X^0 \rightarrow Y^0$ agree as morphisms, not just as maps of sets. – A. Pascal Aug 22 '11 at 9:05
@A.Pascal: but $X$ and $Y$ being reduced, isn't it the same even if they coincide just as maps of sets? – Qfwfq Aug 22 '11 at 11:12
If $X=Y=\mathbb{A}^{1}$ over $k=\mathbb{F}_{2}$, then the identity map $x \mapsto x$ and the the squaring map $x \mapsto x^{2}$ have the same values on $k$-points, but are different as maps of schemes. Of course, you can distinguish them by looking at maps on points over field extensions of $k$. Such points do correspond to closed points of the underlying scheme, but it seems to me a little strange to talk about old-fashioned non-schemey varieties unless you are over an algebraically closed field. I believe Hartshorne presumes his ground field is algebraically closed for classical case. – A. Pascal Aug 22 '11 at 13:34