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I have to annoy you just one further time with these closed subset stories.

I am trying to make rigorous a proof, in which the author tries to show an equality of closed subsets $Y$ and $Z$ of an abelian variety $A$.

He shows that whenever you take a closed (!) point on $A$ which is contained in $Z$, then it is also contained in $Y$. Then he concludes that $Z$ is a subset of $Y$. Why can he do this?

I tried to make it rigorous by thinking locally, but it didn't work because the jacobson radical of an ideal is not the radical of an ideal.

I tried it with general topology, but I didn't see what the essential point should be.

Surely one has to remark that the closed points on $A$ are dense in $A$.

Furthermore, I can remark that the set $Y$ is actually the graph of the inversion morphism $i:A\rightarrow A$. The set $Z$ is the support of a line bundle.

Thanks a lot!

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What is your ground field? algebraically closed? – Soarer Sep 11 '11 at 18:04
Yes, you can assume that, also char zero. – Descartes Sep 11 '11 at 19:38
That is simply Hilbert's Nullstellensatz. – AFK Sep 11 '11 at 19:45
up vote 0 down vote accepted

For any subset $S\subset A \;$ of a scheme $A$ , call $S^{cl}$ the set of closed points of $A$ contained in $S$.
Then :

Let $k$ be a field. If the $k$- scheme $A$ is locally algebraic, then $A^{cl}$ is very dense in $A$.

i) That $A$ is locally algebraic means that $A$ can be covered by open sets isomorphic to $Spec(R)$, with $R$ a finitely generated $k$-algebra .
ii) That $A^{cl}$ is very dense in $A$ means that for all closed subsets $Z\subset A$, the subset $Z^{cl}$ is dense in $Z$, that is $\bar {Z^{cl}}=Z$

Application to your problem The result you want is now immediate: since your hypothesis is $Z^{cl} \subset Y $, you immediately deduce $\bar{ Z^{cl}}=Z\ \subset \bar Y=Y$

As you see the fact that the base field is algebraically closed or of characteristic zero is irrelevant, and similarly the hypothesis that $A$ is abelian or even that $A$ is complete is also supefluous!

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I also tried it this way, Georges, but then I thought: ok, you have $Z^{cl}=Z\cap A^{cl}$, but now in your last argument you are taking the closure of $Z^{cl}$ and $Y$ in $A$ to conclude. But why is the closure of $Z^{cl}$ in $A$ equal to $Z$? – Descartes Sep 12 '11 at 6:33
Dear @Descartes, the closure of $Z^{cl}$ in $A$ is $Z=\bar Z$ because this is what "very dense" means . This equality $\bar { Z^{cl}}=Z$ is stated explicitly at the end of ii). – Georges Elencwajg Sep 12 '11 at 11:41
Dear Georges, yes I registered that, but I wasn't aware that the closure is transitive. With that you get it. Great, thanks for your effort! – Descartes Sep 12 '11 at 13:41

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