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A closed subset of an algebraic group which contains $e$ and is closed under taking products is a subgroup of $G$.

Denote this set as $X$. If the condition of $X$ being closed is dropped, this statement does not hold. The set of nonzero integers in $\mathbb{G}_m$ over $\mathbb{C}$ is a counterexample.

It suffice to prove that for any $x \in X$, $x^{-1}X = X$. As $X$ is closed under taking products, it is clear that $X \subseteq x^{-1}X$. In order to prove the inverse inclusion, the closedness of $X$ (under Zariski topology) must be used.

Let $\phi: G \rightarrow G, y \mapsto x^{-1}y$ is a homoemorphism of $G$ as an algebraic variety. So $x^{-1}X = \phi(X)$ is a closed subset of $G$ containing $X$. But Why are they equal?

Thanks very much.

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I don't know why you're appealing to connectedness in your third paragraph. It's clear that a set satisfying your hypotheses needn't be connected (e.g. $\{\pm 1\} \subseteq \mathbb{G}_m$). –  David Loeffler Sep 30 '11 at 9:41
@David Loeffler: Thank you very much for the comment. That was a type error. I meant to write "closedness". Thank you very much for pointing this out. –  ShinyaSakai Oct 8 '11 at 3:28

2 Answers 2

up vote 6 down vote accepted

As David Loeffler points out, the (attempted) argument with connectedness is invalid. However, if $x \in X$, then $X \supset x X \supset x^2 X \supset \cdots \supset x^n X \supset \cdots$ is a decreasing chain of closed subsets of $G$, which by Noetherian induction must eventually stabilize. Thus in fact $X = x X$, as desired.

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Thank you very much for the answer. That was a type error. I meant to write "closedness". Can I use that fact that $X$ is a notherian topology space and the ascending chain: $X \subsetneq x^{-1}X \subsetneq x^{-2}X \subsetneq \cdots \subsetneq x^{-n}X \subsetneq \cdots$? I think Notherian property means the ascending chain condition. –  ShinyaSakai Oct 8 '11 at 3:48
@ShinyaSakai: Dear Shinya, Noetherianness is about ascending chains of ideals in the ring, and hence descending chains of Zariski closed subsets in the variety. (See for example the discussion of Noetherian schemes in Hartshorne.) Regards, –  Matt E Oct 9 '11 at 2:37
Dear Matt, thank you very much. I was wrong because I took it for granted that Notherianness meant ascending chain condition. On the book I am reading, the Notherianness of a topological space is defined to be the ascending chain condition on open subsets, thus descending chain condition on closed subsets. –  ShinyaSakai Oct 9 '11 at 15:21

Consider a fixed $x\in X$ and the multiplication morphism $m_x:X\to X:y\mapsto xy$. It suffices to prove that $m_x$ is surjective, since then there will exist $y\in X$ with $xy=e$ and so $y=x^{-1}$ will belong to $X$.
And now watch out for the conclusion:

The endomorphism $m_x$ is injective, hence surjective because of the Theorem of Ax-Grothendieck (or here )

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I didn't know about the theorem. Thank you very much for the answer. –  ShinyaSakai Oct 8 '11 at 3:57

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