I have seen a theorem attributed to Chevalley, to the effect that a sub-semi-group in an algebraic group is Zariski dense if and only if the subgroup it generates is Zariski dense. Would anyone happen to have a reference?
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asked Aug 7, 2016 at 1:02
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5$\begingroup$ Presumably the algebraic group $G$ is over an algebraically closed field $k$ and irreducible (to avoid silliness), so it is finite type (not just locally so). The Zariski-closure of a sub-semigroup is a sub-semigroup, so you want that if $M$ is a Zariski-closed subsemigroup and $M(k)$ generates $G(k)$ then $M=G$. It suffices to show $\dim M = \dim G$. By irreducibility of $G$, it suffices to show every element of $G(k)$ is a "word" in a bounded number of letters from $M(k)$. Now see Prop. 2.2 in Ch. I of Borel's textbook on algebraic groups. Chevalley's constructible image theorem is the crux. $\endgroup$– nfdc23Aug 7, 2016 at 1:44
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Here is one proof. It suffices to show a Zariski closed submonoid M of an algebraic group G is an algebraic group. Let x be in M. The left translation by x is an injective polynomial morphism from M to M and hence surjective by Ax-Grothendieck. So xy=1 for some y in M. Since G is a group y is the inverse of x in G.
You should assume here the field is algebraically closed off course.
answered Aug 7, 2016 at 2:54
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$\begingroup$ A very cool argument (though probably not the one Chevalley had in mind :)) $\endgroup$ Aug 7, 2016 at 3:01
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$\begingroup$ I think I learned it from either Putcha or Renner's book on algebraic monoids or from one of Brion's papers. I think there is also in one of those books a direct proof that the inverse of a matrix is in the Zariski closure of its powers. $\endgroup$ Aug 7, 2016 at 3:08
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$\begingroup$ One interesting point is that this seems morally close to the "strong approximation" proof that I had in mind for cases of interest to me (Zariski dense-> modular projections are onto, but that last condition is obviously the same for groups as for semi-groups. Which is somehow very much Ax-Grothendeickian. $\endgroup$ Aug 7, 2016 at 3:23
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$\begingroup$ You can even avoid Ax-Grothendieck by observing that the translation (being an automorphism of $G$ as a variety) induces an isomorphism of $M$ with a closed subset of $G$ (hence of $M$). In other words, it is a closed immersion of $M$ into itself! The rest is an exercise (especially since $M$ is reduced) via dimension arguments. $\endgroup$ Aug 8, 2016 at 15:05
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$\begingroup$ @Laurent Moret-Bailly, I think you can just use Noetherian condition on the descending chain of closed subsets xM > x^2M>... to get stabilization and then use cancellation. $\endgroup$ Aug 8, 2016 at 20:25