I have been trying to understand a sketch of a proof from P. Gabriel's article "Finite representation type is open".

Let $F$ be an algebraically closed field, and let $R$ be a finite dimensional $F$-algebra. Let $\operatorname{Mod}(R,F^n)$ denote the set of all $R$-module structures on $F^n$. Then $\operatorname{Mod}(R,F^n)$ has the structure of a scheme over $F$. Furthermore, the algebraic group $G = GL(F^n)$ acts on $\operatorname{Mod}(R,F^n)$, essentially by conjugation.

Let $M$ be a fixed $R$-module structure on $F^n$. The orbit $G.M$ is locally closed in $\operatorname{Mod}(R,F^n)$, and it is given its reduced scheme structure. Then the natural map $f\colon G \to G.M$, given by multiplication by $M$, is faithfully flat. Gabriel claims that $f$ is smooth. The way he draws this conclusion is by saying that the stabilizer of $M$ is $\operatorname{Aut}_R(M)$, which is an open subset of $\operatorname{Hom}_R(M,M)$, so it is smooth. My question has to do with understanding this.

I am assuming that the stabilizer $\operatorname{Stab}(M)$ is the fibre of $f$ over $M$, which has a natural scheme structure. Since $F$ is algebraically closed, and everything is finite type over $F$, to prove that $f$ is smooth, it suffices to show that this fibre is smooth over $F$. Note that this approach means that I have to take the natural scheme structure on the fibre.

The closed points of this fibre are definitely equivalent to the set of closed points of $\operatorname{Aut}_R(M)$. So, my questions are:

Why are $\operatorname{Stab}(M)$ and $\operatorname{Aut}_R(M)$ isomorphic as schemes?, and why is $\operatorname{Aut}_R(M)$ smooth over $F$? Any help would be appreciated.

  • 1
    $\begingroup$ As suggested by Peter's answer, you should work out the functor of points for both $\mathrm{Aut}_R(M)$ and $\mathrm{Stab}(M)$, reducing the problem to ordinary groups. For the fibre of $f$, it does not matter whether the target is $G.M$ or $\mathrm{Mod}(R,F^n)$. $\endgroup$ – user2035 Aug 9 '11 at 9:10
  • $\begingroup$ This confuses me a bit. I know that the underlying groups for $\operatorname{Aut}_R(M)(S)$ and $\operatorname{Stab}(M)(S)$ are the same. The question is whether their structure sheaves are the same. In particular, I know that the structure sheaf on $\operatorname{Aut}_R(M)$ is reduced, and I need to know whether the same is true for the structure sheaf on $\operatorname{Stab}(M)$. Is this obviously reduced? $\endgroup$ – Sean Sather-Wagstaff Aug 9 '11 at 14:24
  • $\begingroup$ Egads. I think I see it. The point is that the equations defining Stab(M)(S) are also linear: $ABA^{-1}=B$ means that $AB=BA$, and this is a linear system when $B$ is fixed. Ach. $\endgroup$ – Sean Sather-Wagstaff Aug 9 '11 at 14:34
  • $\begingroup$ Giving an isomorphism $X\cong Y$ of schemes is equivalent to giving a compatible collection of isomorphisms $X(S)\to Y(S)$, by Yoneda's lemma. $\endgroup$ – user2035 Aug 9 '11 at 14:35
  • $\begingroup$ Definitely true. But I was only seeing the isomorphisms of the groups, not the schemes. For instance, I was worried about the difference between $\operatorname{Spec}(F)$ and $\operatorname{Spec}(F[x]/(x^2))$. The point is that the defining equations are linear. Thanks again for the help! $\endgroup$ – Sean Sather-Wagstaff Aug 9 '11 at 14:41

Lets write π:R->End(M) for the representation.

Let S be a F-algebra. We first eludicate the S-points of $Aut_R(M)$. We have $$Aut_R(M)(S)=\{ g\in GL_n(S)| g\pi(r)g^{-1}=\pi(r)\forall r\in R \}.$$ We also consider the S-points of Hom(M,M). Hom(M,M) has the same functor of points, except we replace the $g\in GL_n(S)$ condition by $g\in Mat_n(S)$. Note that Hom(M,M) is an affine space, since all conditions are linear. And we read off from their functors of points that $Aut_R(M)$ is the open set in Hom(M,M) where the determinant function is invertible. So $Aut_R(M)$ indeed is open in an affine space, hence smooth.

  • $\begingroup$ Why won't \{ and \} work? $\endgroup$ – Peter McNamara Aug 9 '11 at 5:31
  • $\begingroup$ @Peter: I've edited the TeX to make the brackets show up; you have to put the character ` around the dollar signs. It's a weird quirk of MO. $\endgroup$ – Daniel Litt Aug 9 '11 at 5:33
  • $\begingroup$ @Daniel. Thanks. I knew in theory about the fix to the bug caused by markdown, but had forgotten the particulars. $\endgroup$ – Peter McNamara Aug 9 '11 at 5:36
  • $\begingroup$ Of course. The point is that the equations are linear. I hate it when the answer is that easy. Thanks a lot! $\endgroup$ – Sean Sather-Wagstaff Aug 9 '11 at 14:19

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