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Aug 9, 2011 at 14:41 comment added Sean Sather-Wagstaff 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!
Aug 9, 2011 at 14:35 comment added user2035 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.
Aug 9, 2011 at 14:34 comment added Sean Sather-Wagstaff 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.
Aug 9, 2011 at 14:24 comment added Sean Sather-Wagstaff 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?
Aug 9, 2011 at 14:18 vote accept Sean Sather-Wagstaff
Nov 28, 2011 at 3:42
Aug 9, 2011 at 9:10 comment added user2035 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 $\mathrm{Mod}(R,F^n)$.
Aug 9, 2011 at 5:31 answer added Peter McNamara timeline score: 2
Aug 8, 2011 at 21:06 history asked Sean Sather-Wagstaff CC BY-SA 3.0