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Let $R$ be $p$-torsion free ($p$-prime). We know that $\mathrm{Aut}_{R/pR}((R/pR)[x]) = \mathrm{AL}_1(R/pR)$.

Can someone give me a concise description of $$\mathrm{Aut}_{R/p^nR}((R/p^nR)[x])$$ for $n> 1$? A reference would be nice.

What about multiple variables? What does $$\mathrm{Aut}_{R/p^nR}((R/p^nR)[x_1,\ldots,x_n])$$ look like?

If there isn't an easy description can one describe the normal subgroups/group homomorphisms--- in particular what homomorphisms to $(R/p^nR)$ exist? Do you know of any homomorphism to any algebraic groups? I can only seem to identify the reduction mod $p$ homomorphism to the affine linear group.

References would be stellar. This group makes me feel dumb. I can tell you that an automorphism $n=2$ is induced by a polynomial of the form $a_0x+b_0 + p a(x)$ by using $f(g(x)) =x$ implies $f'(g(x))g'(x)=1$ then studying the multiplicative units in $R[x]/p^nR[x]$ but that hasn't gotten me very far in understanding more about this group.

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about the homomorphism part: A ring homomorphism R[x]-->R' is just given by a ring homomorphism R-->R' and a (free!) choice of an image of $x$. So the remaining question is, when is such a homomorphism a iso. (I hope we are talking about rings and not about objects in some fancy category). –  HenrikRüping Oct 7 '10 at 14:27
    
What is $\rm{AL}_1$ supposed to mean? Does it denote the affine group $\rm{Aff}(1,R/pR)$? –  Qfwfq Oct 7 '10 at 15:13
    
-Yes, for me $\mathrm{AL}_1 = \mathrm{Aff}(1,R/pR)$ the affine linear group. -The application is to $R = \mathcal O(U_{ij})$ where $\mathcal O$ is a sheaf of rings and $U_{ij}=U_i\cap U_j$, the intersection of two affine open sets in a cover. The Automorphism is induced by two local trivializations on special subsets of a (formal) scheme over another scheme. I'm actually interested in the automorphisms of $R[x]^{\wedge}$ which corresponds to the completion of a scheme at a prime $p$. Evaluations maps don't help me too much. –  user8248 Oct 7 '10 at 16:02
    
Wait, I don't think evaluation maps are allowed. $\mathrm{Aut}_R$ is not a functor. –  user8248 Oct 7 '10 at 23:20

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