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(1) Let $R$ be a ring, let $A$ be a finitely presented $R$-algebra, and let $G$ be a finite group of $R$-automorphisms of $A$. Is the algebra of invariant $A^G$ finitely presented over $R$?

I can prove the statement when $R$ is noetherian, but I don't know how to generalize. The previous question is concerned with the following one.

(2) Let $A \to B \to C$ be homomorphisms of rings. If $A \to C$ is of finite presentation and $B \to C$ is finite, then is $A \to B$ of finite presentation?

Can one prove or disprove (1) or (2)? Thanks to all!

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2 Answers 2

Answer to (2): no. Consider $B=A[T]/I$ with $I\subset TA[T]$ and $B$ not finitely presented. Let $C=A$. Then the canonical map $B\to A[T]/(T)=A$ is surjective hence finite, $C$ is of course f.p. over $A$.

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Thanks! I had the suspect that the answer to (2) was no. But my real interest is in (1) and I hope that it has an adfirmative answer. –  Andrea Aug 15 '12 at 13:55
    
@Andrea, I understand. But I don't have any idea on (1). –  user18119 Aug 16 '12 at 14:57
up vote 1 down vote accepted

Maybe the following argument proves (1) under the hypothesis that $\vert G \vert$ is invertible in $R$. Is it correct?

Since $A$ is finitely presented over $R$, there exist a subring $R_\alpha$ of $R$ and an $R_\alpha$-algebra $A_\alpha$ of finite type such that $R_\alpha$ is of finite type over $\mathbb{Z}$ and $A \simeq R \otimes_{R_\alpha} A_\alpha$. Consider the filtrant inductive system $\{R_\lambda \}_{\lambda \geq \alpha}$ of subrings $R_\lambda$ of $R$ which are finitely generated extension of $R_\alpha$. Set $A_\lambda = R_\lambda \otimes_{R_\alpha} A_\alpha$. We are in the situation of [EGA Lemme IV.8.8.2.1], hence the $R$-automorphisms $g_1, \dots g_n$ of $A$ come from $R_\lambda$-homomorphism $g_1^\lambda, \dots, g_n^\lambda$ of $A_\lambda$, for $\lambda \gg \alpha$. The injectivity of the homomorphism (8.8.2.2) implies that, for $\lambda \gg \alpha$, the $g_i^\lambda$ are automorphisms and satisfy the presentation of the group $G$. Besides, if $G$ acts faithfully on $A$, we can require that it acts faithfully also on $A_\lambda$, for some $\lambda \gg \alpha$. Since $\vert G \vert$ is invertible in $R$, then it is invertible in $R_\lambda$, then taking invariants commutes with base change: $A^G = (R \otimes_{R_\lambda} A_\lambda)^G \simeq R \otimes_{R_\lambda} A_\lambda^G$, then $A^G$ is of finite presentation over $R$ because $A_\lambda^G$ is of finite type over the noetherian ring $R_\lambda$.

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I think your proof is correct. –  user18119 Nov 7 '12 at 22:05

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