# A commutative ring which has a group acting locally finitely on itself

The following definitions and proposition are motivated by this question.

All rings are assumed to be commutative and have identity elements.

Definition 1 Let $B$ be a ring. Let $A$ be a subring of $B$. An element $b$ of $B$ is called radical over $A$ if there exists an integer $n > 0$ such that $b^n \in A$. $B$ is called radical over $A$ if every element of $B$ is radical over $A$.

Definition 2 Let $B$ be a ring. Let $A$ be a subring of $B$. Let $Aut(B/A)$ be the group of automorphisms of $B$ fixing every element of $A$. Let $G$ be a group. Let $\psi\colon G \rightarrow Aut(B/A)$ be a homomorphism. Then $G$ acts on $B$ via $\psi$. If every $G$-orbit is finite, we say $G$ acts on $B$ locally finitely over $A$. We denote by $B^G$ the set of elemetns of $B$ which are invariant under $G$. Clearly $B^G$ is a subring of $B$ containing $A$.

Example Let $k$ be a field. Let $\bar k$ be an algebraic closure of $k$. Let $char(k)$ be the characteristic of $k$. We define an integer $p$ as follows.

$p = char(k)$ if $char(k) > 0$.

$p = 1$ if $char(k) = 0$.

Let $k^{p^{-\infty}} = \{x \in \bar k$| there exists an integer $n \ge 0$ such that $x^{p^n} \in k\}$.

Let $A = k[X_1,\dots,X_n]$, $B = \bar k[X_1,\dots,X_n]$ be polynomial rings. Let $G = Aut(\bar k/k)$. Clearly $G$ acts on $B$ locally finitely over $A$ in the obvious way. It is easy to see that $B^G = k^{p^{-\infty}}[X_1,\dots,X_n]$. Hence $B^G$ is radical over $A$.

My question Is the following proposition true? If yes, how do we prove it?

Proposition Let $B$ be a Noetherian ring. Let $A$ be a subring of $B$. Let $G$ be a group acting on $B$ locally finitely over $A$. Suppose $B^G$ is radical over $A$. Let $p$ be a prime ideal of $A$. Let $\Pi$ be the set of prime ideals of $B$ lying over $p$. Then the following assertions hold.

1) $B$ is integral over $A$.

2) $\Pi$ is non-empty and finite and $G$ acts on $\Pi$ transitively.

3) Going down theorem holds on $B/A$.

4) Let $P$ be a prime ideal lying over $p$. Then dim $A_p =$ dim $B_P$.

5) $\Pi$ is the set of minimal prime ideals containing $pB$.

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