The set of prime ideals, whose contraction is a fixed prime in the ring of invariants, is finite.

Question

Let $A$ be a domain and $G$ a finite group of automorphisms of $A$. I define $$A^G=\{a\in A\mid\sigma(a)=a ,\forall\sigma\in G\}.$$

Now take a prime ideal $\mathfrak{p}\subset A^G$ and define $P=\{\mathfrak{P}\subset A \text{ prime ideal}\mid \mathfrak{P}\cap A^G=\mathfrak{p}\}$.

I want to show that $P$ is a finite set.

(Atiyah and Macdonald, Introduction to Commutative Algebra, Exercise 13, Chapter 5.)

I'm a bit at a loss here, how would one go about proving this? So far I have no real results, I only played around with the definitions a bit but don't know where to start this really.

I got a hint though (which I don't understand): For $\mathfrak{P}_1,\mathfrak{P}_2\in P$ I am supposed to show that $\mathfrak{P}_1\subseteq\bigcup_{\sigma\in G}\sigma(\mathfrak{P}_2)$, which in turn apparently would lead to a proof of the claim. I can't make sense of this.

Answer 1

Let me first change some notation. Let $\sum = \{ P \hbox{ prime in } A \mid P \cap A^G = p \}$.

Observation:

a) For $a$ in A, $a$ in $A^G$ iff $\rho(a)$ in $A^G$ for $\rho \in G$.

b) For any $\rho \in G$, $\rho(P \cap A^G) = \rho(P) \cap A^G$.

Proof: b) $\rho(P \cap A^G) \subset \rho(P) \cap \rho(A^G) = \rho(P) \cap A^G$. Suppose $a \in \rho(P) \cap A^G$. Then there exists $b$ in $P$ such that $a = \rho(b)$. By a), $b$ in $A^G$, so $a \in \rho(P \cap A^G)$.

Lemma: Let $P,P'$ prime ideal of $A$ such that $P \cap A^G = P' \cap A^G$, then $P \subset \cup_{\rho \in G} \;\rho (P')$.

Proof: Let $a \in P$. Then $\prod_{\rho \in G} \rho(a)$ is in $P \cap A^G = P' \cap A^G \subset P'$. Since $P'$ is a prime ideal, $\rho(a) \in P'$ for some $\rho \in G$. Equivalently, $a \in \rho^{-1}(P')$.

If $P \cap A^G = P' \cap A^G$, then $P = \rho (P')$ for some $\rho \in G$.

By the lemma, $P \subset \rho(P')$ for some $\rho$ (Prime Avoidance). Applying the lemma to $P'$, there exists $\tau$ such that $\rho (P') \subset \tau (P)$, so $P \subset \tau (P)$. Then $P = \tau(P)$ since $P \subset \tau(P) \subset \tau^2(P) \subset \cdots \subset \tau^{ord \, G}(P) = P$. From $P \subset \rho(P') \subset \tau(P) = P$, the claim follows.

Now, the result follows since the set $\{ \rho(P) \}_{\rho \in G }$ is finite. (By Observation b) every element of the set is again in $\sum$.)