I am working on Exercise 7.5 of Atiyah & Macdonald's Introduction to Commutative Algebra text. The problem is as follows:
Let $A$ be a Noetherian ring, $B$ a finitely generated $A$-algebra, $G$ a finite group of $A$-automorphisms of $B$, and $B^G$ the set of all elements of $B$ which are left fixed by every element of $G$. Show that $B^G$ is a finitely generated $A$-algebra.
A useful result to solve this is Proposition 7.8 in the text: Let $A \subseteq B \subseteq C$ be rings. Suppose that $A$ is Noetherian, that $C$ is finitely generated as an $A$-algebra and that $C$ is either (i) finitely generated as a $B$-module or (ii) integral over $B$. Then $B$ is finitely generated as an $A$-algebra.
With this in mind, here is a proof of this that I've seen in a few different places:
Note that $B^G$ is closed under subtraction and multiplication and contains the multiplicative identity of $B$, and is therefore a subring of $B$ by the subring test. By Proposition 7.8, in order to show that $B^G$ is a finitely generated $A$-algebra, it suffices to show that $B$ is integral over $B^G$. Let $b \in B$ and consider the monic polynomial $f(t) = \prod\limits_{\lambda \in G}(t - \lambda b)$. We have that $f(b) = 0$, and the action of $G$ on $B$ naturally extends to an action of $G$ on $B[t]$, where $G$ acts trivially on $t$ and acts on the coefficients as it does on $B$. For each $\sigma \in G$, we have $\sigma(f(t)) = \prod\limits_{\lambda \in G} (t - \sigma(\lambda b)) = \prod\limits_{\lambda \in G} (t - \lambda b) = f(t)$. Thus, the coefficients of $f(t)$ are $G$-invariant. That is, $f(t)$ has coefficients in $B^G$. It follows that $B$ is integral over $B^G$.
I have a few questions about the proof:
- Is $A$ really contained in $B^G$? The way that Proposition 7.8 is used here suggests that $A \subseteq B^G \subseteq B$.
- Why is it that $f(b) = 0$? This suggests that $b = \lambda b$ for some $\lambda \in G$, but it's not clear to me that every element of $B$ must be fixed by some element of $G$.
- Similarly, why is $\sigma(b) = b$ for each $\sigma \in G$? The equality would certainly follow if $b \in B^G$, but $b$ is just some element of $B$ here.
Thank you!
Q1: Yes. Recall $G$ is a finite group of $A$-automorphisms, so it is the identity on $A$.
Q2: $f(b)=0$ since the identity of the group $G$ (say $e$), $eb=b$, so the term $(t-b)$ occurs in $f(t)$.
Q3: The claim is not that $\sigma(b)=b$ for each $\sigma\in G$, rather all the coefficients of the polynomial $f(t)$ are fixed by $G$.
To see this observe that the coefficients of $f(t)$ are elementary symmetric polynomials in $\lambda b$, as $\lambda$ varies in $G$. Now check that elementary symmetric polynomials in $\lambda b$ are fixed by the action of $G$.