Atiyah Macdonald Exercise 5.12

Question

I am stuck on one part of Exercise 5.12 in Atiyah Macdonald. The entire exercise is quoted below:

Let $G$ be a finite group of automorphisms of a ring $A$, and let $A^G$ denote the subring of $G$-invariants, that is of all $x \in A$ such that $\sigma(x) = x$ for all $\sigma \in G$. Prove that $A$ is integral over $A^G$. [If $x \in A$, observe that $x$ is a root of the polynomial $\prod_{\sigma \in G} (t - \sigma(x))$.] Let $S$ be a multiplicatively closed subset of $A$ such that $\sigma(S) \subset S$ for all $\sigma \in G$, and let $S^G = S \cap A^G$. Show that the action of $G$ on $A$ extends to an action on $S^{-1} A$, and that $(S^G)^{-1} A^G \cong (S^{-1} A)^G$.

The piece that I am stuck on is proving that $(S^G)^{-1} A^G \cong (S^{-1} A)^G$. I have tried two approaches: Using the universal property of localizations, and defining the "obvious" map $\phi : (S^G)^{-1} A^G \to (S^{-1} A)^G$ via $\frac{a}{s} \mapsto \frac{a}{s}$. In both cases, the issue I run into is that I need to show that any $\frac{a}{s} \in (S^{-1} A)^G$, is equal to $\frac{a'}{s'}$, where $a' \in A^G$ and $s' \in S^G$.

My best attempt so far is to start with $\frac{a}{s} \in (S^{-1} A)^G$, and note that $$\frac{a}{s} = \frac{\prod_{\sigma \in G, \sigma \neq 1} \sigma(s) a}{\prod_{\sigma \in G} \sigma(s)}.$$ Clearly the denominator is in $S^G$. For the numerator, if $A$ is an integral domain, then since $\frac{a}{s} \in (S^{-1} A)^G$, we have that $\sigma(a)s - \sigma(s)a = 0$. I can use this to show that the numerator is in $A^G$ by applying $\sigma$ to the numerator, and noting that if $\sigma \neq 1$, then we get $s \sigma(a) \prod_{\tau \in G, \tau \neq \sigma,1} \tau(s) = a\prod_{\tau \in G, \tau \neq 1} \tau(s)$, so the numerator is in $A^G$.

If $A$ is not an integral domain, then we get only that there exists $s_\sigma \in S$ such that $(\sigma(a) s - \sigma(s) a)s_\sigma = 0$, and we have that $(\prod_{\sigma \in G} s_\sigma)(\prod_{\sigma \in G, \sigma \neq 1} \sigma(s))a \in A^G$, but then we need to show that $(\prod_{\sigma \in G} s_\sigma) \in S^G$. Is there a way to do this, or am I missing a simpler approach?

Answer 1

From your work, we may start with $\frac{a}{s}\in (S^{-1}A)^G$ where $s\in A^G$, and $\forall\sigma\in G$, $(a-\sigma(a))ss_{\sigma}=0$ for some $s_\sigma\in S$.

That is we have $a-\sigma(a)$ is annihilated by an element $t_{\sigma}$ fom $S$, and hence it can also be annihilated by $t'_{\sigma}:=\prod_{\tau\in G}\tau(t_{\sigma})\in S^G$. Multiplying all the $t'_{\sigma}$ together, we claim $a-\sigma(a)$ can all be annihilated by a uniform element $s'\in S^G$, no matter what $\sigma$ is, i.e. $$(a-\sigma(a))s'=0\Rightarrow as'=\sigma(a)s'$$

Now we have $\frac{as'}{ss'}\in (S^G)^{-1}A^G$.