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Let $A$ be a domain and $G$ a finite group of automorphisms of $A$. I define $$A^G=\{a\in A\mid\sigma(a)=a ,\forall\sigma\in G\}.$$ Furthermore let $S\subset A$ be multiplicatively closed such that $\sigma(S)\subset S$ for all $\sigma\in G$ and we write $S^G=S\cap A^G$. I want to show that $$(S^G)^{-1}A^G\cong(S^{-1}A)^G.$$ (Atiyah and Macdonald, Introduction to Commutative Algebra, Exercise 12, Chapter 5.)

First off, I'm not really sure what is meant by $(S^{-1}A)^G$ here, so it would be nice if somebody could try to guess what would be the most natural interpretation of this.

Regarding the general procedure: I only know that $A$ is integral over $A^G$, which would give me $(S^G)^{-1}A$ being integral over $(S^G)^{-1}A^G$, though I'm not sure if this would be helpful even if I knew what $(S^{-1}A)^G$ was. So depending on this, maybe a little hint would also be nice.

Thanks in advance.

user26857
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azureai
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  • The elements of $S^{-1}A$ are fractions $a/s$, and since $G$ act on both $A$ and $S$, it act on $S^{-1}A$ by $\sigma \cdot (a/s) = (\sigma a)/(\sigma s)$. – Fredrik Meyer Jun 30 '16 at 13:54
  • @FredrikMeyer Yes, I showed this already. But isn't the set of those elements exactly the left hand expression, $(S^G)^{-1}A^G$ – azureai Jun 30 '16 at 14:20
  • Yes, indeed. So what you must show is that every invariant fraction is a fraction of invariants. – Fredrik Meyer Jun 30 '16 at 14:22
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    @FredrikMeyer Like this: ${\frac{a}{s}|\sigma(a)=a&\sigma(s)=s,\forall\sigma}\cong{\frac{a}{s}|\frac{\sigma(a)}{\sigma(s)}=\frac{a}{s},\forall\sigma}$? So I need to construct an isomorphism between those two sets? – azureai Jun 30 '16 at 14:25

2 Answers2

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The following works without demanding that $A$ is a domain. There is a trick using the finiteness of $G$ that reminds one of the norm in Galois theory: Take $\frac{a}{s} \in (S^{-1}A)^{G}$, i.e., $$\frac{a}{s} = \sigma\left(\frac{a}{s}\right) = \frac{\sigma(a)}{\sigma(s)}$$ for all $\sigma \in G$. We can write $$\frac{a}{s} = \frac{a \prod_{\sigma \in G \setminus \{\rm id\}} \sigma(s)}{\prod_{\sigma \in G} \sigma(s)},$$ having the advantage of the denominator $s'$ being $G$-invariant. Let $a'$ be the numerator. Since now $$\frac{a'}{s'} = \frac{\sigma(a')}{s'}$$ for all $\sigma \in G$, for each $\sigma$ there is some $u_{\sigma} \in S$ such that $$u_{\sigma} s' (a' - \sigma(a')) = 0,$$ and multiplying all those $u_{\sigma}$ gives a $u \in S$ working for all $\sigma \in G$ simultaneously, $$u s' (a'-\sigma(a')) = 0.$$ Finally, using the same trick again, let $u' := \prod_{\sigma \in G} \sigma(u)$. Then $u' \in S^{G}$ and $$u's'(a'-\sigma(a')) = 0$$ for all $\sigma \in G$. Setting $\overline{a} := u's'a'$, for all $\sigma$ we have $$\sigma(\overline{a}) = \sigma(u')\sigma(s')\sigma(a') = u's'\sigma(a') = \overline{a},$$ so $\overline{a} \in A^G$ and hence $$\frac{a}{s} = \frac{a'}{s'} = \frac{u's'a'}{u's'^2} = \frac{\overline{a}}{u's'^2} \in (S^G)^{-1}A^G,$$ as desired.

The injectivity of the canonical map $(S^G)^{-1}A^G \rightarrow (S^{-1}A)^G$ can also be dealt with using this "norm trick".

A. Kret.
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There is a natural injection $(S^G)^{-1} A^G \hookrightarrow (S^{-1}A)^G$. To show that this is an isomorphism, you must show that every invariant fraction is a fraction of invariants.

Here are some hints.

So assume $\sigma \cdot \frac{a}{s}= \frac{a}{s}$. This is equivalent to $s \sigma (a) = a \sigma (s)$.

Since $G$ is a finite group, there is some $n$ such that $\sigma^n=id$. Then note that

$$ \frac{a}{s} = \frac{\sigma^{n-1}(s) \sigma^{n-2}(s)\ldots \sigma(s)a }{\sigma^{n-1}(s) \sigma^{n-2}(s)\ldots \sigma(s)s} = \frac{\sigma^{n-1}(s) \sigma^{n-2}(s)\ldots \sigma(a)s }{\sigma^{n-1}(s) \sigma^{n-2}(s)\ldots \sigma(s)s} $$

By induction one can show that the numerator is $\sigma^{n-1}(a)\sigma^{n-2}(s) \ldots \sigma(s) s$.

Now its not difficult to see that both the numerator and denominator are invariant.

Fredrik Meyer
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  • Doesn't $\sigma\cdot\frac{a}{s}=\frac{a}{s}$ just say that there is some $s'$ such that $s's\sigma(a)=s'a\sigma(s)$? – azureai Jun 30 '16 at 14:41
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    @see Yes, but $A$ is a domain, so you can cancel $s'$. – Fredrik Meyer Jun 30 '16 at 14:42
  • Nevermind. Every $s'$ is invertible because I'm in a domain right. – azureai Jun 30 '16 at 14:42
  • I take $\frac{a}{s}$ as you did above, and now I need $a',s'$ such that for all $\sigma$ I have $\sigma(a')=a'$, $\sigma(s')=s'$ as well as $\frac{a}{s}=\frac{a'}{s'}$. If I take $a'=\sigma^{n-1}(s)...\sigma(s)a$ and $s'=\sigma^{n-1}(s)...\sigma(s)s$, this works for a given $\sigma$, but now I my $a'$ and $s'$ depend on said $\sigma$. I don't know how to proceed from here. – azureai Jun 30 '16 at 18:36
  • @see Use an even larger $n$. Take the maximum of all such $n$ (this is finite since $G$ is finite). – Fredrik Meyer Jun 30 '16 at 19:20
  • I can't come up with the right choice for $a'$ and $s'$. I need to find $a'$ and $s'$ such that $\frac{a}{s}=\frac{a'}{s'}$ and for all $\sigma$ $a'=\sigma(a')$ and $s'=\sigma(s')$ yes? – azureai Jun 30 '16 at 19:23
  • @see The $a'$ and $s'$ I've given works, no? – Fredrik Meyer Jun 30 '16 at 19:52
  • Well if I read your answer right you have $a'=\sigma^{n-1}...\sigma(s)a$ for some $\sigma$. This gives $\sigma(a')=a'$, but by definition it has to work for any element of $G$. So what if I take an $\rho\neq\sigma$. Then I must have $\rho(a')=a'$ for the $a'$ to work, but I don't see how $\rho(a')=\rho(\sigma^{n-1}(s)...\sigma(s)a)=a'$ – azureai Jun 30 '16 at 20:18
  • @see There is nothing special about my $\sigma$. The point is that if $n$ is chosen large enough, the product will be invariant under all of $G$. – Fredrik Meyer Jun 30 '16 at 21:33
  • Well, I tried for a few hours now, but I still don't understand this part. What does the $n$ have to do with the other elements of $G$? – azureai Jun 30 '16 at 21:39
  • There is a natural map $(S^G)^{-1} A^G \rightarrow (S^{-1}A)^G$ (given by $\frac as\mapsto\frac as$) which has to be proved that it is injective! – user26857 Jul 01 '16 at 20:16
  • Moreover, you showed that $\sigma(\frac as)=\frac as\implies\exists u,v$ such that $\frac as=\frac uv$ and $\sigma(u)=u$, $\sigma(v)=v$. But this is not enough to show $\frac as\in (S^G)^{-1} A^G$. – user26857 Jul 01 '16 at 20:19