Action of finite group of automorphisms on Spec A

Question 2.3.20 in Qing Liu's Algebraic Geometry and Arithmetic Curves:

Let $A$ be a ring, $G$ a finite group of automorphisms of $A$. Let $$ p: \mathrm{Spec}\ A \to \mathrm{Spec}\ A^{G} $$ be the morphism of affine schemes induced by the obvious inclusion.

Show that, for $x_{1}$, $x_{2} \in \mathrm{Spec}\ A$, $$ p(x_{1}) = p(x_{2}) \iff \exists \ \sigma \in G : \sigma(x_{1}) = x_{2}.$$

The '$\impliedby$' direction is easy, but I'm having trouble with the other direction (although I have no doubt that it is also simple).

My attempts so far have included assuming that there is no $\sigma \in G$ such that $\sigma(x_{1}) = x_{2}$, so that for each $\sigma$ there is $g_{\sigma} \in x_{1}$ with $\sigma(g_{\sigma}) \notin x_{2}$, and then trying to construct an element of $p(x_{1})\backslash p(x_{2})$ out of these $g_{\sigma}$, for example by taking $$ g = \prod_{\sigma \in G}g_{\sigma} $$ and then $$ h = \prod_{\sigma}\sigma(g) $$ but this appears to fail as I can't ensure that $\sigma(g_{\tau}) \notin p(x_{2})$ for $\sigma \neq \tau$.

Any hints would be much appreciated!

Solution 1:

Note that $A$ is integral over $A^G$.

Suppose $x_1 \cap A^G = x_2\cap A^G$, then for any $g \in x_1$, we have $ \prod_{\sigma \in G} \sigma(g) \in x_1 \cap A^G = x_2 \cap A^G \subset x_2$, thus as $x_2$ is prime, there is some $\sigma \in G$ such that $\sigma(g) \in x_2$ or to put it differently, there is some $\sigma$ such that $g \in \sigma(x_2)$

This shows that $x_1 \subset \displaystyle \bigcup_{\sigma \in G} \sigma(x_2)$.

Now by the prime avoidance lemma, we have for some $\sigma \in G$ $x_1 \subset \sigma(x_2)$. By incomparability, we get $x_1 = \sigma(x_2)$