Given $\pi:X\rightarrow Y$ how to show $X$ is irreducible (resp. normal) $\Rightarrow$ $Y$ is irreducible(resp. normal)?

Question

Let $G$ act on the affine variety $X=\operatorname{Spec}(R)$ such that $R^G$ is a finitely generated $\mathbb C$ - algebrs and let $\pi:X\rightarrow Y=\operatorname{Spec}(R^G)$ be the morphism of affine varieties induced by $R^G\subseteq R$ . Then I need to show the following :

1. If $X$ is irreducible then so is $Y$

2. If $X$ is normal (:= $R$ is integrally closed) then so is $Y$.

Here is the proof from Cox, Little and Shenck -

Part 1 is immediate since $R^G$ is a subring of $R$. For part 2 let $K$ be the field of fractions of $R^G$. If $a\in K$ is integral over $R^G$, then it is also integral over $R$ and hence lies in $R$ since $R$ is normal. It follows that $a\in R\cap K$ which obviously equals $R^G$ since $G$ acts trivially on $K$. Thus $R^G$ is normal. $\square$

Here is what I don't understand -

1. how part 1 follows since $R^G$ is a subring of $R$?

If I write $Y=Y_1\cup Y_2$ where $Y_1$ and $Y-2$ are closed subsets of $Y$ then since $X$ is irreducible, it must be equal to $\pi^{-1}(Y_1)$ or $\pi^{-1}(Y_2)$. But then what?

2. how $R\cap K = R^G$ and how $G$ acts trivially on $K$?

Definitions used -

Let $G$ act on a variety $X$ such that every $g\in G$ defines a morphism $\phi_g:X\rightarrow X$ given by $\phi_g(x)=g\cdot x$. If $X=\operatorname{Spec}(R)$ is affine then $\phi_g$ comes from a map $\phi_g^*:R\rightarrow R$. We define the induced action of $G$ on $R$ by $$g\cdot f=\phi_{g^{-1}}^{*}(f)$$ for $f\in R$

We define the ring of invariants $R^G=\{f\in R:g\cdot f=f \text{ for all }g\in G\}$

Thank you.

EDIT -

Thanks to the comment by msteve my first question is answered. For the second part I have gotten this far -

That $R^G\subseteq R\cap K$ is obvious. If $a\in R\cap K$ then assuming $G$ acts trivially on $K$ and since $a\in K$ we have that $g\cdot a=a$ for all $g\in G \Longrightarrow a\in R^G$. Now all that is left to prove is that $G$ acts trivially on $K$. Any $a\in K$ is of the form $\dfrac{f}{h}$ where $f,h\in R^G$. Then $g\cdot\dfrac{f}{h}=\dfrac{g\cdot f}{g\cdot h}=\dfrac{f}{h}$ . Hence the action is trivial.

But is this the only action of $G$ on $K$? Or can $G$ act in some other way?

Answer 1

I think the proof of the first part really proves that $Y$ is integral if $X$ is integral. You can use msteve's reduction to this case.

Alternatively, if you already know that the map $X \to Y$ is (set-theoretically) surjective, then it follows immediately that $Y$ is irreducible if $X$ is. In fact, the image of an irreducible set is always irreducible.

To answer your last question: it's not the only action of $G$ on $K$. I guess you could prove that it is the only action of $G$ on $K$ such that the restriction to $R$ is the given one. This is straightforward, and most people would not consider it necessary.