I have the following exercise: (This exercise deals with the action of a group $G$ on an affine closed.)
Let $G=\langle w \rangle$ be a cyclic group of order $n$ (in multiplicative notation). Consider the action of $G$ on $\mathbb{A}^2$ defined by $$w(x,y):=(wx,w^{-1}y)$$ Prove that the geometric quotient $\mathbb{A}^2/ G$ is isomorphic to the closed affine set $Z(uv-w^n)\subset \mathbb{A}^3$
I thought about proceeding this way.
I know that a geometric quotient $Y=\mathbb{A}^2/ G$ is such that $$K[Y]=K[\mathbb{A}^2]^G \hookrightarrow^{\psi^*} K[\mathbb{A}^2]$$ where:
- $K[Y]$ is the set of regular functions on $Y$,
- $K[\mathbb{A}^2]^G=\{f\in K[\mathbb{A}^2]: f=f^g \; \forall g \in G\}$ is the set of regular functions on $\mathbb{A}^2$ which are $G$-invariant
- $\psi^*: K[Y] \hookrightarrow K[\mathbb{A}^2]$ is such that for the corresponding $\psi: \mathbb{A}^2 \rightarrow Y$ the following property holds $$\psi(p)=\psi(q) \iff \;\; p \; \textit{and}\; q \;\textit{are in the same orbit of G}$$
So, I'd like to try to figure out who $Y$ is.
I have examined the orbits of $G$ studying the equivalence classes by means of the relation $$p,q \in \mathbb{A}^2, \;\;\;\; p \sim q \iff \exists g \in G: q=p^g$$ in this case, because $G=\langle w\rangle$ of order $n$ $$(x,y),(a,b) \in \mathbb{A}^2 \;\;\; (x,y)\sim (a,b) \iff \exists t\le n : (a,b)=(w^tx,w^{-t}y)$$ so there are four possibilities
- $a=x=b=y=0$ i.e the origin $\{0\}$
- $a=x=0, b,y\neq0$ i.e the $x$ axis
- $b=y=0, a,x\neq0$ i.e the $y$ axis
- $a,x,b,y\neq0$ i.e $\mathbb{A}^2 \setminus \{0\}$
These are the orbits of G, aren't they?
Now, can I assert that Y is made up of these four classes? I believe so because we are in a finite group and therefore the set quotient (and not geometric quotient) has a natural geometric structure and therefore $\psi$ coincides with the application of the set quotient.
Is everything said so far correct?
If everything is correct, then how can I find an isomorphism with $Z(uv-w^n)\subset \mathbb{A}^3$? I can't picture this closed up in my mind. Could someone explain me better?
Why if $w$ generates $G$ of order $n$ we take $w ^ n$? Isn't it unity? If it is the unit, then the origin $(0,0)$ and the points on the axes $(x, 0)$ and $(0, y)$ [which according to my idea are in the quotient $Y$] would not cancel the polynomial $uv-1$ and therefore they would not be in closed $Z(uv-w^n)$. Where am I wrong?
$$K[x,y]^G = \{ \sum_{a,b,\ n\ |\ a-b} c_{a,b} x^ay^b\}= K[x^n,y^n,xy]\cong K[u,v,w]/(w^n-uv)$$ The isomorphism is $G.(x,y)\to (x^n,y^n,xy)$