Let $\mathbb{C}^*=\mathbb C\setminus\{0\}$ act on $\mathbb C^2\setminus \{0\}$ by scalar multiplication, where $\mathbb C^2=\operatorname{Spec}(\mathbb C[x_0,x_1])$. Then $\mathbb C^2\setminus \{0\}=U_0\cup U_1$ , where $$U_0=\mathbb C^2\setminus \mathcal{Z}(x_0)=\operatorname{Spec}(\mathbb C[x_0^{\pm1},x_1])$$ $$U_1=\mathbb C^2\setminus \mathcal{Z}(x_1)=\operatorname{Spec}(\mathbb C[x_0,x_1^{\pm1}])$$ $$U_0\cap U_1=\mathbb C^2\setminus \mathcal{Z}(x_0x_1)=\operatorname{Spec}(\mathbb C[x_0^{\pm1},x_1^{\pm1}])$$
Clearly, $U_0$ , $U_1$ and $U_0\cap U_1$ are invariant under the $\mathbb C^*$ action. So there is an induced action of $\mathbb C^*$ on their respective rings.
I need help finding the rings of invariants - $\mathbb C[x_0^{\pm1},x_1]^{\mathbb C^*}$ $\mathbb C[x_0,x_1^{\pm1}]^{\mathbb C^*}$ and $\mathbb C[x_0^{\pm1},x_1^{\pm1}]^{\mathbb C^*}$
Definition : If $G$ acts on an affine variety $X=\operatorname{Spec} R$ such that every $g\in G$ defines a morphism $\phi_g:X\rightarrow X$ given by $\phi_g(x)=g\cdot x$ , 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 $f\in R$. In other words, $g\cdot f(x)=f(g^{-1}\cdot x)$ for all $x\in X$
We define the ring of invariants $R^G=\{f\in R:g\cdot f=f \text{ for all }g\in G\}$
My attempt :
$\mathbb C[x_0^{\pm1},x_1]^{\mathbb C^*}=\left\{\dfrac{f(x_0,x_1)}{x_0^n} \in \mathbb C[x_0^{\pm1},x_1]: g\cdot\dfrac{f(a_0,a_1)}{a_0^n}=\dfrac{f(g^{-1}a_0,g^{-1}a_1)}{(g^{-1}a_0)^n}=\dfrac{f(a_0,a_1)}{a_0^n}\right\}$ for all $g\in\mathbb C^*$ and $(a_0,a_1)\in U_0$
So I am supposed to find all polynomials $f$ such that $f(g^{-1}a_0,g^{-1}a_1)=(g^{-1})^nf(a_0,a_1)$ for all $g\in\mathbb C^*$ and $(a_0,a_1)\in U_0$. I have no idea how to proceed from here. Is what I have done so far correct?
Thank you.
Let us look at just $\mathbb{C}[x,y,x^{-1}]$ (I have written $x,y$ instead of $x_0,x_1$, for ease of typing). Any element can be written as $f=\sum a_{ij} x^iy^j$, a finite sum, with $i,j\in\mathbb{Z}, j\geq 0, a_{ij}\in\mathbb{C}$. For $g\in\mathbb{C}^*$, we have $gf=\sum a_{ij} g^{i+j} x^iy^j$. So, if we want $gf=f$ for all $g\in\mathbb{C}^*$, we see that $a_{ij}$ must be zero if $i+j\neq 0$. Thus, we see that $f$ is an invariant if and only if $f=\sum_{i+j=0} a_{ij}x^iy^j$, or in other words, $f\in\mathbb{C}[\frac{y}{x}]$. Similar arguments can be made for the other rings in question.