Consider $\mathbb{A}^2$ with $\rho : (x, y) \mapsto (-x, -y)$. Can anyone help me prove that $S = \{f \in \mathbb{C}[x, y] : f \circ \rho = f\}$ is a finitely generated subring? Also, can $S$ be realized as the coordinate ring of some affine variety $V$, and is there a morphism $\mathbb{A}^2 \to V$?
This is what I have so far: The ring $\mathbb{C}[x, y]$ can be viewed as a $\mathbb{C}$-vector space with basis $1, x, y, x^2, xy, y^2, \ldots$ and $\rho^{\ast} : f \to f \circ \rho$ can be viewed as a $\mathbb{C}$-linear map. The generalized eigenvector of the eigenvalue $\lambda \in \mathbb{C}$ is $v \in \mathbb{C}[x, y]$ such that $(\rho^{\ast} - \lambda I)^{m}(v) = 0$ for some integer $m \geq 1$. Suppose $f \circ \rho \to f$. Then $\rho^{\ast}$ is an endomorphism with the minimal polynomial $(x - \lambda_1)^{m_1} \cdots (x - \lambda_n)^{m_n}$, where $\lambda_1, \ldots, \lambda_n$ are district eigenvalues, and $\mathbb{C}[x, y]$ is the direct sum of its generalized eigenspaces: $\mathbb{C}[x,y] = \mathbb{C}[x,y](\lambda_1) \bigoplus \cdots \bigoplus \mathbb{C}[x,y](\lambda_n)$. Moreover, $\mathbb{C}[x,y](\lambda_i) = \ker(\rho^{\ast} - \lambda_i I)^{m_i}$. Because $\mathbb{C}$ is noetherian, $\mathbb{C}[x, y]$ is noetherian by Hilbert's basis theorem, so all ideals in $\mathbb{C}[x, y]$ are finitely generated. Furthermore, the kernel of a ring homomorphism is an ideal, so the generalized eigenspaces are ideals because endomorphisms are homomorphisms. Consequently, they are finitely generated as subrings because ideals are subrings. Since ideals are modules and any finite direct sum of noetherian modules is noetherian, $S$ is noetherian and so it must finitely generated.
This is a special case of finding the ring of invariants of a group action. Here the group is $G = \mathbb{Z}/2\mathbb{Z} = \{1, \sigma\}$ acting on $\mathbb{C}[x,y]$ by $\sigma(x) = -x$, $\sigma(y) = -y$, extended $k$-linearly. The subring $S$ of interest is then $\mathbb{C}[x,y]^G = \{f \in \mathbb{C}[x,y] \mid \sigma(f) = f\}$.
To generate the ring of invariants $S$, we can start by listing obvious invariants. In this case $x^2, xy, y^2$ are invariant, which gives $k[x^2,xy,y^2] \subseteq S$. In fact equality holds: note that $k[x^2,xy,y^2]$ is the subring of $k[x,y]$ consisting of polynomials all of whose terms have even degree. For any $f \in S$, we can write $f = g + h$ with $g \in k[x^2,xy,y^2]$, and every term of $h$ has odd degree. Then $h$ is invariant, which forces $h = 0$ (this works as long as the characteristic of the base field is $\ne 2$).
More generally, the ring of invariants of an affine ring (the coordinate ring of an affine variety) under a finite group action is always finitely generated, by a result of Emmy Noether (one of many in commutative algebra!).
This particular ring $S$ can be realized as an affine ring: namely $S \cong k[u,v,w]/(uw-v^2)$, via $u \mapsto x^2$, $v \mapsto xy$, $w \mapsto y^2$. This is a cone $V = V(uw-v^2) \subseteq \mathbb{A}^3$, and the inclusion $k[x^2,xy,y^2] \subseteq k[x,y]$ gives a dominant map (in fact a quotient map) $\mathbb{A}^2 \to V = \mathbb{A}^2/G$.