Equivariant polynomial maps Let $V$ be a complex vector spaces and assume that a compact group G acts   linearly on $V$. Then look at the  $G$-equivariant polynomial maps from $V$ to $V$. Denote this by $Mor_G(V,V)$. In the case of finite groups,  $Mor_G(V,V)$ is a finitely generated module over the ring of polynomial invariants $\mathbb{C}[V]$. Does such a statement also hold for infinite (reductive) groups? If so what would be a good reference for this?  
 A: The answer should be yes, by the following argument:
With $n := \dim(V)$, consider the polynomial ring $C[x_1..x_n,y_1..y_n] = C[V \oplus V^*]$, so G acts on the $x_i$ as on the variables of $C[V]$, but on the $y_i$ as on the variables of $C[V^*]$ with $V^*$ the dual space.
$C[x_1..x_n,y_1..y_n]^G$ is a bigraded algebra by using the x- and y-degrees. So Mor$_G(V,V)$ is the part of $C[x_1..x_n,y_1..y_n]^G$ where the y-degree is 1.
Since $G$ is reductive, $C[x_1..x_n,y_1..y_n]^G$ has a finite generating set $f_1..f_m$, which we may assume to be bigraded. Renumber the $f_i$ in such a way that $f_1..f_r$ have y-degree 0, $f_{r+1}...f_{r+k}$ have y-degree 1, and the others have higher y-degree.
Now an element from Mor$_G(V,V)$, being an invariant, can be written as a linear combination of power products of the $f_i$. But we can delete all summands whose y-degree is not 1, and so we are left with a linear combination of $f_{r+1}...f_{r+k}$ with coefficients in the subalgebra generated by $f_1..f_r$. These coefficients are all invariants.
This argument shows that $f_{r+1}...f_{r+k}$ generate Mor$_G(V,V)$ as a module over C[V]^G.
The argument generalises to Mor$_G(V,W)$,
