Let $G$ be a complex reductive group acting linearly on $\mathbb{C}^n$ and let $X$ be a $G$-invariant closed subvariety of $\mathbb{C}^n$. Is $X$ the zero-set of finitely many $G$-invariant functions?


In other words, is the ideal $I \subset \mathbb{C}[x\_1, \ldots, x\_n]$ defining $X$ generated by finitely many invariant polynomials?


Since $G$ is reductive, the set $I^G$ of $G$-invariant elements of $I$ is an ideal in the Noetherian ring $\mathbb{C}[x\_1, \ldots, x\_n]^G$, so it suffices to show that $I^G$ generates $I$.



Invariant ideal generated by invariant elements
In general, the answer is no, as Friedrich Knop intimates in a comment. Consider the reductive group $G:=\mathbb{C}^\times = GL\_1(\mathbb{C})$ acting on $\mathbb{C}^2$ by $\alpha\cdot (x,y)\mapsto (\alpha x, \alpha y)$. Then the invariant ring $\mathbb{C}[x,y]^G$ is just $\mathbb{C}$, and the only subvarieties of $\mathbb{C}^2$ it's able to "see" (more precisely the only subvarieties of $\mathbb{C}^2$ cut out by an ideal that is generated by invariants) are the empty set and everything. In particular, $\{0\}$ and any complex line through the origin are invariant subvarieties that are not cut out by invariants.


There is a qualified positive answer in the special case that $G$ is finite, however, or more generally in the situation that the action of $G$ is closed (i.e., every orbit of $G$ is Zariski-closed in $\mathbb{C}^n$).


In this case, the answer to your question about ideals is still no: for example the action of $\mathbb{Z}/2\mathbb{Z}$ on $\mathbb{C}^1$ via the sign representation has no degree $1$ invariants (the invariant ring is $\mathbb{C}[x^2]$), but the ideal $(x)\subset\mathbb{C}[x]$ is $G$-stable, so this is a $G$-stable ideal that is not generated by invariants.


That said, the answer is yes *up to radical*; equivalently, the answer to your question about *varieties* is yes in this situation. This is because when the action of $G$ on $\mathbb{C}^n$ is closed (for example if $G$ is finite), the map


$$\mathbb{A}\_{\mathbb{C}}^n \rightarrow \operatorname{Spec}\mathbb{C}[x\_1,\dots,x\_n]^G$$


induced by the ring inclusion $\mathbb{C}[x\_1,\dots,x\_n]^G\hookrightarrow\mathbb{C}[x\_1,\dots,x\_n]$ is a *geometric quotient* (see Mumford, *GIT*, Chapter 1 Section 2). In particular it is a topological quotient on the underlying topological spaces; we may identify the MaxSpec of $\mathbb{C}[x\_1,\dots,x\_n]^G$ with the topological quotient $\mathbb{C}^n/G$, so a $G$-stable Zariski-closed set in $\mathbb{C}^n$ maps to a closed set in $\mathbb{C}^n/G$, which, because the latter is an affine variety, is cut out by regular functions on this quotient, which are precisely invariant polynomials on the original $\mathbb{C}^n$. So any $G$-stable subvariety of $\mathbb{C}^n$ is indeed cut out by invariant polynomials.


As an illustration, in the situation above with $\mathbb{Z}/2\mathbb{Z}$, the ideal $(x)$ may not be generated by invariants, but the ideal generated by the invariant $x^2$ has the same radical, so it cuts out the same variety $\{0\}$.


(Aside: I did not engage the "finitely many" language in the question because it does not add a new requirement. As you note, reductivity of $G$ implies the ring $\mathbb{C}[x\_1,\dots,x\_n]^G$ is noetherian; thus any variety cut out by invariants will be cut out by finitely many.)


