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Suppose we have a linearly reductive group $G$ acting on $X=\mathbb{A}^n_{\mathbb{C}}$ and we have a closed subvariety $Z=V(f_1,...,f_k)$, where $f_i$ are $G$-invariant functions. If we further know that $Z$ is in fact a divisor, $Z=V(f)$ and $(f)=(f_1,...,f_k)$, can we conclude that $f$ itself can be chosen to be $G$-invariant?

I don't even know if it is possible to rule out the case where $f$ is a relative $G$-invariant, that is $f(gx)=\chi (g) f(x)$ for all $g \in G, x \in X$ and a character $\chi$ of $G$.

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  • $\begingroup$ I'm a bit confused about the question: are you requiring that the ideal generated by $f$ be equal to the ideal generated by $f_1,\dots,f_k$, as in the third line of your question, or are you only asking Supposing $Z$ is a $G$-stable divisor in affine space, is there a $G$-invariant polynomial with zero set $Z$? $\endgroup$
    – Stephen
    Commented Oct 11, 2019 at 13:36
  • $\begingroup$ Now that I think about that, the latter would be good enough for me, because I am really interested in $\mathcal{O}^*(X-Z)$. $\endgroup$
    – Bananeen
    Commented Oct 11, 2019 at 16:04
  • $\begingroup$ I suppose you must also assume that there is some set of $G$-invariant polynomials with common zero set $Z$. $\endgroup$
    – Stephen
    Commented Oct 11, 2019 at 17:02

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