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May assume field $k=\mathbb{C}$. Let $X$ be an affine variety and $G$ be a reductive group (may assume connected). Is the ring of invariants $k[X]^G$ integral closed in $k[X]$? The claim may not true in general, probably we may assume $X$ is normal and the affine quotient $X/G$ also normal. Any reference? |
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Yes, if $G$ is connected. Let $f$ be a function in $k[X]$ which satisfies a polynomial equation over $k[X]^G$. Everything in the $G$-orbit of $f$ must satisfy that polynomial equation. Thus the $G$-orbit is finite, because it is contained in the set of roots of a polynomial, and is connected because it is the orbit of a connected group action, so it is a single point, so $f\in k[X]^G$. In fact, for $G$ not connected with $G_0$ the connected component of the identity and $G/G_0$ finite, the integral closure of $k[X]^G$ in $k[X]$ is $k[X]^{G_0}$ because any element in $k[X]^{G_0}$ has a finite orbit and the coefficients of the monic polynomial that vanishes exactly on the orbit are $G$-invariant. |
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whoops - made a silly mistake. read the question as asking whether k[X]^G is integrally closed in its field of fractions. |
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