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Something went wrong in the 2nd paragraph of Will's answer

whoops - made a silly mistake. Suppose $z$ is an element of $k[X]$ which is integral over $k[X]^G$. Then $z$ is also integral over $k[X]^{G_0}$ since $read the question as asking whether k[X]^G \subseteq k[X]^{G_0}$. Then by Will's first paragraph $z$ lies in $k[X]^{G_0}$. But since $z$ is $G$-invariant $z$ lies in $(k[X]^{G_0})^G = k[X]^G$. Thus $k[X]^G$ is integrally closed in $k[X]$ even when $G$ is not connectedits field of fractions.

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Something went wrong in the 2nd paragraph of Will's answer. Suppose $z$ is an element of $k[X]$ which is integral over $k[X]^G$. Then $z$ is also integral over $k[X]^{G_0}$ since $k[X]^G \subseteq k[X]^{G_0}$. Then by Will's first paragraph $z$ lies in $k[X]^{G_0}$. But since $z$ is $G$-invariant $z$ lies in $(k[X]^{G_0})^G = k[X]^G$. Thus $k[X]^G$ is integrally closed in $k[X]$ even when $G$ is not connected.

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Something went wrong in the 2nd paragraph of Will's answer. Suppose $z$ is an element of $k[X]$ which is integral over $k[X]^G$. Then $z$ is also integral over $k[X]^{G_0}$ since $k[X]^G \subseteq k[X]^{G_0}$. Then by Will's first paragraph $z$ lies in $k[X]^{G_0}$. But since $z$ is $G$-invariant $z$ lies in $(k[X]^{G_0})^G = k[X]^G$. Thus $k[X]^G$ is integrally closed even when $G$ is not connected.