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Let $A$ be a $K$-algebra for some local number field $K$, and denote by $\dim A$ its Krull dimension. Let $G$ be an algebraic group defined over $\text{Spec}K$, and assume $G$ acts on $A$ by $K$-isomorphisms.

My question is: denote the ring of $G$-invariants of $A$ by $A^G$, then, is it true that $\dim A^G \ge \dim A - \dim G$?

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    $\begingroup$ No: fix $n\ge 2$ and take $A=K[x_1,\dots,x_n]$ and $G$ the multiplicative group, acting as $t\cdot P(x)=P(tx)$. Then $A^G$ is reduced to constants, hence has dimension $0$, while $\dim(A)-\dim(G)=n-1$. $\endgroup$
    – YCor
    Commented Oct 2, 2023 at 9:50
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    $\begingroup$ (You mean "acts on $A$ by $K$-automorphisms", not "by $K$-isomorphisms".) $\endgroup$
    – YCor
    Commented Oct 2, 2023 at 9:50

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