Let $R$ be a finitely generated $k$ - algebra and $G$ be a reductive group acting rationally on it. Then a theorem of Nagata says that the invariant ring $R^G$ is also finitely generated.
Here $k$ is assumed to be algebraically closed.
Would this be true if $k=\mathbb R$?
EDIT 1 : As per the comment by Ariyan Javanpeykar I have tried to show that $R^G\otimes_\mathbb R\mathbb C\cong (R\otimes_\mathbb R\mathbb C)^G$
First of all if $G$ acts on $R$ then consider the action on $R\otimes\mathbb C$ given by $$g\cdot \sum_if_i\otimes\lambda_i=\sum_i(g\cdot f_i)\otimes \lambda_i$$ Now since we have $R^G\hookrightarrow R$, we get the map $R^G\otimes\mathbb C\hookrightarrow R\otimes \mathbb C$ which is injective because $\mathbb C$ is flat over $\mathbb R$. Also, by the above action we see that this map takes $R^G\otimes\mathbb C\hookrightarrow(R\otimes\mathbb C)^G$. But I am unable to prove surjectivity.
Also, assuming we have proved $R^G\otimes_\mathbb R\mathbb C\cong (R\otimes_\mathbb R\mathbb C)^G$ then why does $R^G\otimes_\mathbb R\mathbb C$ finitely generated $\mathbb C$ - algebra imply that $R^G$ is a finitely generated $\mathbb R$ - algebra?
EDIT 2 : Thanks to this answer we have that $R^G\otimes_\mathbb R\mathbb C$ being a finitely generated $\mathbb C$ - algebra implies that $R^G$ is a finitely generated $\mathbb R$ - algebra.
So all that is left now is to show that the map $R^G\otimes\mathbb C\hookrightarrow(R\otimes\mathbb C)^G$ is surjective.
But I am still unable to do this.