# Does Nagata theorem hold in a field that is not algebraically closed?

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.

Thank you.

• First of all, isn't this Hilbert's theorem? Also, isn't $R^G\otimes_{\mathbb R} \mathbb C$ the same as the $G$-invariants of $R\otimes_{\mathbb R}\mathbb C$? That shows that $R^G\otimes_{\mathbb R}\mathbb C$ is finitely generated over $\mathbb C$ and implies that $R^G$ is finitely generated over $\mathbb R$. Commented May 2, 2016 at 20:38
• @AriyanJavanpeykar, could you elaborate a little? I am unable to prove these assertions. I can show $R^G\otimes_\mathbb R\mathbb C\subseteq (R\otimes\mathbb C)^G$ but can't do anything else.
– R_D
Commented May 3, 2016 at 5:40