Let $R$ be an $\mathbb R$ - algebra. Suppose $A=R\otimes_\mathbb R\mathbb C$ is a finitely generated $\mathbb C$ - algebra then is $R$ a finitely generated $\mathbb R$ - algebra?
I thought along the following lines - $A$ is finitely generated $\mathbb C$ - algebra. $\mathbb C$ is a finitely generated $\mathbb R$ - algebra. Then $A$ is a finitely generated $\mathbb R$ - algebra and then I can use this answer. However I am not sure if the transitivity I have mentioned above is correct.
Thank you.
Your argument is correct. If $\{a_1,\dots,a_n\}$ generates $A$ as a $\mathbb{C}$-algebra, then $\{a_1,\dots,a_n,i\}$ generates $A$ as an $\mathbb{R}$-algebra. Indeed, any $\mathbb{R}$-subalgebra of $A$ containing $i$ is a $\mathbb{C}$-subalgebra, and then if such a subalgebra contains $a_1,\dots,a_n$ it must be all of $A$. More generally, a similar argument shows that if $A$ is a finitely generated $B$-algebra and $B$ is a finitely generated $C$-algebra, then $A$ is a finitely generated $C$-algebra: just take the union of a generating set for $A$ over $B$ and (the image in $A$ of) a generating set for $B$ over $C$.
Alternatively, this is just a special case of the more general statement I alluded to in my answer to the linked question (notation changed to avoid confusion with your notation):
If $S$ is any base ring and $R$ and $B$ are $S$-algebras such that $B$ is faithfully flat over $S$, then if $R\otimes_S B$ is finitely generated as a $B$-algebra (in particular, if it is finitely generated as an $S$-algebra), then $R$ is finitely generated as an $S$-algebra.
In your case, you want to take $S=\mathbb{R}$ and $B=\mathbb{C}$.
Let me write out the proof of this statement in your case; the proof is almost identical to the proof I gave in the previous answer (in fact, I'm just going to copy-paste that answer and make the appropriate changes). Choose a finite set of generators of $R\otimes_\mathbb{R} \mathbb{C}$ as a $\mathbb{C}$-algebra; each of these is a finite sum of tensors $r\otimes z$. Let $R_0\subseteq R$ be the $\mathbb{R}$-subalgebra generated by all the $r$'s appearing in these tensors. Then $R_0$ is finitely generated, and we see that the natural map $$R_0\otimes_\mathbb{R} \mathbb{C}\to R\otimes_\mathbb{R} \mathbb{C}$$ is surjective (since its image contains all of the tensors $r\otimes z$ in our generators). This implies that $R_0$ is all of $R$ (you can see this by writing $R=R_0\oplus V$ as a $\mathbb{R}$-vector space; then the natural map above will just be the inclusion $R_0\otimes_\mathbb{R}\mathbb{C}\to (R_0\otimes_\mathbb{R}\mathbb{C})\oplus (V\otimes_\mathbb{R}\mathbb{C})$ so $V\otimes_\mathbb{R}\mathbb{C}=0$ and hence $V=0$). Thus $R$ is finitely generated as a $\mathbb{R}$-algebra.
Alternatively to Eric's answer (+1), you might argue as follows: Writing $$A\cong A_{\mathbb R}:=\{a\otimes 1\ |\ a\in A\}\subset A_{\mathbb C}$$ you have $A_{\mathbb C} = A_{\mathbb R}\oplus i A_{\mathbb R}$ as ${\mathbb R}$ vector spaces. Denote $\text{Re}_A, \text{Im}_A: A_{\mathbb C}\to A_{\mathbb R}$ the respective projections. Then, if $A_{\mathbb C}$ is as a ${\mathbb C}$-algebra generated by $a_1,...,a_n$, then $A_{\mathbb R}$ is as an ${\mathbb R}$-algebra generated by $\text{Re}_A(a_i)$ and $\text{Im}_A(a_i)$: namely, rewriting each $a_k=\text{Re}_A(a_k) + i\cdot \text{Im}_A(a_k)$ in some polynomial representation of some $a\in A_{\mathbb R}$ in terms of the $a_k$ results, after simplifying, in a polynomial expression of $a$ in $\text{Re}_A(a_k)$ and $\text{Im}_A(a_k)$ (together with some vanishing polynomial expression for the zero imaginary part of $a$).
