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 $a$).


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.