# Every finitely generated $\mathbb{C}$-algebra is finitely generated $\mathbb{Q}$-algebra.

I want to know if every finitely generated $\mathbb{C}$-algebra is finitely generated $\mathbb{Q}$-algebra or not.

Although by Artin-Tate lemma we can see that $\mathbb{C}[x]/(x^2)$ is finitely generated $\mathbb{C}$-algebra which is not finitely generated $\mathbb{Q}$-algebra since $\mathbb{C}$ is not algebraic over $\mathbb{Q}.$

I want to know is this true in general ? And also want to know what happens if $\mathbb{C}$ is replaced by $\mathbb{R}.$ Thanks.

A non-zero $\mathbb C$-algebra is not denumerable and thus never a finitely generated $\mathbb Q$-algebra.
• Can you please explain why finitely generated $\mathbb{C}$ algebra is not denumerable ? I know that it must be $\mathbb{C}[X_1, \ldots , X_n]/I$ which is isomorphic to $\mathbb{C}[x_1,\ldots,x_n].$ – user371231 Nov 12 '17 at 20:46
• Because $\mathbb{C}$ is already uncountable. – Qiaochu Yuan Nov 12 '17 at 20:52
• @user371231 In general an $A$ algebra $B$ is a ring together with a ring homomorphism $A \to B$. Any $\mathbb{C}$ algebra is therefore a ring $A$ with a map $\mathbb{C} \to A$. Unless $A=0$ this map has to be injective. – leibnewtz Nov 12 '17 at 22:00