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.

  • $\begingroup$ 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 $\mathbb{C}[x_1,\ldots,x_n].$ $\endgroup$ – user371231 Nov 12 '17 at 20:46
  • $\begingroup$ Because $\mathbb{C}$ is already uncountable. $\endgroup$ – Qiaochu Yuan Nov 12 '17 at 20:52
  • $\begingroup$ But it may happen that after quotient it becomes countable $\endgroup$ – user371231 Nov 12 '17 at 20:53
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    $\begingroup$ Every algebra over a field contains that field... $\endgroup$ – MooS Nov 12 '17 at 21:52
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    $\begingroup$ @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. $\endgroup$ – leibnewtz Nov 12 '17 at 22:00

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