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Let $K$ be a field, which is a finitely generated $\mathbb{Z}$-algebra.
Show that $K$ has characteristic $> 0$.

I have this hint but I am not quit sure that I understand it; if $\operatorname{char}(K) = 0$, then $\mathbb{Z} \subset K$, then $\Bbb Q \subset K$, and $K$ is a finitely generated $\Bbb Q$-algebra. Use Zariski's Lemma to show that $K$ is a finitely generated $\Bbb Q$-module, and then Artin-Tate Lemma to get a contradiction

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    $\begingroup$ What don't you understand about the hint? Following the hint, Artin-Tate would imply that $\mathbb{Q}$ is a finitely generated $\mathbb{Z}$ algebra. Is that true? $\endgroup$ – Ragib Zaman Apr 2 '20 at 11:30
  • $\begingroup$ In fact, every such field must be finite: math.stackexchange.com/questions/148745 $\endgroup$ – user26857 Apr 2 '20 at 18:19
  • $\begingroup$ See also here for the same question today. $\endgroup$ – Dietrich Burde Apr 2 '20 at 18:55
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If $\operatorname{char}(K) = 0$, then $\mathbb{Z} \subset K$, then $\Bbb Q \subset K$, and $K$ is a finitely generated $\Bbb Q$-algebra. By Zariski's Lemma $K$ is a finitely generated $\Bbb Q$-module, and by Artin-Tate Lemma we get that $\mathbb Q$ is a finitely generated $\mathbb Z$-algebra, a contradiction.

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