# Show that $K$ has characteristic $> 0$

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

• 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? – Ragib Zaman Apr 2 '20 at 11:30
• In fact, every such field must be finite: math.stackexchange.com/questions/148745 – user26857 Apr 2 '20 at 18:19
• See also here for the same question today. – Dietrich Burde Apr 2 '20 at 18:55

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.