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A ring $R$ is said to be finitely generated if there exists a finite subset $A$ of $R$ such that for all rings $R'\subset R$ with $A\subset R'$ we have $R'=R$. ($A$ is called the generating set of $R$)

Following is the statement that I want to prove:

If $F$ is a field which is finitely generated as a ring, then $F$ is a finite field.

This is supposed to follow as a corollary to the general Nullstellensatz (Eisenbud, Theorem 4.19)

Theorem: Let $R$ be a Jacobson ring. If $S$ is a finitely generated $R$-algebra with the ring homomorphism $\varphi: R\to S$, then $S$ is a Jacobson ring. Further, if $J\subset S$ is a maximal ideal, then $I = \varphi^{-1}(J)$ is a maximal ideal of $R$, and $S/J$ is a finite field extension of $R/I$.

If $F$ is a field of characteristic $p$ then the result follows by taking $R=\mathbb{F}_p$ and $S=F$ in the above theorem. But, how to show that $F$ can't be of characteristic $0$?

Arnaud Mortier
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rationalbeing
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    Replaced Jacobian with Jacobson. – Arnaud Mortier Jun 07 '18 at 17:26
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    @RaviFernando Funny how the accepted answer to the duplicate is a textbook example of "link only" answer. – Arnaud Mortier Jun 07 '18 at 17:27
  • @ArnaudMortier The link at this answer points to a post at this site, which Pierre-Yves Gaillard had written before for this site. Why should he not cite this? "Link only" mostly refers to some document outside of MSE, which may not stay there. – Dietrich Burde Jun 07 '18 at 18:08

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