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Let $R$ be a commutative finitely generated $\mathbb{Z}$-algebra. Then the nilradical is equal to the Jacobson radical.

I am not able to make much traction on this, nor can I find this result in any book I've look at. So far I've reasoned that since $R \cong \mathbb{Z}[x_1,\dots,x_n]/I$ for some ideal $I$, then $\mathrm{nil}(R) = \sqrt{I}/I$ and $J(R) = M/I$, where $M$ is the intersection of all ideals containing $I$. Clearly, the Jacobson radical contains all nilpotent elements, due the to the quasiregular condition, but I can't seem to follow through on the converse.

I already know that if $R$ is a field, then it must be finite, but I can't see how this would help me prove the general case.

Any help would be greatly appreciated.

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    $\begingroup$ This is also proved in Lam's "First Course in Noncommutative Rings". $\endgroup$
    – Pace Nielsen
    Sep 3, 2017 at 23:12

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This follows from the Nullstellensatz (or the version of it), which says that a finitely generated algebra over a Jacobson ring is Jacobson. This can be found in, for example, Eisenbud's book on commutative algebra (I don't have it at hand so I can't be more specific right now, but it should be easy to find in the table of contents).

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    $\begingroup$ This is Theorem 4.19 in Eisenbud. $\endgroup$
    – Zhen Lin
    Apr 23, 2013 at 16:37

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