1
$\begingroup$

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.

Improve this question
$\endgroup$
1
  • 1
    $\begingroup$ This is also proved in Lam's "First Course in Noncommutative Rings". $\endgroup$
    – Pace Nielsen
    Sep 3, 2017 at 23:12

1 Answer 1

Reset to default
3
$\begingroup$

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).

Improve this answer
$\endgroup$
1
  • 1
    $\begingroup$ This is Theorem 4.19 in Eisenbud. $\endgroup$
    – Zhen Lin
    Apr 23, 2013 at 16:37

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.