If a commutative ring $R$ with $1$ is finitely generated over $\mathbb Z$ could one deduce that the Jacobson radical of $R$ is nilpotent?
I am aware of the well-known fact that when $R$ is artinian, the Jacobson radical of $R$ is the unique nilpotent ideal of $R$, but here we are not sure of artinianness of $R$.
Thanks in advance for any cooperation!
Best Answer
Definition: A ring is Jacobson if every prime ideal is an intersection of maximal ideals.
Theorem (Nullstellensatz): Let $R$ be a Jacobson ring. Then every finitely generated $R$-algebra is Jacobson.
Since $\mathbb{Z}$ is Jacobson (as every nonzero prime is maximal, and $0$ is the intersection of any infinite set of maximal ideals), this shows that any finitely generated ring over $\mathbb{Z}$ is Jacobson. But in a Jacobson ring, the Jacobson radical and nilradical coincide (by definition). Finally, any finitely generated ring over $\mathbb{Z}$ is also Noetherian (by Hilbert's Basis Theorem) and in a Noetherian ring, the nilradical is nilpotent.