Is the Jacobson radical of a ring with finite spectrum and nilpotent nilradical nilpotent?

Question

I tried to solve 1.3.3 in Bosch, Algebraic Geometry and Commutative Algebra.

I did not find a way to solve it. But I found this: Finitely many prime ideals ⇒ cartesian product of local rings.

And I am not able to show that in a commutative ring with unit, $R$, which has only finitely many prime ideals and nilpotent nilradical the Jacobson radical is also nilpotent.

I would be happy if someone could give me a hint how to solve it. Thanks. You can find the exercise from Bosch in the link.

Answer 1

This is not true. For instance, consider the localization $R=\mathbb{Z}_{(p)}$, or more generally the localization of any domain at a height $1$ prime. The only primes in $R$ are $0$ and $(p)$, the nilradical is $0$, but the Jacobson radical is $(p)$.