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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.

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migrated from mathoverflow.net Sep 30 '15 at 15:38

This question came from our site for professional mathematicians.

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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 $(p)$.

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  • $\begingroup$ Thank you! So I am not too stupid to solve the exercise, but too stupid to read the other post. $\endgroup$ – user60589 Sep 30 '15 at 15:56

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