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In a preparation question for an exam, I am asked to give an example of a ring $A$ such that the nilradical $\operatorname{Nil}(A)$ is strictly smaller that the Jacobson radical $J(A)$. Here's how I solved the problem:

It is enough to find some ring $A$ with a prime ideal $p$ which is strictly contained in some maximal ideal $m$. Then the localization $A_m$ is a local ring with maximal ideal $m$, and by the correspondence between ideals of $A_m$ and ideals of $A$ contained in $m$ we have that $J(A)=m\supsetneq p\supset\operatorname{Nil}(A)$.

An example is $A=\mathbb{Z}[x]$, $p=(x^2+1)$ and $m=(x^2+1,2)$.

Now I was wondering: are there also any "elementary" examples satisfying the condition above? By this I mean basically examples where I don't have to localize the ring.

Daniel Robert-Nicoud
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    Isn't it enough to construct a ring with no nilpotent elements, but nontrivial Jacobson radical? Say like k[[x]]. Although I suppose that is a localization. – Nate Jan 10 '14 at 21:56

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Local integral domains should give you quite a lot of examples like this. They can arise from localizations (as the one you did above or $\mathbb{Z}_{\mathfrak{p}}$). Or take a ring of power series over a field, e.g. $\mathbb{Q}[[X]]$. These are the easiest examples in my opinion as the Jacobson radical is just the only maximal ideal. (similar to power series rings, you could also look at the $p$-adic integers)

Louis
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