TITLE: Jacobson ring = a ring whose nilradical and Jacobson radical coincide?
QUESTION [8 upvotes]: In Wikipedia it is claimed that "A ring is called a Jacobson ring if the nilradical coincides with the Jacobson radical." Here the word "ring" means a commutative ring.
However, I remember that Jacobson ring is defined to be a ring with the property "every prime ideal is equal to the intersection of some maximal ideals".
Are these two definitions really equivalent?

REPLY [7 votes]: $\newcommand{\mm}{\mathfrak{m}}$The counterexamples given so far seem perhaps more elaborate than necessary.  Here's another:
Let $A$ be a domain which is not a field and which has finitely many maximal ideals $\mm_1,\ldots,\mm_n$.  (So e.g. take $\mathbb{Z}_p$ or $\mathbb{C}[[x_1,\ldots,x_n]]$ or $\mathbb{F}_{17}[[t^2,t^3]]$ or...you have some leeway here.) Let $R = A[t]$.  Then:  
$\bullet$ Like every domain, $R$ has zero nilradical.
$\bullet$ Like every polynomial ring over a domain, $R$ has zero Jacobson radical: if $0 \neq x \in R$, then $1+tx \notin A^{\times} = R^{\times}$.
$\bullet$: The ideal $\mathfrak{p} = \langle t \rangle$ of $R$ is prime.  The maximal ideals of $R$ containing $\mathfrak{p}$ correspond to the maximal ideals $\mm_1,\ldots,\mm_n$ of $R/\mathfrak{p} = A$, so $\bigcap_{i=1}^n \mm_i \supset \mm_1 \cdots \mm_n \supsetneq (0)$.  Thus $\mathfrak{p}$ is not the intersection of the maximal ideals containing it.  
It may be helpful to mention the (easy) fact that a ring $R$ is Jacobson iff for all ideals $I$ of $R$, the nilradical of $R/I$ equals the Jacobson radical of $R/I$.  This simplifies the last bullet point and is really the point of the construction.  In fact for any ring $A$, the Jacobson radical of $A[t]$ is equal to the nilradical of $A[t]$ (Atiyah-Macdonald, Exercise 1.4), so taking any $A$ with nilradical smaller than its Jacobson radical works to give a counterexample: e.g. a domain which not a field and which has only  finitely many maximal ideals.