Equivalent definitions of Jacobson rings We say that a ring $R$ is a Jacobson ring if
$$
J(R/I)=\operatorname{nil}(R/I)
$$
for every proper ideal $I$ of $R$, where
$
J(R)=\bigcap\{M:M \text{ maximal ideal}\}.
$
Then it also says, equivalently, we can define a Jacobson ring as
every prime ideal is an intersection of maximal ideals.

I am wondering how to show the equivalency?
I know that that
$$
\operatorname{nil}(R)=\bigcap\{P:P \text{ prime ideal}\}
$$
Thus the second definition leads to the first one, but how to prove the other direction?
 A: 
$J(R/I)=N(R/I)$ for every proper ideal $I$ of $R$ $\iff$ every prime ideal is an intersection of maximal ideals.

"$\Rightarrow$" Let $P$ be a prime ideal. Then $J(R/P)=N(R/P)$. But $R/P$ is a domain, so $N(R/P)=0$. From $J(R/P)=0$ we get that $P$ is the intersection of all maximal ideals containing it.
"$\Leftarrow$" Let $I$ be a proper ideal of $R$. Then we know that $\sqrt I=\bigcap_{P\supseteq I,\ P\text{ prime}}P$. Let's prove that $$\bigcap_{P\supseteq I,\ P\text{ prime}}P=\bigcap_{M\supseteq I,\ M\text{ maximal}}M.\qquad\qquad (*)$$ Obviously we have "$\subseteq$". For the converse, let $x\in\bigcap_{M\supseteq I,\ M\text{ maximal}}M$. If there is a prime $P\supseteq I$ such that $x\notin P$, then, since $P$ is an intersection of maximal ideals, there is a maximal ideal $M$ containing $P$ (hence containing $I$) such that $x\notin M$, a contradiction. Now, moding out by $I$ in $(*)$ we get $N(R/I)=J(R/I)$.
A: For the other direction, note that $J(R/I) = \mathrm{nil}(R/I) = \mathrm{rad}(I)/I$ for every ideal, $I$, of $R$. In particular, when $I = P$ is prime, we get
$$
P = \mathrm{rad}(P) = \bigcap \{ M \ge I \mid M \text{ maximal} \}
$$
