In [Wikipedia](http://en.wikipedia.org/wiki/Nilradical_of_a_ring) 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?



Jacobson ring = a ring whose nilradical and Jacobson radical coincide?
No. What abx said is exactly correct.


Consider the ring of polynomials in countably many variables over $\mathbb Q$. Consider a surjective homomorphism to the localization of $\mathbb Q[x]$ at $x=0$. The kernel is prime, but is not an intersection of maximal ideals, so the ring is not Jacobson.


However, the Jacobson radical is the zero ideal, which is the nilradical. Indeed, any nonzero element of this ring is a polynomial in finitely many variables. Take a $\mathbb Q$-rational point in affine space that is not a solution to that polynomial equation, and assign those variables to the coordinates of that point. Then assign the other variables to $0$. This gives a homomorphism to $\mathbb Q$ where the polynomial is nonzero, hence a maximal ideal that does not contain that element.


