I'm trying to prove that the intersection of all maximal ideals in a noetherian ring is the nilradical. I know that the nilradical is the intersection of all prime ideals, but don't see why the intersection of maximal ideals should be enough. Also it is clear that the nilradical is in that intersection but the opposite direction stuns me. I think you could use something like the Lasker-Noether theorem, but I'm missing something.
Any help would be appreciated.
Edit: Just for clarity, the nilradical is also the set of nilpotent elements.
Best Answer
This is hopeless: it isn't true. Any Noetherian local domain that isn't a field, (for example, the $2$-adic integers) is a counterexample.
It's true that the nilradical is the intersection of all prime ideals though. You can find that proven on the site and in basically every commutative ring theory book.
For any field $k$, a quotient of $k[t_1,...,t_n]$ is a finitely generated algebra over a Jacobson ring, and so it is itself a Jacobson ring. As such, all the prime ideals are intersections of maximal ideals. Therefore the intersection of all prime ideals is an intersection of many maximal ideals, so it contains the Jacobson radical.
(Of course, the nilradical is always contained in the Jacobson radical in any commutative ring. We can say then that the reverse containment holds for Jacobson rings.)