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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.

Vincent Batens
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    As a concrete counterexample, there exist rings like $\mathbf Z_{(p)}$ with only two primes $0 \subseteq p \mathbf Z_{(p)}$ (noetherian local domains which aren't fields) – Sam Moore May 21 '24 at 23:36
  • I miswrote the question. I'm working in an hopf algebra of a linear algebraic group over an algebraically closed field, so the ring is a quotient of $k[t_1,...,t_n]$ with k alg. closed. Then it should be true, but no idea how. – Vincent Batens May 21 '24 at 23:45
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    Just to give you some key terms to search for: The intersection of all maximal ideals is called the Jacobson radical. A ring in which the nilradical and Jacobson radical coincide is called a Jacobson ring. – Alex Kruckman May 22 '24 at 00:00
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    @AlexKruckman that's not quite the terminology I know. A Jacobson ring has the property that in any quotient, not just the ring itself, the nilradical and Jacobson radical coincide. This can of course be phrased in terms of the ring itself: A ring is Jacobson iff any prime ideal is the intersection of the maximal ideals containing it. – Lukas Heger May 22 '24 at 14:06
  • @LukasHeger Ah, I'm sure your right. I'm far from being a ring theorist, and I must have misremembered the terminology. Thanks for the correction! – Alex Kruckman May 22 '24 at 16:22

1 Answers1

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I'm trying to prove that the intersection of all maximal ideals in a noetherian ring is the nilradical.

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.)

rschwieb
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  • Oh I will reask my question but better than, I'm working in an hopf algebra of a linear algebraic group over an algebraically closed field, so the ring is a quotient of $k[t_1,...,t_n]$ with k alg. closed. Then it should be true, but no idea how. – Vincent Batens May 21 '24 at 23:45
  • @VincentBatens It was not much trouble to add an answer for your refined question. Does this help? – rschwieb May 22 '24 at 14:00