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Let's assume we have a commutative ring with identity. Can the Nil-Radical and the Jacobson Radical be equal in a non-trivial case (i.e. not every nonzero prime ideal in said ring is maximal)? Are there any interesting examples of this case?

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@Natalia S: You tagged the question "commutative algebra", but you might want to be explicit in the question on whether you are considering only commutative rings or not (and while you are at it, whether your rings must have a unity or not). – Arturo Magidin Oct 26 '10 at 19:36
Since the concept of prime ideal is being used, it seems likely that the author of the question has the commutative case in mind. In this case, there is an important class of rings (including in particular the examples of Arturo Magidin's answer) for which the answer is yes, namely Jacobson rings: As that wikipedia entry (and Arturo Magidin's answer) make clear, this question is rather closely bound up with the ideas behind Hilbert's Nullstellensatz. – Matt E Oct 26 '10 at 20:03
@Arturo Magidin: Thanks for your comment, I edited the question. – Pandora Oct 26 '10 at 20:03
@Natalia S: I also added "nonzero"; otherwise, $\mathbb{Z}$ itself would be an example. – Arturo Magidin Oct 26 '10 at 20:06
For any commutative ring R, R[x] has this property. This is one of the first exercises in Atiyah-Macdonald. – user641 Oct 27 '10 at 3:58

3 Answers 3

up vote 11 down vote accepted

There are indeed very many rings in which the nilradical equals the Jacobson radical.

Consider the following property of a ring: every prime ideal is the intersection of the maximal ideals containing it. These rings are called Hilbert-Jacobson rings. In a Hilbert-Jacobson ring, the intersection of all the prime ideals is therefore also the intersection of all maximal ideals, i.e., the nilradical and Jacobson radical coincide.

And indeed there are very many Hilbert-Jacobson rings. One useful result in this direction is:

Theorem: Let $R$ be a Hilbert Jacobson ring, and let $S$ be a ring which is finitely generated as an $R$-algebra. Then $S$ is also a Hilbert-Jacobson ring.

Since any field is trivially a Hilbert-Jacobson ring, it follows that all algebras which are finitely generated over a field are Hilbert-Jacobson, as Arturo Magidin pointed out in his answer.

For some information on this subject, including a proof of the theorem, see these notes.

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Let $k$ be a field, and let $R=k[x_1,\ldots,x_n]/\mathfrak{I}$, where $\mathfrak{I}$ is an ideal of $k[x_1,\ldots,x_n]$.

The nilradical of $R$ is $\sqrt{\mathfrak{I}}/\mathfrak{I}$, and the Jacobson radical of $R$ is $\mathfrak{M}/\mathfrak{I}$, where $\mathfrak{M}$ is the intersection of all maximal ideals of $k[x_1,\ldots,x_n]$ that contain $\mathfrak{I}$.

But by Hilbert's Nullstellensatz, $\sqrt{\mathfrak{I}}$, the radical of $\mathfrak{I}$, is the intersection of all maximal ideals of $k[x_1,\ldots,x_n]$ that contain $\mathfrak{I}$, so the nilradical of $R$ equals the Jacobson radical of $R$.

However, in these rings there are generally prime ideals that are not maximal. For example, if $n\geq 3$ and you take $\mathfrak{I}=(x_1)$, then $(x_1,x_2)+\mathfrak{I}$ is prime but not maximal.

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Theorem 5.1 in T.Y. Lam's book "A First Course in Noncommutative Rings" states that every polynomial ring $R[T]$ over a commutative ring $R$ satisfies $$rad \,\, R[T] = Nil(R[T]) = (Nil \,\,R)[T]$$

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