Nil-Radical equals Jacobson Radical even though not every prime ideal is maximal?

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?

Solution 1:

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.

Solution 2:

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.

Solution 3:

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 $$rad \,\, R[T] = Nil(R[T]) = (Nil \,\,R)[T]$$