rings in which all ideals are radical

Question

If $R$ is a commutative Hilbert ring then each of its prime ideals is an intersection of maximal ideals. Is there a similar class of commutative rings for which every ideal is an intersection of prime ideals (that is, in which all ideals are radical)? Does this class have a name? Have the rings in it been characterized?

Answer 1

As Qiaochu Yuan's answer observed, such a ring must have the property that every finitely generated ideal is generated by an idempotent and therefore every ideal is generated by idempotents. Conversely, I claim that if every principal ideal in a commutative ring $R$ is generated by an idempotent then every ideal in $R$ is radical. To prove this, suppose $I\subseteq R$ is an ideal and $r^n\in I$. Let $e$ be an idempotent that generates the principal ideal $(r)$. Then since $e$ is a multiple of $r$, $e^n=e$ is a multiple of $r^n$ so $e\in I$. But since $(e)=(r)$, this means $r\in I$.

So, all ideals are radical in a commutative ring iff every principal ideal is generated by an idempotent. Such a ring is called a von Neumann regular ring and much is known about them. For instance, one alternative characterization is that a commutative ring is von Neumann regular iff it is $0$-dimensional (i.e., every prime is maximal) and reduced.

Answer 2

This is an extremely rare condition. It implies that every ideal $I$ satisfies $I^2 = I$; that is, every ideal is idempotent. It's known that if $I$ is a finitely generated idempotent ideal then it must be the principal ideal generated by an idempotent element $(e)$, so we conclude that every finitely generated ideal is generated by an idempotent.

The simplest class of rings with this property are the ones in which every element is idempotent; these are the Boolean rings. By a version of Stone's representation theorem the category of Boolean rings is known to be contravariantly equivalent to the category of profinite sets, with the functor to profinite sets given by taking $\text{Spec}$ and the functor in the other direction given by taking continuous functions valued in $\mathbb{F}_2$. It follows that if $I$ is an ideal of a Boolean ring $B$ then the quotient map $B \to B/I$ corresponds via this duality to an injection $\text{Spec}(B/I) \to \text{Spec}(B)$, and the question of whether $I$ is an intersection of prime ideals is dual to the question of whether $\text{Spec}(B/I)$ is a join, in the lattice of subobjects of $\text{Spec}(B)$, of points, which it is. So Boolean rings satisfy the desired condition.

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Some general comments although they don't lead to a full classification: the property that every finitely generated ideal is generated by an idempotent passes to quotients, so if $R$ is such a ring and $P$ is a prime ideal of it then $R/P$ also satisfies this property, and since the only idempotents are $0$ and $1$ it follows that every finitely generated ideal is either zero or the unit ideal, hence that $R/P$ is a field. So every prime ideal of $R$ is maximal, meaning $R$ has Krull dimension zero. It also follows that $R$ is a Jacobson ring and hence its nilradical and Jacobson radical coincide. Since every ideal satisfies $I^2 = I$ no nonzero ideal can be nilpotent, so no nonzero element can be nilpotent either. So $R$ has trivial nilradical, hence trivial Jacobson radical. If $R$ is Noetherian it now follows that $R$ is a finite product of fields, and it's not hard to check that any finite product of fields satisfies the desired condition. I suspect an infinite product of fields also satisfies this condition (Boolean rings are precisely the subrings of arbitrary products of copies of $\mathbb{F}_2$) but I have not checked this.