Jacobson Radical of a Boolean Ring is Zero

Question

I am trying to prove that the Jacobson radical $J(R)$ of a Boolean ring $R$ is $\{0\}$, even when $R$ does not have a unity. In the case where the ring has an identity element, the proof is straightforward. However, I encounter difficulties extending the argument to the non-unital case.

Here's the general idea I'm working with:

Suppose, for contradiction, that there exists some $x \ne 0$ such that $x \in \mathrm{Jac}(B)$. Then by definition, \begin{align*} x \in \bigcap \{ M : M \text{ is a maximal ideal of } B \}, \end{align*} i.e., $x \in M$ for every maximal ideal $M \subseteq B$.

However, since $x \ne 0$, the ideal $\langle x \rangle$ is nontrivial. By Zorn’s Lemma, there exists a maximal ideal $M$ containing $\langle x \rangle$ that is maximal with respect to inclusion. Since $B$ is commutative, the quotient $B/M$ is a field (or at least has no zero divisors). But Boolean rings are of characteristic 2 and satisfy $x^2 = x$ for all $x$, so the only possibility is that $B/M \cong \mathbb{F}_2$.

The issue is that I would now like to construct a maximal ideal not containing $x$, to contradict the assumption that $x$ lies in the intersection of all maximal ideals. But I am unable to construct such a maximal ideal explicitly — this is where my argument stalls. Any help on how to proceed would be greatly appreciated. So

• Is it true that for any Boolean ring $R$ (with or without unity), we have $J(R) = \{0\}$?
• If not, can someone provide a counterexample?

In summary, I want to prove that, for any Boolean ring $R$, \begin{align*} J(R) = \{0\}, \end{align*} without assuming that $R$ has a multiplicative identity.

Thank you in advance for any guidance or references you can provide on this matter.

Answer 1

In a boolean ring, every prime ideal is maximal. Thus, the Jacobson radical of $R$ coincides with the nilradical of $R$, ie. the ideal consisting of all nilpotent elements of $R$. Here, I'm using the fact that the nilradical of a ring coincides with the intersection of all its prime ideals, and this requires the axiom of choice. Now, the nilradical of a boolean ring is obviously trivial, since all elements are idempotents by definition.

The exact same proof actually generalizes to von Neumann regular rings, which are a little bit more general than boolean rings.

Answer 2

The Jacobson radical is characterized as the largest right ideal of $R$ such that all elements are right quasiregular.

An element $x$ is said to be right quasi-regular when there exists a $y$ such that $x+y=xy$. If $x$ is right quasi regular and $x+y=xy$ holds, we also then have this for any $x$ in the radical:

$x+xy = x(x+y)=xxy=xy$, and subtracting $xy$ from both sides, we get $x=0$.

Said another way, any r.q.r idempotent is $0$.

Thus every Boolean ring has trivial Jacobson radical, whether or not it has an identity.

Answer 3

In general the Jacobson radical of a ring can not contain an idempotent (besides zero). For suppose $e$ is an idempotent which belongs to the Jacobson radical then it is well-known that $1-e$ is invertible. But $e(1-e) = 0$ which implies that $e=0$.

Now back to your question, in a Boolean ring all elements are idempotent and therefore the only element that can be in the Jacobson radical is zero.