Do Boolean rings always have a unit element?

Question

Let $(B, +, \cdot)$ be a non-trivial ring with the property that every $x \in B$ satisfies $x \cdot x = x$. How does one prove that such a ring $(B, +, \cdot)$ must have a unit element $1_B$? (Or, in case this is not true in general, what is a counterexample?)

BTW, I'm looking for an elementary proof, not requiring anything more than the definition of a ring, the definition of $(B, +, \cdot)$, and, if necessary, the easily shown facts that $x + x = 0$ and $x\cdot y = y\cdot x,\,\forall\, x,y \in B$.

Answer 1

An example: the family of all finite subsets of a given infinite set.

Answer 2

Here is a positive result: Let $R$ be a commutative ring such that every element can be written as a sum of products of elements of $R$ (i.e. the multiplication map $R \otimes_{\mathbb{Z}}R \to R$ is surjective). If $R$ is finite, then $R$ is unital. In particular, every finite boolean ring is unital (which also can be proven directly, of course).

Proof: Consider the unitalization $R^+$ as an $R$-module. Then we have $R R = R$, hence by Nakayama there is some $e \in R$ with $(1-e) R=0$. But this means that $e$ is a unit of $R$.

Answer 3

In the ring $\prod_{i=1}^\infty\Bbb Z_2$, consider the ideal $\oplus_{i=1}^\infty\Bbb Z_2$. It is a subrng without identity.

(Ultimately I think this is isomorphic to Boris' example, but here the operations are clear.)