Can Boolean ring without unit be embedded into a boolean ring?

Question

While going through a book (Lectures on Boolean algebra, Halmos) I got struck at the following question :
Prove that every Boolean ring without a unit can be embedded in a Boolean ring with a unit.

If possible, give me an elementary proof of the theorem.

Answer 1

Part of modern, abstract algebra is turning imagination into reality by fiat. This is a rather general technique: say you have some concrete object $A$ and you want an element with property $P$: then just adjoin a formal element to $A$ and quotient the result by the collection of all relations that must be satisfied by that element in order for that element to have property $P$. This approach can be used for multiple different elements, and in multiple different contexts (rings, groups, fields, etc.).

Now say $A$ is a Boolean ring (so it is commutative and has characteristic two, remember). We want a new element $\epsilon$ in it that acts as the identity. What relations does the identity need to satisfy; can you turn these relations into an ideal to quotient $A[\epsilon]$ by? Now you simply need to check that the quotient $A[\epsilon]/I$ is Boolean and unital and contains a copy of $A$ (i.e. $A\to A[\epsilon]/I$ is injective).

Answer 2

The most elementary proof is to consider $A'=\mathbf{2}\times A$, where $\mathbf{2}=\{0,1\}$ is the two element boolean ring. The operations are defined by

\begin{gather} (\alpha,a)+(\beta,b)=(\alpha+\beta,a+b)\\ (\alpha,a)(\beta,b)=(\alpha\beta,\alpha b+a\beta+ab) \end{gather} and $0a=0$, $1a=a$ by definition.

Proving that the axioms of rings are satisfied is easy. The map $A\to A'$ defined by $a\mapsto(0,a)$ is a ring homomorphism. Moreover $$ (\alpha,a)^2=(\alpha^2,\alpha a+a\alpha+a^2)=(\alpha,a) $$ so it is a Boolean ring. It has a unit, because $$ (\alpha,a)(1,0)=(\alpha,a). $$

This is the same as the usual Dorroh extension for generic rings (or $\mathbb{Z}$-algebras), only we use the fact that $A$ is a $\mathbf{2}$-algebra.

We are basically adding a complement to each element: indeed $(1,0)+(0,a)=(1,a)$, so $(1,a)$ is the complement of $(0,a)$.