Let $B$ be a boolean ring, that is, for all $x\in B$ we have $x^2 = x$. Then we know a few things about them. First $B$ has characteristic $2$, since taking $x\in B$ we have
$$x+x = (x+x)^2 = x^2 + x^2 + x^2 + x^2 = x + x + x +x$$
which implies that $x + x = 0 $. Secondly, $B$ is commutative, since for $x, y\in B$ we have
$$x+y = (x+y)^2 = x^2 + xy + yx + y^2 = x + xy+yx+y$$
which implies $xy + yx = 0$, but since $\mbox{char}(B) = 2$ then
$$xy + xy + yx = xy \Rightarrow yx = xy$$
So in particular the centre of $B$ is equal to $B$. The way I've shown that $B$ is an associative $\mathbb{Z}/2\mathbb{Z}$-algebra is by defining a ring homomorphism $f:\mathbb{Z}/2\mathbb{Z}\to B$ by $f(x) = x1_B$, that is $f(0) = 0_B$ and $f(1)=1_B$. It's easy to show that this is a homomorphism by just checking all the possible cases. The only case where it could go wrong is
$$f(1 + 1) = 1_B + 1_B = 0_B = f(0)$$
which of course relies on the fact that $\mbox{char}(B) = 2$. The reason I bring this up is because I was reading the Wiki page for Boolean rings and in the properties section it says
The property $x \oplus x = 0$ shows that any Boolean ring is an associative algebra over the field $F_2$ with two elements
As I've said, the fact that $\mbox{char}(B) = 2$ is necessary for $B$ to be a $\mathbb{Z}/2\mathbb{Z}$-algebra, but I get the feeling that the line I've just quoted isn't just referring to the fact that one of the cases wouldn't work without it.
So why does the characteristic being $2$ show that $B$ is an associative $\mathbb{Z}/2\mathbb{Z}$-algebra? Is there something more sophisticated or deeper going on that I'm just not seeing? Or is it really just as simple as it guarantees that the map I defined above is a homomorphism?
If $R$ is any ring, then there is a unique homomorphism $f:\mathbb{Z}\to R$ which maps $n\in\mathbb{Z}$ to the sum of $n$ copies of the multiplicative identity. The image of this homomorphism is always central in $R$, since the multiplicative identity is in the center. So, every ring is a $\mathbb{Z}$-algebra (in a unique way).
If $R$ has characteristic $n$, then $\ker(f)=n\mathbb{Z}$ so $f$ factors through a homomorphism $\mathbb{Z}/n\mathbb{Z}\to R$ whose image is again central. So, any ring of characteristic $n$ is a $\mathbb{Z}/n\mathbb{Z}$-algebra (in a unique way).