# Understanding boolean rings

I'm reading a book "Introduction to Abstract Algebra" by Neal McCoy.

I've come across a few exercises which discuss "Boolean rings".

The text defines Boolean rings as:

A ring $R$ is a Boolean ring if $a^2 = a$ for every element $a$ of $R$.

and from this definition, the only sets that I can think of that could be categorized as Boolean rings are {0, 1} and its subsets.

But then there's a problem that asks:

Give an example of a Boolean ring with 32 elements and an example of a ring with 32 elements which is not a Boolean ring.

and it seems there's something that I'm clearly not understanding.

I'm new to Abstract Algebra, and the posts that come up in the search results are all pretty much over my head.

Even if you can just offer a simple example of Boolean ring with more than two elements, I would find that helpful.

Thanks.

• The ring $\mathbb{F}_2$ with two elements is a Boolean ring. So is any power of it, not necessarily finite. There are also other interesting examples. – Qiaochu Yuan May 26 '17 at 0:25
• Probably what you are thinking is this: $a^2 = a$ implies $a = 0$ or $a=1$. But this is true only if the ring is a domain. – user 1987 May 26 '17 at 0:28
• @user1987 Could you elaborate on what you mean by "domain"? – bt26 May 26 '17 at 0:29
• @bt26 A ring $R$ is a domain if $ab=0$ implies $a = 0$ or $b=0$. The integers, for instance, form a domain. But not $\mathbb{Z}/4\mathbb{Z}$ because $\overline{2}\cdot \overline{2} = 0$. – user 1987 May 26 '17 at 0:32
• @bt26 Remember, that a Ring is equiped with two binary operations $+$, $\cdot$ that satisfy the Ring Axioms. However, these ops need not be what you might, uh, traditionally think of as addition and multiplication. Rather, they are abstractions of the concepts. – Graham Kemp May 26 '17 at 0:52

If I took pairs $(x,y)$ where $x$ and $y$ can be either $0$ or $1$, I get the set $$\{ (0,0), (0,1), (1,0), (1,1) \}.$$ Multiply pairs by multiplying there coordinates, e.g.: $$(0,1) * (1,1) = (0 \cdot 1, 1 \cdot 1) = (0,1).$$ Then $a^2 = a$ for every element of my four element set.
As an added answer, George Boole originally had sets in mind. If $X$ and $Y$ are two sets we can put $$X + Y = X \cup Y, \quad X \cdot Y = X \cap Y.$$ Then $$X^2 = X \cdot X = X \cap X = X.$$ Take the set $\{0,1\}$, then the set of all subsets is $$\{ \emptyset, \{0\}, \{1\}, \{1,2\} \}.$$ This is closed under union and intersection and has four elements, like my previous answer.