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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.

Can someone please help to clarify this subject for me?

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

Thanks.

bt26
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  • 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 00:25
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    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 00:28
  • @user1987 Could you elaborate on what you mean by "domain"? – bt26 May 26 '17 at 00:29
  • @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 00:32
  • @user1987 Sorry, but what does the "two with a bar" mean? – bt26 May 26 '17 at 00:36
  • @user1987 Also, the only set elements that I could think of to test were numbers, vectors, and matrices. Mutliplying coordinate pairs, as Joe Johnson 126 does in an example below, has helped "expand my mind" on this subject, but it's frankly something I've never seen before. – bt26 May 26 '17 at 00:39
  • @bt26 $\overline 2$ is the modulo: the remainder when the integer is divided by (in this case) four. $\Bbb Z/4\Bbb Z = {\overline 0, \overline 1, \overline 2, \overline 3}$ . – Graham Kemp May 26 '17 at 00:45
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    @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 00:52

3 Answers3

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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 their 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.

davidlowryduda
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J126
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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.

J126
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You're probably well beyond this by now, but perhaps this answer will help. I will stick to the commutative case. For any ring $R$, and any field $k$, we can ask for the $k$-valued points of $R$. These are by definition the ring homomorphisms from $R$ to $k$.

For example if $ R = Z[x,y]/(x^2+y^2-1)$ and $k=\mathbb{R}$, this is the unit circle in the Euclidean plane $\mathbb{R}^2$, since any ring homomorphism is determined by the values assigned to $x$ and $y$, and they can be any values of $\mathbb{R}$ satisfying $x^2+y^2 = 1$.

If $\mathbb{F}_2 = \{0,1\}$ is the field with 2 elements and $R = \mathbb{F}_2[x_1, \ldots, x_n]/I$ is a polynomial ring over $\mathbb{F}_2$ modulo some ideal $I$, and we are interested in the $\mathbb{F}_2$ points of $R$, then we might as well enlarge the ideal $I$ by adding the elements $x_i^2 - x_i$, since any ring homomorphism to $\mathbb{F}_2$ must send each $x_i$ to either $0$ or $1$, both of which satisfy $x_i^2 = x_i$. This new ring is a Boolean ring since (a+b)^2 = a^2 + b^2 = a + b, because we are working with $\mathbb{F}_2$ coefficients.

Thus Boolean rings arise naturally whenever we are interested in the $\mathbb{F}_2$ points of any commutative ring $R$.

Robert Bruner
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