Abstract Nonsense

Crushing one theorem at a time

Boolean Rings (Pt. I)

Point of Post: In this post we discuss Boolean rings along with some theorems including prove that they are naturally \mathbb{Z}_2-algebras.

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Often times when dealing with structures which are abstract generalizations of something simple, it is important to break ourselves away from strict intuition and remind ourselves that not all the objects we are dealing with are nice. A good example of this is that one may be naturally inclined to think of rings as objects like \mathbb{C} or \mathbb{Z}, very tame and familiar objects. So, it’s nice to look at weird examples of rings to keep ourselves straight.  A good example of this is encapsulated in the following example. Namely, take a non-empty set X and consider \mathscr{P}(X) with the operations of symmetric difference \Delta (defined as A\Delta B=(A-B)\cup (B-A)) and intersection. A quick check shows that \left(\mathscr{P}(X),\Delta,\cap\right) forms a commutative unital ring. Moreover, we note that every element is an idempotent, i.e. A^2=A\cap A=A for all A\in\mathscr{P}(X) and so we see that every non-zero, non unital element is a zero divisor. This is pretty ‘ugly’ quality, and thus does satisfy our desired quality of being different than the ‘typical’ rings. So, these shall be the rings of study in this post, rings that satisfy x^2=x for all x\in R.

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Definition and Basics

A ring R is called a Boolean ring if it is unital and satisfies r^2=r for all r\in R. Somewhat surprisingly (for those who haven’t seen this before) this idempotency property guarantees that every Boolean ring is commutative. Indeed:

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Theorem: Let R be a Boolean ring, then \text{char}(R)=2 (recall the definition of characteristic).

Proof: We merely note that x+x=(x+x)^2=x^2+x+x+x^2=x+x+x+x and so x+x=0 for all x\in R. \blacksquare

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We see then that:

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Theorem: Every Boolean ring is commutative.

Proof: Let R be a Boolean ring and a,b\in R be arbitrary. Then, a+b=(a+b)^2=a^2+ab+ba+b^2=a+ab+ba+b and so ab+ba=0 and since \text{char}(R)=2 this is equivalent to ab=ba. Since a,b were arbitrary the conclusion follows. \blacksquare

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That said, even though we now know that every Boolean ring is unital and commutative there is, of course, no hope that a Boolean ring is unless it’s isomorphic to \mathbb{Z}^2. Indeed:

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Theorem: Let R be a Boolean ring, then R is a field if and only if R\cong\mathbb{Z}_2.

Proof: Let x\in R then since x(x-1)=0 we evidently have, since R is a field, that x=0 or x=1. Thus, R=\{0,1\} and since evidently there is only one way to define a field structure on a two-element set the conclusion follows. \blacksquare

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So, we know that \left(\mathscr{P}(X),\Delta,\cap\right) is a Boolean ring for every set X, and while currently unjustified, we’d like to think that every Boolean ring is, for all intents and purposes, such a ring, or at least a subring of such a ring. As I said, right now there is no reason to believe that this is entirely true, that said while not proof hopefully the following properties will make the intuition slightly more valid:

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Theorem: Let B be Boolean ring, then any subring S of R is a Boolean ring and any homomorphic image R=\text{im}\left(f:B\to R\right) is a Boolean ring.

Proof: Evidently if x^2=x for every x\in B then x^2=x for every x\in S.

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Let f(x)\in R be arbitrary then f(x)^2=f\left(x^2\right)=f(x). \blacksquare

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Theorem: Let R be a Boolean ring, then the multiplication 0r=0_R and 1r=r turn R into a unital associative \mathbb{Z}_2-algebra.

Proof: Evidently, since R is a ring,  it suffices to prove that R is a \mathbb{Z}_2-space and that a(rs)=(ar)s=r(as) for all a\in\mathbb{Z}_2 and r,s\in R. To prove the first we note that

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\begin{aligned}&\mathbf{(1)}\; 1(r+s)=r+s=(1r)+(1s)\text{ and }0(r+s)=0_R=0r+0s\\ &\mathbf{(2)}(1+0)r=1r=r=1r+0r\text{ and }(1+1)r=0r=0_R=r+r=1r+1r\text{ and }(0+0)r=0r=0_R=0r+0r\\ &\mathbf{(3)}\;1(1r)=1r=r=1r=(11)r\text{ and }1(0r)=10_R=0_R=0r=(10)r\text{ and }0(0r)=00_R=0_R=0r=(00)r\\ &\mathbf{(4)}\; 1r=r\end{aligned}

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since R is an abelian group, the fact that R is a \mathbb{Z}_2-space follows. To prove the second quality we note that

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1(rs)=rs=(1r)s=r(1s)\text{ and }0(rs)=0_R=0_Rs=(0r)s=r0_R=r(0s)

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The conclusion follows. \blacksquare

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Corollary: Let R be a finite Boolean ring, then \#(R)=2^n for some n\in\mathbb{N}.

Proof: We know that R is \mathbb{Z}_2-algebra, and so we know (since evidently R is finitely generated as \mathbb{Z}_2-algebra) that as vector spaces R\cong\mathbb{Z}_2^{\dim_{\mathbb{Z}_2}(R)}. The conclusion follows. \blacksquare

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Remark: Really, if R is a finite ring then we know from the structure theorem for finite abelian groups that R is isomorphic (as groups) to some product of the form  \mathbb{Z}_{p_1^{\alpha_1}}\times\cdots\times\mathbb{Z}_{p_n^{\alpha_n}} but since 2x=0 for all x\in R it evidently follows that p_k=2 for every k=1,\cdots,n. That said, this seems an apt application of a saying a professor of mine is fond of saying “That’s like hunting a butterfly with an elephant gun!”

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1. Dummit, David Steven., and Richard M. Foote. Abstract Algebra. Hoboken, NJ: Wiley, 2004. Print.

2. Bhattacharya, P. B., S. K. Jain, and S. R. Nagpaul. Basic Abstract Algebra. Cambridge [Cambridgeshire: Cambridge UP, 1986. Print.


July 14, 2011 - Posted by | Algebra, Ring Theory | , , , , , ,

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