## Boolean Rings (Pt. I)

**Point of Post: **In this post we discuss Boolean rings along with some theorems including prove that they are naturally -algebras.

**Motivation**

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 or , 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 and consider with the operations of symmetric difference (defined as ) and intersection. A quick check shows that forms a commutative unital ring. Moreover, we note that every element is an idempotent, i.e. for all 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 for all .

*Definition and Basics*

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

**Theorem: ***Let be a Boolean ring, then (recall the definition of characteristic).*

**Proof: **We merely note that and so for all .

We see then that:

**Theorem: ***Every Boolean ring is commutative.*

**Proof: **Let be a Boolean ring and be arbitrary. Then, and so and since this is equivalent to . Since were arbitrary the conclusion follows.

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 . Indeed:

**Theorem: ***Let be a Boolean ring, then is a field if and only if .*

**Proof: **Let then since we evidently have, since is a field, that or . Thus, and since evidently there is only one way to define a field structure on a two-element set the conclusion follows.

So, we know that is a Boolean ring for every set , 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:

**Theorem: ***Let be Boolean ring, then any subring of is a Boolean ring and any homomorphic image is a Boolean ring.*

**Proof: **Evidently if for every then for every .

Let be arbitrary then .

**Theorem: ***Let be a Boolean ring, then the multiplication and turn into a unital associative -algebra.*

**Proof: **Evidently, since is a ring, it suffices to prove that is a -space and that for all and . To prove the first we note that

since is an abelian group, the fact that is a -space follows. To prove the second quality we note that

The conclusion follows.

**Corollary: ***Let be a finite Boolean ring, then for some .*

**Proof: **We know that is -algebra, and so we know (since evidently is finitely generated as -algebra) that as vector spaces . The conclusion follows.

*Remark: *Really, if is a finite ring then we know from the structure theorem for finite abelian groups that is isomorphic (as groups) to some product of the form but since for all it evidently follows that for every . 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!”

**References:**

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.

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