Number Theory: An Introduction to Mathematics

8 Rings and Fields 61

It may seem inconsistent to require that addition is commutative, but not multipli-
cation. However, the commutative law for addition is actually a consequence of the
other axioms for a ring. For, by the first distributive law we have

(a+b)( 1 + 1 )=a( 1 + 1 )+b( 1 + 1 )=a+a+b+b,

and by the second distributive law

(a+b)( 1 + 1 )=(a+b) 1 +(a+b) 1 =a+b+a+b.

Since a ring is a group under addition, by comparing these two relations we obtain first

a+a+b=a+b+a

and thena+b=b+a.
As examples, the setZof all integers is a commutative ring, with the usual defi-
nitions of addition and multiplication, whereas ifn>1, the setMn(Z)of alln×n
matrices with entries fromZis a noncommutative ring, with the usual definitions of
matrix addition and multiplication.
A very different example is the collectionP(X)of all subsets of a given setX.If
we define the sumA+Bof two subsetsA,BofXto be theirsymmetric difference,
i.e. the set of all elements ofXwhich are in eitherAorB, but not in both:

A+B=(A∪B)\(A∩B)=(A∪B)∩(Ac∪Bc),

and the productABto be the set of all elements ofXwhich are in bothAandB:

AB=A∩B,

it is not difficult to verify thatP(X)is a commutative ring, with the empty set∅as
identity element for addition and the whole setXas identity element for multiplication.
For everyA∈P(X),wealsohave

A+A=∅, AA=A.

The set operations are in turn determined by the ring operations:

A∪B=A+B+AB, A∩B=AB, Ac=A+X.

AringRis said to be aBoolean ringifaa=afor everya∈R. It follows that
a+a=0foreverya∈R,since

a+a=(a+a)(a+a)=a+a+a+a.

Moreover, a Boolean ring is commutative, since

a+b=(a+b)(a+b)=a+b+ab+ba

andba=−ba,bywhatwehavealreadyproved.
For an arbitrary setX, any nonempty subset ofP(X)which is closed under union,
intersection and complementation can be given the structure of a Boolean ring in the