Properties Of Rings
Properties of rings: We have some business deferred from earlier to deal with. After that, we prove some basic properties of rings, starting from the axioms.
Uniqueness of zero element: The zero element of a ring is unique. For suppose that there are two zero elements, say z1 and z2. (This means that a z1 = z1 a = a for all a and also
a z2 = z2 a = a for all a.) Then
z1 = z1 z2 = z2.
Exercise: Show that the identity element of a ring, if it exists, is unique.
Uniqueness of additive inverse: The additive inverse of an element a is unique. For suppose that b and c are both additive inverses of a. (This means that a b=b a=0 and a c=c a=0 – we know now that there is a unique zero element, and we call it 0.) Then
b = b 0 = b (a c) = (b a) c = 0 c = c,
where we use the associative law in the third step.
Exercise: Show that the multiplicative inverse of an element of a ring, if it exists, is unique.
Adding more than two elements: The associative law tells us that if we have to add three elements, then the two possible ways of doing it, namely (a b) c and a (b c), give us the same result. For more than three elements, there are many different ways of adding them: we have to put in brackets so that the sum can be worked out by adding two elements at a time. For example, there are five ways of adding four elements:
((a b) c) d, (a (b c)) d, (a b) (c d), a ((b c) d), a (b (c d)).
These are all equal. For the associative law (a b) c) = a (b c) shows that the first and second are equal, while the associative law for b,c,d shows that the fourth and fifth are equal. Also, putting x = a b, we have
((a b) c) d = (x c) d = x (c d) = (a b) (c d),
so the first and third are equal; and similarly the third and fifth are equal. In general we have the following. The proof works for any associative binary operation.
Proposition 2.6 Let be an associative binary operation on a set A, and a1, . . . ,an 2 A. Then the result of evaluating a1 a2 · · · an, by adding brackets in any way to make the expression well-defined, is the same, independent of bracketing.
Proof The proof is by induction on the number of terms. For n = 2 there is nothing to prove; for n=3, the statement is just the associative law; and for n=4, we showed it above. Suppose that the result is true for fewer than n terms. Suppose now that we have two different bracketings of the expression a1 a2 · · · an. The first will have the form (a1* · · · * ai) (ai 1 * · · · * an), with the terms inside the two sets of brackets themselves bracketed in some way. By induction, the result is independent of the bracketing of a1, . . . ,ai and of ai 1*, . . . ,an. Similarly, the second expression will have the form (a1* · · · * aj) (aj 1 * · · · * an), and is independent of the bracketing of a1, . . . ,aj and of aj 1, . . . ,an.
Note that this result applies to both addition and multiplication in a ring. As usual, we denote a1 a2 · · · an by 
Cancellation laws
Proposition 2.7 In a ring R, if a x=b x, then a=b. Similarly, if x a=x b, then a = b.
Proof Suppose that a x = b x, and let y = −x. Then a = a 0 = a (x y) = (a x) y = (b x) y = b (x y) = b 0 = b.
The other law is proved similarly, or by using the commutativity of addition.
These facts are the cancellation laws.
A property of zero: One familiar property of the integers is that 0a = 0 for any integer a. We don’t have to include this as an axiom, since it follows from the other axioms. Here is the proof. We have 0 0 = 0, so 0a 0 = 0a = (0 0)a = 0a 0a, by the distributive law; so the cancellation law gives 0 = 0a. Similarly a0 = 0. It follows that if R has an identity 1, and |R| > 1, then 1 6= 0. For choose any element a 6= 0; then 1a = a and 0a = 0. It also explains why we have to exclude 0 in condition (M3): 0 cannot have a multiplicative inverse.
Commutativity of addition: It turns out that, in a ring with identity, it is not necessary to assume that addition is commutative: axiom (A4) follows from the other ring axioms together with (M2).
For suppose that (A0)–(A3), (M0)–(M2) and (D) all hold. We have to show that a b = b a. Consider the expression (1 1)(a b). We can expand this in two different ways by the two distributive laws:
(1 1)(a b) = 1(a b) 1(a b) = a b a b,
(1 1)(a b) = (1 1)a (1 1)b = a a b b.
Hence a b a b = a a b b, and using the two cancellation laws we conclude that b a = a b.
This argument depends on the existence of a multiplicative identity. If we take a structure with an operation satisfying (A0)–(A3) (we’ll see later that such a structure is known as a group), and apply the “zero ring” construction to it (that is, ab = 0 for all a,b), we obtain a structure satisfying all the ring axioms except (A4).
Boolean rings: We saw that a Boolean ring is a ring R in which xx = x for all x 2 R.
Proposition 2.8 A Boolean ring is commutative and satisfies x x = 0 for all x 2 R.
Proof We have (x y)(x y) = x y. Expanding the left using the distributive laws, we find that
xx xy yx yy = x y.
Now xx = x and yy = y. So we can apply the cancellation laws to get xy yx = 0.
In particular, putting y = x in this equation, we have xx xx = 0, or x x = 0, one of the things we had to prove.
Taking this equation and putting xy in place of x, we have xy xy = 0 = xy yx, and then the cancellation law gives us xy = yx, as required.
We saw that the power set of any set, with the operations of symmetric difference and intersection, is a Boolean ring.