Ring Theory/Properties of rings
We shall now discuss some basic theorems related to rings. We feel that a good way to learn ring theory is to try out proofs of simple theorems on ones own. Hence the reader is encouraged to work out proofs of theorems by him/herserlf before reading the proofs given here. Often we shall provide only a sketch of the proof and the reader is expected to fill in the gaps in that case.
Theorem 1.1: If R is a ring and ; then
1. a+b=a+c implies b=c. (Cancellation Law)
3. The zero element of R is unique.
4. The additive inverse of any element is unique.
1. Clearly adding -a on both sides of a+b=a+c gives us the desired result.
2. It suffices to show that a+(-a)=0 which is obvious from the definition of -a.
3. If there exists two zero elements 0 and 0' in R then 0+0'=0' and 0+0'=0 by definition and so 0=0'.
4. If a' and a'' are two inverses of a then a'=a'+0=a'+a+a''=0+a''=a''.
Theorem 1.2: If R is a ring, then for any ;
If in addition, R has a unit element 1, then
1. a0+0=a0=a(0+0)=a0+a0. By the cancellation law it now follows that a0=0. Similarly 0a=0.
2. It suffices to show that a(-b)=-(ab) or equivalently a(-b)+ab=0. Now a(-b)+ab=a(b-b)=a0=0 by 1. and so the result is proved.
3. (-a)(-b)=-(a(-b)) by 2. Again -(a(-b))=-(-(ab))=-(-ab) by 2. But by Theorem 1.1(2) -(-ab)=ab.
4. a(b-c)=a(b+(-c))=ab+a(-c)=ab-ac by 2.
5. (-1)a+a=(-1)a+1a=(-1+1)a=0a=0 by 1. and so (-1)a=-a.
6. Put a=-1 in 5. and apply Theorem 1.1(2).
Some more resultsEdit
It is strongly recommended that theorems in this section should be treated as exercises by the readers.
Theorem 1.3: Prove that a ring R is commutative if and only if holds for all .
Proof: Suppose R is commutative. Then clearly the result holds. (In fact the binomial theorem: holds in that case. Try to prove it using induction and the Pascal's identity: .) Conversely suppose that for each the given relation is satisfied. Now, on applying the distributive laws to we get and by the cancellation laws we have ab=ba. Hence R is commutative.
Theorem 1.4: If R is a system satisfying all the conditions of a ring with unit element with the possible exception of a+b=b+a, prove that the axiom a+b=b+a must hold in R and that thus R is a ring.
Proof: (a+b)(1+1)=a1+a1+b1+b1=a+a+b+b and (a+b)(1+1)=a1+b1+a1+b1=a+b+a+b by the left and right distributive laws reepectively. Equating the two identities and applying the cancellation laws gives us the result.
Theorem 1.5: Let R be a ring such that for all . Prove that R is commutative.
Note: Such a ring is called a Boolean ring.
Proof: implies . Since and so by the cancellation law. Now as so and so each element in R is its own additive inverse. Hence and so .
Theorem 1.6: If R is a ring with unity satisfying for all , prove that R is commutative.
Proof: By our hypothesis and also by the distributive laws . So equating the two and applying the cancellation laws we have which holds as an identity. Now substituting x+1 for x in the identity we have . This gives and on the application of the distributive laws we have . Cancellation law now gives as required.
Theorem 1.7: Let R be a ring such that for , there exists a unique such that xa=x. Show that ax=x. Hence deduce that if R has a unique right identity e, then e is the unity of R.
Proof: xa=x implies x(a+ax-x)=xa+xax-x2=x. Hence a+ax-x=a or ax=x. If R has a unique right identity e then xe=x implies ex=x and so e is the unity of R.
Theorem 1.8: Let R be a ring with unity . Suppose for a unique such that xyx=x. Prove that xy=yx=1, i.e. x is invertible in R.
Proof: Suppose, if possible xa=0 for some . Now, x(y+a)x=(xy+xa)x=xyx+xax=xyx=x and by the uniqueness of y it follows that y+a=a i.e. a=0. So . Now x(yx-1)=xyx-x=x-x=0 and so yx-1=0. Hence yx=1. Similarly xy=1. So x is invertible.
Theorem 1.9: Show that if 1-ab is invertible in a ring R with unity, then so is 1-ba.
Proof: Let x be the inverse of 1-ab, i.e. let x(1-ab)=(1-ab)x=1. Now (1-ba)(1+bxa)=1+bxa-ba-babxa=1-ba+b(1-ab)xa=1-ba+ba=1. Similarly (1+bxa)(1-ba)=1. So 1-ba is invertible with inverse 1+bxa.
Theorem 1.10: If a,b are any two elements of a ring R and m and n are any two positive integers, then prove that
Proof We shall prove 4. and leave the rest as an exercise for the reader.
4. by repeated application of the distributive law. The RHS is just (nm)(ab).