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.

Basic properties

Theorem 1.1: If R is a ring and $a,b,c\in R$; then

1. a+b=a+c implies b=c. (Cancellation Law)

2. -(-a)=a.

3. The zero element of R is unique.

4. The additive inverse of any element is unique.

Proof:

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 definiton and so 0=0'.

4. If a' and a'' are two inverses of a then a'=a'+0=a'+a+a''=0+a''=a''.$\Box$

Theorem 1.2: If R is a ring, then for any $a,b,c\in R$;

1. a0=0a=0.

2. a(-b)=(-a)b=-(ab).

3. (-a)(-b)=ab.

4. a(b-c)=ab-ac.

If in addition, R has a unit element 1, then

5. (-1)a=-a.

6. (-1)(-1)=1.

Proof:

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).$\Box$

Some more results

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 (a + b)² = a² + 2ab + b² holds for all $a,b\in R$.

Proof: Suppose R is commutative. Then clearly the result holds. (In fact the binomial theorem: $(a+b)^n=\textstyle\sum_{k=0}^n\binom{n}{k}a^{n-k}b^k$ holds in that case. Try to prove it using induction and the Pascal's identity:$\textstyle\binom{n}{k}+\textstyle\binom{n}{k+1}=\textstyle\binom{n+1}{k+1}$.) Conversely suppose that for each $a,b\in R$ the given relation is satisfied. Now, on applying the distributive laws to (a + b)² = (a + b)(a + b) we get = a² + ab + ba + b² = (a + b)² = a² + ab + ab + b² and by the cancellation laws we have ab=ba. Hence R is commutative.$\Box$

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.$\Box$

Theorem 1.5: Let R be a ring such that a² = a for all $a\in R$. Prove that R is commutative.

Note: Such a ring is called a Boolean ring.

Proof: (a + b)² = a + b implies a² + ab + ba + b² = a + b. Since a² = a and b² = b so ab = − ba by the cancellation law. Now as a + a = (a + a)² = a² + 2a² + a² = a + a + a + a so $a+a=0 \ \forall a\in R$ and so each element in R is its own additive inverse. Hence − ba = ba and so ab = ba.$\Box$

Theorem 1.6: If R is a ring with unity satisfying (xy)² = x²y² for all $x,y\in R$, prove that R is commutative.

Proof: By our hypothesis [x(y + 1)]² = x²(y + 1)² = x²(y² + 2y + 1) = x²y² + 2x²y + x² and also by the distributive laws [x(y + 1)]² = [xy + x]² = (xy + x)(xy + x) = x²y² + xyx + x²y + x². So equating the two and applying the cancellation laws we have xyx = x²y which holds as an identity. Now substituting x+1 for x in the identity we have (x + 1)y(x + 1) = (x + 1)²y. This gives (x + 1)(yx + y) = (x + 1)(xy + y) and on the application of the distributive laws we have xyx + xy + yx + y = x²y + xy + xy + y. Cancellation law now gives xy = yx as required.$\Box$

Theorem 1.7: Let R be a ring such that for $x \in R$, there exists a unique $a\in R$ 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.$\Box$

Theorem 1.8: Let R be a ring with unity $1\in R$. Suppose for $x\ne 0\in R \exists$ a unique $y\in R$ such that xyx=x. Prove that xy=yx=1, i.e. x is invertible in R.

Proof: Suppose, if possible xa=0 for some $a\in R$. 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 $xa=0\Rightarrow a=0$. Now x(yx-1)=xyx-x=x-x=0 and so yx-1=0. Hence yx=1. Similarly xy=1. So x is invertible.$\Box$

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.$\Box$

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

1. (m+n)a=ma+na.

2. m(a+b)=ma+mb.

3. m(na)=(mn)a.

4. (na)(mb)=(nm)(ab).

5. a^maⁿ=a^m+n.

6. (a^m)ⁿ=a^mn.

Proof We shall prove 4. and leave the rest as an exercise for the reader.

4. $(na)(mb)=(\underbrace{a+\cdots a}_{n\ times})(\underbrace{b+\cdots b}_{m\ times})=\underbrace{ab+\cdots ab}_{nm\ times}$ by repeated application of the distributive law. The RHS is just (nm)(ab).$\Box$