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 ${\displaystyle a,b,c\in R}$; 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 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''.${\displaystyle \Box }$

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

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).${\displaystyle \Box }$

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 ${\displaystyle (a+b)^{2}=a^{2}+2ab+b^{2}}$ holds for all ${\displaystyle a,b\in R}$.

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

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.${\displaystyle \Box }$

Theorem 1.5: Let R be a ring such that ${\displaystyle a^{2}=a}$ for all ${\displaystyle a\in R}$. Prove that R is commutative.

Note: Such a ring is called a Boolean ring.

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

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

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

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

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

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

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.${\displaystyle \Box }$

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