Show in the definition of a ring, that the axiom that addition be commutative may be omitted if 

a) there is a multiplicative identity ("unit element")

b) not all nonzero elements are zero divisors.

a) Let R have a multiplicative identity and satisfy the ring axioms with a+b=b+a omitted. 

 Then

a+b + a+b

= (a+b)1 + (a+b)1 

= (a+b)(1+1)  ( by left dist.)

= a(1+1) + b(1+1) (by right dist.)

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

= a+a+b+b

Hence a+b+a+b = a+a+b+b.

--> b+a = a+b  (by cancellation prop of +)

b) Suppose that R satisfies the ring axioms with the possible exception of a+b = b+a. Further suppose

R= {0} or ∃c≠0 ∈ R not a zero divisor.  *  (i.e. call this property *).

We'll show a+b= b+a holds ∀ a,b ∈ R. (Thus R is a ring.)

Case: R = {0}. Then a+b = 0+0 = b+a.

Case: R ≠ {0}. Then, ∃c≠0 ∈ R with cx≠0, ∀x≠0 ∈ R.

Further...  cx = cy → c(x-y) = 0 --> x-y = 0 --> x=y.

i.e. c, as a factor, can be canceled.

Now ∀a,b ∈ R...

cb + ca +cb + ca

=  c(b+a) + c(b+a)       (by l. dist.)

=  (c + c)(b+a)             (by r. dist.)

=  (c+c)b + (c+c)a        (by l. dist.)

=  cb +cb +ca +ca        (by r. dist.)

→ ca + cb = cb + ca   (by cancel. prop of +).

→  c(a+b) = c(b+a)    (by left dist.)

→ a+b = b+a  (by cancelation prop. of c).

Corollary: Let 1∈R and R satisfy the ring axioms,

with the possible exception of a+b = b+a.

Then  a+b= b+a holds in R. 

Case:  R = {0}. Then  *  holds.  (Note: here 1 = 0.)

Case R ≠ {0}.  Then 1≠0 and 1 is never a zero divisor (because 1x = 0 --> x = 0), so again  * holds.

Thus a+b = b+a , ∀ a,b ∈ R.

Note: We can't replace a+b = b+a by *.

For example non trivial rings with trivial multiplication are  commutative for + but don't satisfy *.

So axioms with * --> Ring   but Ring -/-> axioms with *.

Suppose R is a system satisfying all the conditions for 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 R is thus a ring. 

PROOF

Let R be as stated above and let a and b be arbitrary members of R. 

a + a + b + b  

=  a(1) + a(1) + b(1) + b(1)   (as R has unit element 1)

=  a(1 + 1) + b(1 + 1)            (by left distributivity)

=  (a + b) (1 + 1)                   (by right distributivity)

=  (a + b)1  + (a + b)1            (by left distributivity)

=  a + b + a + b                     (as 1 is the unit element).

We've established that a + a + b + b = a + b + a + b.

Now, since (R, +) is a group, we may cancel the leading a from each side and the trailing b from each side (Lemma 2.3.2) to obtain

a + b = b +a.

Thus is it established that commutativity holds in (R, +). Since R satisfies all the other axioms of a ring,  R is a ring.

a+b + a+b

= (a+b)1 + (a+b)1       ( since R has unit element 1)

= (a+b)(1+1)               ( by left dist.)

= a(1+1) + b(1+1)       ( by right dist.)

= a(1) +a(1) + b(1) +b(1)   (by left dist.)

= a+a+b+b                       

From a + b + a + b = a + a + b + b we obtain b + 

b) Suppose that R satisfies the ring axioms with the possible exception of a+b = b+a. Further suppose

R= {0} or ∃c≠0 ∈ R not a zero divisor.  *  (i.e. call this property *).

We'll show a+b= b+a holds ∀ a,b ∈ R. (Thus R is a ring.)

Case: R = {0}. Then a+b = 0+0 = b+a.

Case: R ≠ {0}. Then, ∃c≠0 ∈ R with cx≠0, ∀x≠0 ∈ R.

Further...  cx = cy → c(x-y) = 0 --> x-y = 0 --> x=y.

i.e. c, as a factor, can be canceled.

Now ∀a,b ∈ R...

cb + ca +cb + ca

=  c(b+a) + c(b+a)       (by l. dist.)

=  (c + c)(b+a)             (by r. dist.)

=  (c+c)b + (c+c)a        (by l. dist.)

=  cb +cb +ca +ca        (by r. dist.)

→ ca + cb = cb + ca   (by cancel. prop of +).

→  c(a+b) = c(b+a)    (by left dist.)

→ a+b = b+a  (by cancelation prop. of c).

Corollary: Let 1∈R and R satisfy the ring axioms,

with the possible exception of a+b = b+a.

Then  a+b= b+a holds in R. 

Case:  R = {0}. Then  *  holds.  (Note: here 1 = 0.)

Case R ≠ {0}.  Then 1≠0 and 1 is never a zero divisor (because 1x = 0 --> x = 0), so again  * holds.

Thus a+b = b+a , ∀ a,b ∈ R.

Note: We can't replace a+b = b+a by *.

For example non trivial rings with trivial multiplication are  commutative for + but don't satisfy *.

So axioms with * --> Ring   but Ring -/-> axioms with *.

Herstein Problem 12 Page 130

Prove that any field is an integral domain.

Let F be a field.

Then F is commutative ring

and F is nontrivial (since 1 != 0).

To establish that F is an integral domain,

it remains to show

F contains no zero divisors. 

To get a contradiction, suppose that

F contains a zero divisor b. Then

by definition, b is nonzero and

there is a nonzero c in F 

such that bc = 0.

Since F is a field and b is nonzero,

b has a multiplicative inverse b^-1.

Moreover 

0 = b^-1(0)

= b^-1(bc)

= (b^-1b)c

= (1)c

= c.

This contradicts the assumption that c is nonzero.

Therefore F contains no zero divisors.

Hence F is an integral domain.

Herstein Problem 10 Page 130

Show that a commutative ring D 

is an integral domain 

if and only if

for a, b, c, in D with a≠0,

the relation ab = ac implies b = c.

PROOF

First suppose that D is an integral domain.

Let a, b, c be members of D, with a ≠0

and suppose ab = ac.  

Then  0 = ab - ac = a(b - c).

Since D is an integral domain and

and a≠0

we must have b - c = 0, i.e. b = c.

Thus D has the cancellation property.

Now suppose that D is a commutative ring

such that

for a, b, c in D with a≠0, 

the relation ab = ac implies b = c.

Also, I'll assume that D is nontrivial so 

there exists at least one nonzero element.

To establish that D is an integral domain,

consider 0 ≠ a ∈ D and b ∈ D 

such that ab = 0.

Then ab = 0 = a(0).

and by cancellation, b = 0.

Thus D is a commutative ring with no zero divisors.

Thus D is an integral domain.