# Abstract Algebra: Unit question I am stumped!

## Homework Statement

From An Introduction to Abstract Algebra by T. Hungerford
Section 3.2 #29

Let R be a ring with identity and no zero divisors.
If ab is a unit in R prove that a and b are units.

## Homework Equations

c is a unit in R if and only if there exists an element x in R s.t. cx=xc=1
where 1 is the identity element of R.

c is a zero divisor in R if and only if 1)c is not equal to 0 and 2)there exists
and element d in R s.t. either cd=0 or dc=0.

## The Attempt at a Solution

Dick
Homework Helper
In a ring with unit and no zero divisors first prove that you have both left and right cancellation. Then since a*b is a unit you have a d such that (ab)d=d(ab)=1. So a(bd)=1. Now all you need to show is (bd)a=1. Time to multiply both sides by things and use your cancellation properties.

In a ring with unit and no zero divisors first prove that you have both left and right cancellation..

for the left cancellation case, if a is nonzero and ab = ac then ab - ac = 0
hence a(b-c) = 0, since there are no zero divisors, then b-c = 0 which
implies b = c, hence ab = ac -> b = c -> left cancellation

is this correct?

Then since a*b is a unit you have a d such that (ab)d=d(ab)=1. So a(bd)=1. Now all you need to show is (bd)a=1. Time to multiply both sides by things asnd use your cancellation properties.

I imagine that after I have proven that I have left and right cancellation
that it goes along these lines:

Starting off: (ab)d = 1

a(bd) = 1
a(bd)a = 1*a "right" multiplying both sides by a
a(bd)a = a*1 commute the identity.
(bd)a = 1 left cancellation.

which goes to show that a is a unit.

and similarly d(ab) = 1

(da)b = 1
b(da)b = b*1 "left" multiplication both sides by b
b(da)b = 1*b commute the identity is always okay
b(da) = 1 by right cancellation

which shows that b is a unit

is this the correct way?

Dick
Homework Helper
Very nice. But the real test is do you believe the proof?

Thank you Dick for your help. I could not develop on my own the relevance that having no zero divisors had. WHen I worked the question before I posted for help, I encountered a need to use cancellation, yet since I hadn't proven I could use cancellation, I felt I was going in the wrong direction. I didn't know that I should have kept going in that direction, and proven cancellation.

THe lesson I learned here, is the intuitive connection between a lack of zero divisors implies allowable use of cancellation.

Dick