# Abstract Algebra: Prove Unit question

1. Nov 7, 2007

### exprog3

1. The problem statement, all variables and given/known data
Let R be an Integral Domain. Prove that if a,b are elements of R and both a and b are units in R, then prove a*b is a unit of R.

2. Relevant equations
a is a unit in R if and only if there exists an element u in R such that au=1=ua
where 1 is the identity element of R.

We also know that since R is an integral domain, that R is a commutative ring with identity.

Since R is a ring, R is closed under multiplication therefore a*b would still be an element of R.

3. The attempt at a solution

Given that both a and b are elements of R, and both a and b are units. Then by the definition of a unit there exists s,t that are elements of R such that a*s=1=s*a and b*t=1=t*b.

Somehow I need to multiply a and b together. So,

Let a*b be a unit in R, then there exists w that is an element of R such that:
w(a*b)=1=(a*b)w.

I'm lost from this point forward... any help?

2. Nov 7, 2007

### morphism

You were fine until you said

The point is: you want to show that a*b is a unit. This means, you need to find an element u in R such that (a*b)*u = 1. Try using the fact that a*s=b*t=1.

3. Nov 7, 2007

### exprog3

Since a is a unit, then there exist an element u in R such that au=1. Since b is a unit, there exists an element y in R such that by=1.

Therefore

au=1
au(by)=1(by)
au(by)=1(1)
au(by)=1
ab(uy)=1
ab(z)=1

Let z be an element of R such that z=uy.

Since u,y are elements of R, then u*y is an element of R.

Therefore, by the definition of a unit, a*b is a unit because multiplied by some element in R equals 1.