# Show this is Integral Domain

• Dec 9th 2012, 10:05 PM
jzellt
Show this is Integral Domain
A is the set Z (integers), with the following "addition" + and "multiplication" * :

a + b = a + b - 1 and a * b = ab - (a + b) + 2

I've already proven that A is a commutative ring with unity. Now, I have to prove that A is an integral domain.

To my understanding, in order to do this, I assume that a * b = 0 and show that either a = 0 or b = 0. Please correct me if I'm wrong...

So, a * b = ab - (a + b) + 2 = ab - a - b + 2 = 0.

I can't figure out how to proceed and show that either a or b equals zero. Can someone help here? Thanks!
• Dec 16th 2012, 04:40 PM
johng
Re: Show this is Integral Domain
I had to alter your definition slightly, the changed problem has solution:
Attachment 26262
• Dec 16th 2012, 06:22 PM
Deveno
Re: Show this is Integral Domain
i don't think there is any need to alter the definition. if we use # for the "new" plus, we have:

a#b = a + b - 1.

to find an element z such that:

a#z = a, for any a in Z, we solve:

a + z - 1 = a
a + z = a + 1
z = 1, which shows that the "0" of our "modified" integers is 1, not -1.

and indeed:

a*1 = a - a - 1 + 2 = 1.

the trouble you are having, jzellt, is that to prove A is an integral domain, you need to show that if a*b = 0A = 1, then either a = 0A = 1, or b = 0A = 1.

so suppose a*b = 1. assume that a ≠ 1. we need to show that b = 1.

so we have ab -a - b + 2 = 1 (this is an equation in our usual integers)

ab - a - b = -1
ab - b = a - 1
b(a - 1) = a - 1

since a - 1 ≠ 0 (since a ≠ 1), we can use the cancellation property of the integers to cancel the common factor of a - 1 from both sides:

b = 1, QED.