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!
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 = 0_{A} = 1, then either a = 0_{A} = 1, or b = 0_{A} = 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.