# How to define zero divisors if we work on different binary operation systems?

Suppose I have a ring with a binary system in which addition is defined as a^b = a+b-1 and multiplication is defined as a*b = a + b - ab, with a, b are integers.

I got additive identity is 1 and multiplicative identity is 0. Then, how do we define zero divisors in the ring? I know that a and b are zero divisors if a.b is 0, where a and b are not zero. But in this binary operation we have 0 as multiplicative identity. In this case, units are elements, a and b, such that a*b = 0. Am I right?

I need help. Thanks.

• Zero divisors come from non-trivial ways of writing the additive neutral element as a product of two other things. Units come from writing the multiplicative neutral element as a product of two things. Sounds like you got it right. – Jyrki Lahtonen Oct 26 '17 at 5:43
• BTW I didn't check everything, but it looks like $n\mapsto 1-n$ is an isomorphism from $\Bbb{Z}$ to your ring. This should make it easy to locate the zero divisors as well as units. – Jyrki Lahtonen Oct 26 '17 at 5:45
• That makes sense. Thank you. – Nabil Oct 26 '17 at 5:46

## 1 Answer

$a$ is a zero-divisor in the ring if $\exists b\in R$ such that $a*b=1$ for $a,b\neq 1$

Now $a*b=1 \implies b(1-a)=1-a$

• That's what I am thinking too. Thanks. – Nabil Oct 26 '17 at 5:47