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.

  • $\begingroup$ 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. $\endgroup$ – Jyrki Lahtonen Oct 26 '17 at 5:43
  • $\begingroup$ 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. $\endgroup$ – Jyrki Lahtonen Oct 26 '17 at 5:45
  • $\begingroup$ That makes sense. Thank you. $\endgroup$ – Nabil Oct 26 '17 at 5:46

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

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

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

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