Prove whether R is integral domain.

Question

I'm having trouble with figuring out whether a given ring is an integral domain or not. This comes from my confusion about the zero element.

This ring R is a commutative triple $(Z,*,o)$ with identity (which, I suppose, is the additive identity, to which I refer as $1_{(Z,*)}$. Is this correct?). The operations are defined as $a*b=a+b-1$ and $aob=a+b-ab$, for all $a,b \in Z$.

I've figured out the zero element $a*0_{(G,*)} = a$ and the multiplicative identity (unity) $a*1_{(G,o)} = -a$.

So, onto the integral domain: a given ring is an integral domain if it has no zero-divisors such that for any $a,b \in R, ab=0$.

If so, I've tried $aob=a*0_{(G,*)}$. I get to $b-ab=0$, where I should use the cancelation property which is only available to integral domains...

EDIT: I just checked Pierre Grillet's Algebra which is very clear on this. "The zero element of a ring is the identity element 0 for its addition ($0+x=x=x+0$ for all $x \in$ R).

EDIT2: There are some mistakes in my notation and calculations.

Thanks!

Answer 1

Here when we say "addition" we refer to $*$ and when we say "multiplication" we refer to $\circ$.

In this example, confusingly, the additive identity is $1 \in \mathbb{Z}$ because $a * 1 = a + 1 - 1 = a$ for all $a \in \mathbb{Z}$, and the multiplicative identity is $0 \in \mathbb{Z}$ because $a \circ 0 = a + 0 - a\cdot 0 = a$ for all $a \in \mathbb{Z}$.

Now, suppose $a \circ b = 1$. (Remember that $1$ is our additive identity!) This means that $1 = a \circ b = a +b - ab$, which we can rewrite as $(a-1)(b-1) = 0$. But $\mathbb{Z}$ is an integral domain, so we must have $a = 1$ or $b = 1$. Thus your ring is also an integral domain.