Given A=$\Bbb Z$, a commutative ring with element $1$ with $\oplus$ and $\odot$, defined as: $a\oplus b=a+b-1$ and $a\odot b=a+b-ab$, show the ring is an integral domain.
It is known that the identity element in the ring is $1$, so we need to prove $a \odot b=1 \Rightarrow a=1$ or $b=1$; or $a\ne 1$ and $b\ne 1 \Rightarrow a\odot b\ne 1$.
The main problem I find is that I need to prove different cases with each sign of a and b. Is there an easier way?
Suppose $a\odot b=1,$ or more concretely $a+b -ab =1.$
Then we may also write $a-ab= 1-b,$ and $b-ab=1-a,$ factoring gives $$a(1-b)=1-b,$$ or $$b(1-a)=(1-a)$$ must be true. But in $\Bbb{Z}$ this is to say that one of $a$ or $b$ is $1.$
Since $1$ is the additive identity in the ring in question, then we have the desired claim.