Prove a ring is an integral domain given two operations explicitely

Question

We define the following operations $\oplus, \otimes: \mathbb{Z} \times \mathbb{Z} \rightarrow \mathbb{Z}: $

$$a\oplus b:= a+b-1, \,\,\,a\otimes b:=a+b-a \cdot b.$$

I need to show that $(\mathbb{Z}, \oplus, \otimes)$ is an integral domain.

We set: $a\otimes b:=a+b-a \cdot b = 0,\, 0$ is the additive neutral element in $(\mathbb{Z}, +,\cdot).$ We need to check if $a$ or $b$ is $0$. Let's assume, $a \neq0$ and $ b\neq 0.$ We get: $a + b = a\cdot b.$ Division by $a$ leads to $1 + \frac{b}{a} = b.$ Since $a, b \in \mathbb{Z},\,\frac{b}{a} \in \mathbb{Z}.$ Thus,$\,\,\exists \, k \in \mathbb{Z}, \,b = a\cdot k.$ Given that the problem is symmetrical in respect to $b$, we devide $a + b = a\cdot b$ by $b$ and get $a = b \cdot l, $ for some $l \in \mathbb{Z}.$ We combine both equalities to get: $b=b\cdot l \cdot k.$ This is equivalent to $1 = l \cdot k.$ This is only possible if $l = k = \{1,-1\}.$ In terms of $a$ and $b$, it means $a = b.$ Thus, we still can not prove a contradiction. We also get $ a + b - a\cdot b= 0 \implies 2b - b^2 =0 \implies b = 2 \implies a = 2.$ Since the ring structure implies the group structure related to $\oplus, \,\,a\oplus b:= a+b-1 = 0 \implies a$ is the additive inverse of $b$ iff $a+b =1.$ In particular, this must hold for $a = 2$ and $b =2.$ For $a=2, \,\,2\oplus b= 0 \implies 2+b=1 \implies b = -1.$

I can not come to a contradiction. Can somebody provide what is missing or propose another strategy altogether. At the other hand, should one also prove that $(\mathbb{Z}, \oplus, \otimes)$ is a ring ? Thanks.

Answer 1

Hint: The zero element is $1$, since $a\oplus 1 = a+1-1=a$.

Then you need to check $a\otimes b = a+b-ab = 1$ in the ring of integers.

Thus $0 = ab-a - b+1 = (a-1)(b-1)$ in the ring of integers. But this ring has no zero divisors and so $a-1=0$ or $b-1=0$, that is, $a=1$ or $b=1$. Hence, the ring $(\Bbb Z,\oplus,\otimes)$ has no zero divisors.