For the following ring isomorophism question:
Let $\mathbb{Z}^{*}$ be n integral domain defined by the following operations:
$a\oplus b=a+b-1$
$a\odot b=a+b-ab$,
where $a,b\in \mathbb{Z}$
Show that $\mathbb{Z}$ and $\mathbb{Z}^{*}$ are isomorphic.
I need to construct a homorophic bijective function $\phi(x)$ which maps $\mathbb{Z}$ to $\mathbb{Z}^{*}$
$\phi(x)$ is a one variable function but both $a\odot b$ and $a\oplus b$ are binary operation consisting of two element. I am assuming that when either $a\odot b$ or $a\oplus b$ and gets plugged into $\phi(x)$, it is treated as a single variable. But because of the following property of homomorphism: $\phi(ab) = \phi(a)\phi(b)$, I can write $\phi(a \odot b)$ in terms of $\phi(a)$ and $\phi(b)$
Also, I know of the following facts after doing some calculations.
$a\oplus 1=a$
$a\oplus 1-a =0$
$a\odot 1=1$
$a\odot 0=a$
$a\odot \frac{-a}{1-a}=0$
The function $\phi(x)=1-x$ is the supposed function. The supposed function seems to be found by making use of the additive zero element/multiplicative identity $1$ and let $a\odot b=1$ so that $1 - a\odot b=(1-a)(1-b)$. We then obtain $\phi(a\odot b) = \phi(a)\odot \phi(b)=(1-a)(1-b)$. What I want to ask is, is this a general strategy I can use for showing isomorphism between other rings, where one of the rings are defined by a different set of defined addition and multiplication operations. Thank you in advance
