R is commutative ring with identity & define $\circ$ on $R$ by for any $a,b \in R$ $a \circ b=a+b-ab$ Prove the following Let R be a commutative ring with identity. Define a new operation $\circ$ on $R$ by for any $a,b \in R$
$$a \circ b=a+b-ab$$
a) Prove that $\circ$ is associative
b) Prove that R is a field iff the set $\{r \in R: r \ne 1\}$ is an abelian group with respect to the operation $\circ$
Starting with a) we must prove 
$a \circ ( b \circ c)= (a \circ b) \circ c$
Left side: $a \circ ( b \circ c)=a \circ (b+c-bc)=a+b+c-bc-a(b+c-bc)=a+b+c-bc-ab-ac+abc$
Right side: $(a \circ b) \circ c =(a+b-ab) a \circ c=a+b-ab+c-(a+b-ab)c=a+b-ab+c-ac-bc+abc$ and we have thus proved that $\circ$ is associative
As for part b), I am a little confused on what is being asked exactly. Is it asking to go through ALL the properties of a field, closure, associativity, commutativity, identity, inverse, distributive, showing that all hold except for when $r \ne 1$?
 A: Hint $\ $ Any set bijection $\,h\,:\,R'\to R\,$ serves to transport the multiplicative structure of $\,(R,*,1),$ to $\,(R',\circ,1') \,$ by defining the operations in $\,R'\,$ to make $\,h\,$ be an isomorphism
$$\begin{align}  &a \circ b\, =\, h^{-1}(h(a)\, * h(b)),\quad\, 1' = h^{-1}(1)\end{align}\qquad \qquad$$
Yours is the special case $\ h(x) = 1-x$
A: The proof for part (a) is good.
For part (b) you just have to show that any nonzero element in $R$ has a multiplicative inverse and $0\ne1$ if and only if $(R\setminus\{1\},\circ)$ is an abelian group.
The other field properties are already guaranteed by the fact that $R$ is a commutative ring.
If $(R\setminus\{1\},\circ)$ is a group, then it is nonempty, so $0\ne1$ (note that $0$ is the neutral element for the circle operation). Let $a\in R$, $a\ne0$. Then $1-a\ne1$, so $1-a$ has an inverse with respect to the circle operation: $(1-a)\circ b=0$, that is
$$
(1-a)+b-(1-a)b=0
$$
or
$$
1-a+ab=0
$$
and so $a(1-b)=1$. Thus $a^{-1}=1-b$.
Suppose $R$ is a field and consider $a\in R\setminus\{1\}$. You want to find an inverse of $a$ with respect to the circle operation, that is $b$ such that $a+b-ab=0$. This becomes $b(a-1)=a$ or $b=a(a-1)^{-1}$ (note that $a-1\ne0$, so $(a-1)^{-1}$ exists). Also $a(a-1)^{-1}\ne1$ because otherwise $a=a-1$ or $1=0$ which is false in a field.
