A commutative ring R is an integral domain $\iff a,b,c \in R, a \neq0, > ab=ac \implies b=c$.
I am relatively new to rings and I am struggling with shaking off the habit of taking cancellation laws of groups. So, I would like some validation for the following proof which came out to be deceptively straightforward:
($\implies$)
Let $R$ be a commutative ring which is also an integral domain. Let $a,b,c \in R$ be s.t. $a \neq 0, ab=ac$. Then $ab=ac \implies ab-ac=ac+(-ac)\implies ab-ac=0 \implies a(b-c)=0$
By hypothesis $a\neq 0$ so $b-c=0$ since $R$ is an integral domain. Hence, $b=c$.
($\impliedby$)
Let $R$ be a commutative ring s.t. the property $a,b,c \in R$ for $a \neq 0$ be s.t. $ab=ac \implies b=c$ holds.
Let $x,y \in R$ s.t. $xy=0$. $\exists p,q \in R$ s.t. $y=p+(-q)$.
Then $$ \begin{equation} xy = 0 \\ x(p+(-q)) =0 \\ x(p-q)=0 \\ xp-xq =0 \\ xp-xq +xq = 0+xq \\ xp=xq \implies p=q \implies (p-q) =0 \implies y=0 \end{equation} $$ So $R$ has no zero divisors and hence, is an integral domain. $\Box$
For the side $\Leftarrow$ I think that you prove is a little bit too much complicated :
Here is my proof :
$xy=0 \Rightarrow xy+x=x=x(y+1) $ if we assume that x is not zero then :
We can by assumption "eliminate x"
$1=y+1$
Then y is equal to zero