Krdytkyu

$R$ - commutative Ring, and $a$ non-zero-divisor

$R^x$ - multiplicative Monoid? (it's called Einheitsgruppe in German, maybe unitary group?)

I started by stating, that if $amid b$ and $bmid a rightarrow a=b$.

So for $amid b$, $exists xin R$ with $xa=b$ and for $bmid a,exists yin R$ with $yb=a$

Now: $b=xa=xyb=wb$, with $w=xy$.

$stackrel{a=b}{rightarrow} b=wa$

Now the problem that I think I have with my proof, is that R is not a given Ring. For example $a=b$ wouldn't work in $mathbb{Z}$. Is there any other way to do the proof?

I assume you are working in a commutative ring with identity.

The problem is to show that if two elements of a ring divide each other, then they are associates, i.e. the same up to a unit multiple.

Actually this is not true in general. For example, in the ring
$mathbb{Z}[x,y,z]/(xy -z, zy - x)$ we have that $x mid z$ and $z mid x$, but $y$ is not a unit. Note that $x$ and $z$ are both zero divisors, as $z(y^2 - 1) = x(y^2-1) = 0$.

In general,
$a mid b$ means there exists $u in R$ with $au = b$.
$b mid a$ means there exists $v in R$ with $bv = a$.

and putting these equations together...

$$buv = b$$
$$auv = a$$

It follows that if either $a$ or $b$ is not a zero divisor, then $uv = 1$ and $u,v$ are units, therefore $a$ and $b$ are associates.