If $a\mid b$ and $b\mid a$, $a,b\in R$, show that $\exists w \in R^\times$ with $b=wa$

Question

$R$ - 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 $a\mid b$ and $b\mid a \rightarrow a=b$.

So for $a\mid b$, $\exists x\in R$ with $xa=b$ and for $b\mid a,\exists y\in R$ 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?

It's not necessarily true that $a | b$ and $b | a$ implies $a = b$ (even in $\mathbb{Z}$, we have $1 | -1$ and $-1 | 1$). What does the notation $R^x$ denote (in particular, what is $x$)? — platty, Nov 29 '18 at 22:28
Presumably $R^x$ denotes the group of units of $R$ under multiplication — Badam Baplan, Nov 29 '18 at 22:30
Indeed, for any $a, b \in \mathbb Z,$ if $a = -b$, we still have $a \mid b$ and $b \mid a$. — amWhy, Nov 29 '18 at 22:31

Badam Baplan · Accepted Answer · 2018-11-29T23:04:58.913

1

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.

edited Nov 29 '18 at 23:04

answered Nov 29 '18 at 23:01

Badam Baplan

8,688

For the counterexample, it's not sufficient just to show $y$ is not a unit; what's needed is that there is no unit $y'$ with $xy' = z$. (And since, as you mentioned, $x$ is a zero divisor, then for example for any $a$, we have $x[y + a(y^2-1)] = z$. So you would need to make an argument which implies at least that anything of the form $y + a(y^2 - 1)$ is not a unit.) – Daniel Schepler Jan 14 '21 at 21:25
For more counterexamples showing the assumption that $a$ is not a zero divisor is necessary: https://math.stackexchange.com/questions/14270/are-associates-unit-multiples-in-a-commutative-ring-with-1 . I find the example there involving $C[0,3]$ to be the easiest one to check. – Daniel Schepler Jan 14 '21 at 21:29

If $a\mid b$ and $b\mid a$, $a,b\in R$, show that $\exists w \in R^\times$ with $b=wa$

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