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Given a ring $R$ and a ring extension $R'$, if $r=r's$ where $r'\in R'\setminus R$ and $r,s\in R\setminus\{0\}$, does that mean that $s\not\mid r$ in $R$?

I was thinking for example in $\Bbb{Z}$, since $\frac{2}{3} \not\in \Bbb{Z}$ then $3\not\mid 2$ in $\Bbb{Z}$. Or equivalently $2=\frac23 3$, since $\frac23 \in \Bbb{Q}\setminus \Bbb{Z}$ we know $3\not\mid 2$.

So in general I conjecture if "$\frac{r}{s}$" exists only in some larger ring, then $s\not\mid r$. Is this true? Or is there some extreme example where $r=r's$ for some $r'$ in a larger ring and also $r=ts$ for $t$ in the original ring?

Note there are trivial cases where $r$ and $s$ are $0$.

I think this would be a good characterization of divisibility.

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If $s$ is not a right zero divisor in $R'$, a counterexample is obviously impossible, since then $as=bs$ implies $a=b$ so $r'$ is the only element of $R'$ such that $r's=r$. Without this assumption, there are counterexamples. For instance, take $R=\mathbb{Z}$ and $R'=\mathbb{Z}[x]/(2x)$ and let $r=6$, $s=2$, and $r'=3+x$.

Eric Wofsey
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    +1 Nice answer, beat me to it! Pathologies like these are one reason (AFAIK) that notions of divisibility/division are often introduced in the setting of domains, where these kinds of issues don't arise... – Alex Wertheim Jun 22 '16 at 04:48
  • So if you're in a domain, this is true. That's a pretty useful criteria for divisibility. –  Jun 22 '16 at 04:54
  • @ZacharySelk This is a characteristic property of domains - see my answer. – Bill Dubuque Jul 01 '16 at 15:04
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In the commutative case this is a special (linear) case of the well-known result that a ring is a domain iff nonzero polynomials over it have no more roots than their degree. Namely

Lemma $\ R'$ is a domain iff every polynomial $\,s x\! -\! r,\ r,s\neq 0\,$ has at most one root $\,x\in R$

Proof $\ (\Rightarrow)\,\ s x_1\! = r = s x_2 \Rightarrow\, s(x_2\!-x_1) = 0\,$ $\Rightarrow\, x_2 - x_1 = 0\,$ by $\,s\neq 0,\, R'$ domain.

$(\Leftarrow)\ R\,$ not a domain $\Rightarrow$ there are $\,0\neq s,t\in R\,$ with $\,st = 0\,$ so $\,sx\!-\!s\,$ has roots $1\neq 1\!+\!t$

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Bill Dubuque
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