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Is there something similar for division as this Japanese multiplication method?

Let $p$ be a prime element. I need an example of a domain in which $p^n$ divides $ab$ and $p^n$ does not divide $a$ and $p$ does not divide $b$. Obviously, the domain I’m looking for is not a UFD.

Thanks

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In *any* integral domain, prime products behave much as they do in UFDs. For example, their factorizations into atoms are unique. Furthermore products of primes are *primal*, namely, the *prime divisor property* $\rm\ p\ |\ ab\ \Rightarrow\ p\ |\ a\ \ or\ \ p\ |\ b\ $ generalizes from atoms to composites as follows

**Theorem** $\, $ In domain $\cal D\,,\, $ prime product $\rm\, C\mid AB\, \Rightarrow\, C = ab,\ a\mid A,\ b\mid B,\, $ some $\rm\ a,b\in \cal D$

**Proof** $\ $ By induction on product length $\rm\,n.\,$ Trivial if $\rm\ n = 0\!:\,$ then $\rm\ C = 1\ $ so take $\rm\ a = 1 = b\,.\,$

Else $\rm\,n\ge 1\,$ so $\rm\, C = pc,\ p\,$ prime. $\rm\, C = pc\mid AB\, \Rightarrow\, p\mid A\ $ or $\rm\ p\mid B.\,$ W.l.o.g assume $\rm\ p\mid B\,$ so

$\quad \rm pc\mid AB\ \Rightarrow\ c\mid A\,(B/p)\quad $ since $\rm\ p\ne 0\ \Rightarrow\ p$ cancellable, by $\,\cal D\,$ domain

$\rm\qquad\qquad\quad\! \Rightarrow\ \ c\ =\ ab,\ \ a\mid A,\ \ b\mid B/p\quad $ by induction

$\rm\qquad\qquad\ \ \Rightarrow\ \ C = pc\ =\ abp,\ \ a\mid A,\ \ bp\mid B\quad $ **QED**

**Corollary** $\ $ In domain $\cal D\,,\ $ prime power $\rm\ p^n\mid AB,\ \ p\nmid B\ \, \Rightarrow\ p^n\mid A$

You can show by induction that this is impossible. For $n=1$, this is just the definition of $p$ being prime. Now assume that it is impossible for $n$ but possible for $n+1$. Let $p^{n+1}\mid ab$ but $p^{n+1}\nmid a$ and $p\nmid b$. Then $ab=p^{n+1}c=p(p^nc)$, and since $p$ is prime and $p\mid ab$ and $p\nmid b$, we must have $p\mid a$, so $ab=(pd)b=p(db)$. Cancelling $p$ in $p(p^nc)=p(db)$ yields $p^nc=db$ with $p^n\nmid d$ (since $p^{n+1}\nmid a$), but this is impossible by the induction hypothesis.

Contrary to joriki (who gives a perfect answer to his interpretation of the question), I interpreted $p$ in your question as an ordinary prime integer.

In that case, given indeterminates $A,B$ over $\mathbb Z$, the domain $D=\mathbb Z[a,b]=\mathbb Z[A,B]/(A\cdot B-p^n)$ has the properties you require.

[Integrity comes from irreducibility of $A\cdot B-p^n$ and the very construction ensures that $p^n$ divides $ab=p^n$.

That $p^n$ does not divide $a$ and that $p$ does not divide $b$ can be seen by a trick: divide out the domain $D$ by the ideal $(b)$ resp. $(a)$]

The following statements are straightforward and answer immediately (joriki’s interpretation of) your question:

$\bullet$ If an element of your domain is of the form $p^na$ with $p\nmid a$, then the pair $(n,a)$ is unique.

$\bullet$ If two elements are of this form, so is their product.

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