Divisibility question for non UFD rings

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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Which non-UFD domains are you familiar with? Knowing this would help answering the question. – Marc van Leeuwen Jan 15 '12 at 13:50

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 $\rm\: C\ |\ A\:B\ \Rightarrow\ C\: =\: a\:b,\ \: a\ |\ A,\ \: b\ |\ B\:,\:$ 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\ |\ AB\ \Rightarrow\ p\ |\ A\$ or $\rm\ p\ |\ B\:.\:$ W.l.o.g assume $\rm\ p\ |\ B\:.\:$ Then

$\rm\qquad\qquad\quad\ pc\ |\ AB\ \Rightarrow\ c\ |\ A\:(B/p)\qquad$ since $\rm\ p\ne 0\ \Rightarrow\ p$ cancellable, since $\:\cal D\:$ is a domain

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

$\rm\qquad\qquad\qquad\ \ \Rightarrow\ \ C = pc\ =\ abp,\ \ a\ |\ A,\ \ bp\ |\ B\qquad$ QED

COROLLARY $\$ In domain $\cal D\:,\$ prime product $\rm\ p^n\ |\ A\:B,\ \ p\nmid B\ \ \Rightarrow\ p^n\ |\ A\:.$

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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.

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when you assume that this is impossible for $n$ but possible for $n+1$ you are assuming that $p|b$ and $p \nmid b$... Again, how do you conclude that from $p^nc=db$ then necessarily $p^n \nmid d$? – user14174 Jan 15 '12 at 15:03
@Lmn6: I don't understand the first part of that comment. The sentence "Now assume that it is impossible for $n$ but possible for $n+1$" doesn't contain $b$. The next sentence does; it states that $a$, $b$ furnish an example of the possibility for $n+1$. What follows then deduces from this that $d$, $b$ furnish an example of the possibility for $n$, contrary to the induction hypothesis. I don't see where I ever assumed $p|b$. – joriki Jan 15 '12 at 15:15
@Lmn6: I also don't understand what you mean by "again". Regarding the rest of the second part of your comment, I don't conclude $p^n\nmid d$ from $p^nc=db$; as I tried to indicate by the parenthesis "(since $p^{n+1}\nmid a$)", I conclude this from the fact that $a=pd$ and $p^{n+1}\nmid a$: if we had $p^n|d$, then $d=p^nr$, and then $a=p(p^nr)=(pp^n)r=p^{n+1}r$ and thus $p^{n+1}|a$. – joriki Jan 15 '12 at 15:17
I don't understand what exactly you assume that is impossible for $n$ and possible for $n+1$. – user14174 Jan 15 '12 at 15:39
@Lmn6: No, that statement (what you call "the initial statement") doesn't occur anywhere, neither in your question nor in my answer. What I called $A(n)$ in my comment above is $\exists p,a,b:p^n\mid ab\land p^n\nmid a\land p\nmid b$. The negation of this (in case that's what you meant) is $\forall p,a,b:\neg(p^n\mid ab\land p^n\nmid a\land p\nmid b)$, which can be rewritten as $\forall p,a,b:(p^n\mid ab\land p^n\nmid a)\to p\mid b$; note the $\mid$ where you had $\nmid$, and vice versa for your second statement. – joriki Jan 15 '12 at 16:11

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. $(a)$]

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Dear Georges: You have $ab=p^n$ (not $ab=0$), don't you? (+1) – Pierre-Yves Gaillard Jan 15 '12 at 15:29
I don't understand how this is a possible interpretation of the question, even if you read "prime integer" for "prime element": $p$ considered as an element of $D$ is neither a prime element (since $p\mid ab$ but $p\nmid a$ and $p\nmid b$), nor a prime integer, since $D$ consists not of integers but of equivalence classes of polynomials with integer coefficients. In what sense can $p\in D$ be regarded as a prime element? – joriki Jan 15 '12 at 15:30
Dear @Pierre-Yves: as usual you are right. I have just corrected my stupid mistake and am very grateful to you for having spotted it: thanks a lot and Bonne Année! – Georges Elencwajg Jan 15 '12 at 15:33
Dear Georges: Thanks, and Bonne Année to you too! – Pierre-Yves Gaillard Jan 15 '12 at 15:48
Dear @joriki, my interpretation is due to the fact that $p$ was introduced in the question before any mention in the first sentence of a domain. So I didn't think of $p$ as a prime of $D$. But don't worry: I am quite ready to admit that your interpretation is the better one! – Georges Elencwajg Jan 15 '12 at 15:49

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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