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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

Pierre-Yves Gaillard
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4 Answers4

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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.

joriki
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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 $p^n \nmid d$? –  Jan 15 '12 at 15:03
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    @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
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    @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 $n+1$. –  Jan 15 '12 at 15:39
  • @Lmn6: The existence of an example as described in your question. That is, let $A(n)$ be the assertion that there exist $p$ prime and $a$ and $b$ such that $p^n|ab$, $p^n\nmid a$ and $p\nmid b$. Then $A(1)$ is false by definition of a prime element; and my argument shows that $A(n+1)$ is false if $A(n)$ is false. – joriki Jan 15 '12 at 15:44
  • The initial statement is $(p^n \mid ab \wedge p^n \nmid b)\rightarrow p \nmid b$. If $n=1$ it is not true by the def of prime element. So your argument assumes that for, say $m$ it is not valid, i.e. the following is true: $(p^m \mid ab \wedge p^m \nmid b)\wedge p \mid b$ –  Jan 15 '12 at 15:59
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    @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
  • Sorry, just to be clear, in my original question (then edited) I was meaning what I wrote. –  Jan 15 '12 at 18:18
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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 $\rm\ p^n\mid AB,\ \ p\nmid B\ \, \Rightarrow\ p^n\mid A$

Bill Dubuque
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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. $(a)$]

Georges Elencwajg
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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
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    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
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    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
  • @Georges: Ah, I see. Yes, I'd noticed that $p$ was introduced before the domain but just put it down to inexperience in presentation. It seems that people sometimes unorthodoxly use constructions like "let $p$ be a prime element" to express something that would more orthodoxly be expressed as "In the following, the variable name $p$ will stand for a prime element of a domain", i.e. to explain what sort of thing a given letter will be standing for and not to introduce a certain value for a certain variable. – joriki Jan 15 '12 at 15:54
  • Dear @joriki: Any ring $R$ is a $\mathbb Z$-module. Thus, if $n$ is an integer, then $n\cdot1$ is an element of $R$, usually (and abusively) denoted $n$. In Georges's case, we have moreover: $n=0$ in $D$ implies $n=0$ in $\mathbb Z$. – Pierre-Yves Gaillard Jan 15 '12 at 15:56
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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.

Pierre-Yves Gaillard
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