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I would like to prove the following implication: Given any prime $p$, we have $$ a \mid p b \land b \mid p a \implies a \mid b \lor b \mid a $$ I am on a path to proving the prime decomposition theorem in PA, and this popped up a lemma I need in order to finish it. Using prime decompositions it is fairly easy to see that it should hold, but I'm kind of stumped when it comes to a proof idea that does not use it.

Any hints or ideas are welcome!

Bill Dubuque
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Léreau
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3 Answers3

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For some $c,d\in\mathbb Z$ we have $$a\ |\ pb\ \Rightarrow\ pb=ac \tag{1}$$ $$b\ |\ pa\ \Rightarrow\ pa=bd$$ If $b=0$ then $a\ |\ b$, so assume $b\neq0$ and it follows $$p^2b=pac=bcd\ \Rightarrow\ p^2=cd\ \Rightarrow\ c\in\{\pm1,\pm p,\pm p^2\}$$

  • $c=\pm 1\ \Rightarrow\ pb=\pm a\ \text{by (1)}\ \Rightarrow\ b\ |\ a$
  • $c=\pm p\ \Rightarrow\ b=\pm a\ \text{by (1)}\ \Rightarrow\ b\ |\ a$
  • $c=\pm p^2\ \Rightarrow\ b=\pm ap\ \text{by (1)}\ \Rightarrow\ a\ |\ b$
Mengchun Zhang
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    Since you are working in $\Bbb Z$, it should be $c\in{\pm 1, \pm p, \pm p^2 }$ – jjagmath May 31 '24 at 13:09
  • But how do you propose to prove your key claim $,c\mid p^2\Rightarrow c = \pm p^k,, k\le 2,$ without "using prime decompositions" (or equivalent UFD properties)? – Bill Dubuque May 31 '24 at 16:42
  • $${\rm i.e.}\ \ \begin{align}a\mid pb\ b\mid pa\end{align}\Rightarrow \dfrac{pb}a\color{#0a0}{\dfrac{pa}b} = \color{#c00}{p^2}\Rightarrow, \pm\color{#0a0}{\dfrac{pa}b} = \color{#c00}{p^2,p,1}\Rightarrow, \pm\dfrac{a}b = p,1,\color{#0af}{\dfrac{1}p},\Rightarrow,\dfrac{a}b,\ {\rm or}\ ,\color{#0af}{\dfrac{b}a}\in\Bbb Z\qquad\qquad$$ Similarly $,a\mid p^ia,,b\mid p^jb,\Rightarrow, a\mid b,$ or $,b\mid a.,$ The argument boils down to the fundamental fact that the divisors $,d,$ of $,\color{#c00}{p^k}$ are linearly ordered by divisibility, i.e. $,d\mid d'$ or $,d'\mid d.\ \ $ – Bill Dubuque May 31 '24 at 18:01
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I have now found this proof: Let $g$ be the gcd of $a$ and $b$, then $a = g a' \land b = g b'$ with $a', b'$ coprime. We then have $$ \begin{array}{rcl} & {} & a \mid p b \land b \mid p a \\ & \implies & a' \mid p b' \land b' \mid p a' \\ a'\, b' \text{ coprime} & \implies & a' \mid p \land b' \mid p & \\ p \text{ prime} & \implies & (a' = 1 \lor a' = p) \land (b' = 1 \lor b' = p) \\ & \implies & a \mid b \lor b \mid a \end{array} $$

Léreau
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  • This is clearer using fractions - see my answer where I do a more general case that way. – Bill Dubuque Jun 02 '24 at 18:41
  • @BillDubuque Thanks for pointing out that a more general case can actually be covered! The above prove can easily be rewritten for that case as well. Avoiding reasoning in fractions is desirable for me, since I formalise the proofs in first order PA. – Léreau Jun 03 '24 at 09:55
  • Disallowing fractions and formalizing all your proofs in first order PA will be an extremely painful way to learn number theory. – Bill Dubuque Jun 03 '24 at 14:05
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Key idea the divisibilities imply $\,q = a/b\,$ (wlog reduced) can be written in one form where its $\rm\color{#c00}{numerator}$ is a power of $\,p\,$ and another form where its $\rm\color{#0a0}{denominator}$ is a power of $\,p,\,$ hence by $\rm\color{darkorange}{UF}$ = unique fractionization its reduced numerator and denominator are both powers of $\,p,\,$ so $\,q = p^k,\ k\in\Bbb Z,\,$ so $\,q\,$ or $\,1/q\,$ in $\Bbb Z,\,$ i.e. $\,b\mid a\,$ or $\,a\mid b.\,$ More precisely, generalizing we have

$\qquad q=\overbrace{\dfrac{\color{#c00}{p^I}}d =\phantom{a} }^{\textstyle\!\! a\mid p^Ib}\!\!\!\dfrac{a}b\!\!\overbrace{\phantom{a}=\dfrac{c^{\phantom{I}}}{\color{#0a0}{p^J}}}^{\textstyle b\mid ap^J\!\!\!\!\!\!}\,$ $\Rightarrow\, q=\dfrac{\color{#c00}{p^i}}{\color{#0a0}{p^j}}\,\Rightarrow\,\overbrace{q\ \,{\rm or}\,\ \dfrac{1}q\in \Bbb Z}^{\textstyle b\mid a\,\ {\rm or}\,\ a\mid b}\ $ (if $\,i\ge j\,$ or $\,i<j\,$ resp.)

because by $\,\rm\color{darkorange}{UF}$: $\ (a,b)\!=\!1\,\Rightarrow \,a\mid \color{#c00}{p^I}\,$ so $\,a = \color{#c00}{p^i},\,$ and $\,b\mid \color{#0a0}{p^J}\,\Rightarrow\, b = \color{#0a0}{p^j}$

Remark $ $ The proof in your answer is the special case $\,I=1=J\,$ of the above proof. Both proofs use properties of gcds that are equivalent to UPF = uniqueness of prime factorizations, so it would be circular to attempt to use these to prove UPF. The proof in Zhang's answer also implicitly uses a consequence of UPF, that the only divisors of a prime power $\,p^n$ are of the form $\,\pm p^k,\, k\le n.\,$ These properties may fail in domains where UPF fails, i.e. non UFDs.

Proofs like this become clearer when one studies local methods (valuation rings / theory), which helps to algebraically reify the algebraic structure hidden in such divisibility relations.

Bill Dubuque
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