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I read this recently on the web and can't manage to understand it. Not homework -- I haven't done math homework for years.

If $d\mid ab$ then $d=d_1 d_2$, that $d_1\mid a$, that $d_2\mid b$ and, further, if $(a,b)=1$ then $(d_1,d_2)=1$.

I know it's the same question... if $a = a_1 a_2 ... a_n$, why can't $d$ divide some $a_i$ and only $a_i$?

Martin Sleziak
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QED
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1 Answers1

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Hint $ $ by $\, d\mid j,k\!\!\color{#c00}{\overset{\rm U\!\!}\iff} d\mid (j,k)\,$ [gcd Universal property], and $\,\rm\color{#0a0}{DL} =$ gcd Distributive Law

$\,\ d\mid ab\iff d\mid ab,db\!\! \color{#c00}{\overset{\rm U\!\!}\iff} d\mid (ad,db)\overset{\color{#0a0}{\rm D\,L}} = (a,d)b \iff \color{#90f}{d/(a,d)\mid b}$

Hence we conclude $\ d\, =\, (a,d)\, \dfrac{d}{(a,d)} \ $ where $\ (a,d)\mid a,\ $ and $\ \color{#90f}{\dfrac{d}{(a,d)}\mid b},\ $ as sought.

It works fine if $\ d\mid a,\,$ then $\,(a,d)=d\ $ so $\,d/(a,d) = 1,\,$ so $\,d = d\cdot 1\,$ where $\,d\mid a,\ 1\mid b$.

Generalization $ $ The property that is considered in your question may be considered to be a generalization of the prime divisor property from atoms (irreducibles) $\,p\,$ to composites $\,c.\,$

Prime Divisor Property $\quad p\ |\ a\:b\ \Rightarrow\ p\:|\:a\ $ or $\ p\:|\:b\ \ $ [= Euclid's Lemma in $\Bbb Z$]

Primal Divisor Property $\ \ \: c\ |\ a\:b\ \Rightarrow\ c_1\, |\: a\:,\: $ $\ c_2\:|\:b,\ \ c = c_1\:c_2\ $

One easily checks that atoms are primal $\Leftrightarrow$ prime. This leads to various "refinement" views of unique factorizations, e.g. via Schreier refinement and Riesz interpolation, the Euclid-Euler Four Number Theorem (aka Vierzahlensatz), etc, which prove more natural in noncommutative rings - see Paul Cohn's 1973 Monthly survey Unique Factorization Domains.

Bill Dubuque
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  • Thank you. I'm curious about the second item in the equivalence chain in your hint... – QED Feb 08 '15 at 04:14
  • @psoft I elaborated on that. – Bill Dubuque Feb 08 '15 at 04:58
  • I'm sorry, I just don't understand the notation d|x,y. d divides both x and y? – QED Feb 08 '15 at 14:25
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    @psoft Yes $\ d\mid x,y,$ means $,d\mid x,\ d\mid y\ \ $ – Bill Dubuque Feb 08 '15 at 14:28
  • @BillDubuque. I can follow the proof up to the part where you have $d|(a,d)b$ then you divide $d$ by $(a,d)$. Is this division possible in any given gcd domain? wouldn't you have to prove that $(a,d)$ is a unit first? – Mikel Solaguren Jun 03 '22 at 13:03
  • @Mikel $,0\neq c:=(a,d)\mid d,,$ so we can cancel $,c,$ from $,d\mid cb,$ yielding $,d/c\mid ,b,$ by divisibility cancellation (recall elements $\neq 0$ in an integral domain are cancellable - the prior linked post shows how such cancellation extends from equations to divisibility relations). $\ \ $ – Bill Dubuque Jun 03 '22 at 13:27
  • @BillDubuque. I can see this when we have $ac=ab \Rightarrow c=b$. However, when $a$ is not explicitly on say the right hand side wouldn't you need $a$ to have an inverse in order to get $c/a$? – Mikel Solaguren Jun 03 '22 at 14:15
  • @Mikel $,c\mid d,\Rightarrow, cx = d,$ has a root $,x,,$ which is unique: $,cx = d = cx'!\Rightarrow x=x',$ by cancelling $,c\neq 0.,$ This unique quotient $,x,$ is denoted $,d/c,\ $ so $\ c\mid d,\Rightarrow, d = (d/c)c\ \ $ – Bill Dubuque Jun 03 '22 at 14:49
  • @BillDubuque Ahh I see, I wasn't getting the notation. Thanks for the help!! – Mikel Solaguren Jun 03 '22 at 15:10