Divisor of a product of integers is a product of divisors 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 $a_i$?
 A: 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.
