Prove $\,a\mid bc\!\!\iff\!\! a\mid(a,b)c\!\iff\!\!\frac{a}{(a,b)}\!\mid c\ $ [general Euclid's Lemma]

I tried to start by showing that $ \frac{a}{\gcd(a,b)} $ is always an integer, let's call it $d$, because $a$ is always a multiple of $(a,b)$ based on the definition of a g.c.d. I then tried to show that if $a | b$ then $a|bc$ from the theorem that states that $a|b$ implies that $a|bc$ for any integer $c$, but I cant seem to prove that $a | b$ or that $a|bc$ only if $d \mid c$ (where $d = \frac{a}{\gcd(a,b)}$ ). Any help would be much appreciated.

Solution 1:

Note that $x\mid y$ if and only if $(x,y) = x$; then use the properties of the gcd.

\begin{align*} a|bc &\Longleftrightarrow (a,bc) = a\\\ &\Longleftrightarrow \left(\frac{a}{(a,b)}, \frac{bc}{(a,b)}\right) = \frac{a}{(a,b)}\\\ &\Longleftrightarrow \left(\frac{a}{(a,b)}, \frac{b}{(a,b)}c\right) = \frac{a}{(a,b)}\\\ &\Longleftrightarrow \left(\frac{a}{(a,b)}, c\right) = \frac{a}{(a,b)}&&\mbox{(by Euclid's Lemma)}\\\ &\Longleftrightarrow \frac{a}{(a,b)} \Bigm| c, \end{align*} using Euclid's Lemma ("If $(x,y)=1$, then $x|yz\Leftrightarrow x|z$") and $$1 = \frac{1}{(a,b)}(a,b) = \left(\frac{a}{(a,b)},\frac{b}{(a,b)}\right).$$

The manipulations are consequences of $x(y,z) = (xy,xz)$; see Bill Dubuque's proof here.

Solution 2:

It follows quickly by employing basic gcd laws, namely $\rm\color{#c00}{U} = $ gcd (or lcm) Universal Property $\, x\mid y,z\!\!\!\overset{\rm\color{#c00}{U}\!}\iff\! x\mid (y,z)_{\phantom{|_{|_|}}} $ (this is the gcd definition in general domains), and also using the ubiquitous $\,\rm\color{#0a0}D\,$= GCD Distributive Law $\, \ (ac,bc)\, =\, (a,b)\:\!c.\,$ With such

it is easy: $\ \ \ a\mid bc\iff a\mid ac,bc\smash{\overset{\rm\color{#c00}{U}\!}\iff} a\mid (ac,bc) \overset{\rm\color{#0a0}D}= (a,b)\:\!c\iff a/(a,b)\mid c\quad$ QED

Or dually: $\ a\mid bc\iff a,b\mid bc\smash{\overset{\rm\color{#c00}{U}\!}\iff} [a,b]\mid bc\iff [a,b]/b\ |\ c,\ $ $\, [m,n]\, :=\ {\rm lcm}(m,n)$

Combining both yields the GCD $\times$ LCM law: $\ [a,b]/b\, =\, a/(a,b),\ $ i.e. $\ (a,b)\ [a,b] =\, ab\,.$

Alternatively $ $ by cancelling $\ (a,b)\,$ it reduces to Euclid's Lemma $\ (a,b)= 1,\ a\,|\,bc\ \Rightarrow\ a\mid c.\,$ [generally $ $ in homogeneous problems we can assume wlog $\,(a,b)=1\,$ by canceling $(a,b)\,$].

Generally the result is the special case $\ a\,|\,bc\ $ of $\ (a,bc) = (a,(a,b)\,c),\,$ which holds true for both GCDs and ideals. As you can see these basic properties are all intimately connected, so it is useful to master these properties to become proficient with GCD arithmetic. For another example of applying these properties to obtain simple proofs see this proof of the Freshman's Dream $\ (A,B)^n = (A^n,B^n)\ $ for GCDs and invertible ideals.