Use of GCD Distributive Law in proof of Euclid's Lemma

Question

Suppose that c|ab and (b, c) = 1. Then c|a

Proof (ab, ac) =|a|(b, c) = |a|. But by hypothesis, one has c|ab, which implies that c|(ab, ac). We thus conclude that c|a. And the proof is complete.

I am sure it is very simple but I cant seem to understand the stage c|ab implies that c|(ab,ac). I would appreciate if someone could show me this.

Answer 1

Greatest common divisior of $x$ and $y$ is a number with a property that $d \mid x$ and $d \mid y$ implies $d \mid \gcd(x,y)$.

In your case we have $c \mid ab$ from assumption and $c \mid ac$ trivially, because $\frac{ac}{c}=a$. Hence, $c \mid \gcd(ab,ac)$.

I hope this helps $\ddot\smile$

Answer 2

Since $(b,c)=1$ there exist integers $k,l$ with $kb+lc=1$. Now $c\mid ab$ implies $c$ divides $kab=a(1-lc)=a-alc$. Since $c\mid alc$ it follows that $c\mid a$.

Answer 3

It uses basic gcd laws. By the GCD Distributive Law $\,\rm\color{#c00}{DL}\, $ we have

$\qquad\ c\mid ab\iff c\mid ab,ac \color{#0a0}\iff c\mid (ab,ac)\overset{\color{#c00}{\rm D\:\!L}} = a(b,c) \iff c/(b,c)\mid a$

Above we employed $\,\ c\mid j,k\color{#0a0}\iff c\mid (j,k),\,$ the GCD Universal Property.