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

Question

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

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.

because clearly $c$ divides $ac$. And if $x$ is a divisor of $y$ and $z$ it divides their GCD. — Ofir Schnabel, Feb 11 '15 at 13:17

score 3 · Answer 1 · answered Feb 11 '15 at 13:22

3

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 $c\mid a$.

answered Feb 11 '15 at 13:22

Dietrich Burde

130,978

Better to invoke the gcd distributive law by name rather than reprove it every time you use it. – Bill Dubuque Oct 29 '20 at 05:15

score 1 · Answer 2 · answered Feb 11 '15 at 13:19

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 $\ddot\smile$

Bill Dubuque · Answer 3 · 2020-10-29T05:08:40.213

0

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.

edited Oct 29 '20 at 05:08

answered Feb 11 '15 at 17:09

Bill Dubuque

272,048

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

3 Answers3