1
$\begingroup$

If $$\frac{a}{(a,b)}\mid c \;\ \Rightarrow \;\ a\mid b\cdot c$$ $$\frac{a}{(a,b)}\mid c\Rightarrow c=\frac{a}{(a,b)}\cdot k\Rightarrow b\cdot c=\frac{a \cdot b}{(a,b)}\cdot k\Rightarrow b\cdot c=a(\frac{b}{(a,b)}\cdot k)\Rightarrow$$$$a\mid b\cdot c$$ This is correct?

$\endgroup$
2
$\begingroup$

We have: $$a|bc\iff \exists k\in\mathbb Z,\ bc=ka\iff \exists k\in\mathbb Z,\ \frac{b}{(a,b)}c=k\frac{a}{(a,b)}$$

$(\Rightarrow)$ and since $\frac{a}{(a,b)}$ and $\frac{b}{(a,b)}$ are coprime then $\frac{a}{(a,b)}|c$.

$(\Leftarrow)$ $\frac{a}{(a,b)}|c\iff\exists k'\in\mathbb Z,\ c=k'\frac{a}{(a,b)}\Rightarrow \frac{b}{(a,b)}c=\underbrace{\frac{b}{(a,b)}k'}_{=k}\frac{a}{(a,b)}$

$\endgroup$
  • $\begingroup$ that $\frac{a}{(a,b)}$ and $\frac{b}{(a,b)}$ are relatively prime, I get ($(\frac{a}{(a,b)},\frac{b}{(a,b)})=1$), but did not understand why they concluded directly $\frac{a}{(a,b)} \mid c$ Could you explain again with more details? $\endgroup$ – marcelolpjunior Jul 30 '13 at 18:20
  • $\begingroup$ This result is known as the Euclid lemma: if $a|bc$ and $a$ and $b$ are relatively prime then $a|c$. $\endgroup$ – user63181 Jul 30 '13 at 18:24
  • $\begingroup$ But I understand that $\frac{a}{a(a,b)}$ and $\frac{b}{(a,b)}$ are relatively primes, but $a$ and $b$ what too relatively primes? $\endgroup$ – marcelolpjunior Jul 30 '13 at 18:28
  • $\begingroup$ No in your problem $a$ and $b$ aren't relatively prime. $\endgroup$ – user63181 Jul 30 '13 at 18:31
  • $\begingroup$ So why use the motto that says $$a \mid b \cdot c \Rightarrow a \mid c \;\;\ \text{If a and b coprime}$$ Since we have $\frac{a}{(a,b)}$ and $\frac{b}{(a,b)}$ are coprime, and the requirement of the lemma is that $a$ and $b$ are coprime. $\endgroup$ – marcelolpjunior Jul 30 '13 at 18:37

Not the answer you're looking for? Browse other questions tagged or ask your own question.