Divisors in a Euclidean Domain Let $R$ be a Euclidean Domain. If $(a,b)=1$, and $a$ divides $bc$, I would like a hint on how to prove that $a$ divides $c$.
 A: $R$ is a UFD since it is an Euclidean Domain. Now, it follows easily your assertion.
A: HINT $\rm\quad a\ |\ ac,bc\ \Rightarrow \ a\ |\ (ac,bc) = (a,b)\ c = c$ 
Or, in Bezout form $\rm\ a\ |\ (ax+by)c = c $
A: Emiliocba's answer is correct: the desired conclusion holds in any UFD domain (and even a bit more generally, e.g. in a GCD-domain).  
However, the fact that a Euclidean domain is a UFD is really the concatenation of two facts: 
(i) A Euclidean domain is a PID: indeed, every ideal is generated by any element of 
minimal norm.
(ii) A PID is a UFD.
The first fact is quicker to prove: what I said above is the entire idea of the proof, and fleshing it out takes maybe two more lines.  
Thus it may be helpful to give a proof of the desired conclusion in an arbitrary PID.  The method here is familiar from elementary number theory:
since $a$ and $b$ are relatively prime in the PID $R$, there exist elements $x,y \in R$ such that $ax+by = 1$.  Moreover, to say that $a | bc$ is to say that there exists $d \in R$ with $ad = bc$.  Thus
$c = acx + bcy = acx + ady = a(cx+dy)$,
so $a | c$.   
A: Try thinking about ideals and use the fact that if $a|bc$ then $(bc)\subset (a)$. And any ideal is contained in a prime one. 
