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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 $c$.

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$R$ is a UFD since it is an Euclidean Domain. Now, it follows easily your assertion.

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Be careful, using the above could amount to a circular proof. – Bill Dubuque Jan 20 '12 at 3:49

HINT $\rm\quad a\ |\ ac,bc\ \Rightarrow \ a\ |\ (ac,bc) = (a,b)\ c = c$

Or, in Bezout form $\rm\ a\ |\ (ax+by)c = c $

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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 $a | c$.

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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.

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