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This is a somewhat obvious fact that is intuitively obvious to me, but I haven't been able to construct a proof of it. Euclid's lemma states for for $p$ a prime and $ab$ a product of integers (let's take everything to be positive for simplicity), if $p \mid ab$, then $p \mid a$ or $p \mid b$. This is clear, and I know how to prove it. Let's extend it somewhat. Suppose that $a$ and $b$ are two relatively prime integers, and we have $a \mid bc$ for some other integer $c$. Then $a \not \mid b$, so it must divide $c$. This fact is obvious to me, but I can't figure out how to prove it.

Does anyone have any hints or advice? Do I need the assumption of positivity? (For my purposes at the moment, I only need them to be positive, but there is value in having the most general result possible).

EDIT: Updated attempt:

We have that $a,b$ are relatively prime, so there exist $r,s \in \mathbb{Z}$ such that $ar + bs = 1$ by Bézout's lemma. Multiply through by $c$ to get $arc + bsc = c$. Then $a \mid arc$ and $a \mid bsc$, so $a \mid c$.

How is that?

Bill Dubuque
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user861776
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  • This is Gauß' lemma . Your final argument is fine: if a number divides two other numbers, it divides their sum, of course. – Bernard May 18 '21 at 08:57
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    That's the standard Bezout-based proof of Euclid's Lemma. But if fails in higher dimensional UFDs e.g. polynomial rings like $\Bbb Z[x]$ and $,Q[x,y],,$ See the linked dupe for more general proofs. – Bill Dubuque May 18 '21 at 10:41

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