Here is the problem If $c\ |\ ab$ and $\text{gcd}(a,c)=1$ then $c\ |\ b$ Here's my approach. There exists $x,y$ such that $ax+cy=1$, so $c\ |\ axb+cyb=b$ I'm pretty sure my first step is wrong. Any help would be appreciated!
Let $R$ be a UFD and $a,b,c \in R$ be nonzero. If $c \mid ab$ and $\gcd(a, c) = 1$, then $c \mid b$.
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Your proof is fine. – Wuestenfux Jul 26 '19 at 06:10
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Note $c|axb+cyb \Leftrightarrow c|(ab)x$ – Kai Jul 26 '19 at 06:17
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3@Kai Bézout's Identity i.e. $ax + cd = 1$ fails in arbitrary UFDs. You cannot use that result. – balddraz Jul 26 '19 at 06:19
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It is not true that $\gcd(a, c) = 1$ implies that there exist $x, y \in R$ with $ax + cy = 1$. (For example when $R = \mathbb Z[X]$, $a = 2$, $c = X$.)
Instead work with the factorizations of $a, b, c$ into irreducibles.
Bart Michels
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Oh! Yeah, so all irreducible factors of $c$ doesn't divide $a$, so by definition of UFD, they divide $b$. Is that right? – Kai Jul 26 '19 at 06:19
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