# Let $R$ be a UFD and $a,b,c \in R$ be nonzero. If $c \mid ab$ and $\gcd(a, c) = 1$, then $c \mid b$. [duplicate]

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!

## marked as duplicate by Bill Dubuque abstract-algebra StackExchange.ready(function() { if (StackExchange.options.isMobile) return; $('.dupe-hammer-message-hover:not(.hover-bound)').each(function() { var$hover = $(this).addClass('hover-bound'),$msg = $hover.siblings('.dupe-hammer-message');$hover.hover( function() { $hover.showInfoMessage('', { messageElement:$msg.clone().show(), transient: false, position: { my: 'bottom left', at: 'top center', offsetTop: -7 }, dismissable: false, relativeToBody: true }); }, function() { StackExchange.helpers.removeMessages(); } ); }); }); Jul 26 at 14:09

• Note $c|axb+cyb \Leftrightarrow c|(ab)x$ – Kai Jul 26 at 6:17
• @Kai Bézout's Identity i.e. $ax + cd = 1$ fails in arbitrary UFDs. You cannot use that result. – 0XLR Jul 26 at 6:19
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.
• 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 at 6:19
• Indeed. All prime powers dividing $c$ divide $b$. – punctured dusk Jul 26 at 6:21