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Let $R$ be a $\text{UFD}$, and let $a,b,c \in R$ such that $1=\text{gcd}(a,b).$ Suppose $a |c, \ b|c.$ Could anyone advise me on how to prove $ab |c \ ?$ How do I use the fact that every nonzero non unit $r \in R$ can be expressed as finite product of irreducible elements in $R\ ?$

Hints will suffice.

Bill Dubuque
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Alexy Vincenzo
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  • Let $P$ be a set of representatives of the classes of associated primes. Then use that every $x\in R$ has a unique decomposition $$x = \varepsilon\cdot \prod_{p\in P} p^{v_p(x)}.$$ $\gcd(a,b) = 1$ means that for every prime at least one of $v_p(a)$ and $v_p(b)$ is $0$. – Daniel Fischer Mar 02 '14 at 14:35

2 Answers2

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A UFD (or gcd domain) satisfies Euclid's Lemma: $\ (a,b)=1,\,\ a\mid bd\,\color{#c00}{\Rightarrow}\,a\mid d.\,$ In particular

$$ a,b\mid c,\,\ (a,b)=1\,\Rightarrow\, a\mid b(c/b)\,\color{#c00}{\Rightarrow}\,a\mid c/b\,\Rightarrow\, ab\mid c$$

Remark $\, $ In particular this holds true for any domain with a Euclidean algorithm, since this implies all gcds exist, e.g. it is true for $\Bbb Z$ or $\,F[x],\,$ for $F$ a field. Follow the above link for proofs.

See here for the straightforward inductive extension to $n$ arguments.

More conceptually gcd, lcm duality shows $\,{\rm lcm}(a,b) = ab/\gcd(a,b).\,$ OP is case $\,(a,b)=1$

Bill Dubuque
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@Daniel Fischer

Thanks for the reply.

Let $R$ be $\text{UFD}$ and $T$ be a complete set of representatives of associate classes of irreducible elements of $R$ *(so that for each irreducible $r\in R,$ there exists unique $t_r \in T$ such that $r$ is associate to $t_r)$

Then, for each $r \in R^*,$ there exists unique $u_r \in U(R)=$ set of units of $R$ and $n_{r,t} \in \mathbb{Z}_{\geq 0}$ for $t \in T $ such that $r = u_r\cdot \prod_{t\in T} t^{n_{r,t}}.$

Furthermore, if $s=u_s\cdot \prod_{t\in T} t^{n_{s,t}},$ with $u_s \in U(R) $ and $n_{s,t}\in \mathbb{Z}_{\geq 0}$ for all $t\in T,$ then $r|s$ iff $n_{s,t} \leq n_{r,t} $ for all $t \in T.$

Why $t$ is a prime? And if $\text{gcd}(r,s)=1,$ this means $1=\prod_{t\in T} t^{\text{min}(n_{r,t},n_{s,t})}.$ So why there exists $t \in T$ such that $\text{min}(n_{r,t},n_{s,t})=0?$

Alexy Vincenzo
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    In a UFD, we have "irreducible = prime". If $\gcd(r,s) = 1$, then $\min (n_{r,t},n_{s,t}) = 0$ for all $t\in T$, for if we had $\min (n_{r,t},n_{s,t}) > 0$ for some $t$, then that $t$ would be a common divisor of $r$ and $s$, i.e. $t\mid \gcd(r,s)$. And thus we have $n_{r,t} + n_{s,t} = \max(n_{r,t},n_{s,t})$ for all $t$. Oh, and you forgot a 'c' in my name. – Daniel Fischer Mar 02 '14 at 15:36