# equivalence factorial ring , p prime divides $a,…,a_{k}$ is equivalent to $a_{1},…,a_{k}$ are not relatively prime

It will be now shown that this when R is a factorial ring, and $a_{1},…,a_{k}$ in R are not all 0. Then

$a_{1},…,a_{k}$ not relatively prime $\Leftrightarrow$ there exists a prime $p$ so that $p|a_{1},…,p|a_{k}$

1. $(\Rightarrow )$ Assume $a_{1},…,a_{k}$ are not relatively prime, then there is a number p which divides $a_{1},…,a_{k}$. Now it is to show that p must also be a prime. Since R is a factorial ring, that means … stuck.

2. $(\Leftarrow)$ Assume $p|a_{1},…,p|a_{k}$, then $a_{1},…,a_{k}$ are trivially not relatively prime.

Does anybody see a way to use that R is a factorial ring to show that p must also be a prime?

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For $\Rightarrow$, factor the greatest common divisor. Since it is not unit (then the $a_i$'s would be relatively prime), it has at least one prime factor. This prime factor divides every $a_i$.
The gcd itself is not necessarily prime; a simple counterexample is $\mathbb Z$ with $a_1=a_2=6$.