Question

Let $\mathit{R}$ be a UFD. Ifrole="math" localid="1654689077536" $\mathit{a}\mathbf{|}\mathit{b}\mathit{c}$ and if ${\mathbf{1}}_{\mathbf{R}}$is a gcd of role="math" localid="1654689118995" $\mathit{a}$and$\mathit{b}$ , prove thatrole="math" localid="1654689171365" $\mathit{a}\mathbf{|}\mathit{c}$ .

Short answer

We proved that, $a|c$.

Step-by-step solution

Definition of Unique Factorization Domain

Let us see the definition of Unique Factorization Domain (UFD),

An integral domain $R$is a unique factorization domain provided that every nonzero, nonunit element of $R$ is the product of irreducible element and this factorization is unique up to associates.

Let$R$be a UFD and$a,b$and$c$are any elements.

We have to prove that$a|c$.

By Theorem 10.13,

Since$(a,b)=1_R$ , where$(a,b)=$ gcd of $a$and$b$ , we have factorization of $a$as

$a=\prod _{i=1}^m{p}_i^{{\alpha }_i}$

Into irreducible such that $p_i\cancel{|}b$.

To prove  a|c 

Also, by Theorem 10.13, $a|bc$ implies, there is some $d\in R$such that

$bc=d\left(\prod _{i=1}^m{p}_i^{{\alpha }_i}\right)$

Since $p_i\cancel{|}b$, byTheorem 10.15, we have,

$\begin{array}{l}{p}_i^{{\alpha }_i}|c\\ \Rightarrow \prod _{i=1}^m{p}_i^{{\alpha }_i}|c\end{array}$

Hence, $a|c$.