Question

If $\mathit{R}$ is a UFD, if$\mathit{a}\mathbf{,}\mathit{b}$ and$\mathit{c}$ are elements such that$\mathit{a}\mathbf{|}\mathit{c}$ and$\mathit{b}\mathbf{|}\mathit{c}$ , and if${\mathbf{1}}_{\mathbf{R}}$ is a gcd ofrole="math" localid="1654688268521" $\mathit{a}$ and $\mathit{b}$, prove that $\mathit{a}\mathit{b}\mathbf{|}\mathit{c}$.

Short answer

We proved that,$ab|c$ .

Step-by-step solution

Definition of Unique Factorization Domain

Let us see thedefinition 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 role="math" localid="1654688476998" $ab|c$.

ByTheorem 10.13,

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

$a=\prod _{i=1}^m{p}_i^{{\alpha }_i}$and $b=\prod _{j=1}^n{q}_j^{{\beta }_j}$

such that$p_i$ and $q_j$never associates (by definition of UFD).

To prove ab|c

Also, by Theorem 10.13, since $a|c$and $b|c$, there is some $d\in R$such that

role="math" localid="1654688784654" $c=d\left(\prod _{i=1}^m{p}_i^{{{\alpha }^{\text{'}}}_i}\right)\left(\prod _{j=1}^n{q}_j^{{{\beta }^{\text{'}}}_j}\right)$, where ${\alpha }_i\le {{\alpha }_i}^{\text{'}},{\beta }_j\le {{\beta }_j}^{\text{'}}$.

Therefore, the product is given by,

$ab=\left(\prod _{i=1}^m{p}_i^{{\alpha }_i}\right)\left(\prod _{j=1}^n{q}_j^{{\beta }_j}\right)$ is a divisor of $c$.

Hence,$ab|c$ .