Question

Let $\mathit{R}$ be an integral domain in which any two elements (not both ${\mathbf{0}}_{\mathbf{R}}$) have a gcd. With the notation of Exercise 25, prove that if $\mathbf{(}\mathbf{b}\mathbf{,}\mathbf{c}\mathbf{)}\mathbf{~}{\mathbf{1}}_{\mathbf{R}}$ and $\mathbf{(}\mathbf{b}\mathbf{,}\mathbf{d}\mathbf{)}\mathbf{~}{\mathbf{1}}_{\mathbf{R}}$ then $\mathbf{(}\mathbf{b}\mathbf{,}\mathbf{c}\mathbf{d}\mathbf{)}\mathbf{~}{\mathbf{1}}_{\mathbf{R}}$ .

Short answer

If $(b,c)~1_R$ and$(b,d)~1_R$ then$(b,cd)~1_R$ .

Step-by-step solution

Exercise 25

Let $R$be an integral domain in which any two elements (not both $0_R$) have a gcd. Let $\left(r,s\right)$denote any gcd of r and s. Use $~$ to denote associates then for all $r,s,t\in ℝ$:

(a) If $s~t$, then $rs~rt$.

(b) If $s~t$, then $(r,s)~(r,t)$.

(c) $r(s,t)~(rs,rt)$.

(d) $(r,(s,t))~((r,s),t)$ [Hint: show that both are gcd’s of r,s,t.]

(b,cd)~1R

Let $b,c,d\in R$then by (c), $d(b,c)~(bd,cd)$. Given that $(b,c)=1$then $d~(bd,cd)$.

Apply (b) to the above associate and get $(b,d)~(b,(bd,cd))$. It is given that $(b,d)=1$ then $1~(b,(bd,cd))$.

Rewrite $1~(b,(bd,cd))$ by using (a),(b),(c),(d) then,

$\begin{array}{c}1~(b,(bd,cd))\\ ~((b,bd),cd)\\ ~(b(1,d),cd)\end{array}$

Since, $(1,x)=1$ for all $x\in R$ then $1~(b,cd)$.

Therefore, if $(b,c)~1_R$ and $(b,d)~1_R$ then $(b,cd)~1_R$.