Question

Let $R$ be an integral domain and $\mathit{a}\mathbf{,}\mathit{b}\mathbf{\in }\mathit{R}$. Show that $\mathbf{\left(}\mathbf{a}\mathbf{\right)}\mathbf{=}\mathbf{\left(}\mathbf{b}\mathbf{\right)}$ if and only if $\mathit{a}\mathbf{=}\mathit{b}\mathit{u}$ for some unit $\mathit{u}\mathbf{\in }\mathit{R}$ .

Short answer

$\left(a\right)=\left(b\right)$ if and only if $a=bu$ for some unit $\mathit{u}\mathbf{\in }\mathit{R}$.

Step-by-step solution

Theorem 3.7

Cancelation is valid in any integral domain $R$ if role="math" localid="1654240230149" $a\ne 0_R$ and role="math" localid="1654240222828" $ab=ac$ in R, then $b=c$.

a=bu for  u∈R

Let $a,b\in R$ with $\left(a\right)=\left(b\right)$. If $a=0$ then $\left(a\right)=\left\{0\right\}$ implies $b=0$ and so $a=b1$.

If role="math" localid="1654240641633" $a\ne 0$then role="math" localid="1654240648150" $a\in \left(b\right)$ , role="math" localid="1654240651396" $b\in \left(a\right)$ and there exists an elements p and q in R such that $a=bu$and role="math" localid="1654240644533" $b=av$ .

Substitute role="math" localid="1654240722450" $b=av$ in role="math" localid="1654240725676" $a=bu$ then role="math" localid="1654240729308" $a=avu$.

Now by the theorem $vu=1$ and u is an unit. Therefore, $a=bu$ for some unit $u\in R$.

Proof for (a)=(b)

Assume that $a=bu$for some unit $u\in R$. Since a is a multiple of b,$a\in \left(b\right)$ implies

$\left(a\right)\subseteq \left(b\right)$ …… (1)

If role="math" localid="1654241165249" $a{u}^{-1}=b$then b is a multiple of a and $b\in \left(a\right)$ implies $\left(b\right)\subseteq \left(a\right)$…… (2)

Thus, from (1) and (2), $\left(a\right)=\left(b\right)$

Therefore, $\left(a\right)=\left(b\right)$ if and only if $a=bu$for some unit $u\in R$.