Exercise 2.2.15 (Integral Domain Equivalencies)

Question

Let $r$ and $s$ be elements of an integral domain.\
Show that $r | s |r \Longleftrightarrow \langle r angle = \langle s angle \Longleftrightarrow s = ur$ for some unit $u$.

Hassaan Naeem · Accepted Answer

\paragraph{Solution:}
If we have that $r|s|r$ then $\exists\ a \in R: s = ar$ and $ \exists\ b \in R: r = bs$ then:
    \begin{equation*}
        \begin{aligned}
            \frac{s}{a} &= bs\
            b &= \frac{1}{a} \leadsto ab=1 \leadsto b = a^{-1}
        \end{aligned}
    \end{equation*}
Then we have that $s=ar$, and we have just shown that a is a unit, hence $s=ur$. Therefore $r|s|r \implies s = ur$
\\
If we have $\langle r angle = \langle s angle$, then $r=s$. Hence, 
\begin{equation*}
    \begin{aligned}
        r = 1s \quad &\&\quad s = 1r\
        r = as \quad &\&\quad s = ar\quad (	ext{where}\ a=1)\
        \implies s|r  \ &\&\quad r|s
    \end{aligned}
\end{equation*}
Therefore $\langle r angle = \langle s angle \implies r|s|r$
\\
If we have that $s=ur$ for some unit $u$, then also we have that
\begin{equation*}
    \begin{aligned}
        &u^{-1}s = u^{-1}ur \leadsto r = u^{-1}s\
        &	ext{Therefore}\ s \in \langle r angle\ \&\ r \in \langle s angle, \langle s angle \subseteq \langle r angle\ \&\ \langle r angle \subseteq \langle s angle \
        &\implies  \langle r angle = \langle s angle
    \end{aligned}
\end{equation*}
Therefore $s=ur \implies \langle r angle = \langle s angle$

Richard Ganaye · Answer

Let $R$ be an integral domain, and $r,s$ be elements of $R$.

\begin{enumerate} 
\item[$\bullet$]
We know from the text (p.  25) that $r \mid s \iff \langle s angle \subset \langle r angle$. Therefore
$$ r \mid s 	ext{ and  } s \mid r \iff \langle s angle \subset \langle r angle 	ext{ and }  \langle r angle \subset \langle s angle \iff \langle s angle= \langle r angle .$$

\item[$\bullet$] Assume that $ r \mid s$ ans $s \mid r$. Then $s = ur$ and $r = vs$ for some $u,v \in R$. Therefore $r = vu r$ and $r(1 -vu) = 0$. Since $R$ is an integral domain, $r = 0$ or $1 - ba = 0$.

If $r = 0$, then $s = ar = 0$, thus $s= 1\cdot r$.

If $r
e 0$, then $vu = 1$, so $u$ is a unit, and $s = ur$.

In both cases, $s = u r$ for some unit $u$.

Conversely, assume that $s = u r$ for some unit $u$. Since $u$ is a unit, there is some $v \in R$ such that $vu = 1$. Then $vs = vur = 1r = r$. The equalities $s = ur$ and $vs = r$ show that $r \mid s$ and $s \mid r$. This proves
$$ r \mid s 	ext{ and  } s \mid r \iff \exists u \in R^{	imes},\ s = ur.$$
\end{enumerate}

Exercise 2.2.15 (Integral Domain Equivalencies)

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Exercise 2.2.15 (Integral Domain Equivalencies)

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