Associates and Irreducibles

Proposition 1 $a,b$ are non-zero element of an integral domain $R$. $a$ divides $b$ and $b$ divides $a$ if and only if there exist a unit $u$ such that $a=bu$

$\Rightarrow :$
• $\mathrm{\exists }r,s\in R:sa=b$ and $rb=a$. Then, $a=rsa$ and $a(1-rs)=0$. Since $R$ is an integral domain and $a$ is non-zero, $1-rs=0$. Thus, $r,s$ are units and $a=rb$.
$\Leftarrow :$
• $a=bu⟹b\mid a$. Since $u$ is a unit, $\mathrm{\exists }t\in R:ut=1$. $at=b(ut)=b⟹a\mid b$.

Proposition 2 $a|b⟺⟨b⟩\subseteq ⟨a⟩$

Corollary 2.1 $a,b$ are associates $⟺⟨a⟩=⟨b⟩$

Proposition 3 $p$ is irreducible if and only if $⟨p⟩$ is maximal amongst all principal ideals that contain $⟨p⟩$.

$\Rightarrow :$
• For any principal ideal $⟨x⟩$ that contains $⟨p⟩$ and is not the unit ideal, $p\in ⟨x⟩$. So there exists $r\in R:p=xr$. Since $x$ divides $p$ and $p$ is irreducible, $x$ is either an unit or an associate of $p$. If $x$ is an unit, then $⟨x⟩$ is the unit ideal. Thus, $x$ is an associate of $p$. Based on the previous corollary, $⟨x⟩=⟨p⟩$ and $⟨p⟩$ is maximal amongst all principal ideals that contain $⟨p⟩$.
$\Leftarrow :$
• For any $r$ that divides $p$, $⟨p⟩\subseteq ⟨r⟩$. Since $⟨p⟩$ is maximal amongst all principal ideals, $⟨p⟩=⟨r⟩$ or $⟨r⟩$ is the unit ideal. So, $r$ is either an associate of $p$ or an unit. Thus, $p$ is irreducible.

Proposition 4 In an integral domain $R$, every prime is irreducible.

$p$ is a prime in an integral domain $R$. For any $a,b\in R$ such that $p=ab$. Since $p$ is a prime, WLOG, let’s assume $p|a$. Then there exist $t\in R$ such that $pt=abt=a$. So, $a(bt-1)=0$. Since $R$ is an integral domain and $a\ne 0$, $bt=1$ and $b$ is an unit. Therefore, $a$ is an associate of $p$ and $p$ is irreducible.