Let $R$ be an integral domain and $p ∈ R$ be such that $p$ is nonzero and a nonunit. Then $p$ is irreducible if and only if the only divisors of $p$ are the associates of $p$ and the unit elements of $R$.

Proof. Suppose the only divisors of p are the associates of $p$ and the unit elements of $R$. Let $p = ab$ for some $a, b ∈ R$. Suppose $a$ is not a unit. Then $a$ is an associate of $p$. Therefore, $a = pu$ for some unit $u ∈ R$. Now $p = pub$. Since $R$ is an integral domain, it follows that $ub = 1$. Hence, $b$ is a unit and so $p$ is irreducible. We leave the converse as an exercise.

this is a theorem from a book and I have tried to proof the converse in the following way:-

suppose $p$ is irreducible and $a|p$. Then there exist $b\in R$ such that $p=ab$ . Since $p$ is irreducible so one of $a$ & $b$ is unit.Let $a$ is unit.Then we need to show that $b$ is associate of $p$. Since $a$ is unit there exist $u \in R$ such that $au=1$. hence $aup=p \implies aup=ab \implies up=b $.So the result follows.

Am I correct?

  • $\begingroup$ Yes, but notice that you required $R$ to be an integral domain to cancel out $a$ in the step $aup = ab \Rightarrow up = b$. $\endgroup$ – user79202 May 29 '13 at 15:01
  • $\begingroup$ In the question it is mentioned that $R$ is an integral domain. $\endgroup$ – poton May 29 '13 at 15:02
  • $\begingroup$ Yes, I just wanted to make sure you had that in mind... ;-) $\endgroup$ – user79202 May 29 '13 at 15:05
  • $\begingroup$ Isn't that the definition of irreducible element? $\endgroup$ – lhf Nov 7 '18 at 11:27

Yes. The key idea is this: $ $ a factor $\,b\,$ of $\,\color{#c00}{p = ab}\,$ is a unit $\iff$ its cofactor $\,a\,$ is associate to $\,p,\,$ by
$$\,b\ \ {\rm unit} \iff {\overbrace{b\mid 1 \iff\!\!\ p\mid a}^{\Large\color{#c00}{ 1/b\ \ \ \ =\ \ \ \ a/p}}}\!\overset{\color{#c00}{\large a\,\mid\, p}}\iff a\ \ {\rm associate\ to}\ \ p\qquad\ $$


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