Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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?

share|improve this question
    
Yes, but notice that you required $R$ to be an integral domain to cancel out $a$ in the step $aup = ab \Rightarrow up = b$. –  user79202 May 29 '13 at 15:01
    
In the question it is mentioned that $R$ is an integral domain. –  poton May 29 '13 at 15:02
    
Yes, I just wanted to make sure you had that in mind... ;-) –  user79202 May 29 '13 at 15:05

1 Answer 1

up vote 0 down vote accepted

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 $$\,b\ \ {\rm unit} \iff b\mid 1 \stackrel{\color{#c00}{\large 1/b\, =\, a/p}}\iff p\mid a\stackrel{\color{#c00}{\large a\,\mid\, p}}\iff a\ \ {\rm associate\ to}\ \ p$$

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.