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. Show that a non zero, non unit element $c$ of $R$ is NOT reducible iff its only divisors are units of $R$ and elements of $R$ which are associates of $c$.

share|improve this question
    
Welcome to SE Tom. Please note that the way you formatted your question is not reader friendly, nor does it conform to the way questions are typically formatted here. It's generally a bad habit to start asking your question in the title and let it continue into the question text. Please keep in mind, that the quality of the answers you'll get will largely match the quality of the question. –  Ittay Weiss Mar 11 '13 at 23:37
2  
What is your definition of irreducible and associate? –  Math Gems Mar 11 '13 at 23:42
1  
@Tom: So it seems like your statement follows directly from the definition! –  azimut Mar 11 '13 at 23:52
2  
you want a more complicated proof? this is weird –  Integral Mar 11 '13 at 23:57
1  
If the proof is not to apply the definition in a situation where it applies (at the start), just when can the definition be applied? The statement is a tautology by definition. Any other approach seems an unnecessary obfuscation. –  Michael E2 Mar 12 '13 at 12:18
show 1 more comment

1 Answer 1

up vote 1 down vote accepted

Let $a\in R$, $a\neq 0$ and $a$ not a unit.

I will proof your statement by using the following (quite common) definition of the irreducibility of $a$: Whenever $a = bc$ with $b,c\in R$, then at least one element of $b$ and $c$ is a unit in $R$.

Let $A$ be the statement "$a$ is irreducible" and $B$ the statement "Every divisor of $a$ in $R$ is either a unit or an associate of $a$".

We want to show $A\iff B$ or equivalenty, $$\neg A\iff \neg B.$$ We split the proof into the two implications.

,,$\neg B\implies\neg A$'': Assume there exists a divisor $c\in R$ of $a$ which is neither a unit nor associate to $a$. Since $c$ is a divisor of $a$, there a $b\in R$ with $a = bc$. Then $b$ is not a unit (otherwise, $a$ associated to $c$). So $a = bc$ where $b,c$ are no units. Thus, $a$ is reducible.

,,$\neg A\implies \neg B$'': If on the other hand $a$ is reducible, then $a = bc$ with $b,c$ elements of $R$ which are no units. Thus, $c$ is a divisor of $a$ which is neither a unit nor associate to $a$ (otherwise, $b$ is a unit).

share|improve this answer
    
What is that h meant to be? –  Tom Mar 12 '13 at 11:48
    
Thanks for spotting this typo. It's an '$a$'. –  azimut Mar 12 '13 at 11:52
    
If this a proof by contraposition, shouldn't the 3rd para. start by assuming that there are elements $b$, $c$ such that $a=bc$ and $b$, $c$ are neither units nor associates of $a$? –  Michael E2 Mar 12 '13 at 12:23
    
I've edited my answer. Hopefully the logic structure is clearer now. –  azimut Mar 12 '13 at 12:37
    
Wonderful proof, very clear and I understood it well. Thank you. –  Tom Mar 12 '13 at 21:08
show 1 more comment

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.