# Prove that a non-zero, non-unit element $a \in R$ is irreducible [duplicate]

Prove that a non-zero, non-unit element $a \in R$ is irreducible iff its only divisors are units of $R$ and elements of $R$ which are associates of $a$

I have a proof below but i'm not sure if it is correct. I think I need to use $c$ in the second part but I'm not sure how.

Here is my proof

$\Rightarrow$ Assume $a\ne 0$, $a$ is not a unit $\in R$ is irreducible

Then $a=bc$ where at least one of $b$ or $c$ is a unit

Therefore if $b$ is a unit then $c$ is an associate of a and if $c$ is a unit then $b$ is an associate of a

As $a=bc$ then $b|a$ and $c|a$

$b$ and $c$ are either units or associates of $a$ therefore the only divisors of $a$ are units of $R$ or associates of $a\in R$

$\Leftarrow$ Assume $a\in R$ can only be divided by units or associates

Let $b,c\in R$ divide $a$, by assumption $b$ must either be a unit or an associate

If $b$ is a unit then $a$ is irreducible

else $b$ is an associate

This means there exists a unit $u\in R$ where $a=ub$

Therefore $u|a$ so once again a is divided by a unit and is therefor irreducible

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## marked as duplicate by Martin Brandenburg, Amzoti, Micah, Davide Giraudo, Andreas CarantiMar 16 '13 at 22:31

This question was marked as an exact duplicate of an existing question.

What does it mean for a "non-reducible" element to be irreducible? – Erick Wong Mar 16 '13 at 19:04
How do you define "non-reducible" vs. "irreducible". I've never seen them used to mean different things. – Cameron Buie Mar 16 '13 at 19:04
@ErickWong That was meant to say non-unit my bad – Adam Mar 16 '13 at 19:07
@CameronBuie Was meant to say non-unit – Adam Mar 16 '13 at 19:08
@MartinBrandenburg This is not a duplicate as I was trying to see if my own proof was correct which I have been told to put in the question which I shall now do – Adam Mar 16 '13 at 20:38

$\Rightarrow$ Assume $a\ne 0$, $a$ is not a unit $\in R$ is irreducible

Then $a=bc$ where at least one of $b$ or $c$ is a unit

Therefore if $b$ is a unit then $c$ is an associate of a and if $c$ is a unit then $b$ is an associate of a

As $a=bc$ then $b|a$ and $c|a$

$b$ and $c$ are either units or associates of $a$ therefore the only divisors of $a$ are units of $R$ or associates of $a\in R$

$\Leftarrow$ Assume $a\in R$ can only be divided by units or associates

Let $b,c\in R$ divide $a$, by assumption $b$ must either be a unit or an associate

If $b$ is a unit then $a$ is irreducible

else $b$ is an associate

This means there exists a unit $u\in R$ where $a=ub$

Therefore $u|a$ so once again a is divided by a unit and is therefor irreducible

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How did you that in a second – Adi Dani Mar 16 '13 at 19:04
Since this proof is part of your question (this is, you ask if it is correct), it should be included in your question, not presented as answer. – robjohn Mar 16 '13 at 19:31
@robjohn Oh ok, I thought as I was trying to answer it it would be an answer, what if I find out it does work would I then move it to the answers section? – Adam Mar 16 '13 at 19:39
Since your question is about the correctness of the proof, not about what the proof is saying, I would not think the proof would actually answer the question. If you or someone else shows that your proof is correct, then that should be in the answers. – robjohn Mar 16 '13 at 19:51
@robjohn The main question is to find a proof which I have done and I think it is correct I'm just note sure if I need to add something extra into it to make it a more complete proof – Adam Mar 16 '13 at 19:59