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I have to prove the following that every associate of an irreducible element is irreducible.

I am working in an integral domain so I will use the theorem which states that whenever $p$ is a non zero non unit element and when $p=rs$ then either $r$ or $s$ is a unit.

Proof:

Let $p$ be an irreducible element. with $p=rs$ then $r \ or \ s \ is \ a \ unit$

Now let the associate of the irreducible element $p$ be $a$ s.t. $a \in R$ then $a=pu$ where u is a unit. Hence $a=rsu$ with either r or s being units.

If $r$ is a nonunit then $a=s$ and then $s$ must be a unit from theorem.

If $s$ is a nonunit then $a=r$ where r must be a unit from theorem.

So from the theorem $a$ the associate must be irreducible?

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I think you are better off by reformulating everything in terms of divisibility.

Namely, a non-zero, non-unit element $p$ is irreducible if whenever $r \mid p$, then either $r$ is a unit, or $r$ is associate to $p$.

Now recall that $q$ is associate to $p$ iff $p \mid q$ and $q \mid p$.

So let $p$ be irreducible, and consider $r \mid q$. Since $q \mid p$, you have $r \mid p$, so either $r$ is a unit, or it is associate to $p$, and thus to $q$.


If you want to argue your way, it goes like this. Suppose $a = r s$, so that $p = r (s u^{-1})$. Then either $r$ is a unit, or $s u^{-1}$ is a unit, and thus $s = (s u^{-1}) u$ is also a unit, as the product of two units.

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  • $\begingroup$ thanks your method looks more elegant, do you see any flaws so if I fix them my proof would work, or is it already working? $\endgroup$ – user420309 Apr 24 '18 at 13:06
  • $\begingroup$ because it seems like you're using some facts which I do not know about $\endgroup$ – user420309 Apr 24 '18 at 13:27
  • $\begingroup$ @Djhoe, I have a added a proof along your lines. $\endgroup$ – Andreas Caranti Apr 24 '18 at 13:50

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