1
$\begingroup$

Show that, in a domain, every associate of an atom is an atom.

An atom is the same thing as an irreducible element.

I think these two facts will be important to prove this statement:

  1. A nonunit is an atom if and only if it cannot be written as a product of two nonzero nonunits

  2. Two elements are associates if and only if one is a unit multiple of the other.

I just need help with the actual proof writing. I don't know how to convert this information into a nice flowing proof. Any advice would be appreciated!

I'm also a little confused because of this link Irreducible elements are not associates

$\endgroup$
2
$\begingroup$

By $(2)$ an associate of an atom $\,p\,$ must be a unit multiple $\,up.\\$ Ifthis associate were reducible then $\,up = ab\,$ so $\,p = (u^{-1}a) b\,$ is reducible, contradiction.

Generally since the relation of divisibility is preserved by unit scalings, so too are pure divisibility properties such are irreducibility, primality. etc.

$\endgroup$
1
$\begingroup$

Let $p$ be an atom and $u$ an unit. If $pu$ weren't an atom, we'd have $pu=bc$, and $p=b(cu^{-1})$, a contradiction.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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