5
$\begingroup$

In a UFD, primeness and irreducibility are equivalent. In particular, every Euclidean ring which is an integral domain is a UFD.

My question is this: is it possible to prove that "irreducible implies prime" directly (I mean without invoking "Euclidean implies PID implies UFD").

$\endgroup$
  • 1
    $\begingroup$ If you can get to the existence of gcd, it can be finished from there. However from gcd I think is only a small step to PID anyway. $\endgroup$ – coffeemath Nov 8 '14 at 3:51
2
$\begingroup$

It seems most efficient to conclude "irreducible implies prime" after showing ED implies PID. There it is easy to see that a nonprime is properly contained in another proper ideal, whence you can see the nonprime is not irreducible.

$\endgroup$

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.