In a Euclidean ring, could we prove irreducible implies prime directly?

Question

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").

Answer 1

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.