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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").
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.
