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").

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    $\begingroup$ 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, 2014 at 3:51

1 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.


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