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Question: If $R$ is a euclidean ring and $\pi\in R$ is irreducible, prove that $\pi\mid\alpha\beta$ implies $\pi\mid\alpha$ or $\pi\mid\beta$.

A solution is to prove all euclidean rings are PIDs, then prove Euclid's Lemma is true in PIDs. However, is there a solution need not to prove $R$ is a PID and use the definition of euclidean ring directly? And the solution may not use $gcd$ because the existence of $gcd$ doesn't confirm without PID's help.

Forgive my poor English.

Bill Dubuque
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Sailoong
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Hint $ $ If not $\exists$ irred $\,p\!:\ pc = ab,\ p\nmid a,b.\,$ From such counterexamples with min $\,|p|\,$ choose one with min $\,|a|.\,$ Then $\,|a| < |p|,\ $ else $\,a\mapsto \bar a\, =\, a\ {\rm mod}\ p\, =\, a - rp\ne 0\,$ is a counterexample with $\,|\bar a| < |a|.\,$ Now $\,p\nmid b\,\Rightarrow\,a\,$ is a nonunit, so it has an irred factor $\,q\mid a.$ By $\,|p|\,$ minimal and $\,|q|<|p|$ we infer $\,q\mid pc,\ q\nmid p\,\Rightarrow\, q\mid c,\ $ so $\ p(c/q) = (a/q)b\,$ is a counterexample with $\,|a/q| < |a|$ contra hypothesis.

Bill Dubuque
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    I clicked this answer on the front page by curiosity, and it did not take more than 30 seconds to realize that the answer is written in a cryptic shorthand rather than standard mathematical English. – Carl Mummert May 28 '14 at 01:47
  • @CarlMummert So you downvoted because because I used $,\exists,$ instead of "there exists" and abbreviated irreducible? How nice of you. Don't you realize that doing so might mislead students into thinking that there might be something wrong with the answer? – Bill Dubuque May 28 '14 at 02:29
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    I voted because I do think there is something wrong with the answer: I do not think it is useful in its present form. If a student submitted an answer such as this in an undergraduate class, what grade would it receive? I fear this style of presentation might "mislead" students into thinking that this is how mathematics is usually written. Of course, I am certain the idea can be improved into a useful answer. But it is not there yet. I think that is a complete explanation of my reasoning. – Carl Mummert May 28 '14 at 02:35
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    @CarlMummert It is not meant to be an answer submitted by a student. Rather, it is meant to be a hint sketched by a teacher, not unlike a hint one might find in a textbook. In any case, if there is something you do not understand then please tell me precisely what that is and I will be happy to elaborate. – Bill Dubuque May 28 '14 at 03:02