How can we show that any irreducible element in $R[x]$ is prime where R is a unique factorization domain? Note that we should not assume that R[x] is also a unique factorization domain as the question itself is an intermediate step in the proof. A hint is to look at the field of fractions $F$ or $R$ and use the fact that $F[x]$ is a UFD. However I am not sure why this would help us. I think there might be some properties of $F[x]$ that I am unaware of.
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Take a look at "Gauss' Lemma"! – FlipTack Oct 22 '22 at 16:14
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See Nagata's Lemma for a more general perspective. – Bill Dubuque Oct 22 '22 at 16:22
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@FlipTack so gauss lemma gives us that if f is irreducible in R[x] then it is also irreducible in F[x], and since F[x] is a UFD, every irreducible element is prime, now can we conclude that being prime in F[x] implies prime in R[x]? – david h Oct 23 '22 at 13:28
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There's a bit of fiddling around with the definition of a 'primitive polynomial', and proving that a few logical results actually hold, but yes that is the general idea :) – FlipTack Oct 23 '22 at 16:52
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For intuition, you can consider R[X] and Q[X], and it basically comes down to 'clearing denominators' of polynomials in Q[X] to get polynomials in R[X]. – FlipTack Oct 23 '22 at 16:53