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I don't understand the last passage in this proof.

We assume that $P(X)\in R[X]$ is irreducible and want to show that it is prime. So we must prove that $P(X)\mid F(X)G(X)$ implies either $P(X)\mid F(X)$ or $P(X)\mid G(X)$ (in $R[X]$). So we assume that $P(X)\mid F(X)G(X)$, i.e., there is a $Q(X)\in R[X]$ such that $P(X)Q(X)=F(X)G(X)$ in $R[X]$ (and consequently in $K[X]$). How the Gauss lemma implies that either $P(X)\mid F(X)$ or $P(X)\mid G(X)$ in $K[X]$?

user26857
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user557
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    Since $P(X)\mid F(X)G(X)$ in $R[X]$ then $P(X)\mid F(X)G(X)$ in $K[X]$. By Gauss' Lemma $P(X)$ is irreducible in $K[X]$, hence prime. (Recall that $K[X]$ is a UFD.) Then $P(X)\mid F(X)$ or $P(X)\mid G(X)$ in $K[X]$. – user26857 Oct 23 '16 at 22:12
  • See this answer for how to prove a more general polynomial primality criterion. – Bill Dubuque Oct 26 '16 at 15:26

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