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Let $R$ be an integral domain , $p \in R$ be a prime element , then is $p$ also a prime element in $R[x]$ ?

I know it is true if $R$ is a UFD , because then $R[x]$ is a UFD and showing $p$ is prime is equivalent to showing it is irreducible .

But what of $R$ is not a UFD ? Please help . Thanks in advance

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Yes, use the fact that $R[x]/(p) = (R/p)[x]$. Then $R/p$ is an integral domain because $p$ is prime so the polynomial ring $(R/p)[x]$ is as well. And by definition $\mathfrak{p}$ an ideal of a ring $S$ is prime iff $S/\mathfrak{p}$ is an integral domain.

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It is instructive to give a direct proof of $\,p\nmid A,B\,\Rightarrow\,p\nmid AB.\,$ Since $\,p\nmid A,B\,$ when reduced mod $p$ both have lead coefs $\,\color{#0a0}{a,b\not\equiv 0}\,$ so $AB$ has lead coef $\,\color{#c00}{ab\not\equiv 0}\,$ (by $p$ prime), hence $AB\not\equiv 0,\,$ i.e.

$\qquad\qquad{\rm mod}\ p\!: \ \ \ \begin{eqnarray} &&\ 0\ \not\equiv\ A\ \equiv\, \color{#0a0}a\, x^j\! + \:\cdots,\quad\ \ \ \color{#0a0}{a\not\equiv 0}\\ &&\ 0\ \not\equiv\ B\ \equiv\, \color{#0a0}b\, x^k\! + \:\cdots,\quad\ \ \ \color{#0a0}{b\not\equiv 0}\\ \Rightarrow\,\ &&0 \not\equiv AB \equiv \color{#c00}{ab}\ x^{j+k}\! + \:\cdots,\, \color{#c00}{ab\not\equiv 0}\end{eqnarray}$

i.e. primes $\:p\in R\:$ remain prime in $\:R[x]\:$ because the prime divisor property $\,\color{#0a0}{p\nmid a,b}\,\Rightarrow\ \color{#c00}{p\nmid ab}\,$ persists when multiplying leading coefficients. This is one form of Gauss's Lemma.

This is an elementwise view of the proof sketched in Adam's answer (it is simply the common proof of $\,D\,$ domain $\Rightarrow\,D[x]\,$ domain, in the special case that $\,D \cong R/p \cong R\bmod p,\,$ for $p$ prime).

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