# Prime $p\in R$ remains prime in $R[x]$ (Gauss's Lemma)

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 the polynomial ring $$R[x]$$ 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

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.

It's 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.

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

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

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

Beware  The "primitive" form of Gauss's Lemma depends crucially on $$R$$ being UFD. For if $$R$$ has an atom (irreducible) $$\,p\,$$ that is not prime then there exists $$a,b\,$$ such that $$\,p\mid ab,\ p\nmid a,\ p\nmid b\,$$ so $$\,f= px+a,\,g = px+b\,$$ are primitive but $$\,p\mid fg\,$$ so $$\,fg\,$$ is not primitive, so Gauss's Lemma fails. See here for more.

• The above proof immediately extends from a principal prime ideal $(p)$ to any prime ideal $P$ in a commutative ring $R$ - just replace $\,p\mid a\,$ by $\,a\in P\,$ etc. It is an elementwise form of the standard structural proof that reduces to the trivial domain case by factoring out $P[x]$. Apr 14 at 0:10