= Solution
To deduce the final assertion, use the characterization that an <integral> domain is a <unique factorization domain> when it is atomic and every irreducible element is prime. Factoring the contents in $R$ and then the primitive parts by degree shows that $R[X]$ is atomic.
An irreducible constant of $R$ is prime in $R[X]$. Every irreducible <polynomial> of positive degree is, up to a constant unit, primitive. Part (i) makes it irreducible in $F[X]$; because $F[X]$ is a principal <ideal> domain, it is prime there, and part (ii) makes it prime in $R[X]$. Thus every irreducible of $R[X]$ is prime, and
$$
\boxed{R[X]\text{ is a unique factorization domain}.}
$$
This is the <polynomial ring over a unique factorization domain> theorem.
Back to article page
