Can the concept of field extensions be applied equally well to UFDs?

Question

In a nutshell, a field extension is where you take a polynomial $p(x)$ that is irreducible in some field $F$, then define $\alpha$ as a root of $p$, then add $\alpha$ to $F$, then add the minimum number of additional elements to $f$ to maintain its field structure.

Can this still be done if $F$ is merely a UFD rather than a field? Does something fundamental break down in the construction of a "UFD extension" that goes through in a field extension? If not, why do people talk about field extensions when "UFD extension" seems more general?

Answer 1

Given a ring $R$ and an irreducible polynomial $p(x)\in R[x]$, you can construct the ring $R[\alpha]$ in the same way as you would a field extension of the form $F(\alpha)$, namely $R[\alpha]\cong R[x]/(p(x)).$

Even if $R$ is a UFD, there's no guarantee that $R[\alpha]$ will be a UFD. For example, take $R=\mathbb{Z}$ and $p(x) = x^2+5$. Then the resulting ring is $\mathbb{Z}[i\sqrt{5}]$, which is not a UFD: in this ring, $6$ factors as both $2\cdot 3$ and as $(1+i\sqrt{5})(1-i\sqrt{5})$.

Answer 2

A simple example of a non-UFD extension of a UFD is $\, \Bbb Z \subset R =\Bbb Z[x^2,x^3] \subset \Bbb Z[x].\,$ By $\,x\not\in R\,$ both $\,x^2,\, x^3$ are irreducible, but not prime. Indeed, the factorization $\, (x^2)^3\! = (x^3)^2$ is not unique.