Subring of a UFD need not be a UFD

Question

Let $R$ be a UFD with $1$, are the following two statements true?

• If $S$ is a subring of $R$, then $S$ is also a UFD;

• if $S$ is a subring of $R$, and $1\in S$, then $S$ is UFD.

Answer 1

Not in general. For example, note that $R=\Bbb C$ is a field, so trivially a UFD. However, consider $S=\Bbb Z[2i]:=\{x+2iy:x,y\in\Bbb Z\}.$ One can show fairly easily that $S$ is a subring of $R$ with unity. However, while $2$ is irreducible in $S,$ it is not prime in $S$. Hence, while $S$ is an integral domain (as a subring with unity of an integral domain), it is not a UFD. See here for more information on how we might go about proving this (and feel free to ask me questions if you're stuck).

Answer 2

Pick any quadratic number ring UFD $\,R = \Bbb Z[\sqrt n]\,$ and let $\,S = \Bbb Z[2\sqrt n]\subset R.\,$ Notice that $\,w = (2\sqrt{n})/2 = \sqrt{n}\,$ is a fraction over $S$ and $\,w\not\in S\,$ but $w$ is a root of the monic polynomial $\,x^2 - n, \,$ contra the Rational Root Test (which is valid in any UFD). So $S$ is not a UFD.

Answer 3

Generalizing the examples given by Bill Dubuque and Cameron Buie, we can use the fact that a UFD is normal (that is, it is integrally closed in its field of fractions) to give us a strategy for finding counterexamples.

Namely, take some UFD $R$, consider its field of fractions $K$, and then find a (unital) subring $S\subseteq R$ such that the field of fractions of $S$ is still $K$, and $S$ is not integrally closed, i.e., there is $\omega\in K\setminus S$ which satisfies a monic polynomial over $S$.

This gives us the counterexamples from both of the other examples. Cameron Buie's has $R=\Bbb{Z}[i]$, $S=\Bbb{Z}[2i]$, $\omega=i$. Bill Dubuque's has $R=\Bbb{Z}[\sqrt{n}]$, $S=\Bbb{Z}[2\sqrt{n}]$, and $\omega = \sqrt{n}$.

Both of these examples are of rings of integers, however, we can also find a subring of a polynomial ring that is not a UFD via this strategy. Consider $R=k[t]$, $S=k[t^2,t^3]$, $\omega = t$. We have $\omega^2-t^2=0$, so $\omega$ is integral over $S$. Note that $t$ is in the fraction field of $S$, since $t=\frac{t^3}{t^2}$.