Regular local rings are UFDs

Remarks

This is the Auslander–Buchsbaum theorem. Proof idea: $R$ is a domain by regular local rings are domains; one argues by induction on $\dim R$. Pick $x \in \mathfrak m \setminus \mathfrak m^2$; then $x$ is prime, since $R/xR$ is regular, hence a domain. By Nagata's criterion, it is enough to show that $R[1/x]$ is a UFD. Its localizations at primes are regular of smaller dimension, hence UFDs by induction. So each height one prime of $R[1/x]$ is locally principal. Finite free resolutions, which exist by Serre's theorem, then show that it is principal.

Example: the power series ring $k[\![x_1, \ldots, x_n]\!]$ over a field $k$ and the localizations $k[x_1, \ldots, x_n]_{\mathfrak p}$ of a polynomial ring at prime ideals are UFDs.

The regularity hypothesis matters. For a field $k$, the local ring $R = k[x, y, z]_{(x, y, z)}/(xy - z^2)$ is a domain of dimension $2$ and embedding dimension $3$. It is not a UFD, since $xy = z \cdot z$ are two factorizations of the same element into irreducible elements that are not associate.

Going deeper

For a student

The proof sketched above relies on homological algebra, through Serre's theorem and finite free resolutions. In geometry it means that on a smooth variety every hypersurface through a point is defined locally by one equation, so Weil divisors and Cartier divisors agree. It is used with the divisor class group: the class group of a regular local ring is trivial.