Theorem: Every regular local ring $(R, \mathfrak{m})$ is a unique factorization domain.

Step 1: Regular local rings have finite global dimension.

We first show that $\operatorname{gl.dim}(R) = \dim R = d$. It suffices to show $\operatorname{pd}_R(k) = d$ where $k = R/\mathfrak{m}$.

Let $x_1, \ldots, x_d$ be a regular system of parameters. The Koszul complex $K_\bullet(x_1, \ldots, x_d)$ is the chain complex $0 \to R \to R^d \to \binom{d}{2} \cdot R \to \cdots \to R^d \to R \to 0$ For a regular local ring, this complex is exact and provides a free resolution of $R/(x_1, \ldots, x_d) = k$. Therefore $\operatorname{pd}_R(k) \leq d$.

The inequality $\operatorname{pd}_R(k) \geq d$ follows since $\operatorname{Ext}^d_R(k, k) \neq 0$ (computed from the Koszul complex as $\operatorname{Ext}^d_R(k, k) \cong k$).

By the Auslander-Buchsbaum-Serre theorem, every finitely generated $R$-module has finite projective dimension $\leq d$.

Step 2: Every height-one prime is principal (Kaplansky criterion).

By Kaplansky's theorem, a Noetherian domain is a UFD if and only if every height-one prime ideal is principal.

Let $\mathfrak{p}$ be a prime ideal of height $1$. Since $R$ has finite global dimension, $\operatorname{pd}_R(R/\mathfrak{p}) < \infty$. By the Auslander-Buchsbaum formula: $\operatorname{pd}_R(R/\mathfrak{p}) = \operatorname{depth}(R) - \operatorname{depth}(R/\mathfrak{p})$

Since $R$ is Cohen-Macaulay (regular implies Cohen-Macaulay), $\operatorname{depth}(R) = d$. For $R/\mathfrak{p}$: since $\operatorname{ht}(\mathfrak{p}) = 1$ and $R/\mathfrak{p}$ is a domain of dimension $d - 1$ (by the dimension formula for regular rings), and $R/\mathfrak{p}$ is Cohen-Macaulay as a quotient of a regular ring by a height-$1$ prime, we get $\operatorname{depth}(R/\mathfrak{p}) = d - 1$.

Therefore $\operatorname{pd}_R(R/\mathfrak{p}) = d - (d-1) = 1$.

Step 3: Projective dimension $1$ implies principal.

Since $\operatorname{pd}_R(R/\mathfrak{p}) = 1$, there is a free resolution $0 \to R^m \xrightarrow{\varphi} R^n \to R/\mathfrak{p} \to 0$

This means $\mathfrak{p}$ is the image of $\varphi$ in $R^n / \operatorname{im}(\text{surjection onto } R/\mathfrak{p})$. More precisely, applying $\operatorname{Hom}_R(-, R)$ and using the fact that $\operatorname{pd} = 1$:

We have the presentation $0 \to F_1 \to F_0 \to R/\mathfrak{p} \to 0$ with $F_0, F_1$ free. The rank of $F_0$ must be $1$ (since $R/\mathfrak{p}$ is cyclic), giving: $0 \to R^m \to R \to R/\mathfrak{p} \to 0$ This means $\mathfrak{p}$ is the image of $R^m \to R$, i.e., $\mathfrak{p}$ is generated by the images of the standard basis vectors. But $\mathfrak{p}$ is prime of height $1$ in a Noetherian domain, so it is minimal, and the map $R^m \to R$ has image $\mathfrak{p}$.

Since $\mathfrak{p}$ is prime and $R$ is a domain, we can show $m = 1$: the alternating sum of ranks gives $\operatorname{rank}(R/\mathfrak{p}) = 1 - m$ as a rational number, but $R/\mathfrak{p}$ has rank $0$ as an $R$-module (it is a torsion module since $\mathfrak{p} \neq 0$). Wait — let us use a cleaner argument.

Since $R$ is a domain and $\mathfrak{p} \neq 0$, tensoring with the fraction field $K$ gives $K^m \cong K$ (since $R/\mathfrak{p}$ is torsion), so $m = 1$. Therefore $\mathfrak{p} \cong R$ as an $R$-module, meaning $\mathfrak{p}$ is a principal ideal.

Step 4: Conclusion.

Every height-$1$ prime of $R$ is principal. By Kaplansky's criterion, $R$ is a UFD. $\square$