Regular local rings are unique factorization domains

We prove by induction on $d$ that every height-one prime is principal. If $d=0$, $R$ is a field by [F1] and there is nothing to prove. For $d>0$, choose $x\in \mathfrak{m}\setminus {\mathfrak{m}}^2$. By [F1], $R\mathrm{/}(x)$ is a domain, so $x$ is a prime element. A height-one prime containing $x$ equals $(x)$ by [F3]. Fix instead a height-one prime $\mathfrak{p}$ with $x\mathrm{\notin }\mathfrak{p}$.

Assume as induction hypothesis that every height-one prime in a regular local ring of dimension less than $d$ is principal. For every prime $\mathfrak{q}$ of $R$ not containing $x$, $\mathfrak{q}\mathrm{\ne }\mathfrak{m}$ and $\dim R_{\mathfrak{q}}<d$. If $\mathfrak{p}\subset ̸\mathfrak{q}$, the ideal $\mathfrak{p}{R}_{\mathfrak{q}}$ is the unit ideal. Otherwise it is a height-one prime in the regular local ring $R_{\mathfrak{q}}$ and is principal by induction. Thus ${\mathfrak{p}}_x$ is a finite, locally free rank-one $R_x$-module. It is flat by [F3], projective by [F3], and invertible (local multiplication with the dual is an isomorphism, hence so globally). By [F2], ${\mathfrak{p}}_x=(y)$ for some $y\in R_x$. Write $y=x^{-m}f$ with $f\in \mathfrak{p}$, by clearing denominators in the localized ideal.

Every nonzero nonunit in a Noetherian domain factors into irreducibles: a factorization obstruction would give a strict ascending chain of principal ideals by repeatedly splitting a nonirreducible factor, contradicting the ascending chain condition. Factor $f$ and choose an irreducible factor $a\in \mathfrak{p}$. In $R_x$ it is a nonunit dividing $y$, while $y$ generates a prime ideal. Since a prime element is irreducible, $a$ and $y$ are associates in $R_x$. Consequently $(a)R_x$ is prime. The element $a$ is not associated to $x$, since $x\mathrm{\notin }\mathfrak{p}$, so the primeness of $x$ implies $x\nmid a$. If $x^{r}b\in (a)$, write $x^{r}b=ac$; primeness of $x$ forces $x\mid c$, and cancelling $x$ repeatedly gives $b\in (a)$. Hence $(a)R_x\cap R=(a)$. Since contraction preserves prime ideals, $(a)$ is prime in $R$. It is nonzero and contained in the height-one prime $\mathfrak{p}$, so $(a)=\mathfrak{p}$. This completes the induction.

For any irreducible $b\in R$, choose a prime minimal over $(b)$. By [F3] it has height one, and by step 3.1 it is $(c)$. Then $b=ct$, and irreducibility of $b$ implies that $t$ is a unit. Thus $(b)=(c)$ is prime. Factorization exists by the ascending-chain argument of step 3.1, and uniqueness follows by cancelling prime irreducible factors one at a time. This proves that $R$ is a UFD. A smooth finite-type scheme has regular local rings by the geometric-regularity definition of smoothness; applying the result at each point gives local factoriality. AC is used through [F1]–[F3] and the chosen minimal prime.