Under AC, Cartier and Weil divisors agree on a locally factorial Noetherian integral scheme

Irreducible elements of a UFD are prime. Let $R$ be a UFD, let $\pi \in R$ be irreducible, and let $a,b\in R$ with $\pi \mid ab$, say $ab=\pi c$. If $a=0$ or $b=0$ then $\pi \mid a$ or $\pi \mid b$; if $a$ is a unit then $b=a^{-1}ab$ is divisible by $\pi$, and if $b$ is a unit then $a=b^{-1}ab$ is. Otherwise $a,b$ are nonzero nonunits, hence $ab$ is a nonzero nonunit and $c\mathrm{\ne }0$. By [F2] factor $a=p_1\cdots p_m$ and $b=q_1\cdots q_n$ into irreducibles. If $c$ is a unit then $\pi =c^{-1}{p}_1\cdots p_m{q}_1\cdots q_n$ is associate to a product of $m+n$ irreducibles, and since $\pi$ is irreducible uniqueness in [F2] forces $m+n=1$, so $\pi$ is associate to the single factor, which lies among the $p_i$ or the $q_j$; if $c$ is a nonunit, factor $c=r_1\cdots r_k$ and compare the two factorisations $p_1\cdots p_m{q}_1\cdots q_n=\pi r_1\cdots r_k$ of $ab$, so that uniqueness again makes $\pi$ associate to some $p_i$ or $q_j$. In either case $\pi \mid a$ or $\pi \mid b$, so $\pi$ is prime and $(\pi )$ is a nonzero prime ideal.

A UFD is integrally closed. Let $R$ be a UFD with fraction field $K$ and let $x\in K$ be integral over $R$; by [F3] write $x=a\mathrm{/}b$ with $a,b\in R$, $b\mathrm{\ne }0$, subject to a monic relation $x^n+c_{n-1}{x}^{n-1}+\cdots +c_{1}x+c_0=0$ with $c_j\in R$ and $n\ge 1$. For a nonzero $b_0\in R$ let $N(b_0)$ be the number of irreducible factors in a factorisation of $b_0$, and set $N(b_0)=0$ when $b_0$ is a unit; by the uniqueness part of [F2] the number $N(b_0)$ is well defined. Among all representations $x=a^{\mathrm{\prime }}\mathrm{/}{b}^{\mathrm{\prime }}$ with $b^{\mathrm{\prime }}\mathrm{\ne }0$ choose one with $N(b)$ minimal. If $b$ is a unit then $x=a{b}^{-1}\in R$; assume it is not. If $a=0$ then $x=0\in R$, so assume $a\mathrm{\ne }0$. The nonzero nonunit $b$ has an irreducible factor $\pi$ by [F2], say $b=\pi b_1$ with $b_1\mathrm{\ne }0$; multiplying the monic relation by $b^n$ gives $a^n=-b(c_{n-1}{a}^{n-1}+c_{n-2}{a}^{n-2}b+\cdots +c_0{b}^{n-1})$, so $\pi \mid a^n$, and since $\pi$ is prime by step 1.1 we get $\pi \mid a$. Writing $a=\pi a_1$ gives $x=a_1\mathrm{/}{b}_1$, and $N(b_1)=N(b)-1$: if $b_1$ is a unit then $b$ is associate to $\pi$, hence irreducible (a factorisation $u\pi =cd$ gives $\pi =(u^{-1}c)d$, so $u^{-1}c$ or $d$ is a unit), and $N(b)=1$; otherwise appending a factorisation of $b_1$ to $\pi$ factorises $b$. This contradicts the minimality of $N(b)$, so $b$ is a unit and $x\in R$: every UFD is integrally closed.

Height one primes of a UFD are principal. Let $R$ be a UFD and let $\mathfrak{p}\subseteq R$ be a prime ideal of height one. Then $\mathfrak{p}\mathrm{\ne }(0)$, so choose $0\mathrm{\ne }a\in \mathfrak{p}$. The element $a$ is a nonzero nonunit, so by [F2] it factors as $a={\pi }_1\cdots {\pi }_m$ with $m\ge 1$ and all ${\pi }_i$ irreducible; since $\mathfrak{p}$ is prime, some ${\pi }_i$ lies in $\mathfrak{p}$. The ideal $({\pi }_i)$ is then contained in $\mathfrak{p}$, is nonzero, and is prime by step 1.1. Were the inclusion strict, the chain $(0)⊊({\pi }_i)⊊\mathfrak{p}$ of prime ideals would force $\dim R_{\mathfrak{p}}\ge 2$, contradicting $\mathrm{ht}\mathfrak{p}=1$ by [F4]. Hence $\mathfrak{p}=({\pi }_i)$ is principal.

$X$ is normal and the cycle and class maps exist. By [F1] every local ring of $X$ is a UFD, hence integrally closed by step 2.1, and hence $X$ is normal because it is Noetherian: every local ring is an integrally closed domain, as required by [F3]. Since $X$ is in addition integral, the implication of [F13] provides Dependent Choice, so the cycle map $\mathrm{cyc}:\mathrm{CaDiv}(X)\to \mathrm{Div}(X)$ and the canonical homomorphism $\mathrm{Pic}(X)\to \mathrm{Cl}(X)$ are defined by [F10], and $\mathrm{cyc}$ is additive.

Prime divisors are locally cut out by nonzerodivisors. Let $Z\subseteq X$ be a prime divisor and fix $x\in X$. If $x\mathrm{\notin }Z$, then $X\setminus Z$ is an open neighbourhood of $x$ (Z is closed), and picking any affine chart of $X$ through $x$ and applying the distinguished-open refinement of [F14] inside that chart produces an affine open $W=\mathrm{Spec}B$ with $x\in W\subseteq X\setminus Z$; then $Z\cap W=\mathrm{\varnothing }=\mathrm{Spec}(B\mathrm{/}1\cdot B)$ by [F6], and $1$ is a nonzerodivisor. Now suppose $x\in Z$, choose a Noetherian affine chart $U=\mathrm{Spec}A$ containing $x$ from the cover in [F1], let $\mathfrak{q}\subseteq A$ be the prime of $x$ and $\mathfrak{p}=I_Z(U)$ the prime with $Z\cap U=\mathrm{Spec}(A\mathrm{/}\mathfrak{p})$ given by [F6], so that $\mathfrak{p}\subseteq \mathfrak{q}$. The closed subset $Z\cap U$ is a nonempty open subset of the integral scheme $Z$, hence dense by [F14], so its generic point, the point $\xi \in U$ corresponding to $\mathfrak{p}$, is the generic point of $Z$; by [F5] $A_{\mathfrak{p}}={\mathcal{O}}_{X,\xi }$, and therefore $\mathrm{ht}\mathfrak{p}=\dim A_{\mathfrak{p}}=\dim {\mathcal{O}}_{X,\xi }=1$ by [F4] and [F7]. The stalk $A_{\mathfrak{q}}={\mathcal{O}}_{X,x}$ is a UFD by [F1]; the natural map $A_{\mathfrak{p}}\to (A_{\mathfrak{q}}{)}_{\mathfrak{p}{A}_{\mathfrak{q}}}$ is an isomorphism, since both rings are the localisation of $A$ at the multiplicative set $A\setminus \mathfrak{p}$ (every denominator $s\mathrm{\notin }\mathfrak{q}$ also lies outside $\mathfrak{p}$, because $\mathfrak{p}\subseteq \mathfrak{q}$; and an element $a\mathrm{/}s$ of $A_{\mathfrak{q}}$ lies outside $\mathfrak{p}{A}_{\mathfrak{q}}$ exactly when $a\mathrm{\notin }\mathfrak{p}$, so the second localization inverts precisely the remaining numerators outside $\mathfrak{p}$), so $\mathrm{ht}(\mathfrak{p}{A}_{\mathfrak{q}})=\dim (A_{\mathfrak{q}}{)}_{\mathfrak{p}{A}_{\mathfrak{q}}}=\dim A_{\mathfrak{p}}=1$ by [F4]. Applying step 2.2 in the UFD $A_{\mathfrak{q}}$ gives $\mathfrak{p}{A}_{\mathfrak{q}}=f{A}_{\mathfrak{q}}$ for some $f\in A_{\mathfrak{q}}$; write $f=a\mathrm{/}s$ with $a\in A$, $s\mathrm{\notin }\mathfrak{q}$. Then $a=sf\in \mathfrak{p}{A}_{\mathfrak{q}}$ and $a{A}_{\mathfrak{q}}=\mathfrak{p}{A}_{\mathfrak{q}}$. By [F4] the ideal $\mathfrak{p}=(g_1,\dots ,g_m)$ is finitely generated, and each $g_i\in \mathfrak{p}{A}_{\mathfrak{q}}=a{A}_{\mathfrak{q}}$, so there are $s_i\mathrm{\notin }\mathfrak{q}$ with $s_i{g}_i\in aA$; also $a\in \mathfrak{p}{A}_{\mathfrak{q}}$ gives $v\mathrm{\notin }\mathfrak{q}$ with $va\in \mathfrak{p}$. Put $t=v{s}_1\cdots s_m\mathrm{\notin }\mathfrak{q}$, $B=A_t$ and $W=D(t)=\mathrm{Spec}B$, an affine open with $x\in W$. In $B$ one has $a\in \mathfrak{p}B$ and $\mathfrak{p}B=aB$: each $g_i=(s_i{g}_i)\mathrm{/}{s}_i$ lies in $aB$, so $\mathfrak{p}B\subseteq aB$, while $a\in \mathfrak{p}B$ gives the reverse inclusion. Finally $a\mathrm{\ne }0$ because $a{A}_{\mathfrak{q}}=\mathfrak{p}{A}_{\mathfrak{q}}\mathrm{\ne }0$, and $B$ is a domain, so $a$ is a nonzerodivisor; applying [F6] on the affine open $W$ with $I_Z(W)=\mathfrak{p}B=aB$ gives $Z\cap W=\mathrm{Spec}(B\mathrm{/}aB)$.

Every prime divisor is an effective Cartier divisor with $\mathrm{cyc}(D_Z)=[Z]$. Step 3.2 checked the hypothesis of [F9] for the closed subscheme $Z$ (closed by [F7]): every point of $X$ has an affine open neighbourhood $W=\mathrm{Spec}B$ with $Z\cap W=\mathrm{Spec}(B\mathrm{/}aB)$ for a nonzerodivisor $a\in B$. Hence $Z$ carries an effective Cartier divisor $D_Z$ with $I_{D_Z}=I_Z$. To compute $\mathrm{cyc}(D_Z)$, let $Z^{\mathrm{\prime }}$ be a prime divisor with generic point ${\xi }^{\mathrm{\prime }}$. If ${\xi }^{\mathrm{\prime }}\mathrm{\notin }Z$ then $I_Z$ agrees with ${\mathcal{O}}_X$ on the open neighbourhood $X\setminus Z$ of ${\xi }^{\mathrm{\prime }}$ by [F6], so the local equation of $D_Z$ near ${\xi }^{\mathrm{\prime }}$ is a unit of ${\mathcal{O}}_{X,{\xi }^{\mathrm{\prime }}}$, and its order is $0$ by [F11]: the coefficient of $Z^{\mathrm{\prime }}$ in $\mathrm{cyc}(D_Z)$ vanishes. If ${\xi }^{\mathrm{\prime }}\in Z$ then $Z^{\mathrm{\prime }}=\stackrel{‾}{\{{\xi }^{\mathrm{\prime }}\}}\subseteq Z$; to see that $Z^{\mathrm{\prime }}=Z$, take an affine open $U=\mathrm{Spec}A$ meeting $Z^{\mathrm{\prime }}$, write $Z\cap U=\mathrm{Spec}(A\mathrm{/}\mathfrak{p})$ and $Z^{\mathrm{\prime }}\cap U=\mathrm{Spec}(A\mathrm{/}{\mathfrak{p}}^{\mathrm{\prime }})$ with primes $\mathfrak{p}\subseteq {\mathfrak{p}}^{\mathrm{\prime }}$ by [F6], note that the generic points $\xi ,{\xi }^{\mathrm{\prime }}$ both lie in $U$ (each $Z\cap U$ and $Z^{\mathrm{\prime }}\cap U$ is a nonempty open subset of the corresponding integral scheme, hence dense, and contains its generic point), and compute as in step 3.2 that $\mathrm{ht}\mathfrak{p}=\dim {\mathcal{O}}_{X,\xi }=1$ and $\mathrm{ht}{\mathfrak{p}}^{\mathrm{\prime }}=\dim {\mathcal{O}}_{X,{\xi }^{\mathrm{\prime }}}=1$; a strict inclusion $\mathfrak{p}⊊{\mathfrak{p}}^{\mathrm{\prime }}$ with $\mathfrak{p}\mathrm{\ne }0$ would force $\mathrm{ht}{\mathfrak{p}}^{\mathrm{\prime }}\ge 2$, so $\mathfrak{p}={\mathfrak{p}}^{\mathrm{\prime }}$, the two closed subsets $Z\cap U=Z^{\mathrm{\prime }}\cap U$ coincide, and since both $Z$ and $Z^{\mathrm{\prime }}$ are the closure of this common nonempty open subset by [F14], $Z=Z^{\mathrm{\prime }}$. Consequently the only prime divisor whose generic point lies in $Z$ is $Z$ itself. At $\xi$, the stalk $I_{Z,\xi }$ is the kernel of ${\mathcal{O}}_{X,\xi }\to {\mathcal{O}}_{Z,\xi }$ by [F6], and ${\mathcal{O}}_{Z,\xi }$ is a field because $\xi$ is the generic point of the integral scheme $Z$, so $I_{Z,\xi }$ is the maximal ideal of the one-dimensional local domain ${\mathcal{O}}_{X,\xi }$; the germ of any local equation of $D_Z$ at $\xi$ generates this ideal by [F8] and [F9], hence equals a unit times a generator of the maximal ideal, and its $v_{\xi }$-value is $1$ by [F11]. Thus $\mathrm{cyc}(D_Z)$ has coefficient $1$ at $Z$ and $0$ at every other prime divisor, so $\mathrm{cyc}(D_Z)=[Z]$ by [F7].

The cycle map is surjective. Let $D={\sum }_i{n}_i[Z_i]$ be a Weil divisor on $X$; by [F7] the sum has finite support, so it is a finite combination of prime divisors. For each $i$ step 4.1 provides the effective Cartier divisor $D_{Z_i}$ with $\mathrm{cyc}(D_{Z_i})=[Z_i]$, and $D^{\mathrm{\prime }}={\sum }_i{n}_i{D}_{Z_i}$ is a Cartier divisor by [F8]. Since $\mathrm{cyc}$ is a homomorphism of abelian groups by [F10], $\mathrm{cyc}(D^{\mathrm{\prime }})={\sum }_i{n}_i\mathrm{cyc}(D_{Z_i})={\sum }_i{n}_i[Z_i]=D$. Hence every Weil divisor is the associated Weil divisor of a Cartier divisor: the cycle map is surjective, and every Weil divisor is locally Cartier, represented near each point by the local equations of such a Cartier divisor on the members of its representing cover.

The cycle map is an isomorphism. By step 3.1, $X$ is normal and the cycle map is a homomorphism. Its injectivity follows from [F12], and step 5.1 proves surjectivity. Thus it is an isomorphism of divisor groups by [F14].

The canonical map $\mathrm{Pic}(X)\to \mathrm{Cl}(X)$ is an isomorphism. By step 3.1 the canonical homomorphism $\phi :\mathrm{Pic}(X)\to \mathrm{Cl}(X)$ exists, and it is injective by [F12]. For surjectivity let $c\in \mathrm{Cl}(X)=\mathrm{Div}(X)\mathrm{/}P(X)$ be the class of a Weil divisor $D$; by step 5.1 there is a Cartier divisor $D^{\mathrm{\prime }}$ with $\mathrm{cyc}(D^{\mathrm{\prime }})=D$, and by [F10] the class $\phi ([{\mathcal{O}}_X(D^{\mathrm{\prime }})])$ is the class of $\mathrm{cyc}(D^{\mathrm{\prime }})=D$, namely $c$. So $\phi$ is bijective, hence an isomorphism of groups by [F14], and $\mathrm{Pic}(X)\cong \mathrm{Cl}(X)$.