Cartier divisors on a normal Noetherian scheme give Weil divisors

Germs of the local equations at prime divisors are units. Let $Z\subseteq X$ be a prime divisor with generic point $\xi$ and let $i$ with $\xi \in U_i$. By [F3] the stalk ${\mathcal{K}}_{X,\xi }$ is the fraction field of the discrete valuation ring ${\mathcal{O}}_{X,\xi }$. The germ map carries the unit $f_i$ to a unit $f_{i,\xi }$ by [F4], so $f_{i,\xi }\mathrm{\ne }0$ and its normalized valuation $v_{\xi }(f_{i,\xi })$ is defined.

The value is independent of the equation and of the datum. If $\xi \in U_i\cap U_j$, then $u=f_i\mathrm{/}{f}_j$ lies in ${\mathcal{O}}_X^{\times }(U_i\cap U_j)$ by [F1], so its germ $u_{\xi }$ is a unit of ${\mathcal{O}}_{X,\xi }$ and $v_{\xi }(f_{i,\xi })=v_{\xi }(u_{\xi })+v_{\xi }(f_{j,\xi })=v_{\xi }(f_{j,\xi })$. If $\{(V_j,g_j){\}}_{j\in J}$ is any second local-equation datum for $D$, then $f_i\mathrm{/}{g}_j\in {\mathcal{O}}_X^{\times }(U_i\cap V_j)$ for all $i,j$ by [F1]; every generic point $\xi$ lies in some overlap $U_i\cap V_j$, and the same computation gives $v_{\xi }(f_{i,\xi })=v_{\xi }(g_{j,\xi })$. Hence the value ${\mathrm{ord}}_Z(D):=v_{\xi }(f_{i,\xi })$ depends only on $D$ and $Z$, not on the indices or the datum. The additivity of $v_{\xi }$ used in the computation is [F5], and the germ of a unit is again a unit by [F4].

A basic affine neighbourhood carrying a fraction. There are an index $i$, an affine chart $\mathrm{Spec}A$ from the finite cover of [F6], and a basic open $W=D(c)=\mathrm{Spec}B$ with $B=A_c$ such that $x\in W\subseteq U_i\cap \mathrm{Spec}A$, the ring $B$ is Noetherian, and the restriction $f_i{\mathrm{\mid }}_W$ is the image of a fraction $a\mathrm{/}s\in S_X(W{)}^{-1}{\mathcal{O}}_X(W)$. [F2, F6, F7, F8] Choose a chart $\mathrm{Spec}A$ from the finite cover [F6] and an index $i$ with $x\in U_i$, so that $U_i\cap \mathrm{Spec}A$ is an open neighbourhood of $x$. By [F7], choose $c\in A$ with $x\in D(c)\subseteq U_i\cap \mathrm{Spec}A$; then $B=A_c$ is Noetherian by [F8]. The restriction $f_i{\mathrm{\mid }}_{D(c)}$ is a section of the sheafification ${\mathcal{K}}_X=a{\mathcal{P}}_X$, so by [F2] there is a smaller open neighbourhood of $x$ on which it is the image of a presheaf section. Refine that neighbourhood to a basic open $D(d)\subseteq D(c)$ containing $x$, and put $W=D(d)=\mathrm{Spec}{A}_d$. The presheaf section on $W$ is a fraction $a\mathrm{/}s\in {\mathcal{P}}_X(W)=S_X(W{)}^{-1}{\mathcal{O}}_X(W)$. The ring $A_d$ is Noetherian by [F8].

Finitely many supporting prime divisors meet the neighbourhood. In the notation of step 1.3, only finitely many prime divisors $Z$ with $Z\cap W\mathrm{\ne }\mathrm{\varnothing }$ satisfy ${\mathrm{ord}}_Z(D)\mathrm{\ne }0$. Let $Z$ be such a prime divisor and let $\xi$ be its generic point. The scheme $Z$ is integral, so by [F12] its generic point $\xi$ lies in the nonempty open subset $Z\cap W$ of $Z$; let $\mathfrak{p}\subseteq B$ be the prime corresponding to $\xi$, so that ${\mathcal{O}}_{X,\xi }=B_{\mathfrak{p}}$ by [F9]. By [F3] the ring $B_{\mathfrak{p}}$ is a one-dimensional local domain. Write $a_{\xi },s_{\xi }\in B_{\mathfrak{p}}$ for the images of $a$ and $s$. Since $s\in S_X(W)$, its germ at $\xi$ is a nonzerodivisor of the domain $B_{\mathfrak{p}}$, so $s_{\xi }\mathrm{\ne }0$; the germ of the class $a\mathrm{/}s$ at $\xi$ is the fraction $a_{\xi }\mathrm{/}{s}_{\xi }$ and equals $f_{i,\xi }$, which is nonzero by step 1.1, so $a_{\xi }\mathrm{\ne }0$. By step 1.2 and [F3] we have ${\mathrm{ord}}_Z(D)=v_{\xi }(a_{\xi })-v_{\xi }(s_{\xi })$, and $v_{\xi }(a_{\xi })\ge 0$ because $a_{\xi }\in B_{\mathfrak{p}}$ [F5]. Suppose first that $v_{\xi }(a_{\xi })>0$. Then $a_{\xi }\in \mathfrak{p}{B}_{\mathfrak{p}}$, so $a\in \mathfrak{p}$ by [F9]. If a prime $\mathfrak{q}$ satisfies $(a)\subseteq \mathfrak{q}\subseteq \mathfrak{p}$, then $0\mathrm{\ne }{a}_{\xi }\in \mathfrak{q}{B}_{\mathfrak{p}}\subseteq \mathfrak{p}{B}_{\mathfrak{p}}$, and every nonzero prime ideal of the one-dimensional local domain $B_{\mathfrak{p}}$ equals its maximal ideal, so $\mathfrak{q}{B}_{\mathfrak{p}}=\mathfrak{p}{B}_{\mathfrak{p}}$ and hence $\mathfrak{q}=\mathfrak{p}$ by [F9]; thus $\mathfrak{p}$ is a minimal prime of $B\mathrm{/}(a)$. Otherwise $v_{\xi }(a_{\xi })=0$, and ${\mathrm{ord}}_Z(D)\mathrm{\ne }0$ forces $v_{\xi }(s_{\xi })\mathrm{\ne }0$ [F5], so $s\in \mathfrak{p}$ by [F9] and the same argument shows that $\mathfrak{p}$ is a minimal prime of $B\mathrm{/}(s)$. The rings $B\mathrm{/}(a)$ and $B\mathrm{/}(s)$ are Noetherian by [F8] and have finitely many minimal primes by [F10]; distinct prime divisors have distinct generic points and hence distinct primes $\mathfrak{p}$, so the prime divisors meeting $W$ with nonzero coefficient are among the finitely many whose generic point corresponds to a minimal prime of $B\mathrm{/}(a)$ or of $B\mathrm{/}(s)$.

The associated Weil divisor. By steps 1.1 and 1.2 the coefficient ${\mathrm{ord}}_Z(D)=v_{\xi }(f_{i,\xi })$ is a well-defined integer depending only on $D$ and $Z$, and by steps 1.3 and 2.1 the family of prime divisors with nonzero coefficient is locally finite, since every point has a basic affine neighbourhood meeting only finitely many of them; hence the formal sum $\mathrm{cyc}(D):={\sum }_Z{\mathrm{ord}}_Z(D)[Z]$ is a Weil divisor on $X$ by [F11], independent of the charts and local equations used because any two local-equation data give the same coefficients by step 1.2.

The integral case. Suppose that $X$ is integral. Then the global meromorphic units of $X$ are exactly the nonzero elements of $K(X)$ and the Cartier divisor ${\mathrm{div}}_C(f)$ of $f\in K(X{)}^{\times }$ is represented by the single local equation $f$ on the open $X$ [F13]. At the generic point $\xi$ of each prime divisor, [F3] identifies the meromorphic stalk with the fraction field of ${\mathcal{O}}_{X,\xi }$; the germ of the global section $f$ is the element used in the normalized valuation. Thus the coefficient of $[Z]$ in $\mathrm{cyc}({\mathrm{div}}_C(f))$ is $v_{\xi }(f_{\xi })={\mathrm{ord}}_Z(f)$, exactly the coefficient of $[Z]$ in ${\mathrm{div}}_W(f)$ by [F13]. Both sides are Weil divisors, so $\mathrm{cyc}({\mathrm{div}}_C(f))={\mathrm{div}}_W(f)$.