Lemma 31.23.6. Let $X$ be a scheme such that

every weakly associated point of $X$ is a generic point of an irreducible component of $X$, and

any quasi-compact open has a finite number of irreducible components.

Let $X^0$ be the set of generic points of irreducible components of $X$. Then we have

\[ \mathcal{K}_ X = \bigoplus \nolimits _{\eta \in X^0} j_{\eta , *}\mathcal{O}_{X, \eta } = \prod \nolimits _{\eta \in X^0} j_{\eta , *}\mathcal{O}_{X, \eta } \]

where $j_\eta : \mathop{\mathrm{Spec}}(\mathcal{O}_{X, \eta }) \to X$ is the canonical map of Schemes, Section 26.13. Moreover

$\mathcal{K}_ X$ is a quasi-coherent sheaf of $\mathcal{O}_ X$-algebras,

for every quasi-coherent $\mathcal{O}_ X$-module $\mathcal{F}$ the sheaf

\[ \mathcal{K}_ X(\mathcal{F}) = \bigoplus \nolimits _{\eta \in X^0} j_{\eta , *}\mathcal{F}_\eta = \prod \nolimits _{\eta \in X^0} j_{\eta , *}\mathcal{F}_\eta \]

of meromorphic sections of $\mathcal{F}$ is quasi-coherent,

$\mathcal{S}_ x \subset \mathcal{O}_{X, x}$ is the set of nonzerodivisors for any $x \in X$,

$\mathcal{K}_{X, x}$ is the total quotient ring of $\mathcal{O}_{X, x}$ for any $x \in X$,

$\mathcal{K}_ X(U)$ equals the total quotient ring of $\mathcal{O}_ X(U)$ for any affine open $U \subset X$,

the ring of rational functions of $X$ (Morphisms, Definition 29.49.3) is the ring of meromorphic functions on $X$, in a formula: $R(X) = \Gamma (X, \mathcal{K}_ X)$.

**Proof.**
Observe that a locally finite direct sum of sheaves of modules is equal to the product since you can check this on stalks for example. Then since $\mathcal{K}_ X(\mathcal{F}) = \mathcal{F} \otimes _{\mathcal{O}_ X} \mathcal{K}_ X$ we see that (2) follows from the other statements. Also, observe that part (6) follows from the initial statement of the lemma and Morphisms, Lemma 29.49.5 when $X^0$ is finite; the general case of (6) follows from this by glueing (argument omitted).

Let $j : Y = \coprod \nolimits _{\eta \in X^0} \mathop{\mathrm{Spec}}(\mathcal{O}_{X, \eta }) \to X$ be the coproduct of the morphisms $j_\eta $. We have to show that $\mathcal{K}_ X = j_*\mathcal{O}_ Y$. First note that $\mathcal{K}_ Y = \mathcal{O}_ Y$ as $Y$ is a disjoint union of spectra of local rings of dimension $0$: in a local ring of dimension zero any nonzerodivisor is a unit. Next, note that pullbacks of meromorphic functions are defined for $j$ by Lemma 31.23.5. This gives a map

\[ \mathcal{K}_ X \longrightarrow j_*\mathcal{O}_ Y. \]

Let $\mathop{\mathrm{Spec}}(A) = U \subset X$ be an affine open. Then $A$ is a ring with finitely many minimal primes $\mathfrak q_1, \ldots , \mathfrak q_ t$ and every weakly associated prime of $A$ is one of the $\mathfrak q_ i$. We obtain $Q(A) = \prod A_{\mathfrak q_ i}$ by Algebra, Lemmas 10.25.4 and 10.66.7. In other words, already the value of the presheaf $U \mapsto \mathcal{S}(U)^{-1}\mathcal{O}_ X(U)$ agrees with $j_*\mathcal{O}_ Y(U)$ on our affine open $U$. Hence the displayed map is an isomorphism which proves the first displayed equality in the statement of the lemma.

Finally, we prove (1), (3), (4), and (5). Part (5) we saw during the course of the proof that $\mathcal{K}_ X = j_*\mathcal{O}_ Y$. The morphism $j$ is quasi-compact by our assumption that the set of irreducible components of $X$ is locally finite. Hence $j$ is quasi-compact and quasi-separated (as $Y$ is separated). By Schemes, Lemma 26.24.1 $j_*\mathcal{O}_ Y$ is quasi-coherent. This proves (1). Let $x \in X$. We may choose an affine open neighbourhood $U = \mathop{\mathrm{Spec}}(A)$ of $x$ all of whose irreducible components pass through $x$. Then $A \subset A_\mathfrak p$ because every weakly associated prime of $A$ is contained in $\mathfrak p$ hence elements of $A \setminus \mathfrak p$ are nonzerodivisors by Algebra, Lemma 10.66.7. It follows easily that any nonzerodivisor of $A_\mathfrak p$ is the image of a nonzerodivisor on a (possibly smaller) affine open neighbourhood of $x$. This proves (3). Part (4) follows from part (3) by computing stalks.
$\square$

## Comments (2)

Comment #3773 by Laurent Moret-Bailly on

Comment #3903 by Johan on

There are also: