Lemma 31.24.1. Let $X$ be a quasi-compact scheme. Let $h \in \Gamma (X, \mathcal{O}_ X)$ and $f \in \Gamma (X, \mathcal{K}_ X)$ such that $f$ restricts to zero on $X_ h$. Then $h^ n f = 0$ for some $n \gg 0$.

## 31.24 Meromorphic functions and sections; Noetherian case

For locally Noetherian schemes we can prove some results about the sheaf of meromorphic functions. However, there is an example in [misconceptions] showing that $\mathcal{K}_ X$ need not be quasi-coherent for a Noetherian scheme $X$.

**Proof.**
We can find a covering of $X$ by affine opens $U$ such that $f|_ U = s^{-1}a$ with $a \in \mathcal{O}_ X(U)$ and $s \in \mathcal{S}(U)$. Since $X$ is quasi-compact we can cover it by finitely many affine opens of this form. Thus it suffices to prove the lemma when $X = \mathop{\mathrm{Spec}}(A)$ and $f = s^{-1}a$. Note that $s \in A$ is a nonzerodivisor hence it suffices to prove the result when $f = a$. The condition $f|_{X_ h} = 0$ implies that $a$ maps to zero in $A_ h = \mathcal{O}_ X(X_ h)$ as $\mathcal{O}_ X \subset \mathcal{K}_ X$. Thus $h^ na = 0$ for some $n > 0$ as desired.
$\square$

Lemma 31.24.2. Let $X$ be a locally Noetherian scheme.

For any $x \in X$ we have $\mathcal{S}_ x \subset \mathcal{O}_{X, x}$ is the set of nonzerodivisors, and hence $\mathcal{K}_{X, x}$ is the total quotient ring of $\mathcal{O}_{X, x}$.

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

**Proof.**
To prove this lemma we may assume $X$ is the spectrum of a Noetherian ring $A$. Say $x \in X$ corresponds to $\mathfrak p \subset A$.

Proof of (1). It is clear that $\mathcal{S}_ x$ is contained in the set of nonzerodivisors of $\mathcal{O}_{X, x} = A_\mathfrak p$. For the converse, let $f, g \in A$, $g \not\in \mathfrak p$ and assume $f/g$ is a nonzerodivisor in $A_{\mathfrak p}$. Let $I = \{ a \in A \mid af = 0\} $. Then we see that $I_{\mathfrak p} = 0$ by exactness of localization. Since $A$ is Noetherian we see that $I$ is finitely generated and hence that $g'I = 0$ for some $g' \in A$, $g' \not\in \mathfrak p$. Hence $f$ is a nonzerodivisor in $A_{g'}$, i.e., in a Zariski open neighbourhood of $\mathfrak p$. Thus $f/g$ is an element of $\mathcal{S}_ x$.

Proof of (2). Let $f \in \Gamma (X, \mathcal{K}_ X)$ be a meromorphic function. Set $I = \{ a \in A \mid af \in A\} $. Fix a prime $\mathfrak p \subset A$ corresponding to the point $x \in X$. By (1) we can write the image of $f$ in the stalk at $\mathfrak p$ as $a/b$, $a, b \in A_{\mathfrak p}$ with $b \in A_{\mathfrak p}$ not a zerodivisor. Write $b = c/d$ with $c, d \in A$, $d \not\in \mathfrak p$. Then $ad - cf$ is a section of $\mathcal{K}_ X$ which vanishes in an open neighbourhood of $x$. Say it vanishes on $D(e)$ with $e \in A$, $e \not\in \mathfrak p$. Then $e^ n(ad - cf) = 0$ for some $n \gg 0$ by Lemma 31.24.1. Thus $e^ nc \in I$ and $e^ nc$ maps to a nonzerodivisor in $A_{\mathfrak p}$. Let $\text{Ass}(A) = \{ \mathfrak q_1, \ldots , \mathfrak q_ t\} $ be the associated primes of $A$. By looking at $IA_{\mathfrak q_ i}$ and using Algebra, Lemma 10.63.15 the above says that $I \not\subset \mathfrak q_ i$ for each $i$. By Algebra, Lemma 10.15.2 there exists an element $x \in I$, $x \not\in \bigcup \mathfrak q_ i$. By Algebra, Lemma 10.63.9 we see that $x$ is not a zerodivisor on $A$. Hence $f = (xf)/x$ is an element of the total ring of fractions of $A$. This proves (2). $\square$

Lemma 31.24.3. Let $X$ be a locally Noetherian scheme having no embedded points. Let $X^0$ be the set of generic points of irreducible components of $X$. Then we have

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, and

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

**Proof.**
This lemma is a special case of Lemma 31.23.6 because in the locally Noetherian case weakly associated points are the same thing as associated points by Lemma 31.5.8.
$\square$

Lemma 31.24.4. Let $X$ be a locally Noetherian scheme having no embedded points. Let $\mathcal{L}$ be an invertible $\mathcal{O}_ X$-module. Then $\mathcal{L}$ has a regular meromorphic section.

**Proof.**
For each generic point $\eta $ of $X$ pick a generator $s_\eta $ of the free rank $1$ module $\mathcal{L}_\eta $ over the artinian local ring $\mathcal{O}_{X, \eta }$. It follows immediately from the description of $\mathcal{K}_ X$ and $\mathcal{K}_ X(\mathcal{L})$ in Lemma 31.24.3 that $s = \prod s_\eta $ is a regular meromorphic section of $\mathcal{L}$.
$\square$

Lemma 31.24.5. Suppose given

$X$ a locally Noetherian scheme,

$\mathcal{L}$ an invertible $\mathcal{O}_ X$-module,

$s$ a regular meromorphic section of $\mathcal{L}$, and

$\mathcal{F}$ coherent on $X$ without embedded associated points and $\text{Supp}(\mathcal{F}) = X$.

Let $\mathcal{I} \subset \mathcal{O}_ X$ be the ideal of denominators of $s$. Let $T \subset X$ be the union of the supports of $\mathcal{O}_ X/\mathcal{I}$ and $\mathcal{L}/s(\mathcal{I})$ which is a nowhere dense closed subset $T \subset X$ according to Lemma 31.23.9. Then there are canonical injective maps

whose cokernels are supported on $T$.

**Proof.**
Reduce to the affine case with $\mathcal{L} \cong \mathcal{O}_ X$, and $s = a/b$ with $a, b \in A$ both nonzerodivisors. Proof of reduction step omitted. Write $\mathcal{F} = \widetilde{M}$. Let $I = \{ x \in A \mid x(a/b) \in A\} $ so that $\mathcal{I} = \widetilde{I}$ (see proof of Lemma 31.23.9). Note that $T = V(I) \cup V((a/b)I)$. For any $A$-module $M$ consider the map $1 : IM \to M$; this is the map that gives rise to the map $1$ of the lemma. Consider on the other hand the map $\sigma : IM \to M_ b, x \mapsto ax/b$. Since $b$ is not a zerodivisor in $A$, and since $M$ has support $\mathop{\mathrm{Spec}}(A)$ and no embedded primes we see that $b$ is a nonzerodivisor on $M$ also. Hence $M \subset M_ b$. By definition of $I$ we have $\sigma (IM) \subset M$ as submodules of $M_ b$. Hence we get an $A$-module map $s : IM \to M$ (namely the unique map such that $s(z)/1 = \sigma (z)$ in $M_ b$ for all $z \in IM$). It is injective because $a$ is a nonzerodivisor also (on both $A$ and $M$). It is clear that $M/IM$ is annihilated by $I$ and that $M/s(IM)$ is annihilated by $(a/b)I$. Thus the lemma follows.
$\square$

## Post a comment

Your email address will not be published. Required fields are marked.

In your comment you can use Markdown and LaTeX style mathematics (enclose it like `$\pi$`

). A preview option is available if you wish to see how it works out (just click on the eye in the toolbar).

Unfortunately JavaScript is disabled in your browser, so the comment preview function will not work.

All contributions are licensed under the GNU Free Documentation License.

## Comments (2)

Comment #7369 by 代数几何真难 on

Comment #7371 by 代数几何真难 on