The Stacks project

31.15 Effective Cartier divisors on Noetherian schemes

In the locally Noetherian setting most of the discussion of effective Cartier divisors and regular sections simplifies somewhat.

Lemma 31.15.1. Let $X$ be a locally Noetherian scheme. Let $\mathcal{L}$ be an invertible $\mathcal{O}_ X$-module. Let $s \in \Gamma (X, \mathcal{L})$. Then $s$ is a regular section if and only if $s$ does not vanish in the associated points of $X$.

Proof. Omitted. Hint: reduce to the affine case and $\mathcal{L}$ trivial and then use Lemma 31.14.7 and Algebra, Lemma 10.63.9. $\square$

Lemma 31.15.2. Let $X$ be a locally Noetherian scheme. Let $D \subset X$ be a closed subscheme corresponding to the quasi-coherent ideal sheaf $\mathcal{I} \subset \mathcal{O}_ X$.

  1. If for every $x \in D$ the ideal $\mathcal{I}_ x \subset \mathcal{O}_{X, x}$ can be generated by one element, then $D$ is locally principal.

  2. If for every $x \in D$ the ideal $\mathcal{I}_ x \subset \mathcal{O}_{X, x}$ can be generated by a single nonzerodivisor, then $D$ is an effective Cartier divisor.

Proof. Let $\mathop{\mathrm{Spec}}(A)$ be an affine neighbourhood of a point $x \in D$. Let $\mathfrak p \subset A$ be the prime corresponding to $x$. Let $I \subset A$ be the ideal defining the trace of $D$ on $\mathop{\mathrm{Spec}}(A)$. Since $A$ is Noetherian (as $X$ is Noetherian) the ideal $I$ is generated by finitely many elements, say $I = (f_1, \ldots , f_ r)$. Under the assumption of (1) we have $I_\mathfrak p = (f)$ for some $f \in A_\mathfrak p$. Then $f_ i = g_ i f$ for some $g_ i \in A_\mathfrak p$. Write $g_ i = a_ i/h_ i$ and $f = f'/h$ for some $a_ i, h_ i, f', h \in A$, $h_ i, h \not\in \mathfrak p$. Then $I_{h_1 \ldots h_ r h} \subset A_{h_1 \ldots h_ r h}$ is principal, because it is generated by $f'$. This proves (1). For (2) we may assume $I = (f)$. The assumption implies that the image of $f$ in $A_\mathfrak p$ is a nonzerodivisor. Then $f$ is a nonzerodivisor on a neighbourhood of $x$ by Algebra, Lemma 10.68.6. This proves (2). $\square$

Lemma 31.15.3. Let $X$ be a locally Noetherian scheme.

  1. Let $D \subset X$ be a locally principal closed subscheme. Let $\xi \in D$ be a generic point of an irreducible component of $D$. Then $\dim (\mathcal{O}_{X, \xi }) \leq 1$.

  2. Let $D \subset X$ be an effective Cartier divisor. Let $\xi \in D$ be a generic point of an irreducible component of $D$. Then $\dim (\mathcal{O}_{X, \xi }) = 1$.

Proof. Proof of (1). By assumption we may assume $X = \mathop{\mathrm{Spec}}(A)$ and $D = \mathop{\mathrm{Spec}}(A/(f))$ where $A$ is a Noetherian ring and $f \in A$. Let $\xi $ correspond to the prime ideal $\mathfrak p \subset A$. The assumption that $\xi $ is a generic point of an irreducible component of $D$ signifies $\mathfrak p$ is minimal over $(f)$. Thus $\dim (A_\mathfrak p) \leq 1$ by Algebra, Lemma 10.60.11.

Proof of (2). By part (1) we see that $\dim (\mathcal{O}_{X, \xi }) \leq 1$. On the other hand, the local equation $f$ is a nonzerodivisor in $A_\mathfrak p$ by Lemma 31.13.2 which implies the dimension is at least $1$ (because there must be a prime in $A_\mathfrak p$ not containing $f$ by the elementary Algebra, Lemma 10.17.2). $\square$

Lemma 31.15.4. Let $X$ be a Noetherian scheme. Let $D \subset X$ be an integral closed subscheme which is also an effective Cartier divisor. Then the local ring of $X$ at the generic point of $D$ is a discrete valuation ring.

Proof. By Lemma 31.13.2 we may assume $X = \mathop{\mathrm{Spec}}(A)$ and $D = \mathop{\mathrm{Spec}}(A/(f))$ where $A$ is a Noetherian ring and $f \in A$ is a nonzerodivisor. The assumption that $D$ is integral signifies that $(f)$ is prime. Hence the local ring of $X$ at the generic point is $A_{(f)}$ which is a Noetherian local ring whose maximal ideal is generated by a nonzerodivisor. Thus it is a discrete valuation ring by Algebra, Lemma 10.119.7. $\square$

Lemma 31.15.5. Let $X$ be a locally Noetherian scheme. Let $D \subset X$ be an effective Cartier divisor. If $X$ is $(S_ k)$, then $D$ is $(S_{k - 1})$.

Proof. Let $x \in D$. Then $\mathcal{O}_{D, x} = \mathcal{O}_{X, x}/(f)$ where $f \in \mathcal{O}_{X, x}$ is a nonzerodivisor. By assumption we have $\text{depth}(\mathcal{O}_{X, x}) \geq \min (\dim (\mathcal{O}_{X, x}), k)$. By Algebra, Lemma 10.72.7 we have $\text{depth}(\mathcal{O}_{D, x}) = \text{depth}(\mathcal{O}_{X, x}) - 1$ and by Algebra, Lemma 10.60.13 $\dim (\mathcal{O}_{D, x}) = \dim (\mathcal{O}_{X, x}) - 1$. It follows that $\text{depth}(\mathcal{O}_{D, x}) \geq \min (\dim (\mathcal{O}_{D, x}), k - 1)$ as desired. $\square$

Lemma 31.15.6. Let $X$ be a locally Noetherian normal scheme. Let $D \subset X$ be an effective Cartier divisor. Then $D$ is $(S_1)$.

Proof. By Properties, Lemma 28.12.5 we see that $X$ is $(S_2)$. Thus we conclude by Lemma 31.15.5. $\square$

Lemma 31.15.7. Let $X$ be a Noetherian scheme. Let $D \subset X$ be a integral closed subscheme. Assume that

  1. $D$ has codimension $1$ in $X$, and

  2. $\mathcal{O}_{X, x}$ is a UFD for all $x \in D$.

Then $D$ is an effective Cartier divisor.

Proof. Let $x \in D$ and set $A = \mathcal{O}_{X, x}$. Let $\mathfrak p \subset A$ correspond to the generic point of $D$. Then $A_\mathfrak p$ has dimension $1$ by assumption (1). Thus $\mathfrak p$ is a prime ideal of height $1$. Since $A$ is a UFD this implies that $\mathfrak p = (f)$ for some $f \in A$. Of course $f$ is a nonzerodivisor and we conclude by Lemma 31.15.2. $\square$

Lemma 31.15.8. Let $X$ be a Noetherian scheme. Let $Z \subset X$ be a closed subscheme. Assume there exist integral effective Cartier divisors $D_ i \subset X$ and a closed subset $Z' \subset X$ of codimension $\geq 2$ such that $Z \subset Z' \cup \bigcup D_ i$ set-theoretically. Then there exists an effective Cartier divisor of the form

\[ D = \sum a_ i D_ i \subset Z \]

such that $D \to Z$ is an isomorphism away from codimension $2$ in $X$. The existence of the $D_ i$ is guaranteed if $\mathcal{O}_{X, x}$ is a UFD for all $x \in Z$ or if $X$ is regular.

Proof. Let $\xi _ i \in D_ i$ be the generic point and let $\mathcal{O}_ i = \mathcal{O}_{X, \xi _ i}$ be the local ring which is a discrete valuation ring by Lemma 31.15.4. Let $a_ i \geq 0$ be the minimal valuation of an element of $\mathcal{I}_{Z, \xi _ i} \subset \mathcal{O}_ i$. We claim that the effective Cartier divisor $D = \sum a_ i D_ i$ works.

Namely, suppose that $x \in X$. Let $A = \mathcal{O}_{X, x}$. Let $f_ i \in A$ be a local equation for $D_ i$; we only consider those $i$ such that $x \in D_ i$. Then $f_ i$ is a prime element of $A$ and $\mathcal{O}_ i = A_{(f_ i)}$. Let $I = \mathcal{I}_{Z, x} \subset A$. By our choice of $a_ i$ we have $I A_{(f_ i)} = f_ i^{a_ i}A_{(f_ i)}$. It follows that $I \subset (\prod f_ i^{a_ i})$ because the $f_ i$ are prime elements of $A$. This proves that $\mathcal{I}_ Z \subset \mathcal{I}_ D$, i.e., that $D \subset Z$. Moreover, we also see that $D$ and $Z$ agree at the $\xi _ i$, which proves the final assertion.

To see the final statements we argue as follows. A regular local ring is a UFD (More on Algebra, Lemma 15.118.2) hence it suffices to argue in the UFD case. In that case, let $D_ i$ be the irreducible components of $Z$ which have codimension $1$ in $X$. By Lemma 31.15.7 each $D_ i$ is an effective Cartier divisor. $\square$

Lemma 31.15.9. Let $Z \subset X$ be a closed subscheme of a Noetherian scheme. Assume

  1. $Z$ has no embedded points,

  2. every irreducible component of $Z$ has codimension $1$ in $X$,

  3. every local ring $\mathcal{O}_{X, x}$, $x \in Z$ is a UFD or $X$ is regular.

Then $Z$ is an effective Cartier divisor.

Proof. Let $D = \sum a_ i D_ i$ be as in Lemma 31.15.8 where $D_ i \subset Z$ are the irreducible components of $Z$. If $D \to Z$ is not an isomorphism, then $\mathcal{O}_ Z \to \mathcal{O}_ D$ has a nonzero kernel sitting in codimension $\geq 2$. This would mean that $Z$ has embedded points, which is forbidden by assumption (1). Hence $D \cong Z$ as desired. $\square$

Lemma 31.15.10. Let $R$ be a Noetherian UFD. Let $I \subset R$ be an ideal such that $R/I$ has no embedded primes and such that every minimal prime over $I$ has height $1$. Then $I = (f)$ for some $f \in R$.

Proof. By Lemma 31.15.9 the ideal sheaf $\tilde I$ is invertible on $\mathop{\mathrm{Spec}}(R)$. By More on Algebra, Lemma 15.115.3 it is generated by a single element. $\square$

Lemma 31.15.11. Let $X$ be a Noetherian scheme. Let $D \subset X$ be an effective Cartier divisor. Assume that there exist integral effective Cartier divisors $D_ i \subset X$ such that $D \subset \bigcup D_ i$ set theoretically. Then $D = \sum a_ i D_ i$ for some $a_ i \geq 0$. The existence of the $D_ i$ is guaranteed if $\mathcal{O}_{X, x}$ is a UFD for all $x \in D$ or if $X$ is regular.

Proof. Choose $a_ i$ as in Lemma 31.15.8 and set $D' = \sum a_ i D_ i$. Then $D' \to D$ is an inclusion of effective Cartier divisors which is an isomorphism away from codimension $2$ on $X$. Pick $x \in X$. Set $A = \mathcal{O}_{X, x}$ and let $f, f' \in A$ be the nonzerodivisor generating the ideal of $D, D'$ in $A$. Then $f = gf'$ for some $g \in A$. Moreover, for every prime $\mathfrak p$ of height $\leq 1$ of $A$ we see that $g$ maps to a unit of $A_\mathfrak p$. This implies that $g$ is a unit because the minimal primes over $(g)$ have height $1$ (Algebra, Lemma 10.60.11). $\square$

slogan

Lemma 31.15.12. Let $X$ be a Noetherian scheme which has an ample invertible sheaf. Then every invertible $\mathcal{O}_ X$-module is isomorphic to

\[ \mathcal{O}_ X(D - D') = \mathcal{O}_ X(D) \otimes _{\mathcal{O}_ X} \mathcal{O}_ X(D')^{\otimes -1} \]

for some effective Cartier divisors $D, D'$ in $X$. Moreover, given a finite subset $E \subset X$ we may choose $D, D'$ such that $E \cap D = \emptyset $ and $E \cap D' = \emptyset $. If $X$ is quasi-affine, then we may choose $D' = \emptyset $.

Proof. Let $x_1, \ldots , x_ n$ be the associated points of $X$ (Lemma 31.2.5).

If $X$ is quasi-affine and $\mathcal{N}$ is any invertible $\mathcal{O}_ X$-module, then we can pick a section $t$ of $\mathcal{N}$ which does not vanish at any of the points of $E \cup \{ x_1, \ldots , x_ n\} $, see Properties, Lemma 28.29.7. Then $t$ is a regular section of $\mathcal{N}$ by Lemma 31.15.1. Hence $\mathcal{N} \cong \mathcal{O}_ X(D)$ where $D = Z(t)$ is the effective Cartier divisor corresponding to $t$, see Lemma 31.14.10. Since $E \cap D = \emptyset $ by construction we are done in this case.

Returning to the general case, let $\mathcal{L}$ be an ample invertible sheaf on $X$. There exists an $n > 0$ and a section $s \in \Gamma (X, \mathcal{L}^{\otimes n})$ such that $X_ s$ is affine and such that $E \cup \{ x_1, \ldots , x_ n\} \subset X_ s$ (Properties, Lemma 28.29.6).

Let $\mathcal{N}$ be an arbitrary invertible $\mathcal{O}_ X$-module. By the quasi-affine case, we can find a section $t \in \mathcal{N}(X_ s)$ which does not vanish at any point of $E \cup \{ x_1, \ldots , x_ n\} $. By Properties, Lemma 28.17.2 we see that for some $e \geq 0$ the section $s^ e|_{X_ s} t$ extends to a global section $\tau $ of $\mathcal{L}^{\otimes e} \otimes \mathcal{N}$. Thus both $\mathcal{L}^{\otimes e} \otimes \mathcal{N}$ and $\mathcal{L}^{\otimes e}$ are invertible sheaves which have global sections which do not vanish at any point of $E \cup \{ x_1, \ldots , x_ n\} $. Thus these are regular sections by Lemma 31.15.1. Hence $\mathcal{L}^{\otimes e} \otimes \mathcal{N} \cong \mathcal{O}_ X(D)$ and $\mathcal{L}^{\otimes e} \cong \mathcal{O}_ X(D')$ for some effective Cartier divisors $D$ and $D'$, see Lemma 31.14.10. By construction $E \cap D = \emptyset $ and $E \cap D' = \emptyset $ and the proof is complete. $\square$

Lemma 31.15.13. Let $X$ be an integral regular scheme of dimension $2$. Let $i : D \to X$ be the immersion of an effective Cartier divisor. Let $\mathcal{F} \to \mathcal{F}' \to i_*\mathcal{G} \to 0$ be an exact sequence of coherent $\mathcal{O}_ X$-modules. Assume

  1. $\mathcal{F}, \mathcal{F}'$ are locally free of rank $r$ on a nonempty open of $X$,

  2. $D$ is an integral scheme,

  3. $\mathcal{G}$ is a finite locally free $\mathcal{O}_ D$-module of rank $s$.

Then $\mathcal{L} = (\wedge ^ r\mathcal{F})^{**}$ and $\mathcal{L}' = (\wedge ^ r \mathcal{F}')^{**}$ are invertible $\mathcal{O}_ X$-modules and $\mathcal{L}' \cong \mathcal{L}(k D)$ for some $k \in \{ 0, \ldots , \min (s, r)\} $.

Proof. The first statement follows from Lemma 31.12.15 as assumption (1) implies that $\mathcal{L}$ and $\mathcal{L}'$ have rank $1$. Taking $\wedge ^ r$ and double duals are functors, hence we obtain a canonical map $\sigma : \mathcal{L} \to \mathcal{L}'$ which is an isomorphism over the nonempty open of (1), hence nonzero. To finish the proof, it suffices to see that $\sigma $ viewed as a global section of $\mathcal{L}' \otimes \mathcal{L}^{\otimes -1}$ does not vanish at any codimension point of $X$, except at the generic point of $D$ and there with vanishing order at most $\min (s, r)$.

Translated into algebra, we arrive at the following problem: Let $(A, \mathfrak m, \kappa )$ be a discrete valuation ring with fraction field $K$. Let $M \to M' \to N \to 0$ be an exact sequence of finite $A$-modules with $\dim _ K(M \otimes K) = \dim _ K(M' \otimes K) = r$ and with $N \cong \kappa ^{\oplus s}$. Show that the induced map $L = \wedge ^ r(M)^{**} \to L' = \wedge ^ r(M')^{**}$ vanishes to order at most $\min (s, r)$. We will use the structure theorem for modules over $A$, see More on Algebra, Lemma 15.121.3 or 15.121.9. Dividing out a finite $A$-module by a torsion submodule does not change the double dual. Thus we may replace $M$ by $M/M_{tors}$ and $M'$ by $M'/\mathop{\mathrm{Im}}(M_{tors} \to M')$ and assume that $M$ is torsion free. Then $M \to M'$ is injective and $M'_{tors} \to N$ is injective. Hence we may replace $M'$ by $M'/M'_{tors}$ and $N$ by $N/M'_{tors}$. Thus we reduce to the case where $M$ and $M'$ are free of rank $r$ and $N \cong \kappa ^{\oplus s}$. In this case $\sigma $ is the determinant of $M \to M'$ and vanishes to order $s$ for example by Algebra, Lemma 10.121.7. $\square$


Comments (0)


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.




In order to prevent bots from posting comments, we would like you to prove that you are human. You can do this by filling in the name of the current tag in the following input field. As a reminder, this is tag 0B3Q. Beware of the difference between the letter 'O' and the digit '0'.