The Stacks project

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. $\square$

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 locally 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). $\square$

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). $\square$

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. $\square$

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. $\square$

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

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. $\square$

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 $D_1, \ldots , D_ n$ be the pairwise distinct divisors $D_ i$ such that $x \in D_ i$. For $1 \leq i \leq n$ let $f_ i \in A$ be a local equation for $D_ i$. Then $f_ i$ is a prime element of $A$ and $\mathcal{O}_ i = A_{(f_ i)}$. Let $I = \mathcal{I}_{Z, x} \subset A$ be the stalk of the ideal sheaf of $Z$. By our choice of $a_ i$ we have $I A_{(f_ i)} = f_ i^{a_ i}A_{(f_ i)}$. We claim that $I \subset (\prod _{i = 1, \ldots , n} f_ i^{a_ i})$.

Proof of the claim. The localization map $\varphi : A/(f_ i) \to A_{(f_ i)}/f_ iA_{(f_ i)}$ is injective as the prime ideal $(f_ i)$ is the inverse image of the maximal ideal $f_ iA_{(f_ i)}$. By induction on $n$ we deduce that $\varphi _ n : A/(f_ i^ n)\to A_{(f_ i)}/f_ i^ nA_{(f_ i)}$ is also injective. Since $\varphi _{a_ i}(I) = 0$, we have $I \subset (f_ i^{a_ i})$. Thus, for any $x \in I$, we may write $x = f_1^{a_1}x_1$ for some $x_1 \in A$. Since $D_1, \ldots , D_ n$ are pairwise distinct, $f_ i$ is a unit in $A_{(f_ j)}$ for $i \not= j$. Comparing $x$ and $x_1$ at $A_{(f_ i)}$ for $n \geq i > 1$, we still have $x_1 \in (f_ i^{a_ i})$. Repeating the previous process, we inductively write $x_ i = f_{i + 1}^{a_{i + 1}}x_{i + 1}$ for any $n > i \geq 1$. In conclusion, $x \in (\prod _{i = 1, \ldots n} f_ i^{a_ i})$ for any $x \in I$ as desired.

The claim shows 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 that $D \to Z$ is an isomorphism away from codimension $2$ on $X$.

To see the final statements we argue as follows. A regular local ring is a UFD (More on Algebra, Lemma 15.121.2) 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. $\square$

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. $\square$

Proof. By Lemma 31.15.9 the ideal sheaf $\tilde I$ is invertible on $\mathop{\mathrm{Spec}}(R)$. By More on Algebra, Lemma 15.117.3 it is generated by a single element. $\square$

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). $\square$

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. $\square$

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 15.124.3 or 15.124.9. 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. $\square$