Logarithmic transformations of rigid analytic elliptic surfaces

Abstract

We give new examples of algebraic elliptic surfaces and non-algebraic rigid analytic elliptic surfaces by means of logarithmic transformations. In the complex analytic case, it is known that all multiple fibers of elliptic surfaces are obtained by logarithmic transformations. Using rigid analytic geometry, we construct similar transformations of elliptic surfaces over complete non-Archimedean valuation base fields. These operations yield rigid analytic elliptic fibrations with multiple fibers. When the resulting surface admits an ample line bundle, we may algebraize the surface. In the positive characteristic case, we obtain new types of algebraic elliptic surfaces. We also obtain a non-algebraic rigid analytic surface the combination of whose invariants appears neither in the algebraic case nor in the complex analytic case.

Appendix

1.1 Local properties of elliptic fibrations with a section

In this subsection, we show that any two smooth elliptic fibrations with the same $j$-invariant are isomorphic over a sufficiently small admissible open neighborhood of a given point. In particular, we can apply our logarithmic transformations to a fiber of a smooth elliptic fibration whenever the absolute value of the $j$-invariant of the fiber is greater than one.

Proposition 8

Let $(X,C,\pi )$ be a smooth elliptic fibration. Take a point $q$ on $C.$ Then there exists an admissible open subset $U$ of $C$ containing $q$ such that the restriction $\pi |_{\pi ^{-1}(U)}:\pi ^{-1}(U)\rightarrow U$ admits a section.

Proof

Take a point $q^{\prime }$ on the fiber $\pi ^{-1}(q).$ Take an admissible affinoid open subset $U$ of $C$ and an admissible affinoid open subset $V$ of $X$ satisfying the conditions that $q^{\prime }\in V$ and $V\subset \pi ^{-1}(U).$ After shrinking $U$ and $V,$ we may assume that there exists a $U$-isomorphism $V\cong (\pi ^{-1}(q)\cap V)\times U.$ This fact follows from Proposition 1.2 in [22] for the case where $\fancyscript{Z}$ in the proposition is empty. Thus, the restriction $\pi |_V:V\rightarrow U$ admits a section $\sigma .$ The composite $\iota \circ \sigma$ of the section $\sigma$ and the canonical open $U$-immersion $\iota :V\rightarrow \pi ^{-1}(U)$ is a section of the restriction $\pi |_{\pi ^{-1}(U)}:\pi ^{-1}(U)\rightarrow U.$ Thus, the admissible open subset $U$ is a desired one. $\square$

Proposition 9

Let $\nu :B\rightarrow C$ be a finite morphism between smooth affinoid spaces that are curves. Let $(X,C,\pi )$ be an elliptic fibration. Assume that the reductions of all the fibers of $\pi$ are isomorphic to elliptic curves. Suppose that $\nu$ factors through $\pi .$ Then the elliptic fibration is the analytification of an algebraic one.

Proof

We take a morphism $\sigma :B\rightarrow X$ such that the equality $\pi \circ \sigma =\nu$ holds. Since $\sigma$ is proper (Proposition 5 in [6, 9.6.2] and Proposition 4 in [6, 9.6.2]), the proper mapping theorem (Proposition 3 in [6, 9.6.3]) shows that the image $E$ of $\sigma$ is a closed analytic subset of $X.$ We endow $E$ with the reduced structure. Then $E$ is a divisor on $X.$ By the same method as in the proof of Theorem 6, we can show that the line bundle $\fancyscript{O}_X(E)$ is relatively ample for $\pi .$ Thus, the fibration $\pi$ is projective (Theorem 3.2.7 in [9]). Therefore, the proposition follows from the relative GAGA theorems [24]. $\square$

We define the $j$-invariant for an elliptic fibration with a section to be the $j$-invariant for the algebraization of the fibration.

Proposition 10

Let $C$ be a smooth affinoid space that is a curve. Let $(X,C,\pi )$ and $(X^{\prime },C,\pi ^{\prime })$ be smooth elliptic fibrations with sections. Suppose that the two elliptic fibrations have the same $j$-invariant. Then, for any point $q$ on $C,$ there exists an admissible affinoid open neighborhood $U$ of $q$ such that the restrictions $\pi ^{-1}(U)$ and $(\pi ^{\prime })^{-1}(U)$ are isomorphic over $U$ in the category of rigid analytic spaces.

Proof

Proposition 9 shows that the fibrations $(X,C,\pi )$ and $(X^{\prime },C,\pi ^{\prime })$ are the analytifications of algebraic ones. In the following, we use the same letters $X, X^{\prime },$ and $C$ to denote their algebraizations. Put $A:=\fancyscript{O}_C(C).$ To show the proposition, we have only to show that $X\times _A\fancyscript{O}_{C,q}$ is isomorphic to $X^{\prime }\times _A\fancyscript{O}_{C,q}$ over $\fancyscript{O}_{C,q}.$ Since $\fancyscript{O}_{C,q}$ is a strictly Henselian discrete valuation ring (Lemma 2.1.1 (1) in [17]), the proposition follows from the following proposition. $\square$

Proposition 11

Let $R$ be a strictly Henselian discrete valuation ring. Let $X$ and $X^{\prime }$ be elliptic curves over $R.$ Suppose that the two elliptic curves have the same $j$-invariant. Then $X$ and $X^{\prime }$ are isomorphic over $R.$

Proof

Let $L$ be the field of fractions of $R.$ We take a minimal Weierstrass form

$$\begin{aligned} y^2+a_1xy+a_3y=x^3+a_2x^2+a_4x+a_6 \end{aligned}$$

$$\begin{aligned} \big (\text{ resp.}\ (y^{\prime })^2+a_1^{\prime }x^{\prime }y^{\prime }+a_3^{\prime }y^{\prime }=(x^{\prime })^3+a_2^{\prime }(x^{\prime })^2+a_4^{\prime }x^{\prime }+a_6^{\prime }\big ) \end{aligned}$$

of $X$ (resp. $X^{\prime }$) over $R$ (see [36, VII.1]). We define $b_2,b_4,b_6,b_8,c_4\in R$ in the following way:

$$\begin{aligned} b_2&:= a_1^2+4a_2;\\ b_4&:= a_1a_3+2a_4;\\ b_6&:= a_3^2+4a_6;\\ b_8&:= a_1^2a_6-a_1a_3a_4+a_2a_3^2+4a_2a_6-a_4^2;\\ c_4&:= b_2^2-24b_4. \end{aligned}$$

The discriminant $\Delta$ and the $j$-invariant of $X$ are given by

$$\begin{aligned} \Delta :=-b_2^2b_8+9b_2b_4b_6-8b_4^3-27b_6^2 \end{aligned}$$

and

$$\begin{aligned} j:=\frac{c_4^3}{\Delta }, \end{aligned}$$

respectively. We define $b_2^{\prime },b_4^{\prime },b_6^{\prime },b_8^{\prime },c_4^{\prime }\in R$ for $X^{\prime }$ in the same way. The discriminant $\Delta ^{\prime }$ and the $j$-invariant $j^{\prime }$ of $X^{\prime }$ are also given in the same way.

Since $j=j^{\prime },$ there exists a finite separable extension $M/L$ such that $X\times _RM$ and $X^{\prime }\times _RM$ are isomorphic over $M.$ The isomorphism is given by

$$\begin{aligned} (x,y)\longmapsto (x^{\prime },y^{\prime })=(u_0^2x+r_0,u_0^3y+s_0u_0^2x+t_0) \end{aligned}$$

where $r_0,s_0,t_0,u_0\in M$ (Proposition III.3.1 (b) in [36]). We have the following equalities (Table 3.1 in [36, p. 45]):

$$\begin{aligned} u_0a_1&= a_1^{\prime }+2s_0;\end{aligned}$$

(3)

$$\begin{aligned} u_0^2a_2&= a_2^{\prime }-s_0a_1^{\prime }+3r_0-s_0^2;\end{aligned}$$

(4)

$$\begin{aligned} u_0^3a_3&= a_3^{\prime }+r_0a_1^{\prime }+2t_0;\end{aligned}$$

(5)

$$\begin{aligned} u_0^4a_4&= a_4^{\prime }-s_0a_3^{\prime }+2r_0a_2^{\prime }-(t_0+r_0s_0)a_1^{\prime }+3r_0^2-2s_0t_0;\end{aligned}$$

(6)

$$\begin{aligned} u_0^6a_6&= a_6^{\prime }+r_0a_4^{\prime }-t_0a_3^{\prime }+r_0^2a_2^{\prime }-r_0t_0a_1^{\prime }+r_0^3-t_0^2;\end{aligned}$$

(7)

$$\begin{aligned} u_0^2b_2&= b_2^{\prime }+12r_0;\nonumber \\ u_0^4b_4&= b_4^{\prime }+r_0b_2^{\prime }+6r_0^2;\nonumber \\ u_0^6b_6&= b_6^{\prime }+2r_0b_4^{\prime }+r_0^2b_2^{\prime }+4r_0^3;\end{aligned}$$

(8)

$$\begin{aligned} u_0^8b_8&= b_8^{\prime }+3r_0b_6^{\prime }+3r_0^2b_4^{\prime }+r_0^3b_2^{\prime }+3r_0^4;\end{aligned}$$

(9)

$$\begin{aligned} u_0^{12}\Delta&= \Delta ^{\prime }. \end{aligned}$$

(10)

Let $S$ be the integral closure of $R$ in $M.$ Since $\Delta ,\Delta ^{\prime }\in R^\times ,$ the equality (10) shows $u_0\in S^\times .$ The equality (9) shows $3r_0\in S.$ The equality (8) shows $4r_0\in S.$ Then $r_0\in S.$ The equality (4) shows $s_0\in S.$ The equality (7) shows $t_0\in S.$ Therefore, to show the proposition, we have only to show $r_0,s_0,t_0,u_0\in L.$ Let $p$ be the characteristic of the residue field of $R.$

Case (I): $p\not =2,3.$ Since $\Delta ,\Delta ^{\prime }\in R^\times ,$ the equality (10) gives $u_0^{12}\in R^\times .$ Since $R$ is strictly Henselian, we obtain $u_0\in R.$ The equality (3) shows $s_0\in R.$ The equality (4) shows $r_0\in R.$ The equality (5) shows $t_0\in R.$ Therefore, the proposition holds. We remark that this argument works whenever the characteristic of $L$ is different from $2$ and $3.$

Referring to the equalities (3–7), we define $f_i\in R[r,s,t,u]$ in the following way:

$$\begin{aligned} f_1&:= -ua_1+a_1^{\prime }+2s;\\ f_2&:= -u^2a_2+a_2^{\prime }-sa_1^{\prime }+3r-s^2;\\ f_3&:= -u^3a_3+a_3^{\prime }+ra_1^{\prime }+2t;\\ f_4&:= -u^4a_4+a_4^{\prime }-sa_3^{\prime }+2ra_2^{\prime }-(t+rs)a_1^{\prime }+3r^2-2st;\\ f_6&:= -u^6a_6+a_6^{\prime }+ra_4^{\prime }-ta_3^{\prime }+r^2a_2^{\prime }-rta_1^{\prime }+r^3-t^2. \end{aligned}$$

Case (II): $p=3.$ Replacing $y$ by $y-(a_1x+a_3)/2$ and replacing $y^{\prime }$ in the same way, we may assume that the equalities $a_1=a_3=a_1^{\prime }=a_3^{\prime }=0$ hold. The equality (3) shows $s_0=0.$ Thus, the equality (4) shows the following: $a_2\in R^\times$ if and only if $a_2^{\prime }\in R^\times .$ First, we consider the case where $a_2\in R^\times .$ Put $T:=R[r,s,t,u].$ Let $\psi :R\rightarrow T$ be the $R$-algebra structure homomorphism. Let $\phi :T\rightarrow S$ be the $R$-algebra homomorphism given by

$$\begin{aligned} (r,s,t,u)\mapsto (r_0.s_0,t_0,u_0). \end{aligned}$$

Then $(f_1,f_2,f_3,f_4,f_6)\subset \text{ Ker}\phi .$ Let $\mathfrak m$ be the maximal ideal of $R.$ Since $S$ is integral over $R,$ there exists a prime ideal $\mathfrak n$ of $S$ such that $(\phi \circ \psi )^{-1}(\mathfrak n )=\mathfrak m .$ Put $\mathfrak p :=\phi ^{-1}(\mathfrak n ), T^{\prime }:=T_{\mathfrak p }/(f_1,f_2,f_3,f_4),$ and $d:=\det (\partial (f_1,f_2,f_3,f_4)/\partial (r,s,t,u)).$ The equality

$$\begin{aligned} d \equiv -ua_2a_2^{\prime }\text{ mod}(3) \end{aligned}$$

shows $\phi (d)\in S^\times ,$ which implies $d\in T_{\mathfrak p }^\times .$ Thus, the $R$-algebra $T^{\prime }$ is finite and étale over $R.$ Since $R$ is strictly Henselian, the $R$-algebra structure homomorphism $R\rightarrow T^{\prime }$ is an isomorphism. This concludes that $r_0,s_0,t_0,u_0\in R.$ Therefore, the proposition holds. Next, we consider the case where $a_2\not \in R^\times .$ The assumptions $\Delta ,\Delta ^{\prime }\in R^\times$ show $a_4,a_4^{\prime }\in R^\times .$ Since the equality

$$\begin{aligned} \det (\partial (f_1,f_3,f_4,f_6)/\partial (r,s,t,u))\equiv u^3a_4a_4^{\prime } \text{ mod}(3,a_2,a_2^{\prime }) \end{aligned}$$

holds, this case can be shown by a similar argument for $(f_1,f_3,f_4,f_6).$

Case (III): $p=2.$ Replacing $x$ by $x+a_2$ and replacing $x^{\prime }$ in the same way, we may assume that the equalities $a_2=a_2^{\prime }=0$ hold. The equality (3) shows the following: $a_1\in R^\times$ if and only if $a_1^{\prime }\in R^\times .$ First, we consider the case where $a_1\in R^\times .$ Since the equality

$$\begin{aligned} \det (\partial (f_1,f_2,f_3,f_4)/\partial (r,s,t,u))\equiv a_1(a_1^{\prime })^3 \text{ mod} (2) \end{aligned}$$

holds, this case can be shown by a similar argument for $(f_1,f_2,f_3,f_4).$ Next, we consider the case where $a_1\not \in R^\times .$ The assumptions $\Delta ,\Delta ^{\prime }\in R^\times$ show $a_3,a_3^{\prime }\in R^\times .$ Since the equality

$$\begin{aligned} \det (\partial (f_2,f_3,f_4,f_6)/\partial (r,s,t,u))\equiv u^2a_3(a_3^{\prime })^2 \text{ mod} (2,a_1,a_1^{\prime }) \end{aligned}$$

holds, this case can be shown by a similar argument for $(f_2,f_3,f_4,f_6).$ $\square$

1.2 Multiple fibers of elliptic fibrations

In this subsection, we provide steps necessary for the proof of Proposition 1. The analogous studies for elliptic surfaces are given by Proposition 4 in [3]. For completeness, we prove the following lemmas and propositions for algebraic elliptic fibrations.

Let $(X,C,\pi )$ be a relatively minimal algebraic elliptic fibration. By $f_\eta :X_\eta \rightarrow C_\eta$ we denote the generic fiber of $f.$ Take a closed point $q$ on $C.$ Put $X_q:=\pi ^{-1}(q).$ We write $X_q=\sum _{i\in I}d_i\Gamma _i$ where $\{\Gamma _i\}_{i\in I}$ is the set of the irreducible components of the reduction of $X_q.$ Let $m$ be the multiplicity of $X_q.$ We define an effective divisor $D$ on $X$ by $D:=X_q/m.$ Let $\nu$ be the order of the normal bundle $\fancyscript{N}_{D/X}$ of $D$ in $\text{ Pic}D.$ Let $E$ be a divisor on $X$ satisfying $0<E\le X_q.$ We regard $E$ as a scheme over the residue field of $\fancyscript{O}_{C,q}.$ Let $\imath :E\rightarrow X_q$ be the canonical closed immersion. For any line bundle $\fancyscript{L}$ on $E,$ we abbreviate $\imath _*\fancyscript{L}$ to $\fancyscript{L}$ for simplicity. We define the degree of a line bundle $\fancyscript{L}$ on $E$ by the equality

$$\begin{aligned} \deg \fancyscript{L}:=\chi (\fancyscript{L})-\chi (\fancyscript{O}_E) \end{aligned}$$

(Definition 7.3.29 in [26]). The map $\deg :\text{ Pic}E\rightarrow {\mathbb Z }$ is a group homomorphism (Lemma 7.3.30 (b) in [26]). By $E_1\cdot E_2$ we denote the intersection number of two divisors $E_1$ and $E_2$ on $X$ whose supports are contained in the special fiber of $\pi .$ If the inequalities $0<E_2\le X_q$ hold, then the equality

$$\begin{aligned} E_1\cdot E_2=\deg \fancyscript{O}_X(E_1)|_{E_2} \end{aligned}$$

holds (Theorem 9.1.12 in [26]).

Lemma 13

Let $\fancyscript{L}$ be a line bundle on $D.$ Assume that the equality $\deg \fancyscript{L}|_{\Gamma _i}=0$ holds for any $i\in I.$ Then the equality $\deg \fancyscript{L}|_E=0$ holds for any divisor $E$ on $X$ satisfying $0<E\le D.$

Proof

If $E$ is a prime divisor, then there exists an element $i\in I$ such that the equality $E=\Gamma _i$ holds. Thus, we may assume that there exist two non-zero effective divisors $E_1$ and $E_2$ such that the equality $E=E_1+E_2$ holds. Tensoring $\fancyscript{L}$ with the short exact sequence

$$\begin{aligned} 0\longrightarrow \fancyscript{O}_X(-E_1)|_{E_2}\longrightarrow \fancyscript{O}_E\longrightarrow \fancyscript{O}_{E_1}\longrightarrow 0, \end{aligned}$$

we obtain a short exact sequence

$$\begin{aligned} 0\longrightarrow \fancyscript{O}_X(-E_1)|_{E_2}\otimes \fancyscript{L}\longrightarrow \fancyscript{O}_E\otimes \fancyscript{L}\longrightarrow \fancyscript{O}_{E_1}\otimes \fancyscript{L}\longrightarrow 0. \end{aligned}$$

The two short exact sequences give the equalities

$$\begin{aligned} \chi \big (\fancyscript{O}_X(-E_1)|_{E_2}\big )-\chi (\fancyscript{O}_E)+\chi (\fancyscript{O}_{E_1})=0 \end{aligned}$$

and

$$\begin{aligned} \chi \big (\fancyscript{O}_X(-E_1)|_{E_2}\otimes \fancyscript{L}\big )-\chi (\fancyscript{O}_E\otimes \fancyscript{L})+\chi (\fancyscript{O}_{E_1}\otimes \fancyscript{L})=0. \end{aligned}$$

Since the equalities

$$\begin{aligned} \deg \fancyscript{L}|_{E_1}&= \chi (\fancyscript{O}_{E_1}\otimes \fancyscript{L})-\chi (\fancyscript{O}_{E_1}),\\ \deg \fancyscript{L}|_{E_2}&= \deg \big (\fancyscript{O}_X(-E_1)\otimes \fancyscript{L}\big )|_{E_2}-\deg \fancyscript{O}_X(-E_1)|_{E_2}\\&= \chi \big (\fancyscript{O}_X(-E_1)|_{E_2}\otimes \fancyscript{L}\big )-\chi \big (\fancyscript{O}_X(-E_1)|_{E_2}\big ), \end{aligned}$$

and

$$\begin{aligned} \deg \fancyscript{L}|_E=\chi (\fancyscript{O}_E\otimes \fancyscript{L})-\chi (\fancyscript{O}_E) \end{aligned}$$

hold, the above two equalities give the equality

$$\begin{aligned} \deg \fancyscript{L}|_E=\deg \fancyscript{L}|_{E_1}+\deg \fancyscript{L}|_{E_2}. \end{aligned}$$

Thus, we obtain the desired equality by induction. $\square$

Lemma 14

Let $\fancyscript{L}$ be a line bundle on $D.$ Assume that the equality $\deg \fancyscript{L}|_{\Gamma _i}=0$ holds for any $i\in I.$ Then the following conditions are equivalent.

• 1. The line bundle $\fancyscript{L}$ is trivial.

• 2. The equality $h^0(\fancyscript{L})=1$ holds.

• 3. The line bundle $\fancyscript{L}$ admits a non-zero global section.

Proof

The implication Condition 1 $\Rightarrow$ Condition 3 and the implication Condition 2 $\Rightarrow$ Condition 3 are clear. Let us show the implications Condition 3 $\Rightarrow$ Conditions 1 and 2. Assume that Condition 3 is satisfied. Let $h$ be a non-zero global section of $\fancyscript{L}.$ Let $D_1$ be the maximal effective divisor such that $D_1\le D$ and $h|_{D_1}=0.$ Put $D_2:=D-D_1.$ Lemma 13 gives the equality $\deg \fancyscript{L}|_{D_2}=0.$ Thus, the equalities

$$\begin{aligned} -D_1\cdot D_2&= \deg \fancyscript{O}_X(-D_1)|_{D_2}=\deg \big (\fancyscript{O}_X(-D_1)\otimes \fancyscript{L}\big )|_{D_2}\\&= \chi \big (\fancyscript{O}_X(-D_1)|_{D_2}\otimes \fancyscript{L}\big )-\chi (\fancyscript{O}_{D_2}) \end{aligned}$$

hold. Let $\phi :\fancyscript{O}_{D_2}\rightarrow \fancyscript{O}_X(-D_1)|_{D_2}\otimes \fancyscript{L}$ be the canonical homomorphism induced by $h.$ Put $\fancyscript{F}:=\text{ Coker}\phi .$ Then the above equalities and the short exact sequence

$$\begin{aligned} 0\longrightarrow \fancyscript{O}_{D_2}\longrightarrow \fancyscript{O}_X(-D_1)|_{D_2}\otimes \fancyscript{L}\longrightarrow \fancyscript{F}\longrightarrow 0 \end{aligned}$$

give the equality

$$\begin{aligned} -D_1\cdot D_2=\chi (\fancyscript{F}). \end{aligned}$$

Since the support of $\fancyscript{F}$ is at most zero-dimensional, the inequality $\chi (\fancyscript{F})\ge 0$ holds. Since the equality $D_2\not =0$ holds by assumption, Theorem 9.1.23 in [26] gives the equality $D_1=0.$ Thus, we obtain the equality $\chi (\fancyscript{F})=0,$ which implies $\fancyscript{F}=0.$ Therefore, the global section $h$ is nowhere-vanishing. The fact implies that Conditions 1 and 2 are satisfied. $\square$

Lemma 15

The equality $\deg \fancyscript{N}_{D/X}^{\otimes n}|_{\Gamma _i}=0$ holds for any integer $n$ and any $i\in I.$

Proof

The equality follows from the equalities

$$\begin{aligned} \deg \fancyscript{O}_X(nD)|_{\Gamma _i}=nD\cdot \Gamma _i=\frac{n}{m}X_q\cdot \Gamma _i=0. \end{aligned}$$

$\square$

For each integer $n,$ put

$$\begin{aligned} \fancyscript{N}_n:=\fancyscript{O}_X(-nD)|_D. \end{aligned}$$

Lemma 16

For any integer $n$ satisfying $0<n\le m,$ the equality

$$\begin{aligned} h^0(\fancyscript{O}_{nD})=h^1(\fancyscript{O}_{nD}) \end{aligned}$$

holds. For any integer $n,$ the equalities

$$\begin{aligned} h^0(\fancyscript{N}_n)=h^1(\fancyscript{N}_n)= \left\{ \begin{array}{l@{\quad }l} 1,&\nu \mid n,\\ 0,&\nu \mathrel {\not |}n \end{array}\right. \end{aligned}$$

hold.

Proof

Let $n$ be an integer.Since the equalities

$$\begin{aligned} \chi (\fancyscript{N}_n)-\chi (\fancyscript{O}_D)=\deg \fancyscript{O}_X(-nD)|_D=-nD\cdot D=-\frac{n}{m}X_q\cdot D=0 \end{aligned}$$

hold, we obtain the equality

$$\begin{aligned} \chi (\fancyscript{N}_n)=\chi (\fancyscript{O}_D). \end{aligned}$$

If $n>0,$ then there exists a short exact sequence

$$\begin{aligned} 0\longrightarrow \fancyscript{N}_n\longrightarrow \fancyscript{O}_{(n+1)D}\longrightarrow \fancyscript{O}_{nD}\longrightarrow 0. \end{aligned}$$

Thus, if $0<n<m,$ then we obtain the equality

$$\begin{aligned} \chi (\fancyscript{N}_n)-\chi (\fancyscript{O}_{(n+1)D})+\chi (\fancyscript{O}_{nD})=0. \end{aligned}$$

Therefore, by induction on $n,$ we obtain the equality $\chi (\fancyscript{O}_{nD})=n\chi (\fancyscript{O}_D)$ for any integer $n$ satisfying $0<n\le m.$ The constancy of the Euler characteristics of $C$-flat coherent sheaves on a proper family of schemes over $C$ gives the equality $\chi (\fancyscript{O}_{X_q})=0.$ Since the equality $X_q=mD$ holds by definition, the equality $\chi (\fancyscript{O}_{mD})=0$ holds. Thus, the equality $\chi (\fancyscript{O}_{nD})=0$ holds for any integer $n$ satisfying $0<n\le m.$ Therefore, the equality $\chi (\fancyscript{N}_n)=0$ holds for any integer $n.$ Since $D$ is of dimension one, we obtain the first and second equalities. Since there exists an isomorphism

$$\begin{aligned} \fancyscript{N}_n\cong \fancyscript{N}_{D/X}^{\otimes -n} \end{aligned}$$

for any integer $n,$ the last equality follows from Lemmas 15 and 14. $\square$

Lemma 17

The fiber $X_q$ is tamely ramified if and only if the equality $h^1(\fancyscript{O}_{X_q})=1$ holds.

Proof

Since $R^2f_*\fancyscript{O}_X=0,$ the lemma follows from Theorem 5.3.20 (d) in [26]. $\square$

Proposition 12

For each positive integer $n$ satisfying $0<n\le m,$ put

$$\begin{aligned} F(n):=h^1(\fancyscript{O}_{nD}). \end{aligned}$$

Then the following statements hold.

• 1. The equality $F(1)=1$ holds, and the function $F(n)$ is non-decreasing.

• 2. For any positive integer $n,$ the inequality $F(n)>1$ holds if and only if the inequality $n>\nu$ holds.

In particular, the fiber $X_q$ is tamely ramified if and only if the equality $\nu =m$ holds.

Proof

Lemma 16 for $n=0$ gives the equality $F(1)=1.$ Let $n$ be a positive integer. Since $D$ is of dimension one, the short exact sequence in the proof of Lemma 16 induces a long exact sequence

$$\begin{aligned} 0&\longrightarrow&H^0(X,\fancyscript{N}_n)\longrightarrow H^0\big ((n+1)D,\fancyscript{O}_{(n+1)D}\big )\longrightarrow H^0(nD,\fancyscript{O}_{nD})\\&\longrightarrow&H^1(X,\fancyscript{N}_n)\longrightarrow H^1\big ((n+1)D,\fancyscript{O}_{(n+1)D}\big )\longrightarrow H^1(nD,\fancyscript{O}_{nD})\longrightarrow 0. \end{aligned}$$

Thus, the first statement holds. Assume that $n<m$ and $F(n)=1.$ Then Lemma 16 gives the equality $h^0(\fancyscript{O}_{nD})=1.$ Since the pull-back of any non-zero constant function on $(n+1)D$ via the canonical closed immersion $nD\rightarrow (n+1)D$ is non-zero, the third homomorphism in the above long exact sequence is surjective. Thus, the equality

$$\begin{aligned} F(n+1)=1+h^1(\fancyscript{N}_n) \end{aligned}$$

holds. Therefore, the second statement follows from Lemma 16. Since $\nu \mid m,$ the last statement follows from Lemma 17. $\square$

Proposition 13

Let $p$ be the characteristic of the residue field of $\fancyscript{O}_{C,q}.$ Then there exists a non-negative integer $e$ such that the equality $m=\nu p^e$ holds. Here, we put $p^0:=1.$

Proof

For each positive integer $n,$ let $\nu _n$ be the order of the line bundle $\fancyscript{O}_X(D)|_{nD}$ on $nD$ in $\text{ Pic}nD.$ Lemma 6.4.4 in [34] implies that there exists a positive integer $n_0$ such that the equality $\nu _n=m$ holds for any integer $n\ge n_0.$ Let $n$ be a positive integer. Let us denote the connecting homomorphism in the long exact sequence in the proof of Proposition 12 by $\gamma _n.$ The short exact sequence of abelian sheaves on $X$

$$\begin{aligned} 1\longrightarrow 1+\fancyscript{N}_n\longrightarrow \fancyscript{O}_{(n+1)D}^\times \longrightarrow \fancyscript{O}_{nD}^\times \longrightarrow 1 \end{aligned}$$

induces a long exact sequence

$$\begin{aligned} H^0(X,\fancyscript{O}_{nD}^\times )\longrightarrow H^1(X,1+\fancyscript{N}_n)\longrightarrow \text{ Pic}\,(n+1)D\longrightarrow \text{ Pic}nD\longrightarrow 1. \end{aligned}$$

Thus, the relation $\nu _n\mid \nu _{n+1}$ holds. Let us denote the connecting homomorphism in the above long exact sequence by $\gamma _n^\times .$ The isomorphism

$$\begin{aligned} \fancyscript{N}_n\longrightarrow 1+\fancyscript{N}_n,\quad a\longmapsto 1+a \end{aligned}$$

of abelian sheaves induces an isomorphism

$$\begin{aligned} \alpha _n:H^1(X,\fancyscript{N}_n)\longrightarrow H^1(X,1+\fancyscript{N}_n) \end{aligned}$$

of abelian groups. Since the equality $\alpha _n(\text{ Im}\gamma _n)=\text{ Im}\gamma _n^\times$ holds (Proposition in [33, §6, p. 8]) and the abelian group $H^1(X,\fancyscript{N}_n)$ is a $p$-torsion group, the quotient $\nu _{n+1}/\nu _n$ is a power of $p.$ Thus, the equalities $\nu _1=\nu$ and $\nu _{n_0}=m$ show the lemma. $\square$