Proof. Since the intersection matrix of ${\operatorname{Ex}}(f)$ is negative definite [Reference Kollár9, Theorem 10.1], we can apply the same argument as in [Reference Ishii7, Lemmas 7.2.1 and 7.2.3].

Proof. The same argument as in [Reference Ishii7, Proposition 7.2.6] works.

Proof. As for (1), [Reference Wagreich25, the proof of Proposition 2.3] works. In order to show (2), we may replace $f: Y \to X$ by a higher resolution (Proposition 2.3), and hence we may assume that ${\mathfrak{m}}_P{\mathcal{O}}_Y$ is invertible. Then (2) and (3) follow from the same argument as in [Reference Wagreich25, Theorem 2.7].

Let us show (4). We may assume that X is affine and ${\mathcal{O}}_Y(-V)$ is globally generated, i.e. there is a surjective ${\mathcal{O}}_Y$-module homomorphism

\begin{equation*} \theta: {\mathcal{O}}_Y^{\oplus N} \to {\mathcal{O}}_Y(-V), \qquad (a_1, \ldots, a_N) \mapsto a_1\zeta_1+ \cdots a_N\zeta_N, \end{equation*}

\begin{equation*} \text{where} \qquad \zeta_1, \ldots, \zeta_N \in \Gamma(Y, {\mathcal{O}}_Y(-V)). \end{equation*}

Via $\Gamma(Y, {\mathcal{O}}_Y(-V)) \subset \Gamma(Y, {\mathcal{O}}_Y) ={\mathcal{O}}_X$, (a) implies $\zeta_1, \ldots, \zeta_N \in {\mathfrak{m}}_P$. Fix $Q \in Y$ and $\xi \in {\mathcal{O}}_Y(-V)_Q$, where ${\mathcal{O}}_Y(-V)_Q$ denotes the stalk at Q. Since $\theta_Q :{\mathcal{O}}_{Y, Q}^{\oplus N} \to {\mathcal{O}}_Y(-V)_Q$ is surjective, we can write $\xi = \sum_{i=1}^N b_i \zeta_i$ for some $b_i \in {\mathcal{O}}_{Y, Q}$. This implies ${\mathcal{O}}_Y(-V) \subset {\mathfrak{m}}_P {\mathcal{O}}_Y$. Thus (4) holds. The assertion (5) immediately follows from (1) and (4).

Proof. The assertion follows from [Reference Lipman12, Theorem 12.1].

Proof. The implication $(2) \Rightarrow (3)$ is clear.

Let us show $(1) \Rightarrow (2)$. Assume (1). By the exact sequence

\begin{equation*} 0 \to {\mathcal{O}}_Y(-W) \to {\mathcal{O}}_Y \to {\mathcal{O}}_W \to 0, \end{equation*}

we obtain

\begin{equation*} 0 = R^1f_*{\mathcal{O}}_Y \to R^1f_*{\mathcal{O}}_W (\simeq H^1(W, {\mathcal{O}}_W)) \to {R^2f_*{\mathcal{O}}_Y(-Y)=0}, \end{equation*}

where the equality $R^2f_*{\mathcal{O}}_Y(-Y)=0$ follows from the fact that the dimension of every fibre of f is $\leq 1$. Thus (2) holds.

Let us show $(3) \Rightarrow (1)$. Assume (3). Set $Z := Z_f$. For any $n \in {\mathbb{Z}}_{ \gt 0}$, we have

\begin{equation*} 0 \to {\mathcal{O}}_Y(-nZ)|_Z \to {\mathcal{O}}_{(n+1)Z} \to {\mathcal{O}}_{nZ} \to 0. \end{equation*}

It follows from $H^1(Z, {\mathcal{O}}_Y(-nZ)|_Z)=0$ (Proposition 2.5) that we obtain $H^1((n+1)Z, {\mathcal{O}}_{(n+1)Z}) \simeq H^1(nZ, {\mathcal{O}}_{nZ})$. By induction on n, (3) implies $H^1(nZ, {\mathcal{O}}_{nZ})=0$ for every $n \in {\mathbb{Z}}_{ \gt 0}$. By the formal function theorem, we obtain

\begin{equation*} R^1f_*{\mathcal{O}}_Y \otimes_{{\mathcal{O}}_{X,P}} \widehat{{\mathcal{O}}}_{X, P} \simeq \varprojlim_n H^1(nZ, {\mathcal{O}}_{nZ}) =0. \end{equation*}

Since ${\mathcal{O}}_{X, P} \to \widehat{{\mathcal{O}}}_{X, P}$ is faithfully flat, we get $R^1f_*{\mathcal{O}}_Y =0$. Thus (1) holds.

Proof. Note that ${\mathcal{O}}_X(A)$ is ${(m+1)}$-regular with respect to ${\mathcal{O}}_X(A)$, so that

\begin{equation*} H^0(X, {\mathcal{O}}_X(A)) \otimes_k H^0(X, {\mathcal{O}}_X((m+r+1)A)) \to H^0(X, {\mathcal{O}}_X((m+r+2)A)) \end{equation*}

is surjective for every $r \geq 0$ [Reference Fantechi, Göttsche, Illusie, Kleiman, Nitsure and Vistoli4, Section 5.2], [Reference Tanaka21, Definition 11.1, Lemma 11.2].

Proof. The assertions (1) and (2) immediately follow from Proposition 2.5. Let us show (3). By (1), L is an ample globally generated invertible sheaf. Then the assertion follows from $H^1(X, {\mathcal{O}}_X)=0$ and Proposition 2.9. Thus (3) holds.

Let us show (4). Let $\psi : X \to Z \to Y$ be the Stein factorisation of ψ. We obtain $H^1(Z, {\mathcal{O}}_Z) \hookrightarrow H^1(X, {\mathcal{O}}_X)=0$, which implies $H^1(Z, {\mathcal{O}}_Z) =0$. Then the pullback L_Z of ${\mathcal{O}}_{{\mathbb{P}}^N}(1)$ to Z is very ample by (2). By $H^0(X, L) \simeq H^0(Z, L_Z)$, we obtain $Z \simeq Y$. Thus (4) holds.

Proof. The assertion (1) follows from Proposition 3.2. Then (1) and the Riemann–Roch theorem imply (2).

Proof. The assertion immediately follows from the conductor exact sequence (2.8.3).

Proof. Let us show (1). Fix a closed point P satisfying (a) and (b). Take a Cartier divisor D. It suffices to show $D \sim mP$ for some $m \in {\mathbb{Z}}$. Since we have $D = H - H'$ for ample Cartier divisors H and Hʹ, we may assume that D is ample. By Proposition 3.2, $|D|$ is base point free. Hence we may assume that D is an effective Cartier divisor such that X is regular around ${\operatorname{Supp}}\,D$. Then the problem is further reduced to the case when $Q := D$ is a closed point around which X is regular. If $\deg_k Q$ is divisible by $\deg_k P$, then we obtain $Q \sim m P$ by Proposition 3.2.

Suppose that $\deg_k Q$ is not divisible by $\deg_k P$. It suffices to derive a contradiction. By taking the greatest common divisor d of $\deg_k P$ and $\deg_k Q$, we can find an ample Cartier divisor E on X such that $\deg_k E =d$. Automatically, we get $d \lt \deg_k P$. Since $|E|$ is base point free (Proposition 3.2), we may assume that E is effective and X is regular around ${\operatorname{Supp}}\,E$. Pick a closed point $R \in {\operatorname{Supp}}\,E$. Then

\begin{equation*} \deg_k R \leq \deg_k E = d \lt \deg_k P, \end{equation*}

which contradicts the assumption (b) on P. Thus (1) holds.

Let us show (2). We can find a closed point P satisfying (a) and (b). Since $|P|$ is base point free (Proposition 3.2), the assertion (2) follows from $\dim_k H^0(X, {\mathcal{O}}_X(P)) \gt \dim_k H^0(X, {\mathcal{O}}_X)$ (Proposition 3.3).

Proof. Suppose $\Delta(X, D) \lt 0$. By $\Delta(X, D) = 1 + \deg D - h^0(X, D)$, we obtain $h^0(X, D) \gt 1+\deg D \geq 2$. In particular, we may assume that D is effective. By the exact sequence

\begin{equation*} 0 \to {\mathcal{O}}_X \to {\mathcal{O}}_X(D) \to {\mathcal{O}}_X(D)|_D \to 0, \end{equation*}

we get a surjection $H^1(X, {\mathcal{O}}_X) \to H^1(X, {\mathcal{O}}_X(D))$, where $H^1(D, {\mathcal{O}}_X(D)|_D)=0$ holds by $\dim D = 0$. In particular, $h^1(X, {\mathcal{O}}_X) \geq h^1(X, {\mathcal{O}}_X(D))$.

In order to derive a contradiction, it suffices to show $h^1(X, {\mathcal{O}}_X) \lt h^1(X, {\mathcal{O}}_X(D))$. The Riemann–Roch theorem implies

\begin{equation*} h^0(X, D) - h^1(X, D) = \chi(X, D) = \chi(X, {\mathcal{O}}_X) + \deg D = 1 -h^1(X, {\mathcal{O}}_X)+ \deg D. \end{equation*}

Therefore, we get

\begin{equation*} h^1(X, D) - h^1(X, {\mathcal{O}}_X) = h^0(X, D) -(1+\deg D) = -\Delta(X, D) \gt 0, \end{equation*}

as required.

Proof. By the Riemann–Roch theorem and Proposition 3.3, we get (1) $\Rightarrow$ (2). It is clear that (2) $\Rightarrow$ (3).

Let us show that (3) $\Rightarrow$ (1). Assume (3), i.e. there exists an ample Cartier divisor D on X such that $1 + \deg D- h^0(X, D) = \Delta(X, D) =0$. By $h^0(X, D) = 1+ \deg D \gt 0$, we may assume that D is effective. The Riemann–Roch theorem implies

\begin{equation*} h^0(X, D) - h^1(X, D) = \chi(X, D) = \chi(X, {\mathcal{O}}_X) + \deg D = 1 -h^1(X, {\mathcal{O}}_X)+ \deg D. \end{equation*}

This, together with $h^0(X, D) = 1+ \deg D$, implies $h^1(X, D) = h^1(X, {\mathcal{O}}_X)$. Consider the following exact sequence:

\begin{equation*} 0 \to H^0(X, {\mathcal{O}}_X) \to H^0(X, {\mathcal{O}}_X(D)) \xrightarrow{\alpha} H^0(D, {\mathcal{O}}_X(D)|_D) \end{equation*}

\begin{equation*} \to H^1(X, {\mathcal{O}}_X) \xrightarrow{\simeq} H^1(X, {\mathcal{O}}_X(D)) \to 0. \end{equation*}

Then α is surjective and hence $|D|$ is base point free. In particular, we can find an effective Cartier divisor Dʹ such that $D \sim D'$ and ${\operatorname{Supp}}\,D \cap {\operatorname{Supp}}\,D' = \emptyset$. For every $n \in {\mathbb{Z}}_{\geq 0}$, we get the following commutative diagram consisting of the induced maps:

Since α is surjective, so is β_n. Recall that β_n is obtained from the following exact sequence:

\begin{equation*} 0 \to H^0(X, {\mathcal{O}}_X(nD')) \to H^0(X, {\mathcal{O}}_X(D+nD')) \xrightarrow{\beta_n} H^0(D, {\mathcal{O}}_X(D+nD')|_D) \end{equation*}

\begin{equation*} \to H^1(X, {\mathcal{O}}_X(nD')) \xrightarrow{\theta} H^1(X, {\mathcal{O}}_X(D+nD')) \to 0. \end{equation*}

As β_n is surjective, we obtain

\begin{equation*} H^1(X, {\mathcal{O}}_X(nD)) \simeq H^1(X, {\mathcal{O}}_X(nD')) \overset{\theta}{\simeq} H^1(X, {\mathcal{O}}_X(D+nD')) \simeq H^1(X, {\mathcal{O}}_X( (n+1)D). \end{equation*}

Since n is chosen to be an arbitrary non-negative integer, we get

\begin{equation*} H^1(X, {\mathcal{O}}_X) \simeq H^1(X, {\mathcal{O}}_X(D)) \simeq H^1(X, {\mathcal{O}}_X(2D)) \simeq \cdots \simeq H^1(X, {\mathcal{O}}_X(mD)) =0 \end{equation*}

for $m \gg 0$, where the last equality follows from the Serre vanishing theorem. Thus (1) holds.

Proof. Let us show (1). By the Riemann–Roch theorem, we may assume that D is effective. Fix a closed point $P \in {\operatorname{Supp}}\,D$ around which X is regular. It is enough to show $P \not\in {\operatorname{Bs}}\,|D|$. We have

\begin{equation*} H^0(X, D) \xrightarrow{{\rho}} H^0(P, D|_P) \to H^1(X, D-P) \to H^1(X, D) =0, \end{equation*}

where $H^1(X, D)=0$ follows from Serre duality. If D ≠ P, then we get D > P as D is effective, which implies $H^1(X, D-P)=0$ again by Serre duality. Hence the problem is reduced to the case when D = P. Then $h^0(P, D|_P) = \deg P =\deg D \geq 2$, whilst $h^1(X, D-P) = h^1(X, {\mathcal{O}}_X)=1$. Therefore, the restriction map

\begin{equation*} \rho : H^0(X, D) \to H^0(P, D|_P) \end{equation*}

is non-zero. Take an effective divisor D ^ʹ such that the corresponding element $s_{D'} \in H^0(X, D)$ satisfies $\rho(s_{D'}) \neq 0$. We then obtain $P \not\in {\operatorname{Supp}}\,D'$. Therefore, $P \not\in {\operatorname{Bs}}\,|D|$. Thus (1) holds.

Let us show (2). Consider the following exact sequence:

\begin{equation*} 0 \to H^0(X, {\mathcal{O}}_X(D-P)) \to H^0(X, {\mathcal{O}}_X(D)) \to H^0(P, {\mathcal{O}}_X(D)|_P). \end{equation*}

By

\begin{equation*} h^0(X, {\mathcal{O}}_X(D)) =\chi(X, {\mathcal{O}}_X(D)) =\deg D \gt \deg P = h^0(P, {\mathcal{O}}_X(D)|_P), \end{equation*}

we get $H^0(X, {\mathcal{O}}_X(D-P)) \neq 0$. In particular, the problem is reduced to the case when $D = P+D'$ for some effective Cartier divisor Dʹ. By (1), we get

\begin{equation*} {\operatorname{Bs}}\,|D| \subset X_{{\rm non-reg}} \cap {\operatorname{Supp}}(P+D') \subset {\operatorname{Supp}}\,D', \end{equation*}

where the latter inclusion holds by the fact that X is regular around P. Thus it suffices to show that ${\operatorname{Bs}}\,|D| \cap {\operatorname{Supp}}\,D' = \emptyset$. By the exact sequence

\begin{equation*} 0 \to {\mathcal{O}}_X(D -D') \to {\mathcal{O}}_X(D) \to {\mathcal{O}}_X(D)|_{D'} \to 0, \end{equation*}

we obtain another exact sequence

\begin{equation*} H^0(X, {\mathcal{O}}_X(D)) \to H^0(D', {\mathcal{O}}_X(D)|_{D'}) \to H^1(X, {\mathcal{O}}_X(D -D')) = H^1(X, {\mathcal{O}}_X(P))=0. \end{equation*}

Therefore, ${\operatorname{Bs}}\,|D|$ is disjoint from Dʹ, as required. Thus (2) holds

Proof. Taking the base change to the separable closure, the problem is reduced to the case when k is separably closed. If X is geometrically integral, then the assertion follows from [Reference Tanaka21, Lemma 11.10(3)]. Hence we may assume that X is not geometrically integral. In particular, k is an infinite field.

Let $\nu : Y \to X$ be the normalisation. Let $C_X \subset X$ and $C_Y \subset Y$ be the closed subschemes defined by the conductor of ν. We get the following Cartesian diagram:

Set $k_Y := H^0(Y, {\mathcal{O}}_Y)$. Recall that we have

\begin{equation*} K_Y + C_Y \sim \nu^*K_X \sim 0. \end{equation*}

If X is regular, then the assertion follows from Lemma 4.3(1). Then we may assume that $C_Y \neq 0$ and hence Y is a regular curve of genus zero.

Proof of Step 2.

Assume $\# C_X=1$. In this case, $P:=(C_X)_{{\operatorname{red}}}$ is a unique non-regular closed point on X. Fix a regular closed point $Q \in X$ such that $\deg Q = \deg D$, whose existence is guaranteed by Proposition 3.5. We have $Q \neq (C_X)_{{\operatorname{red}}} =P$. For

\begin{equation*} {\operatorname{Spec}}\,A := X \setminus Q, \qquad {\operatorname{Spec}}\,\overline{A} := X \setminus P = X \setminus (C_X)_{{\operatorname{red}}}, \end{equation*}

we get an affine open cover $X = {\operatorname{Spec}}\,A \cup {\operatorname{Spec}}\,\overline{A}$. Take their inverse images to Y:

\begin{equation*} {\operatorname{Spec}}\,B := \nu^{-1}({\operatorname{Spec}}\,A), \qquad {\operatorname{Spec}}\,\overline{B} := \nu^{-1}({\operatorname{Spec}}\,\overline{B}). \end{equation*}

In particular, we have $ Y = {\operatorname{Spec}}\,B \cup {\operatorname{Spec}}\,\overline{B}$ and $\overline{A} = \overline B$. By Step 1, we can find $f \in A$ such that

\begin{equation*} {\mathcal{O}}_X(-D)|_{{\operatorname{Spec}}\,A} = fA, \qquad {\mathcal{O}}_X(-D)|_{{\operatorname{Spec}}\,\overline{A}} = \overline{A} \end{equation*}

\begin{equation*} {\mathcal{O}}_Y(-\nu^*D)|_{{\operatorname{Spec}}\,B} = fB, \qquad {\mathcal{O}}_Y(-\nu^*D)|_{{\operatorname{Spec}}\,\overline{B}}=\overline{B}. \end{equation*}

Set $P_Y := (\nu^{-1}(P))_{{\operatorname{red}}}$ and $Q_Y := \nu^{-1}(Q)$. Both P_Y and Q_Y are closed points (Step 1(d)). By Step 1, we have an equality $\nu^*D|_{{\operatorname{Spec}}\,B} ={\rm div}(f)|_{{\operatorname{Spec}}\,B} = P_Y$. We then have ${\rm div}(f) = P_Y + m Q_Y$ for some $m \in {\mathbb{Z}}$. We obtain $m=-1$, i.e. ${\rm div}(f) = P_Y-Q_Y$, because $\deg Q_Y = \deg Q = \deg D = \deg P_Y$ and $0 = \deg ({\rm div}(f)) = \deg P_Y + m \deg Q_Y$. In particular, $f^{-1} \in \overline B = \overline A$. Set $g := 1+f \in A$. Then

\begin{equation*} \frac{g}{f} = \frac{1+f}{f} = 1 + f^{-1} \in \overline{B} = \overline{A}. \end{equation*}

Therefore, we obtain $\frac{g}{f} \in H^0(X, {\mathcal{O}}_X(D))$. Indeed,

• • $\frac{g}{f} \in f^{-1}A ={\mathcal{O}}_X(D)|_{{\operatorname{Spec}}\,A}$, and

• • $\frac{g}{f} \in \overline{A} ={\mathcal{O}}_X(D)|_{{\operatorname{Spec}}\,\overline{A}}$.

Since $\frac{g}{f}$ does not vanish at the point $P =D_{{\operatorname{red}}}$, $|D|$ is base point free. This completes the proof of Step 2.

Proof of Step 3.

Assume that $\# C_X \geq 2$. By ${\# C_Y \geq \# C_X \geq 2}$, $K_Y +C_Y \sim 0$, and Example 3.9, we obtain $Y \simeq {\mathbb{P}}^1_{k_Y}$ and $C_Y = C_{Y, 1} + C_{Y, 2}$, where $k_Y :=H^0(Y, {\mathcal{O}}_Y)$ and each of $C_{Y, 1}$ and $C_{Y, 2}$ is a k_Y-rational point. In particular, $\# C_X = \# C_Y =2$. For each $i \in \{1, 2\}$,

\begin{equation*} k_i := {\mathcal{O}}_{C_{X, i}} \hookrightarrow {\mathcal{O}}_{C_{Y, i}} =k_Y \end{equation*}

is a field extension, where each $C_{X, i}$ denotes the connected component of C_X containing $\nu(C_{Y, i})_{{\operatorname{red}}}$. We may assume that ${\operatorname{Supp}}\,D = {\operatorname{Supp}}\,C_{X, 1}$.

Set

\begin{equation*} {\operatorname{Spec}}\,B := Y \setminus C_{Y, 2}, \quad {\operatorname{Spec}}\,\overline{B} := Y \setminus C_{Y, 1}, \quad {\operatorname{Spec}}\,A := X \setminus C_{X, 2}, \quad {\operatorname{Spec}}\,\overline{A} := X \setminus C_{X, 1}. \end{equation*}

We introduce affine coordinates

\begin{equation*} B =k_Y[x], \qquad \overline{B} = k_Y[y], \end{equation*}

where the gluing is given by the usual relation xy = 1. We get

\begin{equation*} A = k_1 \oplus xk_Y[x] \subset k_Y[x] =B, \qquad \overline{A} = k_2 \oplus yk_Y[y] \subset k_Y[y] =\overline{B}. \end{equation*}

We can write

\begin{eqnarray*} {{\mathcal{O}}_X(-D)}|_{{\operatorname{Spec}}\,A} &=& (b_1x + b_2 x^2 + \cdots +b_Nx^N)A \quad \text{and}\quad\\ {{\mathcal{O}}_Y(-\nu^*D)}|_{{\operatorname{Spec}}\,B} &=& (b_1x + b_2 x^2 + \cdots +b_Nx^N)B \end{eqnarray*}

for some $b_1, b_2, \ldots, b_N \in k_Y$. By Step 1, we obtain an equality of principal ideals: $(b_1x + b_2 x^2 + \cdots +b_Nx^N)B = xB$. Then $(b_1x + b_2 x^2 + \cdots +b_Nx^N)g(x) = x$ holds for some $g(x) \in k_Y[x]$, so that $b_1x + b_2 x^2 + \cdots +b_Nx^N = b_1x$, i.e. N = 1. Set $b:= b_1 \in k_Y^{\times}$. We get

\begin{equation*} {\mathcal{O}}_X(-D)|_{{\operatorname{Spec}}\,A} = bx A \quad \text{and}\quad {\mathcal{O}}_Y(-\nu^*D)|_{{\operatorname{Spec}}\,B} = bxB (=xB). \end{equation*}

Note that we possibly have $bx A \neq xA$. We obtain

\begin{equation*} \zeta := \frac{b(x+b^{-1})}{bx} \in H^0(X, {\mathcal{O}}_X(D)). \end{equation*}

Indeed, the following hold.

• • $\zeta = \frac{b(x+b^{-1})}{bx} = \frac{1 +bx}{bx} \in (bx)^{-1}(k_1 \oplus xk_Y[x]) = (bx)^{-1}A = {\mathcal{O}}_X(D)|_{{\operatorname{Spec}}\,A}$.

• • $\zeta|_{{\operatorname{Spec}}\,\overline{A}} = \frac{b(y^{-1}+b^{-1})}{by^{-1}} = 1 + b^{-1}y \in k_2 + yk_Y[y] = \overline{A} = {\mathcal{O}}_X(D)|_{{\operatorname{Spec}}\,\overline{A}}$.

Since ζ does not vanish at the point $(C_{X,1})_{{\operatorname{red}}} =D_{{\operatorname{red}}}$, $|D|$ is base point free. This completes the proof of Step 3.

Steps 2 and 3 complete the proof of Theorem 4.4.

Proof. By Theorem 4.4, D is a globally generated ample Cartier divisor. The assertion follows from Proposition 2.9 and $H^1(X, {\mathcal{O}}_X(D))=0$.

Notation 4.6. Let k be a field and let X be a Gorenstein genus-one curve over k. Let D be a Cartier divisor on X with $\deg D \geq 3$. Note that $|D|$ is base point free (Theorem 4.4). Let

\begin{equation*} \varphi_{|D|} : X \xrightarrow{\psi} Y \overset{j}{\hookrightarrow} {\mathbb{P}}^{N}_k \end{equation*}

be the induced morphisms, where $N := h^0(X, D) -1 = \deg D -1$, $Y := \varphi_{|D|}(X)$, and j denotes the induced closed immersion.

Proof. Fix an ample Cartier divisor D_Y such that ${\mathcal{O}}_Y(D_Y) \simeq {\mathcal{O}}_{{\mathbb{P}}^{N}_k}(1)|_Y$. We have $D \sim \psi^*D_Y$. Since $\psi : X \to Y$ is a surjective k-morphism, we obtain

\begin{equation*} k \hookrightarrow H^0(Y, {\mathcal{O}}_Y) \hookrightarrow H^0(X, {\mathcal{O}}_X)=k, \end{equation*}

which implies $H^0(Y, {\mathcal{O}}_Y)=k$. In particular, we get $0 \leq \Delta(Y, D_Y) = \dim Y + \deg D_Y - h^0(Y, D_Y)$ (Lemma 3.7). Then we obtain

\begin{eqnarray*} 0 & \leq & \Delta(Y, D_Y)\\ &=& \dim Y + \deg D_Y - h^0(Y, D_Y)\\ &=& 1 + \frac{\deg D}{\deg \psi } - h^0(X, D)\\ &=& 1 + \frac{\deg D}{\deg \psi } - \deg D\\ &=& 1 + \deg D \left( \frac{1}{\deg \psi } - 1\right)\\ &\leq & 1 + 3 \left( \frac{1}{\deg \psi } - 1\right). \end{eqnarray*}

This implies that $\deg \psi \leq \frac{3}{2}$, i.e. ψ is birational.

Proof. We have the following commutative diagram consisting of the induced maps:

Note that we have $\psi^{-1}(Y \cap \, H) \xrightarrow{\simeq} Y \cap \, H$, because $\psi : X \to Y$ is birational (Lemma 4.7) and H is chosen to be a general hyperplane, so that $Y \cap \, H$ is contained in the isomorphic locus of $\psi : X \to Y$. Here we used the fact that if k is infinite, then a general hyperplane $H \subset {\mathbb{P}}^N_k$ can avoid given finitely many closed points on ${\mathbb{P}}^N_k$. Since $D' := \psi^{-1}(Y \cap H)$ is a member of $|D|$, we obtain an exact sequence:

\begin{equation*} H^0(X, {\mathcal{O}}_X(D)) \xrightarrow{\beta} H^0(D', {\mathcal{O}}_X(D)|_{D'}) \to H^1(X, {\mathcal{O}}_X(D -D')). \end{equation*}

Hence the assertion follows from $\dim_k H^1(X, {\mathcal{O}}_X(D-D')) = \dim_k H^1(X, {\mathcal{O}}_X) =1$.

Proof. We may assume that k is separably closed. In particular, k is an infinite field. If X is geometrically integral, then there is nothing to show [Reference Tanaka21, Proposition 11.11(3)]. Hence we may assume that X is not geometrically integral. Then any regular closed point P of X is not k-rational [Reference Tanaka21, Corollary 2.14], i.e. $\deg P \geq 2$. In particular, $|P|$ is base point free (Theorem 4.4). Fix a general hyperplane H on ${\mathbb{P}}^N_k$, so that Y is regular around $Y \cap H$ and $\psi : X \to Y$ is isomorphic around $Y \cap H$ (Lemma4.7).

Set ${\mathcal{O}}_Y(\ell) := {\mathcal{O}}_{{\mathbb{P}}^N_k}(\ell)|_Y$ for every $\ell \in {\mathbb{Z}}$. It is enough to prove (A)–(E) below.

1. (A) $H^0({\mathbb{P}}^N_k, {\mathcal{O}}_{{\mathbb{P}}^N_k}(n)) \to H^0(Y \cap H, {\mathcal{O}}_{{\mathbb{P}}^N_k}(n)|_{Y \cap H})$ is surjective for every $n \geq 2$.

2. (B) $H^1(Y, {\mathcal{O}}_Y(m))=0$ for every $m \in {\mathbb{Z}}_{ \gt 0}$.

3. (C) $\psi: X \to Y$ is an isomorphism.

4. (D) $\alpha_n : H^0({\mathbb{P}}^N, {\mathcal{O}}_{{\mathbb{P}}^N}(n)) \to H^0(Y, {\mathcal{O}}_{Y}(n))$ is surjective for every $n \geq 0$.

5. (E) $H^0(Y, {\mathcal{O}}_{Y}(m-1)) \otimes_k H^0(Y, {\mathcal{O}}_{Y}(1)) \to H^0(Y, {\mathcal{O}}_{Y}(m))$ is surjective for every $m \geq 1$.

Indeed, (C) and (E) imply that $\bigoplus_{m=0}^{\infty} H^0(X, {\mathcal{O}}_X(mD))$ is generated by $H^0(X, {\mathcal{O}}_X(D))$.

Let us show (A). By Lemmas 4.8 and 4.10,

\begin{equation*} H^0({\mathbb{P}}^N_k, {\mathcal{O}}_{{\mathbb{P}}^N_k}(2)) \to H^0(Y \cap H, {\mathcal{O}}_{{\mathbb{P}}^N_k}(2)|_{Y \cap H}) \end{equation*}

is surjective. In particular,

\begin{equation*} H^0({\mathbb{P}}^N_k, {\mathcal{O}}_{{\mathbb{P}}^N_k}(n)) \to H^0(Y \cap H, {\mathcal{O}}_Y(n)|_{Y \cap H}) \end{equation*}

is surjective for every $n \geq 2$. Thus (A) holds.

Let us show (B). For an integer $n \geq 2$, we have the exact sequence:

\begin{equation*} H^0(Y, {\mathcal{O}}_Y(n)) \to H^0(Y \cap H, {\mathcal{O}}_Y(n)|_{Y \cap H}) \to H^1(Y, {\mathcal{O}}_Y(n-1)) \to H^1(Y, {\mathcal{O}}_Y(n)) \to 0. \end{equation*}

It follows from (A) and the Serre vanishing theorem that $H^1(Y, {\mathcal{O}}_Y(m))=0$ for every $m \in {\mathbb{Z}}_{ \gt 0}$. Thus (B) holds.

Let us show (C). By (B), we have

(4.9.1)\begin{equation} \chi(Y, {\mathcal{O}}_Y(1)) = h^0(Y, {\mathcal{O}}_Y(1)) = h^0(X, {\mathcal{O}}_X(D)) = \chi(X, {\mathcal{O}}_X(D)). \end{equation}

For the closed subschemes $C_X \subset X$ and $C_Y \subset Y$ defined by the conductor of $\psi: X \to Y$, we have the following conductor exact sequence (2.8.3):

(4.9.2)\begin{equation} 0 \to {\mathcal{O}}_Y \to \psi_* {\mathcal{O}}_X \oplus {\mathcal{O}}_{C_Y} \to \psi_*{\mathcal{O}}_{C_X} \to 0. \end{equation}

We then obtain

\begin{equation*} \chi(Y, {\mathcal{O}}_Y) - \chi(X, {\mathcal{O}}_X) = \chi(C_Y, {\mathcal{O}}_{C_Y}) - \chi(C_X, {\mathcal{O}}_{C_X}) \overset{{\rm (i)}}{=} \chi(Y, {\mathcal{O}}_Y(1)) -\chi(X, {\mathcal{O}}_X(D))\overset{{\rm (ii)}}{=} 0, \end{equation*}

where (ii) follows from (4.9.1) and (i) is obtained by applying the tensor product $(-) \otimes {\mathcal{O}}_Y(1)$ to the exact sequence (4.9.2). By $H^0(X, {\mathcal{O}}_X)=H^0(Y, {\mathcal{O}}_Y) =k$, we get $h^1(X, {\mathcal{O}}_X)=h^1(Y, {\mathcal{O}}_Y)$. Then (4.9.2) induces the following exact sequence:

\begin{equation*} 0 \to H^0(Y, {\mathcal{O}}_Y) \to H^0(X, {\mathcal{O}}_X) \oplus H^0(C_Y, {\mathcal{O}}_{C_Y}) \to H^0(C_X, {\mathcal{O}}_{C_X}) \to 0. \end{equation*}

Again by $H^0(X, {\mathcal{O}}_X)=H^0(Y, {\mathcal{O}}_Y) =k$, we get the induced k-linear isomorphism:

\begin{equation*} H^0(C_Y, {\mathcal{O}}_{C_Y}) \xrightarrow{\simeq} H^0(C_X, {\mathcal{O}}_{C_X}), \end{equation*}

which implies ${\mathcal{O}}_{C_Y} \xrightarrow{\simeq} \psi_*{\mathcal{O}}_{C_X}$. Again by the conductor exact sequence (4.9.2), we obtain ${\mathcal{O}}_Y \xrightarrow{\simeq} \psi_*{\mathcal{O}}_X$, i.e. ψ is an isomorphism. Thus (C) holds.

Let us show (D). If $n \in \{0, 1\}$, then α_n is surjective by construction. Fix $n \in {\mathbb{Z}}_{\geq 2}$. We have the following commutative diagram in which each horizontal sequence is exact (note that $H^1({\mathcal{O}}_Y(n-1))=0$ by (B)):

By $n \geq 2$, the restriction map

\begin{equation*} H^0({\mathcal{O}}_{{\mathbb{P}}^{N}}(n)) \to H^0({\mathcal{O}}_{Y \cap H}(n)) \end{equation*}

is surjective by (A). In particular, γ_n is surjective. By the snake lemma, the surjectivity of $\alpha_{n-1}$ implies the surjectivity of α_n. By induction on n, α_n is surjective. Thus (D) holds.

Let us show (E). Fix $m \geq 1$. We have the following commutative diagram:

Since µ and α_m are surjective by (D), also $\mu'$ is surjective. Thus (E) holds.

Proof. Taking the base change to the algebraic closure of k, the problem is reduced to the case when k is algebraically closed. We have

\begin{equation*} Z = Z_1 \amalg \cdots \amalg Z_m, \end{equation*}

where each $(Z_i)_{{\operatorname{red}}}$ is one point.

We introduce a homogeneous coordinate ${\mathbb{P}}^N_k = {\operatorname{Proj}}\,k[X_0, \ldots, X_N]$ and fix an affine space:

\begin{equation*} {\mathbb{A}}^N_k := {\operatorname{Spec}}\,k[x_1, \ldots, x_N] \subset {\mathbb{P}}^N_k, \qquad \text{where}\qquad x_i := X_i/X_0. \end{equation*}

We may assume that $Z \subset {\mathbb{A}}^N_k$. Set $R := {\mathcal{O}}_Z(Z)$, which is an Artinian ring, equipped with the induced surjective k-algebra homomorphism

\begin{equation*} \pi : k[x_1, \ldots, x_N] \to R. \end{equation*}

Set $\overline{x}_i := \pi(x_i)$. We obtain the following commutative diagram:

We now compute under the affine local setting. Via the above right vertical isomorphisms, we identify the following three k-vector spaces:

\begin{equation*} (R=)H^0(Z, {\mathcal{O}}_Z),\qquad H^0(Z, {\mathcal{O}}_{{\mathbb{P}}^N_k}(1)|_Z),\qquad H^0(Z, {\mathcal{O}}_{{\mathbb{P}}^N_k}(2)|_Z). \end{equation*}

Then ρ ₁ can be rewritten as

\begin{equation*} \rho_1^{{\operatorname{aff}}} : k \oplus k x_1 \oplus \cdots \oplus k x_N \to R, \qquad 1 \mapsto 1, \quad x_i \mapsto \overline{x}_i. \end{equation*}

Similarly, ρ ₂ is given by

\begin{equation*} \rho_2^{{\operatorname{aff}}} : k \oplus k x_1 \oplus \cdots \oplus k x_N \oplus \bigoplus_{1 \leq i \lt j \leq N} k x_ix_j \to R, \qquad 1 \mapsto 1, \quad x_i \mapsto \overline{x}_i, \quad x_ix_j \mapsto \overline{x}_i\overline{x}_j. \end{equation*}

Since the induced ring homomorphism $\pi: k[x_1, \ldots, x_N] \to R$ is a surjective k-linear map, R is generated by the monomials $\{\overline{x}_1^{d_1} \cdots \overline{x}_N^{d_N}\}_{d_1, \ldots, d_N}$ as a k-linear space. Let $I_1^{{\operatorname{aff}}}$ and $I_2^{{\operatorname{aff}}}$ be the images of $\rho_1^{{\operatorname{aff}}}$ and $\rho_2^{{\operatorname{aff}}}$, respectively. In particular, $I_1^{{\operatorname{aff}}} \subset I_2^{{\operatorname{aff}}}$. If $I_1^{{\operatorname{aff}}} = R$, then there is nothing to show. Assume $I_1^{{\operatorname{aff}}} \subsetneq R$. It suffices to show that $I_1^{{\operatorname{aff}}} \subsetneq I_2^{{\operatorname{aff}}}$. Since $I_1^{{\operatorname{aff}}}$ contains $1, \overline{x}_1, \ldots, \overline{x}_N$, we can find a unique integer $d \geq 2$ such that

1. (i) $I_1^{{\operatorname{aff}}}$ contains all the monomials $\overline{x}_1^{e_1} \cdots \overline{x}_N^{e_N}$ satisfying $\sum_i e_i \lt d$, and

2. (ii) there exists a monomial $\overline{x}_1^{d_1} \cdots \overline{x}_N^{d_N} \not\in I_1^{{\operatorname{aff}}}$ with $\sum_i d_i =d$.

By (ii), it is enough to show that $I_2^{{\operatorname{aff}}}$ contains $\overline{x}_1^{d_1} \cdots \overline{x}_N^{d_N}$. Since we have d > 0, we may assume, after permuting the indices, that $d_1 \gt 0$. By (i), we can write

\begin{equation*} \rho_1^{{\operatorname{aff}}}( a + b_1x_1 + \cdots + b_Nx_N) = \overline{x}_1^{d_1-1} \overline{x}_2^{d_2}\cdots \overline{x}_N^{d_N} \end{equation*}

for some $a, b_1, \ldots, b_N \in k$. Then we obtain

\begin{equation*} \rho_2^{{\operatorname{aff}}}( x_1(a + b_1x_1 + \cdots + b_Nx_N)) = \overline{x}_1^{d_1} \overline{x}_2^{d_2}\cdots \overline{x}_N^{d_N}, \end{equation*}

as required. This completes the proof.

Proof. Replacing X by a suitable affine open neighbourhood of $P \in X$, we may assume that X is affine and $K_X \sim 0$.

Let us show (1). We have the exact sequence

\begin{equation*} 0 \to {\mathcal{O}}_Y(-Z) \to {\mathcal{O}}_Y \to {\mathcal{O}}_Z \to 0. \end{equation*}

By the Kawamata–Viehweg vanishing theorem for the birational morphism $f : Y \to X$ [Reference Kollár9, Theorem 10.4], we obtain $R^1f_*{\mathcal{O}}_Y(-Z) \simeq R^1f_*{\mathcal{O}}_Y(K_Y) =0$. Hence

\begin{equation*} k \simeq R^1f_*{\mathcal{O}}_Y \xrightarrow{\simeq} H^1(Z, {\mathcal{O}}_Z). \end{equation*}

Since $H^1(Z, {\mathcal{O}}_Z)$ is a $\kappa(P)$-vector space satisfying

\begin{equation*} 1 = \dim_k H^1(Z, {\mathcal{O}}_Z) = [\kappa(P):k] \dim_{\kappa(P)} H^1(Z, {\mathcal{O}}_Z), \end{equation*}

we obtain $[\kappa(P):k]=1$, i.e. P is a k-rational point. Thus (1) holds.

Let us show (2). Take Z ^ʹ as in the statement of (2). For the non-zero effective divisor $Z'' := Z - Z'$, we obtain an exact sequence

\begin{equation*} 0 \to {\mathcal{O}}_Y(K_Y+Z') \xrightarrow{\alpha} {\mathcal{O}}_Y(K_Y+Z) \xrightarrow{\beta} {\mathcal{O}}_Y(K_Y+Z)|_{Z''} \to 0. \end{equation*}

By ${\mathcal{O}}_Y(K_Y+Z) \simeq {\mathcal{O}}_Y$, the map

\begin{equation*} H^0(\beta) : H^0(Y, {\mathcal{O}}_Y(K_Y+Z)) \to H^0(Z'', {\mathcal{O}}_Y(K_Y+Z)|_{Z''}) \end{equation*}

is non-zero. Hence $H^0(\alpha) : H^0({\mathcal{O}}_Y(K_Y+Z')) \hookrightarrow H^0({\mathcal{O}}_Y(K_Y+Z))$ is injective but not bijective. We have the following commutative diagram in which each horizontal sequence is exact:

By $R^1f_*{\mathcal{O}}_Y(K_Y)=0$ and the snake lemma, $H^0(\gamma): H^0(Z', \omega_{Z'}) \to H^0(Z, \omega_Z)$ is injective but not bijective. It follows from Serre duality that $H^1(Z, {\mathcal{O}}_Z) \to H^1(Z', {\mathcal{O}}_{Z'})$ is surjective but not bijective. By (1), we get $H^1(Z', {\mathcal{O}}_{Z'})=0$. Thus (2) holds.

Let us show (3). By $K_Y + Z \sim f^*K_X$, −Z is f-nef. Hence we get $Z_f \leq Z$. Suppose $Z_f \lt Z$. Then (2) implies $H^1(Z_f, {\mathcal{O}}_{Z_f})=0$. However, this leads to a contradiction: $R^1f_*{\mathcal{O}}_Y = 0$ (Proposition 2.6). Thus (3) holds.

Let us show (4). By $R^1f_*{\mathcal{O}}_Y(-Z)=0$, there is the following commutative diagram in which each horizontal sequence is exact:

where ζ is induced by $H^0(X, {\mathfrak{m}}_P) \subset H^0(Y, {\mathfrak{m}}_P {\mathcal{O}}_Y) \subset H^0(Y, {\mathcal{O}}_Y(-Z))$ (Proposition 2.4(1)) and ζ induces ξ. By diagram chase, ξ is surjective. Since $\xi : {\mathcal{O}}_P \to f_*{\mathcal{O}}_Z$ is a ring homomorphism from a field ${\mathcal{O}}_P$, ξ is injective. Thus ξ is bijective. By the snake lemma, also ζ is bijective. Thus (4) holds.

Notation 5.3. Let k be a field and let $P \in X$ be an elliptic singularity over k. Assume that X is affine, $K_X \sim 0$ and $X \setminus P$ is regular. Let $f : Y \to X$ be the minimal resolution of X. Take the canonical model $\overline Y$ of Y over X:

\begin{equation*} f : Y \xrightarrow{g} \overline Y \xrightarrow{\overline f} X. \end{equation*}

More explicitly, $g: Y \to \overline Y$ is the projective birational morphism to a normal surface such that ${\operatorname{Ex}}(g)$ consists of all the f-exceptional prime divisors C with $K_Y \cdot C=0$. Note that the existence of g is guaranteed by [Reference Tanaka20, Theorem 4.2]. In this case, both $\overline Y$ and $\overline f : \overline Y \to X$ are called the canonical model (over X). Set $Z :=Z_f$ and $\overline Z := g_*Z$.

Proof. Let us show (1). If $\overline {Z}=0$, then Y would be canonical, which contradicts $R^1f_*{\mathcal{O}}_Y \neq 0$ [Reference Kollár9, Theorem 10.4]. Hence $\overline Z \neq 0$. By $K_Y + Z \sim 0$ and $\overline Z =g_*Z$, we obtain $K_{\overline Y} + \overline Z =g_*(K_Y + Z) \sim 0$. Since $K_{\overline Y}$ is Cartier, $\overline Z$ is an effective Cartier divisor. Moreover, we obtain $\overline Z = g^*Z$ by the negativity lemma [Reference Tanaka20, Lemma 2.11(1)] and

\begin{equation*} K_{\overline Y} + \overline Z \sim 0 \sim g^*(K_Y + Z) \sim K_{\overline Y} + g^*Z. \end{equation*}

Thus (1) holds.

Let us show (2). We have $g^*\overline Z = Z$ by (1). This means that Z is the scheme-theoretic inverse image of Zʹ, so that the diagram (5.4.1) is Cartesian. Thus (2) holds.

Let us show (3). By $g^*\overline Z = Z$, we get $g_*{\mathcal{O}}_Y(-Z) = {\mathcal{O}}_{\overline Y}(-\overline Z)$. Therefore, we obtain the following commutative diagram in which each horizontal sequence is exact:

where $R^1g_*{\mathcal{O}}_Y(-Z) \simeq R^1g_*{\mathcal{O}}_Y(K_Y)=0$. Therefore, the snake lemma implies $(g|_Z)_*{\mathcal{O}}_Z = {\mathcal{O}}_{\overline Z}$. Thus (3) holds.

Proof. Let us show (1). We have

\begin{equation*} 0 \to L(-Z')|_C \to L \to L|_{Z'} \to 0, \end{equation*}

where $L(-Z') := {\mathcal{O}}_Y(-Z')|_Z \otimes L$. It suffices to show $H^1(C, L(-Z')|_C)=0$. By

\begin{equation*} Z' \cdot C = ( (K_Y+Z) -(K_Y+C)) \cdot C = -\deg \omega_C, \end{equation*}

we obtain

\begin{equation*} \deg (L(-Z')|_C) = L \cdot C -Z' \cdot C \gt \deg \omega_C. \end{equation*}

Hence $H^1(C, L(-Z')|_C)=0$ holds by Serre duality. Thus (1) holds.

Let us show (2). To this end, it suffices to show $H^1(C, L^{\otimes n}(-Z')|_C)=0$ and $H^1(Z', L^{\otimes n}|_{Z'})=0$. The former one follows from

\begin{equation*} \deg (L^{\otimes n}(-Z')|_C) \geq \deg (L(-Z')|_C) \gt \deg \omega_C. \end{equation*}

The latter one holds by Proposition 3.2 and Lemma 5.2. Thus (2) holds.

Proof. By the exact sequence

\begin{equation*} 0 \to {\mathcal{O}}_Y(-Z')|_C \to {\mathcal{O}}_Z \to {\mathcal{O}}_{Z'} \to 0, \end{equation*}

we obtain the following exact sequence:

\begin{equation*} H^0(C, {\mathcal{O}}_Y(-Z')|_C) \to H^0(Z, {\mathcal{O}}_Z) \xrightarrow{\rho} H^0(Z', {\mathcal{O}}_{Z'}) \end{equation*}

\begin{equation*} \to H^1(C, {\mathcal{O}}_Y(-Z')|_C) \to H^1(Z, {\mathcal{O}}_Z) \to H^1(Z', {\mathcal{O}}_{Z'}). \end{equation*}

We have

\begin{equation*} {\mathcal{O}}_Y(-Z')|_C \simeq {\mathcal{O}}_Y(-Z +C)|_C \simeq {\mathcal{O}}_Y(K_Y +C)|_C \simeq \omega_C. \end{equation*}

By Serre duality, we get $h^1(C, {\mathcal{O}}_Y(-Z')|_C) = h^0(C, {\mathcal{O}}_C)$. Recall that $h^0(Z, {\mathcal{O}}_Z)= h^1(Z, {\mathcal{O}}_Z)=1$ (Lemma 5.2(1)(4)). Since ρ is non-zero, $h^0(Z, {\mathcal{O}}_Z)=1$ implies that ρ is injective and $H^0(C, {\mathcal{O}}_Y(-Z')|_C)=0$. We are done by $H^1(Z', {\mathcal{O}}_{Z'})=0$ (Lemma 5.2).

Proof. Let us show (1). If $\overline Z$ is a prime divisor, then the assertion follows from Theorem 4.4. Therefore, the problem is reduced to the case when $Z = C_1 + \cdots +C_r$, where $r \geq 2$, each C_i is a prime divisor, $g^*M \cdot C_1 \gt 0$, and $g^*{M} \cdot C_2 \gt 0$ (we possibly have $C_i = C_j$ even if i ≠ j). Set $Z_i := Z - C_i$ for each $i \in \{1, 2\}$. By Propositions 3.2(1) and 5.5(1), we get ${\operatorname{Bs}}\,|g^*M| \cap Z_i = \emptyset$ for each $i \in \{1, 2\}$. Hence $g^*M$ is globally generated. By $(g|_Z)_*{\mathcal{O}}_Z = {\mathcal{O}}_{\overline Z}$ (Proposition 5.4), also M is globally generated. Thus (1) holds.

Let us show (2). Assume $-Z^2 \geq 2$. By (1), ${\mathcal{O}}_{\overline Y}(- \overline Z)|_{\overline Z}$ is globally generated. Then its pullback ${\mathcal{O}}_Y(-Z)|_Z$ is globally generated. Thus (2) holds.

Proof. Fix a prime divisor C on Y with $-Z \cdot C \gt 0$, whose existence is guaranteed by $Z^2 \lt 0$. We can uniquely write $ Z = aC +Z'$ for some $a \in {\mathbb{Z}}_{ \gt 0}$ and the effective ${\mathbb{Z}}$-divisor Zʹ on Y satisfying $C \not\subset {\operatorname{Supp}}\,Z'$. Since −Z is f-nef, we get

\begin{equation*} 1 = -Z^2 = -Z \cdot (aC +Z') \geq a (-Z) \cdot C \geq a. \end{equation*}

Therefore, we get $a=1, -Z \cdot C = 1$ and $-Z \cdot Z'=0$. Note that

\begin{equation*} H^0(Z, {\mathcal{O}}_Y(-Z)|_Z) \to H^0(Z', {\mathcal{O}}_Y(-Z)|_{Z'}) \end{equation*}

is surjective (Proposition 5.5). Since ${\mathcal{O}}_Y(-Z)|_{Z'}$ is globally generated (Proposition 3.2), the base locus ${\operatorname{Bs}}\,|{\mathcal{O}}_Y(-Z)|_Z|$ is contained in $Z \setminus Z'$. As Z is connected, the restriction map

\begin{equation*} H^0(Z, {\mathcal{O}}_Y(-Z)|_Z) \to H^0(C, {\mathcal{O}}_Y(-Z)|_C) \end{equation*}

is non-zero. Pick a non-zero element $s \in H^0(Z, {\mathcal{O}}_Y(-Z)|_Z)$. We have $h^0(Z, {\mathcal{O}}_Y(-Z)|_Z)= -Z^2 +\chi(Z, {\mathcal{O}}_Z)=1$, where the first equality follows from the Riemann–Roch theorem and the second one holds by $h^0(Z, {\mathcal{O}}_Z) = h^1(Z, {\mathcal{O}}_Z)=1$ (Lemma 5.2(1)(4)). By $h^0(Z, {\mathcal{O}}_Y(-Z)|_Z)=1$ and ${\operatorname{Bs}}\,|{\mathcal{O}}_Y(-Z)|_Z| \subset Z \setminus Z'$, the closed subscheme V(s) of Z defined by s is contained in $Z \setminus Z'$. In particular, V(s) is an effective Cartier divisor on Z. Furthermore, $V(s) = Q$ for some k-rational point $Q \in Z \setminus Z'$, because the restriction $V(s)|_C$ to C, which is the closed subscheme defined by $s|_C$, is an effective Cartier divisor on C with $\deg (V(s)|_C) = \deg ({\mathcal{O}}_Y(-Z)|_C)=-Z \cdot C =1$. Since $V(s) =Q$ is a regular effective Cartier divisor on Z, Z is regular around Q.

Notation 5.9. Let k be a field and let $P \in X$ be an elliptic singularity over k. Assume that X is affine, $K_X\sim 0$ and $X \setminus P$ is regular. Let $f : Y \to X$ be the minimal resolution of X and let $Z \subset Y$ be the fundamental cycle of $P \in X$. Take the canonical model $\overline Y$ of Y over X (cf. Notation 5.3):

\begin{equation*} f : Y \xrightarrow{g} \overline Y \xrightarrow{\overline f} X. \end{equation*}

Set $\overline Z := g_*Z$. Assume that $-Z^2 \geq 3$, so that $L_Z := {\mathcal{O}}_Y(-Z)|_Z$ and $L_{\overline Z} := {\mathcal{O}}_{\overline{Y}}(-\overline{Z})|_{\overline Z}$ are globally generated (Theorem 5.7). Set $N :=h^0(Z, L_Z)-1 =h^0(\overline Z, L_{\overline Z})-1$ and let W be the scheme-theoretic image of $\varphi_{|L_Z|} : Z \to {\mathbb{P}}_k^N$, which coincides with the scheme-theoretic image of $\varphi_{|L_{\overline Z}|} : \overline Z \to {\mathbb{P}}^N_k$ by $(g|_Z)_*{\mathcal{O}}_Z = {\mathcal{O}}_{\overline Z}$. Let

\begin{equation*} \psi : Z \xrightarrow{g|_Z} \overline Z \xrightarrow{\overline \psi} W \end{equation*}

be the induced morphisms. For $L_W := {\mathcal{O}}_{{\mathbb{P}}^N}(1)|_W$, the following isomorphisms hold: $L_{\overline Z} \simeq \overline \psi^*L_W$ and $ L_Z \simeq (g|_Z)^*L_{\overline Z} \simeq \psi^*L_W$.

Proof. Note that ψ is surjective.

In what follows, we assume that $\overline Z$ is not a prime divisor. Fix a prime divisor C on Y such that $L_Z \cdot C = -Z \cdot C \gt 0$. In particular, $C \subset {\operatorname{Supp}}\,Z$. Set $Z' := Z - C$ and let $W' := \psi(Z')$ be the scheme-theoretic image of W ^ʹ by ψ. We have the following commutative diagram in which all the horizontal arrows are closed immersions:

Set $L_{W'} := j_W^*L_W$ and $L_{Z'} := \psi'^*L_{W'} = j_Z^*L_Z$. We have the decomposition $Z' = Z'_{ \gt 0} + Z'_{=0}$ such that

• • $Z'_{ \gt 0}$ and $Z'_{=0}$ are effective ${\mathbb{Z}}$-divisors on Y,

• • $L_Z \cdot \Gamma \gt 0$ for every prime divisor $\Gamma \subset {\operatorname{Supp}}\,Z'_{ \gt 0}$, and

• • $L_Z \cdot \Gamma =0$ for every prime divisor $\Gamma \subset {\operatorname{Supp}}\,Z'_{=0}$.

We have $Z'_{ \gt 0} \neq 0$, because $\overline Z$ is not a prime divisor.

Proof of Step 3

Since the proofs are very similar, we only treat the case when (2) holds. Let $\overline Z = c_1 \overline C_1 + c_2 \overline C_2$ be the irreducible decomposition, where $c_i \in {\mathbb{Z}}_{ \gt 0}$ and $\overline{C}_i$ is a prime divisor. Since $\overline Z$ is not reduced, we may assume that $c_1 \geq 2$. We can write

\begin{equation*} Z = c_1 C_1 + c_2 C_2 + Z_{=0}, \end{equation*}

where each C_i denotes the proper transform of $\overline C_i$ and $Z_{=0} := Z -(c_1 C_1 + c_2 C_2)$. Note that $g_*Z_{=0}=0$.

By applying Step 2(2) for $C := C_1$ and $Z' := Z-C_1$, we see that W has exactly two irreducible components. For the irreducible decomposition $W_{{\operatorname{red}}}= W_1 \cup W_2$, we have $\psi(C_1) = W_1$ and $\psi(C_2) = W_2$ by permuting W ₁ and W ₂ if necessary. Furthermore, $\overline{\psi}:\overline Z \to W$ is birational around the generic point $\xi_{\overline{C}_2}$ of $\overline{C}_2$, i.e. there exists an open neighbourhood U of $\xi_{\overline{C}_2} \in W$ such that $\overline{\psi}|_{\overline{\psi}^{-1}(U)}:\overline{\psi}^{-1}(U) \to U$ is an isomorphism. It suffices to show that $\overline{\psi}:\overline Z \to W$ is birational around the generic point of $\overline{C}_1$. This holds by applying Step 2 for $C := C_2$ and $Z' : Z -C_2$. This completes the proof of Step 3.

By Step 3, we may assume that one of (I)–(III) holds.

1. (I) $\overline Z = \overline{C}_1 + \overline{C}_2$, where $\overline{C}_1$ and $\overline{C}_2$ are distinct prime divisors on $\overline Y$.

2. (II) $\overline Z = 2 \overline C$, where $\overline C$ is a prime divisor on $\overline Y$.

3. (III) $\overline Z = n \overline C$, where $n \in {\mathbb{Z}}_{\geq 3}$ and $\overline C$ is a prime divisor on $\overline Y$.

Proof of Step 4

Assume that (I) or (II) holds. Recall that the prime divisor C on Y has been chosen so that $L_Z \cdot C =-Z \cdot C \gt 0$. If (II) holds, then such a prime divisor is unique. If (I) holds, then there are exactly two such prime divisors, which are the proper transforms C ₁ and C ₂ of $\overline{C}_1$ and $\overline{C}_2$, respectively. Possibly after permuting C ₁ and C ₂, we may assume that $C = C_1$. In order to treat two cases (I) and (II) simultaneously, we set $C_1 := C_2 :=C$ for the case when (II) holds. We get $Z'_{ \gt 0} =C_2$, which is a prime divisor.

Suppose that $(*)$ holds. Let us derive a contradiction. By Step 2(5), Wʹ is irreducible and generically reduced. Then the closed immersion $j_W : W' \hookrightarrow W$ is birational by $(*)$.

We now we prove that

(5.10.1)\begin{equation} \deg L_Z = 2\deg L_W = 2\deg L_{W'}. \end{equation}

Since $j_W : W' \hookrightarrow W$ and $C_2 = Z'_{ \gt 0} \to W'$ are birational, we obtain $\deg L_W = \deg L_{W'}$ and $L_Z \cdot C_2 = \deg L_{W'}$ by [Reference Bădescu1, Definition 1.15 and Lemma 1.18]. By symmetry, we get $L_Z \cdot C_1 = \deg L_{W'}$. Then

\begin{equation*} \deg L_Z = L_Z \cdot C_1 + L_Z \cdot C_2 =2\deg L_{W'}. \end{equation*}

This completes the proof of (5.10.1).

We obtain

\begin{equation*} h^0(Z, L_Z) \overset{\rm (a)}{=} \chi(Z, L_Z) \overset{\rm (b)}{=} \chi(Z, {\mathcal{O}}_Z) + \deg L_Z \overset{\rm (c)}{=} \deg L_Z \overset{\rm (d)}{=} 2 \deg L_W. \end{equation*}

Here (a) follows from Proposition 5.5(2), (b) holds by the Riemann–Roch theorem, (c) is obtained from Lemma 5.2(1)(4) and we get (d) by (5.10.1). For the time being, we finish the proof of Step 4 by assuming Claim 5.11.

Claim 5.11. $H^1(W, {\mathcal{O}}_W)=0$ and $H^1(W, L_W)=0$.

We obtain

\begin{equation*} 2 \deg L_W = h^0(Z, L_Z ) \overset{{\rm (i)}}{=}h^0(W, L_W) \overset{{\rm (ii)}}{=} \chi(W, L_W) \end{equation*}

\begin{equation*} \overset{{\rm (iii)}}{=} \chi(W, {\mathcal{O}}_W) + \deg L_W \overset{{\rm (iv)}}{=} 1 + \deg L_W, \end{equation*}

where (i) follows from the fact that the composite map $H^0({\mathbb{P}}^N, {\mathcal{O}}_{{\mathbb{P}}^n}(1)) \to H^0(W, L_W) \hookrightarrow H^0(Z, L_Z)$ is bijective (and hence both the maps are bijective), (ii) and (iv) hold by Claim 5.11 and (iii) is an obtained by the Riemann–Roch theorem. We then get $\deg L_W = 1$, which implies $\deg L_Z = 2 \deg L_W = 2$ (5.10.1). This contradicts the assumption $\deg L_Z =-Z^2 \geq 3$ (Notation 5.9).

In order to complete the proof of Step 4, it is enough to prove Claim 5.11. Since $H^1(W, {\mathcal{O}}_W)=0$ implies $H^1(W, L_W)=0$ (Proposition 3.2), let us prove $H^1(W, {\mathcal{O}}_W)=0$. We have a birational closed immersion $j_W : W' \to W$ of irreducible one-dimensional projective schemes. Hence we obtain the following exact sequence

\begin{equation*} 0 \to I \to {\mathcal{O}}_{W} \to {\mathcal{O}}_{W'} \to 0 \end{equation*}

for the coherent ideal sheaf I defining W ^ʹ. Then $H^1(W, {\mathcal{O}}_{W}) \simeq H^1(W', {\mathcal{O}}_{W'}) =0$, where the isomorphism follows from $\dim({\operatorname{Supp}}\,I)=0$ and the equality holds by Step 2(4). This completes the proof of Step 4.

In what follows, we assume that $\overline Z$ is irreducible and non-reduced. We can write

• • $Z = nC + \widetilde{Z}$ for some $n \in {\mathbb{Z}}_{\geq 2}$, prime divisor C and effective ${\mathbb{Z}}$-divisor $\widetilde Z$,

• • $\overline Z = n\overline{C}$ for $\overline C := g_*C$.

We set

• • $Z' := (n-1)C + \widetilde{Z}$, and

• • $W' := \psi(Z')$, which denotes the scheme-theoretic image of Zʹ.

Let $\eta_Z, \eta_{\overline{Z}}, \eta_W$ be the generic points of $C, \overline{Z}, W$, respectively. Consider the following local rings at the generic points

\begin{equation*} A := {\mathcal{O}}_{W, \eta_W},\quad B := {\mathcal{O}}_{Z, \eta_Z} = {\mathcal{O}}_{\overline{Z}, \eta_{\overline{Z}}},\quad R := {\mathcal{O}}_{Y, \eta_Z} = {\mathcal{O}}_{\overline{Y}, \eta_{\overline{Z}}}, \end{equation*}

\begin{equation*} A' := {\mathcal{O}}_{W', \eta_{W}},\qquad B' := {\mathcal{O}}_{Z', \eta_{Z}}. \end{equation*}

Note that R is a discrete valuation ring. Pick a generator $t \in R$ of the maximal ideal ${\mathfrak{m}}_R$ of R, i.e. ${\mathfrak{m}}_R = tR$. In particular, $B = R/t^nR$ and $B' = R/ t^{n-1}R$. We have the following commutative diagram consisting of the induced ring homomorphisms:

In order to show that $\psi : \overline Z \to W$ is birational, it suffices to show A = B, because both $\overline Z$ and W are irreducible.

Proof of Step 7

Assume $n \geq 3$. In what follows, we identify Aʹ and Bʹ via the above isomorphism. For $r \in R$, we set $\overline{r} := \pi(r) \in B$. By the above commutative diagram, the following holds:

1. 1. (#) For an element $r \in R$, there exists $s \in R$ such that $\overline{r} - \overline{s}\overline{t}^{n-1} \in A$.

By (#) and $B =R +R\overline{t} + R\overline{t}^2 + \cdots + R \overline{t}^{n-1}$, it suffices to show $(\star)_m$ for every $m \in {\mathbb{Z}}_{ \gt 0}$.

1. 1. $(\star)_m$ $\overline{r}\overline{t}^m \in A$ for every $r \in R$.

Take $r \in R$. By (#), we obtain $\overline{r} =\overline{s}\overline{t}^{n-1} +a$ for some $a \in A$ and $s \in R$. Again by (#), we have $\overline{t} = \overline{s'} \cdot \overline{t}^{n-1} +a'$ for some $a' \in A$ and $s' \in R$. It follows from $\overline{t}^n=0$ and $n-1 \geq 1$ that $\overline{t}^{\ell} = (\overline{t} - \overline{s'} \cdot \overline{t}^{n-1})^{\ell} =a'^{\ell} \in A$ for every integer $\ell \geq 2$. By $n -1 \geq 2$, we get $\overline{t}^{n-1} \in A$, which implies $\overline{r} \overline{t}^{n-1} = (\overline{s}\overline{t}^{n-1} +a)\overline{t}^{n-1} =a \overline{t}^{n-1} \in A$. Thus $(\star)_{n-1}$ holds. Fix $m \in {\mathbb{Z}}_{ \gt 0}$. Let us show $(\star)_m$. By $\overline{t}^n=0$ and $n-1 \geq 1$, we obtain

\begin{equation*} \overline{r} \overline{t}^m = (\overline{s}\overline{t}^{n-1} +a)(\overline{s'}\overline{t}^{n-1} +a')^m = (\overline{s}\overline{t}^{n-1} +a) (m\overline{s'}\overline{t}^{n-1}a'^{m-1} +a'^m) \end{equation*}

\begin{equation*} = \overline{s}\overline{t}^{n-1} \cdot a'^m + a \cdot m\overline{s'}\overline{t}^{n-1}a'^{m-1} + aa'^m. \end{equation*}

By $(\star)_{n-1}$, we obtain $\overline{s}\overline{t}^{n-1} \cdot a'^m \in A$ and $a \cdot m\overline{s'}\overline{t}^{n-1}a'^{m-1} \in A$. Therefore, we get $\overline{r} \overline{t}^m \in A$, i.e. $(\star)_m$ holds. This completes the proof of Step 7.

Steps 3, 5, 6 and 7 complete the proof of Proposition 5.10.

Proof. We may assume that $K_X \sim 0$. We use Notation 5.3. Set $L_{\overline Z} := {\mathcal{O}}_{\overline Y}(-\overline Z)$. By $\bigoplus_{m=0}^{\infty} H^0(Z, L_Z^{\otimes m}) = \bigoplus_{m=0}^{\infty} H^0(\overline Z, L_{\overline Z}^{\otimes m})$ (Proposition 5.4), it is enough to show the corresponding statements for $\bigoplus_{m=0}^{\infty} H^0(\overline Z, L_{\overline Z}^{\otimes m})$.

Recall that we have

\begin{equation*} H^1(\overline Z, L_{\overline Z}) \hookrightarrow H^1(Z, L_Z)=0, \end{equation*}

where $H^1(Z, L_Z)=0$ holds by Proposition 5.5 and the inclusion is obtained by $(g|_Z)_*{\mathcal{O}}_Z = {\mathcal{O}}_{\overline Z}$ (Proposition 5.4(3)) and $g^*L_Z \simeq L_{\overline Z}$. Note that $H^0(Z, {\mathcal{O}}_Z) = H^0(\overline{Z}, {\mathcal{O}}_{\overline Z})=k$ (Lemma 5.2(4)).

Let us show (2). By $\deg L_{\overline Z} \geq 2$, $L_{\overline Z}$ is an ample globally generated invertible sheaf (Theorem 5.7). Then the assertion (2) follows from Proposition 2.9.

Let us show (1). In this case, $L_{\overline Z}^{\otimes 2}$ is an ample globally generated invertible sheaf (Theorem 5.7). By the same argument as in (2), we see that $\bigoplus_{n=0}^{\infty} H^0(\overline Z, L_{\overline Z}^{\otimes 2n})$ is generated by $H^0(\overline Z, L_{\overline Z}^{\otimes 2})$ as a k-algebra. Thus it suffices to show that

\begin{equation*} H^0(\overline Z, L_{\overline Z}^{\otimes 2}) \otimes H^0(\overline Z, L_{\overline Z}^{\otimes (2n+3)}) \to H^0(\overline Z, L_{\overline Z}^{\otimes (2n+5)}) \end{equation*}

is surjective for every $n \geq 0$. By $H^1(\overline Z, L_{\overline Z}^{\otimes 3} \otimes L_{\overline Z}^{\otimes -2}) =0$ (Proposition 5.5), $L^{\otimes 3}_{\overline Z}$ is 0-regular with respect to $L^{\otimes 2}_{\overline Z}$, which implies the required surjectivity [Reference Tanaka21, Definition 11.1 and Proposition 11.2] (cf. [Reference Fantechi, Göttsche, Illusie, Kleiman, Nitsure and Vistoli4, Section 5.2]).

Proof. We have the following commutative diagram consisting of the induced maps:

Recall that $\overline \psi :\overline{Z} \to W$ is birational (Proposition 5.10). Hence we have $\overline{\psi}^{-1}(W \cap H) \xrightarrow{\simeq} W \cap H$, because H is chosen to be a general hyperplane, so that $W \cap H$ is contained in the isomorphic locus of $\overline{\psi} : \overline{Z} \to W$. Since $D := \overline{\psi}^{-1}(W \cap H)$ is a member of $|L_{\overline Z}|$, we obtain an exact sequence:

\begin{equation*} H^0(\overline{Z}, L_{\overline Z}) \xrightarrow{\beta} H^0(D, L_{\overline Z}|_{D}) \to H^1(\overline Z, L_{\overline Z} \otimes {\mathcal{O}}_{\overline Z}(-D)). \end{equation*}

By ${\mathcal{O}}_{\overline Z}(-D) \simeq L_{\overline Z}^{-1}$, the assertion follows from

\begin{equation*} \dim_k H^1(\overline Z, L_{\overline Z} \otimes {\mathcal{O}}_{\overline Z}(-D)) = \dim_k H^1(\overline Z, {\mathcal{O}}_{\overline Z}) \overset{{\rm (i)}}{\leq} \dim_k H^1(Z, {\mathcal{O}}_{Z})\overset{{\rm (ii)}}{\leq}1, \end{equation*}

where (i) and (ii) hold by Proposition 5.4(3) and Lemma 5.2(1), respectively.

Proof. By $\bigoplus_{m=0}^{\infty} H^0(Z, L_Z^{\otimes m}) = \bigoplus_{m=0}^{\infty} H^0(\overline Z, L_{\overline Z}^{\otimes m})$ (Proposition 5.4(3)), it is enough to show that $\bigoplus_{m=0}^{\infty} H^0(\overline Z, L_{\overline Z}^{\otimes m})$ is generated by $H^0(\overline Z, L_{\overline Z})$. We may assume that k is separably closed. In particular, k is an infinite field.

Set ${\mathcal{O}}_W(\ell) := {\mathcal{O}}_{{\mathbb{P}}^N_k}(\ell)|_W =L_W^{\otimes \ell}$ for every $\ell \in {\mathbb{Z}}$. It is enough to prove the following assertions (A)–(E).

1. (A) $H^0({\mathbb{P}}^N_k, {\mathcal{O}}_{{\mathbb{P}}^N_k}(n)) \to H^0(W \cap H, {\mathcal{O}}_{{\mathbb{P}}^N_k}(n)|_{W \cap H})$ is surjective for every $n \geq 2$.

2. (B) $H^1(W, {\mathcal{O}}_W(m))=0$ for every $m \in {\mathbb{Z}}_{ \gt 0}$.

3. (C) $\overline{\psi}: \overline Z \to W$ is an isomorphism.

4. (D) $\alpha_n : H^0({\mathbb{P}}^N, {\mathcal{O}}_{{\mathbb{P}}^N}(n)) \to H^0(W, {\mathcal{O}}_{W}(n))$ is surjective for every $n \geq 0$.

5. (E) $H^0(W, {\mathcal{O}}_{W}(m-1)) \otimes_k H^0(W, {\mathcal{O}}_{W}(1)) \to H^0(W, {\mathcal{O}}_{W}(m))$ is surjective for every $m \geq 1$.

Indeed, (C) and (E) imply that $\bigoplus_{m=0}^{\infty} H^0(\overline Z, {\mathcal{O}}_{\overline Z}(mD))$ is generated by $H^0(\overline Z, {\mathcal{O}}_X(D))$.

Let us show (A). By Lemmas 4.10 and 5.13,

\begin{equation*} H^0({\mathbb{P}}^N_k, {\mathcal{O}}_{{\mathbb{P}}^N_k}(2)) \to H^0(W \cap H, {\mathcal{O}}_{{\mathbb{P}}^N_k}(2)|_{W \cap H}) \end{equation*}

is surjective. Since $W \cap H$ is zero-dimensional, also

\begin{equation*} H^0(W, {\mathcal{O}}_W(n)) \to H^0(W \cap H, {\mathcal{O}}_W(n)|_{W \cap H}) \end{equation*}

is surjective for every $n \geq 2$. Thus (A) holds.

Let us show (B). For an integer $n \geq 2$, we have the exact sequence:

\begin{equation*} H^0(W, {\mathcal{O}}_W(n)) \to H^0(W \cap H, {\mathcal{O}}_W(n)|_{W \cap H}) \to H^1(W, {\mathcal{O}}_W(n-1)) \to H^1(W, {\mathcal{O}}_W(n)) \to 0. \end{equation*}

By (A), it follows from the Serre vanishing theorem that $H^1(W, {\mathcal{O}}_W(m))=0$ for every $m \in {\mathbb{Z}}_{ \gt 0}$. Thus (B) holds.

Let us show (C). By (B) and $H^1(\overline Z, L_{\overline Z})=0$ (Propositions 5.4(3) and 5.5(2)), we have

(5.14.1)\begin{equation} \chi(W, {\mathcal{O}}_W(1)) = h^0(W, {\mathcal{O}}_W(1)) = h^0(\overline Z, L_{\overline Z}) = \chi(\overline Z, L_{\overline Z}). \end{equation}

For the closed subschemes $C_{\overline Z} \subset \overline Z$ and $C_W \subset W$ defined by the conductor of $\overline{\psi}: \overline Z \to W$, we have the following conductor exact sequence (2.8.3):

(5.14.2)\begin{equation} 0 \to {\mathcal{O}}_W \to \overline{\psi}_* {\mathcal{O}}_{\overline Z} \oplus {\mathcal{O}}_{C_W} \to \overline{\psi}_*{\mathcal{O}}_{C_{\overline Z}} \to 0. \end{equation}

We then obtain

\begin{equation*} \chi(W, {\mathcal{O}}_W) - \chi(\overline Z, {\mathcal{O}}_{\overline Z}) = \chi(C_W, {\mathcal{O}}_{C_W}) -\chi(C_{\overline Z}, {\mathcal{O}}_{C_{\overline Z}}) \overset{{\rm (i)}}{=} \chi(W, {\mathcal{O}}_W(1)) - \chi(\overline Z, L_{\overline Z}) \overset{{\rm (ii)}}{=} 0, \end{equation*}

where (ii) follows from (5.14.1) and (i) is obtained by applying the tensor $(-) \otimes {\mathcal{O}}_W(1)$ to the exact sequence (5.14.2). By $H^0(\overline Z, {\mathcal{O}}_{\overline Z})=H^0(W, {\mathcal{O}}_W) =k$ (Lemma 5.2(1), Proposition 5.4(3)), we get $h^1(\overline Z, {\mathcal{O}}_{\overline Z})=h^1(W, {\mathcal{O}}_W)$. Then (5.14.2) induces the following exact sequence:

\begin{equation*} 0 \to H^0(W, {\mathcal{O}}_W) \to H^0(\overline Z, {\mathcal{O}}_{\overline Z}) \oplus H^0(C_W, {\mathcal{O}}_{C_W}) \to H^0(C_{\overline Z}, {\mathcal{O}}_{C_{\overline Z}}) \to 0. \end{equation*}

Again by $H^0(\overline Z, {\mathcal{O}}_{\overline Z})=H^0(W, {\mathcal{O}}_W) =k$, we get the induced k-linear isomorphism:

\begin{equation*} H^0(C_W, {\mathcal{O}}_{C_W}) \xrightarrow{\simeq} H^0(C_{\overline Z}, {\mathcal{O}}_{C_{\overline Z}}), \end{equation*}

which implies ${\mathcal{O}}_{C_W} \xrightarrow{\simeq} \overline{\psi}_*{\mathcal{O}}_{C_{\overline Z}}$. Again by the conductor exact sequence (5.14.2), we obtain ${\mathcal{O}}_W \xrightarrow{\simeq} \overline{\psi}_*{\mathcal{O}}_{\overline Z}$, i.e. $\overline{\psi} : \overline Z \to W$ is an isomorphism. Thus (C) holds.

Let us show (D). If $n \in \{0, 1\}$, then α_n is surjective by construction. Fix $n \in {\mathbb{Z}}_{\geq 2}$. We have the following commutative diagram in which each horizontal sequence is exact (note that $H^1(W, {\mathcal{O}}_W(n-1))=0$ by (B)):

By $n \geq 2$, the composite map

\begin{equation*} H^0({\mathcal{O}}_{{\mathbb{P}}^{N}}(n)) \to H^0({\mathcal{O}}_{W \cap H}(n)) \end{equation*}

is surjective by (A). In particular, γ_n is surjective. By the snake lemma, the surjectivity of $\alpha_{n-1}$ implies the surjectivity of α_n. By induction on n, α_n is surjective. Thus (D) holds.

Let us show (E). Fix $m \geq 1$. We have the following commutative diagram:

Since µ and α_m are surjective by (D), also $\mu'$ is surjective. Thus (E) holds.

Notation 5.15. Let k be a field and let $P \in X$ be an elliptic singularity over k. Assume that $X \setminus P$ is regular, X is affine and $K_X \sim 0$. Let $f : Y \to X$ be the minimal resolution of X and let $Z \subset Y$ be the fundamental cycle of $P \in X$. Take the canonical model $\overline Y$ of Y over X (cf. Notation 5.3):

\begin{equation*} f : Y \xrightarrow{g} \overline Y \xrightarrow{\overline f} X. \end{equation*}

Set $\overline Z := g_*Z$, $L :={\mathcal{O}}_Y(-Z), L_{\overline Y} := {\mathcal{O}}_{\overline Y}(-\overline Z), L_Z := {\mathcal{O}}_Y(-Z)|_Z$ and $L_{\overline Z} := {\mathcal{O}}_{\overline{Y}}(-\overline{Z})|_{\overline Z}$. By $K_Y +Z \sim f^*K_X \sim 0$, we obtain

\begin{equation*} \overline{Y} = {\operatorname{Proj}}\,\bigoplus_{n=0}^{\infty} H^0(Y, nK_Y) \simeq {\operatorname{Proj}}\,\bigoplus_{n=0}^{\infty} H^0(Y, {\mathcal{O}}_Y(-nZ)). \end{equation*}

Let ${\mathfrak{m}}_P$ be the maximal ideal of $H^0(X, {\mathcal{O}}_X)$ corresponding to P. For every $n \in {\mathbb{Z}}_{\geq 0}$, we set

\begin{equation*} I_n :=H^0(Y, {\mathcal{O}}_Y(-nZ)) \subset H^0(Y, {\mathcal{O}}_Y) = H^0(X, {\mathcal{O}}_X), \end{equation*}

which is an ideal of $H^0(X, {\mathcal{O}}_X)$. By

\begin{equation*} 0 \to H^0(Y, {{\mathcal{O}}_Y}(-(n+1)Z)) \to H^0(Y, {{\mathcal{O}}_Y}(-nZ)) \to H^0(Z, {\mathcal{O}}_Y(-nZ)|_Z) \end{equation*}

\begin{equation*} \to H^1(Y, {{\mathcal{O}}_Y}(-(n+1)Z))=0, \end{equation*}

we obtain $I_n/I_{n+1} \simeq H^0(Z, {\mathcal{O}}_Y(-nZ)|_Z) \simeq H^0(Z, L_Z^{\otimes n})$, and hence

\begin{equation*} \bigoplus_{n=0}^{\infty} I_n/I_{n+1} \simeq \bigoplus_{n=0}^{\infty} H^0(Z, L_Z^{\otimes n}). \end{equation*}

Proof. We have ${\mathfrak{m}}_P = f_*{\mathcal{O}}_Y(-Z)$ (Lemma 5.2), which implies $I_1 = {\mathfrak{m}}_P$.

Proof. This map is the same as

\begin{equation*} H^0(\overline Y, L^{\otimes n}_{\overline Y}) \otimes_k H^0(\overline Y, L^{\otimes a}_{\overline Y}) \to H^0(\overline Y, L^{\otimes (n+a)}_{\overline Y}). \end{equation*}

Then the assertion follows from the fact that $L_{\overline Y}$ is (relatively) ample.