Proof. We begin by treating the case where $\delta (K)<\infty$ and claim that, in this case, the Frobenius morphism $R\to R,$ $r\mapsto r^p$ is finite. This is the same as showing that $R$ is finite over $R^p$. As $R$ is integrally closed and integral over $R^p,$ $R$ is the integral closure of $R^p$ in $K$. As $R$ is excellent and $K$ is finite over $K^p,$ this shows that $R$ is finite over $R^p$. The fundamental equation [Sta18, Tag 02MJ] shows that $[\kappa :\kappa ^p]$ is finite and

\[ [K:K^p]=p[\kappa:\kappa^p], \]

as the Frobenius map $R\to R$ clearly has ramification index $p$. This implies the claim from the lemma in this case.

If $\delta (K)=\infty,$ all we must show is that $\delta (\kappa )=\infty$ as well. Let $K^p\subseteq L \subseteq K$ be a finite subextension, and let $R_L$ be the integral closure of $R^p$ in $L$. If $\pi$ denotes a uniformiser of $R^p$ and $\pi _L$ a uniformiserFootnote ² of $R_L,$ then $\pi _L^p=\epsilon \pi ^\nu$ for some $\epsilon \in R^{p,\times }$ and $\nu \in \operatorname {\mathbf {N}}$; moreover, we have $\pi =\eta \pi _L^{e_L}$ for some $\eta \in R_L^\times,$ $e_L\in \operatorname {\mathbf {N}}$. Together these equations imply $\pi _L^p=\epsilon \eta ^\nu \pi _L^{e_L\nu },$ so $e_L\in \{1,p\}$. For each $K^p\subseteq L\subseteq K,$ let $\mathfrak {m}_L$ be the maximal ideal of $R_L$. As above, we have the equation $[L:K^p]=e_L[R_L/\mathfrak {m}_L:\kappa ^p]$. Then

\[ \kappa=\varinjlim_L R_L/\mathfrak{m}_L \]

as an extension of $\kappa ^p$. However, as $[K:K^p]=\infty,$ the integer $[L:K]$ becomes arbitrarily large as $L$ runs through finite subextensions of $K^p \subseteq K$. Since $e_L$ is bounded independently of $L,$ the number $[R_L/\mathfrak {m}_L:\kappa ^p]$ must become arbitrarily large as well, which means that $[\kappa :\kappa ^p]=\infty$.

Proof. (i) $\Rightarrow$ (iii) follows from Lemma 2.1, as does (iii) $\Rightarrow$ (ii) (both by considering $\operatorname {\mathcal {O}}_{S,x}$) and (ii) $\Rightarrow$ (i) is trivial.

Proof. We can write $L$ as a finite separable extension of a purely transcendental extension $T$ of $k$ of transcendence degree $n\geq 2$. Since $L=L^p\otimes _{T^p}T$ (by the primitive element theorem), we see that $\delta (T)=\delta (L)$. Moreover, clearly $\delta (T)=n$ since $k$ is perfect. If $\mathfrak {m}$ denotes the maximal ideal of $R,$ we have $\mathfrak {m}=L\cap \langle t \rangle,$ so we have $k\subseteq R/\mathfrak {m} \subseteq k[\![t]\!]/\langle t\rangle =k$. This shows that $\delta (R/\mathfrak {m})=0$ whereas $\delta (L)\geq 2$ by assumption. Hence, $R$ is not excellent by Lemma 2.1.

Proof. Assume first that $\kappa \subseteq \ell$ is finite. Then we have the towers of extensions $\kappa ^p\subseteq \ell ^p \subseteq \ell$ and $\kappa ^p\subseteq \kappa \subseteq \ell$. The tower law shows that $[\ell :\ell ^p][\ell ^p:\kappa ^p]=[\kappa :\kappa ^p][\ell :\kappa ]$. Since clearly $[\ell :\kappa ]=[\ell ^p:\kappa ^p],$ the desired equality follows in this case. If $\ell$ is separable over $\kappa,$ then for any finite subextension $\kappa \subseteq \ell '\subseteq \ell,$ we have $\ell '=\ell '^p\otimes _{\kappa ^p} \kappa$. This follows from the primitive element theorem. But since $\ell$ is the inductive limit of its finite subextensions and since tensor products commute with inductive limits, we have $\ell =\ell ^p\otimes _{\kappa ^p} \kappa,$ which clearly implies the claim.

Proof. We may assume without loss of generality that $S=\operatorname {\operatorname {Spec}} K$ for some field with $\delta (K)\leq 1$. If $K$ is perfect, all proper morphisms $f\colon X\to \operatorname {\operatorname {Spec}} K$ with $X$ reduced are Stein reduced, so we shall assume $\delta (K)=1$. Moreover, we may assume that $X$ is connected, so that $\Gamma (X, \operatorname {\mathcal {O}}_X)$ is a field, and replace $K$ by $\Gamma (X, \operatorname {\mathcal {O}}_X)$. All we must show is that $X\times _K \operatorname {\operatorname {Spec}} K^{1/p}$ is reduced [Reference SchröerSch10, Proposition 1.2]. Since $K$ has degree of imperfection 1 (and, hence, so does $\Gamma (X, \operatorname {\mathcal {O}}_X)$ by Lemma 2.4), we may pick any $\gamma \in K^{1/p}\backslash K$ and obtain an isomorphism $K^{1/p}\cong K[X]/\langle X^p-\gamma ^p \rangle$. Now let $F$ be the field of rational functions on $X$. If $F\otimes _KK^{1/p}$ were non-reduced, there would have to exist an element $\xi \in F$ such that $\xi ^p=\gamma ^p$. Now cover $X$ by open affine subsets $U_1,\ldots, U_n$. Since $\xi$ is clearly integral over $\Gamma (U_j,\operatorname {\mathcal {O}}_{U_j})$ for all $j$ and since $X$ is normal, we see that $\xi \in \Gamma (U_j, \operatorname {\mathcal {O}}_{U_j})$ for all $j$. Hence, $\xi \in K\subseteq K^{1/p},$ so $\xi =\gamma,$ which contradicts our choice of $\gamma$.

Proof. Let $T$ be any $S$-scheme. Then $X'\times _{S}T=X'\times _{S'}(S'\times _{S} T),$ and a similar identity holds for $Y$. This shows that the map $\Gamma (X'\times _{S}T, \operatorname {\mathcal {O}}_{X'\times _{S}T}) \to \Gamma (Y\times _{S}T, \operatorname {\mathcal {O}}_{Y\times _{S}T})$ is injective. The proof of the second claim is purely formal and left to the reader.

Proof. We shall identify $\mathscr {U}$ and $\psi ^{-1}({\mathscr {U}})$. Let $Z_1, \ldots, Z_n$ be the (reduced) irreducible components of the special fibre of $X'\to S$. By our assumption on $\mathscr {U},$ the intersection $\mathscr {U}\cap Z_j$ is non-empty for all $j=1,\ldots, n$. For each $j,$ we choose a closed point $x_j\in \mathscr {U}\cap Z_j$ with image $\mathfrak {m}_j$ in $S'$ which does not lie on the intersection of two irreducible components and such that $\operatorname {\mathcal {O}}_{(X'\times _{R'} \operatorname {\operatorname {Spec}} R'/\mathfrak {m_j}), x_j}$ is Cohen–Macaulay for all $j=1,\ldots, n$ (see [Reference Demazure and GrothendieckDG70, Exposé $\text {VI}_{\text {A}}$, Lemme 1.1.2]). Raynaud [Reference RaynaudRay70, proof of Proposition 2.2.3(b)] has shown that there exists a rigidificator $Y$ of $\operatorname {\mathrm {Pic}}_{X'/S'}$ whose special fibre is set-theoretically equal to $\{x_1,\ldots, x_n\}$. Because $Y$ is a semilocal scheme, we see that $Y\subseteq \mathscr {U},$ so $Y$ can be regarded as a closed subscheme of $X$. By the preceding lemma, we know that $Y$ is a rigidificator of $\operatorname {\mathrm {Pic}}_{X'/S}$. Because $\psi$ is $S$-universally scheme-theoretically dominant, $Y$ is a rigidificator of $\operatorname {\mathrm {Pic}}_{X/S}$ as well, as desired.

Proof. The exactness of the sequence follows from [Reference OverkampOve24, Proposition 2.30]; the conditions (i)–(v) as well as the standing assumption from [Reference OverkampOve24, Proposition 2.30] can be verified as in the proof of [Reference OverkampOve24, Corollary 3.7]. The remaining claims follow from [Reference Bosch, Lütkebohmert and RaynaudBLR90, Chapter 8.1, Corollary 8] (with the notation from [Reference Bosch, Lütkebohmert and RaynaudBLR90], the scheme $V$ is of finite presentation over $\operatorname {\mathcal {O}}_K$ because so is the $\operatorname {\mathcal {O}}_K$-module $\mathscr {D}$; see [Reference Bosch, Lütkebohmert and RaynaudBLR90, Chapter 8.1, Theorem 7] and [Sta18, Tag 00DO]).

Proof. Put $\mathscr {R}:=\operatorname {\mathrm {Res}}_{\mathcal {A}/\operatorname {\mathcal {O}}_K} \operatorname {\mathbf {G}_m} / \operatorname {\mathrm {Res}}_{\mathcal {B}/\operatorname {\mathcal {O}}_K} \operatorname {\mathbf {G}_m}$ and consider the following commutative diagram.

The rows in this diagram are exact by the exact sequence (2) (the top one being obviously exact); the left vertical column is again clearly exact while the right vertical column is exact by Proposition 3.4. Moreover, the central vertical column is a complex, which implies that it is exact as well.

Proof. If $\mathscr {C}=\widetilde {\mathscr {C}}$ is regular, this follows from [Reference Liu, Lorenzini and RaynaudLLR04, Proposition 3.2] together with Lemma 3.1. Consider the exact sequence

\[ 0 \to \operatorname{\mathrm{Res}}_{\mathcal{A}/\operatorname{\mathcal{O}}_K} \operatorname{\mathbf{G}_m} / \operatorname{\mathrm{Res}}_{\mathcal{B}/\operatorname{\mathcal{O}}_K} \operatorname{\mathbf{G}_m} \to (P_{\mathscr{C}/\operatorname{\mathcal{O}}_K}, Y) \to (P_{\widetilde{\mathscr{C}}/\operatorname{\mathcal{O}}_K}, Y) \to 0 \]

from Lemma 3.6, which induces an exact sequence

\[ 0 \to \operatorname{\mathrm{Res}}_{\mathcal{A}/\operatorname{\mathcal{O}}_K} \operatorname{\mathbf{G}_m} / \operatorname{\mathrm{Res}}_{\mathcal{B}/\operatorname{\mathcal{O}}_K} \operatorname{\mathbf{G}_m} \to \mathscr{G} \to (P_{\widetilde{\mathscr{C}}/\operatorname{\mathcal{O}}_K}, Y)^0 \to 0. \]

Then $\mathscr {G}$ is a group space over $\operatorname {\mathcal {O}}_K,$ and because $(P_{\widetilde {\mathscr {C}}}, Y)^0$ is separated over $\operatorname {\mathcal {O}}_K$ (see [Reference Liu, Lorenzini and RaynaudLLR04, Proposition 3.2]), so is $\mathscr {G}$. In particular, $\mathscr {G}$ is a scheme [Reference AnantharamanAna73, Théorème 4.B]. Clearly, $(P_{\mathscr {C}}, Y)^0$ is an open subspace of $\mathscr {G}$ and, hence, a scheme as well.

Proof. Consider the commutative diagram

where the $Q_j$ are the obvious cokernels. Because $\Gamma (\mathscr {C}, \operatorname {\mathcal {O}}_{\mathscr {C}}),$ $\Gamma (\widetilde {\mathscr {C}}, \operatorname {\mathcal {O}}_{\widetilde {\mathscr {C}}}),$ $\mathcal {A},$ and $\mathcal {B}$ are finite products of discrete valuation rings, and the maps $\Gamma (\mathscr {C}, \operatorname {\mathcal {O}}_{\mathscr {C}}) \to \Gamma (\widetilde {\mathscr {C}}, \operatorname {\mathcal {O}}_{\widetilde {\mathscr {C}}})$ and $\mathcal {B} \to \mathcal {A}$ are finite and faithfully flat, it is easy to see that $q$ is injective and that $\mathrm {coker}\, q_C$ is finitely generated. Using the exact sequence from Proposition 3.4, we see that $T=\ker w$. Hence, the snake lemma gives an injection $T\to \mathrm {coker} \, q_C$.

Proof. The map $V_Y^\times /\operatorname {\mathrm {Res}}_{\Gamma (\widetilde {\mathscr {C}}, \operatorname {\mathcal {O}}_{\widetilde {\mathscr {C}}})/\operatorname {\mathcal {O}}_K}\operatorname {\mathbf {G}_m} \to \widetilde {\mathscr {H}}_{\widetilde {\mathscr {C}}}$ is an open immersion; this follows from [Reference Liu, Lorenzini and RaynaudLLR04, Lemma 3.5(d)] together with Lemma 3.1.

We shall first show the auxiliary claim that the cokernel of the map

\[ (V_Y^\times/\operatorname{\mathrm{Res}}_{\Gamma({\mathscr{C}}, \operatorname{\mathcal{O}}_{{\mathscr{C}}})/\operatorname{\mathcal{O}}_K}\operatorname{\mathbf{G}_m})(\operatorname{\mathcal{O}}_K) \to \widetilde{\mathscr{H}}_{\mathscr{C}}(\operatorname{\mathcal{O}}_K)=\mathscr{H}^1_{\mathscr{C}}(\operatorname{\mathcal{O}}_K) \]

is finitely generated. The cokernel $Q$ of $(V_Y^\times /\operatorname {\mathrm {Res}}_{\Gamma (\widetilde {\mathscr {C}}, \operatorname {\mathcal {O}}_{\widetilde {\mathscr {C}}})/\operatorname {\mathcal {O}}_K}\operatorname {\mathbf {G}_m} )(\operatorname {\mathcal {O}}_K)\to \widetilde {\mathscr {H}}_{\widetilde {\mathscr {C}}}(\operatorname {\mathcal {O}}_K)$ is a finite Abelian group. Now consider the map $\mathscr {H}^1_{\mathscr {C}}(\operatorname {\mathcal {O}}_K) \to Q$ induced by the composition

\[ \mathscr{H}^1_{\mathscr{C}}(\operatorname{\mathcal{O}}_K) \to \mathscr{H}^1_{\widetilde{\mathscr{C}}}(\operatorname{\mathcal{O}}_K) = \widetilde{\mathscr{H}}_{\widetilde{\mathscr{C}}}(\operatorname{\mathcal{O}}_K) \to Q \]

and let $J$ be its kernel. It suffices to show that the cokernel of the induced map

\[ (V_Y^\times/\operatorname{\mathrm{Res}}_{\Gamma({\mathscr{C}}, \operatorname{\mathcal{O}}_{{\mathscr{C}}})/\operatorname{\mathcal{O}}_K}\operatorname{\mathbf{G}_m})(\operatorname{\mathcal{O}}_K) \to J \]

is finitely generated. An element of $J$ is a rigidified line bundle $(\mathscr {L}, \alpha )$ on $\mathscr {C}$ such that $\mathscr {L} \in T$ (Lemma 3.8). An element of the kernel of the induced map $J\to T$ clearly comes from an element of $(V_Y^\times /\operatorname {\mathrm {Res}}_{\Gamma ({\mathscr {C}}, \operatorname {\mathcal {O}}_{{\mathscr {C}}})/\operatorname {\mathcal {O}}_K}\operatorname {\mathbf {G}_m})(\operatorname {\mathcal {O}}_K),$ so the auxiliary claim follows. The remainder of the argument now works as in the proof of [Reference Liu, Lorenzini and RaynaudLLR04, Lemma 3.5(d)]: what we have just shown implies that the cokernel of the induced map

\[ (V_Y^\times/\operatorname{\mathrm{Res}}_{\Gamma({\mathscr{C}}, \operatorname{\mathcal{O}}_{{\mathscr{C}}})/\operatorname{\mathcal{O}}_K}\operatorname{\mathbf{G}_m}) \times_{\operatorname{\mathcal{O}}_K}\operatorname{\operatorname{Spec}} k \to \widetilde{\mathscr{H}}_{\mathscr{C}}\times_{\operatorname{\mathcal{O}}_K} \operatorname{\operatorname{Spec}} k \]

has a finitely generated group of $k$-points. But since $k$ is algebraically closed, this cokernel must be finite over $k,$ so the morphism from the lemma is quasi-finite. Since it is an isomorphism generically and the target is regular, our claim follows from Zariski's main theorem.

Proof. The morphism denoted by $\lambda$ above coincides with the morphism $\lambda \colon H^1(\mathscr {C}, \operatorname {\mathcal {O}}_{\mathscr {C}})\to \operatorname {\mathrm {Lie}} \,P_{\mathscr {C}/\operatorname {\mathcal {O}}_K}^{\mathrm {sep}}$ modulo the identification $\theta _{\mathscr {C}}\colon \operatorname {\mathrm {Lie}} \,P_{\mathscr {C}/\operatorname {\mathcal {O}}_K}\to H^1(\mathscr {C}, \operatorname {\mathcal {O}}_{\mathscr {C}})$ from [Reference Liu, Lorenzini and RaynaudLLR04, Proposition 1.3(b)]. The lemma can now be proven exactly as in [Reference Liu, Lorenzini and RaynaudLLR04, Lemma 3.5(e)]: Lemma 3.9 shows that the map

\[ \Gamma(Y, \operatorname{\mathcal{O}}_Y)/\Gamma(\mathscr{C}, \operatorname{\mathcal{O}}_{\mathscr{C}}) \to \operatorname{\mathrm{Lie}} \,\widetilde{\mathscr{H}}_{\mathscr{C}} \]

is an isomorphism; the top row of the diagram is induced by the exact sequence (3) above. Moreover, we note that the canonical morphism $(P_{\mathscr {C}/\operatorname {\mathcal {O}}_K}, Y)/\mathscr {H}_{\mathscr {C}} \to P_{\mathscr {C}/\operatorname {\mathcal {O}}_K}^{\mathrm {sep}}$ is an isomorphism (cf. [Reference Bosch, Lütkebohmert and RaynaudBLR90, Chapter 9.5, Proposition 3, 3rd case]), which induces the bottom row.

Proof. By definition, there is a unique morphism $\mathscr {G}\to \mathscr {N}$ which extends the identity at the generic fibre. We shall first show that the induced map on identity components $\mathscr {G}^0 \to \mathscr {N}^0$ is an isomorphism; this part of the argument is essentially taken from [Reference Bosch, Lütkebohmert and RaynaudBLR90, p. 269]. By assumption, the map $\mathscr {G}(\operatorname {\mathcal {O}}_K) \to \mathscr {N}(\operatorname {\mathcal {O}}_K)$ is an isomorphism; in particular, the map $\mathscr {G}(k)\to \mathscr {N}(k)$ is surjective. The cokernel of the induced morphism

\[ \nu \colon \mathscr{G}_k^0 \to \mathscr{N}^0_k \]

is a smooth connected algebraic group over $k$. The snake lemma shows that $(\mathrm {coker}\,\nu )(k)$ is a finitely generated Abelian group. But because $k$ is algebraically closed, $(\mathrm {coker}\,\nu )(k)$ is $n$-divisible for any natural number $n$ invertible in $k$. This is only possible if $\mathrm {coker}\,\nu =0,$ so the map $\mathscr {G}^0_k \to \mathscr {N}^0_k$ is surjective. Because the fibres of $\mathscr {G}^0$ and $\mathscr {N}^0$ are of the same dimension, the map $\mathscr {G}^0\to \mathscr {N}^0$ must be quasi-finite. Since it is also an isomorphism generically, Zariski's main theorem shows that it is an isomorphism.

Now let $j\colon \operatorname {\operatorname {Spec}} k \to \operatorname {\operatorname {Spec}} \operatorname {\mathcal {O}}_K$ be the canonical closed immersion and denote by $\Phi _{\mathscr {N}}$ the Abelian group $\mathscr {N}(\operatorname {\mathcal {O}}_K)/\mathscr {N}^0(\operatorname {\mathcal {O}}_K)=\mathscr {N}_k(k)/\mathscr {N}^0_k(k)$. We can view $\Phi _{\mathscr {N}}$ and $\Phi _{\mathscr {G}}$ as group schemes over $k,$ and we have a diagram with exact rows

(the rows are even exact in the Zariski topology). We already know that the left vertical arrow is an isomorphism. This implies that the right vertical arrow is also an isomorphism since $\mathscr {G}(\operatorname {\mathcal {O}}_K)\to \mathscr {N}(\operatorname {\mathcal {O}}_K)$ is an isomorphism by assumption, and we have just seen that it carries $\mathscr {G}^0(\operatorname {\mathcal {O}}_K)$ isomorphically into $\mathscr {N}^0(\operatorname {\mathcal {O}}_K)$. In particular, the central vertical arrow is an isomorphism as well, which implies our claim.

Proof. First note that $P_{\mathscr {C}/K}^{\mathrm {sep}}$ is representable by a smooth separated group scheme over $\operatorname {\mathcal {O}}_K;$ the proof works as in [Reference Bosch, Lütkebohmert and RaynaudBLR90, Chapter 9.5, proof of Proposition 3, 3rd case]. Moreover, by [Reference Bosch, Lütkebohmert and RaynaudBLR90, Chapter 9.2, Corollary 14], the Abelian group

\[ P_{\mathscr{C}/\operatorname{\mathcal{O}}_K}^{\mathrm{sep}}(k)/P_{\mathscr{C}/\operatorname{\mathcal{O}}_K}^{\mathrm{sep},0}(k) =P_{\mathscr{C}/\operatorname{\mathcal{O}}_K}^{\mathrm{sep}}(\operatorname{\mathcal{O}}_K)/P_{\mathscr{C}/\operatorname{\mathcal{O}}_K}^{\mathrm{sep},0}(\operatorname{\mathcal{O}}_K) \]

is finitely generated. Because $k$ is algebraically closed, $\operatorname {\mathrm {Br}} \Gamma (C, \operatorname {\mathcal {O}}_C) =0$ (see [Reference Colliot-Thélène and SkorobogatovCS21, Theorem 1.2.15]), so $\operatorname {\mathrm {Pic}}_{C/K}(K)=\operatorname {\mathrm {Pic}} \,C$. Because $\mathscr {C}$ is semi-factorial, the map $\operatorname {\mathrm {Pic}}_{\mathscr {C}/\operatorname {\mathcal {O}}_K}(\operatorname {\mathcal {O}}_K) \to \operatorname {\mathrm {Pic}}_{C/K}(K)$ is surjective. This map induces a surjection $P_{\mathscr {C}/\operatorname {\mathcal {O}}_K}(\operatorname {\mathcal {O}}_K) \to \operatorname {\mathrm {Pic}}^0_{C/K}(K)$ as $P_{\mathscr {C}/\operatorname {\mathcal {O}}_K}$ is open and closed in $\operatorname {\mathrm {Pic}}_{\mathscr {C}/\operatorname {\mathcal {O}}_K}$. Hence, Proposition 3.11 implies the claim.

Proof. Commutativity of the diagram is clear. The claim on lengths follows from Theorem 3.5 together with Corollary 3.12 (for $\widetilde {\lambda },$ this also follows from [Reference Liu, Lorenzini and RaynaudLLR04, Theorem 3.1]).

Proof. We may suppose without loss of generality that $C$ is connected. Assume that $\operatorname {\mathrm {Pic}}^0_{C/K}$ is semiabelian. Let $C^{\mathrm {sn}}$ be the seminormalisation of $C,$ such that we have a factorisation $\widetilde {C} \overset {\nu }\to C^{\mathrm {sn}} \overset {\varsigma }{\to } C$. If $\varsigma$ is not an isomorphism, [Reference OverkampOve24, Theorem 2.24(i)] shows that we have a factorisation $C^{\mathrm {sn}}=C_1 \to \cdots \to C_n=C$ such that for each $i=1,\ldots, n-1,$ we have a closed point $x_{i+1}$ of $C_{i+1}$ and a closed immersion $\kappa (x_{i+1})[\epsilon ]/\langle \epsilon ^2 \rangle \to C_i$ such that the diagram

is co-Cartesian. Because $C$ is connected, so are $C_i,$ and $C_{i+1},$ and the $K$-algebras $\Gamma (C_i, \operatorname {\mathcal {O}}_{C_i})$ as well as $\Gamma (C_{i+1}, \operatorname {\mathcal {O}}_{C_{i+1}})$ are fields contained in $\kappa (x_{i+1})$. Consider the exact sequence

\begin{align*} 0 &\to \operatorname{\mathrm{Res}}_{\Gamma(C_{i+1}, \operatorname{\mathcal{O}}_{C_{i+1}})/K} \operatorname{\mathbf{G}_m} \to \operatorname{\mathrm{Res}}_{\Gamma(C_{i}, \operatorname{\mathcal{O}}_{C_{i}})/K} \operatorname{\mathbf{G}_m} \to \operatorname{\mathrm{Res}}_{\kappa(x_{i+1})/K} \operatorname{\mathbf{G}_a} \\ & \to \operatorname{\mathrm{Pic}}^0_{C_i/K} \to \operatorname{\mathrm{Pic}}^0_{C_{i+1}/K} \to 0 \end{align*}

[Reference OverkampOve24, Proposition 2.30]. Counting dimensions and using that quotients of split unipotent groups are split unipotent [Reference Conrad, Gabber and PrasadCGP15, Theorem B.3.4], we see that $\operatorname {\mathrm {Pic}}^0_{C/K}$ has a non-trivial split unipotent subquotient, which is impossible. In particular, $C$ is seminormal. By [Reference OverkampOve24, Proposition 2.30], we see that the map $\operatorname {\mathrm {Pic}}^0_{C/K} \to \operatorname {\mathrm {Pic}}^0_{\widetilde {C}/K}$ is surjective, so $\operatorname {\mathrm {Pic}}^0_{\widetilde {C}/K}$ is semiabelian. Hence the same is true for $\operatorname {\mathrm {Pic}}^0_{C'/K},$ where $C'$ is an irreducible component of $\widetilde {C}$. By Proposition 2.6 together with the fact that $\delta (K)=1$ since $k$ is perfect (Lemma 2.1), we conclude that $C'$ is geometrically integral over the field $\Gamma (C', \operatorname {\mathcal {O}}_{C'})$. After replacing $\Gamma (C', \operatorname {\mathcal {O}}_{C'})$ by a finite separable extension if necessary, [Reference Bosch, Lütkebohmert and RaynaudBLR90, Chapter 9.2, Proposition 4] shows that $\operatorname {\mathrm {Pic}}^0_{C'/\Gamma (C', \operatorname {\mathcal {O}}_{C'})}$ contains no algebraic torus. In particular, $\operatorname {\mathrm {Pic}}^0_{C'/K}=\operatorname {\mathrm {Res}}_{\Gamma (C', \operatorname {\mathcal {O}}_{C'})/K} \operatorname {\mathrm {Pic}}^0_{C'/\Gamma (C', \operatorname {\mathcal {O}}_{C'})}$ contains no torus and is therefore an Abelian variety. Hence, $\Gamma (C', \operatorname {\mathcal {O}}_{C'})$ must be separable over $K$ or we must have $\operatorname {\mathrm {Pic}}^0_{C'/\Gamma (C', \operatorname {\mathcal {O}}_{C'})}=0$ (see [Reference Conrad, Gabber and PrasadCGP15, Example A.5.6]), which proves the remaining claim.

Proof. The semi-factorial model $\mathscr {D}$ of $D$ constructed in [Reference OverkampOve24, Theorem 4.3] is naturally endowed with a morphism $\psi \colon \widetilde {\mathscr {D}} \to \mathscr {D},$ which is a strong model pair by construction.

Proof. The Néron lft-model of $\operatorname {\mathbf {G}_m}$ will be denoted by $\operatorname {\mathscr {G}_m};$ the base ring will always be clear from context. Define

\begin{gather*} R_{\widetilde{D}/D} := \operatorname{\mathrm{Res}}_{\Gamma(\widetilde{\mathscr{D}}, \operatorname{\mathcal{O}}_{\widetilde{\mathscr{D}}})/\operatorname{\mathcal{O}}_K}\operatorname{\mathbf{G}_m} / \operatorname{\mathrm{Res}}_{\Gamma({\mathscr{D}}, \operatorname{\mathcal{O}}_{{\mathscr{D}}})/\operatorname{\mathcal{O}}_K}\operatorname{\mathbf{G}_m},\\ \mathscr{R}_{\widetilde{D}/D} := \operatorname{\mathrm{Res}}_{\Gamma(\widetilde{\mathscr{D}}, \operatorname{\mathcal{O}}_{\widetilde{\mathscr{D}}})/\operatorname{\mathcal{O}}_K}\operatorname{\mathscr{G}_m} / \operatorname{\mathrm{Res}}_{\Gamma({\mathscr{D}}, \operatorname{\mathcal{O}}_{{\mathscr{D}}})/\operatorname{\mathcal{O}}_K}\operatorname{\mathscr{G}_m},\\ R_{\mathcal{A}/\mathcal{B}} :=\operatorname{\mathrm{Res}}_{\mathcal{A}/\operatorname{\mathcal{O}}_K} \operatorname{\mathbf{G}_m} / \operatorname{\mathrm{Res}}_{\mathcal{B}/\operatorname{\mathcal{O}}_K} \operatorname{\mathbf{G}_m}, \end{gather*}

and

\[ \mathscr{R}_{\mathcal{A}/\mathcal{B}} :=\operatorname{\mathrm{Res}}_{\mathcal{A}/\operatorname{\mathcal{O}}_K} \operatorname{\mathscr{G}_m} / \operatorname{\mathrm{Res}}_{\mathcal{B}/\operatorname{\mathcal{O}}_K} \operatorname{\mathscr{G}_m}. \]

Note that these group schemes are smooth and separated over $\operatorname {\mathcal {O}}_K$.

Step 1: Since $\mathscr {R}_{\mathcal {A}/\mathcal {B}}$ is smooth over $\operatorname {\mathcal {O}}_K,$ there is a canonical map $\mathscr {R}_{\mathcal {A}/\mathcal {B}} \to \mathscr {T},$ which induces a map $\mathcal {A}/\mathcal {B} \to \operatorname {\mathrm {Lie}} \,\mathscr {T}$ on Lie algebras. The composition $\operatorname {\mathrm {Res}}_{\Gamma (\widetilde {\mathscr {D}}, \operatorname {\mathcal {O}}_{\widetilde {\mathscr {D}}})/\operatorname {\mathcal {O}}_K}\operatorname {\mathbf {G}_m} \to \mathscr {T}$ vanishes generically (and, hence, everywhere), so the induced map $\Gamma (\widetilde {\mathscr {D}}, \operatorname {\mathcal {O}}_{\widetilde {\mathscr {D}}}) \to \mathcal {A}/\mathcal {B} \to \operatorname {\mathrm {Lie}}\,\mathscr {T}$ vanishes. This induces the desired map $\hbar$.

Step 2: We claim that the map $\mathscr {R}_{\mathcal {A}/\mathcal {B}} \to \mathscr {T}$ is surjective on $\operatorname {\mathcal {O}}_K$-points. Since both group schemes are the Néron lft-models of their generic fibres, it suffices to show that the map is surjective on $K$-points. But this follows from the fact that

\begin{align*} &H^1(K, \operatorname{\mathrm{Res}}_{\Gamma(\widetilde{D}, \operatorname{\mathcal{O}}_{\widetilde{D}})/K}\operatorname{\mathbf{G}_m} / \operatorname{\mathrm{Res}}_{\Gamma({D}, \operatorname{\mathcal{O}}_{D})/K}\operatorname{\mathbf{G}_m})\\ &\quad =\ker ( \operatorname{\mathrm{Br}} ( \Gamma({D}, \operatorname{\mathcal{O}}_{D}) ) \to \operatorname{\mathrm{Br}}(\Gamma(\widetilde{D}, \operatorname{\mathcal{O}}_{\widetilde{D}})) )\\ &\quad =0 \end{align*}

[Reference Colliot-Thélène and SkorobogatovCS21, Theorem 1.2.15]. By [Reference Bosch, Lütkebohmert and RaynaudBLR90, Chapter 9.6, Lemma 2], the induced map

\[ R_{\mathcal{A}/\mathcal{B}}(\operatorname{\mathcal{O}}_K) \to \mathscr{T}^0(\operatorname{\mathcal{O}}_K) \]

is surjective as well.

Step 3: Let $\mathscr {K}$ be the kernel of the map $\mathscr {R}_{\mathcal {A}/\mathcal {B}} \to \mathscr {T}$. Now we claim that the induced map $\rho \colon \mathscr {R}_{\widetilde {D}/D} \to \mathscr {K}$ satisfies the universal property of the smoothening (see [Reference Bosch, Lütkebohmert and RaynaudBLR90, p. 174]). This follows immediately from the fact that $\mathscr {R}_{\widetilde {D}/D}$ is the Néron lft-model of the generic fibre of $\mathscr {K}$. Put

\[ \mathscr{R}_{\widetilde{D}/D}':= \rho^{-1}(\mathscr{K}\cap R_{\mathcal{A}/\mathcal{B}}). \]

Then $\mathscr {R}_{\widetilde {D}/D}'$ is the group smoothening of $\mathscr {K}\cap R_{\mathcal {A}/\mathcal {B}};$ this is an immediate consequence of the universal properties of the group smoothening and the Néron lft-model. Note that, moreover, we have an exact sequence

\[ 0 \to \mathscr{K} \cap R_{\mathcal{A}/\mathcal{B}} \to R_{\mathcal{A}/\mathcal{B}} \to \mathscr{T}^0 \]

of group schemes of finite type over $\operatorname {\mathcal {O}}_K$.

Step 4: Consider the commutative diagram

the top row being induced by the sequence (4) above. The remainder of the argument now works as in [Reference Liu, Lorenzini and RaynaudLLR04, p. 480]: The snake lemma shows that $\ker \hbar =\mathrm {coker} \, \alpha,$ we clearly have $\mathrm {coker}\, \hbar = \mathrm {coker}\, u,$ and [Reference Liu, Lorenzini and RaynaudLLR04, Theorem 2.1(a)] shows that

\[ \ell_{\operatorname{\mathcal{O}}_K}( \mathrm{coker}\, \alpha) = \ell_{\operatorname{\mathcal{O}}_K}( \mathrm{coker}\, u).\]

Proof. The surjectivity in part (i) follows from the fact that $P_{\mathscr {C}} \to P_{\widetilde {\mathscr {C}}}$ is surjective (Proposition 3.4) together with the fact that $\mathscr {N}$ and $\widetilde {\mathscr {N}}$ are quotients of $P_{\mathscr {C}}$ and $P_{\widetilde {\mathscr {C}}},$ respectively (Corollary 3.12). This also implies that the map $\mathscr {N} \to \widetilde {\mathscr {N}}$ is fibre-wise flat, so it is faithfully flat by the fibre-wise criterion of flatness.

(ii) (a)$\Rightarrow$(b): If $\mathscr {N} \to \widetilde {\mathscr {N}}$ is smooth, it is elementary to show that its kernel is smooth and satisfies the Néron mapping property.

(b)$\Rightarrow$(c): If the sequence is exact, the map $\mathscr {N} \to \widetilde {\mathscr {N}}$ is smooth. Hence, the map $\operatorname {\mathrm {Lie}} \,\mathscr {N} \to \operatorname {\mathrm {Lie}} \,\widetilde {\mathscr {N}}$ is surjective, which means that the map $H^1(\mathbf {L}(\mathscr {C}))\to H^1(\mathbf {L}(\widetilde {\mathscr {C}}))$ is surjective as well. By Theorem 3.5, the two potentially non-trivial cohomology modules of $\mathbf {L}(\mathscr {C})$ and $\mathbf {L}(\widetilde {\mathscr {C}})$ have the same lengths. Let $t$ (respectively, $\widetilde {t}$) be this common length of the cohomology modules of $\mathbf {L}(\mathscr {C})$ (respectively, $\mathbf {L}(\widetilde {\mathscr {C}})$). We have just shown that $t\geq \widetilde {t}$. However, it is easy to see that the map $H^0(\mathbf {L}(\mathscr {C}))\to H^0(\mathbf {L}(\widetilde {\mathscr {C}}))$ is injective, so that $t\leq \widetilde {t}$. Hence, $t=\widetilde {t}$ and the injectivity and surjectivity we have just seen proves that $\psi ^\ast$ is a quasi-isomorphism.

(c)$\Rightarrow$(a): Let $\mathscr {K}$ be the kernel of the map $\mathscr {N}\to \widetilde {\mathscr {N}}$ and consider the following commutative diagram.

We obtain the exact sequence

\[ H^1(\mathbf{L}(\mathscr{C})) \to H^1(\mathbf{L}(\widetilde{\mathscr{C}})) \to \mathrm{coker} (\operatorname{\mathrm{Lie}} \,\mathscr{N} \to \operatorname{\mathrm{Lie}} \,\widetilde{\mathscr{N}}) \to 0. \]

If $\psi ^\ast$ is a quasi-isomorphism, the map $\operatorname {\mathrm {Lie}} \,\mathscr {N} \to \operatorname {\mathrm {Lie}}\, \widetilde {\mathscr {N}}$ must be surjective, so $\mathscr {N} \to \widetilde {\mathscr {N}}$ is smooth (this can be reduced to the finite-type case, which follows from [Reference Liu, Lorenzini and RaynaudLLR04, Proposition 1.1(e)]).

Proof. By the discussion following the proof of Proposition 3.16, $\mathscr {C}$ is cohomologically flat in dimension zero as well. In particular, $H^0(\mathbf {L}(\mathscr {C}))=H^0(\mathbf {L}(\widetilde {\mathscr {C}}))=0$. Hence, by Theorem 3.5, $H^1(\mathbf {L}(\mathscr {C}))=H^1(\mathbf {L}(\widetilde {\mathscr {C}}))=0$. This shows $\mathbf {L}(\mathscr {C})\cong 0 \cong \mathbf {L}(\widetilde {\mathscr {C}})$. Therefore, the claim follows from the previous proposition.

Proof. We have

\[ \operatorname{\mathrm{Pic}}^0_{\widetilde{C}/\kappa}=\operatorname{\mathrm{Res}}_{\Gamma(\widetilde{C}, \operatorname{\mathcal{O}}_{\widetilde{C}})/\kappa}\operatorname{\mathrm{Pic}}^0_{\widetilde{C}/\Gamma(\widetilde{C}, \operatorname{\mathcal{O}}_{\widetilde{C}})}. \]

Because $\widetilde {C}$ is normal and $\delta (\kappa )\leq 1,$ $\widetilde {C}$ is geometrically reduced over $\Gamma (\widetilde {C}, \operatorname {\mathcal {O}}_{\widetilde {C}})$ (Proposition 2.6). Because the base change of a unirational algebraic group is unirational, the claim now follows from [Reference OverkampOve24, Proposition 2.32].

Proof. That $\ker \nu ^\ast$ is unirational (respectively, equals $\mathrm {uni}(\operatorname {\mathrm {Pic}}^0_{C/\kappa })$ if $\delta (\kappa )\leq 1$) can be shown as in the proof of [Reference OverkampOve24, Theorem 2.33]. For the first claim (respectively, equality), we may assume without loss of generality that $\kappa$ is separably closed ([Reference Conrad, Gabber and PrasadCGP15, Corollary B.3.5] together with [Reference OverkampOve24, Lemma 2.25] and Lemma 2.4 above). The first claim follows as in the proof of [Reference OverkampOve24, Theorem 2.33]. If $\delta (\kappa )\leq 1,$ we can again argue as in [Reference OverkampOve24, Theorem 2.33] as soon as we can show that $\ker \widetilde {\varsigma }^\ast$ contains no closed subgroup isomorphic to $\operatorname {\mathbf {G}_a}$. By [Reference OverkampOve24, proof of Theorem 2.24(ii)] and [Reference OverkampOve24, Proposition 2.30], we see that $\ker \widetilde {\varsigma }^\ast$ is a repeated extension of algebraic $\kappa$-groups of the form $\operatorname {\mathrm {Res}}_{A/\kappa }\operatorname {\mathbf {G}_m}/(\operatorname {\mathrm {Res}}_{B/\kappa }\operatorname {\mathbf {G}_m} \cdot \operatorname {\mathrm {Res}}_{L/\kappa } \operatorname {\mathbf {G}_m}),$ where $A$ is either a field or the product of a finite extension of $\kappa$ with itself, and $B$ and $L$ are subalgebras of $A$. In the first case, both $B$ and $L$ are fields. Since $\delta (\kappa )\leq 1$ and $\kappa$ is separably closed, each finite extension of $\kappa$ is isomorphic to $\kappa ^{1/p^n}$ for some $n$. Hence, either $L\subseteq B$ or $B\subseteq L$ in a canonical way. Denoting the larger field by $M,$ we see that the quotient above is, in fact, isomorphic to $\operatorname {\mathrm {Res}}_{A/\kappa }\operatorname {\mathbf {G}_m}/\operatorname {\mathrm {Res}}_{M/\kappa }\operatorname {\mathbf {G}_m}$. In the second case, we write $A=F\times F$ for a field extension $F\supseteq \kappa$. The argument from [Reference OverkampOve24] shows moreover that we can take $B=F$ with the map $B\to A$ being the diagonal map. Then $L$ is either a field of the product of two field extensions $L_1$ and $L_2$ of $\kappa$. In the first case, the quotient above is isomorphic to $\operatorname {\mathrm {Res}}_{F/\kappa }\operatorname {\mathbf {G}_m};$ in the second case it is isomorphic to $\operatorname {\mathrm {Res}}_{F/\kappa }\operatorname {\mathbf {G}_m}/\operatorname {\mathrm {Res}}_{M/\kappa }\operatorname {\mathbf {G}_m},$ where $M$ is the larger one of $L_1$ and $L_2$. Clearly, we have $\operatorname {\operatorname {Hom}}_{\kappa }(\operatorname {\mathbf {G}_a}, \operatorname {\mathrm {Res}}_{F/\kappa } \operatorname {\mathbf {G}_m})=0$. Hence, all we need to show is that, if $M\subseteq A$ are finite reduced $\kappa$-algebras, we have

\[ \operatorname{\operatorname{Hom}}_{\kappa}(\operatorname{\mathbf{G}_a}, \operatorname{\mathrm{Res}}_{A/\kappa}\operatorname{\mathbf{G}_m}/\operatorname{\mathrm{Res}}_{M/\kappa}\operatorname{\mathbf{G}_m})=0. \]

Any element $\tau$ of this group gives an element $\sigma \in \operatorname {\operatorname {Ext}}^1_{\kappa }(\operatorname {\mathbf {G}_a}, \operatorname {\mathrm {Res}}_{M/\kappa }\operatorname {\mathbf {G}_m})$. However, since $M$ is finite over $\kappa,$ the Grothendieck spectral sequence shows that

\[ \operatorname{\operatorname{Ext}}^1_{\kappa}(\operatorname{\mathbf{G}_a}, \operatorname{\mathrm{Res}}_{M/\kappa}\operatorname{\mathbf{G}_m})=\operatorname{\operatorname{Ext}}^1_{M}(\operatorname{\mathbf{G}_a}, \operatorname{\mathbf{G}_m})=0 \]

[Reference Demazure and GrothendieckDG70, Exposé XVII, Théorème 6.1.1 A) ii)]. Hence, $\sigma =0$ and $\tau$ lifts to a map $\operatorname {\mathbf {G}_a} \to \operatorname {\mathbf {G}_m}$ over $A,$ which must vanish since $A$ is a finite product of fields.

Proof. The first claim follows immediately from Theorem 4.2. Moreover, the proof of Proposition 3.14 shows that, if $C$ is not seminormal, then $\ker \varsigma ^\ast$ is non-trivial. Hence, the second claim also follows.

Proof. We may assume without loss of generality that $C$ is reduced. Indeed, our assumption implies that the map $\operatorname {\mathrm {Pic}}^0_{C/K} \to \operatorname {\mathrm {Pic}}^0_{C_{\mathrm {red}}/K}$ is an isomorphism (see [Reference Bosch, Lütkebohmert and RaynaudBLR90, Chapter 9.2, Proposition 5] together with [Reference OverkampOve24, Lemma 2.3]). Then we know that the map $\operatorname {\mathrm {Pic}}^0_{C/K} \to \operatorname {\mathrm {Pic}}^0_{\widetilde {C}/K}$ is an isomorphism (Corollary 4.3). By Proposition 2.6, we know that $\widetilde {C}$ is geometrically reduced over $\Gamma (\widetilde {C}, \operatorname {\mathcal {O}}_{\widetilde {C}})$. If $S'$ denotes the integral closure of $S$ in $\Gamma (\widetilde {C}, \operatorname {\mathcal {O}}_{\widetilde {C}}),$ then [Reference OverkampOve24, Theorem 3.4] shows that $\operatorname {\mathrm {Pic}}^0_{\widetilde {C}/\Gamma (\widetilde {C}, \operatorname {\mathcal {O}}_{\widetilde {C}})}$ admits a Néron model $\widetilde {\mathscr {N}'}\to S'$. Therefore, the scheme

\[ \widetilde{\mathscr{N}}:=\operatorname{\mathrm{Res}}_{S'/S} \widetilde{\mathscr{N}}' \]

is a Néron model of $\operatorname {\mathrm {Res}}_{\Gamma (\widetilde {C}, \operatorname {\mathcal {O}}_{\widetilde {C}})/K}\operatorname {\mathrm {Pic}}^0_{\widetilde {C}/\Gamma (\widetilde {C}, \operatorname {\mathcal {O}}_{\widetilde {C}})}=\operatorname {\mathrm {Pic}}^0_{\widetilde {C}/K}=\operatorname {\mathrm {Pic}}^0_{C/K}$ over $S$.

Proof. Let $\mathfrak {p}\subseteq A\otimes _R\widehat {R}$ be a prime ideal. We must show that the localisation $(A\otimes _R\widehat {R})_{\mathfrak {p}}$ is regular. If $\mathfrak {p}$ maps to the generic point of $\operatorname {\operatorname {Spec}} \widehat {R},$ this is clear because the extension $\operatorname {\operatorname {Frac}} R \subseteq \operatorname {\operatorname {Frac}} \widehat {R}$ is separable. Otherwise we put $\mathfrak {q}:=\mathfrak {p}\cap A$ and observe that the canonical map $A/\mathfrak {q} \to (A\otimes _R\widehat {R})/\mathfrak {p}$ is an isomorphism since $R\to \widehat {R}$ induces an isomorphism of residue fields. This shows that $\mathfrak {p}=\mathfrak {q}\otimes _R\widehat {R}$ and, hence, $\mathfrak {p}/\mathfrak {p}^2=\mathfrak {q}/\mathfrak {q}^2;$ moreover, we see that the map $\kappa (\mathfrak {q})\to \kappa (\mathfrak {p})$ on residue fields is an isomorphism. Now choose a maximal chain $\mathfrak {p}_0\subset \cdots \subset \mathfrak {p}_n=\mathfrak {p}$ of prime ideals such that $n= \mathrm {ht}_{A\otimes _R\widehat {R}}\, \mathfrak {p}$. Clearly $\mathfrak {p}_0$ is minimal, so $(A\otimes _R\widehat {R})/\mathfrak {p}_0$ is generically of the same dimension as $A\otimes _R\operatorname {\operatorname {Frac}}{R}$. Moreover, we see that

\[ \mathrm{ht}_{(A\otimes_R\widehat{R})/\mathfrak{p}_0} \, \mathfrak{p}/\mathfrak{p}_0 = \mathrm{ht}_{A\otimes_R\widehat{R}} \, \mathfrak{p}. \]

But $R$ is universally catenary, so the dimension formula [Sta18, Tag 02IJ] shows that $\mathrm {ht}_A\, \mathfrak {q} = \mathrm {ht}_{(A\otimes _R\widehat {R})/\mathfrak {p}_0}\,\mathfrak {p}/\mathfrak {p}_0$. From this, we deduce (replacing $\mathfrak {p}$ and $\mathfrak {q}$ with their respective localisations) that

\[ \dim (A\otimes_R\widehat{R})_{\mathfrak{p}} = \dim A_{\mathfrak{q}} = \dim_{\kappa(\mathfrak{q})} \mathfrak{q}/\mathfrak{q}^2 = \dim_{\kappa(\mathfrak{p})} \mathfrak{p}/\mathfrak{p}^2.\]

Proof. We may once again assume that $C$ is reduced. Our assumption implies that $\operatorname {\mathrm {Pic}}^0_{C/K}$ admits Néron lft-models $\mathscr {N}_{\mathfrak {p}}$ at all closed points $\mathfrak {p}$ of $S$ (see [Reference Bosch, Lütkebohmert and RaynaudBLR90, Chapter 10.2, Theorem 2(b’)]). By [Reference Bosch, Lütkebohmert and RaynaudBLR90, Chapter 10.1, Proposition 9], we may replace $S$ by a dense open subscheme (in particular, we may suppose that $S$ is affine). Moreover, [Reference Bosch, Lütkebohmert and RaynaudBLR90] also shows that it suffices to find a finite locally free $\Gamma (S, \operatorname {\mathcal {O}}_S)$-module $\mathscr {L}$ and an isomorphism $\mathscr {L}\otimes _{\Gamma (S, \operatorname {\mathcal {O}}_S)} K \to \operatorname {\mathrm {Lie}} \,\operatorname {\mathrm {Pic}}^0_{C/K}=H^1(C, \operatorname {\mathcal {O}}_C)$ such that, for all closed points $\mathfrak {p}\in S,$ the images of $\operatorname {\mathrm {Lie}}\, \mathscr {N}_{\mathfrak {p}}$ and $\mathscr {L}\otimes _{\Gamma (S, \operatorname {\mathcal {O}}_S)} \operatorname {\mathcal {O}}_{S, \mathfrak {p}}$ inside $\mathscr {L}\otimes _{\Gamma (S, \operatorname {\mathcal {O}}_S)} K$ coincide. Using the method from [Reference OverkampOve24, Corollary 3.7], we construct proper and flat (though a priori not necessarily cohomologically flat) models $\mathscr {C}\to S$ and $\widetilde {\mathscr {C}}\to S$ of $C$ and $\widetilde {C},$ respectively, which fit into a co-cartesian diagram

where $\mathcal {B}\subseteq \mathcal {A}$ are finite regular $\Gamma (S, \operatorname {\mathcal {O}}_S)$-algebras, such that the vertical maps are closed immersions. Replacing $S$ by a dense open subscheme if necessary, we may suppose that $H^1(\mathscr {C}, \operatorname {\mathcal {O}}_{\mathscr {C}})$ is locally free over $\Gamma (S, \operatorname {\mathcal {O}}_S);$ we shall now show that

\[ \mathscr{L}:=H^1(\mathscr{C}, \operatorname{\mathcal{O}}_{\mathscr{C}}) \]

has the required properties. By Lemma 4.5, we see that the scheme $\widetilde {\mathscr {C}}_{\mathfrak {p}}$ is a proper, flat and regular model of $\widetilde {C}\times _K\operatorname {\operatorname {Spec}} \widehat {K^{\mathrm {sh}}}$ over $\operatorname {\operatorname {Spec}} \widehat {\operatorname {\mathcal {O}}^{\mathrm {sh}}_{S, \mathfrak {p}}}$.Footnote ⁷ Using the method from the proofs of [Reference OverkampOve24, Theorem 4.3 and Proposition 4.4], we construct a proper birational morphism $\widetilde {\mathscr {C}}'_{\mathfrak {p}}\to \widetilde {\mathscr {C}}_{\mathfrak {p}}$ such that the push-out

\[ \mathscr{C}'_{\mathfrak{p}}:=\widetilde{\mathscr{C}}'_{\mathfrak{p}}\cup_{\operatorname{\operatorname{Spec}} \mathcal{A}_{\mathfrak{p}}} \operatorname{\operatorname{Spec}} \mathcal{B}_{\mathfrak{p}} \]

is a proper, flat and semi-factorial model of $C\times _K\operatorname {\operatorname {Spec}} \widehat {K^{\mathrm {sh}}}$. Note that we have a canonical map $\mathscr {C}'_{\mathfrak {p}}\to \mathscr {C}_{\mathfrak {p}}$. The lemma of five homomorphisms applied to the exact sequence (4) from § 3.4 shows that the map $H^1(\mathscr {C}_{\mathfrak {p}}, \operatorname {\mathcal {O}}_{\mathscr {C}_{\mathfrak {p}}}) \to H^1(\mathscr {C}'_{\mathfrak {p}}, \operatorname {\mathcal {O}}_{\mathscr {C}'_{\mathfrak {p}}})$ is an isomorphism. Because the residue fields of $S$ are perfect and $H^1(\mathscr {C}, \operatorname {\mathcal {O}}_{\mathscr {C}})$ is torsion-free, Corollary 3.13 (together with [Reference Bosch, Lütkebohmert and RaynaudBLR90, Chapter 10.1, Proposition 3]) shows that the map

\[ H^1(\mathscr{C}'_{\mathfrak{p}}, \operatorname{\mathcal{O}}_{\mathscr{C}'_{\mathfrak{p}}}) \to (\operatorname{\mathrm{Lie}}\, \mathscr{N}_{\mathfrak{p}})\otimes_{\operatorname{\mathcal{O}}_{S, \mathfrak{p}}} \widehat{\operatorname{\mathcal{O}}^{\mathrm{sh}}_{S,{\mathfrak{p}}}} \]

is an isomorphism. Now observe that $\mathscr {C}\times _S\operatorname {\operatorname {Spec}} \operatorname {\mathcal {O}}_{S, \mathfrak {p}}$ is cohomologically flat in dimension zero over $\operatorname {\mathcal {O}}_{S, \mathfrak {p}}$ for all closed points $\mathfrak {p}\in S$ (see [Reference LiuLiu02, Chapter 5.3, Corollary 3.22 and Remark 3.30]), so $\operatorname {\mathrm {Pic}}_{\mathscr {C}\times _S\operatorname {\operatorname {Spec}} \operatorname {\mathcal {O}}_{S, \mathfrak {p}}/\operatorname {\mathcal {O}}_{S, \mathfrak {p}}}$ is a smooth algebraic space over $\operatorname {\mathcal {O}}_{S, \mathfrak {p}}$ ([Reference Bosch, Lütkebohmert and RaynaudBLR90, Chapter 8.3, Theorem 1] and [Reference RaynaudRay70, Corollaire 2.3.2]). It is easy to see that the Néron mapping property remains valid for algebraic spaces, so there is a canonical map $\mathscr {L}\otimes _{\Gamma (S, \operatorname {\mathcal {O}}_S)} \operatorname {\mathcal {O}}_{S, \mathfrak {p}} \to \operatorname {\mathrm {Lie}}\, \mathscr {N}_{\mathfrak {p}}$. Our claim follows once we establish that this map is an isomorphism; however, this can be checked after applying the functor $-\otimes _{\operatorname {\mathcal {O}}_{S, \mathfrak {p}}} \widehat {\operatorname {\mathcal {O}}^{\mathrm {sh}}_{S,{\mathfrak {p}}}}$.

Proof. Throughout the proof, we may assume without loss of generality that $C$ is connected. We begin by observing that $\widetilde {C}_{\widehat {K}}$ is regular. Indeed, the argument from the proof of Lemma 4.5 shows that $\widetilde {\mathscr {C}}_{\widehat {R}}$ is regular at all points on the special fibre. However, since $\widetilde {\mathscr {C}}_{\widehat {R}}$ is excellent, its singular locus is closed. Because this scheme is proper over $\widehat {R},$ the singular locus would have to intersect the special fibre if it were non-empty. Hence, $\widetilde {\mathscr {C}}_{\widehat {R}}$ is already regular (see also [Sta18, Tag 0BG6 (3)]). Since the morphism $\psi _K\colon \widetilde {C} \to C$ is scheme-theoretically dominant, so is $\psi _K\times _K \mathrm {Id}_{\widehat {K}}$. In particular, $C_{\widehat {K}}$ is reduced.

(i) $\iff$ (ii): By [Reference Bosch, Lütkebohmert and RaynaudBLR90, Chapter 10.2, Theorem 2], part (i) is equivalent to $\mathscr {R}_{us, \widehat {K}}(\operatorname {\mathrm {Pic}}^0_{C_{\widehat {K}}/\widehat {K}})=0$. Because we already know that $C_{\widehat {K}}$ is reduced, this is equivalent to part (ii) by Corollary 4.3 (if $\kappa$ is perfect; see Lemma 2.1) and [Reference OverkampOve24, Corollary 2.34] (note that the existence of $K^{\mathrm {sh}}$-points as in the theorem implies geometric reducedness).

(iii) $\Rightarrow$ (ii): If part (iii) holds, then $C$ arises as the push-out of $\operatorname {\operatorname {Spec}} A \to \widetilde {C}$ along $\operatorname {\operatorname {Spec}} A \to \operatorname {\operatorname {Spec}} B$ (see [Reference OverkampOve24, Theorem 2.24(ii)]). Hence, $C_{\widehat {K}}$ arises as the push-out of $\operatorname {\operatorname {Spec}} A\otimes _K\widehat {K}\to \widetilde {C}_{\widehat {K}}$ along $\operatorname {\operatorname {Spec}} A\otimes _K\widehat {K} \to \operatorname {\operatorname {Spec}} B\otimes _K\widehat {K}$. Therefore, our assumptions imply that $C_{\widehat {K}}$ is seminormal.

(i) $\Rightarrow$ (iii): If $\operatorname {\mathrm {Pic}}^0_{C/K}$ admits a Néron lft-model, then $\mathscr {R}_{us,K}(\operatorname {\mathrm {Pic}}^0_{C/K})=0$. This follows from [Reference Bosch, Lütkebohmert and RaynaudBLR90, Chapter 10.2, Theorem 2] together with the fact that $\mathscr {R}_{us, K}(\operatorname {\mathrm {Pic}}^0_{C/K})_{\widehat {K}}\subseteq \mathscr {R}_{us, \widehat {K}}(\operatorname {\mathrm {Pic}}^0_{C_{\widehat {K}}/\widehat {K}})$. However, $\mathscr {R}_{us, K}(\operatorname {\mathrm {Pic}}^0_{C/K})$ vanishes only if $C$ is seminormal (Corollary 4.3). In particular, $C$ arises as the push-out of $\operatorname {\operatorname {Spec}} A \to \widetilde {C}$ along $\operatorname {\operatorname {Spec}} A \to \operatorname {\operatorname {Spec}} B$ (see [Reference OverkampOve24, Theorem 2.24]). Hence, $C_{\widehat {K}}$ is the push-out of $\operatorname {\operatorname {Spec}} A\otimes _K\widehat {K} \to \widetilde {C}_{\widehat {K}}$ along $\operatorname {\operatorname {Spec}} A\otimes _K\widehat {K} \to \operatorname {\operatorname {Spec}} B\otimes _K\widehat {K},$ so if $A\otimes _K\widehat {K}$ is non-reduced, then $C_{\widehat {K}}$ is not seminormal. Indeed, Lemma 4.8 shows that we can find a nilpotent element $y\in A\otimes _K\widehat {K}$ not contained in $B\otimes _K\widehat {K}$. In particular, there exists $N_0\in \operatorname {\mathbf {N}}$ such that, for all $n> N_0,$ we have $y^n\in B\otimes _K\widehat {K}$. Choose $N_0$ minimal with this property. Then $N_0\geq 1$ by assumption. Now replace $y$ by $y^{N_0}$. Then $y\not \in B\otimes _K\widehat {K},$ but $y^2$ and $y^3$ both lie in $B\otimes _K\widehat {K}$. If $U$ is an open affine neighbourhood of $\operatorname {\operatorname {Spec}} A\otimes _K\widehat {K}$ in $\widetilde {C}_{\widehat {K}},$ then we can find $f\in \Gamma (U, \operatorname {\mathcal {O}}_U)$ whose image in $A$ is $y$. The scheme $U':=U\cup _{\operatorname {\operatorname {Spec}} A\otimes _K\widehat {K}} (\operatorname {\operatorname {Spec}} B\otimes _K\widehat {K})$ is an open subscheme of $C_{\widehat {K}}$ by construction. Moreover, $f^2$ and $f^3$ are global functions on $U'$ but $f$ is not. Hence, $U'$ (and therefore $C_{\widehat {K}}$) is not seminormal. This contradicts (i) $\iff$ (ii), which we have already proven.

We construct the morphism $\psi \colon \widetilde {\mathscr {C}} \to \mathscr {C}$ as in [Reference OverkampOve24, Theorem 4.3]. Moreover, we replace the reference to the embedded resolution theorem as formulated in [Reference LiuLiu02, Chapter 9.2, Theorem 2.26], by a reference to [Sta18, Tag 0BIC], in order to avoid the hypothesis on excellence imposed in [Reference LiuLiu02]. Note that the conditions of [Sta18, Tag 0BIC] are satisfied since $A\otimes _K\widehat {K}$ (and, hence, $B\otimes _K\widehat {K}$) are reduced, so the integral closures of $R$ in $A$ and $B$ are finite over $R$. Our hypotheses further imply that the map $\operatorname {\mathrm {Pic}} \,C_{K^{\mathrm {sh}}} \to \operatorname {\mathrm {Pic}}_{C_{K^{\mathrm {sh}}}/K^{\mathrm {sh}}}(K^{\mathrm {sh}})$ is bijective. Since we already know that $\operatorname {\mathrm {Pic}}^0_{C/K}$ admits a Néron lft-model $\mathscr {N},$ the proofs of Proposition 3.11 and Corollary 3.12 can be taken mutatis mutandis to show that $P_{\mathscr {C}_{R^{\mathrm {sh}}}/R^{\mathrm {sh}}}^ \mathrm {sep} \to \mathscr {N}_{R^{\mathrm {sh}}}$ is an isomorphism. Indeed, if $G$ is a smooth connected algebraic group over a separably closed field $\kappa,$ it is still true that $G(\kappa )$ is $n$-divisible for all positive integers $n$ invertible in $\kappa,$ so if $G(\kappa )$ is finitely generated then $G(\kappa )$ is finite. Hence, $G=0$ since $G$ is smooth.

Proof. Let $N_A$ and $N_B$ be the ideals of nilpotent elements in $A\otimes _K\widehat {K}$ and $B\otimes _K\widehat {K},$ respectively. Assume, for the sake of a contradiction, that the inclusion $N_B\subseteq N_A$ is an equality. Then $N_B\not =0$ by assumption. Now choose a $B$-basis $1, \alpha _1,\ldots, \alpha _d$ of $A$ for some $d\in \operatorname {\mathbf {N}}$. Let $\nu \in N_B$ be any non-zero element. Since $1\otimes 1, \alpha _1\otimes 1,\ldots, \alpha _d \otimes 1$ is a $B\otimes _K\widehat {K}$-basis of $A\otimes _K\widehat {K},$ the element $\nu \cdot \alpha _1\otimes 1 \in A\otimes _K\widehat {K}$ is not contained in $B\otimes _K\widehat {K}$. However, it is clearly nilpotent, contradicting the assumption that $N_B=N_A$.

Proof. This follows immediately from the definition together with [Sta18, Tag 0C1S (1)].

Proof. First observe that the cases $d=1$ and $d=2$ are trivial. By splitting the sequence into $0\to G_1\to G_2\to G_2/G_1\to 0$ and $0\to G_2/G_1 \to G_3 \to \cdots \to G_d \to 0,$ as well as using induction on $d,$ we reduce the claim to the case $d=3,$ which is proven in [Reference Halle and NicaiseHN16, Lemma 3.2.2.4].

Proof. Using [Reference Bosch, Lütkebohmert and RaynaudBLR90, Chapter 9.2, Proposition 5] and Lemma 4.10, we may assume that $C$ is reduced. Let $\widetilde {C}$ and $C^{\mathrm {sn}}$ be the normalisation and the seminormalisation of $C,$ respectively. Because the map $C^{\mathrm {sn}} \to C$ is a homeomorphism, Lemma 4.10 shows that $\Gamma (C)$ and $\Gamma (C^{\mathrm {sn}})$ are isomorphic. By Theorem 4.2 and Lemma 4.11, the dimension of the maximal split tori inside $\operatorname {\mathrm {Pic}}^0_{C/\kappa }$ and $\operatorname {\mathrm {Pic}}^0_{C^{\mathrm {sn}}/\kappa }$ coincide. Hence, we may assume that $C$ is seminormal. It follows from the proof of [Reference OverkampOve24, Theorem 2.24(ii)] that the morphism $\widetilde {C}\to C=C^{\mathrm {sn}}$ can be factorised as $\widetilde {C}=C_1 \to C_2 \to \cdots \to C_n=C,$ such that, for each $i=1,\ldots, n-1,$ there is a closed point $x_{i+1}\in C_{i+1},$ a reduced $\kappa (x_{i+1})$-algebra $A_i,$ and a closed immersion $\operatorname {\operatorname {Spec}} A_i\to C_i$ such that the following diagram is co-Cartesian.

Moreover, we may assume that $A_i$ is either isomorphic to $\kappa (x_{i+1}) \times \kappa (x_{i+1})$ or a field extension of $\kappa (x_{i+1})$ (see [Reference OverkampOve24, Corollary 2.23]). We shall prove the theorem by induction on $n$.