Proof. The same argument works with \(q=\ell^e\). If \(\operatorname{div}(f)=qD\), send \([f]\) to the divisor class of \(D\). Changing \(f\) by a \(q\)-th power does not change this class. Every \(q\)-torsion divisor class occurs. The kernel consists of functions \(cu^q\); the constant \(c\in\kappa^\times\) is a \(q\)-th power. This proves the finite-level identification. The \(q\)-torsion of the Jacobian over an algebraically closed field of characteristic prime to \(q\) is \((\mathbb Z/q)^{2g(C)}\) (The Stacks Project Authors 2026, Tag 0C1Z, Lemma 53.17.1). Under reduction from \(\ell^{e+1}\) to \(\ell^e\), the divisor class is multiplied by \(\ell\). Taking the inverse limit gives the exact-order kernel and its rank. ◻

Proof. We give the finite-coefficient argument, keeping track of the infinite rank of \(W_F\). Write \(G=G_F^{(\ell)}\), \(A=G^{\mathrm{ab}}\) and \(R=\mathbb Z/\ell^e\), with trivial action. Kummer duality makes \(A\) a product of copies of \(\mathbb Z_\ell\).

First, a class in \(H^2_{\mathrm{cont}}(A,R)\) is determined by its commutator form. Indeed, a class with zero commutator defines an abelian extension of \(A\) by \(R\). Such an extension splits continuously: the Pontryagin dual of \(A\) is a direct sum of copies of \(\mathbb Q_\ell/\mathbb Z_\ell\), hence is divisible and injective. Conversely every continuous alternating bilinear form \(A\times A\to R\) is the commutator form of a cup-product class. To see this, continuity and compactness give an open subgroup in the radical of the form, so it depends on finitely many coordinates. In coordinate characters \(x_i\) it is a sum of the forms \[x_i(a)x_j(b)-x_j(a)x_i(b),\qquad i<j,\] which come from the cocycles \(x_i\cup x_j\). Consequently cup products generate \(H^2_{\mathrm{cont}}(A,R)\). This proof uses no division by \(2\). In particular, even when \(\ell=2\), a diagonal cup has zero commutator and therefore vanishes. No extra power or Bockstein class survives on the torsion-free group \(A\).

Next, inflation \[H^2_{\mathrm{cont}}(G,R)\longrightarrow H^2(G_F,R)\] is injective. Let \(F(\ell)\) be the maximal Galois pro-\(\ell\) extension of \(F\). An element of \(F(\ell)^\times\) lies in a finite Galois \(\ell\)-extension \(F'/F\). Adjoin to \(F'\) the \(\ell^e\)-th roots of all its \(F\)-conjugates. Since the roots of unity lie in \(F\), the resulting field is again a finite Galois \(\ell\)-extension of \(F\). Thus \(F(\ell)^\times\) is \(\ell^e\)-divisible. Kummer theory gives \(H^1(G_{F(\ell)},R)=0\), and the five-term sequence proves the claimed injectivity.

Fix compatible identifications of the roots of unity with the coefficient groups. The degree-two norm-residue theorem (Merkurjev and Suslin 1983) identifies the kernel of the tensor cup map \[(F^\times/F^{\times\ell^e})\otimes_R (F^\times/F^{\times\ell^e})\longrightarrow H^2(G_F,R)\] with the subgroup generated by the Steinberg tensors \([x]\otimes[1-x]\). Since the tensor cup map onto \(H^2_{\mathrm{cont}}(A,R)\) is surjective, the kernel of inflation from \(A\) to \(G\) is generated by the Steinberg cups \([x]\cup[1-x]\).

Finally, put \(N=[G,G]\). The five-term sequence for \(N\subset G\) identifies this same kernel with \[\operatorname{Hom}_{\mathrm{cont}}(N/[N,G],R) =\operatorname{Hom}_{\mathrm{cont}}(\Delta_F,R).\] The commutator form of the transgression of a character \(\chi\) is \(\chi([\ ,\ ]_F)\), up to the uniform sign determined by the transgression convention. Its value at \(f,g\) therefore vanishes for all \(\chi\) precisely when \[f(x)g(1-x)-f(1-x)g(x)=0\pmod{\ell^e}\] for every \(x\ne0,1\). Continuous characters to finite cyclic \(\ell\)-groups separate points of \(\Delta_F\). Letting \(e\) vary proves the assertion. ◻

Proof. There is an exact sequence of abelian groups \[0\longrightarrow (Fv)^\times/(\kappa v)^\times \longrightarrow F^\times/(\kappa^\times U_v^1) \longrightarrow vF/v\kappa\longrightarrow0.\] Its final term is free, so applying \(\operatorname{Hom}(-,\mathbb Z_\ell)\) gives a surjection \(D_v\to W_{Fv}\) with kernel \(I_v\). This map is continuous for pointwise convergence; compactness makes the induced bijection on the quotient a topological isomorphism.

For the second assertion, use (2). If \(v(x)>0\), then \(1-x\) is a principal unit. If \(v(x)<0\), the identity \(1-x=-x(1-x^{-1})\) makes the characters of \(1-x\) equal those of \(x\). If \(v(x)=0\) and \(v(1-x)>0\), then \(x\) is a principal unit. In each case the alternating determinant is zero. In the remaining case \(x\) and \(1-x\) are units with residues different from \(0,1\), and the determinant is exactly the residual determinant. Conversely every residual element different from \(0,1\) has such a lift. ◻

Proof. For every valuation \(u\), characters kill the divisible constant group, so \[\dim_{\mathbb Q_\ell}\mathcal I_u \leq\operatorname{rrank}(uF/u\kappa).\] Let \(S\) be alternating and let \(\Sigma\) be its valuative subspace, with associated valuation \(u\). If \(S=\Sigma\), the asserted dimension bound follows from (4). Otherwise \(\Sigma\) has codimension one in \(S\). A character in \(S\setminus\Sigma\subset\mathcal D_u\) has nonzero residual character. Thus \(Fu\ne\kappa u\), and \(\operatorname{trdeg}(Fu/\kappa u)\geq1\). Consequently \[\dim S\leq\operatorname{rrank}(uF/u\kappa)+1\leq s.\] This also bounds arbitrary alternating subspaces, by applying the argument to their finite-dimensional subspaces.

There is a full discrete flag at a smooth closed point on a model of \(F/\kappa\). Its inertia space has dimension \(s\) and is alternating by the valuation inequality for \(x\) and \(1-x\). Thus the maximum equals \(s\).

If \(\dim S=s\) and \(S=\Sigma\), all inequalities \[s\leq\dim\mathcal I_u \leq\operatorname{rrank}(uF/u\kappa)\leq s\] are equalities. If \(S\ne\Sigma\), all inequalities \[s-1\leq\dim\mathcal I_u \leq\operatorname{rrank}(uF/u\kappa) \leq s-\operatorname{trdeg}(Fu/\kappa u)\leq s-1\] are equalities. In both cases \(\Sigma=\mathcal I_u\) and \(u\) has no transcendence defect. ◻

Proof. Suppose first that \(v\) is quasi-prime. The residue function field \(Fv/\kappa v\) has two independent discrete valuations of full rank \(s-1\), trivial on the constants. For example, choose two independent prime divisors on a rational subfield on a transcendence basis, complete each to a flag, and prolong the flags to the finite extension \(Fv\). Independent valuation approximation (Anscombe et al. 2020, Theorem 1.2) makes their inertia spaces disjoint. Composing with \(v\) gives two alternating inertia spaces of dimension \(s\) whose intersection is \(\mathcal I_v\).

Conversely, let \(S_1,S_2\) be as stated. By Lemma 8, their valuative parts are \(\Sigma_i=\mathcal I_{u_i}\), where each \(u_i\) has no transcendence defect and \(\dim\Sigma_i\geq s-1\). We first show that \(H\) is valuative. If, say, \(u_1\leq u_2\), then \[\Sigma_1\subset\Sigma_2, \qquad \Sigma_1\subset S_1\cap S_2=H.\] Since \(\dim\Sigma_1\geq1\), this proves the claim. If the valuations are incomparable, let \(w\) be their finest common coarsening. Applying independent valuation approximation on \(Fw\) shows that the principal-unit groups of \(u_1\) and \(u_2\) generate \(U_w\). Indeed, approximate a unit \(a\) by \(b\) for the first residual valuation and approximate \(1\) by \(b\) for the second. Then \(a=(a/b)b\) is a product of the two required principal units. A character in \(H\) annihilates both principal-unit groups, and hence belongs to \(\mathcal I_w\). Again \(H\) is valuative. In either case, \[H=\Sigma_1\cap\Sigma_2.\]

Let \(v=v_H\) be its coarsest associated valuation. It coarsens both \(u_i\), and \[H\subset\mathcal I_v \subset\mathcal I_{u_1}\cap\mathcal I_{u_2}=H.\] A coarsening of a valuation without transcendence defect also has no transcendence defect: relative rational ranks add under composition, and the two valuation inequalities sum to the equality for the finer valuation. Thus \(v\) has no transcendence defect. Its relative value group is finitely generated free, and its rank is \(\dim\mathcal I_v=1\). Hence \(vF/v\kappa\cong\mathbb Z\). A coarsening with the same relative rank would have the same inertia space, contradicting the defining minimality of \(v_H\). Therefore \(v\) is quasi-prime, and the coarsest-valuation description also proves uniqueness.

Every member of \(\mathcal D_v\) alternates with \(h\), by the calculation in Lemma 7. For the converse apply the third part of the local character theorem to the valuative line \(H\). ◻

Proof. Lemma 6 makes the alternating relation intrinsic to the bracket. Lemma 8 then recovers \(d\), and Lemma 9 recovers the rationalized rank-one pairs. Intersecting with the integral modules recovers the exact pairs. Lemma 7 supplies the residual character spaces and their alternating relation. If \(u\) is a valuation of \(Fv\), the inverse images of \(I_u\) and \(D_u\) under \(D_v\to W_{Fv}\) are \(I_{u\circ v}\) and \(D_{u\circ v}\), respectively, directly from the unit and principal-unit definitions. Apply the same characteristic-free recognition argument to each residue function field. Continuing while its transcendence degree is at least two recovers precisely the successive quasi-prime divisors through rank \(d-1\). The construction applies to \(\varphi^{-1}\) as well. ◻

Proof. For a quasi-prime divisor \(u\), an \(\ell\)-divisible convex subgroup \(H\subset uF\) has zero image in \(uF/u\kappa\cong\mathbb Z\). Thus \(H\subset u\kappa\). Coarsening by \(H\) preserves the relative rank one. Since \(u\) attains equality in the valuation inequality, its coarsening does also, so its residual transcendence degree is still \(s-1\). Minimality forces \(H=0\).

Now consider a composite quasi-prime divisor. The value group of its last residual step is a nonzero convex subgroup \(\Gamma\) of the full value group and has the rank-one property just proved. If \(H\) were a nonzero \(\ell\)-divisible convex subgroup of the full group, convexity would make \(H\cap\Gamma\) \(\ell\)-divisible: an \(\ell\)-th part of a positive element remains between zero and that element. Convex subgroups are comparable, so \(H\cap\Gamma\) is nonzero. This contradicts the rank-one case.

For the second hypothesis choose \(x_1,\ldots,x_{s-1}\) whose values are rationally independent modulo \(v\kappa\), and put \(\kappa_1=\kappa(x_1,\ldots,x_{s-1})\). Relative value independence implies algebraic independence. In a polynomial in the \(x_i\), distinct monomials have different values modulo \(v\kappa\), and there is a unique term of least value. A quotient of polynomials with value zero consequently has residue in \(\kappa v\): equality of the two least values forces the same monomial, leaving a quotient of constants. It follows that \(\kappa_1v=\kappa v\). The two asserted transcendence degrees are then one by (3). ◻

Proof. Proposition 10 recovers every group used in (5). A continuous isomorphism preserves the closure of the union, its intersection with decomposition, its quotient image and the maximal procyclic subgroups in that image. Lemma 11 verifies the hypotheses of Pop’s criterion on both sides. The complete-curve-like property therefore holds simultaneously, and the criterion identifies the entire point families when it does. The resulting isomorphism identifies each full procyclic subgroup with its counterpart, so it sends a generator to a generator, differing only by a \(\mathbb Z_\ell\)-unit. ◻

Proof. The classes \([t-a]\) are independent in \(k(t)^\times/\ell\), as evaluation at their distinct zeros shows. The kernel of \(k(t)^\times/\ell\to K^\times/\ell\) is finite dimensional: adjoining roots for any finite independent subset of that kernel gives an elementary abelian \(\ell\)-extension contained in the finite extension \(E/k(t)\). Thus their images span an infinite-dimensional subspace of \(K^\times/\ell\), and the same is true of the \(h_a\) in \(\widehat L/\ell=L^\times/\ell\).

The kernel of divisor evaluation \[L^\times/\ell\longrightarrow \bigoplus_{D\in X^{(1)}}\mathbb F_\ell\] is finite dimensional. Indeed, take a smooth dense open \(U\subset X\). A function whose divisor on \(U\) is divisible by \(\ell\) defines a \(\mu_\ell\)-torsor on \(U\): its divisor divided by \(\ell\) is Cartier. The kernel therefore embeds in \(H^1_{\mathrm{\acute et}}(U,\mu_\ell)\), which is finite dimensional by Deligne’s finiteness theorem (Deligne 1977, Théorèmes de finitude, Theorem 1.1). If \(\mathcal S\) were finite, both the image and the kernel of divisor evaluation on the pencil span would be finite dimensional, a contradiction.

For \(D\in\mathcal S\), let \(w_D\) be its divisorial valuation and let \(v\) be the quasi-prime divisor of \(K/k\) supplied by Proposition 10. Its inertia detects some \([t-a]\). Consequently \(vE/vk\) is nonzero inside the rank-one group \(vK/vk\). The valuation transcendence-degree inequality applied to \(E/k\) gives \(\operatorname{trdeg}(Ev/kv)=0\), so \(Ev=kv\) because \(kv\) is algebraically closed.

If \(\operatorname{ord}_D(\Theta[z])=0\), inertia evaluation implies \(v(z)\in vk\). Multiplying \(z\) first by a constant to give it value zero, and then by a constant with the reciprocal residue, makes it a principal unit. Thus \(D_v\) kills \([z]\), and \(D_{w_D}\) kills \(\Theta[z]\). By Corollary 16, the excluded \(D\) form a finite set. Discard also the finitely many prime divisors in the ordinary divisor of \(g\). For every remaining \(D\), \(g\) is a unit at \(D\), and its residue has zero evaluation modulo \(\ell^j\) against \(D_{w_D}/I_{w_D}=W_{l(D)}\). Finite-level Kummer duality gives (12). ◻

Proof. Place the divisors in a bounded family. Integral subvarieties of fixed dimension and bounded degree in \(\mathbb P^N_l\) have only finitely many Hilbert polynomials (Grothendieck 1961, sec. 2 and Lemma 2.4); see also (Tanaka 2023, Proposition 5.3).

The Hilbert points of \(\mathcal S\) therefore lie in a finite union of projective Hilbert schemes of \(X\), with their universal flat families (Grothendieck 1961, Theorem 3.2 and the Hilbert specialization after Proposition 3.8). Their closure has a positive-dimensional irreducible component on which they are dense. Give it its reduced structure and shrink to an integral variety \(T\) such that the universal family \[Z\subset T\times X\longrightarrow T\] has geometrically integral divisor fibers. This is possible by constructibility of geometric integrality (The Stacks Project Authors 2026, Tags 0579, 055B); the family is flat by construction.

Fix one incidence curve. Choose an integral locally closed curve \(T'\subset T\) now, before making any of the tests in \(\mathcal A\), and put \(Z'=Z\times_T T'\). Both \(Z\) and \(Z'\) are integral. Indeed, flatness over an integral base excludes vertical components and embeds affine coordinate rings in their generic-fiber localizations, while the generic fibers are geometrically integral. Both incidence maps to \(X\) are dominant. Otherwise a proper closed subset of \(X\) would contain infinitely many distinct prime-divisor fibers; each would have to be one of its finitely many components of dimension \(d-1\). Distinct parameter points give distinct divisors because they are distinct Hilbert points.

Set \[M=l(Z'),\qquad P_0=l(T').\] Dominance embeds \(l(X)\) into \(M\). Since \(\dim Z'=d\), this is a finite extension. Geometric integrality of the generic fiber of \(Z'\to T'\) means that \(M/P_0\) is regular.

Figure 1 records the two projections that determine these fields.

Extend each splitting to the fixed curve. It remains to prove all the power assertions without changing \(T'\). Fix one pair \((g,q)\). Let \(U\subset X\) be the unit locus of \(g\), and let \[Y=Z^{\mathrm{sm}/T}\cap\operatorname{pr}_X^{-1}(U).\] Then \(Y\to T\) is smooth. Its fibers over the generic points of \(T\) and of \(T'\) are nonempty and geometrically integral: geometric integrality supplies a dense smooth locus, and dominance of the two incidence maps supplies the unit condition. Over \(Y\) consider the degree-\(q\) finite étale cover \[C_g=\operatorname{Spec}_Y\mathcal O_Y[u]/(u^q-g).\] On every tested fiber satisfying the hypothesis and meeting \(Y\), a rational \(q\)-th root extends to a regular unit by normality of the smooth fiber. Thus the cover splits into \(q\) copies on those fibers. The number of geometric irreducible components is constant on a dense open of \(T\) (The Stacks Project Authors 2026, Tag 055A). That number must be \(q\), by the dense set of tested fibers. A degree-\(q\) finite étale cover of a normal integral scheme with \(q\) components has degree one on each component. It follows that the cover is split on the geometric generic fiber over \(T\).

This last dense open need not meet \(T'\). To pass to the fixed curve, descend the generic splitting to a finite extension of \(l(T)\) and normalize \(T\) in it, obtaining a finite surjective morphism \(\widetilde T\to T\). The scheme \(\widetilde Y=Y\times_T\widetilde T\) is normal because it is smooth over a normal base (The Stacks Project Authors 2026, Tag 034F). It is integral because its generic fiber is geometrically integral and it is flat over the integral base. The pulled-back cover is generically split and finite étale over this normal integral scheme, hence split everywhere: each component is finite birational over the normal target and is an isomorphism (The Stacks Project Authors 2026, Lemma 58.11.2, Tag 0BQL). Surjectivity of \(\widetilde T\to T\) provides a lift of a geometric point above the generic point of \(T'\). Restricting the split cover to that point proves that \(g\) has a \(q\)-th root in \(M\overline{P_0}\). The auxiliary finite extension may depend on \((g,q)\); the already chosen fields \(M\) and \(P_0\) do not. ◻

Proof. A \(q\)-th root \(r\) of \(g\) is separable over \(M\), so it already belongs to \(MP^{\mathrm{sep}}\): the remaining constant extension is purely inseparable. Choose a finite Galois extension \(P'/P\) with \(r\in MP'\). Regularity identifies \(\operatorname{Gal}(MP'/M)\) with \(\operatorname{Gal}(P'/P)\). The ratios \(\sigma(r)/r\in\mu_q\) form a character of this group. By Hilbert’s Theorem 90, there is \(b\in P'^\times\) with these same ratios. Then \(r/b\in M\) and \(b^q\in P\), giving \(g=b^q(r/b)^q\). Finally, if an element of \(P\) has a \(q\)-th root in \(M\), that root is algebraic over \(P\) and therefore belongs to \(P\). This proves uniqueness. ◻

Proof. The compositum \(PQ\) has transcendence degree two over \(l\); otherwise relative algebraic closedness in \(N\) would force \(P=Q\). Let \(C_P\) and \(C_Q\) be the smooth projective curves of these fields. The product is integral, and its function field identifies with \(PQ\): the rational map defined by the two subfields has two-dimensional, hence dense, image in \(C_P\times C_Q\).

For a point \(x\in C_P(l)\), take the valuation of the divisor \(\{x\}\times C_Q\). Extend it by a Gauss valuation over a transcendence basis for \(N/PQ\), then prolong it across the remaining finite extension. Its normalized order \(w\) on \(N\) is zero on \(Q^\times\) and restricts to \(e\operatorname{ord}_x\) on \(P^\times\), for some integer \(e>0\). This construction allows inseparable extensions and all ramification indices.

An element of \(\widehat P\cap\widehat Q\) has \(w\)-order zero by its \(Q\)-representation. Its \(x\)-order is therefore zero, since multiplication by \(e\) is injective on \(\mathbb Z_\ell\). This holds for every \(x\). By Lemma 5, the intersection embeds in the finite-rank Tate module of \(\operatorname{Pic}^0(C_P)\). ◻

Proof of Theorem 17. Let \(E\subset K\) be a curve subfield, and choose \(t\in E\setminus k\). Relative algebraic closedness makes \(E\) the relative algebraic closure of \(k(t)\) in \(K\). Apply Lemmas 18 and 19 to one representative of each \(\Theta[z]\bmod\ell^j\), for \(z\in E^\times\) and \(j\geq1\). We obtain fields \(L\subset M\) and \(P_0\subset M\) as in the incidence lemma.

Put \(H=\Theta(\widehat E)\). We claim that its image in \(\widehat M\) lies in \(\widehat P_0\). At level \(\ell^j\), every element of \(\widehat E\) is represented by an actual \(z\in E^\times\). Lemma 20 puts its image in \(P_0^\times/P_0^{\times\ell^j}\). These classes are unique because \(P_0\) is relatively algebraically closed in \(M\), and their uniqueness makes them compatible as \(j\) varies. Taking inverse limits proves the claim. Completion maps across \(M/L\) are injective by Lemma 3.

Choose a finite normal extension \(N/L\) containing \(M\), allowing an inseparable part, and let \(P_N\) be the relative algebraic closure of \(P_0\) in \(N\). It is a curve subfield of \(N/l\). The image of \(H\) lies in \(\widehat P_N\). Every \(\sigma\in\operatorname{Aut}_L(N)\) fixes \(H\), so \(H\) also lies in \(\widehat{\sigma(P_N)}\). It has infinite rank: \(\widehat E/\ell\) has infinite dimension, as the pencil argument in Lemma 18 shows. Indeed, lifts of any finite mod-\(\ell\) independent set are \(\mathbb Z_\ell\)-independent: a nonzero relation, divided by the smallest \(\ell\)-power in its coefficients, would reduce to a nonzero mod-\(\ell\) relation. Torsion freeness justifies this division. Lemma 21 therefore forces \(\sigma(P_N)=P_N\) for every \(\sigma\).

Set \(P_L=P_N\cap L\). This field has transcendence degree one over \(l\). Indeed, for \(G=\operatorname{Aut}_L(N)\), the field \(P_N^G\) has transcendence degree one and lies in \(N^G\). In characteristic zero \(N^G=L\). In characteristic \(p>0\), \(N^G/L\) is finite purely inseparable, so a common \(p^a\)-th power sends \(P_N^G\) into \(P_L\). Moreover \(P_L\) is relatively algebraically closed in \(L\): an element of \(L\) algebraic over \(P_L\) is algebraic over \(P_N\), hence lies in \(P_N\). Taking the relative algebraic closure of any nonconstant rational subfield shows that \(P_L/l\) is finitely generated.

The norm \(N_{N/L}\) sends \(P_N^\times\) into \(P_L^\times\). In fact, if \(s\) is the inseparable degree of \(N/L\), then \[N_{N/L}(x)=\left(\prod_{\sigma\in G}\sigma(x)\right)^s;\] the right side lies in \(P_N\) because \(P_N\) is \(G\)-stable, and the norm lies in \(L\). On completions, norm composed with inclusion is multiplication by \([N:L]\). We conclude that \([N:L]H\subset\widehat P_L\). The latter is saturated in \(\widehat L\) by Lemma 3, so \(H\subset\widehat P_L\). Here the prime-to-\(\ell\) factor of \([N:L]\) is a unit of \(\mathbb Z_\ell\), and saturation removes its remaining \(\ell\)-power factor.

Apply the same argument to \(\Theta^{-1}\). It gives a curve subfield \(E'\subset K\) with \[\widehat E\subset\Theta^{-1}(\widehat P_L) \subset\widehat E'.\] The infinite-rank intersection forces \(E'=E\) by Lemma 21. Both inclusions are therefore equalities. That lemma also proves uniqueness. Reversing \(K\) and \(L\) proves the asserted bijection. ◻

Proof. Suppose \(v\) is trivial on \(k\). Its residue field has transcendence degree \(d-1\geq1\). Choose \(t\) with transcendental residue, and let \(E\) be the relative algebraic closure of \(k(t)\) in \(K\). The valuation is trivial on \(k(t)\) and hence on its algebraic extension \(E\). Thus \(E\) embeds into \(Kv\). Its relative algebraic closure in \(Kv\) is finite over \(E\), so the induced completion map is injective: factor it through that finite extension and then a relatively algebraically closed inclusion. Residual evaluation by \(D_v/I_v=W_{Kv}\) proves (13).

Conversely, suppose \(v|k\) is nontrivial, and let \(E\) be any curve subfield. Choose \(x\in E\setminus k\), replacing it by \(x^{-1}\) if necessary so that \(v(x)\geq0\), and choose \(c\in k\) with \(v(c)>0\). Then \(1+cx\) is a nonconstant principal unit. Its class in \(\widehat E\) is nonzero and is killed by \(D_v\), so (13) fails. The criterion is invariant under \(\Theta\) by Theorem 17 and the local correspondence. ◻

Proof. A divisorial valuation restricted nontrivially to \(E\) is an integer multiple of a point valuation on its smooth projective curve: its value group is a nonzero subgroup of \(\mathbb Z\), and it is trivial on \(k\). Conversely, extend any point valuation of \(E\) by a Gauss valuation over a transcendence basis for \(K/E\), then prolong it over the remaining finite extension. The resulting valuation is divisorial on \(K/k\). Its restriction is \(e\operatorname{ord}_x\) for some \(e>0\). Saturation of the nonzero image recovers \(\mathbb Z_\ell\operatorname{ord}_x\), including when \(\ell\) divides \(e\).

The dual of \(\Theta|_{\widehat E}\) defines \(\varphi_E\). Evaluation against elements of \(\widehat E\) proves the commutativity of (14). Proposition 22 and the preceding description of point lines show that \(\varphi_E\) matches them. Their generators may differ by units, which do not change the common kernel of point-order evaluations modulo \(\ell\). Its dimension is twice the genus by Lemma 5. Hence genus is preserved. A smooth projective curve over an algebraically closed field has genus zero exactly when its function field is rational.

If \(k(t)\) is relatively algebraically closed in \(K\), the extension \(K/k(t)\) is regular. Only separability needs explanation. In characteristic \(p>0\), relative algebraic closedness implies \(t\notin K^p\). Since \(k\) is perfect, \(t\) can be included in a separating transcendence basis for \(K/k\), proving that \(K/k(t)\) is separably generated. Characteristic zero is immediate. Thus these are exactly the rational subfields defined by general elements. Finally, the residue field at a point of a rational curve is the algebraically closed constant field, whose character group is zero. Point decomposition therefore equals point inertia, and the diagram also carries the required point decomposition data. ◻