The Bogomolov-Pop reconstruction theorem

Proof. Choose a normal integral projective model of F/\kappa. A rational function with zero Weil divisor on this model is a global unit and hence a constant. Thus the divisor map embeds F^\times/\kappa^\times in the free abelian group of prime divisors; a subgroup of a free abelian group is free.

For a basis indexed by a set J, the character group is \Zell^J. Modulo \ell^e, a continuous character on this compact product depends on finitely many coordinates. Compatible such characters are therefore families (c_j)_{j\in J} in \Zell with only finitely many c_j\notin\ell^e\Zell for each e. This is exactly the completion of \bigoplus_{j\in J}\mathbb Z. Reduction modulo \ell^e and coordinate evaluations prove the assertions. Divisibility of \kappa^\times removes the constants at every level.□

Proof. In the first case, a root in F of an element of E is algebraic over E, so injectivity holds at every finite level. Alternatively, on a normal integral projective E-model of F, the divisor map embeds F^\times/E^\times into a free abelian group: its kernel consists of global units, which are algebraic over E and therefore lie in E. Thus F^\times/E^\times is free abelian, and

splits as a sequence of abelian groups. Completion preserves this splitting. The two completed free factors are torsion free, proving closedness and saturation. Every character on E^\times extends along the splitting, proving the asserted surjection; continuity follows from the pointwise topologies.

For a finite extension, the field norm is defined also in the inseparable case and sends an element of E to its [F:E]-th power. It induces the stated composite on completions. Torsion freeness from Lemma 2.1 proves injectivity.□

Proof. Use the given maps to identify the bases of the two central extensions with one abelian pro-\ell group A and their kernels with one group B. Let E_1,E_2 denote these extensions. In the fiber product P=E_1\times_A E_2, quotient by the diagonal central subgroup \{(b,b):b\in B\}. The resulting group Q is abelian: the commutator of two pairs has identical components by bracket compatibility. It fits into an exact sequence of abelian pro-\ell groups

By Lemma 2.1, A is a product of copies of \Zell. Such products are projective in the category of abelian pro-\ell groups. Indeed Pontryagin duality changes A into a direct sum of copies of \mathbb Q_\ell/\Zell, a divisible, hence injective, discrete torsion group. Consequently Q\to A has a continuous section.

The inverse image of this section in P maps isomorphically to each E_i. For fixed first component, the second component can be adjusted by a unique element of B to land over the section; its uniqueness also proves that each projection has trivial kernel. These are continuous bijections between compact Hausdorff groups, hence isomorphisms. The resulting map E_1\to E_2 induces the prescribed base and kernel maps. The proof uses no division by 2 and leaves no additional power datum to specify.□

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)} [TSPA26, 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 \Zell.

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 \Qell/\Zell, 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

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

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 [MS83] identifies the kernel of the tensor cup map

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

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

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

Its final term is free, so applying \operatorname{Hom}(-,\Zell) 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

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

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

are equalities. If S\ne\Sigma, all inequalities

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 [ADF20, 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 3.3, 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

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,

Let v=v_H be its coarsest associated valuation. It coarsens both u_i, and

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 3.2. For the converse apply the third part of the local character theorem to the valuative line H.□

Proof. Lemma 3.1 makes the alternating relation intrinsic to the bracket. Lemma 3.3 then recovers d, and Lemma 3.4 recovers the rationalized rank-one pairs. Intersecting with the integral modules recovers the exact pairs. Lemma 3.2 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 3.5 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 3.6 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 \Zell-unit.□

Proof. We recall the incidence argument for irreducibility; see also [TSPA26, Lemma 37.32.3, Tag 0G4F]. Write B=(\mathbb P^N)^* and

The projection to V is a projective bundle, so I is integral. In I\times_BI, the open set lying over pairs (x,y) with f(x)\ne f(y) is a projective bundle with fiber \mathbb P^{N-2} over an irreducible open subset of V\times V. It is therefore irreducible of dimension 2a+N-2. The locus of pairs with f(x)=f(y) has dimension at most a, since f is quasi-finite; the incidence above it has dimension at most a+N-1. This is strictly smaller than 2a+N-2.

Every component of the generic hyperplane section has dimension a-1: a hyperplane cuts a nonzero nonunit in an integral variety of finite type over a field. Consequently every component of its square has dimension 2a-2. The equal-image locus cannot supply a component of this generic square. The generic square is thus irreducible. Equivalently, the function field of B is separably algebraically closed in that of I, so the generic hyperplane section is geometrically irreducible. Geometric irreducibility spreads to a nonempty open of B [TSPA26, Lemma 37.27.5, Tag 0559]. Finally, a general hyperplane section of the smooth embedded variety U is smooth by Bertini, and its étale inverse image is smooth.□

Proof of Lemma 4.1. Spread the finite data. Let D_1,\ldots,D_s be the prime divisors occurring in the divisors of the g_j. There is a closed subset B_X\subset X of codimension at least two such that X\setminus B_X is smooth, the D_i\setminus B_X are smooth pairwise disjoint Cartier divisors, and, near each of them, every g_j is a power of a local equation times a unit. Off their union the g_j are units. These assertions follow by deleting the singular loci, the pairwise divisor intersections and the proper closed exceptional subsets on which the required local identities fail.

Choose a smooth dense open U\subset X on which all g_j are units. The simultaneous root cover

is finite étale of degree q^r. Let H be the subgroup generated by the [g_j] in L^\times/L^{\times q}. Kummer theory shows that each connected component of V has degree |H| over U. Since U is smooth, these components are integral.

All these data descend to a finitely generated domain R\subset l in which \ell is invertible. Enlarge R and then localize it so that the following properties hold in every geometric fiber:

• •

  the embedded model X_R is projective and flat, with normal integral fibers and the Hilbert polynomial of X;

• •

  the D_{i,R} are distinct geometrically integral divisors with their original degrees, and the local divisor identities and smoothness conditions above hold outside a subset of fiberwise codimension at least two;

• •

  the root cover and its disjoint decomposition into components extend, and every component has geometrically integral fibers and finite locally free degree |H| over U_R.

These are finite spreading conditions. Flatness makes the projective Hilbert polynomials constant. Geometric integrality and normality spread from the generic fiber after shrinking; the latter uses the proper flat finite-presentation form of [Gro66, Theorem 12.2.4(iv),(viii)]. The equations g_j=u\pi^n, the inverses specifying units, the component idempotents, and the finite locally free ranks are all finite algebraic data. Spreading them preserves the exact orders and degrees, rather than just the supports. The dimension bound on the omitted closed set also persists after shrinking.

For later use include a separating transcendence basis x_1,\ldots,x_d of L/l, a primitive element y over l(x_1,\ldots,x_d), its monic minimal polynomial, and common birational opens with X. After another localization, the same presentation and birational identification hold in the geometric fibers, with the reduced polynomial irreducible of its original degree over the rational function field. Preserve as well the nonzero reductions of the finitely many numerators and denominators used for the g_j. This follows either by spreading those common opens and geometric integrality, or by spreading the absolutely irreducible defining polynomial after clearing denominators.

Realize a good fiber as a residue field. Choose a closed point s\in\operatorname{Spec}R. Its residue field is finite and has characteristic different from \ell. A valuation ring of l dominating R_s exists [TSPA26, Lemma 10.50.2, Tag 00IA]. Its residue field is algebraically closed. If that residue field is not algebraic over a finite field, give a transcendence basis independent values in an ordered free abelian group, prolong the resulting valuation to the residue field, and compose. This yields a valuation s_0 of l with the same center s and residue field \lambda algebraic over a finite field. Residues of algebraically closed fields are algebraically closed, so \lambda is an algebraic closure of a finite field. The composition does not change the center because every nonzero element of a finite field has value zero. This construction uses a transcendence basis of arbitrary cardinality; there is no restriction on the size of l.

Extend s_0 to l(x_1,\ldots,x_d) by the Gauss valuation, whose residue field is \lambda(x_1,\ldots,x_d) and whose value group is s_0l. Prolong it to L. The element y is integral and its residue satisfies the reduced irreducible polynomial, of degree [L:l(x_1,\ldots,x_d)]. The fundamental valuation inequality therefore forces the residue degree to be that degree and the value-group index to be one [TSPA26, Section 15.125, Tag 0ASF]. We obtain a valuation s_X with

and with the g_j reducing to their specified rational functions on X_s. This also explains why no extra ramification or defect is hidden in the passage to the good fiber.

Cut down to a curve. Over \lambda, choose a general flag of d-1 hyperplanes in the fixed projective embedding. Make the final section C avoid the omitted codimension-two set and meet the D_{i,s} transversely. The intersections with different D_{i,s} are disjoint. The final section is smooth and integral. Apply Lemma 4.2 successively to the finite list of covering components as well as to the reduced underlying base sections; each restricted covering component remains integral over U_s\cap C. At every step the varieties to which irreducibility is applied have dimension at least two, including the last surface-to-curve step. Smoothness of the restricted cover follows from its étaleness over the smooth curve open.

For completeness, intermediate scheme sections need only have irreducible support and be smooth at the generic points used by the flag. All those points lie in the original smooth locus. Choose the hyperplanes also to avoid the finitely many associated points at each stage, so their equations are non-zero-divisors. The hyperplane exact sequences give

As C is smooth integral, its genus is consequently fixed by the embedded model X; denote it by G_X. All conditions imposed on the flag are finitely many nonempty open conditions. They can be met over the infinite algebraically closed field \lambda, even when it is countable.

Compose s_X with the successive divisorial valuations of this flag. For the first composition, the relative value group is the innermost convex copy of \mathbb Z, and its residue transcendence degree drops by one. Every proper coarsening kills this copy of \mathbb Z and therefore has all its values supplied by constants. The first composition is thus a quasi-prime divisor. The remaining flag valuations are prime divisors over the residue constants, giving a quasi-prime (d-1)-divisor w. All g_j are units at every generic point in the flag, so w(g_j)=0, and Lw=\lambda(C).

Check relations and orders. An old relation \prod_jg_j^{n_j}=b^q specializes: since w(g_j)=0, torsion-freeness of the value group gives w(b)=0. Thus the old relation kernel in (\mathbb Z/q)^r is contained in the new one. On the other hand, every component of the root cover restricted to U_s\cap C is integral and has degree |H|. Kummer theory gives the same order |H| for the subgroup generated by the residue classes. The two finite relation kernels therefore have the same order, so they are equal.

Finally, C meets D_{i,s} in exactly \deg D_i distinct transverse points. At each such point the order of \overline g_j equals the unchanged integer \operatorname{ord}_{D_i}(g_j). There are no other zeros or poles. This proves (6), in fact for every divisor m of q, and completes the construction.□

Proof. Choose five distinct anchors a_1,\ldots,a_5\in k, including 0. In characteristic zero include 1 and \ell as well. Put h_a=\Theta[t-a] and

The N_i are finite before any exact-support assertion is known. Moreover c_\ell\ge1/2, so the constant

is finite and depends only on X, \Theta, t and the fixed anchors. We prove the theorem with B_t=2B.

Fix one additional a\in k, and list the distinct elements among the five anchors and a as a_j. Write u_j=t-a_j. For all sufficiently large e, the classes [u_j] modulo q=\ell^e are nonzero and pairwise distinct and generate a noncyclic subgroup. Indeed the functions t-a_j are independent modulo constants: a multiplicative relation would give a constant rational function of the transcendental element t and hence have all exponents zero. Their images in the free abelian group K^\times/k^\times are therefore independent. Integral evaluations detect nonzero elements and nonzero two-by-two determinants, which remain nonzero modulo a sufficiently large power of \ell. Equivalently, one can prolong two distinct point valuations of k(t) to K to obtain a diagonal evaluation matrix with nonzero integer entries. The necessary lower bound on e may depend on a.

Choose representatives g_j\in L^\times of h_{a_j} modulo q, and apply Lemma 4.1. It gives w, a residual curve C, and the same noncyclic subgroup with distinct nonzero generators in (Lw)^\times/(Lw)^{\times q}. Evaluation on D_w/I_w=W_{Lw} detects these classes. By Proposition 3.5, there is a corresponding quasi-prime (d-1)-divisor v of K/k, and the evaluations of the u_j on D_v modulo q are likewise nonzero, distinct and noncyclic.

Normalize the residual pencil. We first show that

Otherwise v(k(t))/vk is a nonzero subgroup of the finitely generated free group vK/vk. The valuation transcendence-degree inequality for k(t)/k makes it cyclic and gives k(t)v=kv. Every function of value in vk then becomes a principal unit after multiplication by constants. Since D_v kills constants and principal units, evaluation on D_v factors through this cyclic relative value group. Its image modulo q is cyclic, a contradiction.

All the v(u_j) have a common value \gamma\in vk. For if v(u_j)<v(u_h), then u_j/(a_h-a_j)=1+u_h/(a_h-a_j) is a principal unit, forcing the evaluation of u_j to vanish. Also v(a_j-a_h)=\gamma whenever j\ne h: a strictly larger value would make u_j/u_h a principal unit and force two evaluations to coincide. Choose b\in k^\times with v(b)=\gamma, and fix one listed anchor a_0. Set

These expressions are defined, the \beta_j are distinct elements of kv, and

The element \tau_0 is nonconstant: a constant residue would make (t-a_0)/b a constant times a principal unit and give zero evaluation.

The residue characteristic is different from \ell. This is automatic if k has positive characteristic. In characteristic zero, the anchor differences 1-0 and \ell-0 have the same value, so v(\ell)=v(1)=0. Thus \ell is nonzero in kv in that case as well.

Compare the two residual curves. The constants lw are algebraic over a finite field. Proposition 3.7 therefore gives the same assertion for kv and identifies the full families of point inertia lines in W_{Lw} and W_{Kv}. Let C'/kv be the smooth projective curve with function field Kv. By Lemma 2.4, the mod-\ell kernels of the point order maps have dimensions 2g(C) and 2g(C'). The residual isomorphism identifies these kernels, and hence

Generators of matched point inertia lines differ by units of \Zell. Consequently the number of nonzero point orders modulo \ell or q is preserved by the comparison. Because constants have zero Kummer class, evaluation identifies the classes of the functions in (9) with those of \overline g_j modulo q.

Use the five anchors to bound the degree. Write p=\operatorname{char}(kv)>0, using the finite-constant recognition above. Remove inseparability by writing \tau_0=\tau^{p^r} with r\ge0 maximal. Such an r is finite, as a nonzero integer point order of \tau_0 cannot be divisible by arbitrarily high powers of p. The function \tau gives a separable morphism \tau:C'\to\mathbb P^1 of some degree n\ge1. Let \beta'_j be the p^r-th root of \beta_j in kv. These constants are distinct, and

Since p\ne\ell, this power does not affect whether any point order is zero modulo \ell or q.

Over the value \beta'_i corresponding to anchor i, at most N_i points have ramification index not divisible by \ell. Indeed such a point contributes a nonzero order modulo \ell to \tau_0-\beta_i; transport to C and (6) give the bound N_i. Every other point of that fiber has ramification index at least \ell, so there are at most n/\ell of them. If r_i denotes the total number of points in the fiber, then

For a separable morphism of smooth curves, the different exponent at P is at least e_P-1, including in the presence of wild ramification. The five fibers are disjoint. Riemann–Hurwitz therefore yields

Together with (10) this gives c_\ell n\le 2G_X-2+\sum_iN_i, and hence n\le B. This argument uses the different inequality, not the tame Riemann–Hurwitz formula; see [TSPA26, Section 53.12, Tag 0C1B].

Pass from finite tests to exact supports. For every tested j, all zeros and poles of \tau-\beta'_j lie in two fibers, over \beta'_j and infinity. There are at most 2n such points. The number of its nonzero point orders modulo q is therefore at most 2B. Transport this bound to C and apply (6) to obtain

The right side is independent of a, e, the finite descent data and the residual test. For each fixed a, the inequality holds for every sufficiently large e.

Finally S_e(h_a) increases to \operatorname{Supp}_X(h_a). If the exact support had total degree greater than 2B, a finite subset would already have degree greater than 2B; that subset would lie in S_e(h_a) for all sufficiently large e, contradicting (11). Every prime divisor has positive integer degree, so the exact support is finite. This proves the asserted uniform bound.□

Proof. If z\notin k, apply Theorem 5.1 to t=z and a=0. If z\in k^\times, its Kummer class is zero because k^\times is divisible.□

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

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 [Del77, 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 3.5. 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 5.2, 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 [Gro61, Section 2 and Lemma 2.4]; see also [Tan23, 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 [Gro61, 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

has geometrically integral divisor fibers. This is possible by constructibility of geometric integrality [TSPA26, 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

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

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

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 [TSPA26, 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 [TSPA26, 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 [TSPA26, 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 \Zell. This holds for every x. By Lemma 2.4, the intersection embeds in the finite-rank Tate module of \operatorname{Pic}^0(C_P).□

Proof of Theorem 6.1. 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 6.2 and 6.3 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 6.4 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 2.2.

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 6.2 shows. Indeed, lifts of any finite mod-\ell independent set are \Zell-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 6.5 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

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 2.2, so H\subset\widehat P_L. Here the prime-to-\ell factor of [N:L] is a unit of \Zell, and saturation removes its remaining \ell-power factor.

Apply the same argument to \Theta^{-1}. It gives a curve subfield E'\subset K with

The infinite-rank intersection forces E'=E by Lemma 6.5. 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 6.1 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 \Zell\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 6.6 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 2.4. 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.□

Reference and hypotheses. Part (1) is [Pop18, Proposition 2.4]; the surrounding section assumes transcendence degree greater than one. Part (2) is the Isom statement of [Pop12a, Theorem 2.1(2)], restating [Pop12b, Main Theorem]. The stated theorem requires Bertini-type families of rational quotients; all geometric rational quotients form such a family. More explicitly, for algebraically independent x,t with x separating, the general functions

in [Pop12b, Fact/Definition 43] define relatively algebraically closed rational subfields for the prescribed cofinite choices of constants. Every one of these subfields is among the quotients in (2). Thus its hypotheses imply precisely the required Bertini compatibility. No restriction to finite-field closures or to dimension greater than two occurs in this Isom statement.□

Proof of Theorem 1.1. A field isomorphism \alpha:K^i\to L^i induces a contravariant isomorphism of absolute Galois groups, well defined up to an inner automorphism. Purely inseparable extension does not change these groups. Abelianization removes the inner ambiguity, and the induced class-two map preserves the commutator. Thus it gives a member of \Isom^c(\Pi_L^a,\Pi_K^a). In positive characteristic, absolute Frobenius on algebraic closures commutes with field automorphisms; composition with any integer Frobenius power therefore gives the same allowed orbit. This proves well-definedness of \Phi_{K,L}.

Conversely, let \varphi:\Pi_L^a\to\Pi_K^a be bracket-compatible. Lemma 2.3 lifts it to a class-two isomorphism. After choosing Tate identifications, let \Theta be its dual (1). Proposition 3.5 recovers the common relative transcendence degree and the quasi-divisorial pairs. The finite residual tests and the uniform support bound prove the completed curve-subfield correspondence in Theorem 6.1. The recognition statements in Section 6 then show that \varphi preserves true divisorial inertia and every relatively algebraically closed rational quotient, including its point lines.

Proposition 6.7 gives the actual commuting quotient diagrams in (14). Their vertical maps are surjective, and their target isomorphisms \varphi_E match all point inertia lines. For rational targets these are the entire point decomposition data. Thus the same fixed \varphi is compatible with every geometric rational quotient; no independent rescaling of the targets is needed.

Part (1) of Proposition 7.1 gives the total decomposition-graph isomorphism. The quotient squares also commute on these graphs. Indeed, restriction along an actual regular inclusion E=k(t)\subset K is a geometric rational quotient morphism [Pop12a, Section 2, Embeddings and Restrictions]. At a flag vertex its map is obtained by restricting to the decomposition subgroup and passing to the inertia quotient [Pop12b, Definition/Remark 24 and Definition 25]. If the flag restricts trivially to E, its inertia is killed, and the top square descends to the residual square. Otherwise its nonzero inertia image lies in a unique point line of E, where decomposition equals inertia and the residue group is trivial. The same description applies to P\subset L. Local images can have finite index: saturation identifies the point line, but the actual subgroup images and maps are retained. Applying these restriction and quotient operations successively along each flag therefore proves compatibility for the same fixed \varphi and \varphi_E.

Part (2), applied to the recovered rational quotients, gives an isomorphism \alpha:K^i\to L^i with \alpha(k)=l and [\varphi]=[\alpha^*]. Choices of Tate identifications are absorbed by the one allowed global unit. This proves surjectivity. It also proves that different characteristics cannot give a bracket-compatible isomorphism.

Finally, two field isomorphisms with the same Galois-side orbit give the same compatible graph and quotient data after one global unit adjustment. The uniqueness assertion in Proposition 7.1 identifies them modulo Frobenius in positive characteristic, and identifies them outright in characteristic zero. On perfect closures the Frobenius powers are exactly the integer powers appearing in the definition of \Isom_F^i(K,L). This proves injectivity and completes the theorem.□