Reconstruction of Function Fields from Mod-ℓ Milnor K-Theory

Proof. The value group of an algebraically closed field is divisible, and its residue field is algebraically closed. If n\gamma\in v\kappa for \gamma\in vF, choose \delta\in v\kappa with n\delta=n\gamma. Ordered abelian groups are torsion free, so \gamma=\delta. This proves torsion-freeness of the relative group.

Choose elements x_1,\ldots,x_a\in F^\times whose values are rationally independent modulo v\kappa, and units y_1,\ldots,y_b whose residues are algebraically independent over \kappa v. These elements are algebraically independent over \kappa. Indeed, in a putative polynomial relation, distinct monomials in the x_i have distinct values modulo v\kappa. Within a coefficient polynomial in the y_j, its terms of least value cannot cancel, by algebraic independence of their residues after rescaling by a constant. The nonzero summands obtained by grouping according to monomials in the x_i have distinct values, so their sum cannot vanish. This proves (2.2) by taking maximal such families.

Suppose equality holds and choose these families with a+b=s. Then F is finite over F'=\kappa(x_1,\ldots,x_a,y_1,\ldots,y_b). The same least-value calculation gives

The fundamental inequality for the finite extension F/F' shows that [vF:vF'] and [Fv:F'v] are finite. Thus the residue extension is finitely generated, while the relative value group is finitely generated and torsion free, hence free.

For the assertion about composition, write v=w\circ u, meaning that u is a coarsening and w is the induced valuation on Fu. The value groups fit into the exact sequence

Apply (2.2) first to u, and then to w on Fu/\kappa u. The sum of the two inequalities is precisely the inequality for v. Consequently equality for v forces equality in both, and equality in both forces equality for v. This use of the inequality on Fu/\kappa u is legitimate even before finite generation is known: the algebraic-independence argument proving it applies to arbitrary field extensions of finite transcendence degree.□

Proof. There is an exact sequence of abelian groups

For the kernel assertion, rescale by a constant an element whose value lies in v\kappa, and then take its residue. The last group is free, so the sequence splits as a sequence of abstract groups. Dualizing into \Lambda is consequently exact. Characters kill the algebraically closed constants on both sides, giving the asserted surjection \rho_v with kernel I_v. Its induced bijection is a homeomorphism because its source is compact and its target Hausdorff.

For alternation, if v(x)>0, then 1-x\in U_v^1, and the identity (2.1) holds for any pair in D_v. The case v(1-x)>0 is the same. If v(x)<0, then (1-x)/(-x)\in U_v^1, so every character in D_v takes the same value on x and 1-x. The remaining case is v(x)=v(1-x)=0, when the identity is exactly its residue-field counterpart. Conversely, lift any residue element different from zero and one to test the residue identity.

Finally U_v^1 is contained in both U_{w\circ v} and U_{w\circ v}^1. Under reduction of v-units, these two groups map respectively onto U_w and U_w^1. Their annihilators therefore give (2.5).□

Proof. The induced valuations w_1,w_2 on the residue field Fu are independent. For a\in(Fu)^\times, the approximation theorem gives b\in(Fu)^\times with

Thus b\in U_{w_1}^1 and a/b\in U_{w_2}^1, proving U_{w_1}^1U_{w_2}^1=(Fu)^\times. Reduction maps U_{v_i}^1 onto U_{w_i}^1, and both groups U_{v_i}^1 contain U_u^1. Lifting the product identity gives the first equality; taking annihilators gives the second.□

Proof. For any torsion-free abelian group G of finite rational rank,

Indeed, representatives of linearly independent classes modulo \ell are rationally independent: an integral relation can be divided by the greatest common divisor of its coefficients, using torsion-freeness, and the resulting relation has a coefficient nonzero modulo \ell. Since characters of vF kill its divisible subgroup v\kappa, this proves

Now let S be alternating. By Theorem 2.1, its valuative part S_0 is a subspace of codimension at most one and, for its associated valuation v, S_0\subset I_v and S\subset D_v. Moreover S\cap I_v=S_0, since every element of I_v is valuative. If S=S_0, then

If S\ne S_0, an element of S\setminus S_0 has nonzero image under \rho_v, whose kernel is always I_v. Thus A_{Fv}\ne0. Since \kappa v is algebraically closed, this implies \trdeg(Fv/\kappa v)\geq1, and hence

If \dim S=s, equality throughout the applicable chain gives S_0=I_v and equality in (2.2).

To attain the bound, choose a transcendence basis of F/\kappa, take a full coordinate flag on its rational function field, and prolong the associated valuation to F. It is trivial on \kappa, and its value group is a finite-index extension of \Z^s, hence is abstractly \Z^s. Its inertia has dimension s and is alternating by Lemma 2.4. For s=0, the field is \kappa, its character space is zero, and the assertion has the same interpretation.□

Proof. Suppose first that v is quasi-divisorial. On Fv/\kappa v choose two independent full discrete flag valuations w_1,w_2, trivial on \kappa v. To obtain them, start with coordinate flags on a rational subfield of transcendence degree s-1, with distinct first prime divisors, and prolong to the finite extension Fv. Independence survives prolongation. Indeed, a valuation on a finite field extension that restricts trivially to the smaller field is trivial, since its value group is then finite and hence zero. Thus a common nontrivial coarsening of w_1,w_2 would restrict to a common nontrivial coarsening of the original independent flag valuations, which is impossible.

The inertia spaces I_{w_1},I_{w_2} have zero intersection by approximation. Set v_i=w_i\circ v and S_i=I_{v_i}. Each v_i has relative value group \Z^s, so its inertia is alternating of dimension s. By (2.5), the intersection S_1\cap S_2 is I_v.

Conversely, suppose (2.6) holds. Let S_{i,0} be the valuative part of S_i, with associated valuation v_i. Proposition 2.6 gives

and v_i has equality in (2.2). We first show that H is valuative. If, for example, v_1\leq v_2, then

so H=I_{v_1}. The other comparable case is symmetric. If the valuations are incomparable, Lemma 2.5 puts H inside the inertia of their finest common coarsening. In either case every element of H is valuative. Because it lies in each S_i, it then lies in each S_{i,0}, giving

Let u=v_H. By Theorem 2.1, it coarsens each v_i, and therefore

Equality in (2.2) passes to u. Its relative value group is free, so \dim I_u=1 implies uF/u\kappa\cong\Z and \trdeg(Fu/\kappa u)=s-1. If a coarsening u'\leq u still has relative rank one, its one-dimensional inertia is contained in I_u, hence equals H. The defining coarseness of v_H=u then gives u\leq u', so u=u'. Thus u is quasi-divisorial.

For uniqueness, if v is any quasi-divisor with I_v=H, its associated coarsening v_H has the same inertia. Equality in (2.2) passes to that coarsening, so it too has relative rank one and residue transcendence degree s-1. Minimality of v forces v=v_H.

Finally, an element of D_v is alternating with h by Lemma 2.4, since h has zero residue image. Conversely, an element alternating with h belongs to D_{v_H}=D_v by Theorem 2.1.□

Proof. First suppose r=1. An \ell-divisible convex subgroup C of vF maps to zero in vF/v\kappa\cong\Z, and hence lies in v\kappa. Coarsening by C therefore leaves the relative value group unchanged. Lemma 2.2 shows that the residue transcendence degree remains s-1, so minimality of v forces C=0.

For a composite flag, the value group of its final quasi-prime step is a nonzero convex subgroup B\subset vF with the property just proved. If C\subset vF were a nonzero \ell-divisible convex subgroup, then B\cap C\ne0, since convex subgroups are linearly ordered by inclusion. This intersection is \ell-divisible: division by \ell in C stays in B by convexity. It is also convex in B, a contradiction.

Put \Gamma=vF. Since \Gamma/v\kappa\cong\Z^r and v\kappa is divisible, the common kernel in \Gamma of all characters in I_v is

Every convex subgroup contained in \ell\Gamma is itself \ell-divisible: an \ell-th part exists in \Gamma, and convexity places it in the subgroup. The first assertion therefore shows that the largest such convex subgroup is zero. The coarsening description in Theorem 2.1 now gives v=v_{I_v}.□

Proof. The maximum alternating dimension is preserved by \Phi, giving equality of transcendence degrees by Proposition 2.6. Proposition 2.7 then identifies corresponding quasi-prime inertia and decomposition spaces. Quotienting by inertia and using Lemma 2.4 gives the indicated isomorphism on residue characters, with its alternating relation.

As long as the residual transcendence degree is at least two, apply Proposition 2.7 again there. The inverse-image identities (2.5) identify the resulting pairs with those of compositions on the original fields. Lemma 2.8 shows that each such inertia determines its composite valuation, so this matching is independent of any choice of a presentation as a flag. At each step Lemma 2.2 supplies finitely generated residue fields over algebraically closed residue constants. The local results apply even if one of these fields has characteristic \ell. Iteration ends after d-1 steps, with function fields of curves.□

Proof. Reduction of character values gives continuous maps

They carry \widehat I_a into I_a, carry \widehat D_a into D_a, and commute with restriction to units. In particular,

The first inclusion follows from continuity and the definition as a closure of a union.

Fix a c.r. quasi-prime divisor u of F_0/\kappa_0. Characters with either coefficient ring annihilate the divisible group u\kappa_0. Since uF_0/u\kappa_0\cong\Z, reduction \widehat I_u\to I_u is surjective. Given \chi\in I_u, choose a lift \widehat\chi\in\widehat I_u. Theorem 3.1 supplies \widehat\eta\in\widehat S_F^{\mathrm{in}}\cap\widehat D_v with \widehat\rho_v(\widehat\eta)=\widehat\chi. Then r_F(\widehat\eta) belongs to S_F^{\mathrm{in}}\cap D_v and has residual image \chi. Hence \chi\in J_v.□

Proof. Let \mathcal V(F) be the space of valuation rings of F, with its compact patch topology. In A_F\times\mathcal V(F), the condition that a character \chi annihilates the unit group of a valuation ring O is closed. Indeed, for each x\in F^\times, failure of this condition is witnessed by the open conditions

The projection of the closed incidence set to A_F is compact, hence closed. This projection is precisely the set of valuative characters, which contains every I_w and therefore contains S_F^{\mathrm{in}}.

Now let \chi\in I_u\cap D_v for some valuation u. If u\leq v, then \chi\in I_v and its residual image is zero. If v\leq u, then restriction to U_v shows that \rho_v(\chi) annihilates the units of the induced valuation u/v on F_0. In the remaining case let a be the finest common coarsening of u and v. Approximation for the independent induced valuations on Fa gives

For example, a prescribed nonzero residue can be written as an induced u-unit times an induced principal v-unit by choosing an element close to 1 at the first valuation and close to that residue at the second. Lifting gives the displayed identity; the remaining factor in U_a^1 is already a unit for both refinements. Thus \chi annihilates U_a, and \chi\in I_a\subset I_v. Its residual image is again zero.□

Proof. Approximation at distinct point valuations proves that each \delta_x is nonzero and that their lines are distinct. A function on C has only finitely many zeros and poles. Hence every fixed element of B^\times is annihilated by all but finitely many \delta_x, which proves the asserted convergence and defines (3.2), with continuous evaluation at each function.

A family (a_x)_x defines a homomorphism from the divisor group of C to \Lambda, sending the point divisor x to a_x. It lies in the kernel of (3.2) precisely when this homomorphism vanishes on principal divisors, or equivalently factors through \Pic(C). The group \Pic^0(C)(b) is \ell-divisible: multiplication by \ell on the Jacobian is a surjective isogeny, and b is algebraically closed. This remains true in characteristic \ell. Every homomorphism \Pic(C)\to\Lambda therefore factors through the degree map \Pic(C)\to\Z. Since all point divisors have degree one, its coefficient family is constant. Conversely, constant families annihilate principal divisors.

Finally, if all point orders of f\in B^\times are divisible by \ell, write \operatorname{div}(f)=\ell D and associate to [f] the class of D in \Pic(C)[\ell]. This gives an isomorphism from the common kernel to \Pic(C)[\ell]: surjectivity follows from the definition of torsion in \Pic(C); injectivity follows because b^\times is \ell-divisible. When \charac(b)\ne\ell, the \ell-torsion of the Jacobian is isomorphic to (\Z/\ell)^{2g(C)}.□

Proof. Suppose first that \kappa_0 is algebraic over a finite field. Every valuation on \kappa_0 is trivial. A nontrivial valuation of the one-variable field F_0 trivial on \kappa_0 is a point valuation on its smooth projective curve: properness gives a closed center, and the discrete valuation ring at that smooth point is dominated by the valuation ring. Writing each function as a power of a uniformizer times a local unit identifies the two valuations. Thus Lemma 3.3 shows that every line in \mathcal L_v is a point line. The converse inclusion follows from Lemma 3.2, using the trivial constant reduction. Lemma 3.4 now proves the two conditions.

Suppose instead that \kappa_0 is not algebraic over a finite field. There is a nontrivial valuation b on \kappa_0: in characteristic zero extend a nontrivial valuation of the prime field; in positive characteristic use a transcendental element, extend its variable valuation to a rational transcendence basis, and prolong to \kappa_0. Choose a transcendental t\in F_0 with F_0/\kappa_0(t) finite. The Gauss extension of b to \kappa_0(t), followed by a prolongation to F_0, gives a constant reduction c. Indeed the rational residue field has transcendence degree one, and finite prolongation gives a finite residue extension. The value group of c equals its constant-value subgroup: the Gauss extension has this property, and a finite prolongation has finite value-group index, whereas the constant-value group is divisible.

Choose a point valuation q on F_0c/\kappa_0c, and form u=q\circ c. This is a c.r. quasi-prime divisor, whose value group contains the final discrete step as an innermost convex subgroup. In particular uF_0/u\kappa_0\cong\Z, so I_u is a nonzero line, and it belongs to \mathcal L_v. It is different from every true point line. To see this, let p be a point valuation of F_0/\kappa_0. The valuation u is nontrivial on constants, so cannot coarsen p. Conversely, a proper coarsening of u kills its innermost discrete subgroup and retains only values supplied by constants; it cannot equal the nontrivial valuation p that is trivial on constants. Thus p and u are incomparable. Since p has rank one, their finest common coarsening is trivial. Approximation gives I_p\cap I_u=0, proving the assertion even with \Lambda-coefficients.

All point lines already belong to \mathcal L_v, and their generators have the nonzero relation of Lemma 3.4, after rescaling to the chosen generators. If the first condition holds for the full family, extend this relation by zero on the additional lines. It is a nonzero kernel vector with a zero coordinate, contradicting the second condition. If the first condition fails, the family also fails the stated test. This proves the equivalence.□

Proof. Let F' be the relative algebraic closure of F in E. Choose a transcendence basis {\bf x} for E/F such that [E:F({\bf x})]=N<\infty. For every finite extension F_1/F contained in F', the tuple {\bf x} remains algebraically independent over F_1, and

The finite subextensions of F'/F form a directed system with bounded degrees. One of them has maximal degree and contains all the others; thus F'/F is finite.

If r independent classes in V_F vanish in V_E, choose representatives f_1,\ldots,f_r\in F^\times and their \ell-th roots in E. Those roots belong to F', and Kummer theory gives

Hence \ell^r\le [F':F], which bounds the kernel dimension.□

Proof. Smoothness of U\cap H is Bertini’s smoothness theorem for an embedded smooth variety, valid in every characteristic [11, Lemma 33.47.3, Tag 0FD6]. It implies smoothness upstairs because the cover is étale. We give the irreducibility argument, applied to either Y\to\Pj^n or U\to\Pj^n.

Write f:Y\to\Pj^n for the resulting quasi-finite morphism and a=\dim Y\ge2. Let \mathcal H=(\Pj^n)^\vee be the space of hyperplanes. In the incidence of triples (y_1,y_2,H) with f(y_1),f(y_2)\in H, the locus f(y_1)\ne f(y_2) is a projective space bundle with fiber \Pj^{n-2} over a nonempty open of Y\times Y. It is therefore irreducible of dimension 2a+n-2. The locus of pairs with equal images has dimension at most a, by quasi-finiteness; its hyperplane incidence has dimension at most a+n-1<2a+n-2.

For a general hyperplane, f^{-1}(H) is nonempty and has pure dimension a-1. Nonemptiness follows because the image of f contains an open of its a-dimensional closure; purity follows from the principal ideal theorem on the integral variety Y. Thus every component of the square of the generic hyperplane section has dimension 2a-2 over the function field of \mathcal H. No such component can be contained in the equal-image incidence, whose total dimension is too small. The square of the generic section is consequently irreducible.

This implies geometric irreducibility of the generic section. Indeed, if it had more than one geometric component, ordered pairs of points lying on the same component and on different components would give two distinct Galois-invariant unions of components of its square. Its already established generic smoothness excludes nonreducedness. Geometric integrality spreads to a nonempty open of \mathcal H, which proves the assertion for general hyperplanes.□

Proof. We first construct a curve over l, then specialize its finite defining data, and finally realize that specialization by a valuation of L.

A curve preserving orders and independence. Let D_1,\ldots,D_q be the prime divisors occurring in the divisors of the g_j. There is a closed subset B\subset X of codimension at least two such that X\setminus B is smooth, the D_i\setminus B are smooth pairwise disjoint Cartier divisors, and near each D_i\setminus B every g_j is a power of a local equation of D_i times a unit. Outside their union the functions are units. To obtain B, remove the singular locus of the normal variety X, the singular and non-Cartier loci of the D_i, their pairwise intersections, and the exceptional loci of these local expressions. Each has codimension at least two.

On the smooth open U=X\setminus(B\cup D_1\cup\cdots\cup D_q), the simultaneous root cover defined by the equations z_j^\ell=g_j is finite étale. Kummer theory and independence say that it is integral of degree \ell^m. Successively choose d-1 general hyperplanes. Apply Lemma 4.2 to the cover on each successive open section, and Bertini smoothness to the base and its specified divisors. We obtain a smooth integral projective curve C\subset X that avoids B, meets every D_i transversely in exactly \deg D_i distinct points, and has an integral root cover over C\cap U. The latter cover still has degree \ell^m, so the restricted classes of the g_j are independent. At a point of C\cap D_i their orders are exactly their orders along D_i. These are all their zeros and poles on C, proving the required equality of support counts over l.

The genus of this curve is independent of the chosen functions. At each stage we may also require the hyperplane to avoid the associated points of the preceding scheme section. Multiplication by its equation is then injective, so the Hilbert polynomial of a section is the first difference of the preceding Hilbert polynomial. The final scheme section avoids B, and on X\setminus B the successive sections are smooth by Bertini. Hence the final scheme section is reduced and equals C, even if intermediate sections had embedded components supported in B. Its Hilbert polynomial, and thus its genus, depend only on the Hilbert polynomial of the embedded X. Denote this genus by G_X. This argument does not require X to be Cohen–Macaulay.

Specialization of the finite data. Choose a finitely generated subring R\subset l in which \ell is invertible and over which the preceding data are defined. Enlarge R by finitely many elements and localize it as follows. The smooth open of X containing C, the flag sections in that open, and their inclusions descend with their dimensions, smoothness, and geometric integrality preserved in every geometric fiber. The final curve descends as a smooth projective curve of genus G_X. Include the coordinates of its finitely many zeros and poles and the local parameter–unit expressions for the functions. After shrinking \Spec R, they give disjoint point sections with the same orders and no other zeros or poles. The root cover over the complement of those sections remains finite étale of degree \ell^m, with geometrically integral fibers. Thus independence and the support counts persist in every geometric fiber under consideration.

Here the spreading statements have their usual finite-presentation meaning: finitely many schemes, morphisms, functions, and identities descend after adjoining finitely many coefficients [11, Tags 01ZM, 0C0C, 081F]; smoothness and flatness hold after shrinking; and geometric integrality at the generic point persists on an open [11, Tags 0578, 0559]. The Hilbert polynomial in the projective flat family of curves is constant; its constant term is the fiberwise Euler characteristic, locally constant in this proper flat family [11, Lemma 36.32.2, Tag 0B9T]. For the orders, the identities g_j=u\pi^e on finitely many neighborhoods, with u invertible and \pi a parameter for the relevant section, preserve the integer e. On the complement of these neighborhoods the functions and their inverses are regular. Properness of the curve ensures that these finitely many open conditions still cover every fiber after shrinking.

We also include a presentation of the function field, so that the specialized variety will be the residue field of a valuation. Choose a separating transcendence basis x_1,\ldots,x_d for L/l and a primitive element y for the finite separable extension L/l(x_1,\ldots,x_d). Write its monic minimal polynomial as Q(Y)\in l({\bf x})[Y]. After inverting a polynomial in {\bf x}, this gives a finite integral model over an open of affine d-space. Descend that model, a common dense open with the model already chosen, and the rational expressions for all the g_j. By geometric integrality and further localization of R, the reduced polynomial in every geometric fiber is irreducible of the same degree as Q and gives that fiber’s function field. All denominators in the coefficients and in the expressions being used remain nonzero, and the g_j agree with the specified nonzero functions there.

Realization by a valuation. Choose a closed point s\in\Spec R. Its residue field is finite, of characteristic different from \ell. A valuation of \operatorname{Frac}R dominating R_s extends to l; equivalently, one may directly choose a valuation ring of l dominating R_s [11, Lemma 10.50.2, Tag 00IA]. Write its residue field as \lambda_0. Since l is algebraically closed, so is \lambda_0. We may arrange that the ultimate residue field is algebraic over the finite field \kappa(s). Indeed let b be the algebraic closure of \kappa(s) inside \lambda_0, and choose a transcendence basis \mathcal T for \lambda_0/b. On b(\mathcal T) give the basis elements rationally independent values in an ordered free abelian group with basis \mathcal T and give b^\times value zero. Every polynomial has a unique term of least value, so this defines a valuation with residue b. Extend it to the algebraic extension \lambda_0/b(\mathcal T). Its residue extension is algebraic and thus still equals b. Composing valuations gives a valuation u of l with residue \lambda=b. Its center on R remains s, since the second valuation is trivial on the finite field \kappa(s). An ordered free abelian group exists for any cardinality of \mathcal T, so this construction imposes no cardinality condition on l.

Give l({\bf x}) the Gauss extension of u: the values of the x_i are zero and their residues are algebraically independent over \lambda. Prolong it to L, obtaining w_X. The coefficients of Q are integral at the Gauss valuation and its reduction is the specified irreducible polynomial over \lambda(\overline{\bf x}). Since Q is monic, y is integral; its residue therefore has degree \deg Q. The fundamental inequality for a finite extension of valued fields now forces

Thus w_X is a constant reduction with precisely the specialized function field as residue, and each g_j has the prescribed residue.

On Lw_X, take the successive true divisor valuations of the specialized smooth flag, ending at its curve. Compose them with w_X. Write w_1 for the first composition. It is quasi-divisorial: its value group \Gamma has a convex kernel \Z over w_XL, while the constant values map isomorphically onto w_XL=ul. Consequently \Gamma/w_1l\cong\Z. Every nonzero convex subgroup of \Gamma contains this innermost \Z, so any proper coarsening has value group generated by constant values and loses the relative rank-one contribution. This proves the required minimality. The remaining steps are true divisor steps. Their composition w is therefore a quasi-prime (d-1)-divisor with residue the specialized curve field. All g_j are units at the successive generic points. The genus, independence, and support counts already preserved in the specialization give the three conclusions.□

Proof. Put h_a=\Theta([t-a]). The classes [t-a], a\in k, are independent in V_{k(t)}, as is seen from their orders at the finite points of the projective line. Lemma 4.1 shows that the kernel of V_{k(t)}\to V_K is finite dimensional. A basis for its intersection with the span of these classes involves only finitely many indices. After removing those indices, we have an infinite set A\subset k such that the classes [t-a], a\in A, are independent in V_K.

Choose five distinct anchors a_1,\ldots,a_5\in A and put N_i=s_X(h_{a_i}). If \charac k=0, choose the anchors to include b_0,b_0+1,b_0+\ell; this is possible because k\setminus A is finite. Fix one further a\in A distinct from the anchors and put a_6=a. Choose representatives g_1,\ldots,g_6\in L^\times of h_{a_1},\ldots,h_{a_6}, and apply Proposition 4.3 to these representatives. Let w be the resulting valuation and C_w its residual curve. Their residue classes are independent, so their evaluations on D_w are independent by Lemma 2.4. Let v be the quasi-prime (d-1)-divisor of K corresponding to w under Corollary 2.9. The evaluations of the six functions u_j=t-a_j, 1\le j\le6, on D_v are independent.

We first normalize these functions so that their residues are a pencil on Kv. We claim that v(k(t))=vk. Otherwise the nonzero relative value group of k(t)/k has rational rank one, and the valuation transcendence-degree inequality forces k(t)v=kv. Equality in that inequality makes the relative value group cyclic. Any element with value in vk can then be multiplied by a constant to become a unit, and its residue can be removed by a further constant. Its class consequently vanishes on D_v. Thus evaluation of k(t)^\times on D_v factors through the cyclic relative value group and has dimension at most one, a contradiction.

The values of all six u_j are equal. If v(u_i)>v(u_j), then u_j/(a_i-a_j) is a principal unit, contradicting its nonzero evaluation on D_v. Write \gamma for the common value. For distinct indices, v(a_i-a_j)\ge\gamma; a strict inequality would make u_i/u_j a principal unit and give equal evaluations. Hence every difference a_i-a_j has value exactly \gamma. In positive characteristic the residue characteristic equals \charac k\ne\ell. In characteristic zero our three specified anchors give v(1)=v(\ell)=\gamma=0, so again \charac(kv)\ne\ell.

Choose b\in k^\times with v(b)=\gamma and set

The \beta_j\in kv are distinct and \overline{u_j/b}=\tau_0-\beta_j. These residue classes remain independent: the functions u_j/b are units, their evaluations factor through D_v/I_v, and multiplication by b changes no class because constants are \ell-divisible. In particular \tau_0\notin kv. More precisely, the dual of the induced isomorphism D_w/I_w\longrightarrow D_v/I_v carries [\tau_0-\beta_j]\in V_{Kv} to [\overline{g_j}]\in V_{Lw}. Indeed their evaluations on corresponding residual characters are the evaluations of [u_j] and \Theta([u_j])=[g_j] on the original decomposition spaces. This identifies the individual tested classes, as well as preserving their independence.

Proposition 3.5 says that kv, like lw, is algebraic over a finite field, and that the complete point families on the two residual curves correspond. Let C_v be the smooth projective curve with function field Kv. Both residue characteristics are different from \ell; by Lemma 3.4, the dimensions of the common point-order kernels are twice the genera. The residual isomorphism preserves these kernels, so g(C_v)=g(C_w)\le G_X. It also preserves the number of nonzero point orders of each tested class. For the five anchor classes these numbers are N_1,\ldots,N_5.

Write p=\charac(kv)>0 and \tau_0=\tau^{p^r} with r maximal. Such an r exists because a nonconstant function has a nonzero point order, and p^r must divide that fixed nonzero integer. The function \tau defines a separable morphism C_v\to\Pj_{kv}^1; write n' for its degree. Multiplication of orders by p^r does not change their vanishing modulo \ell.

Consider the fiber of this morphism over \beta_i^{1/p^r}. At most N_i points in the fiber have ramification index not divisible by \ell. The other points have index at least \ell, so there are at most n'/\ell of them. If r_i is the total number of points in the fiber, then r_i\le N_i+n'/\ell. Since the sum of the ramification indices in a fiber is n', and the different exponent at a point is at least its index minus one, the fiber contributes at least n'-r_i\ge(1-1/\ell)n'-N_i to the different. The five fibers are disjoint. Riemann–Hurwitz, with its different term in arbitrary characteristic [11, Lemma 53.12.4, Tag 0C1F], therefore yields

or

The coefficient on the left is positive for every prime \ell, including \ell=2. Hence n' has a bound depending only on the five anchors and X.

For the additional parameter, the zeros and poles of \tau_0-\beta_6 lie in two fibers of this same morphism. It has at most 2n' nonzero point orders modulo \ell. The point correspondence and the finite curve test identify this number with s_X(h_a). This proves a uniform bound for all non-anchor members of A. Including the anchors and the finitely many parameters outside A completes the proof.□

Proof. Choose r\in F\overline B with r^\ell=g. The element r already belongs to FB' for a finite Galois extension B'/B. Indeed it is separable over F, so purely inseparable constant extensions are unnecessary, and we may take a finite Galois closure of the remaining constant extension. By regularity, \Gal(FB'/F)=\Gal(B'/B). For \gamma in this group,

These multipliers form a multiplicative cocycle with values in B'^\times. Hilbert’s Theorem 90 gives b'\in B'^\times with \gamma(b')/b'=c_\gamma for every \gamma. Thus r/b'\in F and b'^\ell\in B. The equality g=(r/b')^\ell b'^\ell proves the assertion.□

Proof. The compositum PQ has transcendence degree two over \kappa. Otherwise every element of Q would be algebraic over P, so relative algebraic closedness of P in F would give Q\subset P; reversing their roles would give equality. Let C_P,C_Q be their smooth projective curves. The natural map to C_P\times C_Q is dominant, since its image has dimension two; consequently PQ is the function field of this product.

Inside V_{PQ}, any class coming from both fields has zero order along every divisor \{x\}\times C_Q. Its representative from P therefore lies in the common kernel of all point orders of C_P. This kernel has dimension 2g(C_P) by Lemma 3.4, so the intersection inside V_{PQ} is finite dimensional. The maps from V_P and V_Q to V_{PQ} are injective: each factor field is relatively algebraically closed in the function field of the product.

The kernel of V_{PQ}\to V_F is finite dimensional by Lemma 4.1. To see that passing to F preserves the finite-intersection conclusion, consider pairs (a,b)\in V_P\oplus V_Q whose images in V_F agree. The difference a-b lies in this finite kernel; the kernel of the difference map on such pairs is the intersection already computed in V_{PQ}. The space of these pairs, and hence V_P\cap V_Q in V_F, is finite dimensional.

Finally, choose t\in P\setminus\kappa. The classes [t-a], a\in\kappa, are linearly independent in V_{\kappa(t)}, as their orders at the distinct finite points show. Lemma 4.1 gives only a finite-dimensional kernel on passing to V_P, so V_P is infinite dimensional.□

Proof. A bounded parameter space. We recall why a degree bound gives a parameter space of finite type. Integral subvarieties of fixed dimension and bounded degree in a fixed projective space have only finitely many Hilbert polynomials; see also [10, Proposition 5.3]. One way to establish this boundedness is to cut them out set-theoretically by forms of bounded degree. If an integral subvariety D\subset\Pj^n has dimension r and degree at most b, and x\notin D, choose a linear projection to \Pj^{r+1} whose center avoids the join of x and D. When D is already a hypersurface no projection is necessary. The projection is defined on D and is finite there: a positive-dimensional fiber would contradict ampleness of the pulled-back hyperplane bundle. Its image is a hypersurface of degree at most b, and its equation pulls back to a form vanishing on D but not at x. Multiplication by a form nonzero at x makes the degree exactly b if necessary. It follows that the degree-b forms vanishing on D cut it out as a reduced set.

A fixed number of such forms, namely the dimension of the space of degree-b forms, therefore places all these reduced subvarieties among the geometric reductions of fibers of one projective family of finite type. Such geometric reductions have finitely many Hilbert polynomials. For completeness, over an integral base descend the reduction of the geometric generic fiber to a finite extension of the base function field, and normalize the base in that extension. On a dense open of this finite base change, spread the reduction as a closed subscheme, with nilpotent defining ideal, flat and with geometrically reduced fibers. These properties follow after shrinking from generic flatness and spreading geometric reducedness in the resulting projective flat family [11, Tag 0578]. Thus it gives exactly the geometric reductions there, with constant Hilbert polynomial. The complement upstairs has image in a proper closed subset of the original base because the normalization is finite. Noetherian induction on that closed subset proves the assertion. Applying this to the divisors in \mathcal S gives the required finite union of Hilbert schemes on X.

The incidence fields. In that union take a positive-dimensional irreducible component of the closure of the points corresponding to \mathcal S. After replacing it by an integral locally closed open subset T, the tested points are dense in T, and its universal family

is flat with geometrically integral fibers of dimension d-1. The latter condition is available by constructibility of geometric integrality [11, Tags 0579 and 055B]: the tested fibers are integral over the algebraically closed field l, and their parameter points are dense. Choose an integral locally closed curve T'\subset T, once and for all, and put Z'=Z\times_T T'. Both Z and Z' are integral, by flatness and geometric integrality of their generic fibers. Their maps to X are dominant. Indeed a proper closed subset of X can contain only finitely many distinct prime divisors, whereas the fibers over the infinitely many distinct points of T, or of T', are distinct divisors.

The two projections now give

Since \dim Z'=d, the extension M/L is finite. Geometric integrality of the generic fiber of Z'\to T' says precisely that M/P_0 is regular. The inclusion Z'\subset X\times T' also shows that its function field is generated by the coordinates from the two factors, proving (5.2).

Descent along the fixed curve. We next show that the chosen fields work for every class in H. Fix g\in L^\times with [g]\in H. Let Y be the intersection of the relative smooth locus of Z\to T with the inverse image of the open subset of X on which g is a unit. Every nonempty geometric fiber of Y\to T is integral. Its fibers over the generic points of T and T' are nonempty: the corresponding incidence families dominate X, and the smooth locus is dense in their geometrically integral fibers. On Y consider the finite étale cover defined by adjoining an \ell-th root of g.

This cover splits on a dense set of the tested fibers. Indeed (5.1) gives a rational \ell-th root on each such divisor, and on its smooth unit locus the root and its inverse are regular, by normality. The cover therefore splits on the geometric generic fiber over T as well. Otherwise, since \ell is prime and the constants contain \mu_\ell, its generic Kummer polynomial would be irreducible over that geometric function field. The geometric generic cover would then be integral. Geometric integrality spreads to a nonempty open of T [11, Tags 0578 and 0559], contradicting the dense set of split fibers.

It remains to pass this splitting to our fixed curve T'. Shrinking T separately for each g would not justify this passage: the resulting open could miss T'. Instead, descend the geometric generic splitting to a finite extension of l(T) and let \widetilde T\to T be the finite surjective normalization in that extension. The pullback \widetilde Y=Y\times_T\widetilde T is normal, since it is smooth over the normal scheme \widetilde T [11, Tag 034F]. It is also integral. Its generic fiber is geometrically integral, and every irreducible component dominates \widetilde T: on a normal scheme the components are open, and a smooth map has open image. Thus there is only one component.

The pulled-back finite étale cover of \widetilde Y is generically split, and hence split everywhere. To justify the last implication, each component is finite and birational over the normal integral base; it is therefore isomorphic to that base. Take a geometric point of \widetilde T lying over the generic point of T'. The corresponding fiber of \widetilde Y is a base change of the nonempty fiber of Y over that generic point. Its split cover shows

This reasoning applies to each g separately, while T', M, and P_0 stay fixed. Lemma 5.1, applied to the regular extension M/P_0, now gives the asserted containment in V_M.□

Proof. Fix a normal integral projective model X\subset\Pj_l^n of L/l. Let E\subset K be a curve subfield, choose t\in E\setminus k, and put H=\Theta(V_E). For a\in k, let h_a=\Theta([t-a]), and let \mathcal S be the union of their mod-\ell supports on X. Proposition 4.4 bounds the degrees of these supports, and hence the degree of every prime divisor in \mathcal S.

The set \mathcal S is infinite. To see this, first observe that the kernel of all divisor-order maps on X is finite dimensional. If it contained more than 2G_X independent classes, apply Proposition 4.3 to representatives of a finite independent subset of that size. Their independent residues on the testing curve would all have zero point orders, contradicting Lemma 3.4 and its genus bound. On the other hand, the [t-a] span an infinite-dimensional subspace of V_E, and therefore the h_a span an infinite-dimensional subspace of V_L. If \mathcal S were finite, their order vectors would lie in a finite dimensional space; the finite-dimensional kernel just proved would give a contradiction.

We verify (5.1). For each D\in\mathcal S, some h_a has nonzero order at D. Match the valuation of D to the quasi-divisorial valuation v of K using Proposition 2.7. Its inertia I_v has nonzero restriction to E. Consequently vE/vk\ne0; this group is a subgroup of the cyclic group vK/vk, and the valuation transcendence-degree inequality gives Ev=kv. Restriction of D_v to E therefore factors through vE/vk and has dimension at most one. Explicitly, an element whose value is in vk can be rescaled by a constant to a unit; its residue is in kv, so a further constant rescaling makes it a principal unit. Since inertia already has nonzero restriction, the images of I_v and D_v on E are the same line.

Now let g\in L^\times with [g]\in H. Whenever \ord_D(g)=0, inertia at D annihilates this class. The preceding equality of restriction images, transported through \Phi, implies that decomposition at D annihilates it too. By Lemma 2.4, its residue is an \ell-th power in l(D). Since a fixed rational function has nonzero integral order at only finitely many prime divisors, this proves (5.1). Lemma 5.3 now supplies M/L and P_0\subset M with M=LP_0, M/P_0 regular, and the image of H contained in the image of V_{P_0}.

Choose a finite normal extension N/L containing M, allowing inseparability, and let P_N be the relative algebraic closure of P_0 in N. It is a curve subfield of N/l. The image H_N of H in V_N is infinite dimensional, by Lemma 4.1, and lies in V_{P_N}. Every L-automorphism \gamma of N fixes H_N pointwise. Thus H_N\subset V_{P_N}\cap V_{\gamma(P_N)}, and Lemma 5.2 forces \gamma(P_N)=P_N.

Put P_L=P_N\cap L. This is relatively algebraically closed in L. It also has transcendence degree one over l. Indeed the invariant field P_N^{\Aut(N/L)} has transcendence degree one, and it is contained in the fixed field N^{\Aut(N/L)}. For a finite normal extension the latter is L in characteristic zero and is a finite purely inseparable extension of L in positive characteristic. In the latter case, a uniform p-power of the invariant field lies in P_N\cap L, preserving transcendence degree. Thus P_L is a curve subfield of L.

We now use the compositum identity to obtain containment already in V_L. Since P_L\subset L\subset M and every element of P_L\subset P_N is algebraic over P_0, relative algebraic closedness of P_0 in M gives P_L\subset P_0. The extension P_0/P_L is therefore finite algebraic. Choose the algebraic closures inside a common algebraically closed overfield. Then

Here we used both M=LP_0 and the fact that P_0 and P_L have the same algebraic closure in that overfield.

The extension L/P_L is regular. Relative algebraic closedness is already known, and separability is automatic in characteristic zero. In characteristic p>0, choose a separating variable u for P_L/l. It cannot become a p-th power in L: a p-th root would be algebraic over P_L and hence belong to P_L, contrary to the choice of u. Over the perfect field l, such an element extends to a separating transcendence basis of L/l, by the usual p-basis criterion. Thus L/l(u), and therefore L/P_L, is separable.

For each [g]\in H, Lemma 5.3 and (5.4) give an \ell-th root of g in L\overline{P_L}. Applying Lemma 5.1 once more, now to L/P_L, yields the exact inclusion \Theta(V_E)\subset V_{P_L}.

Apply the same argument to \Theta^{-1} and P_L. For a curve subfield E_1 of K, it gives

Lemma 5.2 and the infinite dimension of V_E force E=E_1. Both inclusions are consequently equalities. The same argument with K and L reversed proves surjectivity of the matching, and Lemma 5.2 proves its uniqueness.□

Proof. Suppose first that v is trivial on \kappa. Since the residue field has transcendence degree d-1\ge1, choose a unit t with transcendental residue, and let E be the relative algebraic closure of \kappa(t) in F. The valuation is trivial on \kappa(t) and therefore on its algebraic extension E. Reduction embeds E into Fv. Its induced map V_E\to V_{Fv} has finite-dimensional kernel by Lemma 4.1. Lemma 2.4 identifies that kernel with V_E\cap D_v^\perp, proving (6.1).

Conversely, suppose that the restriction of v to \kappa is nontrivial, and let E be any curve subfield. Choose x\in E\setminus\kappa with v(x)\ge0, replacing a nonconstant element by its inverse if necessary. There are infinitely many c\in\kappa^\times with v(c)>0, for example the positive powers of one such element. Each 1+cx is a principal unit for v. Their classes are independent in V_{\kappa(x)}: up to constant factors they are [x+c^{-1}], with distinct zeros on the rational curve. The map to V_E has only a finite-dimensional kernel by Lemma 4.1. These principal units therefore span an infinite-dimensional subspace of V_E\cap D_v^\perp. This disproves (6.1) for every E. The final assertion now follows from Theorem 5.4 and the already established matching of quasi-divisorial pairs.□

Proof. For each curve subfield E, let \mathcal Q_E be the collection of nonzero images of true divisorial inertia in A_E. By Proposition 6.1, these collections correspond under the induced character isomorphisms. They are subfamilies of the point lines of E. Choose a nonzero generator of each line. The generators tend to zero outside finite sets, so summation defines a continuous map from a product of copies of \Lambda into A_E.

By Lemma 3.4, when every point occurs, this summation map has one-dimensional kernel, generated by a coefficient family with every coordinate nonzero. A proper subfamily of the point lines has no relation: extending any proposed relation by zero to the missing points would contradict the description of the full relation kernel. Hence the stated kernel property recognizes exactly when \mathcal Q_E contains every point line. Once this holds, the common kernel of their orders on V_E has dimension 2g(E). Its vanishing recognizes genus zero, equivalently rationality over the algebraically closed constants. Both conditions are preserved by the character isomorphisms. Finally, distinct points have distinct inertia lines, so their matching gives the asserted point bijection.□

Proof. Blow up the smooth point q and let v be the valuation of the exceptional divisor. Write e=\trdeg(F/\kappa). Its residue field is the rational field of \Pj_\kappa^{e-1}. The leading linear forms of a and b give independent homogeneous coordinates on the exceptional divisor. Their ratio is the residue of t and is a rational coordinate in Fv. In particular v is trivial on \kappa(t), and \kappa(\overline t) is relatively algebraically closed in Fv. The relative algebraic closure of \kappa(t) in F is a finite algebraic extension on which v is still trivial. It therefore embeds by reduction into Fv, over \kappa(t)\cong\kappa(\overline t). Relative algebraic closedness in the residue field forces this extension to be \kappa(t) itself.

For each c\in\kappa, the function a-cb has nonzero linear term at q and cuts out a prime divisor locally there, with order one. The function b is a unit at its generic point because the linear terms of a and b are independent. The resulting true divisor has \ord(t-c)=1 and detects the point c of \Pj^1_\kappa. The divisor locally cut out by b similarly detects infinity: a is a unit at its generic point and t has a simple pole. Every point line is thus detected, proving that t is good.

Given f\in F^\times, choose the smooth point q in an open set where f is a unit, and choose a,b as above. The functions fa,b also vanish at q with independent differentials, since d(fa)_q=f(q)\,da_q. Hence both ft=(fa)/b and t=a/b are good. The equality f=(ft)/t proves field generation, and [f]=[ft]-[t] proves the spanning assertion for V_F.□

Proof. Choose a smooth closed point q on a model where all the t_i are regular, and choose a,b vanishing at q with independent differentials. Put s=a/b. It is good by Lemma 6.4. For c\ne0, write

At q the numerator has differential t_i(q)\,db_q+c\,da_q, which is independent of db_q. The same lemma therefore makes every indicated sum good.

Let v be the exceptional-divisor valuation at q used in the preceding lemma. The residue of s is transcendental over \kappa. By contrast, the restriction of v to \kappa(t_i) is centered at the finite point t_i(q): the nonzero function t_i-t_i(q) has positive order at q. Its residue field is consequently \kappa. If s were algebraic over \kappa(t_i), relative algebraic closedness of that good subfield in F would imply s\in\kappa(t_i), contradicting its transcendental residue. This proves the required independence.□

Proof. Choose a smooth model open on which all the t_i are regular and all the differentials dt_i are nowhere zero. Such an open exists. In positive characteristic, t_i cannot be a p-th power in K, since its p-th root would be algebraic over the relatively algebraically closed subfield k(t_i). As k is perfect, this gives dt_i\ne0; in characteristic zero the same conclusion is immediate.

At a closed point P of this open, write a_i=t_i(P). The exceptional divisor of the blowup at P defines a true divisorial valuation v with

Indeed dt_i(P)\ne0 says that t_i-a_i has order one in the maximal ideal of the regular local ring at P. Thus restriction of I_v to each k(t_i) is the nonzero point line at a_i.

By Proposition 6.1, the matched valuation on L is a true divisor. Restriction of characters commutes with the correspondence of good subfields, so its center in the t_i'-line is B_{t_i}(a_i) for every i. Specializing the target tuple along this valuation shows that (B_{t_i}(a_i))_i belongs to Y_{\mathbf t}(l): the generic tuple lies in that closed subvariety of the product of projective lines, and so does its specialization.

The image of the chosen model open in X_{\mathbf t} is constructible and dense, hence contains a dense open subset U. Every k-point of U has a closed preimage, since its nonempty fiber is of finite type over the algebraically closed field k. This proves the first inclusion in (7.1). The same argument for \Theta^{-1} proves the second.□

Proof. In characteristic zero, injectivity on a cofinite set implies that a rational map has degree one. In characteristic p, choose N\ge0 such that the given function h satisfies h^{p^N}=r\in l(X). Write r=q^{p^a}, with a\ge0 maximal and q\in l(X) separating. The map q is also injective on a cofinite set, so its degree is one. Indeed a separating map of degree greater than one has more than one distinct point over a general value, and deleting finitely many points cannot change this. Thus q is projective linear, and h=\Frob^{a-N}\circ q on points.

Conjugating a projective linear transformation by Frobenius raises its coefficients to p-th powers. This proves the asserted normal form. No nonzero power of Frobenius is projective linear: for a positive exponent the corresponding map has inseparable degree greater than one, and a negative exponent reduces to this case by inversion. The normal form is therefore unique. Finally, two perfect rational functions agreeing cofinitely are equal, as is seen after clearing Frobenius powers and comparing ordinary rational functions.□

Proof. Consider the relation variety Y for (t,s,z_c). Its projection to the first two factors is dominant. Indeed k(t) and k(s) are distinct curve subfields, so their matched subfields l(t') and l(s') are distinct. Any two distinct curve subfields have compositum of transcendence degree two: if their compositum had transcendence degree one, relative algebraic closedness would make them equal.

On the dense open of Y given by the inverse inclusion in Lemma 7.1, the third coordinate is forced by the first two to be

We restrict to finite coordinates, which is permitted because all the coordinate functions are nonconstant and the B’s fix \infty. If \dim Y=3, then Y=(\Pj^1_l)^3, and a dense open has infinitely many third coordinates above a general pair (x,y), a contradiction. Consequently \dim Y=2.

The projection Y\to(\Pj^1_l)^2 is proper, dominant and generically finite. The complement of the open just used has dimension at most one, so its image is a proper closed subset of the base. After removing that image and restricting to the locus of finite fibers, every closed fiber has precisely one point. Thus the finite extension l(Y)/l(x,y) has separable degree one. Indeed, in positive characteristic write the minimal polynomial of the third coordinate as g(Z^{p^e}), with g separable; in characteristic zero take the minimal polynomial itself. After shrinking the base to preserve the degree and nonzero discriminant of the separable polynomial, the separable degree counts the distinct points of a general fiber, since taking p^e-th roots is a bijection on algebraically closed points. The extension is therefore purely inseparable, and the third coordinate gives a perfect rational function R_c satisfying (7.2).

Fixing a general y in (7.2) gives a perfect rational function of x, agreeing cofinitely with a bijection. A dense open in (\Pj^1_l)^2 has cofinite fibers outside finitely many values of y. Since B_s is a bijection, Lemma 7.2 therefore gives

For two such parameters,

Every element of k is a difference of two members of a cofinite subset: for any prescribed difference, the two required cofinite sets intersect. Hence all translations conjugated by B_t belong to \mathcal H_l. The products D_bD_{b'}^{-1} give the same assertion for B_{z_c}, and then D_b gives B_{z_c}B_t^{-1}\in\mathcal H_l.

Fixing a general x instead yields

Composing on the left successively with

both already in \mathcal H_l, gives

At c=1 this gives B_sB_t^{-1}\in\mathcal H_l. Composing again gives all conjugated multiplications B_t m_cB_t^{-1}. This proves (7.3). Finally, two distinct affine transformations cannot agree outside a finite set; conjugating by a bijection preserves this property. Thus the resulting homomorphism is injective.□

Proof. Choose a good function t, and use Lemma 6.5 to choose a good s satisfying the hypotheses of Lemma 7.3. First consider the embedding B_t\Aff(k)B_t^{-1}\subset\mathcal H_l, with the finite-agreement convention of that lemma.

The affine image is projective linear. In positive target characteristic the Frobenius exponent defines a homomorphism \mathcal H_l\to\Z. Its restriction to the affine image is zero: k^\times is divisible, while the additive group of k is divisible in characteristic zero and torsion in positive characteristic. Both groups therefore have zero image in \Z, and they generate \Aff(k). In characteristic zero for l, there is no exponent to consider. We have in either case an embedding into \PGL_2(l).

Translations yield a field embedding. Let U be the image of the translation subgroup. This is an infinite abelian subgroup of \PGL_2(l), normalized by the whole affine image. It cannot contain a nonidentity semisimple element. To see this, such an element has two fixed points, and every element of U preserves their pair. The subgroup fixing both points has index at most two in U; it is infinite, hence contains an element of order greater than two. The centralizer of that element is the torus fixing the two points, so all of U lies in that torus. Any transformation normalizing U preserves the same pair and acts on U by either identity or inversion. This contradicts the faithful conjugation action of k^\times on its additive group in \Aff(k).

Every nonidentity element of U is consequently unipotent. Conjugate the image in \PGL_2(l) so that one such element is a translation with unique fixed point \infty. Its centralizer consists of translations, and hence so does U. Write this conjugation as postcomposition B=MB_t, with M\in\PGL_2(l). We obtain an injective additive map \chi:k\to l for which

After rescaling the target coordinate, assume \chi(1)=1. The image of m_c normalizes U, hence fixes \infty, and is an affine transformation of some slope \lambda_c. Conjugating translations gives

Setting b=1 gives \lambda_c=\chi(c). Thus \chi is multiplicative as well as additive and is a field embedding. In particular, the characteristics of k and l agree.

The field embedding is onto. Set z=t+s. Lemma 7.3 gives members M_s,M_z\in\mathcal H_l such that

Using R_1 from (7.2), define the perfect rational function

For cofinitely many b\in k, and for each such b for cofinitely many X\in l, equations (7.2) and (7.5) give

Here is the order of the exclusions. First exclude the finitely many values of b at which B(b) is infinite, the equality B_s=M_sB fails, or the corresponding vertical fiber misses the open of (7.2). For each remaining b, that open excludes only finitely many X; the equality B_z=M_zB excludes finitely many more, since its argument is a bijective function of X. Finally (7.5) excludes a finite set of X, which may depend on b.

For each fixed remaining b, (7.6) is therefore an identity of perfect rational functions in X. Choose X_0\in l such that F(X_0,Y) is defined and finite for general Y. This choice is possible by expressing a power of F as a quotient of polynomials and choosing X_0 for which its denominator does not vanish identically as a polynomial in Y. Define

For all but finitely many of the remaining b, the denominator is nonzero at Y=B(b); evaluating the rational identity then gives

In particular, this evaluation does not require X_0 to avoid the original exceptional sets for all b at once.

The injectivity of \chi implies that h is injective on a cofinite subset of \Pj^1(l). By Lemma 7.2, it represents a member of \mathcal H_l, so \chi(k) contains a cofinite subset of l. An infinite proper subfield cannot be cofinite: if a\notin\chi(k), the infinite coset a+\chi(k) is disjoint from it. Thus \chi(k)=l. Taking \sigma=\chi, the equality \sigma=hB=hMB_t outside a finite set proves (7.4) for our initial t.

The same constant isomorphism works for every good function. For any other good function r, Lemma 6.5 supplies one good s_0 that is algebraically independent of both t and r, with t+cs_0 and r+cs_0 good for every c\ne0. Lemma 7.3 gives B_{s_0}B_t^{-1},B_{s_0}B_r^{-1}\in\mathcal H_l, hence B_rB_t^{-1}\in\mathcal H_l. The alignment already obtained for t therefore gives (7.4) for r, with the same \sigma.□

Proof. Let \mathbf t=(t_1,\ldots,t_m) be a finite tuple of good functions, and let Z_{\mathbf t}\subset(\Pj^1_l)^m be the relation variety of (u_{t_1},\ldots,u_{t_m}). We first prove

where \sigma acts coordinatewise.

The point map

is a Zariski homeomorphism. Projective linear maps are isomorphisms; in positive characteristic the coordinate map [X:Y]\mapsto[X^p:Y^p] is finite, radicial and surjective, hence a homeomorphism on point spaces, as is its inverse. Products with different powers in different coordinates have the same property. It follows that

To justify this last equality at the level of relation varieties, place the finite tuple of u_{t_i} in one finite purely inseparable extension of L. On a common model, the tuples (t_i')_i and (u_{t_i})_i are related by \mathcal H wherever they are defined. Their images are dense in their respective relation varieties. Taking closures under the homeomorphism gives the equality.

Apply now the first inclusion of Lemma 7.1. Delete from its source open the finitely many exceptional values in each coordinate where (7.4) fails. Each deleted coordinate fiber is a proper closed subset, since every t_i is nonconstant. The resulting open is still dense, and on it \mathcal H\circ\prod_iB_{t_i}=\sigma. Consequently a dense subset of \sigma(X_{\mathbf t}(k)) lies in Z_{\mathbf t}(l). The former is the point set of the variety obtained from X_{\mathbf t} by applying \sigma to coefficients, so taking closures gives

For the reverse inclusion, use the second inclusion of Lemma 7.1. Delete the finitely many target coordinate values at which B_{t_i}^{-1}=\sigma^{-1}H_{t_i}^{-1} fails. Again this leaves a dense open, now in Y_{\mathbf t}, because each target coordinate is nonconstant. Its image under \mathcal H is dense in Z_{\mathbf t} and lies in \sigma(X_{\mathbf t}(k)). Taking closures proves (7.8).

Thus, for every polynomial P\in k[T_1,\ldots,T_m],

where \sigma(P) is obtained by applying \sigma to the coefficients. Good functions generate K by Lemma 6.4. The displayed equivalence for every finite tuple makes the assignments a\mapsto\sigma(a) for a\in k and t\mapsto u_t for good t well defined on all polynomial expressions, and preserves their nonzero values. Passing to quotients therefore gives an embedding \alpha_0:K\hookrightarrow L^i.

Because L^i is perfect, this embedding extends uniquely to an embedding \alpha:K^i\hookrightarrow L^i. Its image contains l=\sigma(k) and every u_t, and is itself perfect; by (7.7) it contains the perfection of every matched good rational subfield of L. All good rational subfields are matched, and their generators generate L. The image of \alpha therefore contains L^i, proving surjectivity and the final assertion.□

Proof. If v|_E is trivial, reduction embeds E in Kv. The kernel of V_E\to V_{Kv} is finite-dimensional by Lemma 4.1, whereas V_E is infinite-dimensional. Dualizing and using Lemma 2.4 shows that the restriction of D_v to A_E has infinite-dimensional image.

Otherwise v|_E is a positive multiple of the order at a point x of the smooth projective curve of E/k. If f is a unit there, its residue is some c\in k^\times, and f/c is a principal v-unit in K. Every character in D_v therefore kills f. Its restriction to E factors through the value group of v|_E, which is cyclic. The image has dimension at most one. This argument uses decomposition characters; it remains valid if the ramification multiplicity is divisible by \ell.□

Proof. Choose a normal projective model X on which both valuations have codimension-one centers D_1,D_2. We recall why this is possible. For each valuation, lifts of a residue transcendence basis, together with a uniformizer, define a rational subfield over which K is finite. Normalization of a suitable projective model of that subfield realizes the valuation as a prime divisor. A common normal projective model dominating the two models retains codimension-one centers: the residue field of each new center contains that of the old center, of transcendence degree d-1 over k.

Choose two distinct closed points P,Q of D_1\setminus D_2, and a smooth closed point R of X\setminus(D_1\cup D_2). The first choice is possible because d\ge2. There is an affine open containing these points and meeting D_2; write its coordinate ring as A, and let \mathfrak p_2 be the prime ideal of D_2 there. Choose two independent cotangent vectors at R. The Chinese remainder theorem, applied to \mathfrak p_2,\mathfrak m_P,\mathfrak m_Q,\mathfrak m_R^2, produces a,b\in A satisfying

and having value zero and the prescribed independent first-order terms at R. These ideals are pairwise comaximal, so the prescriptions are independent. By Lemma 6.4, t=a/b is good. Its residue on D_1 is nonconstant, since it takes different values at P,Q, and its value along D_2 is positive. Thus v_1 is trivial on k(t) and v_2 is nontrivial there.□

Proof. By Proposition 6.1, \Psi^* permutes the pairs I_v\subset D_v for true prime divisors. Since \Psi fixes each V_E, it preserves the rank of restriction of D_v to A_E. Lemmas 8.1 and 8.2 therefore force this permutation of true divisors to be the identity.

Fix a good rational subfield E. Every point line of A_E is the nonzero restriction of some I_v. The induced dual automorphism \Psi_E^*=(\Psi|_{V_E})^* of A_E therefore fixes every point line. Write

where C_E denotes the smooth projective model of E. The sum of these order characters is zero, and its coefficient families have exactly the diagonal kernel by Lemma 3.4. Continuity permits applying \Psi_E^* to this sum, so all a_x equal one scalar a_E. Since E is rational, its point orders separate V_E, and hence \Psi|_{V_E}=a_E\,\mathrm{id}.

Let E=k(t) and F=k(s) be any two good rational subfields. On a common smooth model open, t,s are regular and both differentials are nonzero. Blow up a closed point of that open. Its exceptional divisor v has

Thus the inertia line I_v restricts nontrivially to a point line of both A_E and A_F. The scalar by which \Psi^* acts on I_v must be both a_E and a_F. All these scalars agree. Good-function classes span V_K by Lemma 6.4, proving the claim.□

Proof. Let E=k(t) be good. Its perfection is relatively algebraically closed in K^i: if an element is algebraic over E^i, clearing Frobenius powers puts an algebraic element over E in K, and relative algebraic closedness then puts it in E. The same applies to \beta(E^i). Choose a nonconstant element of \beta(E^i)\cap K by clearing powers, and let P be the relative algebraic closure of the rational field it generates in K. Relative algebraic closedness of \beta(E^i) gives P^i\subset\beta(E^i). Conversely, every element of \beta(E^i) is algebraic over this rational field; clearing powers puts it in P. Hence \beta(E^i)=P^i. Since \beta_1 is scalar, V_P=V_E inside V_K. Lemma 5.2 forces P=E.

It follows that \beta(t)=h(t) for a perfect rational function h generating k(t)^i over k up to perfection. This implies that h belongs to the group \mathcal H_k of projective transformations and Frobenius powers from Section 7. Indeed, after clearing powers, its separable degree must be one; a larger separable degree could not disappear on passing to perfections. In characteristic zero this says simply that h is projective linear.

Write \rho=\beta|_k. For any two distinct points a,b\in\Pj^1(k), choose a rational function f_{a,b}(t) with divisor [a]-[b]. The point-order vector of \beta_1([f_{a,b}]) is supported exactly at

Here \rho fixes infinity. For negative Frobenius powers, the order vector is interpreted by clearing powers; multiplication by a power of p is invertible in \Lambda. Both displayed coordinates therefore have nonzero orders. Scalar action on V_E implies that the permutation h^{-1}\rho fixes every unordered pair of distinct points. On a set with at least three points this forces every point to be fixed: intersect the fixed pairs \{a,b\} and \{a,c\}. Thus h and \rho agree on all points.

In particular h fixes 0,1,\infty. An element of \mathcal H_k with these three fixed points is a pure Frobenius power, or the identity in characteristic zero. In positive characteristic its exponent is determined by \rho and therefore is the same for every good t. Distinct Frobenius powers act differently on the infinite algebraically closed field k: equality of two powers would force every element to satisfy X^{p^n}-X=0 for some n>0. Good functions generate K, so \beta is this same Frobenius power on K and then on its perfection.□

Proof of Theorem 1.1. A compatible \Theta gives equal transcendence degrees by Proposition 2.6. Theorem 7.5 produces a field isomorphism \alpha:K^i\to L^i respecting constants; in particular, the characteristics agree. Proposition 8.3 shows that \Theta and \alpha_1 differ by a single scalar, proving surjectivity of the stated map. If two field isomorphisms have the same image modulo scalars, their quotient satisfies Proposition 8.4, proving injectivity modulo Frobenius. Conversely, every integral Frobenius power acts on V_L by the corresponding nonzero scalar. The map is therefore well-defined and bijective with exactly the stated equivalence relations.□