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 (3) 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 \[vF'/v\kappa\cong\mathbb Z^a,\qquad F'v=\kappa v(\bar y_1,\ldots,\bar y_b).\] 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 \[0\longrightarrow w(Fu)/w(\kappa u) \longrightarrow vF/v\kappa \longrightarrow uF/u\kappa\longrightarrow0.\] Apply (3) 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 \[1\longrightarrow (Fv)^\times/(\kappa v)^\times \longrightarrow F^\times/(\kappa^\times U_v^1) \longrightarrow vF/v\kappa\longrightarrow0.\] 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) 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 (6). ◻

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 \[w_1(b-1)>0,\qquad w_2(b-a)>w_2(a).\] 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, \[\dim_\Lambda(G/\ell G)\leq\mathop{\mathrm{rrank}}(G).\] 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 \[\dim I_v\leq\mathop{\mathrm{rrank}}(vF/v\kappa).\]

Now let \(S\) be alternating. By Theorem 2, 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 \[\dim S\leq\dim I_v\leq\mathop{\mathrm{rrank}}(vF/v\kappa)\leq s.\] 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 \(\mathop{\mathrm{trdeg}}(Fv/\kappa v)\geq1\), and hence \[\dim S=\dim S_0+1 \leq\mathop{\mathrm{rrank}}(vF/v\kappa)+\mathop{\mathrm{trdeg}}(Fv/\kappa v)\leq s.\] If \(\dim S=s\), equality throughout the applicable chain gives \(S_0=I_v\) and equality in (3).

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 \(\mathbb Z^s\), hence is abstractly \(\mathbb Z^s\). Its inertia has dimension \(s\) and is alternating by Lemma 5. 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 \(\mathbb Z^s\), so its inertia is alternating of dimension \(s\). By (6), the intersection \(S_1\cap S_2\) is \(I_v\).

Conversely, suppose (7) holds. Let \(S_{i,0}\) be the valuative part of \(S_i\), with associated valuation \(v_i\). Proposition 7 gives \[S_{i,0}=I_{v_i},\qquad S_i\subset D_{v_i}, \qquad \dim I_{v_i}\geq s-1\geq1,\] and \(v_i\) has equality in (3). We first show that \(H\) is valuative. If, for example, \(v_1\leq v_2\), then \[0\ne I_{v_1}\subset S_1\cap S_2=H,\] so \(H=I_{v_1}\). The other comparable case is symmetric. If the valuations are incomparable, Lemma 6 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 \[H=I_{v_1}\cap I_{v_2}.\]

Let \(u=v_H\). By Theorem 2, it coarsens each \(v_i\), and therefore \[H\subset I_u\subset I_{v_1}\cap I_{v_2}=H.\] Equality in (3) passes to \(u\). Its relative value group is free, so \(\dim I_u=1\) implies \(uF/u\kappa\cong\mathbb Z\) and \(\mathop{\mathrm{trdeg}}(Fu/\kappa u)=s-1\). 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 (3) 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 5, since \(h\) has zero residue image. Conversely, an element alternating with \(h\) belongs to \(D_{v_H}=D_v\) by Theorem 2. ◻

Proof. First suppose \(r=1\). An \(\ell\)-divisible convex subgroup \(C\) of \(vF\) maps to zero in \(vF/v\kappa\cong\mathbb Z\), and hence lies in \(v\kappa\). Coarsening by \(C\) therefore leaves the relative value group unchanged. Lemma 3 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 \(\Gamma/v\kappa\cong\mathbb Z^r\) and \(v\kappa\) is divisible, the common kernel in \(\Gamma\) of all characters in \(I_v\) is \[v\kappa+\ell\Gamma=\ell\Gamma.\] 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 now gives \(v=v_{I_v}\). ◻

Proof. The maximum alternating dimension is preserved by \(\Phi\), giving equality of transcendence degrees by Proposition 7. Proposition 8 then identifies corresponding quasi-prime inertia and decomposition spaces. Quotienting by inertia and using Lemma 5 gives the indicated isomorphism on residue characters, with its alternating relation.

As long as the residual transcendence degree is at least two, apply Proposition 8 again there. The inverse-image identities (6) identify the resulting pairs with those of compositions on the original fields. Lemma 9 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 3 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 \[r_E:\widehat A_E\longrightarrow A_E.\] They carry \(\widehat I_a\) into \(I_a\), carry \(\widehat D_a\) into \(D_a\), and commute with restriction to units. In particular, \[r_F(\widehat S_F^{\mathrm{in}})\subset S_F^{\mathrm{in}}, \qquad \rho_v r_F=r_{F_0}\widehat\rho_v \quad\hbox{on }\widehat D_v.\] 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 \(uF_0/u\kappa_0\cong\mathbb Z\), reduction \(\widehat I_u\to I_u\) is surjective. Given \(\chi\in I_u\), choose a lift \(\widehat\chi\in\widehat I_u\). Theorem 11 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 \[x,x^{-1}\in O,\qquad \chi(x)\ne0.\] 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 \[U_a=U_u U_v^1.\] 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 (10), 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 (10) precisely when this homomorphism vanishes on principal divisors, or equivalently factors through \(\mathop{\mathrm{Pic}}(C)\). The group \(\mathop{\mathrm{Pic}}^0(C)(b)\) 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 \(\mathop{\mathrm{Pic}}(C)\to\Lambda\) therefore factors through the degree map \(\mathop{\mathrm{Pic}}(C)\to\mathbb Z\). 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 \(\mathop{\mathrm{Pic}}(C)[\ell]\). This gives an isomorphism from the common kernel to \(\mathop{\mathrm{Pic}}(C)[\ell]\): surjectivity follows from the definition of torsion in \(\mathop{\mathrm{Pic}}(C)\); injectivity follows because \(b^\times\) is \(\ell\)-divisible. When \(\mathop{\mathrm{char}}(b)\ne\ell\), the \(\ell\)-torsion of the Jacobian is isomorphic to \((\mathbb Z/\ell)^{2g(C)}\). ◻

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 13 shows that every line in \(\mathcal L_v\) is a point line. The converse inclusion follows from Lemma 12, using the trivial constant reduction. Lemma 14 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 \(uF_0/u\kappa_0\cong\mathbb Z\), 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 14, 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 \[[F_1:F]=[F_1({\bf x}):F({\bf x})]\le N.\] 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 \[[F(f_1^{1/\ell},\ldots,f_r^{1/\ell}):F]=\ell^r.\] 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]. It implies smoothness upstairs because the cover is étale. We give the irreducibility argument, applied to either \(Y\to\mathbb P^n\) or \(U\to\mathbb P^n\).

Write \(f:Y\to\mathbb P^n\) for the resulting quasi-finite morphism and \(a=\dim Y\ge2\). Let \(\mathcal H=(\mathbb P^n)^\vee\) 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 \(\mathbb P^{n-2}\) 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 17 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 \(\mathop{\mathrm{Spec}}R\), 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]; smoothness and flatness hold after shrinking; and geometric integrality at the generic point persists on an open [11]. 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]. 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 \(s\in\mathop{\mathrm{Spec}}R\). 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]. 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 \[w_XL=ul,\qquad Lw_X=\lambda(\overline{\bf x},\overline y).\] 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 \(\mathbb Z\) over \(w_XL\), while the constant values map isomorphically onto \(w_XL=ul\). Consequently \(\Gamma/w_1l\cong\mathbb Z\). Every nonzero convex subgroup of \(\Gamma\) contains this innermost \(\mathbb Z\), 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 16 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 \(\mathop{\mathrm{char}}k=0\), 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 18 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 5. Let \(v\) be the quasi-prime \((d-1)\)-divisor of \(K\) corresponding to \(w\) under Corollary 10. 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 \(\mathop{\mathrm{char}}k\ne\ell\). In characteristic zero our three specified anchors give \(v(1)=v(\ell)=\gamma=0\), so again \(\mathop{\mathrm{char}}(kv)\ne\ell\).

Choose \(b\in k^\times\) with \(v(b)=\gamma\) and set \[\tau_0=\overline{(t-a_1)/b},\qquad \beta_j=\overline{(a_j-a_1)/b}.\] 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 15 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 14, 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 \(p=\mathop{\mathrm{char}}(kv)>0\) 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 \(C_v\to\mathbb P_{kv}^1\); 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], therefore yields \[2g(C_v)-2+2n' \ge 5(1-1/\ell)n'-\sum_{i=1}^5N_i,\] or \[ \bigl(5(1-1/\ell)-2\bigr)n' \le 2G_X-2+\sum_{i=1}^5N_i. \tag{11}\] 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, \(\mathop{\mathrm{Gal}}(FB'/F)=\mathop{\mathrm{Gal}}(B'/B)\). For \(\gamma\) in this group, \[c_\gamma=\frac{\gamma(r)}r\in\mu_\ell.\] 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 14, 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 16. 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 16 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]. One way to establish this boundedness is to cut them out set-theoretically by forms of bounded degree. If an integral subvariety \(D\subset\mathbb P^n\) has dimension \(r\) and degree at most \(b\), and \(x\notin D\), choose a linear projection to \(\mathbb P^{r+1}\) 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]. 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 \[Z\subset X\times T\longrightarrow T\] is flat with geometrically integral fibers of dimension \(d-1\). The latter condition is available by constructibility of geometric integrality [11]: 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 \[\begin{array}{ccccc} && Z' &&\\[-2pt] &\swarrow&&\searrow&\\[-2pt] X&&&&T' \end{array} \qquad\qquad \begin{array}{ccccc} && M=l(Z') &&\\[-2pt] &\nearrow&&\nwarrow&\\[-2pt] L=l(X)&&&&P_0=l(T'). \end{array}\] 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 (13).

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 (12) 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], 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]. 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 \[ g\in(M\overline{P_0})^{\times\ell}. \tag{14}\] This reasoning applies to each \(g\) separately, while \(T'\), \(M\), and \(P_0\) stay fixed. Lemma 20, applied to the regular extension \(M/P_0\), now gives the asserted containment in \(V_M\). ◻

Proof. Fix a normal integral projective model \(X\subset\mathbb P_l^n\) 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 19 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 18 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 14 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 (12). 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 8. 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 \(\mathop{\mathrm{ord}}_D(g)=0\), 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 5, 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 (12). Lemma 22 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 16, 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 21 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 \(P_N^{\mathop{\mathrm{Aut}}(N/L)}\) has transcendence degree one, and it is contained in the fixed field \(N^{\mathop{\mathrm{Aut}}(N/L)}\). 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 \[ M\overline{P_0}=L\overline{P_L}. \tag{15}\] 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 22 and (15) give an \(\ell\)-th root of \(g\) in \(L\overline{P_L}\). Applying Lemma 20 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 \[\Theta(V_E)\subset V_{P_L}\subset\Theta(V_{E_1}).\] Lemma 21 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 21 proves its uniqueness. ◻

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 \[\mathop{\mathrm{ord}}_v(t_i-a_i)=1\qquad\text{for every }i.\] 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 24, 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 (17). 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 \(h=\mathop{\mathrm{Frob}}^{a-N}\circ q\) 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 29, the third coordinate is forced by the first two to be \[B_{z_c}\bigl(B_t^{-1}(x)+cB_s^{-1}(y)\bigr).\] 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 \(Y=(\mathbb P^1_l)^3\), and a dense open has infinitely many third coordinates above a general pair \((x,y)\), a contradiction. Consequently \(\dim Y=2\).

The projection \(Y\to(\mathbb P^1_l)^2\) 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 (18).

Fixing a general \(y\) in (18) gives a perfect rational function of \(x\), agreeing cofinitely with a bijection. A dense open in \((\mathbb P^1_l)^2\) has cofinite fibers outside finitely many values of \(y\). Since \(B_s\) is a bijection, Lemma 30 therefore gives \[D_b:=B_{z_c}T_{cb}B_t^{-1}\in\mathcal H_l \qquad\text{for cofinitely many }b\in k.\] For two such parameters, \[D_b^{-1}D_{b'}=B_tT_{c(b'-b)}B_t^{-1}.\] 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 \[B_{z_c}T_a m_cB_s^{-1}\in\mathcal H_l \qquad\text{for cofinitely many }a\in k.\] Composing on the left successively with \[B_{z_c}T_{-a}B_{z_c}^{-1} \quad\text{and}\quad B_tB_{z_c}^{-1},\] both already in \(\mathcal H_l\), gives \[B_t m_cB_s^{-1}\in\mathcal H_l\qquad(c\ne0).\] 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 (19). 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 28 to choose a good \(s\) satisfying the hypotheses of Lemma 31. First consider the embedding \(B_t\mathop{\mathrm{Aff}}(k)B_t^{-1}\subset\mathcal H_l\), with the finite-agreement convention of that lemma.

The affine image is projective linear. In positive target characteristic the Frobenius exponent defines a homomorphism \(\mathcal H_l\to\mathbb Z\). 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 \(\mathbb Z\), and they generate \(\mathop{\mathrm{Aff}}(k)\). In characteristic zero for \(l\), there is no exponent to consider. We have in either case an embedding into \(\mathop{\mathrm{PGL}}_2(l)\).

Translations yield a field embedding. Let \(U\) be the image of the translation subgroup. This is an infinite abelian subgroup of \(\mathop{\mathrm{PGL}}_2(l)\), 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 \(\mathop{\mathrm{Aff}}(k)\).

Every nonidentity element of \(U\) is consequently unipotent. Conjugate the image in \(\mathop{\mathrm{PGL}}_2(l)\) 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 \(M\in\mathop{\mathrm{PGL}}_2(l)\). We obtain an injective additive map \(\chi:k\to l\) for which \[ BT_bB^{-1}=T_{\chi(b)} \qquad\text{modulo finite disagreement}. \tag{21}\] 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 \[\chi(cb)=\lambda_c\chi(b).\] 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 31 gives members \(M_s,M_z\in\mathcal H_l\) such that \[B_s=M_sB,\qquad B_z=M_zB \qquad\text{modulo finite disagreement}.\] Using \(R_1\) from (18), define the perfect rational function \[F(X,Y)=M_z^{-1}\bigl(R_1(M^{-1}X,M_sY)\bigr).\] For cofinitely many \(b\in k\), and for each such \(b\) for cofinitely many \(X\in l\), equations (18) and (21) give \[ F(X,B(b))=B\bigl(B^{-1}(X)+b\bigr)=X+\chi(b). \tag{22}\] 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 (18). 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 (21) excludes a finite set of \(X\), which may depend on \(b\).

For each fixed remaining \(b\), (22) 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 \[h(Y)=F(X_0,Y)-X_0.\] For all but finitely many of the remaining \(b\), the denominator is nonzero at \(Y=B(b)\); evaluating the rational identity then gives \[\chi(b)=h(B(b)).\] 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 \(\mathbb P^1(l)\). By Lemma 30, 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 (20) for our initial \(t\).

The same constant isomorphism works for every good function. For any other good function \(r\), Lemma 28 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 31 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 (20) for \(r\), with the same \(\sigma\). ◻

Proof. Let \(\mathbf t=(t_1,\ldots,t_m)\) be a finite tuple of good functions, and let \(Z_{\mathbf t}\subset(\mathbb P^1_l)^m\) be the relation variety of \((u_{t_1},\ldots,u_{t_m})\). We first prove \[ \sigma\bigl(X_{\mathbf t}(k)\bigr)=Z_{\mathbf t}(l), \tag{24}\] where \(\sigma\) acts coordinatewise.

The point map \[\mathcal H=\prod_i H_{t_i}^{-1}: (\mathbb P^1(l))^m\longrightarrow(\mathbb P^1(l))^m\] 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 \[\mathcal H\bigl(Y_{\mathbf t}(l)\bigr)=Z_{\mathbf t}(l).\] 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 29. Delete from its source open the finitely many exceptional values in each coordinate where (20) 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 \[\sigma(X_{\mathbf t}(k))\subset Z_{\mathbf t}(l).\]

For the reverse inclusion, use the second inclusion of Lemma 29. 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 (24).

Thus, for every polynomial \(P\in k[T_1,\ldots,T_m]\), \[P(t_1,\ldots,t_m)=0 \quad\Longleftrightarrow\quad \sigma(P)(u_{t_1},\ldots,u_{t_m})=0,\] where \(\sigma(P)\) is obtained by applying \(\sigma\) to the coefficients. Good functions generate \(K\) by Lemma 27. 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 (23) 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 16, whereas \(V_E\) is infinite-dimensional. Dualizing and using Lemma 5 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 \[\begin{array}{c|ccc} & D_2 & P & Q\\ \hline a &0&0&1\\ b &1&1&1 \end{array}\] 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 27, \(t=a/b\) 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 24, \(\Psi^*\) 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 34 and 35 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 \[\Psi_E^*(\mathop{\mathrm{ord}}_x)=a_x\mathop{\mathrm{ord}}_x,\qquad a_x\in\Lambda^\times, \quad x\in C_E(k),\] 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 14. 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 \[\mathop{\mathrm{ord}}_v(t-t(P))=\mathop{\mathrm{ord}}_v(s-s(P))=1.\] 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 27, 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 21 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 \(a,b\in\mathbb P^1(k)\), 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 \[h^{-1}(\rho(a)),\qquad h^{-1}(\rho(b)).\] 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. A compatible \(\Theta\) gives equal transcendence degrees by Proposition 7. Theorem 33 produces a field isomorphism \(\alpha:K^i\to L^i\) respecting constants; in particular, the characteristics agree. Proposition 36 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 37, 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. ◻