On Kato and Kuzumaki's properties for the Milnor $K_2$ of function fields of $p$-adic curves

Let $K$ be the function field of a curve $C$ over a $p$-adic field $k$. We prove that, for each $n, d \geq 1$ and for each hypersurface $Z$ in $\mathbb{P}^n_{K}$ of degree $d$ with $d^2 \leq n$, the second Milnor $K$-theory group of $K$ is spanned by the images of the norms coming from finite extensions $L$ of $K$ over which $Z$ has a rational point. When the curve $C$ has a point in the maximal unramified extension of $k$, we generalize this result to hypersurfaces $Z$ in $\mathbb{P}^n_{K}$ of degree $d$ with $d \leq n$.


Introduction
Kato and Kuzumaki [1986] stated a set of conjectures which aimed at giving a diophantine characterization of cohomological dimension of fields.For this purpose, they introduced some properties of fields which are variants of the classical C i -property and which involve Milnor K -theory and projective hypersurfaces of small degree.They hoped that those properties would characterize fields of small cohomological dimension.
More precisely, fix a field K and two nonnegative integers q and i.Let K q (K ) be the q-th Milnor Kgroup of K .For each finite extension L of K , one can define a norm morphism N L/K : K q (L) → K q (K ); see Section 1.7 of [Kato 1980].Thus, if Z is a scheme of finite type over K , one can introduce the subgroup N q (Z /K ) of K q (K ) generated by the images of the norm morphisms N L/K when L runs through the finite extensions of K such that Z (L) ̸ = ∅.One then says that the field K is C q i if, for each n ≥ 1, for each finite extension L of K and for each hypersurface Z in ‫ސ‬ n L of degree d with d i ≤ n, one has N q (Z /L) = K q (L).For example, the field K is C 0 i if, for each finite extension L of K , every hypersurface Z in ‫ސ‬ n L of degree d with d i ≤ n has a 0-cycle of degree 1.The field K is C q 0 if, for each tower of finite extensions M/L/K , the norm morphism N M/L : K q (M) → K q (L) is surjective.
Kato and Kuzumaki conjectured that, for i ≥ 0 and q ≥ 0, a perfect field is C q i if, and only if, it is of cohomological dimension at most i + q.This conjecture generalizes a question raised by Serre [1965] asking whether the cohomological dimension of a C i -field is at most i.As it was already pointed out at the end of Kato and Kuzumaki's original paper [1986], Kato and Kuzumaki's conjecture for i = 0 follows from the Bloch-Kato conjecture (which has been established by Rost and Voevodsky [2014]); in other words, a perfect field is C 816 Diego Izquierdo and Giancarlo Lucchini Arteche it turns out that the conjectures of Kato and Kuzumaki are wrong in general.For example, Merkur'ev [1991] constructed a field of characteristic 0 and of cohomological dimension 2 which does not satisfy property C 0 2 .Similarly, Colliot-Thélène and Madore [2004] produced a field of characteristic 0 and of cohomological dimension 1 which did not satisfy property C 0 1 .These counterexamples were all constructed by a method using transfinite induction due to Merkurjev and Suslin.The conjecture of Kato and Kuzumaki is therefore still completely open for fields that usually appear in number theory or in algebraic geometry.
Wittenberg [2015] proved that totally imaginary number fields and p-adic fields have the C 1 1 property.Izquierdo [2018] also proved that, given a positive integer n, finite extensions of ‫(ރ‬x 1 , . . ., x n ) and of ‫(ރ‬x 1 , . . ., x n−1 )((t)) are C q i for any i, q ≥ 0 such that i + q = n.These are essentially the only known cases of Kato and Kuzumaki's conjectures.Note however that a variant of the C q 1 -property involving homogeneous spaces under connected linear groups is proved to characterize fields with cohomological dimension at most q + 1 in [Izquierdo and Lucchini Arteche 2022].
In the present article, we are interested in Kato and Kuzumaki's conjectures for the function field K of a smooth projective curve C defined over a p-adic field k.The field K has cohomological dimension 3, and hence it is expected to satisfy the C q i -property for i + q ≥ 3.As already mentioned, the Bloch-Kato conjecture implies this result when q ≥ 3. The cases q = 0 and q = 1 seem out of reach with the current knowledge, since they likely imply the C 0 2 -property for p-adic fields, which is a widely open question.In this article, we make progress in the case q = 2.
Our first main result is the following.
Main Theorem 1. Function fields of p-adic curves satisfy the C 2 2 -property.
steps, that are inspired from the proof of the C 1 1 property for number fields in [Izquierdo 2018] but that require to deal with several new difficulties: (1) Solve the problem locally: For each closed point v of C, prove that x ∈ N 2 (Z K v /K v ).This provides r v finite extensions M (v)  i /K v such that Z (M (v)  i ) ̸ = ∅ and (2) Globalize the extensions M (v)  i /K v : For each closed point v of C and each 1 ≤ i ≤ r v , find a finite extension K (v)   i of K contained in M (v)   i such that Z (K (v)  i ) ̸ = ∅.Then prove that there exists a finite subset of these global extensions, say K 1 , . . ., K r , such that for every closed point v of C, x lies in the subgroup of K 2 (K v ) generated by the norms coming from the (K i ⊗ K K v ).
(3) Establish a local-to-global principle for norm groups: Prove the vanishing of the Tate-Shafarevich group (4) Conclude: By step (2), we have x ∈ X 2 .Hence, step (3) implies that as wished.
Step (1) can be proved by combining some results for p-adic fields due to Wittenberg [2015] and the computation of the groups K 2 (K v ) thanks to the residue maps in Milnor K -theory; see Section 3A3.
In the way it is written above, Step (2) can be easily deduced from Greenberg's approximation theorem.However, as we will see below, we will need a stronger version of that step, that will require a completely different proof.
Step (3) is the hardest part of the proof.The first key tool that we use is a Poitou-Tate duality for motivic cohomology over the field K proved by Izquierdo [2016].This provides a finitely generated free Galois module T over K such that the Pontryagin dual of X 2 is the quotient of by its maximal divisible subgroup.Now, a result of Demarche and Wei [2014] states that, under some technical linear disjointness assumption for the extensions K (v)  i /K , one can find two finite extensions K ′ and K ′′ of K such that the restriction is injective and T is a permutation Galois module over both K ′ and K ′′ .If the groups X 2 (K ′ , T ) and X 2 (K ′′ , T ) were trivial, then we would be done.But that is not the case in our context because the p-adic function field K has finite extensions K ′ such that X 2 (K ′ , ‫)ޚ‬ is not trivial; see for instance the appendix of [Colliot-Thélène et al. 2012].This "failure of Chebotarev's density theorem" makes the computation of X 2 (K , T ) very complicated and technical.By carrying out quite subtle Galois cohomology computations and by using some results of Kato [1980], we prove that, under some technical assumptions on the K (v)   i (see below) and another technical assumption on C (which is trivially satisfied when C(k) ̸ = ∅), the group X 2 (K , T ) is always divisible, even though it might not be trivial; see Section 3A5.This is enough to apply the Poitou-Tate duality and deduce the vanishing of X 2 .Now, in order to ensure that the K (v)   i and C fulfill the conditions required to carry out the previous argument, we have to • add a step (0) in which we reduce to the case where C satisfies a technical assumption close to having a rational point; and • modify the constructions of the K (v) i in Step (2), which cannot be done anymore by using Greenberg's approximation theorem.
The reduction to the case where C satisfies the required conditions uses the Beilinson-Lichtenbaum conjecture for motivic cohomology and a local-to-global principle due to Kato [1980] with respect to the places of K that come from a suitable regular model of the curve C; see Section 3A2.As for Step (2), we want to construct the K (v)   i so that they fulfill two extra conditions: (a) One of the K (v) i has to be of the form k (v)  i K for some finite unramified extension k (v) i /k.This is achieved by observing that Z (k nr (C)) ̸ = ∅ since the field k nr (C) is C 2 and Z is a hypersurface in i have to satisfy some suitable linear disjointness conditions also involving abelian extensions of K that are locally trivial everywhere.This is achieved by an approximation argument that uses the implicit function theorem for Z over the K v , weak approximation and an analogue of Hilbert's irreducibility theorem for the field K , see Section 3A4.
Note that, since we use the implicit function theorem, the previous argument only works when the hypersurface Z is smooth.We thus need to add an extra step to the proof in which we reduce to that case.This uses a dévissage technique that is due to Wittenberg [2015] and that requires to work with all proper varieties over K (instead of only hypersurfaces); see Section 3A7.For that reason, we need to prove a wide generalization of Main Theorem 1 to all proper varieties.This is the object of Theorem 3.1 in the core of the text.Of course, this requires to modify and generalize the proofs of Steps (1), ( 2) and (3) so that they can be applied in that more general setting.
Ideas for the proof of Main Theorem 2. The proof of Main Theorem 2 follows by combining Main Theorem 1 with a result roughly stating that every element of K 2 (K ) can be written as a product of norms coming from extensions of the form k ′ K with k ′ a finite extension of k whose ramification degree is fixed, see Theorem 4.1.The general ideas to prove this last result are similar to (and a bit simpler than) those used in Main Theorem 1.

Notations and preliminaries
In this section we fix the notations that will be used throughout this article.
Milnor K -theory.Let K be any field and let q be a nonnegative integer.The q-th Milnor K-group of K is by definition the group K 0 (K ) = ‫ޚ‬ if q = 0 and if q > 0. For x 1 , . . ., x q ∈ K × , the symbol {x 1 , . . ., x q } denotes the class of More generally, for r and s nonnegative integers such that r + s = q, there is a natural pairing When L is a finite extension of K , one can construct a norm homomorphism satisfying the following properties; see Section 1.7 of [Kato 1980] or Section 7.3 of [Gille and Szamuely 2017]: • For q = 0, the map N L/K : K 0 (L) → K 0 (K ) is given by multiplication by [L : K ].
• If r and s are nonnegative integers such that r + s = q, we have N L/K ({x, y}) = {x, N L/K (y)} for x ∈ K r (K ) and y ∈ K s (L).
• If M is a finite extension of L, we have Recall also that Milnor K -theory is endowed with residue maps; see Section 7.1 of [Gille and Szamuely 2017].Indeed, when K is a henselian discrete valuation field with ring of integers R, maximal ideal m and residue field κ, there exists a unique residue morphism such that, for each uniformizer π and for all units u 2 , . . ., u q ∈ R × whose images in κ are denoted u 2 , . . ., u q , one has ∂({π, u 2 , . . ., u q }) = {u 2 , . . ., u q }.
The kernel of ∂ is the subgroup U q (K ) of K q (K ) generated by symbols of the form {x 1 , . . ., x q } with x 1 , . . ., x q ∈ R × .If U 1 q (K ) stands for the subgroup of K q (K ) generated by those symbols that lie in U q (K ) and that are of the form {x 1 , . . ., x q } with x 1 ∈ 1 + m and x 2 , . . ., x q ∈ K × , then U 1 q (K ) is ℓ-divisible for each prime ℓ different from the characteristic of κ and U q (K )/U 1 q (K ) is canonically isomorphic to K q (κ).Moreover, if L/K is a finite extension with ramification degree e and residue field λ, then the norm map N L/K : K q (L) → K q (K ) sends U q (L) to U q (K ) and U 1 q (L) to U 1 q (K ), and the following diagrams commute: The C q i properties.Let K be a field and let i and q be two nonnegative integers.For each K -scheme Z of finite type, we denote by N q (Z /K ) the subgroup of K q (K ) generated by the images of the maps N L/K : K q (L) → K q (K ) when L runs through the finite extensions of K such that Z (L) ̸ = ∅.The field K is said to have the C q i property if, for each n ≥ 1, for each finite extension L of K and for each hypersurface Z in ‫ސ‬ n L of degree d with d i ≤ n, one has N q (Z /L) = K q (L).
Motivic complexes.Let K be a field.For i ≥ 0, we denote by z i (K , • ) Bloch's cycle complex defined in [Bloch 1986].The étale motivic complex ‫(ޚ‬i ) over K is then defined as the complex of Galois modules . By the Nesterenko-Suslin-Totaro theorem and the Beilinson-Lichtenbaum conjecture, it is known that and Fields of interest.From now on and until the end of the article, p stands for a prime number and k for a p-adic field with ring of integers O k .We let C be a smooth projective geometrically integral curve over k, and we let K be its function field.We denote by C (1) the set of closed points in C. The residual index i res (C) of C is defined to be the g.c.d. of the residual degrees of the k(v)/k with v ∈ C (1) .The ramification index i ram (C) of C is defined to be the g.c.d. of the ramification degrees of the k(v)/k with v ∈ C (1) .
Tate-Shafarevich groups.When M is a complex of Galois modules over K and i ≥ 0 is an integer, we define the i-th Tate-Shafarevich group of M as When a suitable regular model C/O k of C/k is given, we also introduce the following smaller Tate-Shafarevich groups: where C (1) is the set of codimension 1 points of C.
Poitou-Tate duality for motivic cohomology.We recall the Poitou-Tate duality for motivic complexes over the field K ; Theorem 0.1 of [Izquierdo 2016] in the case d = 1.Let T be a finitely generated free Galois module over K .Set Ť := Hom( T , ‫)ޚ‬ and T = Ť ⊗ ‫.)2(ޚ‬ Then there is a perfect pairing of finite groups where A denotes the quotient of A by its maximal divisible subgroup.Note that, in the case T = ‫,ޚ‬ the Beilinson-Lichtenbaum conjecture (2-3) implies the vanishing of X 3 (K , ‫))2(ޚ‬ and hence the group X 2 (K , ‫)ޚ‬ is divisible.By Shapiro's lemma, the same holds for the group X 2 (K , ‫[ޚ‬E/K ]) for every étale K -algebra E.

On the C 2 2 -property for p-adic function fields
The goal of this section is to prove the following theorem: Theorem 3.1.Let l/k be a finite unramified extension and set L := l K .Let Z be a proper K -variety.
Then the quotient Here, χ K (Z , E) denotes the Euler characteristic of E over Z .Main Theorem 1 can be deduced as a very particular case of Theorem 3.1, in which this characteristic is trivial.We explain this at the end of the section.

3A1.
Step 0: Interpreting norms in Milnor K -theory in terms of motivic cohomology.The following lemma, which will be extensively used in the sequel, allows to interpret quotients of K 2 (K ) by norm subgroups as twisted motivic cohomology groups.
Lemma 3.2.Let L be a field and let L 1 , . . ., L r be finite separable extensions of L. Consider the étale L-algebra E := r i=1 L i and let Ť be the Galois module defined by the following exact sequence Proof.Exact sequence (3-1) induces a distinguished triangle By taking cohomology, we get an exact sequence Moreover, as recalled in Section 2, the Nesterenko-Suslin-Totaro theorem and the Beilinson-Lichtenbaum conjecture give the following isomorphisms: We therefore get an exact sequence in which the first map is the product of the norms.□ 3A2.
Step 1: Reducing to curves with residual index 1.In this step, we prove the following proposition, that allows to reduce to the case when the curve C has residual index 1.
Proof.Consider the Galois module Ť defined by the following exact sequence and hence to a distinguished triangle By the Beilinson-Lichtenbaum conjecture, the group H 3 (K ′ , ‫))2(ޚ‬ is trivial.Hence we get an inclusion where C is a fixed regular, proper and flat model of C whose reduced special fiber C 0 is a strict normal crossing divisor.Now, the distinguished triangle and the vanishing of the groups H 3 (K , ‫))2(ޑ‬ and H 4 (K , ‫))2(ޑ‬ = 0 (which follow from Lemma 2.5 and Theorem 2.6.c of [Kahn 2012]) give rise to an isomorphism and by Proposition 5.2 of [Kato 1986], the group on the left is trivial, and hence so is the former group.Now observe that, by Lemma 3.2, we have We claim that the extension K ′ /K totally splits at each place v ∈ C (1) .From this, we deduce that and hence the norm morphism It remains to check the claim.It is obviously satisfied for v ∈ C (1) , so we may and do assume v ∈ C (1)  \ C (1) .If κ and κ ′ denote the residue fields of k and k ′ , we then have to prove that all the irreducible components of C 0 are κ ′ -curves.To do so, consider an infinite sequence of finite unramified field extensions Hence the same is true for all the irreducible components of C 0 .But recall that, by the Lang-Weil estimates, any smooth geometrically integral variety defined over a finite field has a zero-cycle of degree 1.We deduce that the irreducible components of C 0 are κ ′ -curves.□

3A3.
Step 2: Solving the problem locally.In this step, we prove that the analogous statement to Theorem 3.1 over the completions of K holds.For that purpose, we first need to settle a simple lemma.
Lemma 3.4.Let l/k be a finite extension and set K 0 := k((t)) and L 0 := l((t)).The residue map Proof.We have the following commutative diagram from (2-1): For that purpose, recall that we have a commutative diagram with exact lines: But the map N l/k : K 2 (l) → K 2 (k) is surjective since p-adic fields have the C 2 0 -property, and the map ) is also surjective, as wished.□ Proposition 3.5.Let l/k be a finite unramified extension and set K 0 := k((t)) and L 0 := l((t)).Let Z be a proper K 0 -variety.Then the quotient Proof.For each proper K 0 -scheme Z , we denote by n Z the exponent of the quotient group We say that Z satisfies property (P) if it has a model over O K 0 that is irreducible, regular, proper and flat.To prove the proposition, it suffices to check assumptions (1), ( 2) and (3) of Proposition 2.1 of [Wittenberg 2015].
Assumption (1) is obvious.Assumption (3) is a direct consequence of Gabber and de Jong's theorem (Theorem 3 of the introduction of [Illusie et al. 2014]).It remains to check assumption (2).For that purpose, we proceed in the same way as in the proof of Theorem 4.2 of [Wittenberg 2015].Indeed, consider a proper K 0 -scheme X together with a model X that is irreducible, regular, proper and flat and denote by Y its special fiber.Let m be the multiplicity of Y and let D be the effective divisor on X such that Y = m D.
The residue map induces an exact sequence Moreover: (a) Since k satisfies the C 2 0 property, the proof of Lemma 4.4 of [Wittenberg 2015] still holds in our context, and hence the group (c) By Corollary 5.4 of [Wittenberg 2015] applied to the proper k-scheme D ⊔ Spec(l), the group By using exact sequence (3-2), facts (b) and (c) and Lemma 3.4, we deduce that Hence, by fact (a), we get Esnault et al. 2015], and hence the quotient Step 3: Globalizing local field extensions.In the rest of the proof, we will show how one can deduce the global Theorem 3.1 from the local Proposition 3.5.For that purpose, we first need to find a suitable way to globalize local extensions: more precisely, given a place w ∈ C (1) and a finite extension M (w) of K w such that Z (M (w) ) ̸ = ∅, we want to find a suitable finite extension M of K that can be seen as a subfield of M (w) and such that Z (M) ̸ = ∅.For technical reasons related to the failure of Chebotarev's theorem over the field K , we also need M to be linearly disjoint from a given finite extension of K .The following proposition is the key statement allowing to do that.
Proposition 3.6.Let Z be a smooth geometrically integral K -variety.Let T be a finite subset of C (1) .Fix a finite extension L of K and, for each w ∈ T , a finite extension M (w) of K w such that Z (M (w) ) ̸ = ∅.
Then there exists a finite extension M of K satisfying the following properties: (ii) For each w ∈ T , there exists a K -embedding M → M (w) .
(iii) The extensions L/K and M/K are linearly disjoint.
Proof.Before starting the proof, we introduce the following notations for each w ∈ T : so that the integer n := n (w) m (w) is independent of w.We now proceed in three substeps.
Substep 1.By Proposition 4.9 in Chapter I of [Hartshorne 1977], there exists a projective hypersurface Z ′ in ‫ސ‬ m K given by a nonzero equation that is birationally equivalent to Z .Let U and U ′ be nonempty open sub-schemes of Z and Z ′ that are isomorphic.Up to reordering the variables and shrinking U ′ , we may and do assume that the polynomial ∂ f /∂ x 0 is nonzero and that Given an element w ∈ T , the variety Z is smooth, Z (M (w) ) ̸ = ∅ and M (w) is large; for the definition of this notion, please refer to [Pop 2014].Hence the sets U (M (w) ) and U ′ (M (w) ) are nonempty.We can therefore find a nontrivial solution (y (w)  0 , . . ., Substep 2. Given w ∈ T , there exist m (w) elements α 1 , . . ., α m (w) ∈ M (w) whose respective minimal polynomials µ α 1 , . . ., µ α m (w) are pairwise distinct and such that M (w) = K w (α i ) for each 1 ≤ i ≤ m (w) .Recalling that n = n (w) m (w) , introduce the degree n monic polynomial µ (w)  := m (w) i=1 µ α i and consider the set H of n-tuples (a 0 , . . ., a n−1 ) ∈ K n such that the polynomial T n + n−1 i=0 a i T i is irreducible over L. By Corollary 12.2.3 of [Fried and Jarden 2008], the set H contains a Hilbertian subset of K n , and hence, according to Proposition 19.7 of [Jarden 1991], if we fix some ϵ > 0, we can find an n-tuple (b 0 , . . ., b n−1 ) in H such that the polynomial µ : If ϵ is chosen small enough, then there exists a K -embedding w) for each w ∈ T by Krasner's lemma; see Lemma 8.1.6 in [Neukirch et al. 2008].Moreover, since (b 0 , . . ., b n−1 ) ∈ H , the polynomial µ is irreducible over L, and hence the extensions K ′ /K and L/K are linearly disjoint.is coefficient-wise ϵ-close to g (w) for each w ∈ T .Introduce the field M := K ′ [T ]/(g(T )).We check that M satisfies the conditions of the proposition, provided that ϵ is chosen small enough: (i) Fix w ∈ T .By Substep 1, the m-tuple (y (w) 0 , . . ., y (w) m ) lies in U ′ .Hence, for ϵ small enough, if y 0,w stands for the root of g that is closest to y (w)  0 , then the m-tuple (y 0,w , y 1 , . . ., y m ) lies in U ′ .We deduce that U ′ (M) ̸ = ∅, and hence Z (M) ̸ = ∅.
(ii) For each w ∈ T , the polynomial g (w) has a simple root in M (w) , and hence so does g(T ) if ϵ is chosen small enough, again by Krasner's Lemma.The field M can therefore be seen as a subfield of M (w) .
(iii) Since (y 1 , . . ., y m ) ∈ H ′ , the polynomial g(T ) is irreducible over L K ′ .Hence the extensions M/K ′ and L K ′ /K ′ are linearly disjoint.Moreover, by Substep 2, the extensions K ′ /K and L/K are linearly disjoint.We deduce that L/K and M/K are linearly disjoint.□

3A5.
Step 4: Computation of a Tate-Shafarevich group.This step, which is quite technical, consists in computing the Tate-Shafarevich groups of some finitely generated free Galois modules over K .Recall that for each abelian group A, we denote by A the quotient of A by its maximal divisible subgroup.
Proposition 3.7.Let r ≥ 2 be an integer and let L , K 1 , . . ., K r be finite extensions of K contained in K .Consider the composite fields K I := K 1 . . .K r and K î := K 1 . . .K i−1 K i+1 . . .K r for each i, and denote by n the degree of L/K .Consider the Galois module T defined by the following exact sequence Given two positive integers m and m ′ , make the following assumptions: (LD1) The Galois closure of L/K and the extension K I /K are linearly disjoint.
(H1) The restriction map (H2) The restriction map is surjective and its kernel is m-torsion.
(H3) For each i, the restriction maps (H4) For each finite extension L ′ of L contained in the Galois closure of L/K , the kernel of the restriction map Recall that A denotes the quotient of A by its maximal divisible subgroup.
Remark 3.8.In the sequel of the article, we will only use the proposition in the case when L/K is Galois.However, this assumption does not simplify the proof.
Proof.Consider the following sequence: (3-4) It is obviously a complex, and the first arrow is injective by (H1).In order to give further information about the complex (3-4), let us consider the following commutative diagram, in which the first and second rows are obtained in the same way as the third: The second and third columns are exact since the exact sequence (3-3) splits over L, K I and L K I .Moreover, all the lines are complexes, and in the first one, the arrow g 1 is surjective since the restriction map is surjective by (H2).
The next two lemmas constitute the core of the proof of Proposition 3.7.
Lemma 3.9.Let a ∈ X 2 (K , T ) and b and g(b) = 0. Then mb K I comes by restriction from X 2 (K î , ‫[ޚ‬E/K ]) for each i.
Proof.Consider the following commutative diagram, constructed exactly in the same way as diagram (3-5): The last two columns are exact since the exact sequence (3-3) splits over L, K î and L K î , and the restriction morphism then lies in ker(g) ∩ ker(ψ 1 ) and a diagram chase in (3-5) shows that there exists c ∈ X 2 (K I , ‫)ޚ‬ such that By (H2), we have mc = 0, and hence m • (b Before proving the lemma, let us introduce some notation. Notation 3.11.(i) For each i, we can find a family (K i j ) j of finite extensions of K I together with embeddings σ i j : K i → K i j so that K i,1 = K I , the embedding σ i,1 is the natural embedding K i → K I , and the K -algebra homomorphism x ⊗ y → (σ i j (x)y) j is an isomorphism.We denote by σi j : K I → K i j the embedding obtained by tensoring σ i j with the identity of K î .This is well-defined by (LD2).
(ii) For each i, j, we can find a family (L i j j ′ ) j ′ of finite extensions of K i j together with embeddings σ i j j ′ : L → L i j j ′ so that the K -algebra homomorphism is an isomorphism.We denote by σi j j ′ : L K i → L i j j ′ the embedding obtained by tensoring σ i j j ′ with σ i j .
Observe that, when j = 1, the K -algebra homomorphism (3-6) is simply the isomorphism L ⊗ K K I ∼ = L K I , so that σ i,1,1 is none other than the inclusion of L in L K I .
(iii) We can find a family of finite extensions (L α ) α of L together with embeddings τ α : L → L α so that L 1 = L, the embedding τ 1 is the identity of L, and the K -algebra homomorphism is an isomorphism.For each α, we denote by τα : L K I → L α K I the embedding obtained by tensoring τ α with the identity of K I .This is well-defined by (LD1).
Since f 0 is injective, we get µα = µψ 0 ((x α ) α , (y i ) i , z, (t i j ) i, j ), which finishes the proof of the lemma.□ We can now finish the proof of Proposition 3.7.As recalled at the end of Section 2, the group ) is divisible and hence, by Lemma 3.10, In other words, the group On the other hand, using once again the end of Section 2, the group X 2 (L , T ) vanishes.Hence, by restriction-corestriction, X 2 (K , T ) is n-torsion.We deduce that X The following lemma will often allow us to check assumptions (H2) and (H3) of Proposition 3.7: Lemma 3.12.Let l be a finite unramified extension of k of degree n and set L = l K .The restriction map Proof.By restriction-corestriction, ker(Res L/K ) is killed by n.Moreover, since X 2 (K , ‫)ޚ‬ = X 1 (K , ‫,)ޚ/ޑ‬ an element in ker(Res L/K ) corresponds to a subextension K ⊂ L ′ ⊂ L that is locally trivial at every closed point of the curve C. Since L = l K , we can find an extension k ⊂ l ′ ⊂ l such that L ′ = l ′ K .By the local triviality of L ′ /K , the field l ′ has to be contained in the residue field of k(v) for every v ∈ C (1) .In particular, [l ′ : k] and [L ′ : K ] divide i res (C).This shows that ker(Res L/K ) is killed by i res (C), and hence by i res (C) ∧ n.
In order to prove the surjectivity statement, consider an integral, regular, projective model C of C such that its reduced special fiber C 0 is an SNC divisor.Let c be the genus of the reduction graph of C. According to Corollary 2.9 of [Kato 1986], for each m ≥ 1, we have an isomorphism Hence, by Poitou-Tate duality, we also have and the finiteness of the exponent of ker(Res L/K ).□

3A6.
Step 5: Proof of Theorem 3.1 for smooth proper varieties.In this step, we use Poitou-Tate duality to deduce Theorem 3.1 for smooth proper varieties from the previous steps.
Theorem 3.13.Let l/k be a finite unramified extension and set L := l K .Let Z be a smooth proper integral K -variety.Then the quotient Proof.Take x ∈ K 2 (K ).We want to prove that First observe that, if K ′ stands for the algebraic closure of K in the function field of Z , then Z has a structure of a smooth proper K ′ -variety and that χ . Therefore, by restriction-corestriction, we can assume that K = K ′ , and hence that Z is geometrically integral.Moreover, by Proposition 3.3, we may and do assume that C has residual index 1.
Let now S be the (finite) set of places v ∈ C (1) such that ∂ v x ̸ = 0. Given a prime number ℓ, since the curve C has residual index 1 and the field k is large, we can find some point w ℓ ∈ C (1)  \ S such that the residual degree [k(w ℓ ) : k] res of k(w ℓ )/k is prime to ℓ.Moreover, by Proposition 3.5, we have (3-16) Before moving further, we need to prove the following lemma: Lemma 3.14.Let n = [l : k] with l/k as in Theorem 3.13.If v ℓ (n) > v ℓ (χ K (Z , E)), then there exists a finite extension M (w ℓ ) of K w ℓ with residue field m (w ℓ ) such that Z (M (w ℓ ) ) ̸ = ∅ and v ℓ ([m (w ℓ )  : k(w ℓ )] res ) ≤ v ℓ (χ K (Z , E)).
(iv) For each pair (v 0 , i 0 ), the field K (v 0 ) i 0 is linearly disjoint to the composite field over K , where L n stands for the composite of all cyclic extensions of L that are locally trivial everywhere and whose degrees divide n.Note that L n is a finite extension of L since X 1 (L , ‫)ޚ‪/n‬ޚ‬ is finite.
Consider the Galois module T defined by the following exact sequence: where To conclude, we introduce the composite field i and we check the assumptions (LD1), (LD2), (H1), (H2), (H3) and (H4) of Proposition 3.7 with m = χ K (Z , E) and The extension L/K is obviously Galois.The fields L and K I are linearly disjoint over K by (iv).
(H1) By proceeding exactly in the same way as in Lemma 4 of [Demarche and Wei 2014], since we already have (LD1), one gets the injectivity of the restriction map and hence of (H2) Let C I be the smooth projective k-curve with fraction field K I .On the one hand, by (iii), given a prime ℓ such that v ℓ (n) > v ℓ (χ K (Z , E)), the field K I can be seen as a subfield of M (w ℓ ) and the inequality v ℓ ([m (w ℓ ) : k] res ) ≤ v ℓ (χ K (Z , E)) holds by Lemma 3.14.We deduce that v ℓ (i res (C I )) ≤ v ℓ (χ K (Z , E)) for such ℓ.On the other hand, for any other prime number ℓ, we have v ℓ (n) ≤ v ℓ (χ K (Z , E)).We deduce that i res (C I ) ∧ n divides m = χ K (Z , E), and hence (H2) follows from Lemma 3.12.
(H4) Since L/K is Galois, (H4) immediately follows from the choice of m ′ .By Proposition 3.7, we deduce that the group X 2 (K , T ) is ((m ∨ m ′ ) ∧ n)-torsion.But by (iv), the fields K I and L n are linearly disjoint over K , and hence, by the definition of m ′ , we have m ′ ∧ n = 1, so that (m ∨ m ′ ) ∧ n = m ∧ n.The group X 2 (K , T ) is therefore m-torsion.If we set Ť := Hom( T , ‫)ޚ‬ and T := Ť ⊗ ‫,)2(ޚ‬ that is also the case of X 3 (K , T ) according to Poitou-Tate duality.Now, by Lemma 3.2, we may interpret x as an element of H 3 (K , T ).Equations (3-18) and (3-19) together with assertion (ii) imply that mx ∈ X 3 (K , T ), which is m-torsion.Thus m 2 x = 0 ∈ X 3 (K , T ).
3B. Proof of Main Theorem 1.We can now deduce Main Theorem 1 from Theorem 3.1.≤ n, we deduce that there exists a finite unramified extension l of k such that Z (l K ) ̸ = ∅.By Theorem 3.1, the quotient

Proof of
and hence the exact sequence where i : Z → ‫ސ‬ n K stands for the closed immersion, gives 4. On the C 2 1 property for p-adic function fields The goal of this section is to prove Main Theorem 2. Contrary to Main Theorem 1, for which we needed to deal with unramified extensions of k, here we will have to deal with ramified extensions of k.For that purpose, the key statement is given by the following theorem: Theorem 4.1.Assume that C has a rational point, let ℓ be a prime number, and fix a finite Galois totally ramified extension l/k of degree ℓ.Let E 0 l/k be the set of all finite ramified subextensions of l nr /k and let E l/k be the set of finite extensions K ′ of K of the form K Note that, given any two extensions k ′ and k ′′ in E 0 l/k with k ′ ⊂ k ′′ , the extension k ′′ /k ′ is unramified.This observation will be often used in the sequel.
Remark 4.2.We think that the assumption that C has a rational point in Theorem 4.1 cannot be removed.
To check that, we invite the reader to assume that i ram (C) = ℓ.Then, given an integer n ≥ 1, consider the set E 0 n whose elements are extensions of k in E 0 l/k that are contained in the composite l n := lk n , where k n is the degree ℓ n unramified extension of k.Define the set E n of finite extensions K ′ of K contained in L n := l n K that are of the form K ′ = k ′ K for some k ′ ∈ E 0 n and consider the Galois module Tn defined by the exact sequence By following the proof of Proposition 4.5, one can check that, if K 1 and K 2 are two distinct degree ℓ extensions of K in E n , then the Tate-Shafarevich group X 2 (K , Tn ) is the direct sum of the kernel of the map and of a divisible group, given by the kernel of the map In particular The computation of this kernel is a relatively simple (but a bit technical) exercise in the cohomology of finite groups, since it is contained in the group ker In that way, one checks that X 2 (K , Tn ) is an ‫ކ‬ ℓ -vector space of dimension at least nℓ − n − 1.Moreover, the computation being very explicit, one can even check that the morphism X 2 (K , Tn+1 ) → X 2 (K , Tn ) induced by the natural projection Tn+1 → Tn is always surjective.But then, by dualizing thanks to Poitou-Tate duality, this shows that the groups are all nontrivial and that the natural maps Q n → Q n+1 are all injective.We deduce that the nontrivial elements of Q 1 provide nontrivial elements in the quotient 4A. Proof of Theorem 4.1.

4A1.
Step 1: Solving the local problem.The first step to prove Theorem 4.1 consists in settling an analogous statement over the completions of K .We start with the following lemma.
Lemma 4.3.Let ℓ be a prime number and let l/k be a finite Galois totally ramified extension of degree ℓ.
Let m/k be a totally ramified extension such that ml/m is unramified.Then there exists k Proof.If ml/m is trivial, then m contains l and we are done.Therefore we may and do assume that ml/m has degree ℓ.Denote by k ℓ the unramified extension of k with degree ℓ and set l ℓ := l • k ℓ .The extension l ℓ /k is Galois with Galois group ‫)ޚ‪/ℓ‬ޚ(‬ 2 , and since ml is unramified of degree ℓ over m, it contains both k ℓ and l ℓ , so that l ℓ is contained in m ′ l for some finite subextension m ′ of m/k.But Hence the intersection k ′ := m ′ ∩ l ℓ is a degree ℓ totally ramified extension of k, and k ′ ∈ E 0 l/k .□ Proposition 4.4.Let ℓ be a prime number and let l/k be a finite Galois totally ramified extension of degree ℓ.Fix v ∈ C (1) .Then Proof.Three different cases arise: (1) The field k(v) contains l.
Case 1 is trivial, since Let us now consider case 2, and denote by k(v) nr the maximal unramified subextension of k(v)/k.By Lemma 4.3, since lk(v) nr /k(v) nr is a Galois totally ramified extension of degree ℓ and k(v)/k(v) nr is a totally ramified extension such that k(v)l/k(v) is unramified, there exists a finite extension m of k(v) nr such that m ∈ E 0 lk(v) nr /k(v) nr ⊂ E 0 l/k and m ⊂ k(v).By setting M := m K , we get that M ∈ E l/k and that as wished.
Let us finally consider case 3. To do so, fix a uniformizer π of k(v), and as before, let k(v) nr be the maximal unramified subextension of k(v)/k.Denote by k(v) ram π the maximal abelian totally ramified extension of k(v) associated to π by Lubin-Tate theory.Since l/k is abelian, the extension lk(v) ram π /k(v) ram π must be unramified.Hence, by Lemma 4.3, there exists a finite extension m of k(v) nr such that m ∈ E 0 lk(v) nr /k(v) nr ⊂ E 0 l/k and m ⊂ k(v) ram π .We deduce from Corollary 5.12 of [Yoshida 2008] that This being true for every uniformizer π of k(v), we deduce that and hence, by Lemma 3.4, Step 2: Computation of a Tate-Shafarevich group.The second step, which is slightly technical, consists in computing the Tate-Shafarevich groups of some finitely generated free Galois modules over K associated to the fields in E l .Poitou-Tate duality will then allow us to obtain a local-global principle that will let us deduce Theorem 4.1 from Proposition 4.4.
Proposition 4.5.Assume that C has a rational point, and let ℓ be a prime number.Fix a finite Galois totally ramified extension l/k of degree ℓ.Given K 1 , . . ., K r in E l/k so that the fields K 1 and K 2 are linearly disjoint over K , consider the Galois module T defined by the following exact sequence 0 → ‫ޚ‬ → ‫[ޚ‬E/K ] → T → 0, (4-1) where E := K 1 × • • • × K r .Then X 2 (K , T ) is divisible.
Proof.Consider the following complex: (4-2) We start by proving the following lemma: Lemma 4.6.The morphism f 0 is injective.
Proof.Let K I be the Galois closure of the composite of all the K i 's.By inflation-restriction, there is an exact sequence 0 → H 2 (K I /K , T ) → H 2 (K , T ) → H 2 (K I , T )./ / H 2 (K , T ) / / H 2 (K I , T ) in which the first vertical map is an isomorphism.We deduce that the restriction map ker H 2 (K , T ) → H 2 (K v , T ) → ker H 2 (K I , T ) → H 2 (K I,v , T ) is injective.Hence so is the restriction map Res K I /K : X 2 (K , T ) → X 2 (K I , T ) as well as the restriction maps Res K 1 /K : X 2 (K , T ) → X 2 (K 1 , T ), Res K 2 /K X 2 (K , T ) → X 2 (K 2 , T ), since the former factors through these.□ by the multiplicity m of the special fiber Y of X .(b) The proof of Lemma 4.5 of [Wittenberg 2015] also holds in our context, and hence ∂

Substep 3 .
According to Substep 1, for each w ∈ T , y (w) 0 is a simple root of the polynomialg (w) (T ) := f (T, y (w) 1 , . . ., y (w) m ).Let H ′ be the set of m-tuples (z 1 , . . ., z m ) in K ′ such that f (T, z 1 , . . ., z m ) is irreducible over L K ′ .By Corollary 12.2.3 of [Fried and Jarden 2008], the set H ′ contains a Hilbertian subset of K ′m .Hence, by Proposition 19.7 of [Jarden 1991], we can find (y 1 , . . ., y m ) in H ′ such that the polynomial g(T ) := f (T, y 1 , . . ., y m ) Take v ∈ C(k) a rational point.Since the extension K I /K is obtained by base change from an extension k I of k, we have the equalities Gal(K I /K ) = Gal(k I /k) = Gal(K I,v /K v ).The previous inflation-restriction exact sequence therefore induces a commutative diagram with exact lines:0 / / H 2 (K I /K , T ) ∼ = Main Theorem 1. Fix two integers n, d ≥ 1 such that d 2 ≤ n and a hypersurface Z in ‫ސ‬ n K of degree d.By Lang's and Tsen's theorems (Theorem 2a of [Nagata 1957] and Theorem 12 of [Lang 1952]), the field k nr (C) is C 2 .Since d 2