RECONSTRUCTING FUNCTION FIELDS FROM MILNOR K-THEORY
ANNA CADORET AND ALENA PIRUTKA
Abstract. LetF be a finitely generated regular field extension of transcendence degree≥2 over a perfect field k. We show that the multiplicative group F×/k× endowed with the equivalence relation induced by algebraic dependence onF overk determines the isomorphism class ofF in a functorial way. As a special case of this result, we obtain that the isomorphism class of the graded Milnor K-ring K∗M(F) determines the isomorphism class ofF, when k is algebraically closed or finite.
1. Introduction This paper is motivated by the following general question.
Question 1. Does the Milnor K-ring K∗M(F) determine the isomorphism class of the field F? One may also ask whether or not the above holds in a functorial way, that is, whether or not the Mil- nor K-ring functor is fully faithfull from the groupoid1of fields to the groupoid ofZ≥0-graded rings.
Question 1 has a negative answer in general. For instance2, all solvably closed subfields ofQ have isomorphic Milnor K-rings. Indeed, letF1, F2 be two such fields. To show thatK∗M(F1)'K∗M(F2) it is enough to show that KjM(F1) ' KjM(F2), j = 1,2 compatibly with the pairings {−,−} : K1M(Fi)⊗K1M(Fi) →K2M(Fi), i= 1,2 (See Remark 12). Since every polynomial Tn−x∈Fi[T]
is totally split by assumption, K2M(Fi) is uniquely divisible torsion-free [2, I, (1.2)]. On the other hand, since K2M(Fi) is a quotient of the inductive limit of the K2M(F), for Q⊂F ⊂Fi describing the set of all finite field subextensions ofFi|Qand thatK2M(F) is an extension of a torsion abelian group by a finitely generated abelian group [2, II, (1.2)],K2M(Fi) is trivial. So it is enough to show thatF1× 'F2×. SinceFi×is divisible countable,Fi×'(Fi×)tors⊕QZ and the assertion follows from (F1×)tors'(F2×)tors'(Q×)tors.
Also, even when restricted to a class of fields where Question 1 has a positive answer, the naive functorial version of Question 1 still has a negative answer in general, as shown by the example of finite fields F where K∗M(F) has extra automorphisms induced by x→ xu for u prime to |F| −1 on K1M(F). To get a viable functorial version of Question 1, one should at least kill such extra automorphisms.
Since K1M(F) =F×, Question 1 essentially reduces to reconstructing the additive structure of F from the multiplicative groupF×endowed with additional data that can be detected by the Milnor
Date: Article submitted on 2018/09/18. Reports received on 2019/07/19. Article resubmitted on 2019/08/28.
1Here, given a categoryC, the groupoid ofCmeans the category with the same objects asCand with morphisms the isomorphisms inC.
2This counter-example was suggested to us by one of the Referees.
1
K-ring. Our main result (Theorem 4) asserts that for a finitely generated regular field extensionF of transcendence degree ≥2 over a perfect field k, the multiplicative group F×/k× endowed with the equivalence relation induced by algebraic dependence onF overkdetermines the isomorphism class of F in a functorial way. In Section 2, we show (Theorem 7) that for a finitely generated regular field extension of a field k which is either algebraically closed or finite, the Milnor K-ring detects algebraic dependence. This is a consequence of deep K-theoretic results - the n = 2 case of the Bloch-Kato conjecture [12] when k is algebraically closed and of the Bass-Tate conjecture [15] when k is finite. Combined with Theorem 4, this enables us to show that the Milnor K-ring modulo the ideal of divisible elements (resp. of torsion elements) determines in a functorial way finitely generated regular field extensions of transcendence degree ≥ 2 over algebraically closed fields (resp. over finite fields) (see Corollary 10). In particular, this provides a purely K-theoretic description of the group of birational automorphisms of normal projective varieties of dimension
≥2 over algebraically closed or finite fields (Corollary 11).
1.1. Main Result. Recall that a field extensionF|kisregularifkis algebraically closed inF and F|k has a separating transcendence basis. If k is perfect, the latter condition is automatic. Let F|k be a regular field extension.
Definition 2. We say that x, y ∈ F×/k× are algebraically dependent and write x ≡ y if some (equivalenty, every) lifts x, y ∈ F× of x, y ∈ F×/k× are algebraically dependent over k. The relation≡is an equivalence relation on F×/k×.
Note that, as F|k is regular, if x, y∈F×/k× are algebraically dependent then either x=y= 1 or 16=x, y.
LetF|k,F0|k0 be regular field extensions.
Definition 3. We say that a group morphism ψ:F×/k× → F0 ×/k0 × preserves algebraic depen- dence if for everyx, y∈F×/k× the following holds: x≡y if and only ifψ(x)≡ψ(y).
(In particular, a group morphism preserving algebraic dependence is automatically injective).
For a subfield E ⊂ F, write EF ⊂ F for the algebraic closure of E in F. Then a group morphism ψ : F×/k× → F0 ×/k0 × preserves algebraic dependence if and only if k(x)F×/k× = ψ−1(k0(ψ(x))F0/k0 ×) for everyx∈F and some (equivalently, every) liftψ(x)∈F0 of ψ(x).
Let Isom(F, F0) denote the set of field isomorphisms F→F˜ 0 and Isom(F|k, F0|k0)⊂Isom(F, F0)
denote the subset of isomorphisms F→F˜ 0 inducing field isomorphisms k→k˜ 0.
Let Isom(F×/k×, F0 ×/k0 ×) denote the set of group isomorphismsF×/k×→F˜ 0 ×/k0 × and Isom≡(F×/k×, F0 ×/k0 ×)⊂Isom(F×/k×, F0 ×/k0 ×)
the subset of isomorphisms F×/k×→F˜ 0 ×/k0 × preserving algebraic dependence. The group Z/2 acts on the set Isom≡(F×/k×, F0 ×/k0 ×) by ψ→ψ−1. Write
Isom≡(F×/k×, F0 ×/k0 ×) for the resulting quotient.
Theorem 4. Letk, k0 be perfect fields of the same characteristicp≥0, and letF|k,F0|k0 be finitely generated regular field extensions and assume one of them has transcendence degree≥2. Then the canonical map
Isom(F|k, F0|k0)→Isom≡(F×/k×, F0 ×/k0 ×) is bijective.
1.2. Comparison with existing results. Question 1 was considered by Bogomolov and Tschinkel in [3], where they prove (a variant of) Theorem 4 for finitely generated regular extensions of charac- teristic 0 fields ([3, Thm. 2]) and deduce from it Corollary 10 for finitely generated field extensions of algebraically closed fields of characteristic 0 ([3, Thm. 4]).
Variants of our results were also obtained by Topaz from a smaller amount of K-theoretic infor- mation - mod-`MilnorK-rings (for finitely generated field extensions of transcendence degree≥5 over algebraically closed field of characteristic p 6= ` [16, Thm. B]) and rational Milnor K-rings (for finitely generated field extensions of transcendence degree ≥2 over algebraically closed field of characteristic 0 [17, Thm. 6.1]) but enriched with the additional data of the so-called “rational quotients” of F|k. See also [17, Rem. 6.2] for some cases where the additional data of rational quotients can be removed.
Our strategy follows the one of Bogomolov and Tschinkel in [3], where the key idea is to param- etrize lines in F×/k× as intersections of multiplicatively shifted (infinite dimensional) projective subspaces of a specific form arising from relatively algebraically closed subextensions of transcen- dence degree 1. See Subsection 1.3 for details. The strategy of Topaz is more sophisticated and goes through the reconstruction of the quasi-divisorial valuations of F via avatars of the theory of commuting-liftable pairs as developped in the framework of birational anabelian geometry. Though not explicitly stated in the literature, it is likely that Theorem 4 and Corollary 10 for finitely gen- erated field extensions of algebraically closed fields of characteristicp >0 could also be recovered from the techniques of birational anabelian geometry as developed by Bogomolov-Tschinkel [4], Pop (e.g. [14], [13]) and Topaz.
To our knowledge, Theorem 4 for finitely generated regular extensions of perfect fields of charac- teristic p >0 and Corollary 10 for finitely generated field extensions of finite fields are new.
1.3. Strategy of proof. For simplicity, write Fp ⊂F for the subfield generated bykand the xp, x∈F andF×/p:=F×/Fp×.
The proof of Theorem 4 is carried out in Section 3. According to the fundamental theorem of projec- tive geometry (see Lemma 29, for which we give a self-contained proof in the setting of possibly infi- nite field extensions), it would be enough to show that a group isomorphismψ:F×/k×→F˜ 0 ×/k0 × preserving algebraic dependence induces a bijection from lines inF×/k×to lines inF0 ×/k0 ×. This would reduce the problem to describing lines in F×/k× using only ≡and the multiplicative struc- ture ofF×/k×. This classical approach works well ifp= 0. The key observation of Bogomolov and Tschinkel in [3] is that every line can be multiplicatively shifted to a line passing through a “good”
pair of points and that those lines can be uniquely parametrized as intersections of multiplica- tively shifted (infinite dimensional) projective subspaces of a specific form arising from relatively algebraically closed subextensions of transcendence degree 1 [3, Thm. 22]. This is the output of elaborate computations in [3]. Later, Rovinsky suggested an alternative argument using differential
forms; this is sketched in [4, Prop. 9].
When p > 0, the situation is more involved. The original computations of [3] fail due to insep- arability phenomena. Instead, we adjust the notion of “good” for the pair of points (Definition 19) in order to refine the argument of Rovinsky. In particular, we use the field-theoretic notion of “regular” element rather than the group-theoretic notion of “primitive” element used in [3]. To show that every line can be shifted to a line whose image contains a “good” pair of points (Lemma 20), one can invoke Bertini theorems ([7, Cor. 6.11.3] whenk is infinite and [5, Thm. 1.6] whenk is finite); we also give an alternative, more elementary argument due to Akio Tamagawa in Remark 22. This reduces the problem to show thatψ(orψ−1) maps every line inF×/k× whose image con- tains a good pair of points (x1, x2) isomorphically to a line in F0×/k0×. Actually, we cannot prove this directly when p >0. The issue is that, whenp >0, the Bogomolov-Tschinkel parametrization of such line by the setI(x1, x2) introduced in Subsection 3.2 is much rougher than in [3, Thm. 22].
More precisely, when p >0, the setI(x1, x2) only recovers the line passing through (x1, x2) up to prime-to-p powers and certain affine transformations with Fp-coefficients (Lemma 23); this is due to the apparition of constants inFp when one integrates differentials forms. Lemma 23 is however enough to show that there exists a uniquem∈Znormalized as
(1.1) |m|= 1 ifp= 0,2
1≤ |m| ≤ p−12 ifp >2;
such thatψm induces a bijection from lines inF×/pto lines inF0 ×/p(Proposition 27). So Lemma 29 gives a unique field isomorphism φ : F→F˜ 0 such that the resulting isomorphism of groups φ : F×→F˜ 0 × coincides with ψm on F×/p. This concludes the proof if p = 0. But if p > 0, the extensionF/Fpis much smaller (finite-dimensional!) and one has to perform an additional descent step (Section 3.6) to show thatm=±1 andφ coincides withψ±1 onF×/k× (not only onF×/p).
We limited our exposition to function fields, which are those of central interest in algebraic geometry.
However, some of our results extend to more exotic fields provided they behave like function fields.
For instance, Theorem 4 works for the class of regular field extensions F of transcendence degree
≥2 over a perfect fieldk such that for every subfieldk⊂E ⊂F of transcendence degree 2 overk, the algebraic closure of E inF is a finite extension of E. We do not elaborate on this.
1.4. Acknowledgements. We would like to thank heartfully the referees for their thorough and constructive reports. They pointed out several minor but mathematical inacurracies and made suggestions that helped improve a lot the exposition. The first author was partially supported by the I.U.F. and the ANR grant ANR-15-CE40-0002-01. The second author was partially supported by NSF grant DMS-1601680 and by the Laboratory of Mirror Symmetry NRU HSE, RF Govern- ment grant, ag. no. 14.641.31.0001. This project was initiated at the occasion of a working group on geometric birational anabelian geometry at the Courant Institute of Mathematics, NYU. The authors would like to thank F. Bogomolov and Y. Tschinkel for inspiring exchanges, B. Kahn for his explanations concerning the Bass-Tate conjecture and A. Merkurjev, M. Morrow and A. Tamagawa for helpful remarks on a first version of this text. They also thank A. Tamagawa for suggesting the more elementary proof of Lemma 20 presented in Remark 22.
2. Milnor K-rings and algebraic dependence
For a field F, the Milnor K-ring K∗M(F) is the quotient of the Z≥0-graded ring T⊗∗(F×) :=
⊕n≥0(F×)⊗n (with the convention (F×)⊗0 =Z) by the two-sided homogeneous ideal generated by
the degree-2 elementsa⊗(1−a), 16=a∈F×. The map F →K∗M(F) defines a functor from the groupoidsF of fields to the groupoidA of associativeZ≥0-graded anti-commutative rings.
Given a field F and elements x1, . . . , xn ∈ F×, we write {x1, . . . , xn} ∈ KnM(F) for the image of x1⊗ · · · ⊗xn ∈ T⊗n(F×). Recall (e.g. [2, §4]) that given a discrete valuation v : F× → Z on F, with ring of integersOv and residue fieldk(v) there exists a unique group morphism∂v∗:K∗M(F)→ K∗−1M (k(v)) such that for everyx1∈F× andx2, . . . , xn∈ Ov× with imagesx2, . . . , xn ink(v),
∂vn({x1, x2, . . . , xn}) =±v(x1){x2, . . . , xn}=±{xv(x2 1), x3, . . . , xn}.
Forp= 0 or a prime, letFp ⊂ F denote the full subcategory of fields of characteristicp.
2.1. Some geometric observations. LetF|k be a finitely generated regular field extension.
2.1.1. Let x1, . . . , xn∈F be such thatk(x1, . . . , xn) has transcendence degree noverk. Then Lemma 5. There exists a divisorial valuationv:F×→Z of F such that
• v(x1)6= 0, v(x2) =· · ·=v(xn) = 0;
• the images x2, . . . , xn of x2, . . . , xn in the residue field k(v) are algebraically independent over k.
Proof. Fix a normal projective model X|k of F|k. Each xi defines a dominant rational function xi :X99KP1k such that
x= (x1, . . . , xn) :X 99K(P1k)n
is again dominant. Choose any open subscheme U ⊂ X over which the map x : U → (P1k)n is defined. Then, up to replacingX by the normalization of the Zariski closure of the graph ofx|U in X×(P1k)n, one may assume that the mapsxi,i= 1, . . . , nandxare defined overX, and surjective.
Choose an irreducible divisorD∈Div(X) with vD(x1)6= 0. Then (sincex(D) has codimension at most 1 in (P1k)n) the restrictionx|D :D→(P1k)nsurjects onto 0×(P1k)n−1 ⊂(P1k)n; in particular the elements xi :=xi|D ∈k(D),i= 2, . . . , n remain algebraically independent overk and vD(xi) = 0,
i= 2, . . . , n.
2.1.2. The abelian group F×/k× is a free abelian group since it embeds into a free abelian group.
This follows from the exact sequence
0→k×→F×→Div(X),
whereX|kis any normal projective model X|k ofF|k, and the fact thatDiv(X) is a free abelian group.
2.2. Type. We say that a field F is a function field over a fieldk (or with field of constantsk) if F is a finitely generated regular field extension of transcendence degree ≥ 1 over k. We say that a function field F overk is of type 1 (resp. of type 2) if k is algebraically closed (resp. finite), in which casek is uniquely determined byF (see Lemma 6 below).
LetF ⊂ Fe ,Fep⊂ Fp denote the full subcategories of function fields which are of types 1 and 2.
Lemma 6. For F ∈ Fe, the type, the multiplicative group k× of the field of constants and the characteristic p of F are determined byF×=K1M(F) as follows.
Type Torsion subgroup k× p of F×
1 Infinite Divisible 0 or uniquep such that
subgroup of F× (−)p :k×→k× is an isomorphism
2 Finite Torsion
subgroup of F× Unique p such that log(|k×|+ 1)∈Zlog(p) In the following, we sometimes write F|k ∈Fe instead of F ∈Fe implicitly meaning that k is the field of constants ofF.
2.3. Detecting algebraic dependence. ForF ∈ Fp and a prime`6=p, let DK∗M(F)⊂K∗M(F) (resp. T K∗M(F)⊂K∗M(F)) denote the (two-sided) ideal of elements which are infinitely`-divisible (resp. torsion) with respect to the Z-module structure onK∗M(F) and write
K1,M∗ (F) :=K∗M(F)/DK∗M(F);
K2,M∗ (F) :=K∗M(F)/T K∗M(F).
Then K∗M →Ki,M∗ is a morphism of functors fromF toA,i= 1,2.
Theorem 7. For F ∈Fep of type iand every x1, . . . , xn∈F× consider the following assertions.
(a) {x1, . . . , xn}= 0 in Ki,Mn (F);
(b) the transcendence degree of k(x1, . . . , xn) over k is ≤n−1.
Then, (a) ⇒ (b) and (b) ⇒ (a) if i= 1 or if i= 2 andn≤2.
Proof. The assertion forn= 1 follows from 2.1.2.
(a) ⇒ (b): We proceed by induction on n. Assume k(x1, . . . , xn) has transcendence degree n over k and choose a place v of F as in Lemma 5. By induction hypothesis ∂vn({x1, . . . , xn}) =
±{xv(x2 1), x3, . . . , xn} 6= 0 inKi,Mn−1(k(v)) hence,a fortiori,{x1, . . . , xn} 6= 0 inKi,Mn (F).
(b)⇒(a): AssumeE =k(x1, . . . , xn) has transcendence degreed≤n−1 overk. Since{x1, . . . , xn} lies in the image of the restriction mapKi,Mn (E)→Ki,Mn (F), it is enough to show thatKi,Mn (E) = 0.
If F is of type 1, this follows from Tsen’s theorem [9], which ensures that the ´etale cohomology group Hn(E,Z`(n)) is trivial and from the Bloch-Kato conjecture [18, 19]. IfF is of type 2, this follows from the n= 2 case of the Bass-Tate conjecture [15, Thm. 1].
LetF, F0 ∈Fep of the same typei. For a morphismψ∗:Ki,M∗ (F)→Ki,M∗ (F0) ofZ≥0-graded rings and integer n≥1, consider the assertions:
(⊂, n) For every x1, . . . , xn∈F×,ψ1(k(x1, . . . , xn)F×/k×) ⊂ k0(ψ1(x1), . . . , ψ1(xn))F0×/k0× (=, n) For every x1, . . . , xn∈F×,ψ1(k(x1, . . . , xn)F×/k×) = k0(ψ1(x1), . . . , ψ1(xn))F0×/k0× whereψ1 :F×→F0× denotes any set-theoretic lift of ψ1.
Corollary 8. If ψ∗ is injective in degree ≤n+ 1 (resp. an isomorphism), Assertion (⊂, n) (resp.
(=, n)) holds for everyn if i= 1 and for n≤1 if i= 2.
Proof. The second part of Corollary 8 follows from the first part (applied toψ∗andψ−1∗ ). We prove the first part. An element x∈F is algebraic overk(x1, . . . , xn) if and only if there exists a subset I ⊂ {1, . . . , n} such that k(xi|i∈I) andk(xi|i∈I, x) both have transcendence degree|I|overk.
Assume i = 1 and n≥ 1 is arbitrary or i= 2 and n ≤1. Then, from Theorem 7, the condition thatk(xi|i∈I) andk(xi|i∈I, x) both have transcendence degree|I|overkis equivalent to{xi|i∈ I} 6= 0 inKi,M|I| (F) and{xi|i∈I, x}= 0 inKi,M|I|+1(F). Sinceψ∗ :Ki,M∗ (F)→Ki,M∗ (F0) is injective, {xi|i∈ I} 6= 0 in Ki,M|I| (F) and {xi|i∈ I, x} = 0 in Ki,M|I|+1(F) in turn implies {ψ1(xi)|i ∈I} 6= 0 in Ki,M|I| (F0) and {ψ1(xi)|i ∈ I, ψ1(x)} = 0 in Ki,M|I|+1(F0). Applying again Theorem 7 yields the
conclusion.
Remark 9.
• When F is of type 2 the implication (b) ⇒ (a) of Theorem 7 and the second part of Corollary 8 for every n are predicted by the Bass-Tate conjecture in positive characteristic [2, Question, p.390]. See also [8] for the relation between the Bass-Tate conjecture in positive characteristic and classical motivic conjectures.
• For our applications, we only need then= 1 cases of Corollary 8 (viz then= 2 case of Theorem 7). In particular, for function fields of type 1, we only need the n = 2 case of the Bloch-Kato conjecture, an earlier theorem of Merkurjev and Suslin [12].
2.4. Reconstructing function fields. Then= 1 case of Corollary 8 implies that fori= 1,2 and for everyF, F0 ∈Fep of typeiand isomorphismψ∗:Ki,M∗ (F)→Ki,M∗ (F0) ofZ≥0-graded rings the induced isomorphism of multiplicative groupsψ1 preserves algebraic dependence. This implies the following. For i= 1,2, let
Isom(Ki,M∗ (F), KM∗ (F0))
denote the set of isomorphisms of Z-graded ringsKi,M∗ (F) ˜→Ki,M∗ (F0) and Isom(Ki,M∗ (F), Ki,M∗ (F0))
their quotients by the natural action ofZ/2.
Corollary 10. Let F, F0 ∈F. Then˜
(a) Isom(K∗M(F), K∗M(F0))6=∅ only if F, F0 are of the same type i and characteristicp.
(b) If F, F0 ∈ F˜p are of the same type i and have transcendence degree ≥2 over their fields of constants then the natural functorial map
Isom(F, F0)→Isom(Ki,M∗ (F), Ki,M∗ (F0)) is bijective.
Proof. Part (a) follows from Lemma 6. For Part (b), one has a canonical commutative diagram Isom≡(F×/k×, F0× x/k0×)
**
Isom(F, F0)
))
(2) 55
Isom(F×/k×, F0×/k0×)
Isom(Ki,M∗ (F), Ki,M∗ (F0))
' Ki,M
1
44(1)
OO
where the factorization (1) follows from Corollary 8. So the conclusion follows from the fact that (Ki,M1 - hence (1) - is injective and) the map (2) is bijective by Theorem 4.
Corollary 11. Let F ∈F˜ of typei. Then one has the following group isomorphisms Aut(F) ˜→Aut≡(F×/k×) ˜→Aut(Ki,M∗ (F))
Remark 12. Since Ki,M∗ (F) is a quadratic algebra it is entirely determined by Ki,M≤2 (F) (for i = 1, one can even replace K1,M2 (F) with its `-adic completion H2(F,Z`(2))). In particular, in Corollary 10, one could replace Isom(K∗M(F), K∗M(F0)) by the set of group isomorphisms ψ≤2 : Ki,M≤2 (F) ˜→Ki,M≤2 (F0) such thatψ2({x1, x2}) ={ψ1(x1), ψ1(x2)}.
Remark 13. From Lemma 6, one can reconstruct the field of constants kof a function field F in F˜ from K1M(F). In general, one may ask for a relative version of Corollary 10 that is replacing the functor K∗M(−) with the functor sending a finitely generated field extension F of a perfect field k to the morphism K∗M(k)→K∗M(F).
The remaining part of the paper is devoted to the proof of Theorem 4.
3. Proof of Theorem 4
3.1. Recollection on differentials. We will use repeatedly the following classical facts about differentials.
Lemma 14. For x1, . . . , xn∈F, the following are equivalent
• x1, . . . , xn ∈ F is a separating transcendence basis for F|k (i.e. F|k(x1, . . . , xn) is a finite separable field extension);
• dx1, . . . , dxn∈Ω1F|k is an F-basis ofΩ1F|k. If p >0, these are also equivalent to
• x1, . . . , xn∈F is a p-basis ofF|k.
See e.g [11, §27, Thm 59; §38, Thm. 86] for the proof. Recall that since k is a perfect field, the extension F|k always admits a separating transcendence basis, and that if p = 0, every transcen- dence basis is separating.
Corollary 15. ker(d:F →Ω1F|k) =Fp.
In particular for everyx∈F, the following are equivalent:
(a) x∈F\Fp; (b) dx6= 0;
(c) x is a separating transcendence basis fork(x)F|k.
Ifx ∈F verifies the above equivalent conditions (a), (b), (c), for every 06=f ∈k(x)F there exists a unique f0:=f0(x)∈k(x)F such that df =f0dx.
3.2. Notation and overview of the proof of Theorem 4.
Let k be a perfect field of characteristic p ≥ 0 and let F|k be a finitely generated regular field extension of transcendence degree ≥2.
For every subfieldk⊂E ⊂F andx, y∈F×such thatx6=y∈F×/E×, writelE(x, y) :=Ex+Ey⊂ F (which by assumption is a 2-dimensional E-subspace) and
lE(x, y) = ((Ex+Ey)∩F×)/E×⊂F×/E× for the corresponding line inF×/E×.
Definition 16. We say thatx, y∈F×/parep-multiplicatively dependentand writex∼p yif either x=y= 1 or x, y6= 1 and
xZ∩yZ6= 1
We say that x, y∈F×/k× are p-multiplicatively dependentand write again x ∼p y if their images inF×/p arep-multiplicatively dependent. The relations∼p are equivalence relations onF×/p and F×/k×.
Note that, if p >0,x∼p y if and only if xZ=yZ inF×/p.
Forx1, x2, y1, y2 ∈F×/k× and some (equivalently, every) liftsx1, x2, y1, y2 ∈F×, write I(x1, x2, y1, y2) := (k(x1/x2)F×·x2)\
(k(y1/y2)F×·y2) and
I(x1, x2) = [
y1,y2
I(x1, x2, y1, y2) where the union is over allyi ∈k(xi)F×,yi 6∼pxi,1,i= 1,2.
LetI(x1, x2, y1, y2),I(x1, x2) denote the images ofI(x1, x2, y1, y2),I(x1, x2) inF×/k×respectively.
For every I ∈ I(x1, x2) and i = 1,2, set {i, j} = {1,2} and let l◦(I, xi) denote the set of all yi ∈ k(xi)F×, yi 6∼p 1, xi such that for some (equivalently, every) lift I ∈F× of I ∈F×/k×, one hasI ∈k(y1/y2)F×·y2 for someyj ∈k(xj)F,yj 6∼p1, xj. Set also
l(I, xi) =l◦(I, xi)∪ {xi} and letl(I, xi) denote the image ofl(I, xi) in F×/k×.
3.2.1. Roughly,I(x1, x2) andl(I, xi),i= 1,2 have to be regarded as approximations - defined only in terms of the multiplicative structure ofF×/k×and the relation≡of algebraic dependence onF overk - oflk(x1, x2) and lk(1, xi),i= 1,2 respectively. But actually I(x1, x2) and l(I, xi), i= 1,2 do not distinguish lines from their ‘inverse’. More precisely,
Lemma 17. For every x1, x2 ∈ F×/k× such that x1, x2, x1/x2 6∼p= 1 and = ±1, one has lk(x1, x2) ⊂ I(x1, x2) and for every I ∈ lk(x1, x2), I 6= x1, x2, one has lk(1, xi) ⊂ l(I, xi), i= 1,2.
Proof. We perform the proof for = 1; the proof for = −1 is similar. We first observe that for every c ∈ k×, xi−ci 6∼p 1, xi, i = 1,2. Since k is perfect, xi 6∼p 1 forces xi−ci 6∼p 1. If xi−ci∼p xi, there would be nonzero integersai, bi ∈Z and αi ∈k(xi)F such that
(xi−ci)bi−1xaii−1(bixi−ai(xi−ci))
x2ai i dxi=d((xi−ci)bi xaii ) = 0,
which forces (here we use again xi 6∼p 1)ai=bi= 0: a contradiction.
Fix c∈k. On the one hand x1−cx2 = (xx1
2 −c)x2 ∈k(x1/x2)×x2 and on the other hand, writing {i, j}={1,2}, for everyci, cj ∈k× such that c=ci/cj,
(3.1) xi−cxj = (xi−ci
xj−cj −c)(xj −cj)∈k(xi−ci
xj−cj)×(xj −cj).
Since xi−ci 6∼p xi,1, i= 1,2, this ensures that x1−cx2 ∈ I(x1, x2, y1, y2) ⊂I(x1, x2). The last part of the assertion also follows from (3.1) since for everycj ∈k× one can take ci=ccj.
The following picture sums up visually what happens for= 1.
x1 x2
1
l(I, x1) l(I, x2)
I(x1, x2) I
y1 (c1)
y10 (c01)
y2 (c2) y20 (c02)
c=c1/c2=c01/c02
Provided (x1, x2) satisfies certain properties which are encapsuled in the notion of ‘good pair’ (see Definition 19) and p = 0, the indeterminancy = ±1 is the only one (see Remark 26) but when p >0, it is much rougher. More precisely, whenp >0, the setI(x1, x2) only recovers the line pass- ing through (x1, x2) up to prime-to-ppowers and certain affine transformations withFp-coefficients.
The best one can say in whole generality is stated in Lemma 23, which is the technical core of the paper. The proof of Lemma 23 is sketched in Subsection 3.4.2.
For everyx1, x2∈F×/k× such thatx1, x2, x1/x2 6∼p= 1, another key property of Ω◦:=lk(x1, x2)\ {x1, x2} as a subset of I(x1, x2) is that, fori= 1,2 the intersection
\
I∈Ω◦
l(I, xi)
of the ‘parameter spaces’l(I, xi) contains a non-empty subset ∆i(=lk(1, xi)) which is not contained inF×pxZi. This property, which is only set theoretic and multiplicative hence is preserved by any group isomorphism ψ : F×/k×→F˜ 0 ×/k0 ×, will be used crucially in combination with the easy Lemma 28 (which is completely independent of the rest of the proof) in the proof of Proposition 27.
3.2.2. The injectivity of the map
Isom(F|k, F0|k0)→Isom≡(F×/k×, F0 ×/k0 ×)
in Theorem 4 is easy - see the last paragraph of Subsection 3.6. We now summarize the main steps of the proof of the surjectivity in Theorem 4 assuming Lemma 23 (andp >0).
Let ψ:F×/k×→F˜ 0×/k0× be a group isomorphism preserving algebraic independence. The basic idea is to apply the fundamental theorem of projective geometry (Lemma 29). For this, one should show that for =±1, ψ maps lines in F×/k× isomorphically onto lines in F0×/k0×. We are not able to do this directly. Instead, we consider first a very small quotient of ψ:F×/k×→F˜ 0×/k0×, namely
F×/k× '
ψ //
F0×/k0×
F×/Fp× '
ψ
//F0×/F0p×
and show (Proposition 27) that there exists a unique integer m ∈ Z satisfying (1.1) such that ψm maps lines in F×/Fp× isomorphically onto lines in F0×/F0p×. Using symmetry, multi- plicativity and that every line in F0×/k0× can be shifted to a line passing through a good pair (Lemma 20), it is actually enough to show that there exists a unique integer m ∈ Z satisfying (1.1) such that for every x1 6= x2 ∈ F×/p for which (ψ(x1), ψ(x2)) ∈ F0×/k0× is a good pair, ψ(lk(x1, x2))m ⊂ lF0p(ψ(x1)m, ψ(x2)m). From Lemma 17 and the multiplicativity of ψ, one has Ω := ψ(lk(x1, x2))⊂ψ(I(x1, x2)) =I(ψ(x1), ψ((x2)) and, for everyI ∈lk(x1, x2), I 6=x1, x2, one has ∆i :=ψ(lk(1, xi))⊂ψ(l(I, xi)) =l(ψ(I), ψ(xi)),i= 1,2. Write Ω◦ := Ω\ {ψ(x1), ψ(x2)}. From Lemma 23 and Lemma 28, one has a stratification Ω◦=tmΩ◦m, wheremvaries among all integers m∈Z satisfying (1.1) with the property that Ω◦mm ⊂lF0p(ψ(x1)m, ψ(x2)m) and for every ω ∈Ω◦m, l(ω, ψ(xi))m⊂lF0p(1, ψ(xi)m), i= 1,2. In particular, one has the inclusions ∆mi ⊂lF0p(1, ψ(xi)m), i= 1,2 which areindependent of ω. Applying Lemma 28 again, this shows that Ω◦m=∅ for all but onem and then thatm is independent of the line lk(x1, x2).
With Proposition 27 in hands, one can apply the fundamental theorem of projective geometry to ψm :F×/Fp×→F˜ 0×/F0p× to obtain a unique field isomorphism Φ : F→F˜ 0 such that the induced group isomorphism φ: F×/Fp×→F˜ 0×/F0p× coincides with ψm. To conclude, it remains to show that the group isomorphismφ:F×/k×→F˜ 0×/k0×induced by Φ :F→F˜ 0 coincides withψ and that m =±1. We prove both assertions together in Subsection 3.6. It is the consequence of a slightly tricky computation, which relies on the explicit description of ψ(x1+x2)m given by (3.2) and the additivity of Φ.
3.3. Multiplicatively shifting lines to lines passing through good pairs.
Let k be a perfect field of characteristic p ≥ 0 and let F|k be a finitely generated regular field extension of transcendence degree ≥2.
Definition 18. We say that x ∈ F is F|k-regular if x is transcendental over k and F|k(x) is a regular field extension.
Definition 19. We say that (x1, x2)∈F×/k× is a good pairif for some (equivalently, every) lifts x1, x2 ∈F×ofx1, x2 ∈F×/k×,x1 isF|k-regular anddx1, dx2 ∈Ω1F|kare linearly independent over F.
Lemma 20. For every x ∈ F×\Fp×, there exists a F|k-regular y ∈ F× such that dx, dy are linearly independent over F.
In particular, for every x ∈ F×/k×, x 6= 1 in F×/p there exists y ∈ F×/k× such that (y, xy) ∈ F×/k× is a good pair. In particular, for x1 6= x2 ∈ F×/k× and setting x := xx1
2 6= 1, the line lk(x1, x2) can be shifted multiplicatively to a line passing through a good pair (y, xy) as
x−11 ylk(x1, x2) =lk(y, xy).
Proof. This follows from Bertini theorems. Since F has finite transcendence degree r ≥ 2 over k and k is perfect, there exist x2, . . . , xr ∈ F× such that dx, dx2, . . . , dxr are linearly independent overF. WriteE :=k(x2, . . . , xr). Letz∈F× with minimal polynomial
Pz =Td+ X
0≤i≤d−1
aiTi ∈E(x)[T]
overE(x). Thendx, dzare linearly dependent overFif and only if∂ai/∂xj = 0,i= 0, . . . , d−1, j= 2, . . . r, or equivalently, Pz ∈Ep(x)[T]. In particular, anyy ∈E(x)\Ep(x) will have the property thatdx, dyare linearly independent overF. We claim that one can find y∈E(x)\Ep(x) such that y is F|k-regular. Fix a normal quasi-projective model X|kof F|ksuch that x, x2, . . . , xr induce a finite dominant morphism x : X → Prk. As F|k is regular, X is geometrically irreducible over k.
It is enough to show there exists a homogeneous polynomial P ∈ k[x, x1, . . . , xr]\k[x, xp1, . . . , xpr] such that the fiber at 0 of the composite map
y:X→x Prk
→P P1k
is geometrically irreducible. Viewingy as an element inE(x), we deduce thaty is F|k-regular, by [6, 9.7]. If k is infinite, the existence ofP follows directly from [7, Cor. 6.11.3]. If k is finite, this follows from [5, Thm. 1.6] applied toxk :Xk→Prk (note that there is no vertical component since
xk :Xk→Prk is finite) and Lemma 21 below.
Letk be a finite field. Set S0:=k[x, xp2, . . . , xpr]⊂k[x, x2, . . . , xr] and define the density of S0 as δ(S0) = lim
d→+∞
|S0∩Sd|
|Sd| ,
whereSd⊂k[x, x2, . . . , xr] denotes the set of homogeneous polynomials of degreed. Then Lemma 21. δ(S0) = 0.
Proof.3 AsSd+p−1 =⊕p−1i=0xp−1−i(k[xp, x2, . . . , xr]∩Sd+i), one has
dim(Sd+p−1)≥pdim(k[xp, x2, . . . , xr]∩Sd)≥pdim(S0∩Sd), so dim(S0∩Sd)≤ 1pdim(Sd+p−1) and thus, using that dim(Sd) = d+r−1d
, dim(S0∩Sd)−dim(Sd)≤ 1
pdim(Sd+p−1)−dim(Sd)∼ (d+r−1)!
d!(r−1)! (1
p −1)→ −∞.
This meansδ(S0) = lim
d→+∞
|S0∩Sd|
|Sd| = 0.
Remark 22. We resorted to Bertini theorems, which provide a conceptually natural proof of Lemma 20 but one can give more elementary arguments. If k is infinite, the Galois-theoretic Lemma [10, Chap. VIII, Lem. in Proof of Thm. 7] already shows there exists (infinitely many) 06=a∈ksuch thaty:=ax2+xisF|k-regular; by constructiondx, dyare linearly independent over F. Ifkis finite, Akio Tamagawa suggested the following arguments. Fix a smooth (not necessarily
3Our original proof of Lemma 21 was more technical; the following alternative proof was suggested to us by one of the Referees.
proper) model X|k of F|k and a closed point t ∈ X. Let Xe → X denote the blow-up of X at t and Dt = Pn−1k(t) ⊂ Xe the exceptional divisor. Fix a non-empty affine subset U = Spec(R) ⊂ Xe such that Z := U ∩Dt is non-empty and the rational map x : Xe → X 99K A1k given by x is defined over U. We endow Z with its reduced subscheme structure. Since Z is irreducible, one can write Z = Spec(R/P) for some prime ideal P in R. Pick y ∈ R\R∩Fpk(x)F such that fmodP ∈R/P \k(t). Let k[y]F ⊂F denote the normal closure ofk[y] in k(y)F/k(y). Since R is smooth overkhence normal,k[y]F ⊂R. Also, by our choice ofy, the morphismk[y]F ,→RR/I is injective whence the fraction field k(y)F of k[y]F, embeds into the fraction field k(t)(x2, . . . , xn) of R/I. Since k(y)F/k is regular, k(t)⊗k k(y)F ' k(t)·k(y)F ⊂ k(t)(x2, . . . , xn). From Lur¨oth theorem, one thus has k(t)⊗kk(y)F = k(t)(T) for some T ∈ k(t)(x2, . . . , xn) transcendent over k(t). In other words, k(y)F is the function field of ak(t)/k form Cy of P1k. Since k is finite hence perfect with trivial Brauer group, Cy 'P1k. This showsy ∈F is F|k regular. Sincey /∈Fpk(x)F, dx,dyare linearly independent over F.
3.4. Approximating lines passing through good pairs up to powers.
3.4.1. Statement of the main Lemma. Let k be a perfect field of characteristic p ≥0 and let F|k be a finitely generated regular field extension of transcendence degree≥2.
Lemma 23. Let(x1, x2)∈F×/k×be a good pair. Then, for everyI ∈I(x1, x2) there existsm∈Z (depending a priori on I) satisfying (1.1), N ∈Z, α=α(xx1
2)∈k(xx1
2)F× andc∈k× such that
(3.2) Im=α(x1
x2
)p(xm1 −cxpN1 xpN2 xm2 ).
Furthermore, for every yi ∈k(xi)F, yi 6∼pxi,1, i= 1,2 such that I ∈ I(x1, x2, y1, y2), one has for i= 1,2
(3.3) yim=αpi(xmi −cixpNi ) for some αi ∈k(xi)F, ci ∈k× with the condition c=c1/c2.
Remark 24. In particular, for every I ∈ I(x1, x2) there exists m ∈Z (depending a priori on I) satisfying (1.1) such thatIm ∈lFp(xm1 , xm2 ) andl(I, xi)m⊂lFp(1, xmi ),i= 1,2. But let us point out that (3.2) imposes restrictions on the liftsI ∈F×of theI ∈lFp(xm1 , xm2 )m−1 which lie inI(x1, x2).
This will be used crucially in Subsection 3.6.
3.4.2. Main steps of the proof of Lemma 23. The proof when p = 0 is significantly simpler since ker(d) =k (recallF|k is regular). We carry out the proof forp > 0 and just mention the simpli- fications that occur for p = 0. The results for p = 0 are similar (recall that, by our convention, Fp=kwhen p= 0).
We are to determine the possible yi ∈k(xi)F×,yi 6∼p xi,1,i= 1,2 such that I(x1, x2, y1, y2) 6=∅ and for all such yi, i= 1,2 the elements I ∈ I(x1, x2, y1, y2). So assume I(x1, x2, y1, y2)6= ∅ and fixI ∈ I(x1, x2, y1, y2).
Note that, as (x1, x2)∈F×/k× is a good pair and yi 6∼p1,i= 1,2,y1/y26∼p 1. Indeed, otherwise, we would have y01dx1= yy1y02
2 dx2 hence y10 =y02 = 0.
