Milnor K-Groups and Finite Field Extensions
KARIM JOHANNES BECHER
Fachbereich Mathematik und Statistik, D204, Universit¨at Konstanz, D-78457 Konstanz, Germany.
e-mail: becher@maths.ucd.ie
(Received: June 2002)
Abstract. LetE/F be a finite separable field extension and let m denote the integral part of log2[E:F]. David Leep recently showed that if char(F )=2, then fornmthenth power of the fundamental ideal in the Witt ring ofEsatisfies the equalityInE=In−mF·ImE. The aim of this note is to prove the analogous equality for the MilnorK-groups, that isKnE = Kn−mF·KmE fornm. In either of these equalities one may not replacembym−1, as examples of certain m-quadratic extensions indicate.
Mathematics Subject Classifications (2000):11E81, 12F05, 19D45.
Key words:MilnorK-theory, higherK-groups, Witt ring, field extension.
1. Main Result and Consequences
Throughout this paper, letFdenote a field. We recall the definition of the groups KnF (n1) of Milnor K-theory, introduced in [4]: for n1, let KnF be the Abelian group generated by elements{a1, . . . , an}(a1, . . . , an ∈F×), calledsym- bols, which are subject to the only relations that{a1, . . . , an} be zero whenever ai +ai+1 = 1 inF and that {∗, . . . ,∗}viewed as a function (F×)n → KnF be multilinear. The group operation inKnF is written additively. As a consequence of the defining relations, one has, further, fora1, . . . , an ∈F×, that{a1, . . . , an} =0 ifai +ai+1 = 0 for somei < nand that{aσ (1), . . . , aσ (n)} = sgn(σ ){a1, . . . , an} for any permutationσ ∈Snwith signature sgn(σ )∈ {+1,−1}(cf. [4, Lemma 1.3 and Lemma 1.1]).
With the convention K0F := Z, the groupK∗F :=
i0KiF has a natural structure as a gradedZ-algebra. Furthermore, ifL/Fis an arbitrary field extension, then the natural group homomorphismsKnF →KnL (n0)and the induced ring homomorphismK∗F →K∗LturnK∗Linto aK∗F-algebra.
THEOREM 1.1. Let E/F be a finite field extension such that E = F (θ ) for some θ ∈ E. The group KnE is generated by the symbols {f1(θ ), . . . , fn(θ )} where f1, . . . , fn ∈ F[X] are such that fi(θ ) = 0 for 1in and such that deg(fi)12deg(fi+1)for1i < nanddeg(fn)12[E:F].
The proof of this statement will be achieved at the end of Section 2. We first want to discuss its consequences. The following corollary can be considered as the main statement of this article:
COROLLARY 1.2. LetE/F be a finite separable extension and letmdenote the integral part oflog2[E :F]. Fornmone hasKnE=Kn−mF ·KmE.
The equality in the statement can be rephrased by saying thatKnEis generated by those symbols {a1, . . . , an} (a1, . . . , an ∈ E×) where a1, . . . , an−m lie in F. This is actually what we are going to show.
Proof. By the ‘primitive element theorem’ there existsθ ∈ E such that E = F (θ ). Theorem 1.1 then implies that KnE is generated by the symbols {f1(θ ), . . . , fn(θ )} where f1, . . . , fn ∈ F[X] are such that fi(θ ) = 0 and deg(fi)2i−n−1[E:F] for 1in; in particular for in − m, since [E:F]<2m+1,fi must be constant, whencefi(θ )∈F×.
In Section 3 we will see that, in general, the choice of the value for m in Corollary 1.2 is best possible.
At least whenE/Fhas no nontrivial subextension (e.g. when [E:F] is prime), the hypothesis in Corollary 1.2 that the extensionE/F be separable is superfluous, since the existence ofθ such thatE =F (θ )then is evident. In particular one has KnE = Kn−1F ·K1E whenever [E : F]3; this has already been observed by Merkurjev in [3, Lemma 2] (for the crucial casen=2).
COROLLARY 1.3 (Merkurjev). LetE/F be a field extension of degree2or3and letl∈N. IfKnF / l·KnF =0thenKn+1E/ l·Kn+1E=0.
Proof. As just explained, we have Kn+1E = KnF ·K1E. Hence, ifKnF is divisible bylthen so isKn+1E.
Examples of finite field extensions E/F where KnF / l·KnF = 0 while KnE/ l·KnE=0, for givenl, n∈N, are easy to obtain. One way of constructing such examples will be explained in the third section. By this fact together with the last corollary we are lead to
QUESTION 1.4. LetE/F be a finite field extension and l, n ∈ N. Assume that KnF / l·KnF =0. Does it follow thatKn+1E/ l·Kn+1E=0?
Note that the problem can be easily reduced to the case wherelis prime.
In the rest of this section we compare the above corollaries with the results of Leep in [2] which inspired the present investigation.
We assume that char(F )=2. Forn0 one denotes byInF :=(I F )nthenth power of the fundamental ideal in the Witt ring ofF, further byI¯nF the quotient InF /In+1F. The natural homomorphismKnF → ¯InF which sends any symbol
{x1, . . . , xn} to the class of the n-fold Pfister form 1,−x1 ⊗ · · · ⊗ 1,−xn is clearly surjective. (By part of the Milnor conjecture its kernel is precisely 2KnF; in characteristic zero, a proof of this has recently been given in [5, Theorem 4.1].) Elman and Lam have shown by elementary arguments thatKnF /2KnF =0 if and only ifInF =0 [1, Corollary 3.3].
We consider again a finite extensionE/F and denote bymthe integral part of log2[E:F]. Leep showed in [2, Theorem 2.1] that
InE=In−mF ·ImE. (1)
As an immediate consequence we get
I¯nE= ¯In−mF · ¯ImE (2)
in the graded Witt ring ofE. On the other hand, the last equation can also be de- duced from Corollary 1.2 using the natural homomorphismKnF → ¯InF. (Indeed, one applies Corollary 1.2 toE0/F whereE0is the separable closure ofF inEand observes thatI¯nE0→ ¯InEis surjective, as explained in [2, Section 2].)
One can ask whether equalities (1) and (2) are equivalent. If so then Leep’s result would follow from Corollary 1.2. At least under some additional assumption such asIeF =0 or (weaker)IeE =I F ·Ie−1Efor somee ∈N, one can indeed prove that(2)implies(1), regardless of the value ofmn.
Furthermore, sinceKnF /2KnF =0 if and only ifInF =0 [1, Corollary 3.3], it follows from [2, Corollary 3.4] that in the case wherel = 2 and char(F ) = 2, the Question 1.4 has a positive answer at least if [E:F]5.
2. OnKKKnnnof a Rational Function Field
We will obtain Theorem 1.1 at the end of this section as a consequence of an observation which gives generators for certain subgroups Ld ofKnF (X), where F (X)is the rational function field in one variable over F (Proposition 2.3). To carry this information onF (X)down to the finite extension E = F (θ )ofF we use a surjectionKnF (X)→KnE, which has been considered by Milnor.
In order to shrink the set of generators for those groups, we will use a de- gree reduction argument which is based on the division algorithm for polynomials together with the following rule:
LEMMA 2.1. Letf, g, h, t ∈F×be such thatg=f h+t. InK2F one has {f, g} = −{−h, g} + {t, g} − {t, h} − {t, f} =
−h,1 g
+
t, g f h
.
Proof. By hypothesis we have g
f h− t f h =1,
hence
− t f h, g
f h
=0
inK2F. Expanding the last equation yields
0 = {t, g} − {t, f h} − {−f h, g} + {−f h, f h}
= {t, g} − {t, f} − {t, h} − {f, g} − {−h, g}.
This shows the first equality of the statement, the second one is obvious.
We define a partial order on the setN0n in the following way: for twon-tuples d = (d1, . . . , dn) and e = (e1, . . . , en) where d1, . . . , dn, e1, . . . , en ∈ N0, we writedein case we havediei for 1in. Obviously, any nonempty subset ofN0n has a minimal element with respect to this partial order. In the sequel, we implicitly refer to this partial order onN0n when we say that an element ofN0n is minimal with respect to a certain condition.
LEMMA 2.2. LetLbe a proper subgroup ofKnF (X). Letf1, . . . , fn be nonzero polynomials inF[X] such that(deg(f1), . . . ,deg(fn))is minimal with respect to the condition that the symbol{f1, . . . , fn}belong toKnF (X)\L. Then
(a) each of the polynomialsf1, . . . , fnis either a constant or irreducible, (b) for any distinct positivei, jn,ifdeg(fi) deg(fj)then one has actually
deg(fi)12deg(fj).
Proof. (a) Let 1in. Iffi = gh for someg, h ∈ F[X] then the equality {f1, . . . , fi, . . . , fn} = {f1, . . . , g, . . . , fn} + {f1, . . . , h, . . . , fn} shows that at least one of the last two symbols does not belong to L; by minimality of (deg(f1), . . . ,deg(fn)), one of g and h must be of the same degree as fi. This means thatfi is irreducible.
(b) Observe that any change of the order of f1, . . . , fn leaves the symbol {f1, . . . , fn}unchanged up to a sign and therefore does not affect the hypotheses of the lemma. Hence, to prove (b) we may assume without loss of generality that i=1 andj =2.
Suppose now that deg(f1) deg(f2). Using the division algorithm, we may writef2 =f1h+twith polynomialsh, t ∈F[X] where eithert =0 or deg(t) <
deg(f1). If t = 0 then {f1, . . . , fn} = −{−h, f2, . . . , fn}. On the other hand, if t=0 then from Lemma 2.1 we obtain
{f1, . . . , fn} = −{−h, f2, . . . , fn} + {t, f2, . . . , fn} −
− {t, h, f3, . . . , fn} − {t, f1, f3, . . . , fn}
and by the minimality of(deg(f1), . . . ,deg(fn)), the last three symbols lie inL.
In any case we conclude from{f1, . . . , fn} ∈/ L that {−h, f2, . . . , fn} ∈/ L. By
the minimality of (d1, . . . , dn) it follows that deg(h)deg(f1) which leads to deg(f2)=deg(f1)+deg(h)2 deg(f1).
PROPOSITION 2.3. For a given positive integerd,letLd denote the subgroup of KnF (X) generated by the symbols {f1, . . . , fn} where f1, . . . , fn ∈ F[X]
are nonzero polynomials of degree at most d. Ld is already generated by the symbols {g1, . . . , gn} where g1, . . . , gn ∈ F[X] are nonzero polynomials with deg(gi)12deg(gi+1)for1i < nanddeg(gn)d.
Proof. Let L be the subgroup of Ld generated by the symbols {g1, . . . , gn} where g1, . . . , gn ∈ F[X] are nonzero and such that deg(gi)12deg(gi+1) for 1i < nand deg(gn)d. Assume thatLis a proper subgroup of Ld. We may then choose nonzero polynomialsf1, . . . , fn ∈ F[X] of degree less or equal to d such that{f1, . . . , fn} belongs toKnF (X)\L and with(deg(f1), . . . ,deg(fn)) minimal with respect to this condition. Since the symbol{f1, . . . , fn}is invariant up to a sign under any change of order of f1, . . . , fn, we may further assume that deg(f1)· · · deg(fn)d. By the last lemma it follows that deg(fi)
1
2deg(fi+1) for 1i < n. But then {f1, . . . , fn} belongs to L, by definition of L. This contradicts the choice off1, . . . , fn. The conclusion is thatLequals Ld, which gives the statement.
COROLLARY 2.4. The group KnF (X) is generated by the symbols {f1, . . . , fn}wheref1, . . . , fn ∈F[X]are nonzero withdeg(fi)12deg(fi+1)for 1i < n.
Proof. This is clear from the above proposition, sinceKnF (X)is the union of the groupsLd(d1).
The next lemma is [2, Corollary 2.7]. For the convenience of the reader, we include the proof.
LEMMA 2.5. Let E = F (θ ) be a finite field extension of F of degree l. Any nonzero element ofE can be written as a quotientf (θ )/g(θ )wheref andgare nonzero polynomials of degree less or equal to[l/2].
Proof. Let V be an arbitrary F-subspace of E of dimension strictly greater than l/2. We claim that every nonzero element of E is a quotient of two nonzero elements of V. Indeed, for x ∈ E\{0} the F-subspaces V and V x of E both have dimension greater than l/2, hence their intersection is not zero; in other words, there exist v, w ∈ V\{0} such that v = wx, i.e. x = v/w.
We apply this to theF-spaceV generated by 1, θ, θ2, . . . , θ[2l]. SinceV\{0} = {f (θ )∈E |f ∈F[X]\{0},deg(f )[2l]}, the statement follows.
We are now ready to prove Theorem 1.1. Using the identification of E× with K1E, the last lemma actually corresponds to the case in Theorem 1.1 where n=1.
We continue to consider a finite extension E/F such thatE = F (θ )for some θ ∈E. By the ‘primitive element theorem’, the last condition is satisfied whenever E/F is a separable extension. Letπ be the minimal polynomial ofθ overF. Any nonzero element ofF (X)can be written in the formf/g·πz wheref, g ∈F[X]
are polynomials not divisible byπandz∈Z. Associating to an elementf/g·πz∈ F (X)×the elementf (θ )/g(θ )∈Edefines a surjective mapF (X)×−→E×. This map induces a group homomorphism ψn: KnF (X) −→ KnE, uniquely deter- mined by the rules that ψn({f1, . . . , fn}) = {f1(θ ), . . . , fn(θ )} if f1, . . . , fn ∈ F[X] are not divisible by π and thatψn({π,∗, . . . ,∗}) = 0. The homomorphism ψndepends on the choice ofθ, its construction is due to J. Milnor [4, Lemma 2.2].
Let d denote the integral part of 12[E : F] andLd the subgroup of KnF (X) generated by symbols{f1, . . . , fn} where f1, . . . , fn ∈ F[X]\{0} are of degree less or equal tod. From the last lemma we conclude thatψn(Ld) = KnE. Using this, Theorem 1.1 follows immediately from Proposition 2.3.
3. Two Examples
The following proposition is a variant of [2, Proposition 3.1]. It gives an example of a separable field extensionE/F such that the equalityKnE = Kn−mF ·KmE, which holds form = [log2[E : F]] by Corollary 1.2, would be wrong for any lower value form.
PROPOSITION 3.1. Let k be a field of characteristic different from 2 and let mn. Let E = k(X1, . . . , Xn), the rational function field in n variables over k,andF =k(X21, . . . , Xm2, Xm+1, . . . , Xn). Then
(a) E/F is a separable extension of degree2m,that ism=log2[E:F],
(b) the class in the group I¯nE represented by the n-fold Pfister form 1,−X1 ⊗ · · · ⊗ 1,−Xnis not contained inI¯n−m+1F · ¯Im−1E,
(c) InE=In−m+1F ·Im−1E and KnE=Kn−m+1F ·Km−1E.
Proof. Part (a) should be clear. Let k¯ denote an algebraic closure of k. It is clear that then-fold Pfister form1,−X1 ⊗ · · · ⊗ 1,−Xnstays anisotropic over E¯ := ¯k((X1)) . . . ((Xn)) and thus represents a nontrivial class in I¯nE. The field¯ E¯ contains the subfieldF¯ := ¯k((X21)) . . . ((Xm2))((Xm+1)) . . . ((Xn))which in turn containsF.
Consider an anisotropic(n−m+1)-fold Pfister formπdefined overF¯. Since
| ¯F×/F¯×2| =2nandX12, . . . , Xm2 lie inE¯2, it is clear that the image ofF¯×/F¯×2in E¯×/E¯×2has cardinality 2n−m. It follows that in any diagonalization ofπ overE,¯ at least two of the 2n−m+1 entries must lie in the same square class of E¯×. Since
−1∈ ¯E2, it follows thatπbecomes hyperbolic overE.¯
This argument shows that the image ofI¯n−m+1F · ¯Im−1EinI¯nE¯ is zero, while we have seen that the image ofI¯nEinI¯nE¯ contains the class of1,−X1 ⊗ · · · ⊗ 1,−Xn, which is nonzero as a consequence of the Arason–Pfister–Hauptsatz
[6, 4.5.6. Theorem]. This shows (b), while (c) is an immediate consequence of(b).
Remark 3.2. It is not difficult to see that the statements of the proposition hold more generally whenF is a subfield ofk((X¯ 12)) . . . ((X2m))((Xm+1)) . . . ((Xn)) containingk(X12, . . . , X2m, Xm+1, . . . , Xn)andE =F (X1, . . . , Xn).
We finish with a counter-example to a variation of Question 1.4 which might look more natural to ask at first sight.
EXAMPLE 3.3. Letn1 and let p be a prime. Suppose that F1 is a field con- taining a primitivep-root of unity (in particular, not of characteristicp), having no cyclic extension of orderp but having a Galois extension L/F1 whose degree is divisible byp. Hence, there exists anF1-automorphism ofLwhich is of orderp.
LetE1denote the subfield ofLfixed by this automorphism. ThenL/E1is a cyclic extension of degreep, in particularE1is notp-closed whileF1isp-closed. For
F =F1((X1)) . . . ((Xn−1)) and E=E1((X1)) . . . ((Xn−1)) we obtain
KnF /p·KnF =0, KnE/p·KnE=0 and
[E:F]=[E1:F1]<∞.
Note that a Galois extension E1/F1 with the desired properties can be con- structed without difficulties. Take a fieldF0of characteristic different frompwith a Galois extension E0/F0 having as Galois group the alternating group Am for some mmax (5, p). Write E0 ∼= F0[X]/(f ) withf ∈ F0[X]. We choose F1
as a maximal algebraic extension ofF0such thatf is irreducible overF1and put E1 = F1[X]/(f ). ThenE1/F1is a Galois extension with groupAm. SinceAm is simple, this extension is linearily disjoint from any cyclic extension ofF1. Hence, by its choice the fieldF1is perfect and has no cyclic extensions and, in particular, it contains apth root of unity.
Acknowledgements
The author wishes to express his gratitude to David Leep for his interest in this work and many valuable remarks. The author also thanks the referee, whose sug- gestions helped improve an earlier version of the manuscript.
This research was carried out while the author enjoyed the hospitality of the Institut de Math´ematiques de Jussieu in Paris as a postdoctoral fellow, supported by the TMR network ‘AlgebraicK-Theory, Linear Algebraic Groups and Related Structures’ (ERB FMRX CT-97-0107).
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