Milnor K-Theory and Automorphisms

Milnor K-theory and Automorphisms

Daniel Grech Athesispresentedforthedegreeof Doctor of Philosophy

Department of Mathematics University of Warwick September 23, 2018 Contents

1 Introduction 6

2 K-theory, Motivic cohomology and Homology 12 2.1 Milnor K-theory ...... 12 2.1.1 Transfer maps for Milnor K-theory of semi-local rings with inﬁniteresidueﬁelds ...... 14 2.2 Improved Milnor K-theory ...... 18 2.3 Grayson’smotiviccohomology ...... 21 2.4 Grayson’spresentationforQuillenK-theory ...... 23

3 A new presentation for Milnor K-theory of a ﬁeld 27

M M 3.1 The isomorphism for K0 and K1 ...... 30 M 3.2 Transfer maps for Kn ...... 32 3.3 Relations in KM (R)...... 34 n e 3.4 Surjectivity of the map ...... 37 e 3.4.1 Compatibility of the transfers ...... 43 3.5 Injectivity and homotopy invariance ...... 51 3.5.1 Relations in motivic cohomology ...... 51 3.5.2 The map KG(F ) KM (F )...... 59 n ! n

1 CONTENTS

4 Fundamental theorems for Milnor K-theory 69 4.1 Compatibility of the transfers for local rings ...... 69 4.2 Consequences of reciprocity ...... 75 4.3 The additivity theorem ...... 78 4.4 The resolution theorem ...... 81 4.5 Devissage ...... 86

5 The homomorphism to Quillen K-theory 89 5.1 Multilinearity ...... 89 5.2 The coﬁnality theorem ...... 91 5.3 The Steinberg relation for Quillen K-theory ...... 96

6 Further questions 102 6.1 Surjectivity for local rings ...... 102 6.2 ThecaseforDVRs ...... 104 6.3 The map to homology ...... 105

2 Acknowledgements

First and foremost, I’d like to thank my advisor Marco Schlichting for giving me the oppurtunity to study with him and for devoting many hours to discussing mathematics with me. Many of the results in this thesis are a direct result of his ideas, and I would not have been able to write it without his guidance. I’d like to thank the University of Warwick for giving me the oppurtunity to study there. I’d also like to thank EPSRC for funding me throughout these last 4 years. Finally, I’d like to thank my parents for always supporting me throughout my education.

3 Declaration

The work presented in this thesis is my own except where I have stated otherwise. This work has not been presented for a degree at any other university.

4 Abstract

In this thesis, we give a presentation for Milnor K-theory of a field F whose generators are tuples of commuting automorphisms. This is similar to a presentation for Milnor K-theory given by the cohomology groups of Grayson. The main di↵erence is that, in our presentation, we do not use a homotopy invariance relation, which we should not expect to hold for non-regular rings R. We go on to study this presentation for R alocalring.Weconjecturethatit agrees with the usual definition of Milnor K-theory for any local ring. We give some evidence towards this, including showing that the natural map K (R) K (R)is n ! n injective when n =0, 1, 2orwhenR is a regular, local ring containing an infinite e M field. We also show a reciprocity result for Kn (R)anyringR,which,whenR is a field, allows us to deduce surjectivity of the map. e We prove a version of the additivity, resolution, devissage and cofinality theorems

M M for the groups Kn (R). We also construct a comparison homomorphsim from Kn (R) to the presentation of Quillen K-theory given by Grayson. e e

5 Chapter 1

Introduction

M Milnor K-theory Kn (F )ofafieldFisasequenceofabeliangroupswithacertain presentation. It was originally defined, by Milnor, in [14] based on the presentation of K2(F )ofafieldgivenbyMatsumoto[12].Inthispaper,Milnorconjecturesresults connecting Milnor K-theory mod 2 to quadratic forms and Galois cohomology. More precisely, he constructs homomorphisms

M M n hF : Kn (F )/2Kn (F ) H (G; Z/2Z) ! s : KM (F )/2KM (F ) In(F )/In+1(F ), n n n ! where G is the Galois group of the separable closure of F and I(F )isthefundamental ideal of the Witt ring, and conjectures that these maps are isomorphisms. These conjectures became known as the Milnor conjectures and were proven by Voevodsky, Orlov and Vishik in [16], by using methods in motivic cohomology. Milnor K-theory of a local ring was ﬁrst studied in [15] and [7]. In [15], it is shown that the maps

H (GL (R)) H (GL (R)) H (GL (R)) H (GL(R)), n n ! n n+1 ! n n+2 !···! n

6 induced by the natural inclusion

GL (R) GL (R) i ! i+1 A 0 A 7! 0011 @ A are all isomorphisms when R is a local ring with inﬁnite residue ﬁeld. It is also shown that Milnor K-theory occurs as the obstruction to further stability i.e. that the map

M ⇠= Kn (A) Hn(GLn(A))/Hn(GLn 1(A)) ! is an isomorphism. Nowadays, Milnor K-theory is part of motivic cohomology and there are several theorems relating Milnor K-theory of a ﬁeld to various cohomology theories. In [15] and [21], it is also shown that there is an isomorphism

KM (F ) ⇠= CHn(F, n) n ! where CHn(F, n) are Bloch’s higher Chow groups. Another connection with motivic cohomology is with Voevodsky’s motivic cohomology groups; there is an isomorphism

M ⇠= n,n Kn (F ) H (Spec(F ), Z) ! Many of the proofs of the theorems above rely on some of the nice properties of Milnor K-theory. Of particular importance, is the existence of transfer maps

N M : KM (L) KM (F ) L/F n ! n where L/F is a finite field extension. These maps are defined using Milnor’s exact

M sequence from [14], which computes the Milnor K-theory of Kn (F (t)) in terms of M M the groups Kn 1(F [t]/p(t)) and Kn (F )wherep(t)ismonic,irreducible. 7 In [10], Kerz constructs an analogue of this exact sequence for semi-local rings with inﬁnite residue ﬁelds and uses this to construct transfer maps

N : KM (B) KM (A), B/A n ! n where A is a semi-local ring with infinite residue fields and B/A is an etale extension of semi-local rings with infinite residue field. The existence of transfer maps are used to prove the Gersten conjecture for Milnor K-theory in the equi-charactistic case and this is used to show the Bloch formula

n M n Hzar(X, K )=CH (X) for X a regular, excellent scheme over an infinite field. Therefore, if one wishes to generalise some of these results to the realm of local rings with finite residue fields it would seem that the existence transfer maps are important. Unfortunately, the naive generalisation of Milnor K-theory to local rings with finite residue field does not have transfers in general. However, in [11] Kerz gives adefinition,basedonideasofGabber,ofimprovedMilnorK-theoryofalocalring with finite residue field and shows that this definition has transfers. Furthermore, Kerz shows that improved Milnor K-theory agrees with Milnor K-theory when the residue field is suciently large and that this extension. However, this definition is not given by a presentation as Milnor K-theory usually is. The purpose of this thesis is to give a possible presentation of Milnor K-theory of any local ring, motivated by the motivic cohomology groups of Grayson. The idea is to replace the generators in Milnor K-theory, which are n-tuples of units in R⇤,with n-tuples of commuting automorphisms of finitely generated, projective modules. This allows transfers to be naturally defined for any finite, flat extensions of local rings. This is in contrast to Milnor K-theory where the existence of transfers is not obvious, with the construction of these maps dependent on the existence of a certain exact sequence. In fact, replacing tuples of units with tuples of automorphisms

8 allows transfers to be defined for finite, flat extensions of any ring. This presentation is similar to the one given by the motivic cohomology of Grayson [4]. The main di↵erence is that we do not use a homotopy invarience relation which we do not expect to hold when R is not regular. In chapter 1, we review some of the results in Milnor K-theory and related areas of mathematics that we will need. In the first section of the chapter, we review the construction of transfer maps for Milnor K-theory. We do this both for fields and semi-local rings with infinite residue fields. Along the way, we present the residue homomorphisms and the exact sequence necessary to define these transfers. We end the section by stating some of the properties of these transfer maps. In the next section we give the definition of improved Milnor K-theory studied in [11] and state, without proof, some of its properties. In the third section we give the definition of the motivic cohomology groups of Grayson. These groups motivate our goal to give a presentation of Milnor K-theory which has generators n-tuples of commuting automorphisms. In the fourth section, we present the construction of higher algebraic K-theory of Grayson [5] which gives a presentation of the Quillen K-theory of an exact category in terms of binary complexes.

M In chapter 2, we give our definiton of Kn .Thepurposeofthischapteristoshow that KM (F ) = KM (F )whenF is a field. We begin by showing that the groups n ⇠ n e agree when n =0, 1whenF is a local ring. We go on to define transfers for KM , e n and to prove some of the analogous identites that hold in Milnor K-theory. We then e show that the natural map KM (F ) KM (F )issurjective,byshowingKM (F )is n ! m m generated by images of transfers and that the transfers for KM (F )andKM (F )are e n me compatible. In the last section, we prove that the map is injective. To do this we e construct an inverse by first mapping into the cohomlogy groups of Grayson and then constructing an inverse map from these groups to Milnor K-theory.

M In chapter 3, we study some properties of the groups Kn .Webeginbyprovinga

e 9 M reciprocity law for Kn (R). Two immediate corollaries are that the transfer maps for KM (R)andKM (R)arecompatibleandthetransfersforKM (R)satisfynaturality n n e n when R is a semi-local ring with infinite residue fields. The naturality property is e e M M enough to show that if K (R)agreeswithKn (R)whenR is a local ring with infinite residue field then KM (R)willagreewiththeimprovedMilnorK-groupsKˆ M (R)of n e n Gabber-Kerz when R is a local ring with finite reisdue field. The remainder of this e chapter is dedicated to proving some analogues of fundamental theorems, for Quillen K-theory, in our setting. In particular we prove versions of the additivity, resolution and devissage theorems. In chapter 4 we construct a comparison homomorphism KM (R) KQ(R), such n ! n that the standard comparison homomorphism from KM (R)factorsthroughthis n e M map. This provides further evidence that our defintion for Kn is the correct one. To do this we use the presentation of KQ,duetoGrayson[5],whichwereviewedin n e chapter 1. In the first section we review the proof of the bilinearity relation which we take from the thesis of Harris [20]. In the next section we prove the Steinberg relation

Q holds in Kn (R)foranyringR.Beforewedothis,weproveaversionofthecoﬁnality theorem which will allow us to reduce to proving the Steinberg relation for free modules. We then go on to prove the Steinberg relation using homotopy invariance

Q and functorality of Kn .Becausethecomparisonhomomorphismisanisomorphism when n =2thisallowsustoshowthatthemapKM (R) KM (R)isinjective. 2 ! 2 More generally, we can conclude that the kernel of the map KM (R) KM (R)isa n e ! n torsion group annihilated by (n 1)!. e In Chapter 5, we look at some further questions that we were not able to answer.

M We show, using the resolution theorem, that the groups Kn (R)aregeneratedby images of transfers when R is regular, local. We use this to show that KM (R) = e n ⇠ M M M Kn (R), for R a DVR, if the transfers for Kn are compatible with those for Kn . e e

10 In the last section we construct a map

KM (R) H (GL(R)) n ! n which we conjecture to be thee composition KM (R) KQ(R) H (GL(R)). n ! n ! n e

11 Chapter 2

K-theory, Motivic cohomology and Homology

2.1 Milnor K-theory

In this section, we review some facts about Milnor K-theory including the construction of the transfer maps and its properties. We begin by reviewing the deﬁnition of Milnor K-theory.

Deﬁnition 2.1.1. Let A be a commutative ring. We deﬁne Milnor K-theory, de- noted KM (A), of A to be the graded ring ⇤

M K (A) :=TensZ(A⇤)/I ⇤ where I is the two-sided ideal generated by elements of the form a 1 a, for ⌦ M a, 1 a A⇤. We deﬁne the n’th Milnor K-group K (A) to be the abelian subgroup 2 n generated by elements of degree n.

We denote an element a ... a KM (A)by a ,...,a . As noted earlier, 1 ⌦ ⌦ n 2 n { 1 n} this definition is not the correct one in general when A is a local ring with finite residue field.

12 2.1. MILNOR K-THEORY

AfundamentalresultinMilnorK-theoryistheshortexactsequencewhichcal- culates the Milnor K-theory of a rational function ﬁeld. This short exact sequence is used to construct the transfer maps for Milnor K-theory. We now give the deﬁnition of the residue maps in 2.1.2 and use these to give a presentation of the short exact sequence of Milnor in 2.1.3.

Proposition 2.1.2. Let F be a ﬁeld, v be a discrete valuation on F and F (v) be the residue ﬁeld of F. There exists a unique homomorphism

M @v : Kn (F ) Kn 1(F (v)), ! such that

@ ⇡, u ,...,u = u ,...,u v{ 2 n} { 2 n} where ⇡ is any uniformizing element and ui satisfy v(ui)=0.

Of particular importance, is the case when F is a ﬁeld of rational functions. In this case we get a valuation for each monic, irreducible polynomial p(t). We denote

1 the associated residue map by @p(t). We also have a valuation with uniformizer t . We denote the residue map for this valuation as @ . 1 Theorem 2.1.3. Let F be ﬁeld. The sequence

@⇡ M M M 0 Kn (F ) Kn (F (t)) Kn 1(F [t]/⇡) 0 ! ! ! ! ⇡ irreducible, monicM is split exact.

One can use this sequence to define transfer maps for Milnor K-theory of fields. These were originally defined in [1].

Definition 2.1.4. Let F be a field and L := F [t]/p(t) be a simple field extension. We define a map

N M : KM (L) KM (F ) L/F n ! n

13 2.1. MILNOR K-THEORY to be the composition

M M @ K (L) K (F [t]/⇡) K (F (t)) 1 K (F ) n ! n ! n+1 ! n ⇡ irreducible, monicM where the ﬁrst map is inclusion into the appriopriate direct summand and is the splitting map for the exact sequence in 2.1.3.

We can also define transfer maps for an arbitrary finite field extension L/F by writing L as a tower of finite simple extensions. It was shown in [9] that this is independent of the tower of extensions chosen.

2.1.1 Transfer maps for Milnor K-theory of semi-local rings with inﬁnite residue ﬁelds

In this section, we give the definition of transfer maps defined by Kerz in [10] for finite, etale extensions of semi-local rings with infinite residue fields. To do this we first give the analogue of the exact sequence 2.1.3. To give this sequence we only need to define the middle term and the residue maps. We do this in the following definitions:

Deﬁnition 2.1.5. Let A be a semi-local ring. An n-tuple of rational functions

p1 ,..., pn q1 qn ⇣ ⌘ with p ,q A[t] together with a factorization i i 2

i i pi = aip1 ...pni

i i qi = biq1 ...qmi such that a ,b A⇤ and each p ,q is monic irreducible, is called feasible if the i i 2 i i fraction pi is reduced, if every irreducible factor is either equal or coprime and qi

Disc(p ), Disc(q ) A⇤. i i 2 14 2.1. MILNOR K-THEORY

Deﬁnition 2.1.6. Let A be a semi-local ring. We deﬁne

et T (A) := Z (p1,...,pn) (p1,...,pn) feasible, { | p A[t] irreducible or unit /Linear. i 2 } Where Linear denotes the subgroup generated by elements

(p ,...,ap ,...,p ) (p ,...,p,...,p ) (p ,...,a,...,p ) 1 i i n 1 i n 1 i n where a A⇤. i 2 If we have an n-tuple of rational functions together with a choice of factorization

et as in 2.1.5 then we can deﬁne an element in Tn (A)byusingmultilinearfactor- et M ization. We will now deﬁne the group Kn (A)whichwillreplaceKn (F (t)) in the semi-local ring version of the sequence in 2.1.3.

Deﬁnition 2.1.7. Let A be a semi-local ring. We deﬁne

et et et Kn (A)=Tn (A)/St ,

et et where St is the group generated by the elements in Tn (A) which are associated to feasible n-tuples

p p q p1,..., q , q ,...,pn ⇣ ⌘ p ,..., p , p ,...,p 1 q q n ⇣ ⌘ with (p, q)=1and (q p, q)=1. We can now deﬁne the residue maps in the cases we need them.

Proposition 2.1.8. Let A be a semi-local ring with inﬁnite residue ﬁelds. For every monic, irreducible polynomial ⇡ A[t] there exists a homomorphism 2

et M @⇡ : Kn (A) Kn 1(A[t]/⇡) !

15 2.1. MILNOR K-THEORY such that

@ (⇡, u ,...,u )= u ,...,u . ⇡ 2 n { 2 n} where ui are rational functions as in 2.1.5 such that each irreducible factor is in- M vertible in Kn 1(A[t]/⇡). There also exists a homomorphism

et M @ : Kn (A) Kn 1(A) 1 ! such that

1 1 1 @ ( ,p2(t ),...,pn(t )) = (p2(0),...,pn(0)) 1 t where p A[t] are such that p (0) A⇤. i 2 i 2 We can now state the version of the exact sequence 2.1.3 that we need.

Theorem 2.1.9. Let A be a semi-local ring with inﬁnite residue ﬁelds. The sequence

M et M 0 Kn (A) Kn (A) ⇡Kn 1(A[t]/⇡) 0 ! ! ! ! is split exact, where the sum is taken over all monic, irreducible ⇡ A[t] such that 2 Disc(⇡) A⇤. 2 We are now ready to define the transfer maps for finite etale extensions of semi- local rings with infinite residue fields. To do this we use the following proposition taken from [6] Proposition 18.4.5.

Proposition 2.1.10. Let A be a local ring, k its residue field and B be a finite A-algebra. Suppose, moreover, that k is infinite, B is infinite, or that B is a local ring. Let n be the rank of L := B k over k. Then B is etale if and only if there ⌦A exists a monic polynomial f A[t] with Disc(f) A⇤ such that 2 2

B ⇠= A[t]/f.

Moreover, we have that deg(f)=n.

16 2.1. MILNOR K-THEORY

Definition 2.1.11. Let A be a semi-local ring with infinite residue fields. Let B =

A[t]/⇡(t) where ⇡ is an irreducible monic polynomial with Disc(⇡) A⇤. We deﬁne 2 the transfer maps to be the composition

M M et @ K (B) K (A[t]/⇡) K (A) 1 K (A) n ! n ! n+1 ! n M where is the splitting map in 2.1.9 and the sum is taken over all ⇡ which are irreducible, monic and Disc(⇡) A⇤. To deﬁne the splitting we take the retraction 2

et M s : Kn (A) Kn (A) 1 ! which maps a tuple of polynomials to their leading coecients.

Kerz also proves the following compatibility result which we will need.

Proposition 2.1.12. Let i : A A0 be a homomorphism of semi-local rings. Let B ! be as in the previous deﬁnition and let i(⇡)= j ⇡j be a factorization into irreducible polynomials. Let Bj0 = A0[t]/⇡j. Then the followingQ diagram commutes

M M K (B) K (B0 ) n ! j n j N j NB /A B/A L j0 0 M? M? Kn?(A) Kn ?(A0) y ! y The transfer maps for Milnor K-theory satisfy the following properties

M M M 1. The map NB A : K0 (B) K0 (A)isjustmultiplicationby[B : A]. | !

M M M 2. The map NK k : K1 (B) K1 (A)isgievnby | ! b det T , { }7! { b}

where Tb is the A-linear map

T : B B b ! x bx 7! 17 2.2. IMPROVED MILNOR K-THEORY

3. (projection formula) Let B A be a ﬁnite, etale extension ↵ KM (A)and | 2 n KM (B)wehavethat 2 m

M M NB A( ↵B, )= ↵,NB A() | { } { | }

4. (Composition) Given a tower of etale extensions C B A, we have that | |

NC A = NB A NC B | | |

M M 5. Let B A be a ﬁnite, etale extension and i : Kn (A) Kn (B)bethemap | ⇤ ! induced by the inclusion A B.Then !

M NB A i (↵)=[B : A]↵ | ⇤

2.2 Improved Milnor K-theory

In this section we present the generalisation of Milnor K-theory to local rings with finite residue field due to Gabber [2] and studied in [11]. We will present this generalisation, more generally, for certain types of abelian sheaves. Let C be the category of abelian sheaves on the big Zariski site of all schemes. We define NC to be the full subcategory of abelian sheaves in C such that for every finite etale extension of local rings i: A B there are a system of transfers ⇢

[N : F (B0) F (A0)] B0/A0 ! A0 for any A0 which is local A-algebra such that B0 := B A0 is also local. We re- ⌦A quire these transfers to be compatible in the sense that if A0 A00 are both local ! A-algebras with B0 = B A0 and B00 = B A00 also local then the diagram ⌦A ⌦A

18 2.2. IMPROVED MILNOR K-THEORY

F (B0) F (B00)

NB A NB A 0| 0 00| 00

F (A0) F (A00) commutes. We also assume that

NB0 A0 i0 =[B : A]idF (A), | ⇤ where i0 : F (A0) F (B0)isthemapinducedbyi0 : A0 B0. ⇤ ! ! We denote by NC1 the full subcategory of sheaves which have a system of compatible transfers for all finite, etale extensions A B of local rings such that A ⇢ has infinite residue field. Clearly every sheaf in NC gives a sheaf in NC1.The following theorem, proved in [11], proves that every continuous sheaf in NC1 can be extended uniquely to a sheaf in NC.

Theorem 2.2.1. For every continuous F NC1 there exists a continuous Fˆ 2 2 NC together natural transformation F Fˆ, such that for any continuous G ! 2 NC and natural transformation F G there exists a unique natural transforma- ! tion Fˆ G such that the following diagram ! F Fˆ

G commutes. Moreover for a local ring A with inﬁnite residue ﬁeld we have F (A)= Fˆ(A)

ˆ M We therefore deﬁne the improved Milnor K-theory of a local ring A to be Kn (A). Below we give a more explicit, but equivalent, deﬁnition and then we summarize some of the results proved in [11].

Deﬁnition 2.2.2. Let A be a commutative ring. We deﬁne the subset S A[t ,...,t ] 2 1 n

19 2.2. IMPROVED MILNOR K-THEORY to be

i n S := ait A[t1,...,tn] ai i N = A { 2 |h | 2 i } i Nn X2 The set is multiplicatively closed so we can deﬁne the ring of rational functions to be

1 A(t1,...,tn)=S A[t1,...,tn].

We have maps f ,f : A(t) A(t ,t ), where the map f maps t to t . Then we 1 2 ! 1 2 i i ˆ M deﬁne the improved Milnor K-theory Kn (A) to be

M M K (f1) K (f2) Kˆ M (A)=ker(KM (A(t)) n n KM (A(t ,t ))) n n ! n 1 2

Proposition 2.2.3. Let (A, m) be a local ring. Then:

ˆ M 1. K1 (A)=A⇤.

2. Kˆ M (A) has a natural graded commutative ring structure. ⇤

3. For every n 0 there exists a universal natural number M such that if n A/m >M the natural homomorphism | | n

KM (A) Kˆ M (A) n ! n

is an isomorphism.

4. There exists a homomorphism

KQ(A) Kˆ M (A) n ! n

such that the composition

Kˆ M (A) KQ(A) Kˆ M (A) n ! n ! n

20 2.3. GRAYSON’S MOTIVIC COHOMOLOGY

is multiplication by (n 1)! and the composition

KQ(A) Kˆ M (A) KQ(A) n ! n ! n

is the chern class cn,n.

5. Let A be regular and equicharacteristic, F = Q(A) and X = Spec(A). The Gersten conjecture holds for Milnor K-theory, i.e. the Gersten complex

ˆ M M M 0 Kn (A) Kn (F ) x X(1) Kn 1(k(x)) ... ! ! ! 2 !

is exact.

2.3 Grayson’s motivic cohomology

In this section we present certain non standard cohomology groups studied in [4]. These groups serve as the motivation for our new deﬁnition of Milnor K-theory. One of the motivations for the development of motivic cohomology is that these groups should appear as terms in a spectral sequence

pq p q E2 = H (X, Z( q)) = K p q(R) )

Grayson’s approach to this is to study a ﬁltration of the space K(R)

K(R)=W 0 W 1 ...

due to Goodwillie and Lichtenbaum. We can then deﬁne the groups

m t t+1 HG (X, Z(t)) := ⇡2t m(W /W ).

We will first review the construction of W t. Given two rings R and S we let P(R, S)denotedtheexactcategoryofR-S- bimodules which as R-modules are finitely generated and projective. We define the

21 2.3. GRAYSON’S MOTIVIC COHOMOLOGY

K-theory space

K(R, S) := K(P(R, S)).

1 Let Gm :=SpecZ[U, U ]. Note that the category

1 P(R, Gm) := P(R, Z[U, U ]) is isomorphic to the category whose objects are of the form [P,✓]wherePisa ﬁnitely generated, projective module and ✓ is an automorphism of P .Similarlywe can deﬁne, for t 0,

t 1 1 P(R, Gm)=P(R, Z[U1,U1 ,...,Ut,Ut ]).

We deﬁne

t t K0(R, Gm^ ) := K0(R, Gm)/ [P, A1,...,IP ,...,At] . h i

We deﬁne the R-algebra RAd as the algebraic analogue of an n-simplex

d d RA = R[T0,...,Td]/(T0 + + T 1) ··· We can now define the filtration of Goodwillie and Lichtenbaum. We define

t t d t V := K(RA, Gm^ )= d K(RA , Gm^ ) | 7! | t t t W :=⌦ V

Grayson shows that this ﬁltration satisﬁes the required properties when R is a regular noetherian ring and

t t+1 t d t W /W =⌦ d K0(RA , Gm^ ) | 7! | One can show that

t t+1 n+t t ⇡n(W /W ) ⇠= H (K0(RA, Gm^ )).

22 2.4. GRAYSON’S PRESENTATION FOR QUILLEN K-THEORY

In [18] it is shown that these groups are isomorphic to Voevodsky’s groups when X is a smooth variety over a ﬁeld. We also have that Milnor K-theory is isomorphic to certain motivic cohomology groups. So we should have that

M n 0 n Kn (F )=HG(Spec(F ), Z(n)) = H (K0(RA, Gm^ )).

In chapter 2, we shall prove directly that

M 0 n Kn (F )=H (K0(RA, Gm^ )).

M M in order to prove the main result that Kn (F ) ⇠= Kn (F ).

e 2.4 Grayson’s presentation for Quillen K-theory

In this section we present Grayson’s presentation for Quillen K-theory given in [5], and studied in [8] and [20]. We use this presentation in chapter 4 to construct our version of the comparison homomorphism. We will first give the definition of the category of chain complexes and of binary chain complexes. Let N be an exact category. We first look at chain complexes in this category.

Deﬁnition 2.4.1. Let N be an exact category. A chain complex is a sequence

di+1 di ... Ci+1 Ci Ci 1 ... ! ! ! ! where i Z,Ci Ob(N ) and didi+1 =0for all i Z. We denote a chain 2 2 2 complex by C . A map of chain complexes f : C D is a collection of morphisms · · · ! · f : C D such that the diagram i i ! i

di+1 di di 1 ... Ci+1 Ci Ci 1 ...

fi+1 fi fi 1 di0+1 di0 di0 1 ... Di+1 Di Di 1 ...

23 2.4. GRAYSON’S PRESENTATION FOR QUILLEN K-THEORY commutes. We say that a chain complex C is bounded if N Z such that Ci =0 · 9 2 for all i N and i N. We deﬁne the category CN , to be the category whose  objects are bounded chain complexes of N and whose morphisms are maps of chain complexes. We say that a sequence of morphisms of chain complexes

f. g C D · E. · · is exact, if

fi gi Ci Di Ei is exact for all i. This gives CN the structure of an exact category.

Because CN is exact we can inductively deﬁne

n n 1 C N := C(C N ).

We now deﬁne what it means for a chain complex to be acyclic.

Deﬁnition 2.4.2. Let N be an exact category and C be an acyclic chain complex. · We say that C is acyclic if the sequence factors as ·

Zi Zi 2

... Ci+1 Ci Ci 1 ...

Zi+1 Zi 1 where

Zi Ci Zi 1 are exact for each i. We deﬁne Cq(N ) to be the category of bounded, acyclic complexes.

24 2.4. GRAYSON’S PRESENTATION FOR QUILLEN K-THEORY

We now deﬁne binary complexes of an exact category. They will be the generators

Q of the presentation of Kn .

Deﬁnition 2.4.3. A binary chain complex, of objects in some exact category N , is a triple (C ,d,d0) where both (C ,d) and (C ,d0) are chain complexes. We call d · · · the top di↵erential and d0 the bottom di↵erential. A morphism between two binary complexes (C ,d,d0) and (D ,@,@0) is a morphism of the chain complexes · · f :(C ,d) (D ,@) · ! · f :(C ,d0) (D ,@0). · ! · i.e. f must commute with both the top and bottom derivative. We deﬁne BN to be the category of bounded, binary chain complexes. A sequence of morphisms is exact if it is exact on the underlying Z-graded objects.

As with chain complexes, because BN is an exact category we can inductively n n 1 deﬁne B N := B(B N ). Given a chain complex, there is a natural way to get a binary chain complex by taking both the top and bottom di↵erentials to be the di↵erential of the chain complex. Conversely, given a binary chain complex we can deﬁne two chain complexes, one by using the top di↵erential, the other by using the bottom di↵erential. This gives us three functors

:CN BN (C ,d) (C ,d,d) ! · 7! · : BN CN (C ,d,d0) (C ,d) > ! · 7! · : BN CN (C ,d,d0) (C ,d0) ? ! · 7! · We call the diagonal functor, the top functor and the bottom functor. We > ? say that a binary complex is acyclic if its image under both and is acyclic. We > ? deﬁne BqN to be the category of bounded, acyclic binary complexes. One can show q q n q q n 1 that B N is an exact category, so we can deﬁne (B ) (N ) := B ((B ) N ).

25 2.4. GRAYSON’S PRESENTATION FOR QUILLEN K-THEORY

We can describe objects of (Bq)n(N )asZn-graded collections of objects, together with acyclic di↵erentials d1,d10 ,...,dn,dn0

di,di0 : C(x1,...,xi,...,xn) C(x1,...,xi 1...,xn) ! such that di↵erentials in opposite direction commute. We call an object of (Bq)nN an n-dimensional bounded acyclic binary multicomplex. We can extend the functors above to the setting of multicomplexes. If I have an n-dimensional bounded acyclic binary multicomplex I can get a complex of (n 1)- dimensional bounded acyclic binary multicomplexes by forgetting one of the di↵erentials. There are 2n ways to do this which gives us functors

i q n q q n 1 :(B ) N C (B ) N > ! i q n q q n 1 :(B ) N C (B ) N . ? !

We also have a version of the diagonal functor. Given any chain complex of (n 1)- dimensional bounded acyclic binary multicomplexes we can get a n-dimensional bounded acyclic binary multicomplex by duplicating the di↵erential in the i’th direction. This gives us functors

i q q n 1 q n : C (B ) N (B ) N !

If a binary multicomplex is in the image of i,forsomei,thenitiscalleddiagonal. We are now ready to state the main result of [5].

Theorem 2.4.4. Let N be an exact category. We have an isomorphism

Q q n Kn (N ) ⇠= K0((B ) N )/Diag

q n where Diag is the subgroup of K0((B ) generated by the diagonal binary multicomplexes.

26 Chapter 3

A new presentation for Milnor K-theory of a ﬁeld

In this chapter, we give a presentation for Milnor K-theory of ﬁelds in terms of commuting automorphisms. We begin by giving some motivation and proving some of the basic identities for Milnor K-theory in this new setting. We then go on to show that the groups are isomorphic for a ﬁeld F . In section 2.3wesaidthatMilnorK-theoryisisomorphictoGrayon’smotivic cohomology groups. This suggests that a presentation of Milnor K-theory for local rings could be

M Kn (R)=Z [P, A1,...,An] /(some relations). { }

However, the presentatione of Grayson’s cohomology groups includes a homotopy invariance relation which we should not expect to hold when R is not regular. In 3.5 we prove explicitely that these cohomology groups are isomorphic to Milnor K-theory for F a field. In the proof, we need the natural homomorphism to be well-defined and we need an exact sequence relation to hold. For the map to be well-defined we need the multilinearity and Steinberg relations to hold for rank one elements.

27 We also need transfers to exist, so we need the relations to hold for any commuting automorphisms of projective modules. This motivates the following deﬁnition:

M Deﬁnition 3.0.1. Let R be a commutative ring, we deﬁne the groups Kn (R) to be

M e Kn (R) := Z [P, A1,...,An] /(1) (3) { } e where P is a ﬁnitely generated, projective R-module, Ai are automorphisms of P that commute pairwise and relations are (1)-(3) are as follows:

1. [P1,A1,...,An]+[P3,C1,...,Cn]=[P2,B1,...,Bn], if there exists an exact sequence

f g 0 P P P 0 ! 1 ! 2 ! 3 !

such that

f A = B f and g B = C g i i i i

for every i.

2. [P, A1,...,AiAi0 ,...,An]=[P, A1,...,Ai,...,An]+[P, A1,...,Ai0 ,...,An].

3. [P, A1,...,An]=0, if Ai + Ai+1 =IdP for some i.

M More generally we deﬁne Kn E for an exact category E :

Deﬁnition 3.0.2. Let E be ane exact category. We deﬁne Autn(E ) to be the category whose objects are elements of the form [M,⇥ ,...,⇥ ] such that M ob(E ) and 1 n 2 ⇥i are automorphisms of M such that ⇥i⇥j =⇥j⇥i for all i,j. The morphisms between two objects [M1, ⇥1,...,⇥n] and [M2, 1,...,n] are the set of morphisms f : M M in E such that f ⇥ = f for every i. We say that a sequence 1 ! 2 i i

f g [M , ⇥ ,...,⇥ ] [M , ,..., ] [M , ,..., ] 1 1 n ! 2 1 n ! 3 1 n

28 is exact in Autn(E ) if

f g M M M 1 ! 2 ! 3 is exact in E .

This makes Autn(E )intoanexactcategory.WecannowdeﬁnetheMilnor K-groups of an exact category.

M Definition 3.0.3. We define K0 (E ) to be the usual Grothendieck group of an exact category i.e. e M K0 (E ) := Z ob(E ) /short exact sequences. { } We then define KM (eE ) for i 1 as follows: i M M 1 K (eE ) := K (Aut (E ))/ [M,⇥1⇥2]=[M,⇥1]+[M,⇥2] 1 0 h i M M i e e Ki (E ) := K0 (Aut (E ))/H where H is the subgroup generatede by anye element of the two following forms:

[M,⇥ ,...,⇥ ⇥ ,...,⇥ ] [M,⇥ ,...,⇥ ,...,⇥ ] [M,⇥ ,...,⇥ ,...,⇥ ] 0 i i+1 n 0 i n 0 i+1 n

[M,⇥1,...,⇥n] whenever ⇥i +⇥i+1 =IdM for some i

To simplify notation we deﬁne

M M Kn (R) := Kn (ProjR)

M M Gen (R) := Ken (ModR) where ProjR is the category of ﬁnitelye generatede left projective R-modules and ModR is the category of ﬁnitely generated left R-modules. The purpose of this chapter is to show that the natural map

KM (F ) KM (F )(3.1) n ! n a ,...,a [F, a ,...,a ](3.2) { 1 n}7! e 1 n is an isomorphism when F is a ﬁeld.

29 M M 3.1. THE ISOMORPHISM FOR K0 AND K1

M M 3.1 The isomorphism for K0 and K1

In this section, we show that these groups agree with Milnor K-theory when n =0, 1. In fact, we show this for any local ring For n =0,thismapisdeﬁnedas

KM (R) KM (R) 0 ! 0 m [Rm]. 7! e It is simple to show that this map is both injective and surjective. We now deal with the case n =1.

Proposition 3.1.1. Let R be any commutative ring such that every matrix over R can be reduced to a diagonal matrix by elementary row and column operations. Then the map

M g : R⇤ K (R) 1 ! 1 a [R, a] 7! e is an isomorphism.

Proof. To show the map is injective we construct an inverse map. Deﬁne

1 M : K (R) R⇤ 1 ! [Rm,A] det(A). e 7!

M To show the map is well-deﬁned we only need to show that the relations in K1 (R) are satiﬁed. The multilinearity relation follows from the identity e det(AB) = det(A)det(B).

The exact sequence relation follows from the identity

AB det =det(A)det(C) 00 C1 @ A 30 M M 3.1. THE ISOMORPHISM FOR K0 AND K1

To show the map is surjective, we ﬁrst deﬁne e(i,j)()tobethematrix

1 ... 0 ... 0 ... 0 ...... 0...... 1 B C B0 ... 1 ... ... 0C B C B...... C B...... C B C B C B0 ... 0 ... 1 ... 0C B C B...... C B...... C B C B C B0 ... 0 ... 0 ... 1C B C @ A where the is in the i’th row and j’th column. We claim this element is trivial in KM (F ). We prove this by induction on the size of the matrix. For a 1 1matrix 1 ⇥ the result is trivial. Assume it is true for an n n matrix. Take a matrix e () e ⇥ (i,j) and any standard basis vector e with k = j.Thenwehaveanexactsequence k 6

n n 1 0 [F.e , 1] [F ,e ()] [F ,A] 0 ! k ! (i,j) ! ! where A is a matrix of the form e(m,l)(0). By linearity [F.ek, 1] = 0 and by induction n 1 [F ,A]=0. Therefore, using the linearity relation we have that

m m [R ,A]=[R , Ae(i,j)()] for any A GL (R)and R.Sogivenanelement[Rm,A] KM (R)wecanuse 2 m 2 2 1 the above relation to row reduce A to a diagonal matrix. From there we can use the e exact sequence relation to write

m m [R ,A]= [R, ai] i=1 X for some a R⇤. i 2

31 M 3.2. TRANSFER MAPS FOR KN

M e 3.2 Transfer maps for Kn

M In this section, we deﬁne the transfer mapse for Kn . First, we deﬁne multiplication on the graded abelian group e

1 M M K (R) := Ki (R). ⇤ i=0 M e e by the formula

[P ,A ,...,A ] [P ,B ,...,B ] := 1 1 n ⌦ 2 1 m [P P ,A Id ,...,A Id , Id B ,...,Id B ] 1 ⌦ 2 1 ⌦ P2 n ⌦ P2 P1 ⌦ 1 P2 ⌦ m

Deﬁnition/Proposition 3.2.1. Let R S be a ﬁnite map of commutative rings ! such that S is projective as an R-module. The map

N M : KM (S) KM (R) S/R n ! n [M,✓ ,...,✓ ] [M,✓ ,...,✓ ] e1 e n 7! e1 n is well-deﬁned and satisﬁes the following:

1. If R and S are local rings, then the map N M : KM (S) KM (R) is just S/R 0 ! 0 multiplication by [S : R]. e e e

2. If R and S are local rings, the map N M : KM (S) KM (R), is given by S/R 1 ! 1

[V,✓] e[R, dete (✓)] e 7! R

where detR(✓) is the determinant of ✓ as an R-linear map.

3. (Composition) Let R S T be a composition of ﬁnite maps such that S is ! ! a projectve R-module and T is a projective S-module, then

N M = N M N M T/R S/R T/S e 32 e e M 3.2. TRANSFER MAPS FOR KN

M M e 4. (Projection formula) Let i : Kn (R) Kn (S) be the map induced by inclu- ⇤ ! sion, [V,⇥ ,...,⇥ ] KM (R) and [W, ⇥ ,...,⇥ ] KjhM (S). Then 1 n 2 n e en+1 n+m 2 m

M e e NS/R(i ([V,⇥1,...,⇥n]) S [W, ⇥n+1,...,⇥n+m]) = ⇤ ⌦ [V,⇥ ,...,⇥ ] N M ([W, ⇥ ,...,⇥ ]) e 1 n ⌦R S/R n+1 n+m e 5. Let R S be a finite, etale extension of semi-local rings with infinite residue ! M M fields. Let i : Kn (R) Kn (S), be the map induced by inclusion i: R S. ⇤ ! ! Then e e

M NS/R i =[S : R] Id ⇤ ⇥ e Proof. We only prove (4) and leave the other identites to the reader. Note that (5) is a special case of the projection formula.

M NS/R(i ([V,⇥1,...,⇥n]) S [W, ⇥n+1,...,⇥n+m]) = ⇤ ⌦ N M ([V S, ⇥ Id ,...,⇥ Id ] [W, ⇥ ,...,⇥ ]) = e S/R ⌦R 1 ⌦ S n ⌦ S ⌦S n+1 n+m N M ([V W, ⇥ Id ,...,⇥ Id , Id ⇥ ,...,Id ⇥ ]) = S/Re ⌦R 1 ⌦ W n ⌦ W V ⌦ n+1 V ⌦ n+m [V,⇥ ,...,⇥ ] N M ([W, ⇥ ,...,⇥ ]) e 1 n ⌦R S/R n+1 n+m e

We also have transfers of the form

N M : GM (S) GM (R) S/R n ! n for ﬁnite, ﬂat maps R S. e e e !

33 M 3.3. RELATIONS IN KN (R)

M e 3.3 Relations in Kn (R)

In this section, we prove somee of the standard identities for Milnor K-theory for M Kn . Usually the proofs of these theorems only hold for rings with many units, however in these new groups we can get around this by using matrices. This is e M one of the beneﬁts of having more general transfers for Kn (R). We now prove the following useful identity which is used to prove graded commutativity as well as the e reciprocity law.

Proposition 3.3.1. Let E be an exact category. Then

[M,⇥ ,...,⇥ ]=0 KM (E ) 1 n 2 n if ⇥i +⇥i+1 =0for some i. e Proof. We begin by proving the theorem in the case when 1 ⇥ is invertible. For i this we use the identity

1 ⇥i ⇥i = 1 . 1 ⇥ i Using this we can see that

1 ⇥i [⇥1,...,⇥i, ⇥i,...,⇥n]=[⇥1,...,⇥i, 1 ,...,⇥n] 1 ⇥ i =[⇥,...,⇥ , 1 ⇥ ,...,⇥ ] 1 i i n 1 [⇥ ,...,⇥ , 1 ⇥ ,...,⇥ ] 1 i i n =0

We now prove the identity when 1 ⇥ is not invertible. To do this we prove i that 3[M,⇥1,...,⇥n]=0and4[M,⇥1,...,⇥n]=0. Let : M 3 M 3 be the morphism given by the matrix ! 00 IdM 0Id 0 Id ⇥ 1 . M M i B C B 0IdM ⇥i C B C @ A 34 M 3.3. RELATIONS IN KN (R)

Consider the element e

3 [M , ⇥ Id 3 ,...,⇥ , ⇥ ,...,⇥ Id 3 ]. 1 ⇥ M i ⇥ i ⇥ n ⇥ M M We claim that this element is 0 in Kn (E ). To show this we only need to show that

Id 3 ⇥ is invertible which is easy to show. So the element above is trivial M i ⇥ e and using multilinearity we obtain

3 0=[M , ⇥ Id 3 ,...,⇥ Id 3 , ⇥ Id 3 ,...,⇥ Id 3 ] 1 ⇥ M i ⇥ M i ⇥ M n ⇥ M 3 +[M , ⇥ Id 3 ,...,⇥ Id 3 , ,...,⇥ Id 3 ] 1 ⇥ M i ⇥ M n ⇥ M 3 +[M , ⇥ Id 3 ,...,, ⇥ Id 3 ,...,⇥ Id 3 ] 1 ⇥ M i ⇥ M n ⇥ M 3 +[M , ⇥ Id 3 ,...,, ,...,⇥ Id 3 ]. 1 ⇥ M n ⇥ M We claim that the ﬁnal 3 elements in this sum are 0. The last element is 0 because 1 isinvertible.Theothertwoare0becausewecanuseelementaryrowand column operations to reduce to the identity matrix. So we have proved that

3 0=[M , ⇥ Id 3 ,...,⇥ Id 3 , ⇥ Id 3 ,...,⇥ Id 3 ], 1 ⇥ M i ⇥ M i ⇥ M n ⇥ M and using the exact sequence relation we get

3[M,⇥ ,...,⇥ , ⇥ ,...,⇥ ]=0 1 i i n as required. The proof that 4[M,⇥ ,...,⇥ , ⇥ ,...,⇥ ]=0issimilartaking 1 i i n :M 4 M 4 ! to be the morphism given by the matrix

000 Id M 0 1 IdM 00IdM +⇥i . B C B 0IdM 0 ⇥i C B C B C B 00IdM 0 C B C @ A

35 M 3.3. RELATIONS IN KN (R)

We have a few corollaries of this result. It gives us graded-commutativitye of the multiplication deﬁned on KM (R). ⇤

Corollary 3.3.2. Let E bee an exact category. Then the identity

[M,⇥ ,...,⇥ , ⇥ ,...,⇥ ]= [M,⇥ ,...,⇥ , ⇥ ,...,⇥ ] 1 i i+1 n 1 i+1 i n holds in KM (E ), for any [M,⇥ ,...,⇥ , ⇥ ,...,⇥ ] KM (E ). In particular if R n 1 i i+1 n 2 n is a commutative ring we have that e e

[P ,A ,...,A ] [P ,B ,...,B ]=( 1)mn[P ,B ,...,B ] [P ,A ,...,A ] 1 1 n ⌦ 2 1 m 2 1 m ⌦ 1 1 n in KM (R). ⇤

Proof.e The proof is the same as the proof that is given usually for Milnor K-theory.

0=[M,⇥ ,...,⇥ ⇥ , ⇥ ⇥ ,...,⇥ ] 1 i i+1 i i+1 n =[M,⇥ ,...,⇥ , ⇥ ,...,⇥ ]+[M,⇥ ,...,⇥ , ⇥ ,...,⇥ ] 1 i i n 1 i i+1 n +[M,⇥ ,...,⇥ , ⇥ ,...,⇥ ]+[M,⇥ ,...,⇥ , ⇥ ,...,⇥ ] 1 i+1 i+1 n 1 i+1 i n =[M,⇥ ,...,⇥ , ⇥ ,...,⇥ ]+[M,⇥ ,...,⇥ , ⇥ ,...,⇥ ] 1 i i+1 n 1 i+1 i+1 n as required.

Acorollaryof3.3.2isthat[M,⇥ ,...,⇥ ]=0 KM (E )if⇥ +⇥ =1or 1 n 2 n i j ⇥ = ⇥ for any i = j.Beforemovingonweneedoneﬁnalidentity: i j 6 e Corollary 3.3.3. Let E be an exact category, then the identity

⇥i [M,⇥1,...,⇥i, ⇥i+1,...,⇥n]=[M,⇥1,..., , ⇥i +⇥i+1,...,⇥n] ⇥i+1

M holds in Kn (E ), whenever ⇥i +⇥i+1 is invertible.

36 3.4. SURJECTIVITY OF THE MAP

Proof. Using multilinearity we have that

⇥i [M,⇥1,..., , ⇥i +⇥i+1,...,⇥n] ⇥i+1 ⇥i ⇥i ⇥i =[M,⇥1,..., , +1...,⇥n]+[M,⇥1,..., , ⇥i+1,...,⇥n] ⇥i+1 ⇥i+1 ⇥i+1 The ﬁrst element in the sum is trivial by the Steinberg relation. Then using multilinearity on the second term we see that the sum is equal to

[M,⇥ ,..., ⇥ , ⇥ ,...,⇥ ]+[M,⇥ ,...,⇥ , ⇥ ,...,⇥ ] 1 i+1 i+1 n 1 i i+1 n The ﬁrst term is trivial by 3.3.1 and so the sum is equal to

[M,⇥1,...,⇥i, ⇥i+1,...,⇥n] as required.

3.4 Surjectivity of the map

In this section, we will show that the map (3.1)

KM (F ) KM (F ) n ! n

M is surjective. To do this we will first show thate the groups Kn (F )aregeneratedby images of transfer maps. To finish the proof we then show that the transfer maps e M M for Kn (F )arecompatiblewiththemapsforKn (F ). We do the first part, more generally, for the groups K (F, n )definedinchapter2.3becauseitwillbemore 0 Gm e useful later to have this result.

m n For any element [F ,A1,...,An] K0(F, Gm)wedeﬁneaF [t1±,...,tn±]-module 2 m F where multiplication by ti is just multiplication by Ai. Note that this is well- deﬁned because the matrices commute and are invertible. We call an element

m [F ,A1,...,An]simpleifitsassociatedF [t1±,...,tn±]-module is simple. We claim n that the simple elements generate the group K0(F, Gm).

37 3.4. SURJECTIVITY OF THE MAP

m Lemma 3.4.1. Every element [F ,A1,...,An] can be written as a sum of simple n elements in K0(F, Gm).

m Proof. Assume not, then there exists an element [F ,A1,...,An], with m minimal, m which cannot be written as a sum of simple elements. [F ,A1,...,An]cannotbe simple itself so there must be a subspace V F m such that A restricts to an ⇢ i isomorphism on V .Thereforewehaveanexactsequence

0 [V,A ,...,A ] [F m,A ,...,A ] [F m/V, A ,...,A ] 0. ! 1 n ! 1 n ! 1 n ! m Using the exact sequence relation we can write [F ,A1,...,An]asasumoftwo elements each of which have rank less than m. So then each of these elements must

m be a sum of simple elements, hence so is [F ,A1,...,An].

We will now show that the simple elements are images of some rank 1 ele-

m ment under some transfer map. Take any simple element [F ,A1,...,An]. Then, m as explained above, F is naturally a simple F [t1±,...,tn±]-module. The simple

F [t1±,...,tn±]-modules are those of the form F [t1±,...,tn±]/m where m is a maximal m ideal. So there is an F [t1±,...,tn±]-module isomorphism where multiplication on F by Ai corresponds to multiplication on F [t1±,...,tn±]/m by ti.Wethereforehavethe following:

m n Proposition 3.4.2. Let [F ,A1,...,An] K0(F, Gm) be simple and let 2

F [t1±,...,tn±]/m be a ﬁnite extension of F such that

m F [t1±,...,tn±]/m = F as a F [t1±,...,tn±]-module. Then

m NF [t1,...,tn]/m F ([F [t1±,...,tn±]/m, t1,...,tn]]) = [F ,A1,...,An] |

n Hence K0(F, Gm) is generated by images of rank 1 elements under transfer maps.

38 3.4. SURJECTIVITY OF THE MAP

n We now give another proof of the fact that K0(F, Gm)isgeneratedbytheimages of transfer maps in the hope that one of these methods may generalise to the case of local rings considered later.

n Take an element [V,A1,...,An] K0(F, Gm). Take a polynomial, of minimal 2 degree, p(t) F [t]suchthatthenullityofp(A )isgreaterthan0.Thatis,there 2 1 exists a non-zero vector v such that p(A1)v =0.Weclaimthatsuchapolynomial p(t) is irreducible. Assume not then let

p(t)=p1(t)p2(t).

Then we must have that both p1(A1)andp2(A1)havenullity0.Butthenp1(A1) must annihilate p2(A1)v which gives a contradiction. So p(t)mustbeirreducible.

We deﬁne an F -subspace Vp(t) to be the set annihilated by p(A1)i.e.

V = v V p(A )v =0 . p(t) { 2 | 1 }

We claim that Ai restrict to automorphisms on Vp(t).Toshowthis,weonlyneedto show that the map

A : V V i p(t) ! p(t) is well-deﬁned, i.e. that the image of the map is is contained in Vp(t).Thisfollows from the commutativity of A1 and Ai. Hence we have an exact sequence

0 [V ,A ,...,A ] [V,A ,...,A ] [W, B ,...,B ] 0. ! p(t) 1|Vp(t) n|Vp(t) ! 1 n ! 1 n ! Using the exact sequence relation and continuing inductively on W gives that

K0(F, Gm)isgeneratedbyelementsoftheform[Vp(t),A1,...,An]whereeveryvec- tor v V is annihilated by p(A ). We now use a change of basis to put A into 2 p(t) 1 1 rational canonical form which converts A1 into a block diagonal matrix of the form

C1 ... 0 . . . 0 . .. . 1 (3.3) B C B 0 ... ClC B C @ A 39 3.4. SURJECTIVITY OF THE MAP

where Ci is of the form

i 0 ... 0 a0 01 ... 0 ai 1 1 . (3.4) B. .. . . C B. . . . C B C B i C B0 ... 1 ami 1C B C If any of these square matrices@ are of size lessA than deg(p) deg(p)thenthere ⇥ would exist a vector annihilated by a polynomial of smaller degree than p.Thisis impossible by the construction. Alternatively, if any of these blocks are larger than deg(p) deg(p), then p(C )e =0wheree is the ﬁrst standard basis vector. So each ⇥ i 1 6 1 matrix is sqaure of size deg(p) deg(p). We know that the characteristic polynomial ⇥ of each matrix must be p(t), otherwise C (A ) p(A )wouldannihilateanon-zero Ci 1 1 vector. It is known that the characteristic polynomial of matrices of the form 3.4 is

mi i mi 1 i CCi (t)=t ami 1t a0. ···

This shows that Ci = Cj for every i and j.Furthermoreif

m m 1 p(t)=t bm 1t ... b0 then

0 ... 0 b0 0 1 1 ... 0 b1 C = . i B. .. . . C B. . . . C B C B C B0 ... 1 bm 1C B C @ A We have changed A1 into an element which is an image of a transfer. We now look at what this change of basis does to Ai. One useful property of matrices of the form 3.4 is the following:

Lemma 3.4.3. Let R be a commutative ring. Let A GL (R) be a companion 2 n n matrix of the form 3.4 above. If a matrix B commutes with A then B = bnA +...+b0 for some b R. i 2 40 3.4. SURJECTIVITY OF THE MAP

To prove this we use the following:

Lemma 3.4.4. Let R be a commutative ring and A GL (R) be a matrix of the 2 n form 3.4 above. If A commutes with a matrix of the form

0 x(1,2) ... x(1,n 1) x(1,n) 0 1 0 x(2,2) ... x(2,n 1) x(2,n) B = , (3.5) B...... C B. . . . . C B C B C B0 x(n,2) ... x(n,n 1) x(n,n)C B C @ A then x(i,j) =0, for every i, j

Proof. Denote the i’th column of the matrix B above by ci.IfwemultiplyBbyA on both sides, then using commutativity we obtain

0 Ac2 . . . Acn = c2 ... cn a1c2 + + an 1cn . ··· ⇣ ⌘ ⇣ ⌘ In particular, we obtain

c2 = 0 and Aci = ci+1.

So by induction, we obtain that ci =0,foralli.

We can now prove lemma 3.4.3.

Proof. Take an arbitrary matrix

x(1,1) x(1,2) ... x(1,n 1) x(1,n) 0 1 x(2,1) x(2,2) ... x(2,n 1) x(2,n) B = B ...... C B . . . . . C B C B C Bx(n,1) x(n,2) ... x(n,n 1) x(n,n)C B C @ A which commutes with A. Consider the matrix

i 1 n 1 B x I x A x A . (1,1) n ··· (i,1) ··· (n,1)

41 3.4. SURJECTIVITY OF THE MAP

We claim this matrix satisﬁes the conditions of 3.4.4, this is easy to see based on

i the fact that the ﬁrst column of A is ei+1,whereej is the j’th standard unit basis vector. Therefore, the sum above must be equal to 0 and so

i 1 n 1 B = x I + + x A + + x A (1,1) n ··· (i,1) ··· (n,1)

Using lemma 3.4.3 and the discussion above we have the following:

n Lemma 3.4.5. K0(F, Gm) is generated by elements of the form

A... 0 . . . 20 . .. . 1 ,B2(A) ... ,Bn(A)3 6B C 7 6B0 ... AC 7 6B C 7 4@ A 5 where A is a companion matrix with irreducible characteristic polynomial and Bi(t) are matrices of the form

i i p1,1(t) ... pm,1(t) . . . 0 . .. . 1 B i i C Bpm,1(t) ... pm,m(t)C B C @ A where pk F [x]. i,j 2 The symbol in the above lemma is the image under some transfer map. It is equal to

t...0 20. .. .1 3 NF [t]/cA(t) . . . ,B2(t),...,Bn(t) 6B C 7 6B0 ... tC 7 6B C 7 4@ A 5 where cA(t) is the characteristic polynomial of A. Repeating this process with B2(t) n in place of A1 and continuing similarly gives that K0(F, Gm)isgeneratedbyimages of rank one elements of some transfer.

42 3.4. SURJECTIVITY OF THE MAP

3.4.1 Compatibility of the transfers

The aim of this section is to show that the transfer maps commute. The proof we give here is based on the methods in [1] which allows us to reduce to proving the proposition for field extensions K/k with [K : k]=p for some prime p,wherek is afieldwhichhasnofieldextensionswithdegreecoprimetop.Itissimpletoprove the proposition in this case however reducing to this case is dicult. We give a di↵erent proof later which works for semi-local rings and is more elementary.

Proposition 3.4.6. For any ﬁnite extension K k, the diagram | KM (K) KM (K) n ! n N M M K k NK k | e | M? M? e Kn?(k) Kn?(k) y ! y commutes. e

We ﬁrst prove this proposition for the ﬁeld extensions we mentioned above. We need the following lemma to do this:

Lemma 3.4.7. Let K = k(a) be a ﬁeld extension obtained by adjoining an element a of degree d to k. Then KM (K) is generated as a left KM (k)-module by elements ⇤ ⇤ of the form

⇡ (a),...,⇡ (a) { 1 m } where ⇡ are monic irreducible polynomials in k[t] satisfying deg(⇡ ) < < deg(⇡ ) i 1 ··· m  d 1 This allows us to prove proposition 3.4.6 for these certain ﬁeld extensions.

Lemma 3.4.8. Proposition 3.4.6 is true if k is a ﬁeld such that every ﬁnite extension of k is of order pn, for some prime p, and [K : k]=p.

43 3.4. SURJECTIVITY OF THE MAP

Proof. To prove this, we use the properties of the tranfer map and lemma 3.4.7. Take an arbitrary generator given in Lemma 3.4.7. There are no irreducible polynomials

M of degree less than p which have degree greater than 1. So we know that Kn (K)is generated by elements of the form

t + a ,a ,...,a . { 1 2 n} Then using the projection formula and the fact that the transfer maps commute when n =1wearedone.

The following proposition allows us, by induction, to remove the assumption that [K : k]=p in lemma 3.4.8. A proof can be found in [3].

Proposition 3.4.9. Let k be a field such that every finite extension of k has degree pn for some prime p and let K/k be a proper finite extension. Then there exists a subfield k K K such that K /k is a normal extension of degree p. ⇢ 1 ⇢ 1 Using the composition of transfer maps we can deduce that proposition 3.4.6 holds whenever k is a field such that every finite field extension of k has order pn for some prime p. We now begin by trying to reduce the general case to this case. We first need

M the following nice property of the transfer map for Kn .

Proposition 3.4.10. Let F [t]/p(t) F be a simple ﬁelde extension and L/F a ﬁeld | extension. Let

p(t)=p1(t) ...pl(t) be the irreducible factorization of p(t) in L[t]. Then diagram

iF [t]/p(t) L[t]/p (t) KM (F [t]/p(t)) | i KM (L[t]/p (t)) n ! n i N M N M F [t]/p(t) F L L[t]/p (t) L e | e i | P ? e iF L ? e KM?(F ) | KM?(L) ny ! ny commutes. e e 44 3.4. SURJECTIVITY OF THE MAP

M Proof. We ﬁrst compute iF L NF [t]/p(t) F . | |

M e m iF L NF [t]/p(t)([F [t]/p(t),f1(t),...,fn(t)]) = [L ,f1(A),...,fn(A)] | where A is thee companion matrix whose characteristic polynomial is p(t). We claim

1 that we can choose an invertible matrix P such that P AP is a block upper trian- gular matrix, which has companion matrices on the diagonal. Furthermore, we can choose P such that the characteristic polynomials of the matrices on the diagonal are precisely the irreducible factors of p(t)inL[t]. The proof of this was essentially

n done in the second proof of the fact that K0(F, Gm)isgeneratedbyimagesoftrans- fers. Now, using the exact sequence relation to get rid of the elements above the diagonal, we can see that the image of the ﬁrst composition is

deg pi [L ,f1(Ai),...,fn(Ai)], i X where Ai is the companion matrix whose characteristic polynomial is pi. Next we compute the other composition. This is a similar calculation and so we get that the image under the other composition is

deg(pi) [L ,f1(Ai),...,fn(Ai)] i X as required.

An analogous result to the above holds for Milnor K-theory a proof of which can be found in [3].

Remark 3.4.11. We only proved 3.4.10 for rank 1 elements. This is enough to prove that the map g is surjective, which will give that the diagram commutes for ⇤ M all elements in Kn (F ).

We now begine to show Prop 3.4.6 for the general case. We deﬁne to be the

45 3.4. SURJECTIVITY OF THE MAP subgroup

M M := (gF NF [t]/p(t) F NF [t]/p(t) F gF [t]/p(t))( a1,...,an ): h | | { } a ,...,a KM (F [t]/p(t)) e { 1 n}2 n i where gF is the map

KM (F ) KM (F ) n ! n

Our aim is to show this group is trivial. Wee ﬁrst show that it is a torsion group.

Proposition 3.4.12. Let F be a ﬁeld and L be an algebraic extension of F. The kernel of the natural map

KM (F ) KM (L) n ! n is a torsion group. e e

m Proof. Take an arbitrary element [F ,A1,...,An]inthekernel.IfL is finite, the result follows from the projection formula for the transfer map. If L is not finite, it is true that there exists a finite field extension F 0 F such that |

m M [F ,A ,...,A ]=0 K (F 0). 1 n 2 n

e m This is true because only ﬁnitely many relations are needed to reduce [F ,A1,...,An] M to 0 in Kn (L).

So toe show is a torsion group it suces to show that is in the kernel of the map KM (F ) KM (F ). n ! n e e Proposition 3.4.13. For any ﬁeld F with algebraic closure F we have that iF F () = | 0.

46 3.4. SURJECTIVITY OF THE MAP

Proof. Take an arbitrary element

M M (gF NF [t]/p(t) F NF [t]/p(t) F gF [t]/p(t))( f1(t),...,fn(t) ) | | { } in . It is simple to see that e

M M iF F gF NF [t]/p(t) F f1(t),...,fn(t) = gF iF F NF [t]/p(t) F f1(t),...,fn(t) . | | { } | | { } By 3.4.10, we have a commutative diagram

M M M gF iF F NF [t]/p(t) F f1(t),...,fn(t) = gF NF [t]/p(t) F [t]/t a f1(t),...,fn(t) | | { } | i { } X = [F,f1(ai),...,fn(ai)]. i X Where the sum ranges over the roots of p in F .Asimilarcalculationshowsthat

M iF F NF [t]/p(t) F gF [t]/p(t) = i[F,f1(ai),...,fn(ai)]. | | So wee have shown that Pis a torsion group. To continue we need the following proposition:

M M Proposition 3.4.14. Let F be a field, p be a prime and let Gn be Kn or Kn . Then there exists an algebraic extension L of F such that every finite extension of e L has order a power of p and such that the map G (F ) G (L) is injective. n (p) ! n Proof. First we set some notation. We define an ordinal to be an equivalence class of totally ordered set. For any ordinal ↵ and any x ↵ we define x +1tobethe 2 smallest element in the set

y ↵ : y>x . { 2 }

47 3.4. SURJECTIVITY OF THE MAP

Let ⌦be the set of fields contained in F which contain F .Thecardinalityof⌦is less than the cardinality of F so it is a set. We put a partial order on ⌦by saying L K if L is a subfield of K. We define a tower of field extensions to be a function  from an ordinal to ⌦which strictly preserves the ordering and preserves all limits when they exist. We define a p-tower to be a tower f : ↵ ⌦, such that, for every ! x ↵, f(x +1)/f(x) is a finite extension with degree prime to p.Wedefinetheset 2 Tp to be the set of all p-towers. We put a partial order on p-towers by saying that f g,where  f : ↵ ⌦ g : ⌦, ! ! if there is an injective map of sets

i: ↵ , ! such that i(0) = 0, i(x+1) = i(x)+1 and i preserves limits, such that f(x)=g(i(x)). We now use Zorn’s lemma. Take any non-empty chain

C = C : ↵ ⌦: j J T . { j j ! 2 }⇢ p

We can take an upper bound by taking the disjoint union of ↵j and identifying two points if one is the image of the other under the inclusion map. So by Zorn’s Lemma there exists a maximal element f : ↵ ⌦. We deﬁne L to ! be

L := f(x). x ↵ [2 because f is maximal it must be true that L = f(y)forsomey ↵.Wealsohave 2 that L must have no non-trivial, ﬁnite ﬁeld extensions of degree prime to p,elsef would not be maximal. To complete the proof we only need to show that the map

G (F ) G (L) n (p) ! n (p)

48 3.4. SURJECTIVITY OF THE MAP is injective. Assume not, then let z be the minimal element such that the map

G (F ) G (f(z)) n (p) ! n (p) is not injective. We consider two cases.

Assume ﬁrst that there exists z0 ↵ such that z0 +1=z.Byminimalityofz 2 the map

G (F ) G (f(z)) n (p) ! n (p) is injective. Using the projection formula we know that the composition

G (f(z)) G (f(z0)) G (f(z)) n (p) ! n (p) ! n (p) is multiplication by f(z0):f(z) .Becausef(z0):f(z) is coprime to p we deduce | | | | that the composition is an isomorphism and hence the ﬁrst map is injective. By commutativity of the diagram

Gn(F )(p) Gn(f(z0))(p)

Gn(f(z))(p) we can see that the map G (F ) G (f(z)) is injective, giving a contradiction. n (p) ! n (p) Lastly assume that z0 does not exist. In this case we have that

z =lim x ↵: x

f(z)= f(x). x ↵ [2 Because the map G (F ) G (f(z)) is not injective, there exists a non-zero n (p) ! n element s G (F ) that maps to 0. Hence, there exists a ,...,a f(z)such 2 n (p) 1 m 2 49 3.4. SURJECTIVITY OF THE MAP that s =0 G (F (a ,...,a )). Hence, because f(z)isaunionofalltheelements 2 n 1 m less than it, there exists z00

The following Lemma ﬁnally completes the proof of proposition 3.4.6.

Lemma 3.4.15. The p-primary component p is trivial for every prime p. Hence =0.

Proof. We want to ﬁrst show that the following diagram commutes

KM (F [t]/p(t)) KM (F [t]/p(t)) n ! n M i N G iL F NF [t]/p(t) F L F F [t]/p(t) F | | e | | M? e M? Kn?(L) Kn?(L) y ! y By 3.4.10 this is equivalent to showing that e KM (F [t]/p(t)) KM (F [t]/p(t)) n ! n e M ? M ? Kn (L?[t]/pi(t)) Kn (L?[t]/pi(t)) y ! y L L e M? M? Kn?(L) Kn?(L) y ! y commutes. The top square obviously commutes because the vertical maps are just e the maps induced by inclusion. The bottom square commutes because we have already shown 3.4.6 for ﬁeld extensions of this form. This gives that iL F () = 0 | and iL F is injective on p so p =0. | Finally, we can prove the map g is surjective:

Proposition 3.4.16. The map

KM (F ) KM (F ) n ! n is surjective. e

50 3.5. INJECTIVITY AND HOMOTOPY INVARIANCE

M Proof. We have shown that Kn (F )isgeneratedbyelementsoftheform

[Fe[t1±,...,tn±]/m, t1,...,tn].

Hence it suces to show that elements of this form are in the image. We have also shown that the diagram

M M K (F [t±,...,t±]/m) K (F [t±,...,t±]/m) n 1 n ! n 1 n N M N M F [t±,...,t±]/m F F [t ,...,t ]/m F 1 n | e 1± n± | M? M? e Kn?(F ) Kn?(F ) y ! y is commutative. Hence we have that e M [F [t±,...,t±]/m, t ,...,t ]=N g( t ,...,t ) 1 n 1 n F [t±,...,t±]/m F 1 n 1 n | { } = g N M ( t ,...,t ), e F [t±,...,t±]/m F 1 n 1 n | { } as required.

3.5 Injectivity and homotopy invariance

In this section we will complete the proof that the map

KM (F ) KM (F ) n ! n is an isomorphism. To do this, we will constructe an inverse map by ﬁrst mapping n M into HG(Spec(F ), Z(n)) and then mapping to Kn (F ).

3.5.1 Relations in motivic cohomology

In this section we construct a map

M n Kn (F ) HG(Spec(F ), Z(n)). ! n G We denote the group HG(Spec(e F ), Z(n)) by Kn (F ). One can show that these groups G are given by the following presentation, which we take as our deﬁnition of Kn (F ) throughout this chapter.

51 3.5. INJECTIVITY AND HOMOTOPY INVARIANCE

G Definition 3.5.1. Let F be a field. We define the groups Kn (F ), for each n N, 2 to be

G m Kn (F ) := Z[ [F ,A1,...,An]: m N,Ai GLm(F ) { 2 2 and A A = A A for every 1 i, j n ]/(1) (4) i j j i   }

1. [F m1+m2 ,A B ,...,A B ]=[F m1 ,A ,...,A ]+[F m2 ,B ,...,B ] 1 1 n n 1 n 1 n

m m 1 1 2. [F ,A ,...,A ]=[F ,PA P ,...,PA P ] for any P GL (F ). 1 n 1 n 2 m

m 3. [F ,A1,...,An]=0if Ai = Im for some i.

4. [F m,A (1), ,A (1)] = [F m,A (0), ,A (0)] where A (t) GL (F [t]) and 1 ... n 1 ... n i 2 m A (t)A (t)=A (t)A (t) for every 1 i, j n. i j j i   We refer to relation 4 as polynomial homotopy. A simple consequence of this is the following relation:

A1 B1 An Bn A1 0 An 0 ,..., = ,..., 20 1 0 13 20 1 0 13 0 C1 0 Cn 0 C1 0 Cn 4@ A @ A5 4@ A @ A5 which is derived from relation 4 by using the homotopy.

A1 B1t An Bnt ,..., 20 1 0 13 0 C1 0 Cn 4@ A @ A5 This relation is just the exact sequence relation. The above groups ﬁt together to form a graded ring where multiplication is given by

[F m1 ,A ,...,A ] [F m2 ,B ,...,B ] 1 n1 ⇥ 1 n2 =[F m1 F m2 ,A I ,...,A I ,I B ,...,I B ] ⌦ 1 ⌦ m2 n1 ⌦ m2 m1 ⌦ 1 m1 ⌦ n2

G We denote this ring by K (F ). Note that when m1 = m2 =1theabovemultiplica- ⇤ tion is just concatenation of the symbols as it is for Milnor K-theory. We will now prove some useful relations.

52 3.5. INJECTIVITY AND HOMOTOPY INVARIANCE

Proposition 3.5.2. Let F be a ﬁeld. The relation

m m m [F , AB, C2,...,Cn]=[F , A, C2,...,Cn]+[F ,B,C2,...,Cn]

G holds in Kn (F ).

Proof. We ﬁrst show

m m 1 [F ,A1,A2,...,An]+[F ,A1 ,A2,...,An]=0

Using the direct sum relation this is equivalent to showing

A1 0 A2 0 An 0 F 2m, , ,..., (3.6) 2 0 11 0 1 0 13 0 A1 0 A2 0 An 4 @ A @ A @ A5 We use Whitehead’s lemma to give a homotopy

1 10 1(1 A1)t 10 1(1 A1 )t A1(t) := 1 0A t 11 0011 0 t 11 0011 1 @ A @ A @ A @ A Then using the homotopy

A2 0 An 0 F [t]2m,A(t), ,..., 2 1 0 1 0 13 0 A2 0 An 4 @ A @ A5 gives a homotopy between (3.6) and

A 0 A 0 2m 2 n F , Id 2m , ,..., 2 F 0 1 0 13 0 A2 0 An 4 @ A @ A5 So to show the identity

F m, AB, C ,...,C F m, A, C ,...,C F m,B,C ,...,C =0 2 n 2 n 2 n h i h i h i it suces to show

m m 1 m 1 F , AB, C2,...,Cn + F ,A ,C2,...,Cn + F ,B ,C2,...,Cn =0. h i h i h i 53 3.5. INJECTIVITY AND HOMOTOPY INVARIANCE

Using additivity this is equivalent to the relation

AB 00 C2 00 Cn 00 2 3m 0 1 1 0 1 0 13 F , 0 A 0 , 0 C2 0 ,..., 0 Cn 0 =0 6 B 1C B C B C7 6 B 00B C B 00C2C B 00CnC7 6 B C B C B C7 4 @ A @ A @ A5 Note that the ﬁrst matrix can be factored as

AB 00 A 00 B 00 0 1 1 0 1 1 0 1 0 A 0 = 0 A 0 01 0 B 1C B C B 1C B 00B C B001C B 00B C B C B C B C @ A @ A @ A Using Whitehead’s lemma, as in the ﬁrst part of the proof, we can see that each of these factors are homotopic to the identity, hence so is there product.

Next we show anti-commutativity still holds as in Milnor K-theory.

Proposition 3.5.3. The relation

[F m, A, B, C ,...,C ]= [F m, B, A, C ,...,C ] 3 n 3 n

G holds in Kn (F ).

Proof. We ﬁrst show that

m m m [F , AB, AB, C3,...,Cn]=[F , A, A, C3,...,Cn]+[F ,B,B,C3,...,Cn].

To do this we show that

AB 00 AB 00 Cn 00 2 3m 0 1 1 0 1 0 13 F , 0 A 0 , 0 A 0 ,..., 0 Cn 0 6 B 1C B C B C7 6 B 00B C B 00BC B 00CnC7 6 B C B C B C7 4 @ A @ A @ A5 We use the homotopy deﬁned in the previous proof on the ﬁrst matrix in this tuple. The homotopy commutes with the second matrix in the tuple because A and B commute.

54 3.5. INJECTIVITY AND HOMOTOPY INVARIANCE

Using the 3.5.2 we can also show that

m m m [F , AB, AB, C3,...,Cn]=[F , A, A, C3,...,Cn]+[F , A, B, C3,...,Cn]

m m +[F , B, A, C3,...,Cn]+[F ,B,B,C3,...,Cn], which gives the result.

The Steinberg relation

It follows from Proposition 3.5.2 that the map KM (F ) KG(F )iswell-deﬁned n ! n when n =1, 0. In this section, we prove that the map is well-deﬁned for n 2by e proving the Steinberg relation.

Lemma 3.5.4. Let F be a ﬁeld.

1. If ! F , is such that !3 =1and ! =1, then 2[F, a3, 1 a3]=0 KG(F ) 2 6 2 2 3 for every a F ⇤, such that 1 a F ⇤. 2 2

3 3 G 2. If F has no such element !, then 4[F, a , 1 a ]=0 K (F ) for every a F ⇤, 2 2 2 3 such that 1 a F ⇤. 2 Proof. Assume ﬁrst ! F .Considerthehomotopygivenby 2

F 3,A(t), 1 A(t) h i where

00 a3 A(t)=010 t(a3 +1)1 B 3 C B01 t(a +1) C B C @ A

55 3.5. INJECTIVITY AND HOMOTOPY INVARIANCE

Using this we have that

00a3 10 a3 2F 3, 010 01 , 0 11 013 = 6 B C B C7 6 B01 0C B 0 11C7 6 B C B C7 4 @ A @ A5 00 a3 10 a3 2F 3, 010 (a3 +1)1 , 0 11 a3 +1 13 6 B 3 C B 3 C7 6 B01 a +1 C B 0 11 (a +1)C7 6 B C B C7 4 @ A @ A5 Assuming that a3 = ! and a3 = !2 we can diagonalize these matrices to give 6 6

a 00 1 a 00 2F 3, 00 a! 0 1 , 0 01a! 0 13 = 6 B 2C B 2C7 6 B00a! C B 001a! C7 6 B C B C7 4 @ A @ A5 a3 00 1 a3 00 2F 3, 0 0 ! 0 1 , 0 01+! 0 13 6 B 2C B 2C7 6 B 00 ! C B 001+! C7 6 B C B C7 4 @ A @ A5 If a = ! or a = !2 we instead but these matrices in Jordan canonical form, in either case the same argument works. Using the exact sequence relation we have that

F, a, 1 a + F, a!, 1 a! + F, a!2, 1 a!2 = h i h i h i F, a3, 1 a3 + F, !, 1+! + F, !2, 1+!2 h i h i h i expanding the second and third term in the sum gives

F, a, 1 a + F, a, 1 a! + F,!, 1 a! + h i h i h i F, a, 1 a!2 + F,!2, 1 a!2 = h i h i F, a3, 1 a3 + F, !, 1+! + F, !2, 1+!2 h i h i h i

56 3.5. INJECTIVITY AND HOMOTOPY INVARIANCE

Then recombing terms using the multilinear relation gives

[F, a, 1 a3]+[F,!, (1 !a)(1 !2a)2]= [F, a3, 1 a3]+[F, !, 1+!]+[F, !2, 1+!2].

Multiplying both sides by 3 eliminates all terms involving ! because

3[F,!, b]=0 and[F, 1, 1+!]+[F, 1, 1+!2]=0

2 3 3 as (1 + !)(1 + ! )=1Sowehaveshownthat2[F, a , 1 a ]=0when! F ⇤. 2 G G For the case !/F ,weconsidertheﬁeldE := F (!). Let i : K2 (F ) K2 (E) 2 ⇤ ! be the map induced by the inclusion i: F E.Thentheelement !

i [F, a, 1 a]=[E, a, 1 a], satisﬁes ⇤ 2i [F, a, 1 a]=0. ⇤

If we apply the transfer to both sides of this equation and use the projection formula then we obtain

4[F, a, 1 a]=0, as required.

Corollary 3.5.5. For any ﬁeld F, we have that 12[F, a, 1 a]=0 KG(F ) for 2 2 every a F 0, 1 . 2 \{ }

Proof. Using lemma 3.5.4 we know that 4[a3, 1 a3] = 0 for any ﬁeld. If p3 a F 2 then we clearly have 4[a, 1 a]=0.Otherwise,wehavethat4i [F, a, 1 a]=0 ⇤ G 3 3 over K2 F (pa), where i is the map induced by the inclusion i: F F (pa). ⇤ ! Applying the transfer map and using the projection formula gives 12[a, 1 a]=0 as required.

57 3.5. INJECTIVITY AND HOMOTOPY INVARIANCE

G Lemma 3.5.6. Let F be a field and n N.Ifn[K, a, 1 a]=0 K2 (K) for every 2 2 finite field extension K/F and every a K, then [F, a, 1 a]=0 KG(F ) for every 2 2 2 a F 0, 1 . 2 \{ } Proof. Take any a F 0, 1 .Letp be a prime such that n = pm.Let 2 \{ }

i i li 1 li ci(t) := b0 + ...+ bl 1t + t i be the irreducible factors of the polynomial tp a over F [t]. By assumption, we have that mp[F [t]/c (t),t,1 t]=0,sousingproposition3.5.2wehavethat i m[F [t]/c (t), a, 1 t]=0. i Applying the transfer and using the projection formula gives

li m[F ,aId l , 1 A ]=0, F i i where Ai is the matrix

0 ... 0 b 0 0 1 1 ... 0 b1 . B. .. . . C B. . . . C B C B C B0 ... 1 bli 1C B C @ A The determinant of 1 A is c (1) and so i i

m[F, a, ci(1)] = 0.

Because tp a = c (t) ...c (t)wehavethatm[F, a, 1 a] = 0. Continuing inductively 1 n on m gives the result.

We can now show that the Steinberg realtion holds for matrices.

Corollary 3.5.7. Let F be a ﬁeld and [F m,A ,...,A ] KG(F ).IfA + A =1 1 n 2 n i j for some i, j then we have that

[F m,A ,...,A ]=0 KG(F ) 1 n 2 n 58 3.5. INJECTIVITY AND HOMOTOPY INVARIANCE

Proof. Multiplication of rank 1 elements in KG(F )isconcatenationofsymbolsso ⇤ m we have that [F ,A1,...,An]=0whenm =1.Wehavealsoshown,insection3.4 n that K0(F, Gm)isgeneratedbyimagesofrank1elementsundersometransfermap. m Hence, we can write [F ,A1,...,An]asasumofimagesoftransferseachofwhich will be 0.

As a result of proposition 3.5.7 we have that the map from KM to KG is well- deﬁned. e

Corollary 3.5.8. For any ﬁeld F the map

g : KM (F ) KG(F ) ⇤ ⇤ ! ⇤ [F, a ,...,a ] [F, a ,...,a ] 1 e n 7! 1 n is a well-deﬁned homomorphism of graded rings.

3.5.2 The map KG(F ) KM (F ) n ! n In this section, we prove that the map KM KM is injective by constructing an n ! n inverse map ⇥. Our strategy is to deﬁne an inverse map e

KG(F ) KM (F ) n ! n and then compose with the map KM (F ) KG(F ). n ! n Take an element [F m,A ,...,A ] KG(F ). As noted in the previous section we 1 en 2 n can associate to this element a S = F [t1±,...,tn±]-module M.Wethendeﬁne

m M M ⇥([F ,A1,...,An]) := lSm (Mm)NS/m F (t1, ,tn) Kn (F )(3.7) | ··· 2 m S, mmaximalX⇢ We actually show this homomorphism is well-deﬁned on a slightly di↵erent group which we deﬁne in the following.

59 3.5. INJECTIVITY AND HOMOTOPY INVARIANCE

Deﬁnition 3.5.9. Let R be a commutative ring. We deﬁne the group

n n K0(M (R, Gm^ )) := K0(M (R, Gm))/I

n where M (R, Gm) is the category whose objects are of the form

[M,1,...,n]

where M is a ﬁnitely generated R-module, i are commuting automorphisms and I is the subgroup generated by elements of the form [M,1,...,n], with i =IdM for some i.

We now wish to deﬁne a map

n n es : K0(M (R, Gm^ )) K0(M (R/s, Gm^ )) ! for any s R.Thiswillgiveusourhomotopyrelation.Onemightguessthatthe 2 map es might take the form

[M,⇥ ,...,⇥ ] [M R/s, ⇥ Id ,...,⇥ Id ]. 1 n 7! ⌦R 1 ⌦ R/s n ⌦ R/s

However, R/s is not necessarily a ﬂat R-module so this map will not be well-deﬁned because it will not preserve the exact sequence relation. However, given a short exact sequence

0 M M M 0, ! 1 ! 2 ! 3 ! the corresponding exact sequence

M R/s M R/s M R/s 0, 1 ⌦R ! 2 ⌦R ! 3 ⌦R ! can be extended to a long exact sequence involving the Tor functor. More precisely we have an long exact sequence

60 3.5. INJECTIVITY AND HOMOTOPY INVARIANCE

R R ... Tor2 (M2,R/s) Tor2 (M3,R/s)

R R R Tor1 (M1,R/s) Tor1 (M2,R/s) Tor1 (M3,R/s)

M R/s M R/s M R/s 0 1 ⌦R 2 ⌦R 3 ⌦R

If s is a non-zero divisor then R/s has a free resolution of length 1 and so we also have that TorR(M,R/s)=0fori 2. In this case we have that TorR(M,R/s)= i 1 ann (S)andTorR(M,R/s)=M R/s.Thismotivatesthefollowingproposition M 0 ⌦R which holds even when s sazero-divisor.

Proposition 3.5.10. Let R be a commutative ring and s R. The map 2

n n es : K0(M (R, Gm^ )) K0(M (R/s, Gm^ )) ! [M,⇥ ,...,⇥ ] [M R/s, ⇥ Id ,...,⇥ Id ] 1 n 7! ⌦R 1 ⌦ R/s n ⌦ R/s [ann (s), ⇥ ,...,⇥ ], M 1 n where

ann (s)= x M : sx =0 , M { 2 } is well-deﬁned.

Proof. We show ﬁrst that ⇥i restrict to well-deﬁned automorphisms on annM (s). We clearly have that ⇥ (ann (s)) ann (s) because if x ann (s)then i M ⇢ M 2 M

s⇥i(x)=⇥i(sx)=⇥i(0) = 0.

The map ⇥i will obviously still be injective so we only need to show that it is surjective. Take any y ann (s).⇥ is surjective as a map from M to M,sothere 2 M i exists x M such that ⇥ (x)=y.Then 2 i

⇥i(sx)=s⇥i(x)=sy =0,

61 3.5. INJECTIVITY AND HOMOTOPY INVARIANCE

and so sx =0because⇥i is injective. To complete the proof we only need to show the necessary relations hold. If any of the ⇥i are the identity then [M,⇥1,...,⇥n] will clearly map to 0 because the image of ⇥i will still be the identity. To prove the exact sequence relation holds take any exact sequence

g h 0 [M1, ⇥1,...,⇥n] [M2, 1,...,n] [M3, 1,..., n] 0

Now consider the commutative diagram

g h 0 [M1, ⇥1,...,⇥n] [M2, 1,...,n] [M3, 1,..., n] 0

s s s ⇥ ⇥ ⇥ g h 0 [M1, ⇥1,...,⇥n] [M2, 1,...,n] [M3, 1,..., n] 0

The kernel of the vertical maps are precisely annMi (s)andthecokernelofthese maps are M R/s.Then,bythesnakelemma,wehavealongexactsequence i ⌦R

0 annM1 (s) annM2 (s) annM3 (s)

M R/s M R/s M R/s 0 1 ⌦R 2 ⌦R 3 ⌦R

n These maps are also morphisms in M (R, Gm^ )sowehavethatthealternatingsum of elements in the sequence are equal to 0. But this sum is exactly the image of the exact sequence relation and so we are done.

We are now ready to deﬁne groups Hn(F ), which will be the domain of our n inverse map. We deﬁne Hn(F )tobeK0(M (F, Gm^ )) with the extra relation that an element is 0 if it is in the image of et et 1.Thatis

et 1 et n n Hn(F ) :=coker(K0(M (F [t], Gm^ )) K0(M (F, Gm^ ))). !

We will now begin to show that the inverse map, given above, is well-deﬁned on

n Hn(F ). We first show it is well-defined on K0(M (F, Gm^ ))). To check that this gives a homomorphism we must check that the sum on the right hand side is finite and all the relations are satisfied. To check that the sum

62 3.5. INJECTIVITY AND HOMOTOPY INVARIANCE is finite, observe that the maximal ideals for which l (M ) =0arethemaximal Rm m 6 ideals which contain Ann(M). To see this simply note that Mm has length 0, if and only if M =0,ifandonlyifthereexistsr/m such that rm =0.BecauseM is a m 2 finitely generated R-module this is true if and only if there exists an r/m such that 2 rM =0.Weclaimthatthereareonlyfinitelymanymaximalidealswhichcontain

Ann(M). To show this we only need to show that F [t1±,...,tn±]/ Ann(M)isafinitely generated F -module. This is true because for each i,thereexistsamonicpolyno- mial p (t ) ann(M), which has invertible constant term. One such polynomial is i i 2 the characteristic polynomial CAi (ti). This polynomial is clearly monic and has invertible constant term equal to the determinant of Ai.ThenF [t1±,...,tn±]/ Ann(M) has only finitely many maximal ideals because it is artinian. We now begin to show the necessary relations hold for the map to be well-defined

n on K0(M (F, Gm^ )). We ﬁrst show the exact sequence relation holds. Take any exact sequence

1 2 3 [M ,1,...,n] [M , 1,..., n] [M ,✓1,...,✓n].

This gives us an exact sequence of F [t1±,...,tn±]-modules

M 1 M 2 M 3.

Then given any maximal ideal m, we get an exact sequence of F [t1±,...,tn±]m- modules

1 2 3 Mm Mm Mm. because localisation is an exact functor. Then using the exact sequence and the properties of length we get that

l (M 2 )=l (M 1 )+l (M 3 ) F [t1±,...,tn±]m m F [t1±,...,tn±]m m F [t1±,...,tn±]m m which gives the exact sequence relation.

63 3.5. INJECTIVITY AND HOMOTOPY INVARIANCE

We now need to show that an element [M,1,...,n], maps to 0 if i is the identity for some i.Ifm is a maximal ideal such that t 1 m,thent =1 i 2 i 2 M F [t±,...,t±]/m and so N t1,...,tn =0.Ifti 1 / m we claim that 1 n F [t±,...,t±]/m F 1 n | { } 2 l (Mm)=0.ThishappensifandonlyifMm = 0. As mentioned above, this F [t1±,...,tn±]m can only happen if Ann(M) * m which holds in this case because ti 1 Ann(M) 2 and t 1 / m.Thereforewehaveawell-deﬁnedmap i 2

n M K0(M (F, Gm^ )) Kn (F ). ! To complete the proof that the inverse is well-deﬁned, we only need to show that the composition

n n M K0(M (F [t], Gm^ )) K0(M (F, Gm^ )) Kn (F ). (3.8) ! !

n is 0. To do this, we ﬁrst describe a certain set of generators for K0(M (F [t], Gm^ )). Given any element

n [M,1,...,n] K0(M (F [t], Gm^ )), 2 consider the induced F [t, t1±,...,tn±]-module M. Now M is ﬁnitely generated as an

F [t, t1±,...,tn±]-module, so is noetherian. So there exists a series of F [t, t1±,...,tn±]- modules

0=M0 ( M1 ( ( Mt = M, ··· such that each quotient Mi+1/Mi is isomorphic as a F [t, t1±,...,tn±]-module to

F [t, t1±,...,tn±]/p for some prime ideal p.Thenusingtheexactsequencerelationwecandeducethat

n every element in K0(M (F [t], Gm^ )) can be written as a sum of elements of the form

[F [t, t1±,...,tn±]/p , t1,...,tn]

64 3.5. INJECTIVITY AND HOMOTOPY INVARIANCE for some prime ideal p.Soweonlyneedtoshowthattheseelementsmapto0under the composition above. To do this we use a corollary to Weil reciprocity for Milnor K-theory, which we state and use without proof. For a proof of the Weil reciprocity see [3] or [19], for a proof of the following corollary see [13].

Theorem 3.5.11. Suppose L is an algebraic function ﬁeld over k. For each discrete valuation w on L there is a map

: KM (L) KM (k(w)) w n+1 ! n and for every x KM (L): 2 n+1

Nk(w)/kw(x)=0 w X 1 Corollary 3.5.12. Let p : Z AF be a ﬁnite surjective morphism and suppose ! that Z is integral. Let f ,...,f ⇤(Z) and: 1 n 2O

1 0 0 1 1 1 p ( 0 )= n z p ( 1 )= n z { } q i i { } q i i

✏ ✏ ✏ where ni are the multiplicities of the points zi = Spec(Ei )(✏ =0, 1). Deﬁne

0 M 1 M 0 = ni NE0/F ( f1,...,fn E0 ),1 = ni NE1/F ( f1,...,fn E1 ) i { } i i { } i X X Then we have

= KM (F ) 0 1 2 n

We need to show that [F [t, t1±,...,tn±]/p, t1,...,tn]mapsto0underthecomposi- tion for any prime p such that F [t, t1±,...,tn±]/p is a ﬁnitely generated F [t]-module. To do this we consider cases. For the ﬁrst case assume that p F [t] =0.Sop F [t]=(f(t)) for some \ 6 \ irreducible polynomial f(t). We claim in this case that

(et et 1)[F [t, t1±,...,tn±]/p, t1,...,tn]=0

65 3.5. INJECTIVITY AND HOMOTOPY INVARIANCE so clearly the composition is 0. To prove this, ﬁrst assume that f(t) = t and f(t) = t 1. In this case both t 6 6 and t 1areinvertibleinF [t, t±,...,t±]/p.So 1 n

F [t, t±,...,t±]/p F [t]/t =0=F [t, t±,...,t±]/p F [t]/t 1 1 n ⌦F [t] 1 n ⌦F [t]

annF [t,t ,...,t ]/p(t)=0=annF [t,t ,...,t ]/p(t 1) 1± n± 1± n± hence et et 1 =0.Iff(t)=t the same logic as above gives us that et 1 =0.To see that et =0notethat

F [t, t1±,...,tn±]/p F [t] F [t]/t = F [t, t1±,...,tn±]/p =annF [t,t ,...,t ]/p(t) ⌦ 1± n± so et =0.Similarlogicallowsustoconcludethatet et 1 =0whenf(t)=t 1. Hence we can assume that p is such that p F [t]=0.Inthiscasethemap \ F [t] F [t, t±,...,t±]/p is injective. By the going-up theorem, we can conclude ! 1 n that the map

Spec(F [t, t±,...,t±]/p) Spec(F [t]) 1 n ! is surjective. Therefore we can apply corollary 3.5.12 with

Z =Spec(F [t, t1±,...,tn±]/p) to get the following identity in Milnor K-theory:

M l (F [t±,...,t±]/p(1) )N (t ,...,t )= R/p(1)p 1 n p F [t±,...,t±]/p F 1 n 1 1 n | p F [t1±,...,tn±], ⇢pminimalX p(1) p ⇢ M = = l (F [t±,...,t±]/p(0) )N (t ,...,t ) 0 R/p(0)p 1 n p F [t±,...,t±]/p F 1 n 1 n | p F [t1±,...,tn±], ⇢pminimalX p(0) p ⇢ where p(0), p(1) are the ideals p evaluated at 0, 1 respectively.

66 3.5. INJECTIVITY AND HOMOTOPY INVARIANCE

Next we calculate the image of one of these generators under the composition. The image under the map (3.8) is

[F [t±,...,t±]/p(0),t ,...,t ] [F [t±,...,t±]/p(1),t ,...,t ] 1 n 1 n 1 n 1 n Then the image of this element under the inverse map (3.7) is

M l (F [t±,...,t±]/p(0) )N (t , ,t ) Rm 1 n m F [t±,...,t±]/m F 1 n 1 n | ··· m F [t1±,...,tn±], ⇢mmaximalX p(0) m ⇢ M l (F [t±,...,t±]/p(1) )N (t , ,t ) Rm 1 n m F [t±,...,t±]/m F 1 n 1 n | ··· m F [t1±,...,tn±], ⇢mmaximalX p(1) m ⇢ because if a maximal ideal does not contain pi the localisation will be 0. The minimal primes containing p(1),p(0) will be maximal because F [t1±,...,tn±]/p(i)isaﬁnitely generated F -module. So to complete the proof we need only to show that

l (F [t±,...,t±]/p(0)p)=l (F [t±,...,t±]/p(0)p) F [t1±,...,tn±]/p(0)p 1 n F [t1±,...,tn±]p 1 n which is easy to see. So we have constructed the inverse map on the groups Hn(F ).

Lemma 3.5.13. The map

H (F ) KM (F ) n ! n deﬁned in (3.7) is well-deﬁned.

We have a natural homomorphism KG(F ) H (F )sowedeﬁnetheinverse n ! n map to be the composition of this map with the map (3.7). Hence we have shown the following

Theorem 3.5.14. Let F be a ﬁeld. The map

KM (F ) KG(F ) n ! n is an isomorphism.

67 3.5. INJECTIVITY AND HOMOTOPY INVARIANCE

Proof. We showed in section 3.4 that the map is surjective. It only remains to show that the map

M M ⇥([F, a1,...,an]) = lSm (Mm)NS/m F (t1, ,tn)= a1,...,an Kn (F ). | ··· { }2 m S, mmaximalX⇢

If m S is such that t a / m then M =0because(t a )M =0.Sowemust ⇢ i i 2 m i i have t a m for all i. Hence m =(t a ,...,t a )andwearedone. i i 2 1 1 n n In 3.3 we showed that the natural map

KM (F ) KG(F ) n ! n is well-deﬁned. We have shown thate the composition

KM (F ) KM (F ) KG(F )(3.9) n ! n ! n is an ismorphism, hence the ﬁrst map ise injective. We have also shown that the ﬁrst map is surjective. Hence we have shown

Theorem 3.5.15. Let F be a ﬁeld. The map

KM (F ) KM (F ) n ! n is an isomorphism. e

As a result, we have that the second map in 3.9 is an isomorphism. Hence we have the following homotopy invariance relation:

Theorem 3.5.16 (Weak homotopy invariance). Let F be a ﬁeld. The map

evt=1 evt=0 KM (F [t]) KM (F ) n ! n is the zero map. e e

68 Chapter 4

Fundamental theorems for Milnor K-theory

In this chapter, we prove analogues of the additivity, resolution and devisage the-

M M orems from [17] for the groups Kn .WealsoproveareciprocityresultforKn (R) which we use to show compatability of the transfers for semi-local rings. e e

4.1 Compatibility of the transfers for local rings

M M In this section we prove that the transfer maps for Kn and Kn commute. That is we aim to prove the following: e

Theorem 4.1.1. Let A be a semi-local ring with inﬁnite residue ﬁelds and ⇡ A[t] 2 be a monic irreducible polynomial such that Disc(⇡) A⇤. Then the diagram 2 KM (A[t]/⇡) KM (A[t]/⇡) n ! n N M M A[t]/⇡ A NA[t]/⇡ A | e | M? M? e Kn?(A) Kn?(A) y ! y commutes. e

69 4.1. COMPATIBILITY OF THE TRANSFERS FOR LOCAL RINGS

To prove that the diagram commutes it is enough to show that they commute on generators. We will use the following results which gives us generators

M for Kn (A[t]/⇡)aproofcanbefoundin[10].

M Proposition 4.1.2. The group Kn (A[t]/⇡) is generated by elements of the form p (t),...,p (t) , where p (t) are all irreducible in A[t], each p (t) is monic or con- { 1 n } i i stant and

(pi(t),pj(t)) = A[t] for i = j. 6

If any of these pi(t)areinA⇤ then we can show that the diagram above commutes for this element using the projection formula and induction. We therefore only need to show that the diagram commutes for elements with pi(t)non-constant.Recall from chapter 1 that we have a split exact sequence

M et M 0 Kn (A) Kn (A) Kn 1(A[t]/⇡) 0 ! ! ! !

Consider the splitting map

M et ⇡ : Kn 1(A[t]/⇡) Kn (A). (4.1) !

We claim that

p ,...,p =(⇡, p ,...,p )(4.2) ⇡{ 1 n} 1 n n i+1 et + ( 1) pi ⇡, p1,...,pi 1, pˆi,pi+1,...,pn Kn (A)(4.3) { }2 i=1 X To see this, observe that ( p ,...,p )istheuniqueelementsuchthat f { 1 n} 0ifg = f @g(f ( p1,...,pn )) = 6 . { } 8 p ,...,p if g = f. <{ 1 n} : 70 4.1. COMPATIBILITY OF THE TRANSFERS FOR LOCAL RINGS and

s (f ( p1,...,pn )) = 0 1 { } where s is the retraction map which sends an element to its leading coecient. 1 Then we only need to show that the RHS of (4.2) satisﬁes these which is a simple calculation. Now composing 4.2 with @ we can see that 1

M NA[t]/⇡ p1,...,pn = @ (⇡, p1,...,pn)(4.4) { } 1 n i+1 M + ( 1) NA[t]/p ⇡, p1,...,pi 1, pˆi,pi+1,...,pn (4.5) i { } i=1 X We use this identity to prove that the transfer maps commute. We assume, inductively, that the transfer maps commute for A[t]/f where deg(f) < deg(⇡). Then to complete the proof we only need to show the analogous version of (4.4) for

M Kn .Toprovethiswebeginwiththefollowinglemma:

Lemmae 4.1.3. Let A be a semi-local ring and ⇡ A[t] be an irreducible, monic 2 polynomial. Then

M M NA[t]/⇡ A( p1(t),...,pn(t) )=NA[t,x1,...,xn]/(⇡,p1(x1),...,pn(xn) A( t x1,...,t xn ) | { } | { } e e where pi are all monic polynomials.

Proof. We show that

M M NA[t]/⇡ A( p1(t),...,pn(t) )=NA[t,x1]/(⇡,p1) A( t x1,p2(t),...,pn(t) ) | { } | { } and continuee the process inductivelye to obtain the result. To show this, it suces to show that

M NA[t,x1]/(⇡(t),p1(x1)) A[t]/⇡( x1 t, p2(t),...,pn(t) )= p1(t),...,pn(t) (4.6) | { } { } e 71 4.1. COMPATIBILITY OF THE TRANSFERS FOR LOCAL RINGS

M in Kn (A[t]/⇡(t)) because

e M M M NA[t]/⇡ A NA[t,x1]/(⇡,p1) A[t]/⇡ = NA[t,x1]/(⇡,p1) A. | | |

To compute (4.6) wee can use thee projection formulae to get that

M NA[t,x1]/(⇡,p1) A[t]/⇡( t x1,p2(t),...,pn(t) )= d, p2(t) ...,pn(t) | { } { } where d ise the determinant of the A[t]/⇡(t)-linear map

(t x ):A[t, x ]/(⇡, p ) A[t, x ]/(⇡, p ) ⇥ 1 1 1 ! 1 1

We claim that the determinant of this map is p1(t). Let

n 1 n p1(t)=a0 + a1t + + an 1t + t ···

The matrix corresponding to the map above is

t 0 ... 0 a0 0 1 t... 0 a 1 1 B ...... C B . . . . . C (4.7) B C B C B 00... t an 2 C B C B C B 00... 1 t + an 1C B C @ A To calculate the determinant of this matrix we use induction. For a 1 1matrixof ⇥ the form above, the result is trivial. To calculate the determinant of the n n case ⇥ we expand the top row. Doing this we get that the determinant is equal to

t 0 ... 0 a1 1 t...0 0 1 1 t... 0 a2 . . 0 0 1 .. . 1 B ...... C n+1 t det B . . . . . C + ( 1) a0 det . . . ⇥ B C B . . .. t C B C B C B 00... t an 2 C B C B C B 00... 1C B C B C B 00... 1 t + an 1C B C B C @ A @ A 72 4.1. COMPATIBILITY OF THE TRANSFERS FOR LOCAL RINGS

We can calculate the determinant of the ﬁrst matrix using induction. So we get the determinant of (4.7) is

n 2 n 1 n+1 n 1 t (a1 + + an 1t + t )+( 1) ( 1) a0 = p(t) ⇥ ··· ⇥ ⇥ as required.

So if we want to prove the identity (4.4), by (4.1.3), it is enough to show that

M NA[t,x1,...,xn]/(⇡(t),p1(x1),...,pn(xn) A( t x1,...,t xn ) | { } =deg(⇡)deg(p ) ...deg(p ) 1,..., 1 e 1 n { } n i+1 M + ( 1) NA[t,x1,...,xn]/(⇡(t),p1(x1),...,pn(xn)) A( xi t, xi x1 ...,x\i xi,...,xi xn ) | { } i=1 X e (4.8)

To prove this we use the following identity:

Lemma 4.1.4. Let R be a commutative ring and x ,...,x R be such that x 0 n 2 i x R⇤, for all i, j. Then j 2 n i ( 1) [xi x0,...,xi xi 1,xi xi+1,...,xi xn]=[ 1,..., 1] i=0 X M in Kn (R).

Proof.e We prove the result by induction on n.Letn =1,then

x x [x x ] [x x ]=[ 0 1 ]=[ 1] 0 1 1 0 x x 1 0 Now assume the identity holds when n = k.Thenwehavetheidentity

k i ( 1) [xi x0,...,xi xi 1,xi xi+1,...,xi xk]=[ 1,..., 1]. i=0 X

73 4.1. COMPATIBILITY OF THE TRANSFERS FOR LOCAL RINGS

In an attempt to introduce x into the equation we multiply both sides by [x k+1 0 xk+1]togive

k i ( 1) [xi x0,...,xi xi 1,xi xi+1,...,xi xk,x0 xk+1] i=0 X =[ 1,..., 1,x x ]. 0 k+1

Applying the identity [c, d]=[ c ,c+ d]totheﬁrstandlastcoordinatesofthe d elements in sum gives

k i xi x0 ( 1) [ ,...,xi xi 1,xi xi+1,...,xi xk,xi xk+1] x x i=1 0 k+1 X +[x x ,...,x x ]=[ 1,..., 1,x x ]. 0 1 0 k+1 0 k+1

Expanding the ﬁrst term in the sum gives

k i+1 ( 1) [ x0 + xk+1,xi x1,...,xi xi 1,xi xi+1,...,xi xk,xi xk+1] i=1 X k i + ( 1) [xi x0,...,xi xi 1,xi xi+1,...,xi xk,xi xk+1] i=0 X =[ 1,..., 1,x x ]. 0 k+1

The second term is almost the sum we require, so by adding

( 1)k+1[x x ,...,x x ] k+1 0 k+1 k to both sides of the equation we reduce the proof to proving that

k i+1 ( 1) [ x0 + xk+1,xi x1,...,xi xi 1,xi xi+1,...,xi xk,xi xk+1] i=1 X =[ 1,..., 1,x x ]+( 1)k+1[x x ,...,x x ]+[ 1,..., 1]. 0 k+1 k+1 0 k+1 k

74 4.2. CONSEQUENCES OF RECIPROCITY

By rearranging this equation we reduce to showing

k+1 i ( 1) [xk+1 x0,xi x1,...,xi xi 1,xi xi+1,...,xi xk,xi xk+1] i=1 X =[ 1,..., 1,x x ](4.9) k+1 0 The term on the right hand side has order 2 and so by graded commutativity is equal to [x x , 1,..., 1]. So we can see that the identity 4.9 holds by taking k+1 0 the reciprocity formula for x ,...x and multiplying on the left by x x ,and 1 k+1 k+1 0 so by induction we are done.

So we can use this identity, in the ring A[t, x1,...,xn]/(⇡(t),p1(x1),...,pn(xn)), to prove 4.8usingthefactthat

M NA[t,x1,...,xn]/(⇡(t),p1(x1),...,pn(xn)) A([ 1,..., 1]) | =deg(⇡)deg(p ) ...deg(p )[ 1,..., 1] e 1 n

4.2 Consequences of reciprocity

In this section we look at some consequences of reciprocity. In particular, we will

M M show that if Kn is isomorphic to Kn (R)whenR is a local ring with infinite residue field then KM (R)agreeswiththeimprovedMilnorK-groupswhenR has n e finite reisude field. e To do this we only need to show that our system of transfers satisfies the properties stated in 2.2. This is shown in the following proposition.

Proposition 4.2.1. Let A be a local ring with infinite residue field and let A B ⇢ be a finite, etale extension of local rings. Let A0 A00 be a morphism of local ! A-algebras. Assume further that both

B0 := B A0 B00 := B A00 ⌦A ⌦A

75 4.2. CONSEQUENCES OF RECIPROCITY are local. Then we have that

1. The composition

N M i M B0/A0 M Kn (A0) Kn (B0) Kn (A0) ! e !

is just multiplication bye [B : A]. e e

2. The diagram

M M Kn (B0) Kn (B00)

e e M M Kn (A0) Kn (A00)

e M e commutes on rank one elements in Kn (B0).

Proof. Etale morphisms are preserved undere base change so we have that the map

A0 B0 is an etale morphism. By 2.1.10 we can choose a monic ⇡ A[t]with ! 2 Disc(⇡) A⇤ such that 2

B = A[t]/⇡(t).

Furthermore, denoting the image of ⇡ in A0[t]by⇡0, we have

B0 = A0[t]/⇡0(t).

To prove the ﬁrst result, note that the projection formula gives that the composition is equal to multiplication by [B0 : A0]. The result follows from the fact that

[B : A]=deg(⇡)=deg(⇡0)=[B0 : A0]

To prove the second result we need to show that the diagram

76 4.2. CONSEQUENCES OF RECIPROCITY

M M Kn (A0[t]/⇡0(t)) Kn (A00[t]/⇡00(t))

e e M M Kn (A0) Kn (A00)

M commutes on rank 1 elements.e Take a generatore for Kn (A0[t]/⇡0(t)) of the form [A [t]/⇡ ,p (t),...,p (t)] with the p (t)monic,irreducibleandpairwisecoprimewith 0 0 10 n0 i e Disc pi0 A0⇤. Using reciprocity we can write the composition iA A NA [t]/⇡ A as 2 0| 00 0 0| 0 n i+1 e iA A NA [t]/⇡ A [p10 ,...,pn0 ]= iA A NA [t]/p0 A ( 1) [⇡0,p10 ,...,pˆi0 ,...,pn0 ] 0| 00 0 0| 0 00| 0 0 i| 0 i=1 X e e +deg(⇡0)deg(p0 ) ...deg(p0 )[A00, 1,..., 1]. 1 n Using induction we can swap the order of composition in the summation to obtain

iA A NA [t]/⇡ A [p10 ,...,pn0 ] 0| 00 0 0| 0 n i+1 e = NA [t]/p00 A ( 1) [⇡00,p100,...,pˆi00,...,pn00 ] 00 i | 00 i=1 X e +deg(⇡0)deg(p0 ) ...deg(p0 )[A00 1,..., 1]. 1 n

The right hand side of which is NA [t]/⇡ A00 iA [t]/⇡ A [t]/⇡ 00 00 | 00 00| 0 0 We have shown the following:e

Corollary 4.2.2. If the map

KM (R) KM (R) n ! n is an isomorphism when R is a local ring withe inﬁnite residue ﬁeld then there is a unique isomorphism

Kˆ M (R) KM (R) n ! n for R any local ring, such that the diagram e

77 4.3. THE ADDITIVITY THEOREM

ˆ M Kn (R)

M M Kn (R) Kn (R) commutes. e

M Proof. We only need that the transfers are compatible for elements in Kn of rank greater than 1. This follows from the rank one case because we have that the map e KM (R) KM (R)issurjective n ! n e ˆ M Remark 4.2.3. Using the explicit description of Kn (R) as

M M K (f1) K (f2) Kˆ M (R)=ker(KM (R(t)) n n KM (R(t ,t ))), n n ! n 1 2 we can see that there is always a map, regardless of whether the map

KM (R) KM (R) n ! n is an isomorphism for R with inﬁnite residuee ﬁeld. To show this we simply need to show that

M M K (f1) K (f2) M M n n M Kn (R)=ker(Kn (R(t)) Kn (R(t1,t2))), e e ! The proof of thise is identicale to the proof of the analogouse identity for Milnor K- theory [11].

4.3 The additivity theorem

M The aim of this section is to prove a version of the additivity theorem for Kn .The proof is similar to the proof for K we only need to check that the relations are 0 e satisﬁed.

78 4.3. THE ADDITIVITY THEOREM

Deﬁnition 4.3.1. Let A , C be exact subcategories of an exact category B.We deﬁne a category E (A , B, C ), which we call the extension category, whose objects are short exact sequences

0 A B C 0 ! ! ! ! with A A , B B and C C and whose morphisms are commuting diagrams. 2 2 2 Theorem 4.3.2. With notation as in 4.3.1 we have an isomorphism

KM (E (A , B, C )) = KM (A ) KM (C ) n ⇠ n ⇥ n Proof. We ﬁrst deﬁnee maps e e

: K (Autn(E (A , B, C ))) K (Autn(A )) K (Autn(C )) 0 ! 0 ⇥ 0 : K (Autn(A )) K (Autn(C )) K (Autn(E (A , B, C ))) e 0 ⇥ 0 e ! 0 e and then showe these maps satisfye the necessary relations.e Take an element [E,✓ ,...,✓ ] K (Autn(E (A , B, C ))) where 1 n 2 0 E =0 A eB C 0and✓ is ! ! ! ! i 0 A B C 0 ! ! ! !

✓i,A ✓i,B ✓i,C ? ? ? 0 A? B? C? 0 ! y ! y ! y ! We deﬁne ([E,✓1,...,✓n]) = ([A,✓1,A,...,✓n,A], [C,✓1,C ,...,✓n,C ]). Given an element ([A,✓1,A,...,✓n,A], [C,✓1,C ,...,✓n,C ]) we deﬁne the map to be

([A,✓1,A,...,✓n,A], [C,✓1,C ,...,✓n,C ]) = [E,✓1,A C ,...,✓n,A C ]where

E =0 A A C C 0and✓i,A C ! ! ! ! E =0 A B C 0and✓ is ! ! ! ! i 0 A A C C 0 ! ! ! ! ✓ ✓ ✓ ✓ i,A i,A i,C i,C ? ? ? 0 A? A ? C C? 0 ! y ! y ! y ! 79 4.3. THE ADDITIVITY THEOREM

It is a simple calculation to show that the exact sequence relation is satisﬁed so these maps are well-deﬁned. To show that the composition is the identity is suces to show that

[0 A B C 0,✓ ,...,✓ ] ! ! ! ! 1 n =[0 A A C C 0,✓ ✓ ,...,✓ ✓ ] ! ! ! ! 1,A 1,C n,A n,C

n in K0(Aut (E (A , B, C ))). This follows by using the exact sequence relation on the following exact sequence in E (A , B, C ) 000

? ? ? 0 A? A? ?0 0 ! y !IdA y ! y !

IdA f ? ? ? 0 A? B? C? 0 ! y !f y !g y !

g IdC ? ? ? 0 ?0 C? C? 0 ! y ! y !IdC y !

? ? ? ?000? ? y y y Each column is exact so this is an exact sequence of elements in E (A , B, C ). We have to show that this gives an exact sequence in Autn(E (A , B, C )). We show that the morphism between 0 A A 0 0and0 A B C 0givesan ! ! ! ! ! ! ! ! morphism between

[0 A B C 0,✓ ,...,✓ ]and[0 A A 0 0,✓ ,...,✓ ] ! ! ! ! 1 n ! ! ! ! 1,A n,A

To show this we observe that the diagram

80 4.4. THE RESOLUTION THEOREM

Id A A A 0

IdA f f g A B C ✓i,A ✓i,A ✓i,B IdA ✓i,A A A 0

IdA f f g A B C commutes. Hence the maps and are inverse to each other. It can also be shown that the necessary relations are satisﬁed, this implies that and induce maps on ˜ M Kn which are mutually inverse.

4.4 The resolution theorem

The main result of this section is the following:

Theorem 4.4.1. Let R be a regular local ring. Then the natural map

KM (P) KM (M ) i ! i [P, ⇥ ,...,⇥ ] [P, ⇥ ,...,⇥ ] 1 e i 7! e 1 i is an isomorphism.

To prove this we will construct an inverse map

KM (M ) KM (P) i ! i We will ﬁrst show that there is ane map e

K (Auti(P)) K (Auti(M )) 0 ! 0 and then that this map preserves the necessary relations. The ﬁrst thing to show is that every element in Auti(M )hasaresolutionwith elements in Auti(P). This is clear for i =0becauseRisregular.Forthegeneral case we need the following lemma.

81 4.4. THE RESOLUTION THEOREM

Lemma 4.4.2. Let R be any commutative ring and M a ﬁnitely generated R-module. Let ⇥:M M be an automorphism of M as an R-module. Then there exists a ! polynomial r(t) R[t] such that r is monic, r(0) = 1 and r(⇥) = 0. 2 Proof. MisﬁnitelygeneratedasanR-modulesothereexistsasurjectiveR-module homomorphism

n f : R ⇣ M n (r ,...,r ) r m 1 n 7! i i i=1 X 1 n because ⇥is invertible we can lift the maps ⇥and ⇥ to maps on R so that we have commutative diagrams.

B Rn A Rn Rn Rn ! ! f f f f

1 ? ⇥ ? ? ⇥ ? M? M? M? M? y ! y y ! y 1 So we must have monic polynomials p and q of degree n, such that p(⇥) = q(⇥ )=0 (take, for example, the characteristic polynomials of A and B). Then deﬁne r(t)to be

n 1 r(t):=t (p(t)+q(t )).

One easily checks that r(t)satisﬁestherequiredproperties.

i We use this to construct a resolution of [M,⇥1,...,⇥i]withelementsinAut (P).

Proposition 4.4.3. Let [M,⇥ ,...,⇥ ] Auti(M ). Then there exists a long exact 1 i 2 sequence

0 [P ,A ,...,A ] [P ,A ,...,A ] [M,⇥ ,...,⇥ ] 0 ! n (n,1) (n,i) !···! 0 (0,1) (0,i) ! 1 i ! such that P P for every i. i 2 82 4.4. THE RESOLUTION THEOREM

Proof. We show that there is a surjective map

[P ,A ,...,A] [M,⇥ ,...,⇥ ] 0 1 i ! 1 i with P0 projective. We then proceed by induction. M is finitely generated so we have a homomorphism f : Rn M defined by ! n f(r1,...,rn)= j=1 rjmj.By2.2.therearemonicpolynomialsrj(t)ofdegree2n with rj(0) = 1 andP rj(⇥j)=0.WedefineP0 to be the R[T1,...,Ti]-module

P := (R[T ,...,T]/ r (T ),...,r(T ) )n 0 1 i h 1 1 i i i

The rj(Tj)aremonicsoP0 is a free R-module. We deﬁne an R[T1,...,Tn]-module homomorphism

f˜ : P M 0 ! n

f˜(q1(T1,...,Ti),...,qi(T1,...,Ti)) = qj(⇥1,...,⇥i)mj j=1 X

This map is surjective because f is surjective and is well-deﬁned because rj(⇥j)=0.

We deﬁne the maps Aj to be multiplication by Tj. These maps clearly commute and are invertible because rj(Tj)hasconstantterm1sowecanﬁndaninverseofTj in R[T ,...,T]/ r (T ),...r(T ) .Tocompletetheproofoftheclaimweonlyneed 1 i h 1 1 i i i to show that f˜ gives a homomorphism from [P , T ,..., T ]to[M,⇥ ,...,⇥ ], 0 ⇥ 1 ⇥ n 1 n i.e. that the following square commutes

T i P ⇥ P 0 ! 0 f f

? ⇥i ? M? M? y ! y which is simple to show. Hence we have an exact sequence

0 [ker(f˜),A ,...,A] [P ,A ,...,A] [M,⇥ ,...,⇥ ] 0 ! 1 i ! 0 1 i ! 1 i !

83 4.4. THE RESOLUTION THEOREM

Continuing this process with [ker(f˜),A1,...,Ai]replacing[M,⇥1,...,⇥i]givesa projective resolution for [M,⇥1,...,⇥i]. The process must terminate because R is regular.

Note that the above proposition gives us that the map in Theorem 4.4.1issurjec- tive for any regular ring because the resolution allows us to write each element as an alternating sum of the elements in its projective resolution. To show it is injective, we shall define an inverse map to be the alternating sum of the elements in its resolution and show that this is independent of the choice of resolution. We know that the map is well defined because Autn(P)andAutn(M )satisfytheconditionsforthereso- lution theorem for K .Therefore,wehaveamapfromKM (Auti(M )) KM (P) 0 0 ! i which takes an element to the alternating sum of the elements in its projective resolution. We need to show it satisfies linearity and the steinberg relation.

Proposition 4.4.4. The map KM (Auti(M )) KM (P) factors through a map 0 ! i KM (M ) KM (P). i ! i Proof. We show the steinberg relation ﬁrst because this is easy. We do it for the case i =2tosimplifynotation.Take[M,⇥, 1 ⇥]. Take a projective resolution f f f 0 [P ,A ,B ] n ... 1 [P ,A ,B ] 0 [M,⇥, 1 ⇥] 0 ! n n n ! ! 0 0 0 ! ! However, because the diagram on the left commutes the diagram on the right commutes A 1 A P j P P j P j ! j j ! j fj fj fj fj (4.10)

? Aj 1 ? ? 1 Aj 1 ? P?j 1 P?j 1 P?j 1 P?j 1 y ! y y ! y So we can choose another projective for [M,⇥, 1 ⇥] similar to the one above but with B =1 A . i i f f f 0 [P ,A , 1 A ] n ... 1 [P ,A , 1 A ] 0 [M,⇥, 1 ⇥] 0 ! n n n ! ! 0 0 0 ! !

84 4.4. THE RESOLUTION THEOREM

To prove the linear relation we make the following claim

Lemma 4.4.5. Let R be a regular local ring and M a ﬁnitely generated R-module.

M i Given two elements of [M,⇥0, ⇥2,...,⇥i] and [M,⇥1, ⇥2,...,⇥i] of K0 (Aut (M )) there exists projective resolutions

f 0 [P ,A ,A ,...,A ] n ... ! n (n,0) (n,2) (n,i) ! f f ... 1 [P ,A ,A ,...,A ] 0 [M,⇥ , ⇥ ,...,⇥ ] 0 ! 0 (0,0) (0,2) (0,i) ! 0 2 i !

f 0 [P ,A ,A ,...,A ] n ... ! n (n,1) (n,2) (n,i) ! f f ... 1 [P ,A ,A ,...,A ] 0 [M,⇥ , ⇥ ,...,⇥ ] 0 ! 0 (0,1) (0,2) (0,i) ! 1 2 i !

Proof. We construct the ﬁrst term and then we can continue similarly.

Deﬁne the polynomials rj(t)asin2.2.Misﬁnitelygeneratedsowehavea homomorphism f : Rn M where f(r ,...,r )= n r m .Wechoosethismap ! 1 n j=1 j j f so that n is minimal. Using Nakayama’s LemmaP we can show that there exist automorphisms A ,A : Rn Rn which make the following diagram commute. 0 1 ! Rn A0 Rn Rn A1 Rn ! ! f f f f (4.11)

? ⇥0 ? ? ⇥1 ? M? M? M? M? y ! y y ! y We deﬁne

S := R[t±,...,t±]/ r (t ),...,r(t ) . 2 i h 2 2 i i i

We tensor S with the diagrams (4.11) and compose with the maps

g : M S M ⌦R ! m q(t ,...,t ) q(⇥ ,...,⇥ ) m ⌦ 2 n 7! 2 n ⇤

85 4.5. DEVISSAGE to obtain the diagram

A0 Id A1 Id Rn S ⌦R S Rn S Rn S ⌦R S Rn S ⌦R ! ⌦R ⌦R ! ⌦R f Id f Id f Id f Id ⌦R S ⌦R S ⌦R S ⌦R S ? ⇥0 RIdS ? ? ⇥1 RIdS ? (4.12) M ?R S ⌦ M ?R S M ?R S ⌦ M ?R S ⌦y ! ⌦y ⌦y ! ⌦y g g g g

? ⇥0 ? ? ⇥1 ? M? M? M? M? y ! y y ! y The diagrams (4.12) commute so we can deﬁne maps

n g(f IdS ) [R S, A Id , Id n t ,...,Id n t ] ⌦ [M,⇥ , ⇥ ,...,⇥ ] ⌦R 0 ⌦R S R ⌦R 2 R ⌦R n ! 0 2 n n g(f IdS ) [R S, A Id , Id n t ,...,Id n t ] ⌦ [M,⇥ , ⇥ ,...,⇥ ] ⌦R 1 ⌦R S R ⌦R 2 R ⌦R n ! 1 2 n

S is a free R-module hence so is Rn S.Themapissurjectivebecausef is, and ⌦R so we can take the kernel and cotinue inductively.

To ﬁnish the proof, we take resolutions for

[M,⇥0, ⇥2,...,⇥i]and[M,⇥1, ⇥2,...,⇥i]

of the form in Lemma 2.5. Then the following is a resolution for [M,⇥0⇥1, ⇥2,...,⇥i]

f 0 [P ,A A ,A ,...,A ] n ... ! n n,0 (n,1) (n,2) (n,i) ! f f ... 1 [P ,A A ,A ,...,A ] 0 [M,⇥ ⇥ , ⇥ ,...,⇥ ] 0 ! 0 (0,0) (0,1) (0,2) (0,i) ! 0 1 2 i !

M Using linearity in Ki (P) gives the result.

e 4.5 Devissage

M In this section, we prove a Devissage theorem for Kn . To do this we mimic the proof for K . To ﬁnish the proof we only need to show that the necessary relations 0 e are satisﬁed.

86 4.5. DEVISSAGE

Theorem 4.5.1. Let I be an ideal of a noetherian ring R. Let ModI (R) be the abelian subcategory of Mod(R) whose objects are ﬁnitely generated modules M, such that InM =0for some M. Then

M M Kn (ModI (R)) ⇠= Kn (Mod(R/I))

Proof. Given an R/I-modulee M,wecan,byrestrictionofscalars,obtainane R- module M such that IM =0.Wethereforehaveaninclusionofabeliancategories

Mod(R/I) Mod (R). ⇢ I This gives us an inclusion of abelian categories

Autn(Mod(R/I)) Autn(Mod (R)). ⇢ I

This induces a homomorphisms on K0

f : K (Autn(Mod(R/I))) K (Autn(Mod (R))). 0 ! 0 I To show this map is an isomorphism we only need to show that each object of n n Aut (ModI (R)) has a ﬁltration with quotients in Aut (Mod(R/I)). Take any object n [M,⇥1,...,⇥n] in Aut (ModI (R)). Then [M,⇥1,...,⇥n]hasaﬁltration

m 1 [M,⇥ ,...,⇥ ] [IM,⇥ ,...,⇥ ] [I M,⇥ ,...,⇥ ] 0. 1 n 1 n ··· 1 n

Therefore, we can apply Devissage for K0 to conclude that the map f is an isomorphism with inverse

1 n n f : K (Aut (Mod (R))) K (Aut (Mod(R/I))) 0 I ! 0 m 1 [M,⇥ ,...,⇥ ] [IiM/Ii+1M,⇥ ,...,⇥ ]. 1 n 7! 1 n i=0 X ˜ M To get two mutually inverse maps on Kn it remains to show that the multilinearity and Steinberg relations are satisﬁed under the maps

K (Autn(Mod(R/I))) K˜ M (Mod (R)) 0 ! n I K (Autn(Mod (R))) K˜ M (Mod(R/I)). 0 I ! n

87 4.5. DEVISSAGE

Both relations hold trivially and so we are done.

We now give a few special cases of the above theorem.

Corollary 4.5.2. Let I be a nilpotent ideal of a noetherian ring R. Then the inclusion Mod(R/I) Mod(R) induces an isomorphism ⇢

M M Gn (R/I) ⇠= Gn (R)

Corollary 4.5.3. Let R be an artiniane locale ring. Then

GM (R) = KM (R/m). ⇤ ⇠ ⇤ e e Corollary 4.5.4. Let R be a local noetherian ring and Modfl(R) be the category of modules of ﬁnite length. Then

M M Kn (Modfl(R)) ⇠= Kn (R/m)

Proof. This follows from thee fact that a modulee M over a local noetherian ring has ﬁnite length i↵it is annihilated by a power of m.

88 Chapter 5

The homomorphism to Quillen K-theory

In this chapter we will construct a homomorphism to Grayson’s deﬁniton of higher K-theory. One consequence of this is that the kernel of the map KM (R) KM (R) n ! n is annihilated by (n 1)!. In particular, this shows the map is injective when n =2. e More precisely, we will show that the map which sends [P, ⇥1,...,⇥n]tothen- dimensional cube whose top di↵erential di := Ai and whose bottom is the identity, is well-deﬁned.

5.1 Multilinearity

In this section, we will give a sktech of a proof of the multilinear relation which we take from [8]. The proof uses the identity in 5.1.2, which is an analogue of an identiy of Nenashev.

Deﬁnition 5.1.1. A bounded binary double complex N.. is a pair of bounded double complexes which have the same objects in each position.

89 5.1. MULTILINEARITY

Proposition 5.1.2. . Let N.. be a bounded binary double complex of objects in q 1 (B) N that is supported on [0,m] [0,n], and whose rows and columns are acyclic. ⇥ th th q n Let N.,j be the j row and Ni,. the i row considered as objects in (B ) N . Then the equation

n m ( 1)j[N ]= ( 1)i[N ] .,j i,. j=0 i=0 X X Q holds in Kn (N ).

Proof. Let [P, ⇥1,...,⇥n]denotethen-dimensional cube whose top di↵erential di is ⇥i and whose bottom is the identity. To prove the multilinearity we wish to prove

[P, ⇥0⇥1,...,⇥n]=[P, ⇥0,...,⇥n]+[P, ⇥1,...,⇥n]

Let Q =[P, ⇥2,...,⇥n]. Consider the binary double complex 0 0 0 0 0 0

0 Q ⇥1 Q 0 Q 1 Q

1 ⇥0 1 1 0 Q ⇥0⇥1 Q 0 Q 1 Q Using the relation 5.1.2 we get that

1 ⇥0 ⇥0⇥1 ⇥1 -[ Q Q ]+[ Q Q ]=[ Q Q ]-[ Q Q ] 1 1 1 1 The ﬁrst term in the sum is diagonal so is trivial. So

[P, ⇥0⇥1, ⇥2,...,⇥n]=[P, ⇥0, ⇥2,...,⇥n]+[P, ⇥1, ⇥2,...,⇥n], as required.

90 5.2. THE COFINALITY THEOREM

5.2 The coﬁnality theorem

In section 5.1 we proved that the multilinearity relation holds in Grayson’s deﬁnition of higher K-theory. In the next section we will show that the Steinberg relation holds. The purpose of this section is to prove the following theorem, which will reduce proving the Steinberg relation for projective modules to proving it just for free modules.

Theorem 5.2.1. Let R be a ring and F be the category of ﬁnitely-generated, free left R-modules. Then the map

KM (F ) KM (P) n ! n is an isomorphism when n 1 e e It is easy to see the map is surjective; take an element [P, ⇥ ,...,⇥ ] KM (R). 1 n 2 n Now P is projective so there exists Q such that P Q is free. Because n 1we e have that

[Q, IdQ,...,IdQ] is trivial, so

[P, ⇥1,...,⇥n]= P Q, ⇥ Id ,...,⇥ Id 1 Q n Q h i which is in the image. To show the map is injective we construct an inverse map. We deﬁne the inverse map s to be

s: KM (P) KM (F ) n ! n

[P, ⇥1,...,⇥n] P Q, ⇥ Id ,...,⇥ Id 7! e 1 e Q n Q h i

91 5.2. THE COFINALITY THEOREM

We ﬁrst show that the choice of Q is irrelevant. Let Q1 and Q2 be two left R-modules such that P Q and P Q are free. Then 1 2

P Q , ⇥ Id ,...,⇥ Id = 1 1 Q1 n Q1 h i P Q P Q , ⇥ Id Id Id ,...,⇥ Id Id Id 1 2 1 Q1 P Q2 n Q1 P Q2 h i

P Q , ⇥ Id ,...,⇥ Id = 2 1 Q2 n Q2 h i P Q P Q , ⇥ Id Id Id ,...,⇥ Id Id Id 2 1 1 Q2 P Q1 n Q2 P Q1 h i These two terms are obviously equal. Next we show the exact sequence relation. Take any exact sequence

f g 0 [P , ,..., ] [P , ,..., ] [P , ⇥ ,...,⇥ ] 0 ! 1 1 n ! 2 1 n ! 3 1 n !

Let Q and Q be ﬁnitely generated modules such that P Q and P Q are 1 3 1 1 3 3 free. Then there is an exact sequence of free modules

f 0 [P Q , Id ,..., Id ] ! 1 1 1 Q1 n Q1 ! g [P Q Q , Id Id ,..., Id Id ] 2 1 3 1 Q1 Q3 n Q1 Q3 ! [P Q , ⇥ Id ,...,⇥ Id ] 0. 3 3 1 Q3 n Q3 !

Where P Q Q is free because P = P P . 2 1 3 2 ⇠ 1 3 The multilinearity is simple to show. Take an element

[P, ⇥0⇥1, ⇥2,...,⇥n] this elements maps to an element of the form

[P Q, ⇥ ⇥ Id , ⇥ Id ,...,⇥ Id ] 0 1 Q 2 Q n Q

92 5.2. THE COFINALITY THEOREM

M M in Kn (F ). Using the multilinearity relation in Kn (F )thisisequalto

e e [P Q, ⇥ Id , ⇥ Id ,...,⇥ Id ]+ 0 Q 2 Q n Q [P Q, ⇥ Id , ⇥ Id ,...,⇥ Id ] 1 Q 2 Q n Q

The Steinberg relation is more dicult. We ﬁrst show the image of a Steinberg symbol is independent of the automorphisms of the module.

Lemma 5.2.2. Let P be a finitely-generated projective module for which there exists an autmorphism of P such that 1 is invertible. Then there exists a finitely generated module Q such that there is an automorphism ⇥ of Q with 1 ⇥ invertible and P Q is free. Proof. Because P is projective there obviously exists a Q such that P Q is free. If Q satisfies the necessary properties then we are done. Otherwise we replace Q with P Q Q and let 0 0 0 1 ⇥ := 00IdQ B C B 0IdQ IdQC B C @ A

Lemma 5.2.3. Let P be a projective module and let ⇥1, ⇥10 be automorphisms of P such that 1 ⇥0 and 1 ⇥ are both invertible. Then 1 1

s[P, ⇥ , 1 ⇥ , ⇥ ,...,⇥ ]=s[P, ⇥0 , 1 ⇥0 , ⇥0 ,...,⇥0 ] 1 1 3 n 1 1 3 n

Proof. We take Q to be a projective as in lemma 5.2.2. Then

s[P, ⇥ , 1 ⇥ , ⇥ ,...,⇥ ]=s[P, ⇥ , 1 ⇥ , ⇥ ,...,⇥ ] 1 1 3 n 1 1 3 n +[P Q,⇥0 , (1 ⇥0 ) (1 ), ⇥0 Id,...,⇥0 Id] 1 1 3 n

93 5.2. THE COFINALITY THEOREM

Combining these two terms using the exact sequence relation gives

[P Q P Q, ⇥ Id ⇥0 , (1 ⇥ ) Id (1 ⇥0 ) (1 ), 1 1 1 1 ⇥ Id ⇥0 Id,...,⇥ Id ⇥0 Id] (5.1) 2 2 n n One can get the result from this by taking an exact sequence whose middle term is 5.1andwhoseﬁrsttermisjusttheinclusionoftheﬁrstandlastcoordinate.

Lemma 5.2.3actuallycompletestheproofthattheSteinbergrelationholdswhen n 3becausewecanjustchoose⇥0 =Id.Theonlycaseleftisthecasen =2.In 3 this case we have shown that

s[P, ⇥, 1 ⇥] = s[P, ⇥0, 1 ⇥0] whenever this makes sense. We denote an element s[P, ⇥, 1 ⇥] by s(P ) Note that for projective modules M,N we have that s(M N)=s(M) s(N) providing both s(M)ands(N)exist. Our aim now is to show that s(P )=0wheneveritexists.Webeginwiththe following lemma.

Lemma 5.2.4. Let Q be a projective R-module. If there is an automorphism ✓ of Q such that 1 ✓2 are invertible then 3s(Q)=0 KM (F ) 2 n Proof. We have that e

s(Q)=[Q,✓2, 1 ✓2] =[Q,✓2, (1 ✓)(1 + ✓)] =[Q, ( ✓)2, 1+✓]+[Q,✓2, 1 ✓] =2[Q, ✓2, 1+✓]+2[Q,✓, 1 ✓]=2s(Q)+2s(Q) which gives the result.

94 5.2. THE COFINALITY THEOREM

From this we can show that for any projective module P we have that 3s(P 2)=0 and 3s(P 3)=0.Toprovetheﬁrstoftheseidentitiestake

0 IP 0 IP ✓ = , 1 ✓2 = 0I I 1 0 I I 1 P P P P @ A @ A and for the second identity take

00I I I 0 P P P ✓ = 0I 0 I 1 , 1 ✓2 = 0 00I 1 P P P B C B C B 0 IP 0 C B IP 00C B C B C @ A @ A Then the two identities above give us the following

Lemma 5.2.5. Let P be any projective R-module. We have that

3s(P )=0 KM (F ) 2 n To ﬁnish the proof we will show that 4es(P )=0.Wedothisbypickingan explicit representation of s(P 4). We take this to be

000 I I 00I P P P 20 1 0 13 IP 00IP IP IP 0 IP s(P 4)= , 6B C B C7 6B 0 IP 0 IP C B 0 IP IP IP C7 6B C B C7 6B C B C7 6B 00IP IP C B 00IP 0 C7 6B C B C7 4@ A @ A5 One can check that both these maps are invertible. Furthermore, it is true that

10 000 I I 000 P P 0 1 0 1 IP 00IP 0 IP 00 = B C B C B 0 IP 0 IP C B 00IP 0 C B C B C B C B C B 00IP IP C B 000IP C B C B C @ A @ A So we have that 10s(P 4)=0,butwealsohavethat3s(P 4)=0bytheprevious lemma so s(P 4)=0,hence4s(P )=0.

95 5.3. THE STEINBERG RELATION FOR QUILLEN K-THEORY

5.3 The Steinberg relation for Quillen K-theory

In this section we prove the Steinberg relation for Grayson’s deﬁnition of higher K-theory of a ring.

Lemma 5.3.1. Let R be any ring. Denote elements of the form

x R R 1 1 y 1 y x R R 1

Q in K2 (R) by [x, y]. Then we have that

4[a3, 1 a3]=0 KQ(R) 2 2

3 3 for all a , 1 a R⇤. 2

1 1 Proof. We show that this relation holds when R is the ring Z[t, t , (1 t) ]. R is Q aregularringsoweknowthatK2 (R) is homotopy invariant. Using this we may show that

00t3 10 t3 2010 01 , 0 11 013 = 6B C B C7 6B01 0C B 0 11C7 6B C B C7 4@ A @ A5 00 t3 10 t3 2010 (t3 +1)1 , 0 11 t3 +1 13 6B 3 C B 3 C7 6B01 t +1 C B 0 11 (t +1)C7 6B C B C7 4@ A @ A5 Using the homotopy

00 t3 10 t3 2010 x(t3 +1)1 , 0 11 x(t3 +1) 13 6B 3 C B 3 C7 6B01 x(t +1) C B 0 11 x(t +1)C7 6B C B C7 4@ A @ A5 96 5.3. THE STEINBERG RELATION FOR QUILLEN K-THEORY

We reduce these matrices to 1 1matricesintheringR[!]/(!2 + ! +1).Weﬁrst ⇥ use following change of bases matrices on the matrices above

t2 01 100 0 t 101 , 0 1101 B C B C B100C B 101C B C B C The ﬁrst column of each is@ an eigenvectorA changing@ basesA gives the following:

t 10 1 t 10 200 t 11 , 0 01+t 113 = 6B 2 C B 2 C7 6B0 t 0C B 0 t 1 C7 6B C B C7 4@ A @ A5 t3 0 t3 1 t3 0 t3 2000 11 , 0 01113 6B C B C7 6B01 1C B 0 10C7 6B C B C7 Using the exact sequence relation we get4 that@ A @ A5

t 1 1+t 1 [t, 1 t]+ , = 20 t2 01 0 t2 1 13 4@ A @ A5 0 1 11 [t3, 1 t3]+ , 20111 0 1013 Using the change of bases matrices 4@ A @ A5

10 10 0 !2t 11 0! 11 we get that @ A @ A

!t 1 1 !t 1 [t, 1 t]+ , = 20 0 !2t1 0 01!2t13 4@ A @ A5 ! 1 1+! 1 [t3, 1 t3]+ , 20 0 !21 0 01+!213 4@ A @ A5 97 5.3. THE STEINBERG RELATION FOR QUILLEN K-THEORY

Then using the exact sequence relation we get that

[t, 1 t]+[!t, 1 !t]+[!2t, 1 !2t]=[t3, 1 t3]+[ !, 1+!]+[ !2, 1+!2] Using linearity we can get that

[t, 1 t3]+[!, (1 !t)(1 !2t)2]=[t3, 1 t3]+[ !, 1+!]+[ !2, 1+!2] Multiplying both sides by 3 eliminates all terms involving ! because 3[!, b]=0 and [ 1, 1+!]+[ 1, 1+!2]=0sowehaveshownthat2[t3, 1 t3]=0.We 3 3 Q 1 3 1 use the transfer map to get that 4[t , 1 t ]=0 K2 (Z[t, t , (1 t ) ]). To 2 get the result for a general ring R we use the fact we can take a homomorphism

1 3 1 Z[t, t , (1 t ) ] R with t a. ! 7!

Corollary 5.3.2. Let R be any ring, and a, 1 a R⇤. Then 12[a, 1 a]=0 2 2 Q K2 (R).

1 1 Proof. We prove this for the ring R = Z[t, t , (1 t) ]andtheelement[t, 1 t]. 1 1 3 Consider the ring S = Z[t, t , (1 t) ][x]/(x t). Then we know, by 5.3.1, that 4[x3, 1 x3]=0 KQ(S). 2 2 Hence taking the image under the transfer map we have that 12[t, 1 t]=0 2 Q K2 (R).

We are ﬁnally able to prove the Steinberg relation

Proposition 5.3.3. Let R be any ring and a, 1 a R⇤. Then 2 [R, a, 1 a]=0 KQ(R) 2 2

Q 1 1 1 Proof. We show that [t, 1 t]=0 K2 (Z[t, t , (1 t) ]). Let R = Z[t, t , (1 2 1 12 t) ][x]/(x t). Then we know, by 5.3.2, that [t, 1 x]=12[x, 1 x]=0 KQ(R) 2 2

98 5.3. THE STEINBERG RELATION FOR QUILLEN K-THEORY

Applying the transfer map to this element gives us the following 12 12 matrices ⇥ 1 ... 0 t t...0 . 2 0 1 .. 0013 0. .. .1 Q 1 1 . . . =0 K2 (Z[t, t , (1 t) ]) 6 B ...... C7 2 6 B . . . . C7 6B0 ... tC B C7 6B C B C7 6B C B 0 ... 11C7 6@ A B C7 4 @ A5 We can now use elementary row and column operations to reduce the matrix on the right to

1 ... 00 t...0 2 0. .. . . 13 ...... 0. .. .1 , 6 B C7 6 B0 ... 10C7 6B0 ... tC B C7 6B C B C7 6B C B0 ... 01 tC7 6@ A B C7 4 @ A5 The result follows.

To ﬁnish the proof that the map KM (R) KQ(R)iswell-deﬁnedweneedto n ! n show that the identity holds for free modules. e Q First note the Steinberg relation holds for free modules of rank 1 in Kn (R) because

[a , 1 a ,...,a ]=[a , 1 a ] [a ,...,a ] 1 1 n 1 1 ⌦ 3 n We need to show that

Rm,A , 1 A ,A ,...,A =0 KQ(R). 1 1 3 n 2 n h i Let S be the commutative subring of Mn(R)generatedbyA1,A3,...,An. We know that [S, A , 1 A ,A ,...,A ]=0 KQ(S). We deﬁne a functor 1 1 3 n 2 n

F :Proj Proj S ! R P Rm P 7! S O

99 5.3. THE STEINBERG RELATION FOR QUILLEN K-THEORY

and given a morphism f : P Q we deﬁne F (f)=Id m f.Thisfunctorinduces ! R ⌦ amaponK-theoryKQ(S) KQ(R). One can show that the image of S under this n ! n m functor is R and the image of the homomorphism Ai is the matrix Ai. We have shown the following:

Theorem 5.3.4. Let R be a ring. There exists a homomorphism

: KM (R) KQ(R) n ! n such that when R is a local ring withe inﬁnite residue ﬁeld the comparison homomorphism from Milnor K-theory to Quillen K-theory is equal to the composition

KM (R) KM (R) KQ(R) n ! n ! n M Q We know that for a local ring withe infinite residue field K2 (R) ⇠= K2 (R). We conjecture the map defined above is an isomorphism more generally.

Conjecture 5.3.5. Let R be any ring. The map

KM (R) KQ(R) 2 ! 2 is an isomorphism. e

We know that this map is an isomorphism for R a ﬁeld. We also know, because, by [15], the composition

KM (R) KM (R) KQ(R) 2 ! 2 ! 2 is an isomorphism for R alocalringwithinﬁniteresidue,thate Kˆ M (R) KQ(R)is 2 ! 2 surjective.

Corollary 5.3.6. Let R be a regular, local ring with inﬁnite residue ﬁeld, then the map

KM (R[t ,...,t ]) KQ(R[t ,...,t ]) n 1 n ! 2 1 n is surjective. e

100 5.3. THE STEINBERG RELATION FOR QUILLEN K-THEORY

Proof. We do this by induction on n. The case n = 0 is considered above. Assume that this holds for n = k.Considerthecommutativediagram

KM (R[t ,...,t ]) KQ(R[t ,...,t ]) n 1 k+1 ! n 1 k+1 e M ? Q ? Kn (R[t1,...,t? k,tk+1]) Kn (R[t1,...,t? k,tk+1]) y ! y by induction thee top map is surjective and by homotopy invariance the right map is surjective. Hence, the bottom map is surjective.

We can also use the map to Quillen K-theory to prove the following:

Corollary 5.3.7. Let R be a local ring with inﬁnite residue ﬁeld. Then the kernel of the map

KM (R) KM (R) n ! n is annihilated by (n 1)!. In particular, whene n =2the map is injective. Proof. There is a map

KQ(R) KM (R) n ! n such that the composition

KM (R) KM (R) KQ(R) KM (R) n ! n ! n ! n is multiplication by (n 1)!. e

101 Chapter 6

Further questions

6.1 Surjectivity for local rings

In section 4.1 we showed that the transfers for Milnor K-theory are compatiable

M with the transfers for Kn .InthecasewhenR is a ﬁeld we can use compatibility of the transfer or the reciprocity laws to prove that the map is surjective. To do this e M we need that every element in Kn (F )istheimageoftransfersofrankoneelements. In this section, we show that when R is a regular local ring KM (R)isimagesofrank e n one elements under a transfer map. Unfortunetly, we do not have a reciprocity law e to manipulate these elements nor do we have the corresponding transfers we need for Milnor K-theory. Let R be a regular, local ring. We have shown, in section 4.4.1, that

KM (R) GM (R) n ! n is an isomorphism. Take an elemente e

[M,⇥ ,...,⇥ ] GM (R). 1 n 2 n e Like in the ﬁeld case, we can consider M as a R[t1±,...,tn±]-module. We can take a

102 6.1. SURJECTIVITY FOR LOCAL RINGS

ﬁltration of M where each qoteient is of the form

R[t1±,...,tn±]/p

M where p is prime. Hence every element in Gn (R)canbewrittenasasumofelements of the form e

NR[t±,...,t±]/p R[R[t1±,...,tn±]/p, t1,...,tn] 1 n |

To complete the proofe as in the ﬁeld case we either need a more general version of reciprocity or transfers for Milnor K-theory.

M In the proof of surjectivity for ﬁelds we gave an alternative proof that Kn (F )is generated by the image of rank one transfers. This also carries over, in some way, e to the realm of regular local rings. Take an element

m [R , ⇥1,...,⇥n]

Let c⇥1 (t)bethecharacteristicpolynomialof⇥1.Let

c⇥1 (t)=p1(t) ...pl(t) be the factorizations into irreducibles in the field of fractions. As in the field case we can define M to be the subspace annihilated by some monic polynomial.

0 [M,⇥ ,...,⇥ ] [[Rm, ⇥ ,...,⇥ ]] [Rm/M, ⇥ ,...,⇥ ] 0 ! 1 n ! 1 n ! 1 n !

Now M is a R[t]/p(t)-module where t M =⇥ M. ⇥ 1 ⇥ M Hence, we have that Kn (R)isgeneratedbytransfersoftheform

[R[t]/p(t),t] [M,⇥ ,...,⇥ ] GM (R[t]/p(t)) ⌦R[t]/p(t) 2 n 2 n where p(t)isanirreducible,monicpolynomial. e

103 6.2. THE CASE FOR DVRS

6.2 The case for DVRs

M In the previous section, we showed that Kn (R)isgeneratedbyimagesofrank 1elementsundersometransfer.Inthissection,weshowthatifR is a discrete e valuation ring we can deﬁne the necessary transfers for Milnor K-theory. However we do not know whether these transfers commute.

M Let R be a discrete valuation ring. We know that the group Gn (R)isgenerated by elements of the form e

[R[t1±,...,tn±]/p, t1±,...,tn±] where p is prime. Consider the map

R R[t±,...,t±]/p. ! 1 n

First assume that the map is not injective. Then the kernel is a non-trivial prime ideal, so must be ⇡. Hence the map factors as

R R/⇡ R[t±,...,t±]/p. ! ! 1 n

Hence the element [R[t1±,...,tn±]/p, t1,...,tn]isintheimageofthetransfer

M M M NR/m R : Gn (R/⇡) Gn (R) | !

e M e e R/⇡ is a ﬁeld so we know that Gn (R/⇡)isageneratedbyelementsoftheform

e [R/⇡, a1,...,an].

M We claim that these elements are equal to 0 in Gn (R). Let ai be any lifting of ai in R.Wehaveanexactsequence e b ⇡ 0 [R, a ,...,a ] ⇥ [R, a ,...,a ] [R/⇡, a ,...,a ] 0 ! 1 n ! 1 n ! 1 n !

b b b b 104 6.3. THE MAP TO HOMOLOGY

So using the exact sequence relation we have that

[R/⇡, a ,...,a ]=[R, a ,...,a ] [R, a ,...,a ]=0 1 n 1 n 1 n

So we are left with the case R R[t±b,...,t±b]/p is injective.b b ! 1 n R[t1±,...,tn±]/p is finite over R and so is 1-dimensional. Let S denote the integral closure of R in the field of fractions of R[t1±,...,tn±]/p.WeknowthatS is a finite R- module which contains R[t1±,...,tn±]/p.WealsoknowthatS is a Dedekind domain. Consider the exact sequence

⇡ 0 [R[t±,...,t±]/p, t ,...,t ] ⇥ [S, t ,...,t ] [M,t ,...,t ] 0 ! 1 n 1 n ! 1 n ! 1 n ! where M is a ﬁnitely generated R/⇡-module. Using a similar argument to above we can deduce that [M,t1,...,tn]=0. M So we have shown that Gn (R)isgeneratedbyelementsoftheform[S, a1,...,an] where S is a Dedekind domain. Consider the diagram e

@⇡ 0 KM (S) KM (L) i KM (S/⇡ ) 0 n n ⇡i n i M M NL/F NS/⇡ R/⇡ i| M M @⇡ M P 0 Kn (R) Kn (F ) Kn (R/⇡) 0

The diagram commutes and each of the rows are exact when R contains an inﬁnite ﬁeld. So we have constructed transfers

KM (S) KM (R) n ! n

M If these transfers are compatible with those for Kn then we are done.

e 6.3 The map to homology

M In this section, we give a possible map from Kn (R)toHn(GL(R))/Hn(GLn 1(R)) which agrees with map from Milnor K-theory. e 105 6.3. THE MAP TO HOMOLOGY

We begin by deﬁning a map

m : Z [R ,A1,...,An] Hn(GLn(R)) { }!

m Given an element [R ,A1,...,An]wedeﬁne

m ([R ,A1,...,An]) = sgn()[A(1),...,A(n)]

Sn X2 We need to show this map is well-deﬁned. We ﬁrst show that

@i( sgn()[A(1),...,A(n)]) = 0

Sn X2 when 0

@i( sgn()[A(1),...,A(n)]) =

Sn X2 [A ,...,A A ,...,A ] [A ,...,A A ,...,A ] (1) (i) (i+1) (n) (1) (i+1) (i) (n) Sn,+ 2X this is 0 because all matrices commute. We claim that

@ +( 1)n@ =0. 0 n

To show this we need to show that

n sgn()[A(2),...,A(n)]+( 1) sgn(0)[A (1),...,A (n 1)]=0 0 0 Sn (1)=i Sn (n)=i 2 X| 02 X| 0 right multiplication by (1,...,n)sendselementsofSn which send 1 to i to elements which send n to i.

sgn()[A(2),...,A(n)]+

Sn (1)=i 2 X| n ( 1) sgn(0(1,...,n))[A (1,...,n)(1),A (1,...,n)(2) ...,A (1,...,n)(n 1)]=0 0 0 0 Sn (1,...,n)(n)=i 02 | 0X

106 6.3. THE MAP TO HOMOLOGY

So we have constructed a symbol in Hn(GL(R)). We denote this symbol by

µ(A1,...,An). We show that this symbol satisﬁes the multlilinear relation. This means we have to show that

µ([A A ,A ,...,A ]) µ([A ,A ,...,A ]) µ([A ,A ,...,A ]) 1 2 3 n 1 3 n 2 3 n is the image of some boundary map. We claim this is

@( sgn()[A(1),A(2),...,A(n)]) 1 1 Sn (1)< (2) 2 | X First take 1 i n 1. Then  

@i( sgn()[A(1),A(2),...,A(n)]) = 1 1 Sn (1)< (2) 2 | X

sgn()[A(1),...,A(i)A(i+1),...,A(n)] 1 1 Sn (1)< (2) 2 | X Using a similar argument to above (apply (i, i +1))wecanseethateveryelement in the sum cancels unless (i)=1or(i) = 2. Again using similar argument as above we see that the only elements in the image of @0 that do not cancel with some element in @n are elements such that (1) = 1. Conversely, the only elements of @n that do not cancel with some element of @0 are elements of the form (n)=2. So we have that µ is a multilinear symbol. However µ is likely not additive unless n =1.Forµ to be additive when n =2wewouldneedthat

A 0 C 0 µ( , )=µ([A, C]) + µ([B,D]) 00 B1 0 0 D1 @ A @ A Using linearity we can see that this is equivalent to

A 0 10 µ( , )=0 0011 00 D1 @ A @ A

107 6.3. THE MAP TO HOMOLOGY

For any A, D. This is not true in general. However, given a multilinear symbol we can define a additive symbol. We first do this for the case n =2.Wedefine

A 0 10 c(A, B) := µ(A, B) µ( , ) 0011 00 B1 @ A @ A c(A, B)isalsobilinearsoweonlyneedtoseetheidentityholdsabove.

A 000) 1000) A 0 10 A 0 10 001001 0010 01 c( )=µ( ) µ( ) 0 1 0 1 0 1 0 1 B C B C 01 0 D 01 0 D B0010C B001 0C B C B C @ A @ A @ A @ A B C B C B0001C B000DC B C B C @ A @ A Changing basis and using the fact we are working in GL(R)givestheresult. We can rewrite this formula as

A 00 B 00 0 1 1 0 1 c(A, B)=µ( 0 A 0 , 01 0 ) B C B 1C B001C B 00B C B C B C @ A @ A We outline how to construct an additive symbol generally and then present this map in the case n =3. Let A ,...,A be commuting automorphisms of Rm = P and let f : 1,...,n 1 n { }! 1,...,n be a function. Deﬁne f[A ,...,A ]=[B ,...,B ]whereB is an auto- { } 1 n 1 n i morphism of P n of the form

1 . 0 .. 1 B C B 1 C B C B C B Ai C B C B C B 1 C B C B .. C B . C B C B C B 1C B C @ A 108 6.3. THE MAP TO HOMOLOGY

where Ai is in the (f(i),f(i)) position. Using the same argument as above we can see that a symbol is additive providing that c(f[A1,...,An]) = 0 whenever f with exactly 2 elements in the image. We will construct a symbol such c(f[A1,...,An]) = 0wheneverf is not constant. Take any multilinear symbol µ.Deﬁne

c (A ,...,A )=µ([A ,...,A ]) µ(f[A ,...,A ]) n 1 n 1 n 1 n where f is the identity on 1,...,n .Itiseasytoseethatc (f[A ,...,A ]) = 0 { } n 1 n for any bijective f.Inductively,wedeﬁne

ci 1([A1,...,An]) = ci([A1,...,An]) ci(f[A1,...,An]) f : 1,...,n 1,...,n s.t. im(f)=i 1 { }!{ X} | | The sum is over functions f : 1,...,n 1,...,n such that f(1) = 1, the image { }!{ } has precisely i 1elementsandf(j) max f(1),...,f(n) +1forallj  { } Example 6.3.1. We do the above computation when n =3. First let

A 1 1

c3([A1,...,A3]) := µ([A1,A2,A3]) µ(0 1 1 , 0 A 1 , 0 1 1) 2 B C B C B C B 1C B 1C B A3C B C B C B C @ A @ A @ A Next we deﬁne c2 to be

A1 0 A2 0 10 c ([A ,...,A ]) := c ([A ,A ,A ]) c ([ , , ]) 2 1 3 3 1 2 3 3 0 1 0 1 0 1 01 01 0 A3 @ A @ A @ A A1 0 10 10 A1 0 10 A3 0 c ([ , , ]) c ([ , , ]) 3 0 1 0 1 0 1 3 0 1 0 1 0 1 01 0 A2 0 A3 01 0 A2 01 @ A @ A @ A @ A @ A @ A

109 6.3. THE MAP TO HOMOLOGY

Writing this in terms of µ gives

A1 0 A2 0 10 c([A ,A ,A ]) := µ([A ,A ,A ]) µ([ , , ]) 1 2 3 1 2 3 0 1 0 1 0 1 01 01 0 A3 @ A @ A @ A A1 0 10 10 A1 0 10 A3 0 µ([ , , ]) µ([ , , ]) 0 1 0 1 0 1 0 1 0 1 0 1 01 0 A2 0 A3 01 0 A2 01 @ A @ A @ A @ A @ A @ A A1 1 1 0 1 0 1 0 1 +2µ( 1 , A2 , 1 ) B C B C B C B 1C B 1C B A3C B C B C B C @ A @ A @ A This map gives an additive multilinear symbol. We conjecture that the steinberg relation and the exact sequence relation hold under this map. One may be able to prove that they do by using homotopy invariance as was done in Grayson’s deﬁnition of higher K-theory. Therefore we conjecture that the map

Kn(R) Hn(GL(R))/Hn(GLn 1(R)) ! [Rn,A ,...,A ] c(A ,...,A ) e 1 n 7! 1 n is well-deﬁned.

M Recall that the map Kn (R) Hn(GL(R))/Hn(GLn 1(R)) is an isomorphism. ! One can see that the composition

M M Kn (R) Kn (R) Hn(GL(R))/Hn(GLn 1(R)) ! ! is equal to a constant multiplee of the above map.This constant should be (n 1)! but we have no proof of this. It should also be true that the map we have deﬁned above factors as

KM (R) KQ(R) H (GL (R)). n ! n ! n n e

110 Bibliography

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