Algebraic K-theory
Math 600D - Fall 2011
Sujatha Ramdorai
Lecture notes by Pooya Ronagh
Department of mathematics, University of British Columbia
Room 121 - 1984 Mathematics Road, BC, Canada V6T 1Z2
E-mail address: pooya@math.ubc.ca
Chapter 1. Zeroeth K-groups
1. K-theory of an abelian category
2. K0 of local rings
3. K0 of some quotient rings
4. K-theory of Dedekind domains
5. K-theory of rings
6. K-theory of rings of polynomials
7. K-theory of topological spaces
8. K-theory of schemes
3
3
5
6
7
8
10
12
13
Chapter 2. K1 of a ring
1. Stable range results
2. K1 and topology
3. Description of K1 using group cohomology
4. Baby K-theory long exact sequence
14
17
18
19
19
Chapter 3. K2 groups
1. Milnor’s K2
2. K2 as an obstruction element
3. Constructing elements of K2
4. Case of fields
22
22
23
26
29
Chapter 4. Galois Cohomology
1. Lower cohomology groups
2. Computations
3. Maps in the level of cohomology
4. Profinite groups
30
31
32
33
36
Contents
CHAPTER 1
Zeroeth K-groups
1. K-theory of an abelian category
Recall that if R is a ring, an R-module P is said to be projective if any of the following
equivalent conditions are satisfied:
(1) P is a direct summand of free modules,
(2) 0 Ð
→M Ð
→N Ð
→P Ð
→ 0 splits for any sequence of R-modules,
(3) the functor HomR (P, −) is exact,
(4) the following diagram always completes
X
~~
~
~
~
P .
// Y
The projective dimension, pdR (M ), of the R-module M is the minimum length of all
projective resolutions. In particular pdR M = 0 whenever M is projective. The global
dimension (or homological dimension) of R is the supremum over pdR (M ) of all R-modules
M . A good example to recall is that if R is a regular local ring, then gl. dimR < ∞ and
is equal to its Krull dimension. We will use the notations R − Mod , R − mod and PR
respectively for the categories of all, finitely generated, and finitely-generated projective
R-modules.
By an abelian category we mean a small preadditive category (i.e. Hom(A, B) has the
structure of an abelian group, compatible with category structure), that has finite (co)products, has a zero object (i.e. both initial and terminal), is (co-)normal (i.e. every
mono(epi-)morphism is the (co-)kernel of some morphism), and finally has epi-mono factorization:
MF
F
f
FF
FF
FF
"
/N
y<
y
yy
yy
y
y
im(f )
3
1. K-THEORY OF AN ABELIAN CATEGORY
4
In an abelian category, A , the abelian group K0 (A ) is one generated by isomorphism classes
of objects of the category with the relations
[B] = [A] + [C], for any short exact sequence 0 Ð
→AÐ
→BÐ
→CÐ
→ 0.
This group is universal in the sense that for any abelian group G, and group homomorphism
f ∶ ob(A ) Ð
→ G respecting the relations f (B) = f (A) + f (C) for any short exact sequence as
above, then there is a unique group morphism completing
/ K0 (A) .
JJ
JJ
JJ
J
JJ !
f
J% ob(A )
G
Lemma 1. If A has countable direct sums then K0 (A ) = 0.
Proof. Let B = ∐∞ A for any object A. Then A ⊕ B = B implying [A] = 0.
If A and B are two abelian categories and T ∶ A Ð
→ B is an exact functor we get an induced
group homomorphism K0 (A ) Ð
→ K0 (B ). One therefore concludes that K0 is a co-variant
functor from the category of abelian categories and exact functors between them to the
category of abelian groups.
Corollary 1. If A and B are equivalent abelian categories, then K0 (A ) = K0 (B ).
It is immediate that K0 (ab ), that of the category of abelian groups is isomorphic to Z and
the isomorphism is given by the rank of the abelian group. If A is however the category
of finite abelian groups then K0 (A) is the free abelian group with basis [Z/p]’s. A more
general observation is the following
Lemma 2. If A is an abelian category with Jordan-Holder filtrations, i.e.
A = An ⊇ An−1 ⊇ ⋯ ⊇ A1 ⊇ A0 = 0
with simple factors, Ai /Ai−1 , then
K0 (A ) =
⊕ [S].
simple S
Lemma 3. Two objects A, B in the abelian category A have the same class in K0 (A ) if and
only if there is C in A such that A ⊕ C = B ⊕ C.
In particular, this motivates the jargon stably isomorphic for R-modules: M and N are
stably isomorphic R-modules if
M ⊕ Rk ≅ N ⊕ Rk
for some integer k > 0. We say M is stably free if
M ⊕ Rk ≅ Rn
for some integers k, n > 0.
2. K0 OF LOCAL RINGS
5
Proof. One way is obvious. For the other direction, let [A] = [B]. Then we can
rewrite [A] − [B] = 0 as
[A] − [B] = ∑([Ai ⊕ Bi ] − [Ai ] − [Bi ]) − ∑([A′j ⊕ Bj′ ] − [A′j ] − [Bj′ ])
i
So
j
[A] + ∑[A′i ⊕ Bi′ ] + ∑([Aj ] + [Bj ]) = [B] + ∑[Ai ⊕ Bi ] + ∑([A′j ⊕ Bj′ ]).
i
Therefore C =
j
(⊕i Ai ⊕ Bi ) ⊕ (⊕ A′j
⊕ Bj′ )
i
we prove our claim.
j
2. K0 of local rings
In what follows we do not need R to be necessarily a commutative ring. Our goal is to
study the Grothendieck group,
K0 (R) ∶= K0 (PR ),
of the category of projective R-modules. It is for instance immediate that if R is any PID,
then rank induces K0 (R) ≅ Z.
Jacobson radical is the intersection of all maximal ideals as in the commutative case. Then
the following version of Nakayama lemma is handy:
Theorem 2.1 (Nakayama lemma). let I be an ideal of the ring R containing the Jacobson
radical. Suppose M is a finitely-generated R-module. Then M /IM = 0 implies M = 0.
Theorem 2.2 (Kaplansky). Let R be a local ring and P a projective R-module. Then P is
free.
Proof. Let J = J(R) be the Jacobson radical of R. Since R is assumed to be a local
ring, this is the unique maximal ideal of R. Hence D = R/J(R) is a division ring. Let M
be a finitely generated projective R-module. In particular say
M ⊕ N ≅ Rn .
Then M /JM and N /JN are D-vector spaces. We now fix e1 , ⋯, em and e′1 , ⋯, e′s to form
bases of the former vector spaces. By Nakayama’s lemma these lift to generators e1 , ⋯, em
and e′1 , ⋯, e′s of M and N . It now suffices to show that
{x1 , ⋯, xn } = {e1 , ⋯, em , e′1 , ⋯, e′s }
form a basis of Rn . Let f1 , ⋯, em+s be the standard basis of M ⊕ N ≅ Rn . So we already
have
fi = ∑ aij xi , xi = ∑ bij fj
for all i = 1, ⋯, n. By the notation A = (aij ) and B = (bij ) we have AB = In . Recall now
that
J(R) = {x ∈ R ∶ 1 − ax is invertible for any a ∈ R}.
It follows now that BA − I ∈ Mn (J(R)) proving our claim.
3. K0 OF SOME QUOTIENT RINGS
6
3. K0 of some quotient rings
Another interesting variant of K0 is
G0 (R) ∶= K0 (R − mod ).
Remark. If f ∶ R Ð
→ R′ is a ring homomorphism we get an induced group homomorphism
f∗ ∶ K0 (R) Ð
→ K0 (R′ )
if moreover R′ is R-flat then we have an induced homomorphism
f∗ ∶ G0 (R) Ð
→ G0 (R′ )
as well. If R′ is a finitely-generated R-module, then the forgetful functor R′ − mod Ð
→ R− mod
induced
res ∶ G0 (R′ ) Ð
→ G0 (R)
′
and if R is moreover R-projective, we get an induced
res ∶ K0 (R′ ) Ð
→ K0 (R).
Our final remark is that the natural inclusion PR ⊂ R − mod induces the so called Cartan
homomorphism
c0 ∶ K0 (R) Ð
→ G0 (R).
Theorem 3.1 (Idempotent lifting). Let R be a ring and J a two-sided ideal of R contained
in the Jacobson radical. Let R = R/J. Then the quotient map induces a functor
PR Ð
→ PR
which is full, additive and satisfies the following: if f ∶ P Ð
→ Q is a morphism in PR such
that f ∶ P Ð
→ Q is an isomorphism then f is also an isomorphism.
This is easy to prove. In particular we have
Theorem 3.2. If R is J-adicaly complete (i.e. lim R/J n ≅ R), then the above functor
←Ð
PR Ð
→ PR is bijective.
Proof. Given Q ∈ PR we look for a lift of it P ∈ PR . Since Q is projective we may
think of it as the image of an idempotent:
n
Q = im(ρ), ρ ∈ Mn R = EndR R , ρ2 = ρ.
All we have to do is to lift ρ to idempotent ρ ∈ Mn (R). Let A = Mn (R), A = Mn (R/J) =
Mn (R)/Mn (J). Given any a ∈ A, n > 0 we have
2n
2n
1 = (a + (1 − a))2n = ∑ ( )a2n−j (1 − a)j .
j
0
But
n
2n
fn (a) = ∑ ( )a2n−j (1 − a)j ≡ a mod an A.
j
0
4. K-THEORY OF DEDEKIND DOMAINS
7
therefore fn (a) ≡ a mod a(1 − a)A. Now choose a2 − a = 0, then fn (a)2 = fn (a). If R is
J-adically complete, then A is Mn (J)-adically complete. Thus we can form a compatible
system of idempotents in R/J k ’s, constructing an element ρ ∈ lim A/Mn (J) ≅ A such that
←Ð
ρ ↦ ρ.
Corollary 2. K0 (R) = K0 (R/J) if R is J-adically complete and J ⊆ J(R).
4. K-theory of Dedekind domains
Recall that a Dedekind domain R is an intergrally closed noetherian domain of Krull dimension 1.
Example 4.1. Examples are Z, k[x], ring of integers in an algebraic number field. Let
K be a finite extension of Q and
OK = {θ ∈ K ∶ f (θ) = 0 for some monic polynomial f (x) ∈ Z[x]}
be the integral closure of Z in K
OK
Z
K
⊆
Q
Then OK is a Dedekind domain. If R is a noetherian, integrally closed ring and p is a prime
ideal of height one, then Rp is also a Dedekind domain.
⌟
So let R be a Dedekind domaion and K its function field.
Definition 1. A fractional ideal of K is a nonzero R-submodule I of F such that there is
an element a ∈ I with aI ⊆ R.
Fractional ideals form an abelian monoid under multiplication with (1) the identity element.
If I is a fractional ideal , there exists a fractional ideal
I −1 = {a ∈ K ∶ aI ⊆ R}
and that II −1 = R. So the fractional ideals of R form a group and the principal fractional
ideals form a subgroup.
Definition 2. The class group C(R) for a Dedekind domaion R is defined to be
C(R) =
the group of fractional ideals
subgroup of principal fractional ideals
Theorem 4.1. If R is a Dedekind domain, then every fractional ideal is finitely generated
and projective.
5. K-THEORY OF RINGS
8
Proof. Let I be a fractional ideal, since I −1I = R there are elements x1 , ⋯, xn ∈ I −1
and y1 , ⋯yn ∈ I such that
n
∑ x i yi = 1
i=1
If b ∈ I, then b = ∑(bxi )yi with bxi ∈ II
−1
= R. Hence y! , ⋯, yn generate I.
Consider the homomorphism Rn Ð
→ I via ei ↦ yi . This has a splitting with right inverse
b ↦ (bx1 , ⋯, bxn )
Hence I is a direct summand of a free modli and therefore projective.
Corollary 3. If R is a Dedekind domain, then every finitely generated projective Rmodulie is isomorphic to a direct sum of ideals. In particular the isomorphism classes of
ideals generate K0 (R).
Proof. Let M ∈ PR . Then M ⊆ Rn . Proof is by induction on n. Let π ∶ Rn Ð
→ R be the
projection to the last factor. Then π(M ) ⊆ R, hence π(M ) is an ideal in R. If π(M ) = 0
we are done otherwise, we may assume that
0 ≠ π(M ) = I ⊆ R.
As I is projective we get M ≅ ker π∣M ⊕ I. But ker π ⊆ Rn−1 and we are done by induction
hypothesis.
It is now easy to prove
Theorem 4.2. If R is a Dedekind domain, then K0 (R) ≅ Z ⊕ C(R).
Proof idea. Any finitely-generated projective module, P , of rankn can be expressed
as Rn−1 ⊕ I for a fractional ideal I. The mapping K0 (R) Ð
→ Z is just the rank map.
5. K-theory of rings
Recall that K0 (R) = K0 (PR ) and G0 (R) = K0 (R − mod ).
Theorem 5.1 (Devisage, Hellor). Let A be an abelian category, C and B full subcategories
with C ⊆ B and such that
(1) C is closed in A with respect to subobjects and quotient objects.
(2) Eery object of B has afinite filtration with all quotients in C .
Then the canonical map K0 (C ) Ð
→ K0 (B ) is an isomorphism.
5. K-THEORY OF RINGS
9
Proof. The inverse to the natural map K0 (C ) Ð
→ K0 (B is defined by f ∶ K0 (B ) Ð
→
K0 (C ) as follows. Let B ∈ B and
B = B0 ⊇ B1 ⊃ ⋯ ⊃ Bn = 0
such that all factors are in C . Check that [B] ↦ ∑[Bi /Bi+1 ] works.
Corollary 4. Let A be an abelian category in which each object has finite length (with
respect to simple objects). Then K0 (A ) is the free abelian group on [S] where S varies over
a set of representatives of the simple objects of A .
Theorem 5.2. Let A be an abelian category and P be the full subcategory of projective
objects (i.e. those for which Hom(P, −) is exact). If pdA (via resolution by projectives) is
finite for every object A ∈ A , the the natural map
I ∶ K0 (P ) Ð
→ K0 (A )
is an isomorphism.
Proof. Check that A ↦ ∑(−1)i Pi works. For this we need Schanuel’s lemma that we
recall here.
Lemma 4 (Schanuel’s). Let A be an abelian category and
0Ð
→BÐ
→P Ð
→AÐ
→ 0 and 0 Ð
→ B′ Ð
→ P′ Ð
→ A′ Ð
→0
are exact sequences with A ≅ A′ and P, P ′ projective. Then B ⊕ P ′ ≅ B ′ ⊕ P .
This proves independence of the projective resolution once we pass to long exact sequences
from this. In fact let P ● Ð
→ A and P ′● Ð
→ A are two resolutions. By Schanuel’s lemma
′
Pn ⊕ Pn−1 ⊕ ⋯ is isomorphic to Pn′ ⊕ Pn−1 ⊕ ⋯. Thus
′
′
∑ Pi ⊕ ∑ Pk ≅ ∑ Pi ⊕ ∑ Pk .
odd
even
even
odd
Next step is to show that this map is independant of the choice of representative for the
class of A.
Corollary 5. If R is regular (i.e. every finitely generated R-module has a finite projective
resolution) then K0 (R) Ð
→ G0 (R) is an isomorphism.
Corollary 6. If R is a regular commutative ring, then K0 (R) Ð
→ G0 (R) is an isomorphism.
When working with commutative rings the notion of regularity is given equivalently by
Remark. A local ring (A, m) is regular whenever krull dim A = dimk m/m2 . A ring is
regular if Ap is regular for all prime ideals in A (equivalently if so for maximal ideals).
6. K-THEORY OF RINGS OF POLYNOMIALS
10
In general if R is a ring (domain), we can define Pic(R) to be the group generated by the
set of all invertible ideals of R. Then we always have a surjection
K0 (R) ↠ Pic(R).
It is not always the case that K0 (R) = G0 (R). For instance let R = k[t2 , t3 ] ⊂ k[t]. Then
Pic(A) ≅ k+ , the additive group of k.
Remark. The above remarks motivate a general definition of Krull dimension for an Rmodule. And this is given as
dimR M ∶= dim(R/AnnR M ).
Remark. G0 (A) = K0 (A − mod ) has two broad classes: that of the free modules (a Zcontribution) and those with finite length (giving a free group on simple modules).
6. K-theory of rings of polynomials
The question we are going to consider is conditions under which K0 (R[t]) ≅ K0 (R). This is
not always the case but, we will see that this holds when R has finite homological dimension.
We will need the following tools:
Theorem 6.1 (Hilbert’s Syzygy). If R is a ring such that gl. dim R ≤ n, then R[t] has
gl. dim R[t] ≤ n + 1.
Proposition 1. If 0 Ð
→ M1 Ð
→ M2 Ð
→ M3 Ð
→ 0 is an exact sequence of R-modules, then
pd M2 ≤ max{pd M1 , pd M2 }
and equality holds if and only if pd M3 ≠ pd M1 + 1.
Proposition 2. pdk M ≥ n then ToriR (M, N ) = 0 for all i ≥ n + 1 and for all modules N .
This proves
Lemma 5. Let x be a (central) nonzero divisor in a ring S. If M is a nonzero S/x-module
with pdS/x M < ∞ then pdX M = 1 + pdS/x M .
Theorem 6.2. If R is a ring with gl dimR ≥ n then R[t] has gl dim ≤ n + 1.
Proof. Step 1. Let S = R[x]. As S is R-flat, hence pdS M [x] ≤ pdR M . For projective
resolution of R-modules
P● ↠ M
we tensor by R[x] and get
R[x] ⊗R P ● ↠ R[x] ⊗R M
which is a projective resolution of R[x] ⊗R M ≅ M [x]. Check that the previous lemma
implies that
gl dim R[x] ≥ 1 + gl dim R.
6. K-THEORY OF RINGS OF POLYNOMIALS
11
Step 2. Let M be an S-module and consider MR . Consider the S-modules
β
µ
→ S ⊗ R MR Ð
→M Ð
→0
0Ð
→ S ⊗ R Mr Ð
where µ ∶ s ⊗ m Ð
→ sm and β ∶ s ⊗ m ↦ s(x ⊗ m − 1 ⊗ xm). (Check that this is exact).
Step 3. Use comparison statement for projective dimension on exact sequence to conclude
that pdS M ≤ 1 + pdR M . So
gl dim S ≤ 1 + gl dim R
and we are done.
Theorem 6.3. For noetherian R with finite global dimension, we have
K0 (R[t]) ≅ K0 (R) ≅ K0 (R[t, t−1]).
Proof. We have split short exact sequences
0Ð
→ tR[t] Ð
→ R[t] Ð
→RÐ
→0
and
0Ð
→ (t − 1)R[t, t−1] Ð
→ R[t, t−1] Ð
→RÐ
→0
of R -modules. Hence K0 (R[t]) contains K0 (R) as a direct summand.
As S = R[t] is R-flat, we have f∗ ∶ G0 (R) Ð
→ G0 (S). Also hdR[t] R = 1. Now take a short
exact sequence
0Ð
→ M1 Ð
→ M2 Ð
→ M3 Ð
→0
of R[t]-modules and we derive with tensoring with R as an R[t]-module. Because of
projective dimension of R higher Tors vanish and our ong exact sequence reduces to
R[t]
R[t]
R[t]
0Ð
→ Tor1 (R, M1 ) Ð
→ Tor1 (R, M2 ) Ð
→ Tor1 (R, M3 )
Ð
→ R ⊗R[t] M1 Ð
→ R ⊗R[t] M2 Ð
→ R ⊗[R[t] M3 Ð
→0
We will show that f∗ has a right inverse, ϕ and then prove that f∗ is also injective: ϕ ∶
G0 (R[t]) Ð
→ G0 (R) is define by
R[t]
[M ] ↦ [R ⊗R[t] M ] − [Tor1
´¹¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¸ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹¶
(R, M )]
M /tM
and is well-defined by our long exact sequence. Now
f∗
ϕ∗
R[t]
M Ð→ [R[t] ⊗R M ] Ð→ [R ⊗R[t] M [t]] − [Tor1 (R, R[t] ⊗R M )]
´¹¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹¸ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¶
´¹¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹¸ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¶ ´¹¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¸¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹¶
[M [t]]
[M ]
so f∗ is a split surjection. It remains to show it is injective.
0
7. K-THEORY OF TOPOLOGICAL SPACES
12
For this part we use the following observation of Grothendieck: When R is increasing-filtered
R = inj lim Rn , we get a grading on R by factors of the filtration.
G0 (R) ⊗Z Z[t]
≅
Ggr
0 (R)
≅
K0 (grR − mod)
amd we jave
Ggr
0 (R)
forget
G0 (R)

θ
/ Ggr (R[t, s])
0
ψ∗
/ G0 (R[t])
So the completion of the proof is by showing that this is a commutative diagram, ψ∗ is
sujrective and therefore that θ is an isomorphism.
Corollary 7. If gl dim R < ∞, so is gl dim R[x1 , ⋯, xn ].
Corollary 8. For a field k, K0 (k[x1 , ⋯, xn ]) ≅ G0 (k[x1 , ⋯, xn ]) ≅ Z.
So this shows that every projective k[x1 , ⋯, xn ]-module P is stably free, i.e.
P ⊕ Rk ≅ Rk+t .
Can we do a cancellation then? The ideas to come are
Quillen/Suslin: Stably free k[x1 , ⋯, xn ]-projective moduels are free.
Bass cancellation: Let R bea commutative noetherian domain of Krull dimension d. The
every stably free R-module of rank > d is a free module over R.
7. K-theory of topological spaces
Let X be a compact, Hausdorff topological space and VectF (X) be the commutative monoid
of isomorphism classes of F -vector bundles over X, under the Whitney sum operation and
X being the unit. Here F is the field R or C. The topological K-theory of X is defined
by
KF0 (X) = K0 (VectF (X)).
Then X ↦ VectF (X) is a contravariant functor from the category of compact Hausdorff
spaces to the category of abelian groups.
Theorem 7.1 (Swan). Let R = C F (X) be the ring of continuous F -valued functions on X
nad Γ(X, E) be the R-module of sections of E, then
(1) Γ(X, E) is finitely generated and projective over R.
(2) Conversely, every finitely-generated projective R-module arises this way.
8. K-THEORY OF SCHEMES
13
(3) The map E ↦ Γ(X, E) induces an isomorphism of categories betwenn VectF (X)
and PR , hence an isomorphism
KF0 (X) Ð
→ K0 (R).
8. K-theory of schemes
Let Vect(X) be the category of algebraic vector bundles on the ringed space (X, OX ) (for
us, locally free OX -module with finite rank at every point). In the case (X, OX ) is the
affine scheme Spec R, with R a commutative noetherian ring, then
Vect(X) = PR
since free OX -modules are precisely the projective ones.
Proposition 3. For every scheme X and OX -module F, the following are equivalent:
(1) F is a vector bundle on X.
(2) F is a coherent sheaf and the stalks Fx are free OX,x -modules of finite rank.
(3) For every affine open U = Spec R in X, F∣U is a finitely-generated projective Rmodule.
So we have a dictionary
sheaf R-module
coherent sheaf finitely-generated R-modules
locally free of finite rank (vector bundles) finitely-generated projective modules
So we have analogously K 0 (Vect(X) = K0 (R) at least in the affine case. In general we
have a map
K 0 (Vect(X) Ð
→ Pic(X)
by highest exterior powers.
CHAPTER 2
K1 of a ring
Let R be a noetherian ring, not necessarily commutative. We already saw that matrices
over R play a role in K0 (R). Let A be the full subcategory of an abelian category which
contains 0, is closed under ⊕, and is equivalent to a small category.
Notation: A as above, A [x, x−1] is defined to be the following category: objects are pairs
(A, f ) where A is an object of A and f ∶ A Ð
→ A is an isomorphism. Morphisms are
A
f
/A .
ϕ
ϕ
B
g
/B
A sequence 0 Ð
→ (A′ , f ′ ) Ð
→ (A, f ) Ð
→ (A′′ , f ′′ ) Ð
→ 0 is then exact if and only 0 Ð
→ A′ Ð
→AÐ
→
′′
A Ð
→ 0 is exact.
Definition 3. K1 (A ) is the abelian group whose generators are objects [(A, f )] ∈ ob(A [x, x−1]).
with relations
(a) If 0 Ð
→ [(A′ , f ′ )] Ð
→ [(A, f )] Ð
→ [(A′′ , f ′′ )] Ð
→ 0 is exact then
[(A, f )] = [(A′ , f ′ )] + [(A′′ , f ′′ )].
(b) [(A, f g)] = [(A, f )] + [(A, g)].
Clearly if B is a category like A and T ∶ A Ð
→ B is a functor that preserves exact sequence
then we have an induced homomorphism K1 (A ) Ð
→ K1 (B ).
Remark. For C a compact Hausdorff topological space K 1 (X) = K 0 (SX).
Definition 4. If R is a ring, then K1 (R) ∶= K1 (PR ).
If ϕ ∶ R Ð
→ R′ is a ring homomorphism, then the functor T (M ) = R′ ⊗R M preserves exact
sequence of projective modules hence we get a functor from the induce morphisms
ϕ∗ ∶ K1 (R) Ð
→ K1 (R′ ).
Proposition 4. Let R be a ring and let A = PR . Let F be the category of finitely-generated
free R-modules. The inclusion F ⊂ PR induces an isomorphism K1 (F ) Ð
→ K1 (R).
14
2. K1 OF A RING
15
Proof. Clearly by relation (b) we have [(A, 1A )] = 0 for any A ∈ A . Now let [(P, f )] ∈
K1 (R), then there is a projective module Q such that P ⊕ Q ≅ Rn =∶ F ∈ F . Let g ∶ F Ð
→F
be defined by g = (f ⊕ 1Q ). Define j ∶ K1 (R) Ð
→ K1 (F ) via [(P, f )] ↦ [(P ⊕ Q, f ⊕ 1Q )].
Then
ij[(P, f )] = [(F, g)] = [(P, f )] + [(Q, 1∣Q )] = [(P, f )].
Note that j is well-defined since if P ⊕ Q′ = F ′ , g ′ = (f ⊕ 1∣Q′ ). To show that [(F, g)] =
[(F ′ , g ′ )] consider
[(P ⊕ Q ⊕ P ⊕ Q′ , f ⊕ 1 ⊕ 1 ⊕ 1)] ∈ K1 (F )
we have
[(P ⊕ Q ⊕ P ⊕ Q′ , f ⊕ 1 ⊕ 1 ⊕ 1)] = [(P ⊕ Q ⊕ P ⊕ Q′ , 1 ⊕ 1 ⊕ f ⊕ 1)]
So [(P ⊕ Q, f ⊕ q)] = [(P ⊕ Q′ , f ⊕ 1)].
LATEX2e
Let GL(n, R) be the group of n×n matices over R that are invertible. The embeddings
GL(n, R) ⊂ GL(n + 1, R), A ↦ (
A 0
)
0 1
give the definition of the infinite general linear group
GL(R) = lim GL(n, R).
Ð→
n
If f ∈ Gl(n, R), consider [(Rn , f )] ∈ K1 (F ). We get a homomorphism
GL(n, R) Ð
→ K1 (F )
and also Gl(R) Ð
→ K1 (F ). Now let the matrices Eij (a) = id + aeij be the elementary
matrices (i ≠ j) and E(n, R) ⊂ GL(n, R) be the subgroup generated by elementary matrices
E(R) = lim E(n, R).
Ð→
Theorem 0.1 (Whitehead). Let R be any ring. Then E(R) = [GL(R), GL(R)].
Proof. Observe that Eij (a)−1 = (I − aeij ). To show the converse inclusion for A ∈
GL(n, R),
A 0
(
) ∈ E(2n, R).
0 A−1
One can go from the above matrix to I2n by a series of steps sonsiting of multiplication by
elements of E(2n, R). Then observe that
(
[A, B] 0
) ∈ E(2n, R).
0
I
2. K1 OF A RING
16
By the next result, we have an equivalent definition for K1 (R):
K1 (R) ≅
GL(R)
.
E(R)
Note that as E(R) = [GL(R), GL(R)], we immediately have that K1 (R) is the abelianization of GL(R). Note that universal property
/ / K1 (R)
KK
KK
KK
KK
KK %
GL(R)
G
for any abelian group G.
Theorem 0.2. Let R be any ring. Then the map GL(R)/E(R) Ð
→ K1 (R) = K1 (F ) is an
isomorphism.
Proof. We already know that GL(R) ↠ K1 (R). This gives surjectivity. We construct
j ∶ K1 (F ) Ð
→ GL(R)/E(R) which is an inverse to the map. Let [(F, f )] ∈ K1 (F where F
is a rank n free module. Take (e1 , ⋯, en ) to be a basis for F . Let A be the corresponding
matrix for f . Then j[(F, f )] = [A].
If A′ is another matrix obtained as above, there is an isomorphism
−1
ϕ ∶ (F ⊕ F, f ⊕ 1) Ð
→ (F ⊕ F, 1 ⊕ f )
in F (x, x ) making
F ⊕F
/ F ⊕F
/ F ⊕F
F ⊕F
A 0
I 0
commute. Therefore (
) and (
) give the same element in GL(R)/E(R). This
0 I
0 A
shows that j is well-defined.
It also preserves relations of K1 ; for the relation
0Ð
→ (F ′ , f ′ ) Ð
→ (F, f ) Ð
→ (F ′′ , f ′′ ) Ð
→0
pick a basis e′1 , ⋯, e′n for F ′ , extend it to a basis e′1 , ⋯, e′ n, e′′1 , ⋯, e′′m for F so that the images
of e′′i form a basis for F ′′ . If A′ , A′′ correspond respectively to f ′ and f ′′ , we get
A′ B
A=(
)
0 A′′
for the matrix corresponding to f . But
A′ 0 I 0
I B
A=(
)(
)
′′ ) (
0 I 0 A
0 I
1. STABLE RANGE RESULTS
and therefore j[(F, f )] = j[(F ′ , f ′ )] + j[(F ′′ , f ′′ )].
17
Corollary 9. If F is a field, then K1 (F ) ≅ F × via the determinant map.
1. Stable range results
If R is commutative then
det ∶ GL(R) Ð
→ R∗
is split. Let SK1 (R) = ker(det) so K1 (R) ≅ R∗ ⊕ SK1 (R). Let
SL(n, R) = ker(GL(n, R) Ð
→ R∗ ), SL(R) = lim SL(n, R).
Ð→
So we have
E(R) ⊂ SL(R) ⊂ GL(R).
For a general ring R, define
SK1 (R) ∶= Sl(R)/E(R).
IF R is a field, then SL(n, R) = E(n, R), hence SK1 (R) = 0 and K1 (R) ≅ R∗ .
Now the questions are:
(1) When are the following maps surjective/isomorphisms?
GL(n, R)/E(n, R) Ð
→ K1 (R).
(2) Is GL(n, R)/E(n, R) a group? an abelian group?
Definition 5. The integer n defines a stable range for GL(R) if whenever r > n, and
(a1 , ⋯, ar ) is a unimodular row, then there exist b1 , ⋯, br ∈ R such that
(a1 + ar b1 , ⋯, ar−1 ar br−1 )
is a unimodular row.
Lemma 6. If n defines a stable range for Gl(R), and if r > n and (a1 , ⋯, ar ) is a unimodular
row, then there is A ∈ E(r, R) such that
(a1 , ⋯, ar )A = (1, 0, ⋯, 0).
Theorem 1.1. IF n defines a stable range for R, then
(1) Gl(m, R)/E(m, R) Ð
→ GL(R)/E(R) is surjective for all m ≥ n + 1.
(2) E(m, R) is a normal subgroup of Gl(m, R), if m ≥ n + 1.
(3) Gl(r, R)/E(r, R) is an abelian group if r ≥ 2n.
Remark. Bass, Milnor and Serre, proved that if R is commutative, the stable range for R
is 1. If R is the ring of integers of an algebraic number field, then st.r(R) = {0}.
2. K1 AND TOPOLOGY
18
Proposition 5. If D is a division ring, then the inclusion D∗ ↪ GL(1, D) ↪ GL(D)
induces a surjection
D∗ /[D∗ , D∗ ] ↠ K1 (D).
Remark. IF (R, m) is a local ring, then the inclusion R∗ ↪ GL(1, R) Ð
→ Gl(R) induces a
surjection
∗
Rab
↠ K1 (R).
One can in fact define a non-commutative determinant for the local rings to get an isomorphism
≅
×
Ð
→ K1 (R).
Eab
For semi-local rings, again there is a surjection
×
Rab
↠ K1 (R).
The idea in defining a non-commutative determinant for local rings is
(1) In any n × n matrix in Gl(n, R), any given row has at least on element which is
not in m and is therefore a unit.
(2) Use induction to define the usual determinant along with the last step to define
determinant.
2. K1 and topology
These are mainly works of Milnor and C. T. C. Wall. Suppose (K, L) is a pair, consisting
of a finite connected CW complex K and a subcomplex L which is a deformation retract
of K. Then
G ∶= π1 (K) = π1 (L)
as we know. Milnor defined an element τ (K, L) called the Whitehead torsion, in the
Whitehead group
W h(G) ∶= K1 (Z[G])/⟨{±g}⟩.
Whitehead torsion has applications in surgery theory, Poincare conjecture, etc. Milnor
showed that the Whitehead torsion of a homotopy equivalence between finite CW-complexes
can be used to distinguish between homotopy and simple homotopy types. Some of the
computations done where for G = Z/5 when W h(G) is infinite and if G ≅ Z/2 the W h(G) =
{1}. For R = k[x, y]/(x2 + y 2 − 1) we get
SK1 (R) ≠ 0, SK1 (R) ≅ Z/2Z.
Then Wall constructs the Wall obstruction: for a topological space X, dominated by a
finite complex, he defines a generalized Euler characteristic χ(X) ∈ K1 (Z[G]) when G =
π1 (X). He then shows that X has the homotopy type of a finite complex if and only if
χ(X) ∈ Z.
4. BABY K-THEORY LONG EXACT SEQUENCE
19
3. Description of K1 using group cohomology
Recall the definition of a G-module for an arbitrary group G.
Example 3.1. If K is a field and L/K is a Galois extension, then the Galois group
G = Gal(L/K) acts on L× .
⌟
If G is any group. Z is the trivial G-module. We have definitions
H 0 (G, M ) = M G = {m ∈ M ∶ g.m = m, ∀g ∈ G}
and the dual notion is
H0 (G, M ) = MG = M /IG M.
Here IG is the augmentation ideal defined to be the kernel of the augmentation map
e
Z[G] Ð
→ Z, via ∑ ag g ↦ ∑ ag.
As an example H0 (G, Z) = Z.
Define
Z(G, M ) = {f ∶ G Ð
→ M ∶ f (g1 g2 ) = g1 f (g2 ) + g2 f (g1 )}
to be the set of Crossed homomorphisms. For m ∈ M , fm ∶ G Ð
→ M via g ↦ gm − m is a
crossed homomorphism. Then let
B 1 (G, M )
be the gorup of boundaries. Then the group cohomology of M is defined as
H 1 (G, m) = Z(G, M )/B 1 (G, M ).
In particular
H 1 (G, Z) = Hom(G, Z) = Hom(Gab , Z).
Once defined systematically, one gets
H1 (G, Z) = H1 (G, Z)∨ = Gab .
A consequence of this is a description
K1 (R) = H1 (GL(R), Z).
4. Baby K-theory long exact sequence
Theorem 4.1 (Resolution theorem). Suppose M and P are categories with exact sequences
both contained in an abelian category A and P ⊂ M is a full subcategory. Assume the
following
(1) For each object M of M , there is an epimorphism P ↠ M in A with P ∈ ob(P )
such that every endomorphism of M lifts to an endomorphism of P .
dn
(2) If ⋯ Ð
→ Pn Ð→ Pn−1 Ð
→⋯Ð
→ P0 Ð
→M Ð
→ 0 is exact in M with Pi ∈ ob(P ) then
ker dn ∈ ob(P ) for n >> 0.
4. BABY K-THEORY LONG EXACT SEQUENCE
20
(3) If 0 Ð
→ M1 Ð
→ M2 Ð
→ M3 Ð
→ 0 is a short exact sequence in A with M2 , M3 ∈ ob(M )
then M1 ∈ ob(M ).
≅
Then the inclusion P ⊂ M induces an isomorphism K1 (P ) Ð
→ K1 (M ).
Proof. Let [M, α] be an object in K1 (M ), α ∶ M Ð
→ M an automorphism of M .
−1
−1
Consider α ⊕ α ∈ Aut(M ⊕ M ). Then α ⊕ α is a product of matrices
(
A 0
I
0 I A 0 −I
I A
) ( −1 ) (
)(
)
−1) = (
0 I −A
0 A
1 0 I I 0
so it can be written as a product of elementary automorphisms of the form
1
( M
0
β
1
),( M
M
γ
0
),
1M
β, γ ∈ End(M ).
Lift β, γ to elements in End(P ) by (1).
// M
P
̃γ
β,̃
β,γ
// M
P
Hence we can lift α ⊕ α−1 to an automorphism of P ⊕ P . Since the kernel is in M by (3) we
keep repeating this process and take the corresponding alternating sums of the resolution
obtained in this manner.
We have similar corollaries from this theorem as in case of K0 :
(1) K1 (PR ) ≅ K1 (R − mod ) ∶= G1 (R) if the global dimension of R is finite.
(2) K1 (R[t]) = K1 (R) if gl. dim R < ∞.
(3) K1 (R[t, t−1]) = K1 (R) ⊕ K0 (R).
Let f ∶ R Ð
→ R′ be a homomorphism of rings. This gives a homomorphism K1 (R) Ð
→ K1 (R′ ).
We define a category Pf with objects (A, g, B) where A, B are projective left R-modules and
g ∶ R′ ⊗R A Ð
→ R′ ⊗R B is an isomorphism. A morphism between (A, g, B) and (A′ , g ′ , B ′ )
is a pair (α, β) where α ∶ A Ð
→ A′ , β ∶ B Ð
→ B ′ such that
R ′ ⊗R A
1⊗α
/ R ′ ⊗ A′
R
g
R ′ ⊗R B
1⊗β
g′
/ R′ ⊗ B ′
R
is commutative and
0Ð
→ (A′ , g ′ , B ′ ) Ð
→ (A, g, B) Ð
→ (A′′ , g ′′ , B ′′ ) Ð
→0
is exact (meaning 0 Ð
→ A′ Ð
→AÐ
→ A′′ Ð
→ 0 and 0 Ð
→ B′ Ð
→BÐ
→ B ′′ Ð
→ 0 are exact).
4. BABY K-THEORY LONG EXACT SEQUENCE
21
Definition 6. Let K0 (R, f ) be an abelian group defined similar to K0 and K1 with
generators (A, g, B) ∈ ob(Pf ) with the scissor relations and an extra relation
(A, gh, B) = (A, h, C) + (C, g, B).
One can check that if we associate to θ ∈ K1 (R′ ) the triple (Rn , θ, Rn ) ( viewing θ as an
n × n matrix in GL(n, R′ ) ) we get a well-defined mapping
K1 (R′ ) Ð
→ K0 (R, f ).
It is straightforward to check that
Theorem 4.2. Let f ∶ R Ð
→ R′ be a homomorphism of rings. The the sequence
K1 (R) Ð
→ K0 (R′′ ) Ð
→ K0 (R, f ) Ð
→ K0 (R) Ð
→ K0 (R′ )
is exact.
4.1. Localization sequence. Let R be commutative and S ⊂ R be a multiplicative
closed set and suppose gl. dim R < ∞. Let S − tor be the subcategory of R − mod consisting
of objects M ∈ R − mod such that S −1M = 0. S − tor is closed with respect to exact sequence:
If
0Ð
→ M1 Ð
→ M2 Ð
→ M3 Ð
→0
−1
is an exact sequence of R-modules, then if S Mi = 0 for any two for the i’s, S −1Mi = 0 for
the other one. One gets an exact sequence
K0 (S − tor ) Ð
→ K0 (R − mod ) Ð
→ K0 (S −1R − mod ) Ð
→0
called the localization sequence.
Remark. If fact K0 (S − tor ) ≅ K0 (R, f ) for the inclusion mapping f ∶ R Ð
→ S −1R.
CHAPTER 3
K2 groups
1. Milnor’s K2
Let R be a ring and E(R) be the set of elementary matrices. E(R) is generated by
Eij (λ), λ ∈ R with relations
(1)
Eij (r).Eij (s) = Eij (r + s)
(2)
⎧
E
j = k, i ≠ `
⎪
⎪
⎪ i`
j ≠ k, i ≠ ` .
[Eij (s), Ek` (t)] = ⎨ id
⎪
⎪
⎪
E
(−ts)
j ≠ k, i = `
ikj
⎩
The Steinberg group, St(R), is on the other hand defined as a non-abelian group with
generators xij (t), t ∈ R where i ≠ j ranges over positive integers with the following relations:
(1)
xij (s)xij (t) = xij (s + t).
(2)
⎧
x (st)
j = k, i ≠ `
⎪
⎪
⎪ i`
i ≠ `, r ≠ k .
[xr (s), xk` (t)] = ⎨ 1
⎪
⎪
⎪
x
(−ts)
j ≠ k, i = `
⎩ kj
We can view St as a functor form the category of rings to the category of groups.
Remark. Since [a, b]−1 = [b, a] the relation
[xij (s) , xki (t)] = xkj (−ts)
holds. We clearly have a group homomorphism f ∶ St(R) ↠ E(R).
Generalizing the above notion with define Stn (R) as the quotient of the free group on
(n)
symbols xij (a) for 1 ≤ i, j ≤ n, i ≠ j and a ∈ R. The relations will be as before. Then we
take the direct limit
St(R) ∶= lim Stn (R).
Ð→
22
2. K2 AS AN OBSTRUCTION ELEMENT
23
Definition 7. K2 (R) is defined to be the kernel of the surjection:
K2 (R) ∶= ker(f ∶ St(R) ↠ E(R)).
Lemma 7. K2 (R) is the center of St(R).
Proof. Let x ∈ Z(St(R)). Then f (x) ∈ Z(E(R)) as f is surjective. So it suffices
to show that Z(E(R)) is trivial. This is clear since if A ∈ GL(n, R) is in the center,
A.Eij (1) = Eij (1).A hence A is a diagonal matrix with all entries on the diagonal equal.
Therefore A = id.
For the other direction let α ∈ K2 (R) ∩ im(Stn−1 (R) Ð
→ St(R)). So we can write α as a
word in xij (s) with i, j < n. Let An be the subgroup of St(R):
An = ⟨{xin (t) ∶ 1 ≤ i ≤ n − 1, t ∈ R}⟩.
By the relation
[xij (s), xk` (r)] = 1, if i ≠ `, j ≠ k, r, s ∈ R
we see that An is commutative. From the relation xij (s)xij (t) = xij (s+t) we have xij (0) = 1.
Hence each element of An has a unique expression of the form
x1n (a1 )⋯xn−1,n (an−1 ).
Thus f ∣An ∶ An Ð
→ E(R) is an isomorphism
form
⎛1 0
⎜0 1
⎜
⎜
⎝0 0
Further if i, j < n then
onto the group of matrices in SLn (R) of the
⋯ a1 ⎞
⋯ a2 ⎟
⎟
⋱ ⋮⎟
⋯ 1⎠
xij (a)xkn (b)xij (−a) = {
xkn (b)
j≠k
.
xin (ab)xkn (b) j = k
Therefore α normalizes An . But f ∣An is injective and f (α) = 1 as α ∈ K2 (R) = ker(St(R) Ð
→
E(R)). Hence α centralizes An :
[α, xin (a)] = 1,
∀1 ≤ i ≤ n − 1, a ∈ R.
Similarly [α, xij (a)] = 1 for any j ∈ {1, ⋯, n − 1} and a ∈ R. Thus α commtues with xij (ab)
for all a, b ∈ R and 1 ≤ j ≠ i ≤ n − 1. This holds for all large n completing the proof.
2. K2 as an obstruction element
If G acts on M , the second group cohomology H 2 (G, M ) coincides with the (group of )
central extensions:
0Ð
→M Ð
→EÐ
→GÐ
→ 0, M ⊂ Z(E).
So we suspect a relation between K2 (R) and central extensions. We will explore this
relation in what follows.
2. K2 AS AN OBSTRUCTION ELEMENT
24
Definition 8. An extension 1 Ð
→AÐ
→EÐ
→GÐ
→ 1 is called a universal central extension if
(1) it is central,
(2) given any other central extension 1 Ð
→ A′ Ð
→ E′ Ð
→GÐ
→ 1 then there is a unique
surjective ϕ ∶ E Ð
→ G such that
ϕ
1
/A
/E
1
/ A′
/ E′
/G
/1
/G
/1
commutes.
Theorem 2.1. A group G has a universal central extension if and only if G is perfect (i.e.
G = [G, G]). In this case, a central extension (E, ϕ) is universal if and only if the following
two conditions hold:
(i) E is perfect,
(ii) all central extensions of E are trivial.
Proof. For necessity suppose G has a nontrivial abelian quotient A. Let ψ ∶ G Ð
→A
be the quotient map. Then if
ϕ
1Ð
→ A′ Ð
→EÐ
→GÐ
→1
p1
is a central extension of G we have two distinct morphisms from (E, ϕ) to (G × A Ð→ G1 ):
ϕ
E
(ϕ,1)
/G
G×A
p1
/G
and
E
(ϕ,ψ○ϕ)
ϕ
G×A
p1
/G
/G
so (E, ϕ) cannot be universal.
For sufficiency, consider a presentation of G with a free group F and a subgroup R of
relations:
1Ð
→RÐ
→F Ð
→GÐ
→ 1.
Step 1. We have a surjection
[F, F ]
[F, F ]
(*)
↠
= [F /R, F /R] = [G, G] = G.
[F, R]
R
2. K2 AS AN OBSTRUCTION ELEMENT
25
We will show that this is in fact a central extension of G.
Step 2. Show that any central extension of G satisfying (i) and (ii) is universal. Let (E, ϕ)
be a central extension of G satisfying (i) and (ii). Let (E ′ , ϕ′ ) be any other central extension
of G. If
E
ψ,ψ ′
ϕ
E′
ϕ′
/G
/1
/G
/1
are two morphisms, for x ∈ E we have ϕ′ ○ ψ(x) = ϕ′ ○ ψ ′ (x) = ϕ(x). So ψ(x) = cx ψ ′ (x)
where cx ∈ A′ = ker ϕ′ . If y ∈ E, then ψ(y) = cy ψ ′ (y) for cy ∈ A′ = ker ϕ′ and cx , cy are
central. Hence
ψ([x, y]) = [ψ(x), ψ(y) = [ψ ′ (x), ψ ′ (y)].
Thus ψ = ψ ′ on [E, E] but E = [E, E] showing uniqueness.
We still need to construct a morphism (E, ϕ) Ð
→ (E ′ , ϕ′ ) to show existence. Let E ′′ =
′
′
′′
E×G E . Since ϕ, ϕ are surjective, π ∶ E Ð
→ E is surjective. Note that ker(π1 ) ≅ A′ ∶= ker ϕ′ .
′′
Therefore π ∶ E Ð
→ E is a central extension of E. Check the remaining details.
Step 3. Show that (*) satisfies (i) and (ii). Let E =
[F,F ]
[F,R] ,
ϕ ∶ E ↠ G. We have a diagram
E _ PPP
PPP
PPϕP
PPP
P'
ϕ
/
/
F /R = G
E1 = F /[F, R]
oo
o
o
oo
ooo
ooo
F /R
So ker ϕ1 ⊂ R/[F, R] and hence [ker ϕ1 , E1 ] ⊂ [R/[F, R], F /[F, R]], So ker ϕ1 is central in
E1 and therefore ϕ, ϕ1 are central extensions of G. We check (i) and (ii) for E.
(i) Clearly E1 has the property that [E1 , E1 ] = E. Let e1 ∈ E1 then there is e ∈ E such that
−1
ϕ(e) = ϕ1 (e1 ) therefore ϕ1 (ee−1
1 ) = 1. Therefore ee1 ∈ ker ϕ1 . Hence
E = [E1 , E1 ] = [E ker ϕ1 , E ker ϕ1 ] = [E, E]
as ker ϕ1 is central. We conclude that E is perfect.
ψ
(ii) Let 1 Ð
→AÐ
→ E1 Ð
→EÐ
→ 1 be any central extension. Define E3 = E1 ×G E2 . Consider
π1 ∶ E3 Ð
→ E1 . The claim is that (E3 , π1 ) is a central extension.
ker(π1 ) ≅ ker(ϕ ○ ψ ∶ E2 Ð
→ G).
But E = [E, E] hence ψ([E2 , E2 ]) = [ψ(E2 ), ψ(E2 )] = [E, E]. Hence E2 = [E2 , E2 ]A. Also
ψ([E2 , ker ϕ ○ ψ]) ⊂ [E, ker ϕ] = 1
3. CONSTRUCTING ELEMENTS OF K2
26
as ker ϕ is central. We conclude from all this that if x ∈ ker ϕ ○ ψ then x commutes with
[E2 , E2 ] and A (as A is central). Therefore x commutes with all of E2 . Hence E3 is central.
Consider
qF
η qq
q
q
xq q
/ / E1 = F /[F, R]
E3
π1
Use η to get a homomorphism θ ∶ F Ð
→ E2 such that if x ∈ F , then ϕ ○ ψ(θ(x)) is the image
of x in G = F /R. Therefore θ descends to
θ ∶ F /[F, R] = E1 Ð
→ E2 .
and together with the identity map on E1 this gives a splitting E1 Ð
→ E1 × E2 = E3 . Then
we restrict this to E gives the desired trivialization proving (ii).
Corollary 10. K2 (R) is the kernel of a universal central extension of E(R).
3. Constructing elements of K2
Let R be any ring and consider two matrices A, B ∈ E(R) which commute. Choose representatives a, b ∈ St(R) under f ∶ St(R) Ð
→ E(R). Then we define
A ∗ B ∶= aba−1b−1 ∈ K2 (R)
which is independent of the liftings as K2 R is central.
Lemma 8. The following statements hold:
(1) Star is skew-symmetric: (A ∗ B)−1 = B ∗ A.
(2) Star is bimultiplicative: A1 A2 ∗ B = (A1 ∗ B)(A2 ∗ B).
(3) (P AP −1) ∗ (P BP −1) = A ∗ B.
At least when R is commutative, given u, v ∈ R∗ then
⎛u 0 0⎞
Du = ⎜ 0 u−1 0⎟ ,
⎝ 0 0 1⎠
⎛v 0 0 ⎞
Dv′ = ⎜0 1 0 ⎟
⎝0 0 v −1⎠
then Du and Dv′ commute giving {u, v} = [u, v] ∈ K2 (R).
Remark. We haven’t shown that the elements constructed this way are nontrivial. We do
not even know yet that K2 (R) is nontrivial. In case of fields for instance, a highly nontrivial
fact is that {−1, −1} ≠ 0, and more generally {a, 1 − a} ≠ 0.
A central simple algebra is a finite dimensional algebra over its center with no proper
nontrivial two sided ideals. Examples are Mn (F ), the quaternion algebra H, Mn (D) if D
is a division algebra.
3. CONSTRUCTING ELEMENTS OF K2
27
Theorem 3.1. The Steinberg group St(R) is the universal central extension of K2 (R).
Proof. It is easy to see that St(R) is perfect. All we have to do is to construct a
section for a central extension
ϕ
1Ð
→CÐ
→Y Ð
→ St(R) Ð
→ 1.
For xij (a) ∈ St(R), choose an index k distinct from i and j and let
y = ϕ−1(xik (`)), y ′ = ϕ−1(xkj (a)).
Then we let sij (a) = [y, y ′ ]. Note that ϕ(sij (a)) = xij (a). One shall show now that sij (a) is
independent of the choice of index k, and satisfies Steinberg relations. For this we consider
any central extension
1Ð
→CÐ
→Y Ð
→ St(n, R) Ð
→ 1(n ≥ 0)
then for x, x′ ∈ St(n, R) the symbol [ϕ−1(x), ϕ−1(x′ )] ∈ Y will denote the commutator [y, y ′ ]
where y ∈ ϕ−1(x) and y ′ ∈ ϕ−1(x′ ). Using the Steinberg relations one makes the following
Observation: If j ≠ k and i ≠ ` and a, b ∈ R then [ϕ−1(xij (a)), ϕ−1(xk` (b))] = 1 in Y .
So if we choose four distinct indices h, i, j, k and u ∈ ϕ−1(xki (1)), v ∈ ϕ−1(xij (a)) and w ∈
ϕ−1(xjk (b) then [u, v] = 1. Let G be the subgroup of Y generated by u, v, w. Then G′ =
[G, G] is generated by elements in ϕ−1(xhj (a)), ϕ−1(xik (ab)) and ϕ−1(xhk (ab)). This implies
that G′′ = 1 by Steinberg relations. Therefore
[ϕ−1xhj (a), ϕ−1xjk (b)] = [ϕ−1xhi (a), ϕ−1xik (ab)].
We set a = 1 and conclude
shk (b) = [ϕ−1xhj (1), ϕ−1xjk (b)]
i.e. that sij is independent of the chosen index h. Also we have shown that sij satisfies
the first Steinberg relation. For the second relation apply [u, v][v, w] = [u, vw][v, [w, u]] to
u ∈ ϕ−1(xij (1)), v ∈ ϕ−1(xhi (a)), w ∈ ϕ−1(xjk (b)).
Definition 9. A monomial matrix is one in GL(n, R) that can be expressed as a product
P D where P is a permuation matrix and D is a diagonal matrix.
Let W ⊂ St(n, R) be the subgroup generated by all the wij s. Then and important fact is
that ff R is commutative, then ϕ(W ) ⊂ GL(n, R) via ϕ ∶ St(R) Ð
→ E(R) ⊂ Gl(R). In fact
ϕ(W ) is athe set of all monomial matrices of determinant 1.
Our goal is to prove
{,}
Theorem 3.2. If R is a commutative ring, the Steinberg symbols map R× × R× Ð→ K2 (R)
satisfying {u, 1 − u} = 1 for u ∈ R× and 1 − u ∈ R× .
3. CONSTRUCTING ELEMENTS OF K2
28
We define two new symbols:
wij (u) = xij (u)xji (−u−1)xij (u)
hij (u) = wij (u)wij (−1).
We see that there are relations such as [h12 (u), h13 (v)] = h13 (uv)h13 (u)−1h13 (v)−1. The
following facts are just tedious computations using definitions:
Lemma 9. We have wij (u)−1 = wij (−u), hij (1) = 1 and wij (u) = wji (−u−1). In addition if
u, v ∈ R× and i ≠ j and k ≠ ` then
⎧
wij (−v)
⎪
⎪
⎪
⎪
⎪
w (−u−1v)
wk` (u)wij (v)wk` (u)−1 = ⎨ ij
wi` (−v)
⎪
⎪
⎪
⎪
−1 −1
⎪
⎩ wji (−u vu )
Take u = v and we get
if
if
if
if
j, k, ` are distinct
k = i and j, i, ` are distinct
k = j and j, i, ` are distinct
k = i and j = `.
wk` (u)wij (u)wk` (k)−1 = wji (−u−1).
If R is commutative, {u, v} = {v, u}−1, {u, u2 , v} = {u, v}{u2 , v} and also
Lemma 10. If R is a commutative ring and u, v ∈ R× then h12 (uv) = h12 (u)h12 (v){u, v}−1.
Proof of the theorem. For {u, −u} = 1 we have to show that
h12 (u)h12 (−u) = h12 (−u2 ).
Which is the case since the left hand side can be written as
w12 (u)w12 (−1)w12 (−u)w12 (−1) = w21 (u−2 )w12 (−1) = w12 (−u2 )w12 (−1) = h12 (−u2 ).
For {u, 1 − u} = 1 we will be showing that
h12 (u − u2 ) = h12 (u)h12 (1 − u)
and this can be shown starting from the right hand sind and rewriting everything in terms
of w’s and x’s.
Finally we have some nonzero elements
Proposition 6.
(1) {a, 1} = 1 = {1, a}.
(2) {a−1, b} = {a, b−1} = {a, b}−1.
(3) {a, b} = {b, a}−1.
4. CASE OF FIELDS
29
4. Case of fields
Let F be a field, and let T (F × ) = ⊕n Tn (F × ) = ⊕n (F × )⊕n be the torus. Then K∗M (F ) =
T (F × )/I where I is the 2-sided ideal generated by all elements of the form a ⊗ (1 − a) for
all 1 ≠ a ∈ F × .
Theorem 4.1 (Matsumoto). If F is a field, then K2 (F ) has a presentation as the free
abelian group on the symbols {a, b} with a, b ∈ F × and subject to the relations
(1) (a1 a2 , b) = (a1 , b)(a2 , b),
(2) (a, b) = (b, a)−1,
(3) (a, 1 − a) = 1.
Example 4.1 (Symbols on a field). A symbol on F with values in an abelian group
G is a map (, ) ∶ F × × F × Ð
→ G such that it satisfies the above three relations. From
Matsumoto’s theorem there is a bijection between symbols on F with values on G and
homomorphism from K2 (F ) to G. One type of tame symbols are the tame symbols for
a field (F, ν) with a discrete valuation ν ∶ F × Ð
→ Z on it. Let Oν be the valuation ring
→ k(ν)× we have
{x ∈ F × ∶ ν(x) ≥ 0} which is a local ring. Then form the natural ψ ∶ P×ν Ð
Tν ∶ F × × F × Ð
→ k(ν)×
ν(b)
via (a, b) = ψ((−1)ν(a)+ν(b) abν(a) .
The differential symbols are useful in the context of algebraic geometry. Let F be a field
again and Ω1F = Ω1F /Z be the F -vector space of Kahler differentials of F . As an F -vector
space Ω1F is spanned by the symbols da for all a ∈ F × given via
d(ab) = adb + bda.
Then
Ω2F
=∧
2
Ω1F .
The differential symbol F × × F × Ð
→ Ω2F is
da db
∧ .
(a, b) ↦
a
b
⌟
Let L/F be a field extension. Then F × ↪ L× induces L2 (F ) Ð
→ K2 (L). If L/F is finite,
then there exists homomorphisms tr ∶ Ki (L) Ð
→ Ki (F ) induced by the norm. For instance
in the case of i = 2 if (λ, θ) ∈ K2 (l) such that λ ∈ F, θ ∈ L then
tr(λ, θ) = (λ, NL/F θ).
CHAPTER 4
Galois Cohomology
The setup is this: G is a group. M is a G-module, i.e. M is a module over Z[G]. And
of course this need not be commutative, so whenever we neglect saying, we assume a left
action. We say M is a trivial module if g.m = m for all g ∈ G. For G-modules, M and N ,
Hom(M, N ) as groups, has a G-module structure via
(g.f )(m) = gf (g −1m).
On M ⊗ N we put also the G-module structure g(m ⊗ n) = gm ⊗ gn. If M and N are
G-modules then
HomG (M, N ) ∶= HomZ[[G]] (M, N ) = {f ∶ M Ð
→ N, f (gm) = g.f (m)}.
Then we defined
M G = H 0 (G, M ) ∶= {m ∈ M ∶ gm = m, ∀g ∈ G} ≅ HomZ[G] (Z, M ).
where we consider Z as a trivial G-module. The last isomorphism if given by f ↦ f (1) in
the reverse direction. The cohomology groups of G with coefficients in M are a sequence
fo abelina groups
H q (G, A) q = 0, 1, 2, ⋯
such that
(1) H 0 (G, M ) = M G .
(2) For q ≥ 0 the assigment M ↦ H q (G, M ) is a covariant functor.
(3) Given an exact sequence 0 Ð
→ M′ Ð
→M Ð
→ M ′′ Ð
→ 0 of G-module there exists a
long exact sequence constructed from connecting homomorphisms
δq ∶ H q (G, M ′′ ) Ð
→ H q+1 (G, M ′ )
that fits into a long exact sequence
0Ð
→ H 0 (G, M ′ ) Ð
→ H 0 (G, M ) Ð
→ H 0 (G, M ′′ ) Ð
→ H 1 (G, M ′ ) Ð
→⋯
and that furthermore δ is functorial .
(4) Let H ⊂ G and define N = HomZ[H] (Z[G], M ) (such modules are called co-induced
modules from H to G). Then H q (G, N ) = H q (H, M ) for all q ≥ 1.
Theorem 0.2. For any group G and any G-module M , the cohomology groups H q (G, M )
exists for q ≥ 0 and are unique up to functorial isomorphisms.
30
1. LOWER COHOMOLOGY GROUPS
31
The idea is to left-derive M Ð
→ M G.
̃ From exact sequence
Proof. Suppose we have two cohomology theories H and H.
0Ð
→ M ↪ M∗ Ð
→ M′ Ð
→ 0 where M ∗ = HomZ (Z[G], M ), we have
0
/ MG
/ (M ∗ )G
/ (M ′ )G
0
/ MG
/ (M ∗ )G
/ (M ′ )G
/ H 1 (G, M )
/H
̃ 1 (G, M )
/0
/0
̃ 1 (G, M ) such that f1 δ0 = ̃
we get f1 ∶ H 1 (G, M ) Ð
→H
δ0 . As δ0 and ̃
δ0 are functorial in M
we get that f1 is also functorial. The general proof follows by induction.
For existence we define
H q (G, M ) = ExtqZ[G] (Z, M ).
To compute ExtqZ[G] (Z, M ) we find a Z[G]-projective resolution of Z. Let Pi be the free
Z[G]-module G⊕i+1 given the module structure
g.(g0 , ⋯, gi ) = (gg0 , ⋯, ggi ).
Pi is a free Z[G]-module with basis {(1, g1 , ⋯, go ) ∶ gk ∈ G}. The differentials di ∶ Pi Ð
→ Pi−1
are defined via
(g0 , ⋯, gi ) ↦ ∑(−1)i (g0 , ⋯, ̂
gi , ⋯, gi ).
This gives an exact sequence of Z[G]-modules P ● Ð
→ Z.
For connecting homomorphisms the maps δi ∶ HomZ[G] (Pi , M ) Ð
→ HomZ[G] (Pi+1 , M ) work:
δi (θ)(x1 , ⋯, xi+1 ) = x1 δ1 (x2 , ⋯, xi+1 )
+ ⋯ + ∑ (−1)j θ(x1 , ⋯, xj xj+1 , ⋯, xi+1 )
1≤j≤i
+ (−1)i+1 θ(x1 , ⋯, xn ).
1. Lower cohomology groups
So we have expressions
H 1 (G, M ) = B 1 (G, M ) = {fm ∶ m ∈ M, fm ∶ G Ð
→ M such that fm (g) = gm − m}.
For the second cohomology we have H 2 (G, M ) = Z 2 (G, M )/B 2 (G, M ) where
Z 2 (G, M ) = {f ∶ G × G Ð
→ M ∶ x1 f (x2 , x3 ) − f (x1 , x2 , x3 ) + f (x1 , x2 , x3 ) − f (x1 , x2 ) = 0}
B 2 (G, M ) = {δh ∶ G1 × G1 Ð
→M ∶h∶GÐ
→ M and δh(x1 , x2 ) = x1 h(x2 ) + h(x1 ) − h(x1 x2 )}.
If f is a co-cycle, for (x, 1, 1) ∈ G3 we have xf (1, 1) = f (x, 1). The map f ∗ ∶ G2 Ð
→ M given
by f ∗ (x1 , x2 ) = f (x1 , x2 ) − f (x, 1) is verified to be a 2-cocyle. In fact f ∗ = f − δh where
2. COMPUTATIONS
32
h∶GÐ
→ M is given via h(x) = f (1, 1). Hence f and f ∗ are cohomologous. So every 2-cocyle
is cohomologous to a normalized 2-cocyle (i.e. a 2-cocyle f̃ such that f̃(x, 1) = f̃(1, x) = 0
for any x ∈ G).
2. Computations
Let L/K be a Galois extension and G = Gal(L, K). Then we have the famous
Theorem 2.1 (Hilbert’s theorem 90). H 1 (G, L∗ ) = (e).
Proof. Let f ∈ Z 1 (G, L∗ ). By Dedekind’s theorem elements of G are linearly independent over L. Hence there are elements a, b ∈ L∗ such that
∑ f (τ )τ (b) = a.
τ ∈G
By cocyle condition we have f (στ ) = σf (τ )f (σ) so for any σ ∈ G
σ(a) = ∑ σf (τ )στ (b) = ∑ f (στ )f (σ)−1στ (b) = af (σ)−1.
τ ∈G
τ ∈G
Therefor ef (σ) = σ(a−1).a and hence f is a coboundary.
Corollary 11. Let L/K be a finite cyclic extension and σ be a generator of G. Let a ∈ L×
then NL/K (a) = 1 if and only if there is b ∈ L× such that a = σb/b.
Proof. If a = σb/b then
NL/K (a) = a.σa.⋯.σ n−1 a = 1
trivially. Conversely suppose a ∈ L× with norm 1. Check that the map σ ↦ a can be
extended to a 1-cocyle on G. But by Hilbert’s theorem 90 then this is a coboundary. Hence
there is b ∈ L∗ such that f (σ) = σb/b. Therefore a = (σb)b−1.
Proposition 7. If L/K is Galois and G = Gal(L/K) then for all n ≥ 1 we have H n (G, L) =
0.
Proof. There is a normal basis for L/K, i.e. there is a ∈ L such that {a, σa, ⋯} forms
a basis. Any b ∈ L can hence be written as
b = ∑ bσ σ(a).
Then the map L Ð
→ HomZ (Z[G], K) via b ↦ (σ ↦ bσ−1) is an isomorphism of G-modules
(check this). Hence L is coinduced. Therefore H n (G, L) = 0 for all n ≥ 1.
3. MAPS IN THE LEVEL OF COHOMOLOGY
33
3. Maps in the level of cohomology
3.1. Inflation and restriction. Let G, G′ be groups, M a G-module and M ′ is a
G -module. Let f ∶ G′ Ð
→ G be a group homomorphism. Suppose ϕ ∶ M Ð
→ M ′ is a G′ homomorphism. Then (M, M ′ ) is said to be (f, ϕ)-compatible and we get a homomorphism
H n (G, M ) Ð
→ H n (G′ , M ′ ) for all n ≥ 0 from
′
(G′ )n Ð→ Gn Ð
→M Ð
→ M ′.
fn
ϕ
Example 3.1. For H ↪ G this is called the restriction homomorphism H n (G, M ) Ð
→
H (H, M ).
⌟
n
Example 3.2. If H ⊂ G is a normal subgroup we have the quotient mapping G ↠ G/H.
The action of G on M induces an action of G/H on M H and we get homomorphisms
H n (G/M, M H ) Ð
→ H n (G, M ).
There are called the inflation maps denoted by inf n .
⌟
The inflation and restriction maps are functorial and commute with the connecting homomorphism.
3.2. Corestriction. Let M be a G-module and let N be co-induced as
N = HomZ[H] (Z[G], M ).
Then H (f ) ∶ H (G, M ) Ð
→ H (G, N ) ≅ H n (H, M ) is the restriction map.
n
n
n
If H ⊂ G is a subgroup of finite index, let {xi ∶ i ∈ I} be a set of right coset representatives
of H in G. We have a G-linear map
HomZ[H] (Z[G], M ) Ð
→M
via f ↦ ∑i∈I x−1
i f (xi ). One can observe that this homomorphism is independent of the
choice of representatives. It is moreover functorial and we get induced homomorphims
/ H n (G, M )
5
j
j
jjjj
j
j
j
jjjcores
jjjj
H n (G, HomZ[H] (Z[G], M ))
H n (H, M )
called the corestriction map.
The corestriction commutes with the connecting homomorphisms. For n = 0 this is just the
averaging map
MH Ð
→ MG
m ↦ ∑ xi m.
3. MAPS IN THE LEVEL OF COHOMOLOGY
34
Proposition 8. Let G be a group and H ⊂ G a subgroup. Let M be a G-module. The
composite
res
cores
H n (G, M ) Ð→ H n (H, M ) ÐÐ→ H n (H, M )
is multiplication by k = [G ∶ H].
Example 3.3 (Kummer theory). Let L/F be a Galois field extension with Galois group
Gal(L/F ) = G. Take L = F in characteristic zero and let G = Gal(F /F ). Let µn ⊂ F be the
Galois module consisting of n-th roots of 1. We have
×
×
0Ð
→ µn Ð
→F Ð
→F Ð
→1
where the sujrection is x ↦ xn . So
×
×
×
→ H 0 (G, F ) Ð
→ H 1 (G, µn ) Ð
→ H 1 (F, F ) = 0
0Ð
→ H 0 (G, µn ) Ð
→ H 0 (G, F ) Ð
is the result of taking the long exact sequence. So we get
→ F× Ð
→ F × /(F × )n .
0Ð
→ µn (F ) Ð
→ F× Ð
n
Taking n = 2 we have
H 1 (GF , µ2 ) = F × /(F × )2 .
⌟
3.3. Cup product. Suppose G is a group and M, N are two G-modules. The M ⊗Z N
is again a G-module. Then for all p, q ≥ 0 there exists unique homomorphisms
∪ ∶ H p (G, M ) ⊗ H q (G, N ) Ð
→ H p+q (G, M ⊗ N ).
such that
(1) are functorial in the sense that
H p (G, M ) ⊗ H q (G, N )
f ⊗g
∪
H p (G, M ′ ) ⊗ H q (G, N ′ )
/ H p+q (G, M ⊗ N )
∪
f ⊗g
/ H p+q (G, M ′ ⊗ N ′ )
is commutative.
(2) commutes with boundary, restriction, corestriction and inflation maps.
In fact explicitly the cup product can be expressed as follows: if a ∶ Gp Ð
→ M and b ∶ Gq Ð
→N
p
q
are elements are H (G, M ) and H (G, N ) respectively then
(a ∪ b)(g1 , ⋯, gp+q ) = a(g1 , ⋯, gp ) ⊗ b(gp+1 , ⋯, gp+q ).
3. MAPS IN THE LEVEL OF COHOMOLOGY
35
Example 3.4. Recall that H 1 (GF , µn ) = F × /(F × )n . The special elements in H 2 (GF , µ2 )
are of the form
{λ1 , λ2 }, λi ∈ F × /(F × )2 .
These are called symbols in Galois cohomology. We have
×
K1M (F ) = F × = H 0 (GF , F ).
→ H 1 (GF , µ2 ) which is a homomorphism: λ ↦ [λ] ∈ F × /(F × )2 .
Thus we get a map K1M (F ) Ð
It follows that since KnM (F ) is generated by symbols {λ1 , ⋯, λn } then H n (GF , µ2 ) is generated by
[λ1 ] ∪ ⋯ ∪ [λn ].
There is a theorem of Merkurjev showing that there is an isomorphism K2M (F ) Ð
→ H 2 (G, µ2 )
induced as such and called the norm residue homomorphism.
⌟
Proposition 9. Let H ⊂ G be a normal subgroup and A a G-module. If H i (H, A) = 0 for
1 ≤ i ≤ n − 1 (no condition if n = 1) then the sequence
inf
res
0Ð
→ H n (G/H, AH ) Ð→ H n (G, A) Ð→ H n (H, A)
is exact.
Proof. By induction on n. Let n = 1 and let [f ] ∈ H 1 (G/H, AH ) such that inf[f ] = 0.
Then the cocyle inff becaomes a coboundary, i.e. there is a ∈ A such taht f (σ) = σa − a
for all σ ∈ G. Since f ∣H = 0 then a ∈ Ah then f ∈ B 1 (G/H, AH ) and therefore the inflation
is injective.
The composition res ○ inf = 0. since H Ð
→GÐ
→ G/H is the zero map. Let f ∈ Z 1 (G, A)
be such that res(f ) = 0. Then f (σ) = σa − a for all σ ∈ H and a ∈ A. Let fa ∶ G Ð
→ A be
fa (g) = ga − a. Thus [f ] and [f ′ ] are equal in H 1 (G, A) where f ′ = f − fa . But f ′ ∣H = 0
and hence f ′ induces a 1-cocyle: f ′′ ∶ G/H Ð
→ AH . Then
inf[f ′′ ] = [f ]
proving the exactness at H 1 (G, A). Suppose n > 1 and assume the proposition is true for
n − 1. Consider the exact sequence
0Ð
→AÐ
→ A× Ð
→ A′ Ð
→0
where A× = HomZ (Z[G], A). Taking the long exact sequence we get the short exact sequence
0Ð
→ AH Ð
→ (A× )H Ð
→ (A′ )H Ð
→0
of G/H-modules.
Now A× is coinduced also as an A-module and (A× )H is coinduced as a G/H-module. Taking
the long exact sequence associated to the above short exact sequence and comparing it to
that of the quotient 0 Ð
→GÐ
→H Ð
→ G/H Ð
→ 0 the isomorphism for n ≥ 2 are all proved. 4. PROFINITE GROUPS
36
Corollary 12. Let A = L× and G = Gal(L/K) be finite. Let H ⊆ G be a normal subgroup.
We get
0Ð
→ H 2 (G/H, F × ) Ð→ H 2 (G, L× ) Ð→ H 2 (H, L× )
inf
res
is exact.
Remark. Just as in more general setups we have a Hochschild-Serre spectral sequence with
page two
E2p,q = H p (G/H, H q (H, A))
which converges to H n (G, A). The construction is the same as any Grothendieck spectral
sequence.
4. Profinite groups
Let K be a field and L a Galois extension, not necessarily finite. Let G = Gal(L/K). In
this case G comes with a topology called the pro-finite topology. The opens are the normal
subgroups corresponding to subfields of L that are finite and Galois over K. The fact
that
L = ∪Kα = lim Kα
Ð→
where Kα ranges over all intermediate field extensions is equivalence to
G = lim G/Gα .
←Ð
Also note that G is a profinite group in the sense that it is the inverse limit of an inverse
system of finite groups. Then
G ⊂ ∏ G/Gα
α
is a compact, totally disconnected, Hausdorff topological group if Gα ’s have discrete topology.
Definition 10. Let G be a profinite gorup and A a G-module. Then A is a discrete
G-module if G × A Ð
→ A is continuous when A is given the discrete topology.
′
This is equivalently the case when A = ∪AG for G′ ranging over open subgroups of G.
Hcn (G, A) = Hcn (lim G/Hα , A) = lim H n (G/Hα , A) = lim H n (G/Hα , AHα ).
←Ð
Ð→
Ð→