If you find a mistake, omission, etc., please let me know by e-mail.
The orange balls mark our current location in the course, and the current problem set.
Review of some basic notations concerning field extensions etc.
h1.ps: Overview of Galois theory (also in PDF format)
Jacobson 4.1: Preliminaries on field extensions, etc.
Theorem 2.16:
Let K/F be a field extension and u some element of K.
Then u is algebraic over F if and only if F(u)=F[u].
(When u is algebraic, inverses in F[u] constructed by linear algebra
on the finite-dimensional vector space F[u]/F)
Theorem 4.1:
Under the hypotheses of Thm. 2.16, u is algebraic over F
if and only if [F(u):F] is finite, in which case [F(u):F]
equals the degree of the minimal polynomial of u over F.
Theorem 4.2:
If K/E and E/F are field extensions, then
[F:K] is finite if and only if [K:E] and [E:F] are both finite,
in which case [K:F] equals the product of [K:E] and [E:F].
Corollary:
If [K:F] is finite then [E:F] is a factor of [K:F]
for every subfield E of K containing F.
Theorem 4.23 [sic]:
Let A be a subring of some ring B.
Then the integral closure of A in B is a ring,
and is integrally closed in B.
(In particular, if A,B are fields we recover the analogous results
for algebraic closure.
Proof sketch: see the
closure
notes.)
Definitions and basic results concerning algebraic and integral closure.
The map taking a to the multiplication-by-a operator M_{a} is surprisingly important and useful. We have used it already to show that F[u]=F(u) when u is algebraic over F, and to show that the integral closure of any subring of a ring A is itself a subring of A. In all contexts in which we're likely use that map, it is a homomorphism: it satisfies the identities M_{a+b}=M_{a}+M_{b} and M_{ab}=M_{a}M_{b}. So, for instance, if [K:F]=n we get a copy of K in the (noncommutative) ``endomorphism ring'' End_{K}F, isomorphic with the algebra of n*n matrices over F. As an application, when F (and thus also K) is finite, we can use a generator of the multiplicative group K^{*} to obtain a linear transformation of F^{n} that cyclically permutes the nonzero vectors of F^{n}.
Jacobson 4.3: Splitting fields
Motivation and definitionTheorem 4.3: Every monic polynomial in F[X] has a splitting field K/F.
This already lets us construct, for each prime power q, a finite field of q elements -- namely, the splitting field of X^{q}-X over the prime field -- and prove it is unique up to isomorphism. We shall come back to this; see for instance Chapter 4, section 13 in Jacobson. Note that X^{q}-X has distinct roots because its derivative is the nonzero constant polynomial -1 (see ``multiple roots'').These fields were discovered by Galois, and the old notation GF(q) ["GF"="Galois Field"] for the q-element field persists in the engineering/CS literature. In modern mathematical writing, the more efficient notation F_{q} is preferred. Under either terminology, these fields are fundamental to number theory, and play important roles in other branches of mathematics, pure (e.g., finite simple groups) as well as applied (e.g., error-correcting codes).
Jacobson 4.4: Dealing with multiple roots
Since all splitting fields are isomorphic, it makes sense to say
that a polynomial f in F[X] has a multiple root, or that
two polynomials f,g in F[X] have common roots, in a splitting field
of f or fg respectively. The next two results give criteria
for these two properties that do not require field extensions:
Lemma:
Let f,g be monic polynomials in F[X]. Then f,g have no common roots
in a splitting field of fg if and only if f and g are relatively prime.
(If h=gcd(f,g) has positive degree, then each root of h
in the splitting field is a common root of f and g.
If not, then use a,b in F[X] such that af+bg=1.)
Theorem 4.5:
A monic polynomial f in F[X] has no repeated factors
if and only if f is relatively prime with its derivative.
In particular, if f is irreducible then it has no repeated factors
unless f'=0, that is, unless F has positive characteristic p and
f(X)=g(X^{p}) for some g in F[X].
An algebraic introduction to derivatives and derivationsLemma: If F is a field of positive characteristic p, then a polynomial in F[X] of the form x^{p}-a is either: irreducible, when a is not a p-th power in F; or splits completely as a p-th power (x-b)^{p}, if a=b^{p} in F.
We already have all we need to give the basic results about finite fields. Let F be a finite field of q_{0} elements, and q some power of q_{0}, say q=q_{0}^{n}. Then the splitting field K of X^{q}-X over F consists of the roots of that polynomial and is isomorphic with every q-element field containing F. In particular (taking F=Z/pZ), for each prime power q there exists a finite field K of q elements, and that field is unique up to isomorphism. Moreover, K has a q_{0}-element subfield F if and only if q=q_{0}^{n} for some n>0. In this case, the group G of isomorphisms of K that act trivially on F is the cyclic group of order n generated by the ``Frobenius map'' taking any x to x^{q0}. This also gives us a special case of the Galois correspondence: there is an inclusion-reversing bijection from subgroups H of G and subfields K' of K that contain F, namely the map that takes H to the subfield K^{H} of K consisting of elements fixed by H.
Jacobson 4.5: The Galois group and fundamental Galois correspondence
Lemmata:
Theorem 4.7: Let E/F be any field extension. Then TFAE:
Example 1: symmetric functions.
Let E be the field k(x_{1},...,x_{n})
of rational functions in n variables over some ``ground field'' k.
Let G be the symmetric group of permutations of n letters,
acting on E by permuting the x_{i}. Then F=E^{G}
is the field of symmetric rational functions. We claim that
F=k(s_{1},...,s_{n}) where s_{j} is
the j-th elementary symmetric function, that is,
(-1)^{j} times the X^{n-j} coefficient of
g(x)=(X-x_{1})...(X-x_{n}).
[Jacobson uses p_{j} for our s_{j};
but nowadays p_{j} is generally used
for the j-th power sum of the x_{i},
and the first n of these generate F only in characteristic 0 or >n.]
Call that field K as long as we haven't proved that K=F.
Clearly K is in F, and E is the splitting field of g over K;
moreover, g is separable, so E/K is normal.
Finally, an element of Gal(E/K) permutes the roots x_{i} of g,
and is determined by this permutation.
Thus by the fundamental theorem K=E^{G}, so K=F as claimed.
[In fact the G-invariant subring of
k[x_{1},...,x_{n}] is likewise
k[s_{1},...,s_{n}] (see Jacobson 2.13);
this generalizes to the theory of finite reflection groups.]
This example yields
Theorem 4.15 (Jacobson 4.9): Let F be any field,
and t_{1},...,t_{n} indeterminates over F.
The Galois group of
x^{n}-t_{1}x^{n-1}+t_{2}x^{n-2}-...+(-1)^{n}t_{n}
over F(t_{1},...,t_{n}) is the symmetric group
S_{n}.
Corollary (Jacobson 4.9):
Every finite group is the Galois group of some field extension.
(Proof: embed the group in some symmetric group
and apply the fundamental theorem to the Galois extension
of Theorem 4.15.)
It is conjectured (E.Noether), but not yet proved,
that in fact every finite group is the Galois group
of some field extension of the rational numbers.
For some idea of the work that Noether's conjecture inspired,
see J.-P. Serre's Topics in Galois Theory
(Boston: Jones & Bartlett, 1992), QA214.S47 in Cabot.
Example 2: doubly periodic functions
(a.k.a. isogenies between elliptic curves over C;
cf. the fields C(e^{z}),
C(e^{z/n})
of singly-periodic functions).
A ``lattice'' in C (or in any finite-dimensional
real vector space) is the Z-span
of an R-basis.
Let L,L' be lattices in C with L' containing L.
Let F and F' be the associated fields of doubly periodic functions,
that is, meromorphic functions on C invariant
under translation by each element of L or L' respectively.
Let G=L'/L. Clearly G acts on F by translation, and F'=F^{G}.
Moreover, the action of G is faithful, since it is already faithful
on the orbit of the Weierstrass p-function associated with L.
Therefore G=Gal(F/F') and [F:F']=[L':L]. The intermediate fields
are precisely the fields of meromorphic functions for some intermediate
lattice L"; since G is abelian, all its subgroups are normal, and thus
all these subfields are normal over F'.
(C/L and C/L'
are ``elliptic curves'' over C, and
the natural map from C/L to and C/L'
is an ``isogeny''. See also problem 5 on the second problem set.)
The ``Galois group of a polynomial'' over some field F is the Galois group of the polynomial's splitting field K over F (provided that K/F is separable). Some standard examples with F=Q:
We didn't say much about algebraically closed fields and algebraic closure. A field K is algebraically closed if every monic polynomial over K splits completely over K; equivalently, if K is the only algebraic extension of K. Likewise, K is separably closed if every separable monic polynomial over K splits completely over K; equivalently, if K is the only separable algebraic extension of K. If such K is an algebraic (or separable) extension of some subfield F, we say that K is an algebraic (or a separable) closure of F. For instance, C is algebraically closed (the ``Fundamental Theorem of Algebra''), and is an algebraic closure of R; the field of algebraic numbers in C is an algebraic closure of Q. In general, we need the Axiom of Choice (a.k.a. Zorn's Lemma) to construct an algebraic or separable closure of a given field F, and to show that any two algebraic closures are isomorphic over F. The group of isomorphisms of a separable closure, or more generally of any infinite-dimensional separable extension, can be studied by a generalization of Galois theory to infinite Galois groups. Besides the Axiom of Choice, this theory requires also a topological structure on the Galois group. (This topology can be defined for finite normal extensions too, but is discrete in this case, so gives no new information.) For instance, if F is the cyclotomic field obtained from Q by adjoining the roots X^{p}-1 (p some odd prime), and L is the infinite algebraic extension of F obtained by adjoining p^{m}-th roots of unity for all m, then L/F is normal with Galois group isomorphic with Z_{p}, the p-adic integers! Note that Gal(L/F) can be uncountable even when L is countable. See the third problem set for the details of this theory and more examples.
``Lemma 3'' in Jacobson 4.7 generalizes to normal extensions with a cyclic Galois group of arbitrary order n (not necessarily prime) over a field F containing the n-th roots of unity (that is, F such that the polynomial X^{n}-1 is separable and split over F). Besides Jacobson's application to solvability in radicals -- for which the case of prime n is sufficient -- this generalization also leads to Kummer theory of any finite normal extension K/F of such a field F such that the Galois group G is abelian and of exponent dividing n. (This last condition means that g^{n}=1 for all g in G.) Let M_{1} be the subgroup of F^{*} consisting of elements c with an n-th root d in K, and let M be the quotient of M_{1} by F^{*n} [note that M_{1} must contain F^{*n}]. For g in G and c in M_{1} define (g,c)=g(d)/d where d^{n}=c. This is an n-th root of unity that does not depend on the choice of n-th root d and is constant when c varies over a coset of F^{*n} in M_{1}. Therefore (.,.) descends to a well-defined pairing between G and M taking values in the n-th roots of unity. (``Pairing'' means that for each g the map taking c to (g,c) is a homomorphism from M to the n-th roots of unity, and likewise for each c the map taking g to (g,c) is a homomorphism from G to the n-th roots of unity. This particular pairing is known as the ``Kummer pairing''.) The key fact is that this pairing is perfect: if for some g we have (g,c)=1 for all c in M, then g=1, and if for some c we have (g,c)=1 for all g in G, then c=1. As a corollary, G and M are isomorphic abelian groups. For instance (taking n=2 and F=Q), the extension of Q generated by the square roots of the first N primes has Galois group (Z/2Z)^{N}.
Jacobson 4.15: Traces and norms, continued
In a Galois extension E/F, the norm and trace of any element x
are the product and sum of its conjugates;
more precisely, they are the product and sum of g(x),
g ranging over Gal(E/F). This is because the g(x) are
the eigenvalues of the F-linear operator M_{x} on E
introduced in the first problem of Problem Set 1.
More generally, any separable extension E/F is K^{H}
for some Galois extension K/F and some subgroup H of G=Gal(K/F);
then the eigenvalues of M_{x} are g(x) for g ranging
over representatives of the cosets of H in G.
It follows by ``independence of characters''
that the trace is surjective onto F.
If E/F is inseparable than the eigenvalues occur
with multiplicities p^{m} for some m>0,
so in particular the trace is identically zero.
Theorem 4.28 [Hilbert's ``Theorem 90'',
also known by the equivalent German name ``Satz 90'']
Let E/F be a cyclic extension, that is, a normal extension
whose Galois group G is cyclic. Let e be a generator of G.
Then an element u of E has norm 1 in F if and only if
u=v/e(v)[=``v^{1-e}''] for some nonzero v in E.
Jacobson shows this by proving the following more general theorem
of E.Noether that is sometimes also called ``Satz/Theorem 90'' because
(as we shall see later this term) it can also be deduced from it:
Theorem 4.29
Let E/F be a normal extension with (finite) Galois group G.
Then a |G|-tuple {u_{e}} with coordinates in E indexed by G
is of the form u_{e}=v/e(v) for some nonzero v in E
if and only if u_{ee'}=e(u_{e'})u_{e}
for all e,e' in G.
To recover 4.28 from this, let
u_{ei}=ue(u)e^{2}(u)...e^{i-1}(u).
To prove 4.29, let v(w) (for any w in E) be the sum of
u_{e}e(w) over e in G, and check that
u_{e'}e'(v(w))=v(w) for all e' in G.
So we may take v=v(w) for any w that makes v(w) nonzero,
and such w must exist because the elements of G
are linearly independent in the E-vector space End_{F}(E).
The following generalization of our ``Lemma 3'' above quickly follows
from Satz 90:
Theorem 4.32
If E/F is a cyclic extension of degree n and F contains the n-th roots
of unity then E=F(u) where u^{n} is in F.
Another application of Satz 90 is the description of all Pythagorean
triples, since (x,y,z) is such a triple if and only if (x+iy)/z
is an element of norm 1 in the quadratic extension
Q(i)/Q.
To be sure, there are many other ways to obtain the general solution
of x^{2}+y^{2}=z^{2}...
The additive analogue of 4.29 is also true:
Theorem 4.30
Let E/F be a normal extension with (finite) Galois group G.
Then a |G|-tuple {d_{e}} with coordinates in E indexed by G
is of the form d_{e}=c-e(c) for some c in E
if and only if d_{ee'}=e(d_{e'})+d_{e}
for all e,e' in G.
This is proved by taking c=Sum_{e}d_{e}e(u)/Tr(u)
for any u in E whose trace Tr(u) is nonzero.
Theorem 4.30 yields the additive analogue of Hilbert 90:
Theorem 4.31
Let E/F be a cyclic extension, with Galois group G generated by e.
Then an element d of E has trace 0 in F if and only if
d=c-e(c)[=``(1-e)c''] for some c in E.
[Simpler proof of 4.31: dimension count on kernel and image of 1-e.]
This in turn yields the additive analogue of Theorem 4.32:
Theorem 4.33
Let F be a field of positive characteristic p,
and E/F a normal extension with [E:F]=p.
Then E=F(c) where c^{p}-c is in F.
[Compare this with the last problem of Problem Set 2.
An extension E/F in characteristic p obtained by adjoining a root of
an irreducible polynomial of the form X^{p}-X-a
is called an Artin-Schreier extension.]
The representation ring of G can also be defined as
the group of all pairs (V,W) of representations
modulo the following equivalence relation:
(V,W)=(V',W') if the direct sums V+W', V'+W
are isomorphic representations. The idea is
that (V,W) stands for the virtual representation V-W
(so an honest representation is one equivalent to (U,0)
for some representation U).
One must then check that this is in fact an equivalence
relation and that addition and tensor product are well-defined
(by (V,W)+(V',W')=(V+V',W+W') and (V,W)*(V',W')=(V*W+V'*W',V*W'+V'*W)).
This generalizes a familiar construction
of the integers from the whole numbers.
As is true in that case, the transitivity of the equivalence
relation requires a cancellation law: if U+V is isomorphic with U+V'
then V is isomorphic with V'. In the absence of this law,
we still have a notion of ``stable equivalence'',
and can define (V,W)=(V'+W') to mean that there exists
a representation U such that U+V+W' and U+V'+W are isomorphic.
This approach is due to Witt, and the resulting group or ring
is called the ``Witt group'' or ``Witt ring''.
See Serre's A Course in Arithmetic for an example,
involving quadratic forms, in which the cancellation law fails.
Theorems 4.29 and 4.30 in Jacobson say that the first Galois cohomology group H^{1}(Gal(E/F),A) vanishes when A is either the additive or the multiplicative group of E. This is usually written "H^{1}(Gal(E/F),G_{a})=0" and "H^{1}(Gal(E/F),G_{m})=0", or even "H^{1}(E/F,G_{a})=0" and "H^{1}(E/F,G_{m})=0"; here G_{a} (G_{m}) is the additive (resp. multiplicative) group. See Tate's notes on Galois cohomology, the source of most of the material for the next few lectures. [Note the casual identification of Noether's Theorem 4.29 and Hilbert's Satz 90 starting on page 3.] We will then explain in detail the remark on H^{2}(K/k,G_{m}) and Brauer groups in the next-to-last paragraph of page 3 of Tate's notes. We won't cover everything in Tate; many of the topics we omit would make good final projects for the class.
Tate uses the notion of an ``algebraic group'' C, which you may not have previously encountered. All the algebraic groups we'll use can be found inside the group GL_{n} of invertible matrices of order n. (The only groups Tate uses that are not of this form are elliptic curves.) An algebraic subgroup of GL_{n} is one obtained by imposing finitely many polynomial conditions on the entries of the matrix. Examples of such algebraic groups are: GL_{n} itself; its subgroup SL_{n}, with polynomial condition det=1; the multiplicative group G_{m}, which is just GL_{1}; the group mu_{n}, which is the subgroup of GL_{1} defined by a^{n}=1; and the additive group G_{a}, isomorphic with the subgroup of GL_{2} consisting of upper triangular matrices with unit diagonal entries. The ``ax+b'' group is the subgroup of GL_{2} consisting of matrices with bottom row (0,1); it has a normal subgroup G_{a}, consisting of elements with a=1; the quotient is G_{m}. Any finite group G with trivial Galois action can be represented as a group of permutation matrices of order |G|, and thus (with some unrewarding effort) exhibited as an algebraic group. We may also use the group PGL_{n}, which is the quotient of GL_{n} by its (algebraic) subgroup of scalar matrices. As it stands, this construction of PGL_{n} does not exhibit it as an algebraic group; we do not need to worry about this for now, but we remark that PGL_{n} it can be obtained as an algebraic subgroup of GL_{N} where N=n^{2}, using the action of GL_{n} on the tensor product of the defining representation and its dual.
The formulas that define group cohomology may look a bit less strange under the following change of variables (Eckmann, c.1950; see PS6): let x_{i}=a_{i-1}^{-1}a_{i} for some a_{0},...,a_{r} in G. A different (r+1)-tuple (a_{0}',...,a_{r}') yields the same x_{i} if and only if there is an element g of G such that a'_{i}=ga_{i} for all i. We may associate to an arbitrary r-cochain f, which is a function from G^{r} to A, the ``homogeneous function'' f_{homog} from G^{r+1} to A defined by
Eckmann uses these homogeneous cochains to give an explicit (albeit mysterious-looking) one-line formula for the corestriction; see Theorem 7 on page 490 of his ``Cohomology of groups and transfer'' in Annals of Math. 58 #3 (11/1953), 481-493.
Serre points out that the proof of Tate's Corollary 4.1
(H^{r}(G,A) is a |G|-torsion group for r>0)
can be translated to the following computation:
Suppose f: G^{r} --> A is an r-cocycle for some r>0.
Fix y in G, and let g be the (r-1) cochain whose value at any
(x_{1},x_{2},...,x_{r-1}) is
f(x_{1},x_{2},...,x_{r-1},y).
Then the coboundary of g is the r-cochain whose value at any
(x_{1},x_{2},...,x_{r}) is
The ``p-primary component'' of a finite abelian group (mentioned in Exercise 4.2) is the subgroup consisting of elements of order p^{f} for some integer f.
Here are some basic definitions and examples concerning
algebras
over a field.
Theorem 1 (Wedderburn):
A finite-dimensional k-algebra is simple if and only if it is
isomorphic with M_{n}(k') for some positive integer n
and some (possibly) skew field k' containing k.
The ``only if'' is proved by taking any finite-dimensional
irreducible representation E of the simple k-algebra A
and identifying A with its ``bicommutant''.
Here E may be taken to be a minimal left ideal.
The ``bicommutant'' is the commutant of the commutant.
The ``commutant'' of any subalgebra A of End_{k}(E)
is {x in End_{k}(E) : ax=xa for all a in A}.
This is a subalgebra, and clearly its commutant in turn contains A.
The hard part is showing that in our case it is no larger than A.
Note that our A is indeed a subalgebra of End_{k}(E):
since A is simple, its representation to End_{k}E
is automatically faithful. It also follows that its commutant
is a skew field K, and thus that the bicommutant is a matrix algebra
over the opposite skew field K'=K^{o}.
In the process of identifying A with its bicommutant,
Serre quotes a result about ``semisimple modules''
that also figures in the representation theory of finite groups.
Here
are the relevant definitions and proofs.
Theorem 2:
Every representation of a simple algebra is isomorphic with
a direct sum of isomorphic simple representations.
This is proved first for the left regular representation,
using Wedderburn's theorem;
then deduced from this for an arbitrary representation
using the general facts about simple and semisimple modules.
Corollaries:
Every representation of a simple algebra is completely irreducible;
two such representations are isomorphic if and only if they are
of the same dimension.
The center C(A) of any k-algebra A is automatically a commutative k-algebra. If A is simple, then C(A) is in fact a field. Serre obtains this as a consequence of Lemma 4, but we can also show it directly. For any nonzero x in C(A), consider the set xA=Ax. This is a nonzero two-sided ideal, so equals A. In particular, it contains 1, so we obtain a two-sided inverse y of x. Moreover, y commutes with all elements of A: any such element b can be written as xa=ax for some a in A; then by=axy=a=yxa=yb. So y is also in C(A) and we're done.
Since C(A) is a field, we can regard A as a C(A)-algebra, indeed a central C(A)-algebra. This means that any simple algebra over a field k_{0} can be obtained by:
The punchline of the first Serre handout is that the set of these equivalence classes has a natural group structure. More precisely, it is an abelian group, whose identity is the class of k itself, and with the inverse of any central skew field K being its opposite K^{o}. The addition law comes from the tensor product of k-algebras. The punchline of the second handout will be that this abelian group, called the Brauer group of k, is none other than the Galois cohomology group H^{2}(k^{sep}/k,G_{m})!
Lemma 4:
Let B,B' be subalgebras of k-algebras A,A'.
Let C,C' be their commutants.
Let A'' be the tensor product of A with A',
and likewise B'' and C''.
Then C'' is the commutant of B'' in A''.
In particular,
the tensor product of two central algebras is again central.
Theorem 3: Let A,A' be finite-dimensional simple k-algebras, at least one of which is central. Then the tensor product of A with A' is simple.
Suppose A' is central. Then it is a matrix algebra over some central skew field K'. It is enough to show that the tensor product of A with K' is simple; then by Wedderburn it is a matrix algebra over some skew field K'', and then so is the tensor product of A with A'. (Matrix algebras over skew fields are known to be simple.) We obtain this from the following more general result (in which neither A nor K' is assumed finite over K):
Theorem 4:
Let K' be a central skew field over k,
and A any associative k-algebra with unit.
Then every two-sided ideal N of the tensor product of A with K'
is generated as a left K'-vector space by its intersection with A.
Here K' operates by x(a.tensor.y)=a.tensor.xy,
and A is imbedded in the tensor product as {a.tensor.1}.
Corollary:
Let A be a central simple k-algebra, of dimension n over k.
Then the tensor product of A with its opposite A^{o}
is isomorphic with M_{n}(k).
Indeed, the left and right actions of A on itself yield a homomorphism
from that tensor product to End_{k}(A). Once we know that the
product is simple, this homomorphism must be injective.
By comparing dimensions we then see that it is bijective,
and are done.
In the proof of Theorem 4, Serre uses the notion of a ``primordial element'' of a vector subspace N of V relative to a basis {e_{i}} of V. Every vector v in V can be written as a sum of c_{j}e_{j}, with j ranging over a finite subset J=J(v) of the i's. A vector w in N is said to be ``primordial'' if:Combining Theorem 4, its corollary above, and general properties of tensor products, we define the Brauer group of an arbitrary (commutative) field k and show that it is abelian (Theorem 5). Serre denotes this group G_{k}; we'll use the notation Br(k), which has become more-or-less standard (see for instance the Tate handout).The key property of this notion is that every vector subspace of V is generated by its primordial elements. This is used in much the same way that we used ``weights'' in our proof of Artin's Lemma; and indeed we could have used primordial vectors there instead of vectors of minimal weight. In general, though, the minimal-weight vectors in a subspace need not generate it, so we cannot conversely use minimal-weight vectors to prove Theorem 4. For example, in the subspace of k^{5} with basis v_{1}=(1,1,0,0,0) and v_{2}=(0,0,1,1,1), the primordial vectors are v_{1} and v_{2}, and the minimal-weight vectors are nonzero multiples of v_{1}.
- w is a nonzero vector such that the only vector w' in N with J(w') strictly contained in J(w) is the zero vector; and
- at least one of the coefficients c_{j} equals 1.
Theorem 6:
Let A be a central simple k-algebra,
and L a commutative field containing k.
Then the tensor product of A with L is a simple L-algebra.
Note that we do not require that [L:k] be finite.
In particular, if we take L to be an algebraic closure of k,
then the tensor product must be a matrix algebra over L
(since Br(L) is then trivial). Since it has the same dimension
over L as did A over k, we conclude:
Corollary:
dim_{k}(A) is a perfect square
for every central simple k-algebra A.
Theorem 6 follows readily from Wedderburn and Theorem 4. It yields a homomorphism from Br(k) to Br(L). If L/k is Galois then this homomorphism can be interpreted as restriction from H^{2}(k^{sep}/k,G_{m}) to H^{2}(k^{sep}/L,G_{m})^{Gal(L/k)}, and we'll identify its kernel with H^{2}(Gal(L/k),K^{*}), cf. page 9 of the Tate notes. In general, the kernel of the map from Br(k) to Br(L) is called Br(L/k) [Serre calls it H_{k,L}]; it consists of (equivalence classes of) central simple k-algebras that become matrix algebras when we extend the field of scalars from k to L. Such algebras are said to admit L as a ``decomposition field''. Thus k itself is a decomposition field for A if and only if A is a matrix algebra over k; and we've seen that an algebraic closure L of k is a decomposition field for every A.
In fact, each A admits a decomposition field K with [K:k] finite. Serre shows this by letting K be generated by the coefficients of the linear map used to decompose A over L. Alternatively, we could do this in stages. Let A be a central division algebra over k. if A is k itself then we are done; else pick an element x of A not in k, and adjoin a root of the minimal polynomial of x. Then x becomes a zero divisor, and A decomposes as a matrix algebra of order at least 2 over a central division algebra of strictly smaller dimension over k. After finitely many such extensions we are done.
Either way, we have obtained Theorem 7: Br(k) is the union of its subgroups Br(K/k) as K ranges over finite extensions of k.
We shall see in the second Serre handout that it is enough to take K/k separable. If K' contains K then Br(K'/k) contains Br(K/k). In particular, if K is separable, we may take K'/k to be a normal closure of K/k, and conclude that Br(k) is the union of the subgroups Br(K/k) as K ranges over finite Galois extensions of k.
Here's another source for central simple algebras and the Brauer group (and also for group and Galois cohomology): Milne's notes on class field theory.From Br(k) to H^{2}(k^{sep}/k,G_{m}):
Construction of the reduced characteristic polynomial, and in particular of the reduced norm of an element of a central simple algebra
Theorem 8 (Skolem-Noether):
Let A be a central simple k-algebra of finite dimension;
and let f,g be k-algebra homomorphisms from some simple algebra B to A.
Then A contains an invertible element x such that
f(b)=xg(b)x^{-1} for all b in B.
Note: Serre writes that f and g are "k-isomorphisms",
but means isomorphisms to the images of B under f,g.
Since B is assumed simple, this holds automatically
for any linear map from B to A that respects the algebra structure
(including the unit element).
In particular, taking B=A we obtain:
Corollary:
Every automorphism of a central simple k-algebra of finite dimension
is inner.
For instance, if K is a skew field of finite dimension over its center k
then Aut_{k}M_{n}(K)=GL_{n}(K)/k^{*}.
Theorem 9:
Let A be a central simple k-algebra of finite dimension,
and B a simple subalgebra with commutant C.
Then C is simple, its commutant is B, and
dim_{k}A = dim_{k}B dim_{k}C.
Corollary 1:
If moreover B is central over k then the intersection of B with C is k
and A is the tensor product of B with C.
Also, C is central.
Corollary 2:
Let L be a commutative subfield of a simple central k-algebra
of finite dimension. Then the following are equivalent:
Theorem 10:
Let W be an element of the Brauer group of a field k,
and L a commutative field containing k with [L:k] finite.
Then L is a decomposition field for W if and only if
W is represented by a central simple algebra A containing L
with dim_{k}A=[L:k]^{2}.
Corollary 1:
Let D be a skew field with center k.
Every maximal commutative subfield of D/k
is a decomposition field for D.
Corollary 2:
Let D be a skew field with center k, and let
dim_{k}D=r^{2}.
Then for each decomposition field L of D,
its degree [L:k] is a multiple of r.
Theorem 11:
Every skew field D finite over its center k contains a maximal
commutative subfield that is separable over k.
Corollary: Every element of Br(k)
has a decomposition field that is a Galois extension of k.
The key step in the proof of Theorem 11 is given by
Lemma:
Every skew field D finite over its center k is either k itself
or contains a commutative subfield M properly containing k,
with M/k a separable extension.
If not, then each x in D would be ``radical'' over k,
that is, would be a p^{e}-th root of an element of k
for some e=1,2,3,...
Since p^{e}=[k[x]:k], which is bounded above by
dim_{k}(D), there is a single integer e that works for all of D.
Since finite fields are perfect, k can be assumed infinite.
Then the fact that x^{pe} is in k for all x in D
can be translated to polynomial identities that remain true over
an algebraic closure -- including the inseparable closure
that decomposes D. But in a matrix algebra of order greater than 1,
it is not possible for each element to be radical over the scalar
matrices. For instance, such an algebra contains nontrivial
``idempotents'' (elements other than 0 or 1 satisfying x^{2}=x).
So we're done.
We can now map Br(L/k) to H^{2}(Gal(L/k),L^{*}) for any Galois extension L/k of finite dimension. By Theorem 10, each element of Br(L/k) is represented by a central simple algebra A of dimension [L:k]^{2}, with an embedding of L. This representative is unique up to k-isomorphism, and so is the embedding of L up to conjugation in A^{*} (Skolem-Noether). Let E be the group of such conjugations; that is, of x in A^{*} such that xL=Lx. For each such x we get a k-homomorphism of L taking any field element c to xcx^{-1}; this gives a map from E to Gal(L/k). This map is surjective by Skolem-Noether. Its kernel is the unit group of the commutant of L. The commutant is L itself by Theorem 9 (Cor.2). So, we have an extension of Gal(L/k) by the abelian group L^{*}. Moreover, the map of Gal(L/k) on L^{*} via conjugation in E coincides with the Galois action on L^{*}. Thus we obtain an element of H^{2}(Gal(L/k),L^{*}) with that Galois action.
To check that we actually get an isomorphism with H^{2}(Gal(L/k),L^{*}), we must show that the map is a homomorphism relative to the abelian group laws we have defined on Br(L/k) and H^{2}(Gal(L/k),L^{*}), and that each element of H^{2}(Gal(L/k),L^{*}) is actually realized by some central simple algebra, uniquely determined up to k-isomorphism. To further show that Br(k) is isomorphic with H^{2}(Gal(L/k),L^{*}), we need more than the corollary of Theorem 11, which shows that each element of Br(k) is in Br(L/k) for some Galois L/k: we must also show that different choices of L yield the same element of H^{2}(k^{sep}/k,G_{m}). Since the compositum of any two Galois extensions of finite dimension is again Galois of finite dimension, it is enough to check that our maps to H^{2}(Gal(L/k),L^{*}) commute with inflation between different fields of decomposition L that are finite Galois extensions of k.
We recover A from E as a ``crossed product'' algebra. For each g in Gal(L/k), let I_{g} be the preimage of g under the map from E to Gal(L/k). This is a coset of L^{*} in E consisting of those x such that xcx^{-1}=g(c) for all c in L^{*}. Let N_{g} be the union of I_{g} with {0}; this is an L-vector space of dimension 1. Now let A_{E} be the direct sum of all the N_{g}'s. This is an L-vector space of dimension |Gal(L/k)|=[L:k]. We give A_{E} the structure of an associative algebra by using the multiplicative structure on E and extending linearly to A_{E}. Then A_{E} is a k-algebra, with unit element the identity of E (contained in I_{1}).
Theorem 13 [sic]:
A_{E} is a central simple k-algebra.
[Serre's ``Thm.13'' looks different but is tantamount
to the same thing.]
Now there's a natural homomorphism of k-algebras
from A_{E} to our A.
Since A_{E} is simple, this homomorphism is injective.
It is therefore an isomorphism, because A_{E} and A have
the same dimension [L:k]^{2} over k. We have thus
associated to each extension E of Gal(L/k) by L^{*}
a unique central simple algebra A_{E} in the preimage of E
under the map from Br(L/k) to H^{2}(Gal(L/k),L^{*}).
It remains to show that the map is a group homomorphism. We shall use the description of the group law on H^{2}(G,A) as ``Baer multiplication''. Let E,E' be extensions of G by A with the same action of G on A, corresponding to two classes in H^{2}(G,A). Then the sum of these two classes is represented by an extension E'' constructed as a quotient (E,E')/Q. Here (E,E') is the fiber product of E with E' with respect to the projection maps to G; that is, the subgroup of E*E' consisting of pairs (e,e') that map to the same element of G. This (E,E') is an extension of G by A*A. Its normal subgroup A*A contains the subgroup Q={(a,a'):aa'=1}, which is normal in (E,E'). The quotient group is an extension of G by A, which we may regard again as an element of H^{2}(G,A). We readily check that it is the sum of the elements corresponding to E and E'.
Using this description and the construction in Theorem 10, Serre shows:
Lemma:
Our map from Br(L/k) to H^{2}(Gal(L/k),L^{*})
is a homomorphism.
Combining this with Theorem 13, Serre finally obtains:
Theorem 12: Our map from Br(L/k) to H^{2}(Gal(L/k),L^{*}) is an isomorphism.
Corollaries:
Second problem set: Galois theory II (PS, PDF)
A consequence of the first two problems is the following generalization
of Gauss's Lemma on factoring integer monic polynomials:
Proposition: Let A be a subring of a field F
that is integrally closed in F. If g,h are monic polynomials in F[X]
whose product is in A[X] then g and h are in A[X].
(Gauss proved the special case
F=Q, A=Z of this Proposition.)
Corollary (Eisenstein):
Let f be a monic polynomial in Z[X]
all of whose coefficients except the leading coefficient
are multiples of a prime p. If the constant coefficient f(0)
is not a multiple of p^{2} then f is irreducible
in Q[X].
Proof sketch:
By Gauss, any putative factorization f=gh in Q[X]
already works in Z[X]. We may then reduce it mod p
and use unique factorization of polynomials over the field
F_{p}=Z/pZ
to show that g,h are congruent mod p to powers of X. Evaluating at
X=0 yields p^{2}|P(0), contradiction.
Example: The polynomial
f(X)=((X+1)^{p}-1)/X) satisfies the Eisenstein criterion,
and hence is irreducible.
Translating X to X-1, we find that the polynomial
(X^{p}-1)/(X-1), whose roots are
the nontrivial p-th roots of unity,
is irreducible in Q[X].
It easily follows that the splitting field of this polynomial
is obtained by adjoining one root,
and that its Galois group over Q
is the group of units of
F_{p}=Z/pZ,
which is known to be a cyclic group of order p-1.
Above we show more generally that the Galois group of
X^{N}-1 over Q is the group
(Z/NZ)^{*}
of units in Z/NZ.
Third problem set: infinite Galois theory, and a bit of Kummer theory (PS, PDF)
Fourth problem set: A last bit of Galois theory, and a first bit of representations of finite groups (PS, PDF)
Fifth problem set: Representations of finite groups (PS, PDF)
Sixth problem set: Group and Galois cohomology (PS, PDF)
Seventh problem set: Simple algebras, etc. (PS, PDF)
Eighth problem set: tensor products; Chevalley's theorem (PS, PDF)