Home page for Math 250: Higher Algebra (Fall 2004)

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.


h0.ps: Initial Math 250a handout (also in PDF format)

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 Ma 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 Ma+b=Ma+Mb and Mab=MaMb. So, for instance, if [K:F]=n we get a copy of K in the (noncommutative) ``endomorphism ring'' EndKF, 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 Fn that cyclically permutes the nonzero vectors of Fn.

Jacobson 4.3: Splitting fields

Motivation and definition
Theorem 4.3: Every monic polynomial in F[X] has a splitting field K/F.
[K:F] is finite, indeed at most n! where n=deg(f).
Theorem 4.4: The splitting field of a polynomial is unique up to isomorphism.
More precisely, if K and K' are both splitting fields for the same polynomial f in F[X], there exists an isomorphism from K to K' that acts as the identity on F. The number of such isomorphisms is at most [K:F], and exactly equals [K:F] when f has distinct roots in K.
In particular, if K=K' we have obtained important information about the isomorphisms of K that act as the identity on F, which we call isomorphisms of K over F. For a general field extension K/F, the isomorphisms of K over F constitute a group, called the Galois group of K over F and denoted Gal(K/F).

This already lets us construct, for each prime power q, a finite field of q elements -- namely, the splitting field of Xq-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 Xq-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 Fq 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(Xp) for some g in F[X].

An algebraic introduction to derivatives and derivations
Lemma: If F is a field of positive characteristic p, then a polynomial in F[X] of the form xp-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=bp in F.
Simplification of the end of Jacobson's proof: once g(x)=(x-b)k with coefficients in F, we know b must be in F because the xk-1 coefficient is -kb.
The map taking any element b of a field of characteristic p to bp is in fact a field endomorphism, due to the identities (1/b)p=1/bp, (ab)p=apbp, and crucially (a+b)p=ap+bp. The image of this ``Frobenius'' map is denoted Fp.
We shall say that a nonzero polynomial over a field is separable if it has no repeated roots in a splitting field. Note that this is a more stringent definition than Jacobson's, in which only the irreducible factors are required to have distinct roots. Therefore we must restate his Theorem 4.6 thus:
Theorem 4.6: A field F of positive characteristic p equals Fp if and only if every irreducible polynomial in F[X] is separable.
(A field F is said to be perfect if and only if every irreducible polynomial in F[X] is separable; using Thm. 4.6, we conclude that F is perfect if and only if either F has characteristic zero or F has positive characteristic p and F=Fp.)
Corollary: Every finite field is perfect.
We already have all we need to give the basic results about finite fields. Let F be a finite field of q0 elements, and q some power of q0, say q=q0n. Then the splitting field K of Xq-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 q0-element subfield F if and only if q=q0n 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 xq0. 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 KH of K consisting of elements fixed by H.

Jacobson 4.5: The Galois group and fundamental Galois correspondence

Lemmata:

  1. Let E/F be the splitting field of some separable polynomial in F[X]. Then |Gal(E/F)|=[E:F].
  2. (Artin) Let G be a finite group of automorphisms of a field E, and F=EG [Jacobson uses the notation "Inv G"]. Then [E:F] is finite and no larger than |G|. Proof

Theorem 4.7: Let E/F be any field extension. Then TFAE:

  1. E/F is a splitting field of a separable polynomial in F[X];
  2. F = EG for some finite group of automorphisms of E;
  3. E/F is finite dimensional, normal, and separable.
Moreover, F=EGal(E/F) if (1) holds, and G=Gal(E/F) if (2) holds.
FUNDAMENTAL THEOREM OF GALOIS THEORY: Let K/F be a finite-dimensional, normal, separable field extension with Galois group G. Every subfield E of K that contains F is of the form KH for some subgroup H of G, namely the group consisting of the isomorphisms that fix each element of E. Moreover, E/F is normal if and only if H is a normal subgroup of G, in which case Gal(E/F) is canonically isomorphic with G/H.
(Only the last sentence requires a new idea. This idea is the observation that KgHg-1=g(KH) for all g in G and all subgroups H of G.)
It follows that if E,E' are the subfields KH, KH' then E is contained in E' if and only if H contains H'. That is, the lattice of subfields is isomorphic with the lattice of subgroups, with inclusions reversed.

Example 1: symmetric functions. Let E be the field k(x1,...,xn) 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 xi. Then F=EG is the field of symmetric rational functions. We claim that F=k(s1,...,sn) where sj is the j-th elementary symmetric function, that is, (-1)j times the Xn-j coefficient of g(x)=(X-x1)...(X-xn).
[Jacobson uses pj for our sj; but nowadays pj is generally used for the j-th power sum of the xi, 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 xi of g, and is determined by this permutation. Thus by the fundamental theorem K=EG, so K=F as claimed.
[In fact the G-invariant subring of k[x1,...,xn] is likewise k[s1,...,sn] (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 t1,...,tn indeterminates over F. The Galois group of xn-t1xn-1+t2xn-2-...+(-1)ntn over F(t1,...,tn) is the symmetric group Sn.
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(ez), C(ez/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'=FG. 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:

In both cases we in effect found elements of G by embedding Q in a field F' (namely R or Qp) and using the fact that in any embedding of a field F in F' the Galois group of any separable polynomial P over F contains the Galois group of P over F'. For a general polynomial over Q, we can often use this technique (and refinements of it) to guess G, but except in special cases (such as G=Sn or G=An) it can take some effort to prove rigorously that G is what it seems to be.

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 Xp-1 (p some odd prime), and L is the infinite algebraic extension of F obtained by adjoining pm-th roots of unity for all m, then L/F is normal with Galois group isomorphic with Zp, 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 Xn-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 gn=1 for all g in G.) Let M1 be the subgroup of F* consisting of elements c with an n-th root d in K, and let M be the quotient of M1 by F*n [note that M1 must contain F*n]. For g in G and c in M1 define (g,c)=g(d)/d where dn=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 M1. 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 Mx on E introduced in the first problem of Problem Set 1. More generally, any separable extension E/F is KH for some Galois extension K/F and some subgroup H of G=Gal(K/F); then the eigenvalues of Mx 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 pm 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)[=``v1-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 {ue} with coordinates in E indexed by G is of the form ue=v/e(v) for some nonzero v in E if and only if uee'=e(ue')ue for all e,e' in G.
To recover 4.28 from this, let uei=ue(u)e2(u)...ei-1(u).
To prove 4.29, let v(w) (for any w in E) be the sum of uee(w) over e in G, and check that ue'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 EndF(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 un 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 x2+y2=z2...

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 {de} with coordinates in E indexed by G is of the form de=c-e(c) for some c in E if and only if dee'=e(de')+de for all e,e' in G.
This is proved by taking c=Sumedee(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 cp-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 Xp-X-a is called an Artin-Schreier extension.]


An introduction to representation theory of finite groups G over fields of characteristic not dividing |G| (also in PDF);
Some further examples and exercises on representations of finite groups

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 H1(Gal(E/F),A) vanishes when A is either the additive or the multiplicative group of E. This is usually written "H1(Gal(E/F),Ga)=0" and "H1(Gal(E/F),Gm)=0", or even "H1(E/F,Ga)=0" and "H1(E/F,Gm)=0"; here Ga (Gm) 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 H2(K/k,Gm) 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 GLn of invertible matrices of order n. (The only groups Tate uses that are not of this form are elliptic curves.) An algebraic subgroup of GLn is one obtained by imposing finitely many polynomial conditions on the entries of the matrix. Examples of such algebraic groups are: GLn itself; its subgroup SLn, with polynomial condition det=1; the multiplicative group Gm, which is just GL1; the group mun, which is the subgroup of GL1 defined by an=1; and the additive group Ga, isomorphic with the subgroup of GL2 consisting of upper triangular matrices with unit diagonal entries. The ``ax+b'' group is the subgroup of GL2 consisting of matrices with bottom row (0,1); it has a normal subgroup Ga, consisting of elements with a=1; the quotient is Gm. 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 PGLn, which is the quotient of GLn by its (algebraic) subgroup of scalar matrices. As it stands, this construction of PGLn does not exhibit it as an algebraic group; we do not need to worry about this for now, but we remark that PGLn it can be obtained as an algebraic subgroup of GLN where N=n2, using the action of GLn 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 xi=ai-1-1ai for some a0,...,ar in G. A different (r+1)-tuple (a0',...,ar') yields the same xi if and only if there is an element g of G such that a'i=gai for all i. We may associate to an arbitrary r-cochain f, which is a function from Gr to A, the ``homogeneous function'' fhomog from Gr+1 to A defined by

fhomog(a0,...,ar) = a0 f(x1,...,xr).
By saying that fhomog is ``homogeneous'', we mean that
g fhomog(a0,...,ar) = fhomog(ga0,...,gar).
Conversely, every homogeneous map from Gr+1 to A is fhomog for a unique r-cochain f, namely
f(x1,...,xr) = fhomog(1, x1, x1x2, ..., x1x2x3...xr).
Now if the function F from Gr+1 to A is homogeneous then so is the function dF on Gr+2 defined by the more familiar coboundary rule
(dF)(a0,a1,...,ar+1) = F(a1,a2,a3,...,ar+1) - F(a0,a2,a3,...,ar+1) + F(a0,a1,a3,...,ar+1) ... + (-1)r+1 F(a0,a1,a2,...,ar).
This makes it easy to check that dd=0. On the other hand, being homogeneous, dF must correspond to some (r+1)-cochain. If F=fhomog, then dF corresponds to exactly what we defined as the coboundary of f!

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 (Hr(G,A) is a |G|-torsion group for r>0) can be translated to the following computation:
Suppose f: Gr --> 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 (x1,x2,...,xr-1) is f(x1,x2,...,xr-1,y). Then the coboundary of g is the r-cochain whose value at any (x1,x2,...,xr) is

(-1)i ( f(x1,x2,...,xr-1,xr) - f(x1,x2,...,xr-1,xry) + f(x1,x2,...,xr-1,y) )
Now sum over y in G. The first terms add to (-1)i |G| f(x1,...,xr-1,xr), that is, to (-1)i |G| times our cocycle f. The remaining terms cancel out. Therefore |G|f is a coboundary, as claimed.

The ``p-primary component'' of a finite abelian group (mentioned in Exercise 4.2) is the subgroup consisting of elements of order pf 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 Mn(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 Endk(E) is {x in Endk(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 Endk(E): since A is simple, its representation to EndkE 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'=Ko.
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 k0 can be obtained by:

Now the center of Mn(A) is readily seen to equal the center of A. Thus by Wedderburn the second step is equivalent to choosing a central skew field K over k and an integer n, and then A is Mn(K). We shall say that two central simple algebras over k are ``equivalent'' if they come from isomorphic skew fields K. Thus the description of central simple algebras comes down to describing central skew fields, or (what's the same) equivalence classes of central simple algebras, over a given commutative field k.

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 Ko. 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 H2(ksep/k,Gm)!

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 Ao is isomorphic with Mn(k).
Indeed, the left and right actions of A on itself yield a homomorphism from that tensor product to Endk(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 {ei} of V. Every vector v in V can be written as a sum of cjej, 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: 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 k5 with basis v1=(1,1,0,0,0) and v2=(0,0,1,1,1), the primordial vectors are v1 and v2, and the minimal-weight vectors are nonzero multiples of v1.
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 Gk; we'll use the notation Br(k), which has become more-or-less standard (see for instance the Tate handout).

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: dimk(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 H2(ksep/k,Gm) to H2(ksep/L,Gm)Gal(L/k), and we'll identify its kernel with H2(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 Hk,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.
Construction of the reduced characteristic polynomial, and in particular of the reduced norm of an element of a central simple algebra
From Br(k) to H2(ksep/k,Gm):

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 AutkMn(K)=GLn(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 dimkA = dimkB dimkC.
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:

Corollary 3: Let D be a skew field with center k. Then for any maximal commutative subfield L of D such that L contains k, dimkD=[L:k]2.

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 dimkA=[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 dimkD=r2. 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 pe-th root of an element of k for some e=1,2,3,... Since pe=[k[x]:k], which is bounded above by dimk(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 xpe 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 x2=x). So we're done.

We can now map Br(L/k) to H2(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 H2(Gal(L/k),L*) with that Galois action.

To check that we actually get an isomorphism with H2(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 H2(Gal(L/k),L*), and that each element of H2(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 H2(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 H2(ksep/k,Gm). 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 H2(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 Ig 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 Ng be the union of Ig with {0}; this is an L-vector space of dimension 1. Now let AE be the direct sum of all the Ng's. This is an L-vector space of dimension |Gal(L/k)|=[L:k]. We give AE the structure of an associative algebra by using the multiplicative structure on E and extending linearly to AE. Then AE is a k-algebra, with unit element the identity of E (contained in I1).

Theorem 13 [sic]: AE 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 AE to our A. Since AE is simple, this homomorphism is injective. It is therefore an isomorphism, because AE 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 AE in the preimage of E under the map from Br(L/k) to H2(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 H2(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 H2(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 H2(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 H2(Gal(L/k),L*) is a homomorphism.
Combining this with Theorem 13, Serre finally obtains:

Theorem 12: Our map from Br(L/k) to H2(Gal(L/k),L*) is an isomorphism.

Corollaries:

  1. Br(L/k) is a torsion group, i.e., consists only of elements of finite order. (Indeed we have shown that [L:k]H2(Gal(L/k),L*)=0.)
  2. Br(k) is a torsion group. (Use Theorem 11 and its Corollary.)
  3. If k is a finite field then the groups Br(k) and H2(ksep/k,Gm) are trivial. (Serre proves this using facts about H2(G,A) when G is cyclic; we gave an alternative proof using the Br(k) interpretation and Chevalley's theorem in the 8th problem set. Milne gives yet another proof on page 107 of his notes.)
Note that Serre does not complete the proof of Br(k)=H2(ksep/k,Gm) by showing that our maps from Br(L/k) to H2(Gal(L/k),L*) are compatible with the inflation maps from H2(Gal(K/k),L*) to H2(Gal(L/k),L' *), for any Galois extension L'/k containing L. Serre shows how to do this in Local Fields, Chapter X, section 5. Once this is done, one also recovers the injectivity of the inflation maps. (This appears at the beginning of page 9 of the Tate notes, as part of the inflation-restriction sequence with vanishing H1's.)
First problem set: Galois theory I (
PS, PDF)

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 p2 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 Fp=Z/pZ to show that g,h are congruent mod p to powers of X. Evaluating at X=0 yields p2|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 (Xp-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 Fp=Z/pZ, which is known to be a cyclic group of order p-1. Above we show more generally that the Galois group of XN-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)

Ninth and last problem set: Quaternion algebras (PS, PDF)