Home page for Math 250: Higher Algebra (2001-2002)

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.

To Math 250b

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.

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).

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.
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.

Jacobson 4.5: The Galois group and fundamental Galois correspondence


  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 EgHg-1=g(EH) 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: FINITE FIELDS (see Jacobson 4.13). Suppose F is the prime finite field Z/pZ, and [E:F]=n. Then E is a finite field of size q:=pn. In fact, for each prime power q there exists a field E of size q, unique up to isomorphism, namely the splitting field of f(X):=Xq-X over Z/pZ. Indeed f is separable (since f'=-1), and its q roots in a splitting field are closed under the field operations (cf. also Problem 6 on the second problem set); and conversely in any q-element field every element is a root of the polynomial f (for instance, because the exponent of the finite group E* divides |E*|=q-1). Moreover, Gal(E/F) is the cyclic group of order n generated by the Frobenius map taking x to xp; and E contains a copy of the finite field E1 of order q1 if and only if q=q1e for some integer e, in which case Gal(E/E1) is cyclic of order e. The key to the identification of Gal(E/F) is that if g is the m-th power of Frobenius then E<g> consists of roots of the polynomial Xpm-X of degree pm, and thus cannot be all of E if 0<m<n. Thus Frobenius has exponent n in Gal(E/F), and hence generates Gal(E/F) since n=[E:F].

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).

Example 2: symmetric functions. Now 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 >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); we shall say more about this in M250b.]

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.

Example 3: doubly periodic functions (a.k.a. isogenies between elliptic curves over C). 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.)

Jacobson 4.6: Some results on finite groups

Theorem 4.8: Any finite group of prime power order is solvable.
The key lemma is that a nontrivial such group has nontrivial center. Proof
Definition: The commutator [g,h] of elements g,h of the same group is g-1h-1gh. The commutator subgroup G' of a group G is the subgroup of G generated by all the commutators in G.
Lemma: The commutator subgroup G' of any group G is normal; the quotient group G/G' is abelian; and G' is the minimal such group: it is contained in every normal subgroup K of G such that G/K is abelian.
Theorem 4.9: A group G is solvable if and only if its k-th derived group G(k) is trivial for some k.
Theorem 4.10: i) Any subgroup of a normal group is solvable.
ii) Let K be a normal subgroup of a group G. Then G is solvable if and only if K and the quotient group G/K are both solvable.
Theorem 4.11 (Galois): The alternating group An is simple as long as n is at least 5.
(The key tool here is that if H is a normal subgroup of G then H contains all the commutators [g,h]=g-1h-1gh with g,h in G,H respectively. If G=An and H is a nontrivial normal subgroup then we can repeatedly use such commutators to show that H contains a 3-cycle; since the 3-cycles generate An and are all conjugate in An, it follows that H=An.)
Corollary: The symmetric group Sn is not solvable for any n>4.

Definition: A composition series for a group G is a normal series in which each term Gi+1 is a maximal normal subgroup of Gi (excluding Gi itself). Necessarily each quotient Gi/Gi+1 is simple; these quotients are called the composition factors of G determined by the series.
JORDAN-HÖLDER THEOREM: The composition factors of a finite group G are determined by G up to permutation.
Theorem 4.12: A finite group is solvable if and only if its composition factors are all abelian (hence cyclic of prime order).

Jacobson 4.7: Galois' criterion for solvability by radicals

Definition 4.2: A root tower for a field extension K/F is a tower of subfields F=F1 in F2 in ... in Fs+1=K such that each Fi+1 is Fi(di) for some di such that ai=dini is in Fi. A monic polynomial of positive degree in F[X] is solvable by radicals over F if its splitting field is contained in an extension K/F possessing a root tower.

GALOIS SOLVABILITY CRITERION: A monic polynomial over a field of characteristic zero is solvable by radicals if and only if its Galois group is solvable.


  1. Let n be a positive integer that is not a multiple of the characteristic of a field F. Then Xn-1 has abelian Galois group over F.
    (The key is the the roots in a splitting field K constitute a finite subgroup of K*, and such a group is necessarily cyclic! Else it would have at least p2 elements of order p for some p, contradicting the upper bound of p on the number of roots of Xp-1 in K.)
    The Galois group is necessarily contained in the group of units of Z/nZ. We shall see that over Q the Galois group is exactly equal to (Z/nZ)* (this is essentially Thm. 4.17 in Jacobson). That is among other things the basis for the criterion (Thm. 4.18) for Euclidean constructibility of the regular n-gon. For constructions with means beyond Euclid, see A.M.Gleason's ``Angle trisection, the heptagon, and the triskaidecagon'' [13-gon] in the American Math. Monthly 85 (1988) #3, 185-194 (addenda: same volume, #10, 911).
  2. If the field F contains n distinct n-th roots of unity, then for each c in F the Galois group of Xn-c over F is cyclic of order a divisor of n.
  3. If p is prime, the field F contains p distinct p-th roots of unity, and E/F is a normal extension of degree p, then E=F(d) with dp in F.
    (Fix a generator e of Gal(E/F). For each c,z in E with zp=1, let R(z,c) be the Lagrange resolvent c+ze(c)+z2e2(c)+...+zp-1ep-1(c). This is an element d of E such that d=ze(d). This is enough provided that d is nonzero and z is not 1. There are several ways to show that this is possible; perhaps the easiest is to fix c not in F and sum R(z,c) over the p choices of z. The result is pc, which is not in F; the summand R(1,c) is in F, so one of the other summands must be nonzero.
    This is already enough to explain the solution by radicals of the general quadratic and (with some more work) equations.
  4. Let f be a monic polynomial in F[X], and K/F any extension. Then the Galois group of f over K is isomorphic to a subgroup of the Galois group of f over F.
  5. Let E/F be a separable extension with a root tower. Then the normal closure K/F of E/F has a root tower that requires only the extraction of roots of orders among the ni for a root tower of E/F.

In Theorem 4.4 (existence and uniqueness up to isomorphism of splitting fields), we already in effect identified the Galois group of a polynomial with a permutation group of its roots. In particular, we have
Theorem 4.14: Let f in F[X] be a monic polynomial without repeated roots. Then f is irreducible in F[X] if and only its Galois group acts transitively on its roots.
This is one of many tools for actually computing the Galois group of a given polynomial or extension. Another is Theorem 4.17 (the Galois group of XN-1 over Q), already mentioned earlier.
In another direction: if p is prime then the symmetric group on p letters is generated by any p-cycle and any simple transposition. If f is an irreducible polynomial of prime degree then its Galois group is a transitive group on p letters (Thm. 4.14), so automatically contains a p-cycle (Sylow). If it also can be shown to contain a simple transposition then it must be the full symmetric group Sp. For instance, this holds if it is a polynomial over Q (more generally, over any subfield of R) with exactly one complex conjugate pair of roots (Thm. 4.16). We thus easily construct polynomials over Q with Galois group Sp.
In fact, ``most'' polynomials over Q have the full symmetric group as Galois group. What other finite groups can arise? Since every finite group G is a subgroup of some symmetric group Sn (for instance, n=|G| [Cayley]), it is the Galois group of some normal extension K/KG of your favorite Sn extension. But if we specify the ground field, the ``inverse Galois problem'' can be much harder. For instance, Emmy Noether conjectured that G is always the Galois group of some polynomial over Q. This is likely to be a very difficult problem. For an idea of some 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.

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 even for finite normal extensions, 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! See the fourth problem set for the details of this theory and more examples.

Further directions (and potential paper topics) in Galois theory
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.
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.
We show this by proving the following more general theorem of E.Noether that is sometimes also called ``Satz/Theorem 90'' because 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.]

Theorems 4.29 and 4.30 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 shall see in 250b that it is an algebraic subgroup of GLN for some N>n, but we do not need to worry about this for now.

The defining formulas for cohomology may look a bit less strange under the following change of variables (Eckmann, c.1950): 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 we'll need again to study group representations. 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.
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.


  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.)
Further topics II: potential paper topics in group/Galois cohomology and central simple algebras
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
p1.ps: First problem set: Galois theory I, corrected
(thanks to Chris Luhrs for these typo corrections to Problems 1(i) and 6, and to Bridget Tenner and Alex Healy for pointing out the typo in Problem 3)
Here it is in PDF format.
David Jao's comments on this problem set, in PS and PDF

p2.ps: Second problem set: Galois theory II (also in PDF).
David Jao's comments on this problem set, in PS and 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. We shall show more generally that the Galois group of XN-1 over Q is the group (Z/NZ)* of units in Z/NZ.

p3.ps: Third problem set: Galois theory III (also in PDF).
David Jao's comments on this problem set, in PS and PDF
In problem 1, the subfield of k(t) fixed by PGL2(k) is k(u) where u=(tq2-t)q+1/(tq-t)q2+1, a rational function of degree q3-q=|PGL2(k)| [the numerator and denominator have a common factor (tq-t)q+1] fixed by PGL2(k). This can be obtained from the ``q-polynomial'' with roots at+b (a,b in k).
The group of order 2nn! arising in problem 4 is known as the ``hyperoctahedral group'', since it is the symmetry group of the regular hyperoctahedron in n-space, spanned by the unit vectors and their negatives.

p4.ps: Fourth problem set: Infinite Galois theory (also in PDF), incorporating Joe Rabinoff's correction to Problem 2.
David Jao's comments on this problem set, in PS and PDF

p5.ps: Fifth problem set: More about trace, norm, and cyclic extensions, and a taste of group cohomology (also in PDF). Problem 3 corrected thanks to Chris Luhrs and Joe Rabinoff.
David Jao's comments on this problem set, in PS and PDF

p6.ps: Sixth problem set: Group and Galois cohomology, continued (also in PDF). Problem 1 corrected thanks to Hoe Teck Wee.
David Jao's comments on this problem set, in PS and PDF
One application of noncommutative H1(Gal(K/k),M) is to the description of ``twists'' of objects over k with symmetries by M. See for instance Chapter X, especially section 2, of J.H.Silverman's The Arithmetic of Elliptic Curves (Springer, 1985; GTM 106).

p7.ps: Seventh problem set: Simple algebras, etc. (also in PDF; typo in Problem 2 corrected thanks to Alex Healy)
David Jao's comments on this problem set, in PS and PDF

p8.ps: Eighth problem set: More about k-algebras: tensor products, generalized quaternions, and Chevalley (also in PDF)
David Jao's comments on this problem set, in PS and PDF

p9.ps: Ninth problem set: Baer multiplication and more about generalized quaternions (also in PDF)
David Jao's comments on this problem set, in PS and PDF
Re problem 3: it is even easy to show directly that there are no nontrivial 2-adic solutions of r2+s2+t2+u2=0, and thus that c=d=1 yields a division algebra over Q2.

h0b.ps: Initial Math 250b handout (also in PDF format)

In Chapter 1 of the Fulton-Harris textbook, we'll develop representation theory over R and induced representations from 1.3, and the combinatorial theory of representations of the symmetric group Sn from 1.4. You should already be familiar with, or willing to quickly learn on your own, the material in 1.1 (basic definitions, Schur's lemma, etc.), 1.2 (characters and their orthogonality), and 1.3.4 (the group algebra, introduced in 250a). Note in particular questions (ii) and (iii) on page 8. Be sure to notice Exercise 1.14* -- and the fact that in the F-H text the star indicates not unusual difficulty but the presence of a solution and/or other edification in the back of the book, starting at page 516. We'll omit 1.5 (except possibly 1.5.1, representations of the alternating group An), and also omit 1.6 or at least postpone it until much later in the term. We won't cover in class the direct determination of the character tables of the symmetric and alternating groups of orders at most 5 (see 1.1.3, 1.2.3, 1.3.1, and note that we did these through order 4 last term in 250a), but you might find it interesting to read through those sections of the textbook on your own.

In the proof of Lemma 3.35 (page 40), it might not be obvious that lambda is a positive real. To show this, recall that H may be assumed positive definite, and let x be any nonzero vector. Then H(phi2x, x) is the complex conjugate of H(phi(x),phi(x)) [plug in (phi(x), x) for (x,y) in the displayed identity]. But that is a positive real number, as is H(x, x). Hence lambda, the ratio H(phi2x, x) / H(x, x) of these two positive reals, is also a positive real number.

We have shown that an element g of a finite group G is conjugate with its inverse g-1 if and only if chi(g) is real for every character chi of G. Here is a similar characterization of group elements each of whose characters is an integer:
Theorem: Let g be an element of order n in a finite group G. Then chi(g) is an integer for every character chi of G if and only if gd is conjugate with g for each integer d coprime to n.

(Equivalently: if and only if g is conjugate with every generator of the cyclic group <g>.)

Proof (sketch): It is enough to show that the criterion is equivalent to the rationality of each chi(g), because character values are automatically algebraic integers. Diagonalize the action of g on any representation. The eigenvalues are n-th roots of unity. Conclude that chi(gd) is the image of chi(g) under the Galois automorphism sending each nth root of unity to its d-th power. This equals chi(g) for each chi if and only if gd is conjugate to g.

Example: Every character of the symmetric group Sd takes integer values at all permutations of d letters.

We shall see that Sd actually satisfies a stronger condition: all its representations are defined over Q (and hence over Z).

Recall that for any finite group G we have the formula |G|=sum(dim(V))2 where V ranges over the (isomorphism classes of) irreducible representations of G. In our case of G=Sd , each dim(V) is the number of standard tableaux in the Young diagram corresponding to dim(V), so the sum of (dim(V))2 is the number of pairs of standard tableaux of order d with the same shape. Can the fact that this number equals d! be proved combinatorially, by exhibiting a bijection between such pairs and permutations of d letters? Yes, but nontrivially: that's the (Robinson-)Schensted correspondence.

A nice example of the hook-length formula (4.12, p.50) for the dimension of an irreducible representation of Sd is the case of a partition of the form (m,m) where d=2m. The Young diagram for this partition is a rectangle of height 2 and width m. The dimension obtained from the hook-length formula is the m-th Catalan number (2m)! / (m! (m+1))!. For m=2 and 3, we obtain representations of dimension 2 and 5 of S4 and S6 respectively, coming from the exceptional homomorphisms from those groups onto S3 and S6. These two (and the representation of S6 associated with the partition (2,2,2) conjugate to (3,3)) are the only representations of Sd of dimension less than d other than the trivial, sign, standard, and standard-tensor-sign representations (Exercise 4.14*). [The Catalan numbers also appear with some regularity as the solutions of various combinatorial enumeration problems, such as the number of triangulations of an (m+2)-gon.]

The final step in the proof of Lemma 4.25 (p.54), and thus of Theorem 4.3 (the description of irr.reps. of Sd in terms of Young symmetrizers), hinges on the fact that if W is a (one-sided) ideal of the group algebra A then W=0 if and only if W2=0. Here's another way to see this, which works in any semisimple algebra A. Since A is a direct sum of matrix algebras End(V), it is enough to prove the result for A=End(V). But we know the ideals of End(V) are just Hom(V,V') as V' ranges over subspaces of V. This ideal is zero if and only if V' is. But Hom(V,V') contains projections from V to V'. A projection is idempotent (equal to its own square); thus if W2=0 then the projection itself is zero, whence V' is the zero subspace and W=0 as claimed.

In the course of proving that [X,Y] = XY-YX in GL(V) [p.107], Fulton and Harris implicitly use the following formula: let g be a differentiable map from an interval in R to GL(V), with derivative g'; then the derivative of g-1 (the composition of g with the inverse map on GL(V)) equals -g-1g'g-1. This generalizes the familiar formula d(1/u)=-du/u2 of elementary calculus, and can be proved in much the same way: differentiate the identity gg-1=e and use the product rule.

A couple of warnings about the exponential map exp from g to G:

Here's the example showing that for large n there are lots of Lie algebras of dimension n, even two-step nilpotent ones. This is basically the same construction that produces p2n3/27-O(n2) finite groups of order pn. Fix vector spaces V, W of dimensions 2m, m respectively. Let (.,.) be any alternating bilinear pairing from V to W. Then the direct sum V+W can be made into a Lie algebra of dimension n:=3m by taking [(X,Y),(X',Y')]=(0,(X,X')). Now (.,.) varies in a vector space of dimension 2m3-m2. Different choices of (.,.) may yield isomorphic Lie algebras, but the available isomorphisms are contained in the group GL(V+W) of dimension 9m2. Hence there is a space of 2m3-O(m2) distinct Lie algebras of dimension 3m, as claimed. Of course it follows that there are lots of n-dimensional Lie groups as well.
In the opening paragraph of the proof of Engel's Theorem 9.9 it is shown that if X is a nilpotent element of End(V) then ad(X) is a nilpotent linear transformation of gl(V), and hence on any Lie subalgebra of gl(V). Another, possibly simpler, way to see this is to note that for any Y in gl(V) we may write (adm(X))(Y) as a linear combination of the endomorphisms XaYXb of V with a+b=m. [Digression: what are the coefficients of this linear combination?] If X is nilpotent then Xn=0 for some n; hence (adm(X))(Y)=0 once m>2n-2, so ad(X) is nilpotent as claimed.
In Lie's Theorem 9.11, the hypothesis that the solvable Lie subalgebra of gl(V) be complex is necessary: over R, the span of the 2*2 matrix [0,1;-1,0] (a.k.a. so2) in gl(R2) would be a counterexample. For the same reason, this hypothesis is needed (though not stated) for Exercise 9.16. Also, while the paragraph preceding the statement of Lie's theorem suggests that it depends on Engel's Theorem 9.9, the proof seems not to require Engel (though it proceeds along similar lines).
Proposition 9.17 may remind you of an analogous description of the irreducible representations of a semidirect product of finite groups. There is good reason for this similarity: the short exact sequence 0->Rad(g)->g->gss->0 always splits (Levi's Theorem, see Appendix E.1 in the textbook), and this is the condition on the Lie-algebra level that corresponds to a semidirect product of Lie groups.
At the end of the proof of this Proposition, it's not enough to extend lambda to an arbitrary linear functional: such a functional yields a one-dimensional representation of g if and only if it vanishes on [g,g]. We can find an extension of lambda satisfying this condition provided that lambda vanishes on the intersection of [g,g] with Rad(g). Let X be contained in this intersection. Then X acts on V by scalar multiplication by lambda(X). On the other hand, being a commutator, X must have trace zero in any finite-dimensional representation of g. Hence lambda(X)=0 as desired. [Your copy of the textbook has this correction if it's the ``corrected fifth printing, 1999''; see the bottom of page iv.]
Finally, note that L* is the contragredient (dual) of L as a Lie-algebra representation; that is, any X in g acts on L* by multiplication by (-lambda(X)). In general, the contragredient of a representation into gl(V) is obtained by composing the map to gl(V) with the automorphism of gl(V) taking any square matrix X to (-tX) -- this being the action on gl(V) of the outer automorphism of GL(V) taking every invertible matrix to its inverse transpose.
A nice companion to the result of Exercise 11.22 is the following geometrical description of the map from the exterior square of Sym2V to Sym2V. The generic element of Sym2V can be described both as a pair of points in the Riemann sphere P1(C) and as the involution in SL2(C) whose fixed points in P1(C) are that pair. [Any pair is equivalent to {0,inf}, which is the fixed pair of the involution z<-->-z and of no other involution in SL2(C).] Now for any two disjoint pairs of points there exists a unique involution switching both pairs. But this involution is itself identified with a pair of points. This extends to a map from Sym2*Sym2 to Sym2, which is none other than our identification of Sym2V with its exterior square! That is, if P,Q are quadrics without repeated or common roots then so is their Jacobian R=J(P,Q), and the roots of R are the fixed points of the involution that switches the roots of P and switches the roots of Q. To prove this, it is enough to check the case that this involution is z<-->-z. But then P,Q are polynomials of the form x2+cy2 for distinct choices of c, and we readily check that their Jacobian is a multiple of xy as desired. (See also Proposition A in the Appendix of my ANTS-III paper where this result is used.)
Here are some examples of computations concerning invariant polynomial rings (corrected; also in PDF)
Here's Remark 2 on page 21 of Serre's Complex Semisimple Lie Algebras (Springer, 1987):
Here is an example of an application of Theorems 3 and 4 [description of the representations of g=sl2], independent of the interpretation of sl2 as the Lie algebra of SL2:
Let U be a compact Kähler variety of complex dimension n, and let V be the cohomology algebra H*(V,C). Hodge theory associates endomorphisms Lambda and L of V with the kählerian structure of U (cf. A. Weil, Variétés kähleriennes, Chap. IV); let us take X and Y to be these endomorphisms, and define H by the relation Hx=(n-p)x if x is in Hp(V,C). Then one can check (Weil, loc. cit.) that V becomes a g-module. By applying Theorems 3 and 4 to this module, one retrieves Hodge's theorems on ``primitive'' cohomology classes.

p1b.ps: First problem set: Representations of finite groups (also in PDF)
p2b.ps: Second problem set: More representations of finite groups (especially Sd), and a bit about Lie groups (corrected; also in PDF)
David Jao's comments on this problem set, in PS and PDF
Problem set #3 consists of the following problems from the textbook: 8.10* (page 109), 8.27 (p.113), 8.31 (p.115), 8.44 (p.119-120), 9.1 and 9.2 (p.121-122), and 9.8 (p.125). [For 9.2, as with 9.1, you'll use the ideas of Chapter 8 to recover information about G from its Lie algebra.] These are due in class Wednesday, March 13.
David Jao's comments on this problem set, in PS and PDF
p4b.ps: Fourth problem set: More basics about Lie groups and algebras, and some SL2 ``plethysm'' (also in PDF)
p5b.ps: Fifth problem set: More about invariant rings and their Hilbert functions; the Weyl character formula for SL3; Killing / Cartan stuff (also in PDF)