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 definitionTheorem 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.

Jacobson 4.4: Dealing with __multiple roots__

Since all splitting fields are isomorphic, it makes sense to say
that a polynomial f in F[X] has a multiple root, or that
two polynomials f,g in F[X] have common roots, in a splitting field
of f or fg respectively. The next two results give criteria
for these two properties that do not require field extensions:

Lemma:
Let f,g be monic polynomials in F[X]. Then f,g have no common roots
in a splitting field of fg if and only if f and g are relatively prime.

(If h=gcd(f,g) has positive degree, then each root of h
in the splitting field is a common root of f and g.
If not, then use a,b in F[X] such that af+bg=1.)

Theorem 4.5:
A monic polynomial f in F[X] has no repeated factors
if and only if f is relatively prime with its derivative.

In particular, if f is irreducible then it has no repeated factors
unless f'=0, that is, unless F has positive characteristic p and
f(X)=g(X^{p}) for some g in F[X].

An algebraic introduction to derivatives and derivationsLemma: If F is a field of positive characteristic p, then a polynomial in F[X] of the form x

The map taking any element b of a field of characteristic p to b

We shall say that a nonzero polynomial over a field is

Theorem 4.6: A field F of positive characteristic p equals F

(A field F is said to be

Corollary: Every finite field is perfect.

Jacobson 4.5:
The __Galois group__ and fundamental Galois correspondence

Lemmata:

- Let E/F be the splitting field of some
*separable*polynomial in F[X]. Then |Gal(E/F)|=[E:F]. - (Artin) Let G be a finite group of automorphisms of a field E,
and F=E
^{G}[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:

- E/F is a splitting field of a separable polynomial in F[X];
- F = E
^{G}for some finite group of automorphisms of E; - E/F is finite dimensional, normal, and separable.

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 K

(Only the last sentence requires a new idea. This idea is the observation that E

Example 1: FINITE FIELDS (see Jacobson 4.13).
Suppose F is the prime finite field
**Z**/p**Z**,
and [E:F]=n. Then E is a finite field of size q:=p^{n}.
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):=X^{q}-X over **Z**/p**Z**.
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 x^{p};
and E contains a copy of the finite field
E_{1} of order q_{1}
if and only if q=q_{1}^{e} for some integer e,
in which case Gal(E/E_{1}) 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 X^{pm}-X
of degree p^{m}, 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 **F**_{q} is preferred.
Under either terminology, these fields are fundamental to number theory,
and play important roles in other branches of mathematics,
pure (e.g., finite simple groups) as well as applied
(e.g., error-correcting codes).

Example 2: symmetric functions.
Now let E be the field k(x_{1},...,x_{n})
of rational functions in n variables over some ``ground field'' k.
Let G be the symmetric group of permutations of n letters,
acting on E by permuting the x_{i}. Then F=E^{G}
is the field of symmetric rational functions. We claim that
F=k(s_{1},...,s_{n}) where s_{j} is
the j-th *elementary symmetric function*, that is,
(-1)^{j} times the X^{n-j} coefficient of
g(x)=(X-x_{1})...(X-x_{n}).

[Jacobson uses p_{j} for our s_{j};
but nowadays p_{j} is generally used
for the j-th power sum of the x_{i},
and the first n of these generate F only in characteristic >n.]

Call that field K as long as we haven't proved that K=F.
Clearly K is in F, and E is the splitting field of g over K;
moreover, g is separable, so E/K is normal.
Finally, an element of Gal(E/K) permutes the roots x_{i} of g,
and is determined by this permutation.
Thus by the fundamental theorem K=E^{G}, so K=F as claimed.

[In fact the G-invariant sub*ring* of
k[x_{1},...,x_{n}] is likewise
k[s_{1},...,s_{n}] (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 t_{1},...,t_{n} indeterminates over F.
The Galois group of
x^{n}-t_{1}x^{n-1}+t_{2}x^{n-2}-...+(-1)^{n}t_{n}
over F(t_{1},...,t_{n}) is the symmetric group
S_{n}.

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'=F^{G}.
Moreover, the action of G is faithful, since it is already faithful
on the orbit of the Weierstrass *p*-function associated with L.
Therefore G=Gal(F/F') and [F:F']=[L':L]. The intermediate fields
are precisely the fields of meromorphic functions for some intermediate
lattice L"; since G is abelian, all its subgroups are normal, and thus
all these subfields are normal over F'.

(**C**/L and **C**/L'
are ``elliptic curves'' over **C**, and
the natural map from **C**/L to and **C**/L'
is an ``isogeny''. See also problem 5 on the second problem set.)

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^{-1}h^{-1}gh.
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 A_{n} 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^{-1}h^{-1}gh
with g,h in G,H respectively. If G=A_{n} 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 A_{n}
and are all conjugate in A_{n}, it follows that
H=A_{n}.)

Corollary: The symmetric group S_{n}
is not solvable for any n>4.

Definition: A __composition series__
for a group G is a normal series in which each term G_{i+1}
is a *maximal* normal subgroup of G_{i}
(excluding G_{i} itself).
Necessarily each quotient G_{i}/G_{i+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=F_{1} in F_{2} in ... in F_{s+1}=K
such that each F_{i+1} is F_{i}(d_{i})
for some d_{i} such that
a_{i}=d_{i}^{ni}
is in F_{i}. 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.

Lemmata:

- Let n be a positive integer that is not a multiple of the
characteristic of a field F. Then X
^{n}-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 p^{2}elements of order p for some p, contradicting the upper bound of p on the number of roots of X^{p}-1 in K.)The Galois group is necessarily contained in the group of units of

**Z**/n**Z**. We shall see that over**Q**the Galois group is exactly equal to (**Z**/n**Z**)^{*}(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). - If the field F contains n distinct n-th roots of unity, then
for each c in F the Galois group of X
^{n}-c over F is cyclic of order a divisor of n. - 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 d
^{p}in F.

(Fix a generator e of Gal(E/F). For each c,z in E with z^{p}=1, let R(z,c) be the*Lagrange resolvent*c+ze(c)+z^{2}e^{2}(c)+...+z^{p-1}e^{p-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.

- 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.
- 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 n
_{i}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
X^{N}-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 S_{p}.
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 S_{p}.

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
S_{n} (for instance, n=|G| [Cayley]), it is the Galois group
of some normal extension K/K^{G} of your favorite
S_{n} 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

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 X^{p}-1
(p some odd prime), and L is the infinite algebraic extension of F
obtained by adjoining p^{m}-th roots of unity for all m,
then L/F is normal with Galois group isomorphic with
**Z**_{p}, the p-adic integers!
See the fourth problem set for the details of this theory
and more examples.

Further directions (and potential paper topics) in Galois theoryJacobson 4.15: Traces and norms, continued

In a Galois extension E/F, the norm and trace of any element x
are the product and sum of its conjugates;
more precisely, they are the product and sum of g(x),
g ranging over Gal(E/F). This is because the g(x) are
the eigenvalues of the F-linear operator M_{x} on E
introduced in the first problem of Problem Set 1.

Theorem 4.28 [Hilbert's ``Theorem 90'',
also known by the equivalent German name ``Satz 90'']
Let E/F be a cyclic extension, that is, a normal extension
whose Galois group G is cyclic. Let e be a generator of G.
Then an element u of E has norm 1 in F if and only if
u=v/e(v)[=``v^{1-e}''] for some nonzero v in E.

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 {u_{e}} with coordinates in E indexed by G
is of the form u_{e}=v/e(v) for some nonzero v in E
if and only if u_{ee'}=e(u_{e'})u_{e}
for all e,e' in G.

To recover 4.28 from this, let
u_{ei}=ue(u)e^{2}(u)...e^{i-1}(u).

To prove 4.29, let v(w) (for any w in E) be the sum of
u_{e}e(w) over e in g, and check that
u_{e'}e'(v(w))=v(w) for all e' in G.
So we may take v=v(w) for any w that makes v(w) nonzero,
and such w must exist because the elements of G
are linearly independent in the E-vector space End_{F}(E).

The following generalization of our ``Lemma 3'' above quickly follows
from Satz 90:

Theorem 4.32
If E/F is a cyclic extension of degree n and F contains the n-th roots
of unity then E=F(u) where u^{n} is in F.

Another application of Satz 90 is the description of all Pythagorean
triples, since (x,y,z) is such a triple if and only if (x+iy)/z
is an element of norm 1 in the quadratic extension
**Q**(i)/**Q**.
To be sure, there are many other ways to obtain the general solution
of x^{2}+y^{2}=z^{2}...

The additive analogue of 4.29 is also true:

Theorem 4.30
Let E/F be a normal extension with (finite) Galois group G.
Then a |G|-tuple {d_{e}} with coordinates in E indexed by G
is of the form d_{e}=c-e(c) for some c in E
if and only if d_{ee'}=e(d_{e'})+d_{e}
for all e,e' in G.

This is proved by taking c=Sum_{e}d_{e}e(u)/Tr(u)
for any u in E whose trace Tr(u) is nonzero.

Theorem 4.30 yields the additive analogue of Hilbert 90:

Theorem 4.31
Let E/F be a cyclic extension, with Galois group G generated by e.
Then an element d of E has trace 0 in F if and only if
d=c-e(c)[=``(1-e)c''] for some c in E.

[Simpler proof of 4.31: dimension count on kernel and image of 1-e.]
This in turn yields the additive analogue of Theorem 4.32:

Theorem 4.33
Let F be a field of positive characteristic p,
and E/F a normal extension with [E:F]=p.
Then E=F(c) where c^{p}-c is in F.

[Compare this with the last problem of Problem Set 2.
An extension E/F in characteristic p obtained by adjoining a root of
an irreducible polynomial of the form X^{p}-X-a
is called an *Artin-Schreier extension*.]

Theorems 4.29 and 4.30 say that the first
__Galois cohomology group__ H^{1}(Gal(E/F),A) vanishes
when A is either the additive or the multiplicative group of E.
This is usually written
"H^{1}(Gal(E/F),**G**_{a})=0" and
"H^{1}(Gal(E/F),**G**_{m})=0", or even
"H^{1}(E/F,**G**_{a})=0" and
"H^{1}(E/F,**G**_{m})=0"; here
**G**_{a} (**G**_{m})
is the additive (resp. multiplicative) group. See
Tate's
notes on Galois cohomology, the source of most of the material
for the next few lectures. [Note the casual identification of
Noether's Theorem 4.29 and Hilbert's Satz 90 starting on page 3.]
We will then explain in detail the remark on
H^{2}(K/k,**G**_{m}) and Brauer groups
in the next-to-last paragraph of page 3 of Tate's notes.
We won't cover everything in Tate; many of the topics we omit
would make good final projects for the class.

Tate uses the notion of an ``algebraic group'' C,
which you may not have previously encountered.
All the algebraic groups we'll use can be found
inside the group GL_{n} of invertible matrices of order n.
(The only groups Tate uses that are not of this form
are elliptic curves.)
An algebraic subgroup of GL_{n} is one obtained
by imposing finitely many polynomial conditions
on the entries of the matrix.
Examples of such algebraic groups are:
GL_{n} itself;
its subgroup SL_{n}, with polynomial condition det=1;
the multiplicative group **G**_{m},
which is just GL_{1}; the group **mu**_{n},
which is the subgroup of GL_{1} defined by a^{n}=1;
and the additive group **G**_{a},
isomorphic with the subgroup of GL_{2}
consisting of upper triangular matrices with unit diagonal entries.
The ``ax+b'' group is the subgroup of GL_{2}
consisting of matrices with bottom row (0,1);
it has a normal subgroup **G**_{a},
consisting of elements with a=1;
the quotient is **G**_{m}.
Any finite group G with trivial Galois action can be represented
as a group of permutation matrices of order |G|, and thus
(with some unrewarding effort) exhibited as an algebraic group.
We may also use the group PGL_{n},
which is the quotient of GL_{n}
by its (algebraic) subgroup of scalar matrices.
As it stands, this construction of PGL_{n}
does not exhibit it as an algebraic group;
we shall see in 250b that it is an algebraic subgroup of GL_{N}
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
x_{i}=a_{i-1}^{-1}a_{i}
for some a_{0},...,a_{r} in G.
A different (r+1)-tuple (a_{0}',...,a_{r}')
yields the same x_{i} if and only if there is an element
g of G such that a'_{i}=ga_{i} for all i.
We may associate to an arbitrary r-cochain f, which is a function
from G^{r} to A, the ``homogeneous function''
f_{homog} from G^{r+1} to A defined by

Eckmann uses these homogeneous cochains to give an explicit
(albeit mysterious-looking) one-line formula for the corestriction;
see Theorem 7 on page 490 of his ``Cohomology of groups and transfer''
in *Annals of Math.* **58** #3 (11/1953), 481-493.

Serre points out that the proof of Tate's Corollary 4.1
(H^{r}(G,A) is a |G|-torsion group for r>0)
can be translated to the following computation:

Suppose f: G^{r} -> A is an r-cocycle for some r>0.
Fix y in G, and let g be the (r-1) cochain whose value at any
(x_{1},x_{2},...,x_{r-1}) is
f(x_{1},x_{2},...,x_{r-1},y).
Then the coboundary of g is the r-cochain whose value at any
(x_{1},x_{2},...,x_{r}) is

The ``p-primary component'' of a finite abelian group (mentioned
in Exercise 4.2) is the subgroup consisting of elements
of order p^{f} for some integer f.

Here are some basic definitions and examples concerning
algebras
over a field.

Theorem 1 (Wedderburn):
A finite-dimensional k-algebra is simple if and only if it is
isomorphic with M_{n}(k') for some positive integer n
and some (possibly) skew field k' containing k.

The ``only if'' is proved by taking any finite-dimensional
irreducible representation E of the simple k-algebra A
and identifying A with its ``bicommutant''.
Here E may be taken to be a minimal left ideal.
The ``bicommutant'' is the commutant of the commutant.
The ``commutant'' of any subalgebra A of End_{k}(E)
is {x in End_{k}(E) : ax=xa for all a in A}.
This is a subalgebra, and clearly its commutant in turn contains A.
The hard part is showing that in our case it is no larger than A.
Note that our A is indeed a subalgebra of End_{k}(E):
since A is simple, its representation to End_{k}E
is automatically faithful. It also follows that its commutant
is a skew field K, and thus that the bicommutant is a matrix algebra
over the opposite skew field k'=K^{o}.

In the process of identifying A with its bicommutant,
Serre quotes a result about ``semisimple modules''
that 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 k_{0}
can be obtained by:

- Choosing a field extension k with [k:k
_{0}] finite; - Choosing a
__central__simple algebra A over 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 K^{o}. The addition law comes from the
*tensor product* of k-algebras. The punchline of the second
handout will be that this abelian group, called the __Brauer group__
of k, is none other than the Galois cohomology group
H^{2}(k^{sep}/k,**G**_{m})!

Lemma 4:
Let B,B' be subalgebras of k-algebras A,A'.
Let C,C' be their commutants.
Let A'' be the tensor product of A with A',
and likewise B'' and C''.
Then C'' is the commutant of B'' in A''.

In particular,
the tensor product of two central algebras is again central.

Theorem 3: Let A,A' be finite-dimensional simple k-algebras, at least one of which is central. Then the tensor product of A with A' is simple.

Suppose A' is central. Then it is a matrix algebra over some central skew field K'. It is enough to show that the tensor product of A with K' is simple; then by Wedderburn it is a matrix algebra over some skew field K'', and then so is the tensor product of A with A'. (Matrix algebras over skew fields are known to be simple.) We obtain this from the following more general result (in which neither A nor K' is assumed finite over K):

Theorem 4:
Let K' be a central skew field over k,
and A any associative k-algebra with unit.
Then every two-sided ideal N of the tensor product of A with K'
is generated as a left K'-vector space by its intersection with A.

Here K' operates by x(a.tensor.y)=a.tensor.xy,
and A is imbedded in the tensor product as {a.tensor.1}.

Corollary:
Let A be a central simple k-algebra, of dimension n over k.
Then the tensor product of A with its opposite A^{o}
is isomorphic with M_{n}(k).

Indeed, the left and right actions of A on itself yield a homomorphism
from that tensor product to End_{k}(A). Once we know that the
product is simple, this homomorphism must be injective.
By comparing dimensions we then see that it is bijective,
and are done.

In the proof of Theorem 4, Serre uses the notion of a ``primordial element'' of a vector subspace N of V relative to a basis {eCombining Theorem 4, its corollary above, and general properties of tensor products, we define the Brauer group of an arbitrary (commutative) field k and show that it is abelian (Theorem 5). Serre denotes this group G_{i}} of V. Every vector v in V can be written as a sum of c_{j}e_{j}, with j ranging over a finite subset J=J(v) of the i's. A vector w in N is said to be ``primordial'' if:The key property of this notion is that every vector subspace of V is generated by its primordial elements. This is used in much the same way that we used ``weights'' in our proof of Artin's Lemma; and indeed we could have used primordial vectors there instead of vectors of minimal weight. In general, though, the minimal-weight vectors in a subspace need not generate it, so we cannot conversely use minimal-weight vectors to prove Theorem 4. For example, in the subspace of k

- w is a nonzero vector such that the only vector w' in N with J(w') strictly contained in J(w) is the zero vector; and
- at least one of the coefficients c
_{j}equals 1.^{5}with basis v_{1}=(1,1,0,0,0) and v_{2}=(0,0,1,1,1), the primordial vectors are v_{1}and v_{2}, and the minimal-weight vectors are nonzero multiples of v_{1}.

Theorem 6:
Let A be a central simple k-algebra,
and L a commutative field containing k.
Then the tensor product of A with L is a simple L-algebra.

Note that we do not require that [L:k] be finite.
In particular, if we take L to be an algebraic closure of k,
then the tensor product must be a matrix algebra over L
(since Br(L) is then trivial). Since it has the same dimension
over L as did A over k, we conclude:

Corollary:
dim_{k}(A) is a perfect square
for every central simple k-algebra A.

Theorem 6 follows readily from Wedderburn and Theorem 4.
It yields a homomorphism from Br(k) to Br(L).
If L/k is Galois then this homomorphism can be interpreted
as restriction from
H^{2}(k^{sep}/k,**G**_{m})
to
H^{2}(k^{sep}/L,**G**_{m})^{Gal(L/k)},
and we'll identify its kernel with
H^{2}(Gal(L/k),K^{*}), cf. page 9 of the Tate notes.
In general, the kernel of the map from Br(k) to Br(L) is called Br(L/k)
[Serre calls it H_{k,L}];
it consists of (equivalence classes of)
central simple k-algebras that become matrix algebras when we extend
the field of scalars from k to L. Such algebras are said to admit L
as a ``decomposition field''. Thus k itself is a decomposition field
for A if and only if A is a matrix algebra over k; and we've seen that
an algebraic closure L of k is a decomposition field for every A.

In fact, each A admits a decomposition field K with [K:k] finite. Serre shows this by letting K be generated by the coefficients of the linear map used to decompose A over L. Alternatively, we could do this in stages. Let A be a central division algebra over k. if A is k itself then we are done; else pick an element x of A not in k, and adjoin a root of the minimal polynomial of x. Then x becomes a zero divisor, and A decomposes as a matrix algebra of order at least 2 over a central division algebra of strictly smaller dimension over k. After finitely many such extensions we are done.

Either way, we have obtained Theorem 7: Br(k) is the union of its subgroups Br(K/k) as K ranges over finite extensions of k.

We shall see in the second Serre handout
that it is enough to take K/k separable.
If K' contains K then Br(K'/k) contains Br(K/k).
In particular, if K is separable, we may take K'/k
to be a normal closure of K/k, and conclude that
Br(k) is the union of the subgroups Br(K/k)
as K ranges over finite *Galois* extensions of k.

Here's another source for central simple algebras and the Brauer group (and also for group and Galois cohomology): Milne's notes on class field theory.From Br(k) to H

Construction of the reduced characteristic polynomial, and in particular of the reduced norm of an element of a central simple algebra

Theorem 8 (Skolem-Noether):
Let A be a central simple k-algebra of finite dimension;
and let f,g be k-algebra homomorphisms from some simple algebra B to A.
Then A contains an invertible element x such that
f(b)=xg(b)x^{-1} for all b in B.

Note: Serre writes that f and g are "k-isomorphisms",
but means isomorphisms __to the images__ of B under f,g.
Since B is assumed simple, this holds automatically
for any linear map from B to A that respects the algebra structure
(including the unit element).

In particular, taking B=A we obtain:

Corollary:
Every automorphism of a central simple k-algebra of finite dimension
is inner.

For instance, if K is a skew field of finite dimension over its center k
then Aut_{k}M_{n}(K)=GL_{n}(K)/k^{*}.

Theorem 9:
Let A be a central simple k-algebra of finite dimension,
and B a simple subalgebra with commutant C.

Then C is simple, its commutant is B, and
dim_{k}A = dim_{k}B dim_{k}C.

Corollary 1:
If moreover B is central over k then the intersection of B with C is k
and A is the tensor product of B with C.
Also, C is central.

Corollary 2:
Let L be a commutative subfield of a simple central k-algebra
of finite dimension. Then the following are equivalent:

- L is its own commutant in A;
- dim
_{k}A = [L:k]^{2}; - L is a maximal commutative subring of A.

Theorem 10:
Let W be an element of the Brauer group of a field k,
and L a commutative field containing k with [L:k] finite.
Then L is a decomposition field for W if and only if
W is represented by a central simple algebra A containing L
with dim_{k}A=[L:k]^{2}.

Corollary 1:
Let D be a skew field with center k.
Every maximal commutative subfield of D/k
is a decomposition field for D.

Corollary 2:
Let D be a skew field with center k, and let
dim_{k}D=r^{2}.
Then for each decomposition field L of D,
its degree [L:k] is a multiple of r.

Theorem 11:
Every skew field D finite over its center k contains a maximal
commutative subfield that is separable over k.

Corollary: Every element of Br(k)
has a decomposition field that is a Galois extension of k.

The key step in the proof of Theorem 11 is given by

Lemma:
Every skew field D finite over its center k is either k itself
or contains a commutative subfield M properly containing k.

If not, then each x in D would be ``radical'' over k,
that is, would be a p^{e}-th root of an element of k
for some e=1,2,3,...
Since p^{e}=[k[x]:k], which is bounded above by
dim_{k}(D), there is a single integer e that works for all of D.

Since finite fields are perfect, k can be assumed infinite.
Then the fact that x^{pe} is in k for all x in D
can be translated to polynomial identities that remain true over
an algebraic closure -- including the inseparable closure
that decomposes D. But in a matrix algebra of order greater than 1,
it is not possible for each element to be radical over the scalar
matrices. For instance, such an algebra contains nontrivial
``idempotents'' (elements other than 0 or 1 satisfying x^{2}=x).
So we're done.

We can now map Br(L/k) to H^{2}(Gal(L/k),L^{*})
for any Galois extension L/k of finite dimension.
By Theorem 10, each element of Br(L/k) is represented by
a central simple algebra A of dimension [L:k]^{2},
with an embedding of L.
This representative is unique up to k-isomorphism,
and so is the embedding of L up to conjugation in A^{*}
(Skolem-Noether).
Let E be the group of such conjugations; that is, of
x in A^{*} such that xL=Lx. For each such x
we get a k-homomorphism of L taking any field element c
to xcx^{-1}; this gives a map from E to Gal(L/k).
This map is surjective by Skolem-Noether.
Its kernel is the unit group of the commutant of L.
The commutant is L itself by Theorem 9 (Cor.2).
So, we have an extension of Gal(L/k) by the abelian group L^{*}.
Moreover, the map of Gal(L/k) on L^{*} via conjugation in E
coincides with the Galois action on L^{*}.
Thus we obtain an element of H^{2}(Gal(L/k),L^{*})
with that Galois action.

To check that we actually get an isomorphism with
H^{2}(Gal(L/k),L^{*}),
we must show that the map is a homomorphism
relative to the abelian group laws we have defined on Br(L/k)
and H^{2}(Gal(L/k),L^{*}),
and that each element of H^{2}(Gal(L/k),L^{*})
is actually realized by some central simple algebra,
uniquely determined up to k-isomorphism.
To further show that Br(k) is isomorphic with
H^{2}(Gal(L/k),L^{*}),
we need more than the corollary of Theorem 11,
which shows that each element of Br(k) is in Br(L/k)
for some Galois L/k:
we must also show that different choices of L yield the same element of
H^{2}(k^{sep}/k,**G**_{m}).
Since the compositum of any two Galois extensions of finite dimension
is again Galois of finite dimension, it is enough to check that
our maps to H^{2}(Gal(L/k),L^{*})
commute with inflation between different fields of decomposition L
that are finite Galois extensions of k.

We recover A from E as a ``crossed product'' algebra.
For each g in Gal(L/k), let
I_{g} be the preimage of g under the map from E to Gal(L/k).
This is a coset of L^{*} in E consisting of those x such that
xcx^{-1}=g(c) for all c in L^{*}. Let
N_{g} be the union of I_{g} with {0};
this is an L-vector space of dimension 1.
Now let A_{E} be the direct sum of all the N_{g}'s.
This is an L-vector space of dimension |Gal(L/k)|=[L:k].
We give A_{E} the structure of an associative algebra
by using the multiplicative structure on E and extending linearly
to A_{E}. Then A_{E} is a k-algebra, with unit element
the identity of E (contained in I_{1}).

Theorem 13 [sic]:
A_{E} is a central simple k-algebra.

[Serre's ``Thm.13'' looks different but is tantamount
to the same thing.]

Now there's a natural homomorphism of k-algebras
from A_{E} to our A.
Since A_{E} is simple, this homomorphism is injective.
It is therefore an isomorphism, because A_{E} and A have
the same dimension [L:k]^{2} over k. We have thus
associated to each extension E of Gal(L/k) by L^{*}
a unique central simple algebra A_{E} in the preimage of E
under the map from Br(L/k) to H^{2}(Gal(L/k),L^{*}).

It remains to show that the map is a group homomorphism.
We shall use the description of the group law on
H^{2}(G,A) as ``Baer multiplication''.
Let E,E' be extensions of G by A with the same action of G on A,
corresponding to two classes in H^{2}(G,A).
Then the sum of these two classes is represented by an extension E''
constructed as a quotient (E,E')/Q.
Here (E,E') is the __fiber product__ of E with E'
with respect to the projection maps to G;
that is, the subgroup of E*E' consisting of pairs (e,e')
that map to the same element of G.
This (E,E') is an extension of G by A*A.
Its normal subgroup A*A contains the subgroup
Q={(a,a'):aa'=1}, which is normal in (E,E').
The quotient group is an extension of G by A,
which we may regard again as an element of H^{2}(G,A).
We readily check that it is the sum of the elements corresponding
to E and E'.

Using this description and the construction in Theorem 10, Serre shows:

Lemma:
Our map from Br(L/k) to H^{2}(Gal(L/k),L^{*})
is a homomorphism.

Combining this with Theorem 13, Serre finally obtains:

Theorem 12:
Our map from Br(L/k) to H^{2}(Gal(L/k),L^{*})
is an isomorphism.

Corollaries:

- Br(L/k) is a torsion group, i.e., consists only of elements
of finite order. (Indeed we have shown that
[L:k]H
^{2}(Gal(L/k),L^{*})=0.) - Br(k) is a torsion group. (Use Theorem 11 and its Corollary.)
- If k is a finite field then the groups Br(k) and
H
^{2}(k^{sep}/k,**G**_{m}) are trivial. (Serre proves this using facts about H^{2}(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.)

Further topics II: potential paper topics in group/Galois cohomology and central simple algebrasAn 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 p^{2} then f is irreducible
in **Q**[X].

*Proof sketch*:
By Gauss, any putative factorization f=gh in **Q**[X]
already works in **Z**[X]. We may then reduce it mod p
and use unique factorization of polynomials over the field
**F**_{p}=**Z**/p**Z**
to show that g,h are congruent mod p to powers of X. Evaluating at
X=0 yields p^{2}|P(0), contradiction.)

Example: The polynomial
f(X)=((X+1)^{p}-1)/X) satisfies the Eisenstein criterion,
and hence is irreducible.
Translating X to X-1, we find that the polynomial
(X^{p}-1)/(X-1), whose roots are
the nontrivial p-th roots of unity,
is irreducible in **Q**[X].
It easily follows that the splitting field of this polynomial
is obtained by adjoining one root,
and that its Galois group over **Q**
is the group of units of
**F**_{p}=**Z**/p**Z**,
which is known to be a cyclic group of order p-1.
We shall show more generally that the Galois group of
X^{N}-1 over **Q** is the group
(**Z**/N**Z**)^{*}
of units in **Z**/N**Z**.

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 PGL_{2}(k) is
k(u) where
u=(t^{q2}-t)^{q+1}/(t^{q}-t)^{q2+1},
a rational function of degree q^{3}-q=|PGL_{2}(k)|
[the numerator and denominator have a common factor
(t^{q}-t)^{q+1}] fixed by PGL_{2}(k).
This can be obtained from the ``q-polynomial''
with roots at+b (a,b in k).

The group of order 2^{n}n! 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 H^{1}(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
r^{2}+s^{2}+t^{2}+u^{2}=0,
and thus that c=d=1 yields a division algebra over
**Q**_{2}.

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 S_{n} 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 A_{n}),
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*(phi^{2}*x*, *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*(phi^{2}*x*, *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 g ^{d}
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__(*g ^{d}*) is the image of

*Example*: Every character of the symmetric group
S* _{d}*
takes integer values at all permutations of

We shall see that S* _{d}* actually satisfies
a stronger condition: all its representations are defined
over

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*=S* _{d}* ,
each dim(

A nice example of the hook-length formula (4.12, p.50)
for the dimension of an irreducible representation of
S* _{d}* is the case of a partition of the form
(

The final step in the proof of Lemma 4.25 (p.54), and thus of
Theorem 4.3 (the description of irr.reps. of S* _{d}*
in terms of Young symmetrizers), hinges on the fact that if

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*^{-1}*g'g*^{-1}.
This generalizes the familiar formula
*d*(1/*u*)=-*du*/*u*^{2}
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*:

- exp is in general
*not*a group homomorphism! We shall see from the Campbell-Hausdorff formula that exp is a homomorphism__iff__the connected component of the identity in*G*is abelian__iff__[*X,Y*]=0 for all*X,Y*in__g__. The restriction of exp to any one-dimensional subspace of__g____is__a homomorphism, though - The operation
*X***Y*:= log(exp(*X*)exp(*Y*)) is*not*bilinear -- it's claimed to be bilinear in p.117 of the textbook, but that ``bilinear'' must be a typo for ``binary''. Also, in general*X***Y*is not defined for all*X,Y*in__g__, only for*X,Y*sufficiently small.

In the opening paragraph of the proof of Engel's Theorem 9.9 it is shown that if

In Lie's Theorem 9.11, the hypothesis that the solvable Lie subalgebra of

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(

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

Finally, note that

A nice companion to the result of Exercise 11.22 is the following geometrical description of the map from the exterior square of Sym

Here are some examples of computations concerning invariant polynomial rings (corrected; also in PDF)

Here's Remark 2 on page 21 of Serre's

Here is an example of an application of Theorems 3 and 4 [description of the representations ofg=sl_{2}], independent of the interpretation ofsl_{2}as the Lie algebra of SL_{2}:

LetUbe a compact Kähler variety of complex dimensionn, and letVbe the cohomology algebraH^{*}(V,C). Hodge theory associates endomorphisms Lambda andLofVwith the kählerian structure ofU(cf. A. Weil,Variétés kähleriennes, Chap. IV); let us takeXandYto be these endomorphisms, and defineHby the relationHx=(n-p)xifxis inH(^{p}V,C). Then one can check (Weil,loc. cit.) thatVbecomes ag-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 S

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

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 SL

p5b.ps: Fifth problem set: More about invariant rings and their Hilbert functions; the Weyl character formula for SL