Lectures on Abstract Algebra for Graduate Students

267 Pages·2003·1.11 MB·English

Lectures on Abstract Algebra for Graduate
Students
Alexander Kleshchev

Contents
Introduction page 1
1 Groups 2
1.1 Things known 2
1.2 Cyclic groups 2
1.3 Simplicity of A 3
n
1.4 Isomorphism and Correspondence Theorems 5
1.5 Group Actions and First Applications 8
1.6 Direct and Semidirect Products 15
1.7 Sylow Theorems 18
1.8 Jordan-H¨older Theorem 23
1.9 Solvable and Nilpotent Subgroups 25
1.10 More on simple groups: projective unimodular groups 32
1.11 Generators and Relations 38
1.12 Problems on Groups 47
2 Fields 55
2.1 Things known 55
2.2 More on irreducible polynomials 56
2.3 First steps in fields 58
2.4 Ruler and compass 63
2.5 What is Galois theory? 66
2.6 Normality and separability 67
2.7 The Fundamental Theorem 72
2.8 Galois group of a polynomial 80
2.9 Discriminant 83
2.10 Finite fields 88
2.11 Cyclotomic Polynomials 90
2.12 The theorem of the primitive element 94
2.13 Solution of equations by radicals 95
iii
iv Contents
2.14 Transcendental extensions 100
2.15 Symmetric functions and generic polynomials 104
2.16 The algebraic closure of a field 106
2.17 Problems on Fields 107
3 Modules 122
3.1 Definition and the first properties 122
3.2 Direct sums and products 125
3.3 Simple and Semisimple modules 129
3.4 Finiteness conditions 131
3.5 Jordan-H¨older and Krull-Schmidt 133
3.6 Free modules 135
3.7 Modules over PID’s 137
3.8 Normal forms of a matrix over a field 144
3.9 Algebras and Modules over Algebras 146
3.10 Endomorphism Ring of a Module 147
3.11 The Wedderburn-Artin Theorem 150
3.12 The Jacobson Radical 157
3.13 Artinian Rings 159
3.14 Projective and Injective Modules 161
3.15 Hom and Duality 170
3.16 Tensor Products 174
3.17 Problems on Modules 185
4 Categories and Functors 205
4.1 Categories 205
4.2 Functors 209
4.3 Adjoint functors 214
4.4 Problems on Categories and Functors 217
5 Commutative algebra 219
5.1 Noetherian rings 219
5.2 Rings of Quotients and Localization 225
5.3 Ring extensions 231
5.4 Krull Theorems on Noetherian Rings 239
5.5 Introduction to Algebraic Geometry 246
5.6 Problems on Commutative Algebra 254
Bibliography 264
Introduction
These are lecture notes for a year long graduate course in abstract alge-
bragivenattheUniversityofOregonin2002-2003. ThetextisAdvanced
Modern Algebra by J. Rotman. I will greatly appreciate if you will let
me know of any misprints or errors you can find in these lecture notes.
Thisisadifficultcoursetotakeandtoteach: thereisalot ofmaterial
to cover. As a result, it is difficult to get into things in a deep way, so
the course might sometimes even seem boring.
The homework assignments will be given weekly on Mondays and
collected also on Mondays. Only part of the problems will be graded.
The assignment will usually include sections from the textbook or these
lecture notes to read. This part of the assignment should never be
ignored! Sometimes I might assign sections which were not explained in
class.
The midterm will be on Wednesday of the 6th week of each term,
from 6 to 8:15 p.m., if possible. Final during the finals week according
to schedule.
Iwill assumethatthematerialusuallycoveredin500Algebracourses
has been mastered.
Finally, never fear! I am there to help.
1
1
Groups
1.1 Things known
Throughout the course I will assume some maturity in linear algebra
(vector spaces, bases, linear transformations, bilinear forms, duals, ...)
As far as groups are concerned we assume that sections 2.1-2.5 and
partsofsection2.6inRotmanarewellunderstood,asthesemattersare
covered in detail in 500 Algebra.
Among other things, these sections discuss the following important
topics (which I will skip):
2.2 Symmetricgroupasaninterestingexampleofafinitegroup. You
need to understand symmetric and alternating groups very well,
as these will occur throughout the course as main examples of
finite groups.
2.3 Formal definition of a group and more examples: cyclic groups,
dihedralgroups,generallineargroups,etc. Allofthemshouldbe
your friends, just like symmetric groups.
2.4 Subgroups,cosets,Lagrange’sTheorem,Fermat’sTheorem,Sub-
groups generated by subsets.
2.5 Homomorphisms and automorphisms, kernels, images, conjuga-
tion, normal subgroups, center.
2.6 Quotient groups.
1.2 Cyclic groups
We will review some useful properties of cyclic groups. Most of the
proofsareomitted. ThecyclicgroupofordernisdenotedbyC ,andthe
n
infinitecyclicgroupisdenotedbyC (orZifweuseadditivenotation).
∞
2
1.3 Simplicity of A 3
n
Lemma 1.2.1 Let C = hgi. For each divisor d of n, C has exactly
n n
one subgroup of order d, namely hgn/di={h∈C |hd =1}.
n
Lemma 1.2.2 Let C = hgi. Every automorphism of C has form
n n
α(gi) = gki for a fixed k with (k,n) = 1. Hence Aut(C ) ∼= (Z/nZ)×,
n
the multiplicative group of the residue ring Z/nZ.
Remark 1.2.3 Note that Aut(C ) does not have to be cyclic. What is
n
Aut(C )?
8
Lemma 1.2.4 If an abelian group G has elements of orders k and l,
then it has an element of order LCM(k,l)
Proof Let g and h have orders k and l, respectively. If (k,l) = 1 it
is easy to see that the order of gh is kl. Otherwise let d = (k,l), and
consider the elements g and hl/d.
The following nice result is used very often.
Lemma 1.2.5 Let F be a field, and G be a finite subgroup of the mul-
tiplicative group F×. Then G is cyclic.
Proof Firstobservethat,foreveryd∈Z ,thereareatmostdsolutions
>0
oftheequationxd =1inF×—indeed,thepolynomialxd−1hasatmost
d roots.
Now let g be an element of G having a maximal possible order n.
We claim that G = hgi. Otherwise, pick an element h ∈ G \ hgi.
By the choice of g, the order k of h is at most n. If k = n, then
1,g,g2,...,gn−1,h are n+1 solutions of xn =1, giving a contradiction.
Sowehavek <n. Now,kdividesn,forotherwiseLemma1.2.4yieldsan
element of order LCM(k,n)>n, giving a contradiction. Finally, we get
a contradiction anyway, as now 1,gn/k,g2n/k,...,g(k−1)n/k,h are k+1
solutions of xk =1.
1.3 Simplicity of A
n
Recall that a group is called simple if it has exactly two normal sub-
groups (which then have to be {1} and G itself). Just to start our
course somewhere, in this section we will prove the classical result that
the alternating group A is simple for n ≥ 5. Simple groups are very
n
4 Groups
important in group theory, and A was historically the first example of
n
a simple group.
Lemma 1.3.1 Let n≥5. Then all 3-cycles are conjugate in A .
n
Proof Let i,j,k,l,m,,... be arbitrary (distinct) numbers from
{1,2,...,n}.
Wecanconjugate(1,2,3)to(i,j,k)usingthepermutationσwhichmaps
1 to i, 2 to j, and 3 to k, and leaves 4,5,... invariant. If σ happens to
be odd, then use (l,m)σ instead.
Lemma 1.3.2 Let n≥3. Then A is generated by the 3-cycles.
n
Proof Any element of A is a product of an even number of transposi-
n
tions. Consideraproductoftwotranspositions(i,j)(k,l). Ifallnumbers
i,j,k,l are distinct, then
(i,j)(k,l)=(i,j)(j,k)(j,k)(k,l)=(i,j,k)(j,k,l).
Otherwise the product looks like (i,j)(j,k), which equals (i,j,k). The
lemma follows.
If g,h are two elements of a group G, we write [g,h] for the element
ghg−1h−1, called the commutator of g and h. This terminology comes
fromthefactthatg andhcommuteifandonlyiftheircommutatoris1.
Theorem 1.3.3 Let n≥5. Then A is simple.
n
Proof Let {1} =6 H E A . By the lemmas above, it suffices to show
n
that H contains a 3-cycle. Take σ ∈H \{1}.
Let us suppose first that n = 5. Then either σ = (i,j)(k,l) or σ =
(i,j,k,l,m). In the former case take τ = (i,j)(k,m). Then (m,l,k) =
[τ,σ] ∈ H. In the latter case, take τ = (i,j,k). Then (i,j,l) = [τ,σ] ∈
H. We are done in both cases.
Now, let n = 6. If σ fixes some i ∈ {1,2,...,6}, then σ belongs
to a subgroup A < A permuting the remaining 5 symbols. Thus,
5 6
σ ∈ H ∩ A . Therefore H ∩ A is a non-trivial normal subgroup of
5 5
A . As we already know that A is simple, this implies H ≥ A . In
5 5 5
particular,H containsa3-cycle,andwearedone. Soassumeσ doesnot
fix any element. Then either σ =(i,j)(k,l,m,r) or σ =(i,j,k)(l,m,r).
1.4 Isomorphism and Correspondence Theorems 5
In the first case, 16=σ2 ∈H fixes i (and j), and we are reduced to the
case already considered above. In the second case take τ =(j,k,l). We
have again (k,i,m)(j,l,k)=[σ,τ]∈H fixes r.
Finally, let n ≥ 7. There exist i 6= j with σ(i) = j. Chose a 3-cycle
α which fixes i but moves j. Then ασ 6=σα, as the two elements differ
on i. Hence γ := [α,σ] is a non-trivial element of H. But σα−1σ−1 is
a 3-cycle. So γ is a product of two 3-cycles. Hence it moves at most
6 elements, say, i ,...,i . Let F ∼= A be the alternating group on
1 6 6
{i ,...,i } considered as a subgroup of A . Then γ ∈ H ∩F, whence
1 6 n
H ≥F by simplicity of F, and so H contains a 3-cycle.
Example1.3.4Apermutationoftheset{1,2,...}iscalledfinitary ifit
fixesallbutfinitelymanypoints. DenotebyA thefinitaryalternating
∞
group, i.e. the group of all even finitary permutations. Using the fact
that A = ∪ A , it is easy to see that A is simple, giving us an
∞ n≥1 n ∞
example of an infinite simple group.
1.4 Isomorphism and Correspondence Theorems
Theorem 1.4.1 (First Isomorphism Theorem) If f : G → H is a
homomorphism of groups, then K :=kerf /G, and the map
f¯:G/K →imf, gK 7→f(g)
is an isomorphism.
Proof It is routine to check that K is normal, that f¯is a well-defined
homomorphism(themostimportantpart, somakesureyouunderstand
this), and that f¯is surjective and injective.
Example1.4.2ThecyclicgroupC ofordermisisomorphictoZ/mZ.
m
Example 1.4.3 Let S1 be the group of all complex numbers of abso-
lute value 1. Then considering the map R → S1, x 7→ e2πix, gives an
isomorphism R/Z∼=S1.
Example 1.4.4 The determinant map det : GL (F) → F× yields an
n
isomorphism GL (F)/SL (F)∼=F×.
n n
Example 1.4.5 The sign map S →{±1} shows that S /A ∼=C .
n n n 2
6 Groups
Example 1.4.6 If d|n then C /C ∼=C .
n d n/d
Example 1.4.7 For any group G, we have G/Z(G) ∼= Inn(G), where
Z(G) is the center of G, and Inn(G) is the group of the inner automor-
phisms of G.
Example 1.4.8 Let p be a prime, and C < C× be the group of all
p∞
pnth roots of 1 for all n ≥ 0. Considering the map z 7→ zp yields an
isomorphism C /C ∼= C . It can be proved that actually any non-
p∞ p p∞
trivial quotient of C is isomorphic to C . Another curious property
p∞ p∞
of this group is that any finitely generated subgroup of it is cyclic, even
though it is not cyclic itself.
Example 1.4.9 Let p be a prime Q be a subgroup of (Q,+) which
(p)
consists of all numbers of the form m/pn for m,n∈Z. Considering the
map Q → Z , m/pn 7→ e2πim/pn yields an isomorphism Q /Z ∼=
(p) p∞ (p)
Z .
p∞
Theorem 1.4.10 (Second Isomorphism Theorem) Let H E G,
K ≤G. Then H EHK ≤G, H ∩K EK, and the map
ϕ:K/(H ∩K)→HK/H, k(H ∩K)7→kH
is an isomorphism of groups.
Proof Things to check: (1) HK < G, (2) H E HK, (3) H ∩K E K,
(4) ϕ is a well-defined homomorphism, (5) ϕ is surjective and injective.
All are routine. For example, for (1): given elements h ,h ∈ H and
1 2
k ,k ∈K,
1 2
h k (h k )−1 =h k k−1h−1 =h h k k−1 ∈HK,
1 1 2 2 1 1 2 2 1 3 1 2
where h =k k−1h−1(k k−1)−1 ∈H, as H is normal.
3 1 2 2 1 2