Math 901 Notes
3 January 2012
1 of 33
Math 901 ∼ Fall 2011 ∼ Course Notes
Information: These are class notes for the second year graduate level algebra course at UNL (Math 901) as taken
in class and later typed by Kat Shultis. The notes are from the Fall of 2011, and that semester the course was taught
by Brian Harbourne. During the fall semester we covered the following topics:
B Category Theory, Functors, Natural Transformations
B Group Theory: Free (Abelian) Groups, Group Actions, Semi-Direct Products, Sylow Theorems, Nilpotent Groups,
Solvable Groups
B Field Theory: Algebraic Extensions, Separable Extensions, Separable Degree, Galois Theory, Galois Extensions,
Finite Fields, Transcendence Degree, Transcendence Bases
For each class day, I’ve indicated the topic at the top of that day’s notes.
Disclaimer: I created these notes in order to help me study, and so I’ve expanded proofs where I find it useful,
shortened things I was comfortable with before this course, and changed at least one proof to one that I like better
than the one presented in class. These notes are not meant to be a substitute for your own notes. They have been
proof-read, but are not guaranteed to be without errors. If you find errors, please email Kat at the following address:
s-kshulti1 “at” math.unl.edu
22 August 2011
Categories
B Comment: The purpose of categories and functors is to describe structure that is common in many parts of
mathematics.
B Comment: Intuitively, a category is a structure consisting of objects and arrows (morphisms), and the arrows are
maps between the objects.
B Examples:
Category
Objects
Arrows
Typical Notation
Sets
sets
set maps
S
Groups
groups
group homomorphisms G
Abelian Groups
abelian groups
group homomorphisms Ab
Rings
rings
ring homomorphisms
Vector Spaces
finite dimensional R-vector spaces linear transformations
V
Manifolds
topological manifolds
continuous maps
differentiable manifolds
differentiable maps
Group, G
?
elements of G
B Definition: A category, A, consists of a collection of objects, Ob(A), and for each pair of objects, A, B ∈ Ob(A),
there is a set of morphisms, Mor(A, B), and a composition law defined for morphisms ϕ ∈ Mor(A, B) and
ψ ∈ Mor(B, C) when A, B, C ∈ Ob(A), such that ψ ◦ ϕ ∈ Mor(A, C), which satisfies the following:
1. Composition is associative when defined.
2. For all A ∈ Ob(A), there exists a unique identity morphism, which is a left and right identity whenever composition
is defined. That is, there is a morphism 1A ∈ Mor(A, A) such that if ϕ ∈ Mor(A, B) and ψ ∈ Mor(B, A), then
ϕ ◦ 1A = ϕ and 1A ◦ ψ = ψ.
3. Mor(A, B) ∩ Mor(A0 , B 0 ) = ∅, unless A = A0 and B = B 0 ,1 in which case Mor(A, B) = Mor(A0 , B 0 ).
24 August 2011
More Categories & Universal Properties
B Example: Let X be a topological space. We’ll use TX as the collection of open sets in X. We can turn this
into a category, by setting Ob(X) = TX , and if U, V ∈ TX , we have either Mor(U, V ) = ∅ when U * V or
Mor(U, V ) = {U ⊆ V } when U ⊆ V .
B Definition: A morphism f : A → B in a category C is an isomorphism if and only if there is a morphism
g : B → A such that g ◦ f = idA and f ◦ g = idB .
B Comment: We discussed the idea of direct products, and how a direct product is really just a function. This was a
bit confusing, but the general idea was to generalize n-tuples. I’ll only put it here in the most general context, not in
the more specific cases we discussed.
1 These are actually equalities and not isomorphisms. In the category of groups, for example, my copy of Z is isomorphic, but not
necessarily equal to, Sarah’s copy of Z, and the morphisms from my copy to Caitlyn’s copy are hence fundamentally different from the
morphisms from Sarah’s copy to Caitlyn’s copy.
3 January 2012
Math 901 Notes
2 of 33
B Definition: Let {Ai : Y
i ∈ I} be a family of objects in a category C. We say P ∈ Ob(C) is a product of the objects
Ai , sometimes written
Ai , if P was given together with morphisms πi : P → Ai called projections such that they
i∈I
satisfy
the universal property. That is, given any B ∈ Ob(C) and morphisms fi : B → Ai there is a unique morphism
Y
fi = ϕ ∈ Mor(B, P ) such that the diagram below commutes for each i ∈ I.
i∈I
ϕ
B
P
πi
fi
Ai
B Aside: Suppose P and P 0 are both products, that is we have projections πi : P → Ai and πi0 : P 0 → Ai . Then by
the universal property we have morphisms f : P → P 0 and g : P 0 → P , and these are inverse morphisms because the
diagram below commutes.
f
P0
g
P0
P
πi
πi0
πi0
Ai
In other words πi0 ◦ g ◦ f = πi0 and so we must have that g ◦ f = 1P 0 and similarly 1P = f ◦ g. Hence P and P 0
are isomorphic. However, note that we can actually only use this diagram for one direction, that is we can only
show that g ◦ f = 1P 0 . However, a similar argument will show the other direction. Once things are more formal,
we’ll see that objects satisfying universal properties are unique.2
26 August 2011
Monomorphisms & Epimorphisms
B Review: From last time, we discussed the categorical product, and we’ve discussed the category of a topological
space, X. The product of U and V in this category is their intersection. For homework, we’ll ask more sophisticated
questions about the product.
B Example: Let C be the category whose objects are all the elements in the set R ∩ [0, 1], and Mor(x, y) =
∅,
if x > y
. Let {xi : i ∈ I} be a collection of objects in this category, then the categorical product is
{x ≤ y}, if x ≤ y.
p = inf{xi : i ∈ I}. This is easily seen to be what you want if you do the case of two objects and their product, so
that the product of x and y is min(x, y).
B Definition: Let C be a category and let A, B ∈ Ob(C), with ϕ ∈ Mor(A, B). We say ϕ is a monomorphism if
given any γ, δ ∈ Mor(C, A) with ϕ ◦ γ = ϕ ◦ δ then γ = δ.
B Proposition: In the category of sets, ϕ is a monomorphism if and only if it is injective.
B Definition: (Again, using pictures) A map ϕ is a monomorphism if whenever we have
γ
C
ϕ
A
B
δ
and it commutes, then γ = δ.
B Comment: An epimorphism is the categorical dual of a monomorphism, and also a generalization of a surjective map.
B Definition: A map ϕ is an epimorphism if whenever we have
γ
C
A
B
ϕ
δ
and it commutes, then γ = δ.
2 Also,
the argument is frequently almost identical to this one.
3 January 2012
Math 901 Notes
3 of 33
B Example: Back to the order relations. Here, all morphisms are both monomorphisms and epimorphisms, but only
the identity morphisms are isomorphisms. This indicates that the mono- and epi-morphisms are related to injective
and surjective maps, but do not always act the same.
B Comment: The categorical dual of a product is a coproduct. Also, in most cases, categorical duals can be defined
by simply reversing the directions of the arrows.
29 August 2011
Coproducts; Initial & Terminal Objects
B Comment: We’ll repeat the definition of a product here, so that we can compare it to the dual, the coproduct.
B Definition: Let {Ai : i ∈ I} be a family
Y of objects in a category C.
Y
The product (if it exists) is an object
Ai with a family of morphisms πj :
Ai → Aj such that given any object
i∈I
i∈I
B and family of morphisms pj : B → Aj , there is a unique morphism ϕ : B →
Y
Ai such that the diagram below
i∈I
commutes for each j ∈ I.
ϕ
Y
B
Ai
i∈I
πj
pj
Aj
The coproduct (if it exists) is an object
a
Ai with a family of morphisms π
ej : Aj →
i∈I
a
Ai such that given any
i∈I
object B and family of morphisms pej : Aj → B, there is a unique morphism ϕ :
a
Ai → B such that the diagram
i∈I
below commutes for each j ∈ I.
ϕ
a
B
Ai
i∈I
pej
π
ej
Aj
B Example: The category of abelian groups. In this category, the
B Example: The category of sets. In this example, the coproduct is
more rigorous: Let {Ai : i ∈ I} be our collection of sets. Then set
coproduct is the direct sum.
the disjoint union. Let’s
[ make that concept a bit
ei = {i} × Ai ⊆ I ×
ei ∩ A
ej = ∅
A
Ai , so that A
i∈I
if i 6= j. Then P =
[
ei works and π
A
ej : Aj → P is given by x 7→ (j, x).
i∈I
B Definition: An object, I in a category C is said to be initial (aka universally repelling) if there is a unique
morphism I → A for every object A ∈ Ob(C).
B Definition: An object T in a category C is said to be terminal (aka universally attracting) if there is a unique
morphism A → T for every object A ∈ Ob(C).
B Example: The initial object in the category of abelian groups is the trivial group, I = (0). In the category
corresponding to a topological space, the empty set is an initial object, and the whole space is a terminal object.
B Theorem: If A and B are initial (respectively terminal) objects in a category C, then there is a unique isomorphism
ϕ : A → B.
31 August 2011
Topology Review; (co)Products as (Initial)Terminal Objects
B Comment: We discussed a variety of possible definitions of continuity, and how these arose in history. We also
discussed the idea that topology was in some ways a natural abstraction of other areas of mathematics. We then
proceeded to define a topology (defined using open sets), closed sets, the closure and interior, the topological definition
of continuous, and a basis. We then proceeded to continue our discussion of initial and terminal objects. The purpose
Math 901 Notes
3 January 2012
4 of 33
of this discussion was to introduce ideas that show up in the first homework assignment due Friday. I omit the details
here as I was already comfortable with these definitions.
B Definition: Let C be a category, and let F = {Ai : i ∈ I} be a family of objects in C. We define a new category
CF . In this category, the objects are the objects of C together with a family of morphisms fi ∈ Mor(C, Ai ) for
each i ∈ I. We’ll denote this object as C{fi } . Given two objects in this category C{fi } and D{gi } , a morphism
ϕ ∈ Mor(C{fi } , D{gi } ) is a morphism ϕ ∈ Mor(C, D) in the original category such that the diagram (existing in the
category C) below commutes for all i ∈ I.
ϕ
C
D
gi
fi
Ai
B Comment: The product in C is precisely a terminal object in CF .3
B Definition: Let C be a category, and let F = {Ai : i ∈ I} be a family of objects in C. We define a new category
C F . In this category, the objects are the objects of C together with a family of morphisms fi ∈ Mor(Ai , C) for
each i ∈ I. We’ll denote this object as C {fi } . Given two objects in this category C {fi } and D{gi } , a morphism
ϕ ∈ Mor(D{gi } , C {fi } ) is a morphism ϕ ∈ Mor(D, C) in the original category such that the diagram (existing in the
category C) below commutes for all i ∈ I.
ϕ
C
D
gi
fi
Ai
B Comment: The coproduct in C is precisely an initial object in C F .4
2 September 2011
Fiber Products
B Comment: For intuition, we discuss the fiber product in the category of sets. Given X, Y, Z sets with f : X → Z
and g : Y → Z the fiber product is the product of the fibers, that is, X ×Z Y = {f −1 (z) × g −1 (z) : z ∈ Z}, or more
traditionally though of as X ×Z Y = {(x, y) ∈ X × Y : f (x) = g(y)}.
B Definition: Let A, B ∈ Ob(C) for C some category, and let f ∈ Mor(A, C) and g ∈ Mor(B, C) be morphisms. We
say an object F together with morphisms α ∈ Mor(F, A) and β ∈ Mor(F, B) is a fiber product of f and g if
– the diagram below commutes:
F
α
β
A
B
g
f
C
– and whenever we have morphisms ϕ ∈ Mor(G, A) and ψ ∈ Mor(G, B) such that the diagram below commutes
ϕ
G
ψ
A
B
g
f
C
3 This
4 This
is “obvious” if you go through the definitions of a product and a terminal object.
is “obvious” if you go through the definitions of a product and a terminal object.
Math 901 Notes
3 January 2012
5 of 33
then there is a unique morphism θ ∈ Mor(G, F ) such that the third diagram below commutes.
G
θ
ϕ
F
α
ψ
β
A
B
g
f
C
B Comment: The categorical fiber product of f ∈ Mor(A, C) and g ∈ Mor(B, C) in a category C is just a categorical
product of objects A{f } and B{g} in the category C{C} .5
B Comment: We next started to think about functors, and discussed the idea of a subcategory, but did not complete
the definition of a functor.
B Definition: We say a category C is a subcategory of a category D if Ob(C) ⊆ Ob(D) and if given A, B ∈ Ob(C),
then MorC (A, B) ⊆ MorD (A, B) and if moreover an identity in C is an identity in D, and the composition law in C
agrees with the composition law in D.
B Example: The category of abelian groups is a subcategory of the category of groups. The inclusion of the category
of abelian groups in the category of groups is an example of a covariant functor.
7 September 2011
Functors
B Definition: Let C, D be categories.
A covariant functor F : C → D is an assignment of an object F (A) in D for each object A in C, and for each
morphism ϕ : A → B of objects in C, a morphism F (ϕ) : F (A) → F (B) such that
– F (1A ) = 1F (A) , and
– if we have morphisms f : A → B and g : B → C in C, then F (g ◦ f ) = F (g) ◦ F (f ).
A contravariant functor F : C → D is an assignment of an object F (A) in D for each object A in C, and for each
morphism ϕ : A → B of objects in C, a morphism F (ϕ) : F (B) → F (A) such that
– F (1A ) = 1F (A) , and
– if we have morphisms f : A → B and g : B → C in C, then F (g ◦ f ) = F (f ) ◦ F (g).
B Examples:
– The inclusion of abelian groups into groups is a covariant functor.
– The forgetful functor is a covariant functor. For example, we could have this functor from the category of groups
to the category of sets, where the functor sends a group to its underlying set and a group homomorphism is
sent to itself (as a set map).
B Comment: The next few items here are intended as a review of things necessary to understand the double dual
as a covariant functor, and the dual as a contravariant functor.
B Definition: Given a vector space V , the dual of V is V ∗ = Mor(V, R).
B Definition: Given vector spaces V, W and a linear map f : V → W , we define the dual of the map as f ∗ : W ∗ → V ∗
via f ∗ (ϕ) = ϕ ◦ f where ϕ : W → R is any linear transformation.
B Recall: If V is a real finite dimensional vector space, then Mor(V, R) is as well; moreover, they have the same dimension, so there exists an isomorphism between V and V ∗∗ (and with V ∗ but we care significantly less about that here).
B Definition: Let V be the category whose objects are real finite dimensional vector spaces, with morphisms as linear
transformations.
B Example: We now define the functor F : V → V as follows. Given an object V ∈ Ob(V), then F (V ) = V ∗∗ . Also,
if V, W ∈ Ob(V), with f ∈ Mor(V, W ), then F (f ) is a map from F (V ) to F (W ), that is from V ∗∗ to W ∗∗ given
by F (f ) = (f ∗ )∗ = f ∗∗ .
B Lemma: Given f : V → W and g : W → Z, where V, W, Z are finite dimensional real vector spaces, and f, g are
linear transformations, then (g ◦ f )∗ = f ∗ ◦ g ∗ .
Proof. Let ϕ ∈ Z ∗ , that is ϕ : Z → R. Then it is clear that (g ◦ f )∗ (ϕ) = ϕ ◦ (g ◦ f ). Also, we have that
(f ∗ ◦ g ∗ )(ϕ) = f ∗ (g ∗ (ϕ)) = f ∗ (ϕ ◦ g) = (ϕ ◦ g) ◦ f . Now, as composition of morphisms is associative, we have that
the action of (g ◦ f )∗ and of f ∗ ◦ g ∗ is the same, and hence, they are the same function.
5 See
31 August 2011 for a reminder of what this category is.
Math 901 Notes
3 January 2012
6 of 33
B Proposition: As defined F : V → V is a covariant functor.
Proof. First, we’ll check that F (1V ) = 1F (V ) or equivalently, that (1V )∗∗ = 1V ∗∗ . So let ϕ ∈ V ∗∗ so that
ϕ : Mor(V, R) → R. Then clearly, 1V ∗∗ (ϕ) = ϕ. Also, 1V ∗∗ (ϕ) = ϕ ◦ 1V ∗ . Both of these are maps from
V ∗ to V ∗ , and we want to check they are the same map. So let λ ∈ V ∗ . Then 1V ∗∗ (ϕ)(λ) = ϕ ◦ λ. Also,
1V ∗∗ (ϕ)(λ) = (ϕ ◦ 1V ∗ )(λ) = ϕ(1V ∗ (λ)) = ϕ ◦ λ ◦ 1V = ϕ ◦ λ. Hence, we have F (1V ) = 1F (V ) .
Now, let f : V → W , and g : W → Z where V, W, Z ∈ Ob(V) and f, g are morphisms. Then we’ll check that
F (g ◦ f ) = F (g) ◦ F (f ). From the Lemma, we know that (g ◦ f )∗ = f ∗ ◦ g ∗ . So applying the lemma to f ∗ ◦ g ∗ = (g ◦ f )∗
then gives that F (g ◦ f ) = ((g ◦ f )∗ )∗ = (f ∗ ◦ g ∗ )∗ = g ∗∗ ◦ f ∗∗ = F (g) ◦ F (f ) as required.
9 September 2011
More on Functors
B Last Time: The double dual gave us a covariant functor. Note here that we didn’t need finite dimensional or the
field we were working over to be R to get a covariant functor, but we’ll discuss this functor again later, and then we’ll
make use of those properties.
B Example: We can get a contravariant functor D : V → V by taking duals. On objects, the functor is defined by
D(W ) = W ∗ = {λ : W → R|λ is a linear transformation}. On morphisms, f : V → W , D(f ) = f ∗ known as the
“pull back” morphism and is obtained by composing with f .6 We checked in class that this was a contravariant
functor, but Sarah and I did that work already, and the explanation is already incorporated into the notes for the
previous class (7 Sept 2011).
B Comment: We get the covariant functor from last time by composing D with itself, or equivalently, using “push
forwards.” That is, we can define F : V → V by F (W ) = W ∗∗ and F (f ) = f∗ . We now describe this push forward.
B Definition: Given f : V → W we want to define f∗ : V ∗∗ → W ∗∗ . Let ϕ ∈ V ∗∗ . We have a morphism, ϕ : V ∗ → R
and we know what f ∗ is, and so we define f∗ (ϕ) = ϕ ◦ f ∗ .
B Alternate Viewpoint: Given f : V → W and ϕ : V ∗ → R, we want f∗ (ϕ) to be an element of W ∗∗ , so we should
be able to evaluate f∗ (ϕ) at elements λ ∈ W ∗ . We then define (f∗ (ϕ))(λ) = ϕ(λ ◦ f ).7
B Definition: Let F be a covariant functor F : C → D. Then for any two objects C1 , C2 in C, there is a map
MorC (C1 , C2 ) → MorD (F (C1 ), F (C2 )) which is given by f 7→ F (f ).
– If the map is injective for all C1 , C2 in C, then we say F is faithful.
– If the map is surjective for all C1 , C2 in C, then we say F is full.
– If the map is bijective for all C1 , C2 in C, then we say F is fully faithful.
B Examples:
– The inclusion functor from the category of abelian groups to the category of groups is fully faithful.
– The forgetful functor from the category of groups to the category of sets is faithful but not full.
– The functor from the category of sets to the category with a single object, and single morphism is full, but not
faithful.
12 September 2011
Functor Categories & Representability of Functors
B Motivation: We discussed the motivation for functor categories. The basic idea was to stick in things that we want
to have in our categories, and then maybe they were already there, but we have them now. We then defined two
representation functors.
B Definition: Let A be a category, and let A be an object in A. Then we define a covariant functor, MA from A
to the category of sets, S. On objects, this functor is MA (B) = MorA (A, B), and if f ∈ MorA (C, D), then MA (f )
should be a morphism from MorA (A, C) to MorA (A, D), so we set MA (f ) = f∗ where f∗ is the push forward map,
f∗ (g) = f ◦ g.
B Proposition: The above defines a covariant functor.8
B Definition: Similarly, we define a contravariant functor, M A from A to S. On objects, the functor is M A (B) =
MorA (B, A), and if f ∈ MorA (C, D), then M A (f ) should be a map from MorA (D, A) to MorA (C, A), so we set
M A (f ) = f ∗ where f ∗ is the pull back map, f ∗ (g) = g ◦ f .
B Proposition: The above defines a contravariant functor.9
6 and
was also defined on 7 Sept 2011
not sure how this helps, it may make more sense if we had other examples of a push forward, right now, it seems like a pull back
still and to be honest, I’m a bit confused.
8 The proof is not difficult, but it can be confusing to think about.
9 The proof is not difficult, but it can be confusing to think about.
7 I’m
Math 901 Notes
3 January 2012
7 of 33
B Definition: Let C and D be categories. Also, let F and G be covariant functors from C to D. A natural
transformation from F to G is a rule, Φ which assigns to every object C of C a morphism ΦC ∈ MorD (F (C), G(C))
such that given any morphism f ∈ Mor(C1 , C2 ) in C, the diagram below commutes.10
F (C1 )
F (f )
F (C2 )
ΦC1
G(C1 )
ΦC2
G(f )
G(C2 )
B Fact: Let C and D be categories. Let C D be the collection of covariant functors from C to D. This becomes a category
if given two covariant functors F and G from C to D we define MorC D (F, G) to be the natural transformations from
F to G.11
B Definition: A natural equivalence is a natural transformation which is an isomorphism in the functor category.
B Definition: A covariant functor, F , from a category C to the category S is said to be representable if F is
naturally equivalent to MA for some object A in C.
A contravariant functor, G from C to the category S is representable if G is naturally equivalent to M A for some
object A in C.
B Comment: We’re now ready to discuss the vector space example again. Next time, we’re going to show that the
double dual functor is naturally equivalent to the identity functor on the category V as defined on 7 September 2011.
14 September 2011
Natural Equivalence Example
B Example: We spent all of this day explaining why the identity functor on V, which we’ll denote 1V , is naturally
equivalent to the double dual functor, F (as defined on 7 September 2011).
∗
– Step 1: First we defined an evaluation map eW
w : W → R for any fixed object W in V and any fixed element
w ∈ W , by λ 7→ λ(w). We showed this is a linear map, and hence is an element of W ∗∗ .
– Step 2: We next defined a map eW : W → W ∗∗ for any finite dimensional real vector space, W via w 7→ eW
w .
We again showed12 this is a linear map, and hence an object of MorV (W, W ∗∗ ).
– Step 3: We next defined a natural transformation Φ from 1V to F . For each object W of V we need a morphism
ΦW : 1V (W ) → F (W ). Note that 1V (W ) = W and F (W ) = W ∗∗ . We’ll take ΦW = eW . We must show that
this is natural. That is, given any morphism L : W1 → W2 of real finite dimensional vector spaces, we must show
that the following diagram commutes:
1V (W1 )
L
eW1
F (W1 )
1V (W2 )
eW2
L∗∗ = L∗
F (W2 )
We’ll rewrite the diagram with “easier” notation:
W1
L
eW1
W1∗∗
W2
eW2
L∗∗ = L∗
W2∗∗
We checked that the diagram commutes in class, and it was not difficult.
B Aside: For a natural transformation Φ to be a natural equivalence, we just need for each object W , that ΦW is an
isomorphism. We can then define an inverse natural transformation Ψ as ΨW = (ΦW )−1 and then we just need to
check Ψ is a a natural transformation and that it is an inverse to Φ. However, I believe Brian Harbourne said this will
ALWAYS happen if each ΦW is an isomorphism.
10 You
can similarly define a natural transformation of contravariant functors by changing the direction of the horizontal arrows.
believe that we call the category C D the functor category.
12 We actually just claimed this, and didn’t show it; but it took me all of 60 seconds to write it down including finding a working marker.
11 I
3 January 2012
Math 901 Notes
8 of 33
B Example: Back to our example, it is enough then to show that eW is an isomorphism for each W . We’ll assume
without proof that dim(W ) = dim(W ∗ ) = dim(W ∗∗ ). However, this fails unless dim(W ) < ∞. We’ll also use the
fact that if dim(W ) = dim(W ∗∗ ) and eW is injective, then eW is an isomorphism.13 We then made a clever choice in
order to show that ker(eW ) = 0.
B Comment: “People” often say that W and W ∗∗ are canonically isomorphic. They’re referring to the isomorphism
eW because it defines a natural equivalence from 1V to F .
16 September 2011
Examples of Representability
B Example: Let F be the forgetful functor from V to S. Note that F is a covariant functor. We showed that F
is representable. That is, there is a vector space U such that F is naturally equivalent to MU = MorV (U, −). Or
restated again, we showed there is a vector space U such that for all objects W in V, there is a natural bijection
between W = F (W ) and MU (W ) = MorV (U, W ). In fact, we can say that R represents F , i.e., that F is naturally
equivalent to MR . We define the natural transformation Φ : MR → F via ΦW : MorV (R, W ) → W is given by
L 7→ L(1). It’ll be a homework problem to actually check that ΦW is a bijection for all objects W in V and also that
Φ is natural.
B Example: Now, let W1 and W2 be fixed objects in V. We’ll define F to be the contravariant functor from V to
S which is given by F (W ) = MorV (W, W1 ) × MorV (W, W2 ), and F (f ) will be given by (λ01 , λ02 ) 7→ (λ01 ◦ f, λ02 ◦ f ).
This functor is also representable. In this case, it is represented by U = W1 × W2 . Again, we didn’t actually show the
natural transformation, and I don’t feel like typing the details of the setup.
B Example: Let F = {Wi : i ∈ I} be a family of objects in V. We define
a contravariant functor F from V to
Q
S similarly to the previous example. That is, on objects, F (W ) = i∈I MorV (W, Wi ) and given a morphism
L : W → W 0 , we define F (f ) to be the map from F (W 0 ) to F (W ) which is given by
Y
Y
λ0i : i ∈ I 7−→
λ0i ◦ L : i ∈ I.
i∈I
i∈I
P
However, this functor is not representable
if
i∈I dim Wi = ∞, as the vector space we would want it to be
Q
representable by (namely U = i∈I Wi is not an object in V).
B Comment: Grothendieck had a solution to this issue of non-representability. The solution basically boils down to
embedding the category into a larger category where the object you want exists. We’ll start explaining this with the
Yoneda Lemma next time.
19 September 2011
Yoneda Lemma
S
B Yoneda Lemma Setup: Let C be a category. We can define a covariant functor F from C to C where C S is the
category of contravariant functors C to S.14 On objects, we define F (C) = M C .15 Given a morphism f ∈ MorC (C, D)
we define F (f ) to be the natural transformation F (f ) : M C → M D . In particular, if B is an arbitrary object in C,
then we have (F (f ))(B) : M C (B) → M D (B) or equivalently (F (f ))(B) : MorC (B, C) → MorC (B, D) is given by
g 7→ f ◦ g.
B Yoneda Lemma: F as defined above is a fully faithful functor.
Proof. We need to check four things here. They are:
1. F (f ) is a natural transformation:
Let A, B be objects in C with a morphism g : B → A. We must show the following diagram commutes:
M C (A)
M C (g)
F (f )(A)
M D (A)
13 This
14 See
M C (B)
F (f )(B)
M D (g)
M D (B)
is because the dimensions are equal, so a basis must map to a basis to get injectivity.
Fact on 12 September.
that M C is a contravariant functor.
15 Recall
3 January 2012
Math 901 Notes
9 of 33
So let ϕ ∈ M C (A) = MorC (A, C). Then we have (F (f )(B) ◦ M C (g))(ϕ) = F (f )(B)(ϕ ◦ g) = f ◦ ϕ ◦ g. Similarly,
we get (M D (g) ◦ F (f )(A))(ϕ) = M D (g)(f ◦ ϕ) = f ◦ ϕ ◦ g, so that the diagram commutes. However, this is
“obvious” as M C (g) and M D (g) act by composing with g on the right, and F (f )(A) and F (f )(B) act by composing
with f on the left, and composition is associative.
2. F is a covariant functor: Let C be an object in C, and let 1C be the identity morphism on C in C. Then,
F (1C ) is a natural transformation from M C to M C . So let B ∈ Ob(C), then F (1C )(B) : M C (B) → M C (B) is
given by g 7→ g ◦ 1C = g. Thus, F (1C ) = 1F (C) . We now need to look at compositions. So let f ∈ MorC (A, B) and
g ∈ MorC (B, C). Then we want to show that F (g ◦ f ) = F (g) ◦ F (f ). For any object D in the category C, then
F (g◦f )(D) : M A (D) → M C (D) is given by composition with g◦f on the left, so we get that h 7→ (g◦f )◦h. Similarly,
F (f )(D) : M B (D) → M C (D) is given by composition with f on the left, so h 7→ f ◦ h, and F (g)(D) : M A (D) →
M B (D) is given by composition with g on the left, so h 7→ g ◦ h. Now, if we have h ∈ M A (D), then (F (g) ◦
F (f ))(h) = F (g)(f ◦h) = g◦(f ◦h). However, composition is associative, and so we have that F (g◦f ) = F (g)◦F (f ).
3. F is faithful: Let C, D be objects in C and consider the map MorC (C, D) → MorC S (M C , M D ) which is given
by f 7→ F (f ). We must show that this map is injective. So let g, h ∈ MorC (C, D) such that g 6= h. We must
show that F (g) 6= F (h). We’ll show this on C. So consider F (g)(C), F (h)(C) : M C (C) → M D (C). Then,
F (g)(C)(1C ) = g ◦ 1C = g 6= h = h ◦ 1C = F (h)(C)(1C ), so that F (g)(C) 6= F (h)(C) and hence F (g) 6= F (h).
4. F is full: Again, Let C, D be objects in C and consider the map MorC (C, D) → MorC S (M C , M D ) which is given by
f 7→ F (f ). We must show that this map is surjective. So let η ∈ MorC S (M C , M D ). We’ll set α = η(C)(1C ). Note
first that η(C) : M C (C) → M D (C) so that α ∈ MorC (C, D). We now claim that F (α) = η. To show this, let A be
an object in C and ϕ ∈ MorC (A, C). Then as η is a natural transformation, we know the following diagram commutes:
M C (C)
η(C)
M D (C)
ϕ∗ = M C (ϕ)
M C (A)
M D (ϕ) = ϕ∗
η(A)
M D (A)
Now, tracing the identity morphism on C will give us what we want. Note here that the red equality is due to the definition of α, the blue equality is due to the diagram commuting, and the green equality is due to the definition of F (α).
idC
idC ◦ ϕ = ϕ
η(C)(idC )= α
η(A)(ϕ)= α ◦ ϕ =F (α)(A)(ϕ)
Thus the diagram tells us that η(A)(ϕ) = F (α)(A)(ϕ) and so we have that η = F (α) as desired so that the map
is surjective.
21 September 2011
Representability and Free Groups
B Example: Let X be a set. We’ll define a covariant functor FX from Ab to S. On objects, FX (A) = MorS (X, A)
and on morphisms, if g : A → B, then FX (g) : F (A) → F (B) is given by ϕ 7→ g ◦ ϕ. We wonder now if FX is
representable. It is, and I’ll show how we got there in class.
Proof. We started by assuming that FX is representable and seeing what that meant about the functor and about
the object that represents it.
– So we’ll assume that B represents FX , i.e. that FX is naturally equivalent to MB . Let Φ : MB → FX be the
natural transformation.
– Then in particular we’d have ΦB : MB (B) → FX (B) which is given by 1B 7→ ΦB (1B ) = ϕ, where ϕ is simply
defined to be the image of ΦB applied to the identity on B.
– However, we really want to know what Φ is in general. So let D be an abelian group, and g : B → D a group
homomorphism. We create the following commutative diagram to help us figure out what ΦD (B) is.
3 January 2012
Math 901 Notes
10 of 33
MB (g) = g∗
MB (B)
ΦB
MB (D)
ΦD
FX (g) = g∗
FX (B)
FX (D)
– Tracing the element 1B tells us what we need, namely, that ΦD (g) = g ◦ ϕ = g∗ (ϕ). See the below diagram where
the red equality is what will make the diagram commute.
g∗ (1B ) = g
1B
ϕ
g∗ (ϕ) = g ◦ ϕ = ΦD (g)
– Claim: The pair (B, ϕ : X → B) satisfies the following universal property: for any abelian group D and any
set map ψ : X → D there is a unique homomorphism hψ : B → D such that the following diagram commutes:
hψ
B
D
ϕ
ψ
X
Sub-Proof. This proof was fairly simple and was based entirely on the fact that we’re assuming that Φ is a natural
equivalence, i.e., that ΦD is an isomorphism of sets (which is just a bijection).X
B Comment: The pair (B, ϕ) exists, and is referred to as the free abelian group on the set X.
B Comment: The above shows that the free abelian group on the set X represents the functor FX .
B Construction/Definition: A much more explicit construction
M of a free abelian group can also be given. Here,
we did this as the free abelian group on the set X is B =
Z, with the map ϕ : X → B. More generally, if
x∈X
{Bx : x ∈ X} is a family of abelian groups such that Bx = Z for all x ∈ X, then B =
M
Bx and ϕ : X → B is
x∈X
1, x = y
0, x 6= y.
With all this, we denote the canonical inclusions by π
ex : Bx → B.
B Claim: We claim now that B is a free abelian group, i.e. that it satisfies the appropriate universal property. The
idea is to either:
– Show directly that (B, ϕ) satisfies the universal property OR
– Use the universal property of B as a categorical coproduct (sum) to show that (B, ϕ) satisfies the universal
property of a free abelian group.
given by x 7→ ex where ex : X → Z is the map ex (y) = δxy =
23 September 2011
Free Groups & Coproducts
B Claim: (B, ϕ) as defined in the last class is a free abelian group, that is, it satisfies the following universal property:
given any set map Ψ : X → D, where D is an abelian group, there is a unique homomorphism hΨ : B → D such that
the following diagram commutes:
hΨ
B
D
ϕ
Ψ
X
Proof. We did most of the details of this in class. However, we used the fact that B is a categorical coproduct,
and that basically gave us what we wanted, so I’m going to not rewrite down the details since neither existence
nor uniqueness seemed at all tricky.
Math 901 Notes
3 January 2012
11 of 33
B Definition: Let X be a set, F a group, and ϕ : X → F a set map. We say (F, ϕ) is a free group on the set X
if for every group G and every map Ψ : X → G, there exists a unique homomorphism hΨ : F → G such that the
diagram below commutes.
hΨ
F
G
ϕ
Ψ
X
B Fact: Coproducts and free groups exist in the category of groups.
B Proposition: Let A and B be groups with A ∩ B = ∅ (for simplicity). Define the free product A ◦ B = A ∗ B to be
the set of sequences (a1 , . . . , am ) for m ≥ 0 such that ai ∈ (A \ {1A }) ∪ (B \ {1B }) and such that if ai ∈ A, then
ai+1 ∈
/ A and if ai ∈ B then ai+1 ∈
/ B. Then A ◦ B is a group under the following:
– 1A◦B is the unique sequence of length 0, namely (),
−1
– (a1 , . . . , am )−1 = (a−1
and
m , . . . , a1 ), 
if am , b1 are in different groups, A, B
(a1 , . . . , am , b1 , . . . , bn )
– ab = (a1 , . . . , am ) · (b1 , . . . , bn ) = (a1 , . . . , am b1 , . . . , bn )
if am , b1 ∈ A (or B), and am b1 6= 1A or 1B .

(a1 , . . . , am−1 ) · (b2 , . . . , bn ) if am b1 = 1A or 1B
Proof. We discussed the ideas behind this proof in the next class, but I don’t think that any of them are really worth
typing up. Basically, these things make sense. The only difficulty arose in showing associativity,16 and a sort of
clever induction argument made it not so bad.
26 September 2011
More on Coproducts & Free Groups
B Comments:
– A◦B =B◦A
– We have the obvious injective homomorphisms:17 π
eA : A → A ◦ B, and π
eB : B → A ◦ B where x 7→ (x) for
x ∈ A (or x ∈ B as is appropriate), and obviously in each case the identity maps to the identity in A ◦ B which
is the sequence of length 0.
– Given any group homomorphisms, ϕ∗ : ∗ → G for ∗ = A, B, then we claim there is a unique18 group homomorphism
ϕ : A ◦ B → G such that the following diagrams commute for ∗ = A, B:
ϕ
A◦B
G
π
e∗
ϕ∗
∗
In fact, we define ϕ as follows: Clearly we set ϕ(()) = eG . Also, ϕ((x)) = ϕ∗ (x) where x ∈ ∗ and ∗ = A, B, and
in general, if x = (x1 , . . . , xn ), then ϕ(x) = ϕ((x1 )) · . . . · ϕ((xn )). There are a bunch of cases to check it is a
homomorphism, but it seems clear enough without checking them. Also, the fact that the diagrams commute
seems clear, so I won’t write down any of the details of either.19
B Corollary: Let A and B be groups, which we regard as having disjoint sets of elements. Then (A ◦ B, π
eA , π
eB ) is a
coproduct of A and B in the category of groups.
Proof. It only remains to show uniqueness of the map ϕ. However, this is easy to show since the diagrams must
commute and ϕ must be a homomorphism, so I’ll omit the details from these typed notes.
B Examples: The coproduct of A = Z/2Z = {0, 1A } and B = Z/2Z = {1B , −1} contains the elements: A ◦ B =
{(), (1A ), (−1), (1A , −1), (−1, 1A ), (1A , −1, 1A ), (−1, 1A , −1), . . .}. Note here that there are exactly two elements of
each length. However, in the category of abelian groups, the coproduct is the direct sum, namely A⊕B = Z/2Z⊕Z/2Z
which is the Klein Vier Group.20
B Lemma: Let X1 , and X2 be disjoint sets, and assume (Fi , ϕi ) is a free group on Xi for i = 1, 2. Then F1 ◦ F2 is the
free group on the set X = X1 ∪ X2 with respect to the map ϕ : X → F1 ◦ F2 defined by x 7→ π
ei (ϕi (x)).
16 as
always!
they are injective and homomorphisms is “clear.”
18 “I haven’t asserted uniqueness, but it will be unique. It’s the freeness that makes it unique.” –B.H.
19 I’m aware that this silly ∗ notation sucks, but I’m lazy and don’t want to do everything twice for both A and B.
20 Vier is 4 in German.
17 that
3 January 2012
Math 901 Notes
12 of 33
28 September 2011
Free (Abelian) Groups
B Lemma: If X1 , X2 are disjoint sets, and (Fi , ϕi ) is a free group on Xi for i = 1, 2, then F1 ◦ F2 is a free group on
X = X1 ∪ X2 .
Proof. Let π
ei : Fi → F1 ◦ F2 be the coproduct canonical homomorphisms, and define ϕ : X → F1 ◦ F2 via
x 7→ π
ei (ϕi (x)) for x ∈ Xi . We need to show that (F1 ◦ F2 , ϕ) is a free group on X. So let ψ : X → G be any set
map to any group G. We must show there is a unique homomorphism hψ : F1 ◦ F2 → G such that hψ ◦ ϕ = ψ.
In order to work the coproduct into the proof, we’ll need homomorphisms Γi : Fi → G. Let Γi be the unique
homomorphism guaranteed by the universal property of Fi free from the set map γi : Xi → G, where γi = ψ|Xi .
The maps Γi then induce the map Γ : F1 ◦ F2 → G since F1 ◦ F2 is a coproduct. Set hψ = Γ. We must show that
hψ ◦ ϕ = ψ and that this map is unique. So, let x ∈ X = X1 ∪ X2 . Without loss of generality (WLOG), let x ∈ X1 .
Then,
(hψ ◦ ϕ)(x) = Γ ◦ π
ei ◦ ϕi (x) = Γ1 ◦ ϕ1 (x) = γ1 (x) = ψ|X1 (x) = ψ(x).21
B Examples:
1. The free group on the empty set is the trivial group.
2. The free group on X = {1} is isomorphic to Z.
3. Inductively, using the lemma, we have the the free group on a set X = {1, . . . , n} is Z
. . ◦ Z}.
| ◦ .{z
n times
4. It is possible to define a free group on an infinite set as well, and Lang describes this. It is similar in idea to
what we’ve done so far, but is a bit messier.22
B Lemma 1: Let (F, ϕ : X → F ) be a free group. Then ϕ is injective, and ϕ(X) generates F .
Proof. Notice that
the injectivity is trivial if |X| < 2, so we may let x1 , x2 be distinct elements of X. Define ψ : X →
1, x 6= x1
Z/2Z as ψ(x) =
. Then by the universal property, there is a unique homomorphism hψ : F → Z/2Z
0, x = x1
such that hψ ◦ ϕ = ψ. However, ψ(x1 ) 6= ψ(x2 ), so we must have that ϕ(x1 ) 6= ϕ(x2 ) due to the fact that hψ ◦ ϕ = ψ.
Next, let G be the subgroup of F generated by ϕ(X). We’ll show G = F . Consider the diagram below:
F
hϕe
ϕ
G
ϕ
e
i
F
ϕ
X
Notice that by definition of F as a free group, there is a unique homomorphism hϕ : F → F such that hϕ ◦ ϕ = ϕ. As
1F works for hϕ , then we must have that 1F = i ◦ hϕe. Since 1F is a bijection, then we must have that i is surjective,
meaning that F ⊆ G, and hence F = G.
B Lemma 2: Let (A, ϕ : X → A) be a free abelian group. Then ϕ is injective and ϕ(X) generates A.
Proof. The proof is the same as the proof of Lemma 1, mutatis mutandis.23
30 September 2011
B Example: If X is a set, then A =
Free (Abelian) Groups
M
Z together with e : X → A given by x 7→ ex is a free abelian group, where
x∈X
ex : X → Z is given by ex (y) = 0 for y 6= x, and ex (x) = 1.
B Lemma
X 3: Let (A, ϕ : X → A) be a free abelian group on X. Then every element a ∈ A has a unique expression
a=
mx ϕ(x), where mx ∈ Z.
x∈X
21 This
is “obvious” once you draw the pictures, but there are too many to be worth putting in here.
22 obviously!
23 See
Hungerford page xv. This is latin for (roughly) “by changing the things which (obviously) must be changed (in order that the
argument will carry over and make sense in the present situation).”
3 January 2012
Math 901 Notes
13 of 33
Proof. Notice that existence here is due to!Lemma 2 since ϕ(X) generates A as an abelian group. So we really only
M
Z, e is a free abelian group, then we have homomorphisms he and hϕ such
need to show uniqueness. Since
x∈X
that he ◦ ϕ = e and hϕ ◦ e = ϕ. Now, suppose
X
mx ϕ(x) =
x∈X
X
mx ex =
X
mx he (ϕ(x)) = he
X
X
nx ϕ(x). We’ll apply he to get the following:
x∈X
X
X
X
mx ϕ(x) = he
nx ϕ(x) =
nx he (ϕ(x)) =
nx ex .
We can apply these functions to any element y ∈ X to get that
X
X
my =
mx ex (y) =
nx ex (y) = ny
for all y ∈ X. Hence, the two expansions are equivalent, which gives us uniqueness.
B Definition: A subset, B, of an abelian group, A, is a basis for A if B generates A in a unique way.
B Definition: If B is a basis for A, then we say |B| is the rank of A.
B Lemma 4: If (F, ψ : X → F ) is a free abelian group and we have a commutative diagram:
'
F
G
ϕ
ψ
X
then (G, ϕ) is free abelian on X as well.
B Proposition: If B is a basis for an abelian group A, then (A, ϕ) is a free abelian group where ϕ : B ,→ A is the
inclusion of B into A.
Proof. Let (F, ψ) be a free abelian group on the set B. Then we have the induced commutative diagram:
hϕ
F
A
ϕ
ψ
B
We now claim that in fact hϕ is an isomorphism. We’ll first do surjectivity. As B = ϕ(B) generates A, and
ϕ(B) = ψ(hϕ (B)), then hϕ (F ) contains B, so that hϕ (F ) =X
A since B is a basis for A. Next, we’ll do injectivity.
Let f ∈ ker(hϕ ). Since ψ(B) generates F , then we get f =
mb ψ(b) for some mb ∈ Z. Now,
b∈B
0 = hϕ (f ) = hϕ
X
X
X
mb ψ(b) =
mb hϕ (ψ(b)) =
mb ϕ(b),
which means that mb = 0 for all b ∈ B, and hence f = 0, so that hϕ is injective. Applying Lemma 4 now completes
the proof.
B Lemma 5: If we have a free abelian group (A, ϕ : X → A) and a set Y with a bijection β : X → Y , then (A, ϕ ◦ β −1 )
is a free abelian group on Y .24
B Corollary: Let A be an abelian group. Let X be a set with a map ϕ : X → A. then (A, ϕ) is a free abelian group
if and only if ϕ(X) is a basis for A and ϕ is injective.25
B Fact: If X is an infinite set, and Fin(X) is the set of finite subsets of X, then |X| = | Fin(X)|.
3 October 2011
Free Groups, Group Actions
B Theorem: Let (Ai , ϕi : Xi → Ai ) be free abelian groups for i = 1, 2. Then A1 ∼
= A2 if and only if |X1 | = |X2 |.
24 The
25 The
proof here is “easy” if you draw the pictures.
proof here is an application of the lemmas and the proposition.
3 January 2012
Math 901 Notes
14 of 33
Proof. (⇐) Let ψ : X1 → X2 be a bijection. Since (A2 , ϕ2 ) is free abelian on X2 , then (A2 , ϕ2 ◦ ψ) is free abelian
on X1 by Lemma 5. Thus, A1 ∼
= A2 by the universal property of their both being free on X1 .
(⇒) Now, assume A1 ∼
= A2 , and let h : A1 → A2 be an isomorphism. Set 2Ai = {2a|a ∈ Ai } and note that 2Ai C Ai .
Set Ai = Ai /2Ai . We claim now that h induces an isomorphism h : A1 → A2 . So set h(a) = h(a). We first show this
is well defined. Suppose a = b ∈ A1 . Then a − b ∈ 2A1 . Note that h(2A1 ) = 2h(A1 ) ⊆ 2A2 so that h(a) − h(b) ∈ 2A2
which gives h(a) = h(b). Next, we must show that h is an isomorphism. In fact, we have the following commutative
diagram:
A1
h
A2
−
−
A1
◦h
−
A2
h
Since h and − are surjective, then so is − ◦ h. We claim then that ker(− ◦h) = 2A1 and apply the first isomorphism
theorem.26 Note that (− ◦h)(2a) = 2h(a) = 0 in A2 so that 2A1 ⊆ ker(− ◦h). Now, suppose (− ◦h)(a) = 0. Then
we have h(a) = 0 so that h(a) ∈ 2A2 . So let h(a) = 2b for some b ∈ A2 . But h is an isomorphism, so there exists
some β ∈ A1 such that h(β) = b. Thus, h(a) = 2h(β) = h(2β) so that a = 2β ∈ 2A1 since h is an isomorphism.
Thus, h is an isomorphism.
Now, for i = 1, 2 we have the following:
M
M
Z
Z
M
x∈Xi
x∈Xi
∼
∼
∼
Z/2Z,
Ai = M = M
=
2Z x∈Xi
2
Z
x∈Xi
which gives the isomorphism
M
Z/2Z ∼
=
x∈X1
M
x∈Xi
Z/2Z. Now, if |X1 | < ∞, then we get
x∈X2
2|X1 |
M
M
=
Z/2Z = Z/2Z = 2|X2 |
x∈X1
x∈X2
which means that |X1 | = |X2 |. If, on the other hand, X1 (and hence also X2 ) is infinite, then we have
M
M
|X1 | = | Fin(X1 )| = Z/2Z = Z/2Z = | Fin(X2 )| = |X2 |
x∈X1
x∈X2
which completes the proof.
B Comment: We’re now going to move on from free groups and start discussing group actions. We’ll also briefly cover
the Sylow Theorems, and will probably cover semi-direct products.
B Definition: Let G be a group and X a set. An action of G on X is a map α : G × X → X (usually we write g · x
or simply gx for the action of g on the set element x) such that
1. 1G · x = x for all x ∈ X, and
2. g1 · (g2 · x) = (g1 g2 ) · x for all x ∈ X and for all g1 , g2 ∈ G.
B Definition: We say the orbit of an element x ∈ X is the set G · x = {g · x|g ∈ G}.
B Definition: Let x ∈ X. Then the stabilizer of x is StabG (x) = {g ∈ G|g · x = x}. Notice here that the stabilizer
is a subgroup of G.
B Comment: Sometimes the notation Gx is used for the stabilizer of x, but that notation sucks given our notation for
the orbits!
B Definition: We say an action is transitive if it has a single orbit. Equivalently, if for some x ∈ X (but really it
will be true for all x ∈ X), we have G · x = X.
26 Apparently
the isomorphism theorems are due to Emmy Noether?
3 January 2012
Math 901 Notes
15 of 33
B Comment: Giving an action of G on X is equivalent to giving a homomorphism from G to the group of permutations
on the set X. Note here that this group is really the group of bijections X → X where the group law is composition.
We’ll call this group of permutations Perms(X).
5 October 2011
More Group Actions; Starting Semi-Direct Products
B Recall: We have two equivalent definitions of an action being transitive. The first is that there is a unique orbit.
The second is that given any two elements x, y ∈ X, there is a group element g ∈ G such that g · x = y.
B Example: We discussed the action of S 1 on C where S 1 acts on C via regular multiplication in C. Here, this action
is not transitive, but each orbit is a circle centered about the origin.
B Theorem: (In particular a version of Cayley’s Theorem) Let ϕ : G × X → X be an action of a group G on a set
X, let x ∈ X, and H = StabG (x).
1. Then ϕ induces a homomorphism Φ : G → Perms(X) which is given\by g 7→ πg where πg (y) = g · y.
2. If the action is transitive, then ker Φ ⊆ H and moreover, ker Φ =
gHg −1 and lastly, ker Φ contains every
g∈G
subgroup N ≤ G such that N C G and N ⊆ H.
Proof. We’ll show each part:
1. We first need to show that πg is actually a permutation for every g ∈ G. So note that given g ∈ G, then
(πg ◦ πg−1 )(x) = πg (g −1 · x) = g(g −1 x) = eG x = x and (πg−1 ◦ πg )(x) = g −1 (gx) = eG x = x so that πg is a
bijection for each g and hence an element of Perms(X).
Next, we need to show that Φ is a homomorphism. That is, we need to show that πg1 g2 = Φ(g1 g2 ) =
Φ(g1 ) ◦ Φ(g2 ) = πg1 ◦ πg2 . So let x ∈ X. We’ll show that both permutations send x to the same place. Now,
πg1 g2 (x) = (g1 g2 )x = g1 (g2 x) = πg1 (g2 x) = πg1 (πg2 (x)) = πg1 ◦ πg2 (x). Thus, Φ is a homomorphism.
2. First, we’ll show that ker Φ ⊆ H. So let a ∈ ker Φ. Then πa = 1X so that πa (y) = a · y = y for any y ∈ X. Hence,
a · x = x so that a ∈ StabG (x) = H.\
Next, we need to show that ker Φ ⊆
gHg −1 . So let a ∈ ker Φ and let g ∈ G. It is clearly enough to show that
g∈G
a ∈ gHg −1 . If a ∈ ker(Φ) then a(gx) = gx so\
that g −1 agx = x. this\
means that g −1 ag ∈ H and so a ∈ gHg −1 .
−1
Next, we’ll show the other inclusion ker Φ ⊇
gHg . So let b ∈
gHg −1 , and let y ∈ X. Then y = γx for
g∈G
g∈G
some γ ∈ G since the action is transitive. Also, b = γhγ −1 for some h ∈ H by choice of b. Thus, multiplying both
sides of these two equations together gives by = γhγ −1 γx. Simplifying the right hand side gives by = γhx = γx = y.
Thus, Φ(b) = πb = 1X so that b ∈ ker(Φ).
Lastly, we must show that ker(Φ) is the largest normal subgroup of G which \
is contained in H. So let N C G with
N ⊆ H, and choose any g ∈ G. Then N = gN g −1 ⊆ gHg −1 . Thus, N ⊆
gHg −1 ⊆ ker(Φ) which completes
g∈G
the proof.
B Corollary: If a group G acts transitively on a finite set X and if |G| > n! where n = |X| > 1, then G has a
nontrivial normal subgroup, N .
Proof. We didn’t prove this in class, but I’ll do it now. Since |G| > |X|! = | Perms(X)|, then the homomorphism
Φ from part 1 of the theorem cannot be injective. This means that ker(Φ) {1G }. By part 2 of the theorem above,
we know that H = StabG (x) contains ker(Φ) for any x ∈ X. Now, since the action is transitive, and X has more
than one element, then there must be at least one g ∈ G such that gx 6= x. Hence, H G, so that ker(Φ) is a proper
subgroup of G as we now have {1G } ker(Φ) ⊆ H G. Since the kernel of any group homomorphism is a normal
subgroup, then we have that ker(Φ) is a non-trivial normal subgroup.
B Semi-Direct Products: We can do this in a more general setting, but we’ll give an example first to illustrate the
idea.
B Example: Let G be a group, with subgroups N, H such that N ∩ H = {1G } and N C G. Then, N H = {nh : n ∈
N, h ∈ H} is a subgroup of G. We let H act on N by conjugation, and can refer to N H as an internal semi-direct
product.
3 January 2012
Math 901 Notes
16 of 33
7 October 2011
Semi-Direct Products & Back to Group Actions
B Definition: Let N, H be groups and Φ : H → Aut(N ) where Aut(N ) is the set of automorphisms of N . Note that
Aut(N ) ≤ Perms(N ). We define the semi-direct product, N oΦ H, as follows. As a set N o H is just the set
N × H. We define the group operation27 as
(n1 , h1 ) · (n2 , h2 ) = n1 Φ(h1 )(n2 ) , h1 h2 .
B Claim: N ∼
= N × {1H } C N o H and H ∼
= {1N } × H ≤ N o H.28
B Recall: If N = Z/pZ where p is a prime, then Aut(N ) ∼
= Z/(p − 1)Z, and in general, Aut(Z/nZ) is of order φ(n)
where φ is the Euler φ function.
B Example: Let N = Z/7Z and H = Z/3Z. Then, Aut(N ) ∼
= Z/6Z. Let Φ : H → Aut(N ) be an injective
homomorphism given by [1]3 7→ [2]6 . Now, N o H has order 21, but is not an abelian group. You can see this
by considering the two elements ([1]7 , [0]3 ) and ([0]7 , [1]3 ) which do not commute since Φ([1]3 ) is not the identity
automorphism. On the other hand if we were to try and construct a semi-direct product of the groups in the other
order, H o N , then we must use the zero map Φ : N → Z/2Z and so H o N = H × N .
B Lemma: Let G be a finite group, acting on a finite set X and choose any x ∈ X. Then |G| = |Gx | · |G · x|.
Proof. This is mostly just an application of Lagrange’s theorem, and noting that there is a bijection between elements
of the orbit G · x and the left cosets of StabG (x) = Gx in G.
B Orbit Decomposition Theorem: Let G be a finite group acting on a finite set X. Let X1 , . . . , Xr be the distinct
orbits in X for the action. Choose xi ∈ Xi for each i. Then
|X| =
r
X
|G/Gxi | =
i=1
r
X
[G : Gxi ].
i=1
Proof. This proof is a simple application of the lemma and Lagrange’s theorem.
10 October 2011
More Group Actions
B Proposition: Let G be a finite group acting on a finite set X. Let n be the number of orbits. LetX
f : G → Z be a
map where f (g) is the number of fixed points of g, i.e. f (g) = |{x ∈ X : gx = x}|. Then n|G| =
f (g).29
g∈G
Proof. Let X1 , . . . , Xn be the distinct orbits, and set fi (g) = |{x ∈ Xi : gx = x}|. Note then that f (g) =
n
X
fi (g).
i=1
Let Γi be a “correspondence” in G × Xi . In particular, let Γi = {(g, x) ∈ G × Xi |gx = x}. We then have the
following diagram where π1 and π2 are the usual projection maps:
Γi ⊆ G × Xi
π1
G
π2
Xi
Consider π2 |Γi : Γi → Xi . Note that this map is surjective as (eG , x) ∈ Γi for all x ∈ Xi . Also, the fibers are
(π2 |Γi )−1 ({x}) = {(g, x) : gx = x} = StabG (x) × {x}, and
(π1 |Γi )−1 ({g}) = {(g, x) : x ∈ Xi and gx = x},
27 We didn’t actually show this operation turned N × H into a group, but it isn’t hard to show, and some people even did it as homework.
The only tricky part is determining that (n, h)−1 = (Φ(h−1 )(n−1 ), h−1 ).
28 This seems not too hard to show and part of it may have been done by some people for homework.
29 This was actually stated the previous day, but proved this day, so it’s listed here.
3 January 2012
Math 901 Notes
17 of 33
−1
which is the set of fixed points of G in X
({g})| = fi (g). The fibers of a map are always disjoint,
i and so |(π1 |Γi ) X
X
and their union is the domain; hence,
fi (g) = |Γi | =
| StabG (x)|. Recall that |G| = | StabG (x)| · |G · x|.
g∈G
Hence,
X
g∈G
fi (g) = |Γi | =
X
x∈Xi
|G|/|Xi | = |G|. Now, we add these up over i to get
x∈Xi
X
g∈G
f (g) =
n
XX
fi (g) =
g∈G i=1
n X
X
fi (g) =
i=1 g∈G
n
X
|G| = n|G|.
i=1
B Corollary: Say G is a finite group acting transitively on a finite set X with |X| > 1. Then there exists an element
g ∈ G which is fixed-point free, that is, so that gx 6= x for all x ∈ X. In the language of the previous proposition,
this says there is an element g ∈ G such that f (g) = 0.
Proof. Suppose not, then for each g ∈ G there existsX
an xg ∈ X such that g · xg = xg . Or equivalently, f (g) ≥ 1
for every g ∈ G. Since the action is transitive, |G| =
f (g). This means that f (g) = 1 for every g ∈ G, but this
g∈G
is a contradiction since f (1G ) = |X| ≥ 2.
B Corollary: Let G be a group, H a subgroup such that H G. If G is a finite group, then
[
gHg −1 ( G.
g∈G
Proof. We know G acts on the left cosets of H by left multiplication. Set X = {gH}g∈G . Since H G, then we
have that |X| > 1. Also, note that this action is transitive, and choose any a ∈ G. Then a ∈ gHg −1 for some g ∈ G
if and only if g −1 ag ∈ H, which happens if and only if ag ∈ gH,
[ or equivalently if and only if a is in the stabilizer of
gH under the group action, i.e. a · gH = gH. Now, let S =
gHg −1 . Then a ∈ S if and only if f (a) ≥ 1, where
g∈G
as before, f (a) is the number of fixed points of a under the action, or equivalently, f (a) is the number of conjugates
gHg −1 which contain a. However, by the previous corollary, there is some b ∈ G such that f (b) = 0. Hence, b ∈
/ S.
12 October 2011
More Group Actions & the Class Equation
B We now provide an alternate proof for the Corollary proved last time:
Proof. Alternate Proof: Let G act on the set of conjugates of H by conjugation, and set X = {gHg −1 }.30
Note that StabG (H) = {g ∈ G : gHg −1 = H} = NG (H).31 Recall that |gHg −1 | = |H| for all g ∈ G, and that
1G ∈ gHg −1 for all g ∈ G. Let n be the number of conjugates of H, i.e. n =[|X| = |G|/| StabG (H)| = |G|/|NG (H)|.
If n = 1, then H C G and for all g ∈ G, we have gHg −1 = H, so clearly
gHg −1 G. So suppose n 6= 1. We
g∈G
have the following computation:
[
|G|
|G|
|G|
−1 gHg
≤ n (|H| − 1) + 1 = |NG (H)| (|H| − 1) + 1 = |NG (H)| |H| − |NG (H)| + 1.
g∈G
Now,
NG (H)
G, |H|/|NG (H)| ≤ 1, and |G|/|NG (H)| = n ≥ 2, and so from the computation we get that
[
−1 gHg ≤ |G| − 2 + 1 = |G| − 1 < |G|.
g∈G
B Class Equation: Let G act on itself via conjugation g · h = ghg −1 . This gives us a homomorphism Φ : G → Aut(G)
with kernel the center of the group, which we recall is denoted Z(G). The image of Φ is a subgroup of Aut(G) which
is known as the group of inner automorphisms, and is sometimes denoted Inn(G). For h ∈ G, we have that StabG (h)
is {g ∈ G : ghg −1 = h} = CG (h), the centralizer of h in G. More generally, the centralizer of a subset S ⊆ G is
CG (S) = {g ∈ G|gs = sg for all s ∈ S}. Note here that CG (G) = Z(G).
We now assume that G is a finite group, and that X1 , . . . , Xn are the disjoint orbits of the group under the conjugation
action. For each i, choose any xi ∈ Xi . Then we have the class equation:
|G| =
n
X
i=1
30 This
IS an action.
that NG (H) is the normalizer of H in G.
31 Recall
|Xi | =
n
X
[G : CG (xi )].
i=1
Math 901 Notes
3 January 2012
18 of 33
However, this is more commonly written as follows and separates out the singleton orbits:
X
|G| = |Z(G)| +
[G : CG (xi )]
where now the sum is taken over those i’s such that CG (xi ) G.
B Proposition: Let G be a finite group, and H a subgroup of index 2. Then H C G.
Proof. Let g ∈ G. If g ∈ H, then obviously gH = Hg. However, if g ∈
/ H, then gH and Hg are both disjoint from H,
and since the index is 2, we must have that gH = Hg. Hence, gH = Hg for all g ∈ G, and so H is normal in G.32
14 October 2011
Group Actions & Sylow Theorems & Nilpotent Groups33
B Corollary: If G is a finite group and has an element with exactly two conjugates, then G has a non-trivial, proper,
normal subgroup.34
Proof. Let g ∈ G be an element with exactly two conjugates, then under conjugation of the group on itself, StabG (g)
has index 2. So by the proposition, StabG (g) is a proper normal subgroup of G. It remains to show that StabG (g)
is a non-trivial, proper subgroup. However, this is easy since g ∈ StabG (g) and g 6= 1G since 1G has only itself as
a conjugate, and so StabG (g) is a non-trivial, proper, normal subgroup.
B Proposition: Let G be a finite group, and let p be the smallest prime such that p divides the order of the group. If
G has a subgroup, H, of index p, then H C G.
Proof. We didn’t prove this in class, but I’ll do it here anyways. Let X be the set of left cosets of H in G. Then, G
acts on H by left multiplication, and this action is clearly transitive. Then the homomorphism, Φ, given by Cayley’s
theorem has kernel ker(Φ) ⊆ H as we have that H = StabG (H) since gH = H if and only if g ∈ H.35 Our goal
is to get that ker(Φ) = H. Now, we know that Perms(X) is a group of size p!, and so we must have that |G/ ker(Φ)|
divides p!. However, if r divides the index of ker(Φ) in G, then r divides the order of G since the index divides the
order of the group. Since p is the smallest prime that divides the order of the group, and no number larger than
p divides G/ ker(Φ), then we must have that |G/ ker(Φ)| is either 1 or p. We now do the following computation:
|G/ ker(Φ)| = [G : ker(Φ)] = [G : H][H : ker(Φ)] = p[H : ker(Φ)] ≥ p.
Thus, we must have that [G : ker(Φ)] = p, so that ker(Φ) = H since we already knew that ker(Φ) ⊆ H. Hence H
is normal in G.
B Definition: Let p be a prime. A finite group, G, is called a p-group if |G| = pn for some n ∈ N.
B “Trivial” Example: The trivial group is a p-group for all primes p.36
B Lemma: Let G be a non-trivial p-group for some prime p. Then |Z(G)| > 1.
Proof. We apply the class equation. The size of the group G is the size of its center plus the sum of the order of
the orbits of elements having conjugates other than themselves. Note that p divides the order of the group, and
p divides the order of each orbit of elements having conjugates other than themselves. Thus, p must divide the order
of the center of G, and so |Z(G)| is at least p, and hence greater than 1.
B Cauchy’s Theorem: Let G be a finite group, and p a prime such that p divides the order of G. Then G has an
element of order p.
Proof. From class notes: “A beautiful proof can be given using Z/pZ acting via cyclic permutations on a subset
of G × . . . × G.” –B.H.37
|
{z
}
p times
B Definition: Let p be a prime. A subgroup of order pn of a finite group, G is called a p-subgroup.
B Definition: Let G be a finite group of order pn · m where p is prime and (m, p) = 1. Then a p-subgroup of G that
has order pn is called a Sylow p-subgroup.
B Sylow’s Theorem(s): Let G be a finite group of order pn · m where p is prime and p - m.
1. G has a Sylow p-subgroup.
32 This
proof is different from the one given in class, but makes more sense to me.
My!
with no non-trivial, proper, normal subgroup are called simple, so we could just say that then G is not a simple group.
35 The subgroup H and the stabilizer H from the theorem are now the same thing.
36 Hahaha!
37 This seems to indicate that we don’t need to know the proof, but at some point, I think I should understand and type up a real proof.
33 Oh
34 Groups
3 January 2012
Math 901 Notes
19 of 33
2. If Q is a p-subgroup of G and P is a Sylow p-subgroup, then Q ⊆ gP g −1 for some g ∈ G. In particular, any
two Sylow p-subgroups of a group are conjugate.
3. If np denotes the number of Sylow p-subgroups, then np ≡ 1 (mod p) and np |m.38
B Definition: A sequence of subgroups G1 ≤ G2 ≤ . . . ≤ Gr = G of a group G is called an abelian tower if
Gi C Gi+1 and Gi+1 /Gi is abelian and both statements hold for all i.
B Comment: Nilpotent and Solvable groups be defined based upon special abelian towers.
19 October 2011
Sylow Theorems & Nilpotent Groups
B Useful Fact: Let P be a Sylow p-subgroup of a finite group G. Then NG (NG (P )) = NG (P ).
Proof. We have that P C NG (P ) C NG (NG (P )) but remember that normality is not a transitive relation. So, we
know that P is the unique Sylow p-subgroup of NG (P ) by part 2 of the Sylow Theorem(s). We easily get the inclusion
NG (P ) ⊆ NG (NG (P )), so we need to show the other. Let g ∈ NG (NG (P )). Then gP g −1 C gNG (P )g −1 = NG (P )
by choice of g. Thus, P = gP g −1 since P is the unique Sylow p-subgroup of NG (P ). In particular, this means that
g ∈ NG (P ) and so we get the other inclusion.
B Lemma: Let G be a finite group. Assume the Sylow subgroup of G are all normal. Then G is the direct product of
its Sylow subgroups.
Proof. Let |G| = pr11 · . . . · prss where pi are distinct primes and assume ri > 0 for all i. By part 1 of Sylow’s
Theorem(s), there exists a Sylow pi -subgroup, Pi for each i. Then we need to show that G ∼
= P1 × . . . × Ps . If
s = 1, then the statement is trivial since G = P1 is a p1 -group, so assume s > 1. Define θ : P1 × . . . × Ps → G by
(g1 , . . . , gs ) 7→ g1 · . . . · gs . We need to check three things:
θ is a Homomorphism: Let (g1 , . . . , gs ) and (h1 , . . . , hs ) be elements of P1 × . . . × Ps . Then,
θ((g1 , . . . , gs ) · (h1 , . . . , hs )) = θ((g1 h1 , . . . , gs hs )) = g1 h1 · . . . · gs hs .
Also, θ((g1 , . . . , gs )) · θ((h1 , . . . , hs )) = (g1 · . . . · gs ) · (h1 · . . . · hs ). We get that this is a homomorphism once we
show that gh = hg whenever g ∈ Pi , h ∈ Pj , and i 6= j. Note that ghg −1 h−1 = (ghg −1 )h−1 ∈ Pj since h ∈ Pj C G.
Also, ghg −1 h−1 = g(hg −1 h−1 ) ∈ Pi since g ∈ Pi C G. However, we have that the orders of the subgroups Pi and
Pj are coprime, and so we must have that the order of the element ghg −1 h−1 is 1, and hence ghg −1 h−1 = eG = e
so that gh = hg.
Injectivity: Suppose that (g1 , . . . , gs ) ∈ ker(θ). Then we have g1 · . . . · gs = eG = e. Let mi = |G|/pri i so that
mi and pri i are coprime. Thus, there exist x, y ∈ Z such that 1 = xmi + ypri i . Now, e = (e)xmi = (g1 · . . . · gs )xmi =
g1xmi · . . . · gsxmi = gixmi since if i 6= j, then the order of gj divides mi . Thus, we get
ri
e = gixmi = gi1−yp
ri −y
p
= gi · gi i
= gi · (1G )−y = gi .
This means that (g1 , . . . , gs ) = (e, . . . , e) and so θ is injective.
Surjectivity: This is clear since the orders of the two groups are the same and θ is injective.
B Definition(s): Let G be a group. Consider the following sequence of quotients:
G = G0 G1 G2 . . .
where Gi+1 = Gi /Z(Gi ). We get a sequence then of surjective homomorphisms σj : G → Gj . Let
C0 (G) := (eG ), and
Cj (G) := ker(σj ).
T hen C0 (G) < C1 (G) < C2 (G) < . . . is called the ascending central series. We’ll show next time that it is
an abelian tower.39 The group G is nilpotent if G = Ci (G) for some i ∈ N. Equivalently, we can say that G is
nilpotent if |Gi | = 1 for some i ∈ N.
B Example: Any abelian group is nilpotent. This is obvious since G/Z(G) = G1 is the trivial group.
38 This second bit is usually (in my oh-so-humble experience) stated that n divides the order of the group, but given the first part,
p
the two are equivalent.
39 We did a terrible job doing it this day, but next time it actually makes sense!
3 January 2012
Math 901 Notes
20 of 33
21 October 2011
Nilpotent Groups
B Note: The following are fairly easy facts to show about the set of group homomorphisms f : A → B and g : B → C.
1. f (A) ∼
= A/ ker(f )
2. ker(f ) ⊆ ker(g ◦ f )
3. f (ker(g ◦ f )) ∼
= ker(g ◦ f )/ ker(f )
4. ker(g ◦ f ) = f −1 (ker(g))
5. If f is surjective, then f (ker(g ◦ f )) = f (f −1 (ker(g))) = ker(g) by applying fact 4 from this list. Applying this
to the 3rd fact gives us
ker(g) ∼
(1)
= ker(g ◦ f )/ ker(f )
B Proposition: The ascending central series defined in the last class is an abelian tower.
Proof. Let f : G → Gi , and g : Gi → Gi+1 be the given surjections. Then ker(f ) = Ci (G), ker(g) = Z(Gi ),
and ker(g ◦ f ) = Ci+1 (G) all by definition or simple observation. Plugging these into equation (1) gives us
Z(Gi ) ∼
= Ci+1 (G)/Ci (G). In particular, this means that the ascending central series is an abelian tower.
B Facts:
– If p is prime, then any finite p-group, G, is nilpotent.
Proof. If |Gi | > 1, then since Gi is a p-group, we have that |Z(Gi )| > 1 so |Gi+1 | = |Gi /Z(Gi )| = |Gi |/|Z(Gi )|
is less than |Gi |. Hence, we muse eventually get that Gt is trivial.
– A product of nilpotent groups is nilpotent.
Proof. Let B and C be nilpotent groups, and set G = B × C. Note that Z(G) = Z(B) × Z(C), so that we have
B×C
∼
G/Z(G) =
= B/Z(B) × C/Z(C). In general, we get Gi ∼
= Bi × Ci and so eventually we get both
Z(B) × Z(C)
Bi and Ci are trivial, and hence Gi is trivial as well.
– We can easily extend this to finite products, but it is not true for infinite products.
– A group G = B(1) × . . . × B(r) where each B(i) is a finite pi -group for primes pi is nilpotent.
B Lemma: Let G be a nilpotent group, and let H G. Then H NG (H).
Proof. Let n be the largest index such that Cn (G) ⊆ H. We know such an n exists since G is nilpotent. Then
choose any b ∈ Cn+1 (G) \ H. Since Cn+1 (G)/Cn (G) ∼
= Z(Gn ) ∼
= Z(G/Cn (G)), then bCn (G) ∈ Z(G/Cn (G)). Let
h ∈ H. Then bhCn (G) = (bCn (G))(hCn (G)) = (hCn (G))(bCn (G)) = hbCn (G) and this all takes place in the group
G/Cn (G). Thus, there is some a ∈ Cn (G) ≤ H so that bha = hb. Thus, as h and a are both elements of H, we
get, b−1 hb = ha ∈ H. Thus, we get b ∈ NG (H) but we also had b ∈
/ H, and so H NG (H) as desired.40
B Theorem: A finite group G is nilpotent if and only if it is the product of its Sylow subgroups.
Proof. (⇐) If G is a product of its Sylow p-subgroups for primes p which divide the order of the group, then G
is a finite product of nilpotent groups, and hence is nilpotent.
(⇒) Say G is a finite nilpotent group. It is enough to show that each Sylow subgroup is normal in G, because if
so, then G is a product of its Sylow subgroups (see the first Lemma from 19 Oct). If G is a p−group, then we’re
clearly done. So, we assume otherwise, and let P G be a Sylow p−subgroup of G. From the Useful Fact at
the beginning of the 19th of October, we know that NG (NG (P )) = NG (P ). Also, by the Lemma just previous to
this, we know that if NG (P ) G, then we’d have NG (P ) NG (NG (P )) which would be a contradiction, and so
we must have that NG (P ) = G. Thus, P C G and that completes the proof.
24 October 2011
B Definition: A group G is solvable if it has an abelian tower of the following form:
Solvable Groups
(eG ) = Gm C Gm−1 C . . . C G0 = G,
where Gi /Gi+1 is abelian for all i. A series of this form is called a solvable series for G.
B Note: It is clear that nilpotent groups are all solvable. We’ll see later that not all solvable groups are nilpotent.
40 I’ve
modified this proof a bit from the one given in class so that I understand it better.
Math 901 Notes
3 January 2012
21 of 33
B Feit-Thompson Theorem: (Conjectured by Burnside) Every finite group of odd order is solvable.41
B Derived Series: Let G(1) (also written G0 ) be the commutator subgroup of G, [G, G], and call it the first derived
subgroup of G. Recursively, G(i+1) is called the i + 1st derived subgroup, and is the first derived subgroup of
the ith derived subgroup, that is, G(i+1) = (G(i) )0 . Note that G ⊇ G(1) ⊇ G(2) ⊇ . . ., and that G(i) /G(i+1) is abelian
(we know this from the homework). This sequence of subgroups is an abelian tower known as the derived series.
B Note: If there is an m such that G(m) = (eG ), then G is solvable.
B Proposition: Let G be a group. Then G is solvable if and only if the derived series terminates at the identity.
Proof. (⇐) This is clear!
(⇒) Assume G is solvable. Then there is an abelian tower Gm = (eG ) C Gm−1 C . . . C G0 = G. Since, G0 /G1 is
abelian, then G1 ⊇ G(1) . We can inductively show that G(i) ⊆ Gi . Then we get G(m) ⊆ Gm = (eG ), which completes
the proof since Gi /Gi+1 being abelian means that G0i ⊆ Gi+1 for all i. Thus, G(i+1) = (G(i) )0 ⊆ G0i ⊆ Gi+1 .
B Useful Fact: Let G be a group, and N C G, then G is solvable if and only if N and G/N are.
Proof. (⇒) Assume G is solvable, then N (i) ⊆ G(i) , so if G(m) = (1G ), then so does N (m) , and so G being solvable
implies that N is solvable. Next, if f : G → H is any group homomorphism, then f (G0 ) ⊆ H 0 , and if f is surjective,
then f (G0 ) = H 0 . Thus, under the quotient homomorphism q : G → G/N , we get q(G(i) ) = (G/N )(i) , so that G
being solvable implies that G/N is also solvable.
(⇐) We’ll build an abelian tower for G out of towers for N and G/N . First, let Nm = (eG ) C Nm−1 C . . . C N0 = N ,
be a solvable series for N , and let Gs = (eG/N ) C Gs−1 C . . . C G0 = G/N be a solvable series for G/N . Set
Gi = q −1 (Gi ). Then Gi /Gi+1 ∼
= (Gi /N )/(Gi+1 /N ) ∼
= Gi /Gi+1 is abelian. Note that G0 = G, and Gs = N . Then
we have the following as a solvable series for G, which completes the proof:
(eG ) = Nm C . . . C N0 = N = Gs C . . . C G0 = G.
B Corollary: If G is solvable, then any subgroup of G and every homomorphic image of G is solvable.42
B Example: Not every solvable group is nilpotent. The group S3 is solvable. This is because A3 C S3 , and A3 is
abelian, and every abelian group is solvable. Also, S3 /A3 ∼
= Z/2Z which is abelian and hence also solvable. Thus by
the Useful Fact we’ve got that S3 is solvable. However, S3 is not nilpotent! Note here that Z(S3 ) = (1S3 ), and so
the ascending central series is constant. Alternately, the sylow subgroups of S3 are of order 3 and 8, but S3 has order
6, and so it cannot be the direct product of its Sylow subgroups.
26 October 2011
Finish Solvable Groups & Start Rings (Fields)
B Corollary: Sn is not solvable for n ≥ 5.
B Corollary: An is not solvable for n ≥ 5.
B Definition: A ring is an abelian group (R, +) with a binary operation ∗ called multiplication such that
– ∗ is associative
– R contains a multiplicative identity, denoted 1R or 1.43
– Multiplication distributes over addition on both the left and the right. That is, for a, b, c ∈ R, we have
a(b + c) = ab + ac and (a + b)c = ac + bc.
B Note: If we also have the ∗ is commutative, then we say that R is a commutative ring.
B Example: If 0 = 1, then R = {0}.
B Facts: Let R be a ring, and b ∈ R. Then the following are easy to show from the properties.
– For any b ∈ R, we get −b = (−1)b = b(−1).
– For any b ∈ R, we get b0 = 0b = 0.
B Definition: We say a ring R is a division ring if 1 6= 0 and for any b ∈ R \ {0}, there exists b−1 ∈ R such that
bb=1 = b−1 b = 1.
B Definition: A commutative division ring is a field.
B Example: The real quaternions are a division ring. As a set they are Q = {a + bi + cj + dk : a, b, c, d ∈ R}, where 1
is a multiplicative identity for Q, addition works like a vector space, and ij = k, jk = i, ki = j, ji = −k, kj = −i,
ik = −j, and i2 = j 2 = k 2 = −1. We also have that multiplication is associative and distributes over addition. It is
possible in this ring to find inverses of any non-zero element in a manner similar to the field C, but in this ring,
multiplication is not commutative, so it is not a field.
41 Thompson
got a fields medal for this!
proof of this is embedded in the proof of the Useful Fact.
43 This is not always part of the definition, but it will be for our class.
42 The
Math 901 Notes
3 January 2012
22 of 33
B Wedderburn’s Little Theorem: A finite division ring is a field.
B Definition: Let R be a commutative ring, and let a ∈ R be nonzero. We say a is a zero-divisor of R if there is a
nonzero b ∈ R such that ab = 0.
B Definition: A commutative ring R where 1 6= 0 which has no zero-divisors is called an integral domain, or simply
a domain.
B Example: Z is a domain, as is any field.
B Proposition: Let R be a finite domain. Then R is a field.
Proof. Let b ∈ R be a nonzero element. Then {b, b2 , . . .} is a finite subset of R, which does not contain 0. Thus,
there exists i < j such that bi = bj . We can then cancel to get 1 = bj−i = b · bj−i−1 .
B Fact: Let F be a finite field, and let P be the prime field, that is, the field generated by 1 or equivalently the
smallest subfield of F which contains 1. Then |P | = p for some prime p, and the additive order in (F, +) of every
non-zero element is p.
B Definition: Let F be a field. The smallest positive integer such that mb = 0 for all b ∈ F is called the characteristic
of F , and is denoted char(F ) = m if it exists. If no such integer exists, we say char(F ) = 0.
B Example: Let F = Z/pZ, where p is a prime, then P = F , and char(F ) = p. If F is a subfield of C, then charF = 0.
B Fact: If F is a field and char(F ) = p > 0, then p is a prime, and the prime subfield of F is isomorphic to Z/pZ.
28 October 2011
Field Extensions
B Terminology: Let F ⊆ E be fields. We say F is a subfield of E or that E is an extension field of F .
B Note: If F ⊆ E is a field extension, then E is an F −vector space.
B Definition: We say F ⊆ E is a finite extension if the vector space dimension of E over F is finite. Otherwise, we
say E is an infinite extension. Either way, we write [E : F ] = dimF (E) and call it the degree of the extension.
B Fact: If F ⊆ E ⊆ K are fields, then [K : F ] = [K : E][E : F ].
B Definition: Let R, S be rings. We say a map h : R → S is a ring homomorphism if the following hold:
1. h(0R ) = 0S
2. h(1R ) = 1S
3. h(a + b) = h(a) + h(b) for all a, b ∈ R
4. h(ab) = h(a)h(b) for all a, b ∈ R
B Aside: The composition of homomorphisms is also a homomorphism. The only homomorphism out of the stupid
ring, {0} is the trivial map to the stupid ring.
B Example: Let F ⊆ E be a field extension and let α ∈ E. Consider the polynomial ring F [x], and let evα : F [x] → E
be the map given by evaluating the polynomial at α. Then evα is a homomorphism.
B Notation: If ker(evα ) 6= 0, then Lang denotes the unique monic generator of the ideal by Irr(α, F, x) and calls it the
irreducible polynomial for α over F .
B Note: If ker(evα ) 6= 0, then Irr(α, F, x) is an irreducible polynomial since F [x]/(Irr(α, F, x)) is isomorphic to a
subring of E, but E is a domain, and hence F [x]/(Irr(α, F, x)) is also a domain, so Irr(α, F, x) cannot have non-trivial
factors.
B Definition: Let F ⊆ E be fields, and let α ∈ E. We say α is algebraic over F if ker(evα ) 6= 0. That is, if there is
a non-trivial polynomial f ∈ F [x] such that f (α) = 0. We say α is transcendental if it is not algebraic over F . We
say a field extension F ⊆ E is algebraic if every element of E is algebraic over F .
B Criterion: Let F ⊆ E be a field extension. If [E : F ] < ∞, then the extension is algebraic. The converse is false
since Q is algebraic over Q but has infinite degree.
Proof. If [E : F ] < ∞, then dimF (F [x]) = ∞ so any homomorphism h : F [x] → E has a nontrivial kernel. In
particular, for any α ∈ E, we get that ker(evα ) 6= 0. Thus, α is a root of some non-trivial polynomial, Irr(α, F, x).
B Comment: Let F ⊆ E be a field extension, and let α ∈ E. If α is algebraic, then F (α) is the field generated by F
and α inside E, and is also the image of evα , or equivalently, F [α]. That is, when α is algebraic over F , we get
F [α] ∼
= F (α). This is false if α is transcendental.
31 October 2011
B Note: We started today by proving the comment from the previous day.
Algebraic Extensions
Math 901 Notes
3 January 2012
23 of 33
B Fact: If α is algebraic over a field F , then
[F (α) : F ] = [F [α] : F ] = dimF (F [α]) = dimF
F [x]
(Irr(α, F, x))
= deg(Irr(α, F, x)) < ∞.
However, if α is transcendental, then [F (α) : F ] = ∞.
B Question: Why is the set of elements in a field E which are algebraic over a subfield F actually a subfield of E?44
B Theorem: Let k be a field, L an algebraically closed field such that there is an embedding σ : k ,→ L. Let E be an
algebraic extension field of k. Then there is a homomorphism σ 0 : E → L such that the diagram below commutes.45
E
k
σ0
σ
L
B Corollary: Any two algebraic closures, A, B, of a field k are isomorphic over k; that is, there exists an isomorphism
φ : A → B such that φ|k = idk .46
B Notation: Lang denotes the algebraic closure of a field k by k a . Note that k a can technically be any algebraically
closed field containing k which is algebraic over k. The point of the corollary is that the algebraic closure of a field is
unique up to isomorphism.
2 November 2011
More Algebraic Extensions
B Recall: A field K is algebraically closed if every non-constant polynomial f ∈ K[x] has a root in K. We say an
algebraically closed field K which contains k is an algebraic closure of k if K is algebraic over k.47
B Comments:
– The only algebraic extension of an algebraically closed field, k, is the field k itself.
– Every field has an algebraic closure.
B Definition: We say a field extension k ⊆ K is an algebraic closure if K is algebraically closed, and the extension
is algebraic.
B Example: C is algebraically closed, but is not the algebraic closure of Q. Instead, Qa = {z ∈ C : z is algebraic over Q}.
B Definition: Let S ⊆ k[x] be a set of non-constant polynomials. Then K is a splitting field for S if every element
of S splits (factors) into linear factors in K[x] and if K is generated over k by the roots of elements of S.
B Examples:
1. C √
is a splitting field for x2 + 1 over R.
2. Q[ √2] is a splitting field over Q for x2 − 2.
3. Q[ 3 2] is not a splitting field over Q for x3 − 2 since it’s missing the complex roots.
4. For any field k, k a is the splitting field over k for the set S = {all positive degree polynomials in k[x]}.
B Definition: A field extension k ⊆ K is normal if K is a splitting field of some set of polynomials of positive degree
in k[x].
B Theorem:48 If A and B are extension fields of k, and A, B are splitting fields for the same set S of polynomials,
then A ∼
= B over k. That is, there is an isomorphism φ : A → B such that φ|k = idk .
B Theorem:49 Consider fields k ⊆ K ⊆ k a . The following three statements are equivalent:
1. K is a normal extension of k.
2. Every homomorphism h : K → k a such that h|k = 1k is an automorphism of K.
3. Every irreducible polynomial in k[x] which has a root in K actually splits (factors linearly) in K.
4 November 2011
Field Homomorphisms & Separability
√
3
B Example: The extension
Q ⊆ Q( √
2) is a finite algebraic extension, but it is not a normal
extension
since not
all of
√
√
√
√
iπ
2iπ
3
3
3
the roots of Irr( 3 2, Q, x) are in Q( 3 2). We also explained that √
Q[x]/(x3 − 2) ∼
= Q[ 2] ∼
= Q[ 2e 3 ] ∼
= Q[ 2e 3 ].
Hence, there are at least three homomorphisms over Q from Q[ 3 2] to Qa .
44 We
justified this, but I won’t here.
as review.
partially proved the corollary given the theorem, but it is one of those non-interesting proofs where you just follow your nose.
47 The algebraic closure of a field k is frequently denoted k.
48 This was cited from Lang, and is review.
49 Also cited and review.
45 Cited
46 We
Math 901 Notes
3 January 2012
24 of 33
B Proposition: Let k be a field, and let α ∈ k a . Let f (x) = Irr(α, k, x). Then the number of distinct homomorphisms
from k[α] to k a over k is equal to the number of distinct roots of f (x) in k a .
Proof. This is just a sketch! It is easy to show that if h is such a homomorphism, then f (h(α)) = 0 so that h(α)
is also a root of f (x). Since h is determined by its value on α, then it is clear that the number of homomorphisms
over k is no larger than the number of distinct roots of f in k a . Also, you can define a homomorphism by sending
α to any of the other roots, so there are exactly as many homomorphisms over k as there are distinct roots of the
irreducible polynomial.
B Corollary: The number of k−homomorphisms of k[α] → k a is no larger than deg Irr(α, k, x) if α ∈ k a .
B Example: Let k be a field, and
then x2 − y ∈ (k(y))[x] is irreducible. If k has characteristic 2,
√ y 2and indeterminate,
2
2
a
then x − y = x + y = (x + y) ∈ (k(y)) [x] and so x2 − y only has a single root in k(y)a .
B Definition: Let k be a field. We define a linear operator (known as the derivative) by
d
n
n−1
and by extending linearly.
dx ax = anx
B Comments:
– The derivative satisfies the Liebnitz rule:
d
dx
: k[x] → k[x] via
d
df
dg
(f g) =
g+f
dx
dx
dx
d
d
– If k ⊆ E is an algebraic field extension, then dx
on k[x] is the restriction of dx
on E[x].
– We can use derivatives to detect polynomials with multiple roots since the linear factors aren’t unique and due
to the Liebnitz rule! In particular, α is a multiple root of f (x) ∈ k[x] if and only if the power of (x − α) in the
factorization of f (x) is strictly greater than one.
d p
– If k is a field of characteristic p > 0, then dx
x = pxp−1 = 0.
B Definition: Let k ⊆ E be a field extension, and let α ∈ E be algebraic over k. We say α is separable over k if
Irr(α, k, x) does not have any multiple roots in k a . An extension k ⊆ E is separable if E is algebraic over k and
every element of E is separable over k.
B Definition: A finite algebraic extension is Galois if it is both normal and separable.
7 November 2011
Characterizing Separable Extensions
B Frobenius: Let k be a field of characteristic p > 0. We get a homomorphism F : k → k given by a 7→ ap , which is
called the Frobenius map. This is because in a field of characteristic p, we have (a + b)p = ap + bp and in any
commutative ring (ab)p = ap bp .
B Example: Let k be a field, f ∈ k[x] where d = deg f ≥ 1. If f 0 (x) = 0 (since d > 1, this means k is a field of
positive characteristic, say p), then f is actually a polynomial in xp and p|d. If we work over k a , then we can take
pth roots of elements of k, and so f 0 (x) = 0 gives that f (x) = (h(x))p where h ∈ k a [x].
B Corollary: Let k be a field. Let α be algebraic over k, and let f = Irr(α, k, x). The following are equivalent:
– α is not separable over k,
– f 0 (α) = 0, and
– f 0 (x) = 0.50
B Corollary: Let k ⊆ K be an algebraic field extension, and assume that the characteristic of k is zero, (and hence
so is the characteristic of K). Then K is a separable extension of k.51
B Proposition: Let k ⊆ K be finite fields. Then the extension is separable and normal.
Proof. Let p > 0 be the characteristic of both fields. Then the prime field is Z/pZ. This means K is a Z/pZ
vector space of dimension dimZ/pZ K = m. That is, K ∼
= Z/pZ ⊕ . . . ⊕ Z/pZ. This gives that |K| = pm , and
|
{z
}
m times (copies)
[K : Z/pZ] = m. Let K ∗ = K \ {0} be the multiplicative group of K. Then |K ∗ | = pm − 1 and by Lagrange’s
∗
m
Theorem, mα|K | = 1K for mall α ∈ K ∗ . Thus, xp −1 − 1 is a polynomial of degree pm − 1 with pm − 1 distinct roots.
Hence, xp −1 − 1 and xp − x are separable and the roots of m
the second polynomial are the pm distinct elements
of K. Now, for any α ∈ K, we get that Irr(α, k, x) divides xp − x, and so Irr(α, k, x) cannot have any multiple
roots, so that α is separable over k so that the extension is normal (since α was arbitrary). Moreover, the extension is
m
m
normal since xp − x splits in K[x] and the set of roots of xp − x is precisely K, and so it generates K over k.
50 The
51 This
proof is not really difficult and I’ve seen it before, so I’m omitting it here.
is almost entirely a direct application of the previous corollary.
3 January 2012
Math 901 Notes
25 of 33
B Comment: If K is an algebraic extension of k, then if E is the set of all elements of K which are separable over k,
then E is an intermediate field, k ⊆ E ⊆ K.52
9 November 2011
Separable Degree
a
B Definition: Let K be a finite algebraic extension of a field k so that k ⊆ K ⊆ k . We define the separable
degree of K over k, denoted [K : k]s to be the cardinality of the set of k−homomorphisms K → k a .
B Note/Example: For simple algebraic extensions, the separable degree [k(α) : k]s is the number of distinct roots of
Irr(α, k, x). If α is separable, then the separable degree and the degree [k(α) : k] are equal. If α is not separable, then
we know char(k) = p > 0, and if f (x) = Irr(α, k, x), then f 0 (x) = 0, so that f (x) = (h(x))p for some h(x) ∈ k a [x]. In
m
fact, there is an m > 0 such that f (x) = (g(x))p where g has no multiple roots. In this case, [k(α) : k]s = deg g(x)
[k(α):k]
as the inseparable degree and
and [k(α) : k] = deg f (x) = (deg g(x)) · pm = pm [k(α) : k]s . We refer to [k(α):k]
s
denote it [k(α) : k]i .
B Theorem: Let k ⊆ E ⊆ K ⊆ k a be fields and assume K is a finite extension of k. Then the separable degree is
multiplicative and finite. Moreover, [K : k]s divides [K : k]. Here, we call [K : k]/[K : k]s the inseparable degree
and denote it by [K : k]i .
Proof. Let SK/k be the set of k−homomorphisms K → k a . So |SK/k | = [K : k]S . Similarly, SK/E is the set of
E−homomorphisms K → k a and SE/k is the set of k−homomorphisms E → k a . Thus, |SK/E | = [K : E]s and
|SE/k | = [E : k]s . For each k−homomorphism γ : E → k a pick an extension γ
e : k a → k a so that γ
e|E = γ. We’ll
now define a map
ϕ : SK/k → SK/E × SE/k
via
h 7→
g
h|
E
−1
◦ h, h|E .
Showing that this map is bijective will complete the proof. We’ll do this by finding an inverse to ϕ. Let
λ : SK/E × SE/k → SK/k
be given by (h, g) 7→ ge ◦ h. Note that h : K → k a is a E−homomorphism (and hence a k−homomorphism) and
ge : k a → k a is a k−homomorphism and so their composition exists as written and is a k−homomorphism from K
to k a , and hence an element of SK/k as desired. We now must show that compositing these two maps is the identity
in both directions. We have
−1
−1
g
g
g
(λ ◦ ϕ)(h) = λ (ϕ(h)) = λ
h|E
◦ h, h|E
= h|E ◦
h|E
◦h =h
and similarly since h|E = idE and ge|E = g, we get
(ϕ ◦ λ)((h, g))
= ϕ(λ((h, g)))
= ϕ(e
g ◦ h)
−1
=
(e
g^
◦ h) |E
◦ (e
g ◦ h), ge ◦ h|E
−1
g
=
ge
|E
◦ ge ◦ h, ge|E
=
(h, g).
Thus, we have that |SK/k | = |SK/E | · |SE/k | as desired to give that separable degree is multiplicative.
Now, since K is a finite extension of k, we can pick a finite set of elements α1 , . . . , αr so that
k ⊆ k(α1 ) ⊆ k(α1 , α2 ) ⊆ . . . ⊆ k(α1 , . . . , αr ) = K.
Using what we learned about simple extensions in the Note/Example, we have that
|SK/k | = |SK/k(α1 ,...,αr−1 ) | · |Sk(α1 ,...,αr−1 )/k(α1 ,...,αr−2 ) | · . . . · |Sk(α1 )/k |
=
52 We’ll
[K : k(α1 , . . . , αr−1 )] · [k(α1 , . . . , αr−1 ) : k(α1 , . . . , αr−2 )] · . . . · [k(α1 ) : k].
justify this first thing on the 14th of November using the idea of separable degree.
3 January 2012
Math 901 Notes
26 of 33
As each of the factors53 if finite, then so is their product, |SK/k |. That is, [K : k]s < ∞. The fact that separable
degree in general divides the regular degree is a combination of the facts that degree is multiplicative, separable
degree is multiplicative, and that for a simple extension separable degree divides the regular degree.
11 November 2011
Separability & Characteristic p
B Definition: Let k ⊆ E be a field extension. We define Autk (E) := {f ∈ Aut(E) : f |k = idk }.
B Note: Every field homomorphism is injective!5455 Also note that for any field extension k ⊆ E, the set Autk (E) is
actually a group under composition.
B Comment: Given any field k, the group Autk (k a ) acts on the roots of any polynomial f ∈ k[x]. By letting
σ ∈ Autk (k a ), we get the following commutative diagram where the two vertical(ish) maps are the obvious inclusion
map(s):
ka
σ
ka
k
This map σ can then be extended to an automorphism of k a [x] so that the similar diagram commutes where again
the vertical(ish) maps are the obvious inclusion map(s):
k a [x]
σ
e
k a [x]
k[x]
By definition if f =
n
X
i=1
e(f ) =
ai xi ∈ k a [x], then σ
n
X
e is a homomorphism, we have σ
e(f + g) =
σ(ai )xi . Because σ
i=1
σ
e(f )+e
σ (g) and σ
e(f g) = σ
e(f )e
σ (g). So if f ∈ k[x] is monic, then in k a [x] we have that f (x) = (x−α1 )·. . .·(x−αm ) and
so f = σ
e (f (x)) = (x − σ(α1 ))·. . .·(x − σ(αm )) which means that σ permutes the roots of f , and this gives the action.
B Note: The action need not be transitive; however, if f ∈ k[x] is irreducible, then the action will be transitive. We
show that now: If α1 , α2 are roots in k a of an irreducible polynomial f , then k[α1 ] ∼
= k[x]/(f ) ∼
= k[α2 ] where the
maps are given by x 7→ αi and called τ and σ respectively. So there is a k−isomorphism στ −1 : k[α1 ] → k[α2 ] given
by α1 7→ α2 , and this can be extended to all of k a to get a k−homomoprhism γ : k a → k a such that γ|k[α1 ] = στ −1 .
Thus, γ(α1 ) = α2 and the action is transitive as desired.
B Corollary: If f ∈ k[x] is a (monic),56 irreducible polynomial that factors over k a as f (x) = (x−α1 )m1 ·. . .·(x−αr )mr .
Then m1 = . . . = mr .
Proof. Choose any αi , αj roots of the polynomial f . For simplicity, make them α1 and α2 in the expansion of f
in k a [x]. Then, there is some k−homomorphism γ : k a → k a such that γ(α1 ) = α2 by the transitivity of the action
as explained above. Then γ
e(f ) = f as f ∈ k[x] and γ is a k−homomorphism. However, we also see that
(x − α1 )m1 · . . . · (x − αr )mr = f = γ
e(f ) = (x − α2 )m1 (x − γ(α2 ))m2 · . . . · (x − γ(αr ))mr .
As the power of (x − α2 ) is the same in each expansion of f we must have that m1 = m2 since γ is bijective and
k[x] is a UFD.
Note: If char(k) = p > 0, then p divides the multiplicity of each root (and they’re all the same by the corollary).
In fact, each multiplicity must be a power of p. We’ll show that now: Let f be a monic, irreducible polynomial
in k[x]. Then over k a [x], f factors as f (x) = (x − α1 )m · . . . · (x − αr )m by the above corollary. Now, Let m = ps n
53 The
correct word is not productands...idiot!
is because fields have no nontrivial ideals, the kernel must be an ideal, and not everything can be mapped to 0.
55 This means that we never need to show that field homomorphisms are injective on the homework (or Friday night activities)! Ever!
56 Monic just makes the proof easier, but isn’t necessary!
54 This
3 January 2012
Math 901 Notes
27 of 33
where p - n. Then using the fact that the Frobenius map is in fact a homomorphism we get
n
ps
f (x) =
((x − α1 ) · . . . · (x − αr ))
s
n
s
s
s
=
(xp − α1p ) · . . . · (xp − αrp )
=
s
g(xp )
s
p
where
g is some polynomial
n in k[x]. Moreover, g is irreducible with roots αi for each i. Since g(x) =
s
s
(x − α1p ) · . . . · (x − αrp )
then if n > 1 we get that g is irreducible with multiple roots, but p - n which
is a contradiction, and so we must have n = 1 so that m = ps .
B Corollary: Let f ∈ k[x] be irreducible with a field of characteristic p > 0. Then there exists an s ≥ 0 such that
f (x) = g(xs ) and g is an irreducible, separable polynomial.
s
B Corollary: Let α be algebraic over a field k, with char k = p > 0. Then αp is separable for some s ≥ 0.
B Fact: Let k ⊆ K be a field extension, and let α1 , . . . , αr ∈ K be algebraic and separable over k. Then
E = k(α1 , . . . , αr ) is a separable extension of k.
Proof. Let β ∈ E and assume that β is not separable over k. Then [k(β) : k]s < [k(β) : k]. Thus, [E : k]s = [E :
k(β)]s · [k(β) : k]s < [E : k(β)] · [k(β) : k] = [E : k]. However, we also have that because simple algebraic extensions
are separable, then
[E : k]s
=
=
[E : k(α1 , . . . , αr−1 )]s · . . . · [k(α1 ) : k]s
[E : k(α1 , . . . , αr−1 )] · . . . · [k(α1 ) : k] = [E : k]
This is a contradiction, and so we must have that β is separable over k.
14 November 2011
Theorem of the Primitive Element
B Corollary: Let k ⊆ K be an algebraic extension of fields. Also, let E = {α ∈ K : α is separable over k}. Then E
is an intermediate field k ⊆ E ⊆ K.
Proof. It is enough to show that k(E) = E. It is clear that E ⊆ k(E), so let β ∈ k(E). Then there exist
α1 , . . . , αr ∈ E such that β ∈ k(α1 , . . . , αr ) ⊆ k(α). But we saw last time that k(α1 , . . . , αr ) is separable over k.
Thus, β is separable over k and so β ∈ E.
B Definition: Let k ⊆ F be an algebraic extension of fields, and let E = {α ∈ F : α is separable over k}. Then we
call E the separable closure of k in F .
B Comment: If F 6= E in the above definition, then char k = p > 0, and for each β ∈ F , there is an n (which depends
n
n
on β) so that β p ∈ E. This is because Irr(β, k, x) = f (xp ) where f is separable. Moreover, Irr(β, k, x) divides
n
n
n
xp − β p = (x − β)p and so we have that Irr(β, k, x) = (x − β)m , and as we saw before, m must be a power of p.
B Definition: We say an algebraic extension k ⊆ F (where char k = p > 0) is purely inseparable if for each
n
element β ∈ F there is an n such that β p ∈ k. If k = F , then regardless of characteristic, we also say F is a purely
inseparable extension of k.
B Corollary: Let k ⊆ F be an algebraic field extension. Let E be the separable closure of k in F . Then k ⊆ E is
separable and E ⊆ F is purely inseparable.
B Definition: We say an extension k ⊆ F is simple if there is an α ∈ F such that k(α) = F . If k(α) = F , we say
that α is a primitive element.
B The Theorem of the Primitive Element: Let k ⊆ E be a finite extension of fields.
1. If E is separable over k, then E is a simple extension, that is, there is an element δ ∈ E such that E = k(δ).
2. The extension is simple if and only if there are only finitely many intermediate fields.
Proof. 1. First, suppose that |k| < ∞. Then E is also a finite field. But E ∗ = E \ {0} is a cyclic group (under
multiplication), so let E ∗ = (α). Then we get E = k(α). Next, suppose that k is an infinite field. It is enough
to show that k(α, β) = k(δ) for some δ ∈ k(α, β). Let σ1 , . . . , σn be the distinct k−homomorphisms of E → k a .
Note n = [E : k]s = [E : k]. So consider the polynomial
Y
P (x) =
(σi (α) + σi (β)x − (σj (α) + σj (β)x)) .
i6=j
Math 901 Notes
3 January 2012
28 of 33
In particular, since σi 6= σj , then we have that P (x) 6= 0. Since |k| = ∞, we can find a c ∈ k so that P (c) 6= 0.57
Thus, σi (α) + σi (β)c is different for all i = 1, . . . , n. Thus, σi |k(α+βc) is different for each i = 1, . . . , n. Thus,
[k(α + βc) : k]s ≥ n. This gives then that n ≤ [k(α + βc) : k]s = [k(α + βc) : k] ≤ [k(α, β) : k] = n. Hence,
k(α + βc) = k(α, β).
2. First, suppose that |k| < ∞. Then for any finite extension k ⊆ E, we have that |E| < ∞ as well, and k ⊆ E
is simple, and so there are only finitely many intermediate fields. Next, suppose that k is an infinite field. We’ll
start by assuming there are finitely many intermediate fields and showing that the extension is simple. Again,
it is enough to show that for α, β ∈ K that k(α, β) is actually a simple extension. So there exist c1 , c2 ∈ k so
that c1 6= c2 6= 0 and k(α + c1 β) = k(α + c2 β). This gives that (c1 − c2 )β = (α + c1 β) − (α + c2 β) ∈ k(α + c2 β).
Thus, β ∈ k(α + c2 β). Now using the fact that β ∈ k(α + c2 β), we get that, α = α + c2 β − c2 β ∈ k(α + c2 β).
Thus, k(α, β) ⊆ k(α + c2 β) ⊆ k(α, β) and hence k(α + c2 β) = k(α, β).
Now, assume k ⊆ E is a simple extension. In particular, let E = k(δ) for some δ ∈ k a = E a . Our goal is to show
that there are only finitely many intermediate fields between k and E. We’ll show there is an injection from the
set of intermediate fields to the set of monic factors of f = Irr(δ, k, x). Since the monic factors is clearly a finite
set, then the other set will be finite as well, and that will complete the proof. So let F be an intermediate field, i.e.
k ⊆ F ⊆ E. Then F (δ) = E. So consider gF (x) = Irr(δ, F, x). Then clearly gF |f . We’re now ready to define a map
ϕ : {intermediate fields} → {monic factors of f }
by setting ϕ(F ) = gF . Instead of directly showing that ϕ is injective, we’ll actually show that we can recover F from
gF . Let KgF = k(a1 , . . . , am ) where gf = xm + a1 xm−1 + . . . + am . It is clear that KgF ⊆ F as each ai ∈ F . So
we must show the other inclusion. We’ll do this by looking at degrees. It is clear that [E : F ] = [F (δ) : F ] = deg gF
as E = F (δ) and by the definition of gF . Similarly, it is clear that [E : KgF ] = deg(Irr(δ, KgF , x)) = deg(gF ) since
E = KgF (δ) and Irr(δ, KgF , x) = gF . Thus, we have [E : F ] = [E : KgF ] = [E : F ]·[F : KgF ] so that [F : KgF ] = 1
and hence F = KgF so that there is an inverse map, an hence ϕ is injective (and hence completing the proof).
16 November 2011
Proof Completion
B Today was short and we just finished the proof that I’ve put completely above for the sake of simplicity.
18 November 2011
B No class today. Brian Harbourne was out of town.
None
21 November 2011
Separable Degree & Galois Theory
B Recall: Let k ⊆ F be fields, and assume the extension is algebraic. Then we have k ⊆ E ⊆ F where E := {α ∈ F :
α is separable} and E ⊆ F is purely inseparable.
B Recall: If k ⊆ F is a finite extension and E is the separable closure of k in F , then
[F : k] = [F : E] · [E : k],
[F : k]s = [F : E]s · [E : k]s , and
[F : k]i = [F : E]i · [E : k]i
since each type of degree is multiplicative (i.e. this holds for any intermediate field k ⊆ E ⊆ F ).
B Proposition: Let k ⊆ F be a finite field extension, and let E be the separable closure of k in F . Then
[E : k]s = [E : k] = [F : k]s , [F : E]s = 1, [F : E] = [F : E]i , and [E : k]i = 1.
Proof. We’ll start by showing that [E : k]s = [E : k]. By the theorem of the primitive element, there is an α ∈ E
such that E = k(α) (since E is a separable extension of k). However, [E : k] = deg(Irr(α, k, x)) and Irr(α, k, x)
is separable. For each root, β, of Irr(α, k, x) we get a k−homomorphism k(α) → k a given by α 7→ β. Thus, the
number of k−homomorphisms k(α) → k a is at least the number of distinct roots of Irr(α, k, x), but that’s equal to
deg Irr(α, k, x) = [E : k] since α is separable. Hence, [E : k]s ≥ [E : k] ≥ [E : k]s and so we get that [E : k]s = [E : k].
Next, we’ll show that [F : E]s = 1. Once we’ve shown that, then we get [F : k]s = [F : E]s · [E : k]s = [F : E]s · [E :
k] = [E : k]. So, let σ be an E−homomorphism σ : F → E a = F a . Thus, σ|E = idE . Let β ∈ F . If char(E) = 0,
then F is automatically a separable extension of k so that F = E and σ(β) = β. However, if char(E) = p > 0, then
57 This
is because the polynomial can have only finitely many roots.
3 January 2012
Math 901 Notes
n
n
n
n
29 of 33
n
n
n
β p ∈ E for some n. Thus, σ(β)p = σ(β p ) = β p . This then gives that (σ(β) − β)p = σ(β)p − β p = 0 and
hence σ(β) = β. This gives that [F : E]s = 1.
Now, we have that [F : E] = [F : E]s · [F : E]i by definition of the inseparable degree, since [F : E]s = 1, then we
clearly have [F : E] = [F : E]i . Similarly, we have [E : k] = [E : k]s · [E : k]i = [E : k] · [E : k]i and so [E : k]i = 1.
B Note: We’ll now start Galois Theory!!!
B Remark: Let K be a field. We’ll associate a group to K, namely, we associate Aut(K) to it. For a field extension
k ⊆ K, we associate the group Autk (K) ≤ Aut(K). Similarly, given G ⊆ Aut(K), we can associate a subfield,
K G := {α ∈ K : ∀g ∈ G, g(α) = α}. This is called the fixed field of the subgroup.
B Comment: These associations can both be described as contravariant functors. In particular, let K be a field. Let G
be the functor from subfields of K to subgroups of Aut(K) where G(E) = AutE (K) for any subfield E ⊆ K. These
categories are equipped with morphisms of inclusion in both cases. Similarly, let F be the functor from subgroups of
Aut(K) to subfields of K given by F(G) = K G for any G ≤ Aut(K). We should discuss the contravariance of these
functors. If H < G < Aut(K), then K G ⊆ K H since fewer things are fixed by all of G than are fixed by all of H.
Similarly, if k ⊆ E ⊆ K, then AutE (K) < Autk (K) since if E is fixed then k is automatically fixed.
B Motivation: The idea behind Galois theory is to understand the composition of these two functors. In general
they’re not inverse functors, but it would be awfully nice if they were.
√
√
B Example: Consider the√field extension Q ⊆ Q( 3 2).
Then we√want to figure out what G(Q) = AutQ Q( 3 2) is. It is
√
3
3
3
√
3
clear that AutQ( √
2) Q( 2) = {1Q( 3 2) }. Let σ : Q( 2) → Q( 2) be a Q−homomorphism. Then we can extend this
√
√
to a Q−homomorphism
σ
e : Qa → Qa . Here σ
e permutes √
the roots
of Irr( 3 2, Q, x) = x3 − 2. However, Q( 3 2) ⊆ R
√
√
√
3
and so Q( 3 2) contains only one root of x3 − 2. Thus, σ( 3 2) = 3 2 so that σ = 1Q( √
2) and hence G(Q) = {1Q( 3 2) }
as well.
28 November 2011
Artin’s Theorem
B Definition: Let k ⊆ K be an algebraic extension of fields. We say it is a Galois extension if it is normal and
separable.
B Fundamental Theorem of Galois Theory: Let k ⊆ K be a finite Galois extension of fields. Then F and G give a
bijection from the set of intermediate fields k ⊆ E ⊆ K to the set of subgroups of Autk (K).58 Moreover, an intermediate
field E is Galois over k if and only if AutE (K) C Autk (K), and in this case, Autk (E) ∼
= Autk (K)/ AutE (K).
B Artin’s Theorem: Let K be any field, and let G be a finite subgroup of Aut(K). Let k = K G . Then k ⊆ K is a
finite Galois extension with Galois group G = Autk (K) and [K : k] = |G|.
Proof. We start by claiming that k ⊆ K is separable. So let α ∈ K. It is enough to show that f (α) = 0 for some
polynomial f ∈ k[x] which does not have multiple roots. So let {α1 , . . . , αr } = Gα be the orbit of α under the action of
G. Also, do this so that αi 6= αj whenever i 6= j. Then set f (x) = (x − α1 ) · . . . · (x − αr ). Note here that f ∈ K[x], and
that G acts on K, so that we have an action of G on K[x]. In particular, if g ∈ G, then gf = (x−g(α1 ))·. . .·(x−g(αr )).
But we know that {g(α1 ), . . . , g(αr )} = {α1 , . . . , αr } since G just permutes the elements of an orbit. Thus, gf = f .
Hence the coefficients of f as a (non-factored) polynomial are fixed by the action of G, and so they’re elements of
k and f ∈ k[x]. But f is separably by construction, and so k ⊆ K is separable and algebraic.
We now claim that k ⊆ K is a normal extension. Let α ∈ K and let fα be the f as defined in the previous part.
Note that fα splits over K and the roots are all in K. Thus, K is the splitting field of {fα : α ∈ K}. Thus, k ⊆ K
is Galois and G ≤ Autk (K).
Now, if [K : k] < ∞, then K G = k, and of course K Autk (K) = k. Thus, under the Galois bijection, both G and
Autk (K) correspond to k. Thus, G = Autk (K). Then by the usual statement of the Fundamental Theorem of Galois
Theory, we have that |G| = | Autk (K)| = [K : k]. So consider the case when [K : k] = ∞. Then there must be an
intermediate field k ⊆ E ⊆ K so that |G| < [E : k] < ∞.59 By choosing such an intermediate field E, we know that
E will be a finite extension of k. Then the theorem of the primitive element says that E = k(γ) for some γ ∈ E. Now,
we claim that fγ where fγ is as before is actually the irreducible polynomial of γ over k. That is, fγ = Irr(γ, k, x).
It is clear that fγ ∈ k[x] from our previous reasoning, and that fγ (γ) = 0 by definition. So thus, we get that
Irr(γ, k, x)|fγ . On the other hand, the group Autk (K) acts transitively on the roots of fγ , but preserves the set
of roots of Irr(γ, k, x). Since g(Irr(γ, k, x)) = Irr(γ, k, x) for all g ∈ Autk (K), we see that if δ is a root of Irr(γ, k, x)
then so is gδ for all g ∈ Autk (K). Thus, the set of roots of Irr(γ, k, x) are exactly (Autk (K)) · γ and so contains G · γ.
Since Irr(γ, k, x)|fγ , then we get the other inclusion, and so the sets of roots are equal, and hence fγ = Irr(γ, k, x)
as they’re both monic and separable. Now, we have |G| ≥ t = deg Irr(γ, k, x) = [k(γ) : k] = [E : k] > |G|. This is
a contradiction, and so we can’t ever have that [K : k] = ∞, and that completes the proof.
58 We
defined these functors on 21 November 2011.
can be done by adding elements one at a time until you get that the index is sufficiently large.
59 This
3 January 2012
Math 901 Notes
30 of 33
B Aside: If k ⊆ K is a Galois extension, then | Autk (K)| = [K : k].
B Aside: Here, we see how to get extensions k ⊆ K with any Galois group, as long as G ≤ Aut(K).
B Noether’s Problem: What groups occur as Galois groups of finite Galois extensions of Q?60
30 November 2011
Galois Extensions & Finite Fields
B Aside: If k ⊆ F is a finite field extension, then what can we say about | Autk (F )| and [F : k]? If F/k is Galois, then
they’re equal. More generally, for any finite extension F/k, then [F : k]s is equal to the number of k−homomorphisms
F → F a . Thus, | Autk (F )| ≤ [F : k]s since for each σ ∈ Autk (F ) we have a k−homomorphism F → F a given by
composition with the inclusion morphism F → F ,→ F a . Here the image of the composition is all of F , but there can
be k−homomorphisms from F → F a whose image is not F . If σ is the k−homomoprhism of F → F , then we call
the composition ϕσ .
√
√
B Example: Consider the homomorphism
ϕ : Q( 3 √
2) → Qa which is given by sending Q to √
itself identically and 3 2 to
√
a nonreal root of x3 − 2. Then ϕ(Q( 3 2)) 6= Q( 3 2) since the nonreal root is not in Q( 3 2).
B Comments:
– An extension F/k is normal if and only if for all α ∈ F , then all roots of Irr(α, k, x) are in F as well.
– When F/k is normal, then for any k−homomorhism ϕ : F → F a , then ϕ(F ) = F .
– Thus, ϕ = ϕσ for some σ ∈ Autk (F ).
– So when F/k is finite and normal (but not necessarily separable) then [F : k]s = | Autk (F )|.
– If F/k is not Galois, then [F : k] > [F : k]s and so [F : k] > | Autk (F )|.’
B In Conclusion: If F/k is finite, we have
| Autk (F )| ≤ [F : k]s ≤ [F : k]
where if the extension is normal then the first is an equality and if it is also Galois, then they’re both equalities. (i.e.
the first is an equality when the extension is normal and the second is an equality when the extension is separable).
B Finite Fields: Let K be a finite field, and set p = char(K) > 0. Note that |K| = pn for some n ≥ 1. Also, let k be
the prime field, so k ∼
= Z/pZ. Thus, [K : k] = n. We have the Frobenius automorphism, which we’ll call ϕ that sends
elements of K to their pth powers. In particular if α ∈ k then ϕ(α) = αp = α so that ϕ ∈ Autk (K) ≤ Aut(K).
B Claim: Autk (K) is a cyclic group generated by the Frobenius map ϕ for any finite field K where k is the prime field,
and hence isomorphic to Z/pZ for some prime p.
Proof. It is enough to show that the order of ϕ is n.61 So recall thatm K ∗ is a cyclic group under multiplication,
and let K ∗ = (α). But the order of ϕ is the least m ≥ 1 such that β p = ϕm (β) = β for all β ∈ K. However, the
m
number of roots of xp − β is at most
pm so the order of ϕ is at least n since we need pn roots to getnall elements
pm
of K as
roots
of
the
polynomial
x
−
β. On the other hand, the order of α in K ∗ is pn − 1. So αp −1 = 1 and
pn −1
pn
so (α
)α = α or equivalently α = α. Hence, ϕn (α) = α so that ϕn (αi ) = αi for all i. Since α generates K ∗
n
then ϕ (β) = β for all β 6= 0 and so ϕn (β) = β for all β ∈ K. Thus, n is the order of ϕ.
B Comment: If k ⊆ E ⊆ K, then Autk (K) ⊇ AutE (K), so AutE (K) is generated by a power of ϕ.
2 December 2011
Finite Fields
B Last Time: We saw that if F is a finite field with prime field k = Z/pZ, then Autk (K) = (ϕ) where ϕ is the
Frobenius homomorphism. If |F | = pn for some n ≥ 1 then the order of ϕ is n.
B Note: If F1 and F2 are finite fields with |F1 | = |F2 |, then F1 ∼
= F2 .62
B Galois Correspondence for Finite Fields: Say k is the prime field, and F is an extension field with |F | = pn .
Consider an intermediate field k ⊆ E ⊆ F . Then these correspond to (respectively)
Autk (F ) ⊇ AutE (F ) ⊇ AutF (F ) = {idF }.
n
Here, |k| = p, |F | = p and n = [F : k] = [F : E][E : k] so if we let [F : E] = r and [E : k] = s, then n = rs, and
|E| = ps . Let ϕF be the Frobenius map on the field F . As a vector space E ∼
= k s . Also Autk (E) is a cyclic group of
∼
order s, and Autk (E) = Autk (F )/ AutE (F ). Since | AutE (F )| = [F : E] = r, it is clear that AutE (F ) is a group of
order r, but it is not clear which group of order r it is. Since | Autk (F )| = n and Autk (F ) = (ϕF ) and | AutE (F )| = r,
[k:k]
[E:k]
[F :k]
then AutE (F ) = (ϕsF ). Thus, we have k ⊆ E ⊆ F correspond to (ϕF ) ⊇ (ϕF ) ⊇ (ϕF ) = (1F ) respectively.
60 This
problem is still open and interesting.
is sufficient because we know that the group has order n = [K : k] = | Autk (K)| so if it has an element of order n then it will
be cyclic of order n.
62 This is true because of the uniqueness of splitting fields, since if |F | = pn , then F is the splitting field of xpn − x over k ∼ Z/pZ.
=
61 This
Math 901 Notes
3 January 2012
31 of 33
B Example(s):
– For every integer
n ≥ 1, there is a finite field, F of order pn where p is any prime. Namely, F is the splitting
pn
field for x − x over k = Z/pZ. If n is prime, then the only intermediate fields are F and k. (Due to the Galois
correspondence!)
– Let p = 31991. Note that this is prime. (We could use any other prime, but this one is large enough to be awkward
to raise to powers, and yet still relatively small in computer terms). We want to find all intermediate fields of the
field of order p12 . Note here that the lattice of subfields only depends on the exponent (and not on the prime itself).
This is because Autk (F ) depends only on n when k = Z/pZ and |F | = pn . First, we’ll find the lattice of subgroups
of the cyclic subgroup of order 12. For notational simplicity, we’ll denote the cyclic group of order n by Cn :
C1
C2
C3
C4
C6
C12
Thus, there are 6 intermediate fields, and their lattice is the same as for the lattice of subgroups, except the
bigger fields are at the top, rather than at the bottom of the picture.
– How many elements of F generate F over k if F and k are fields, p is prime, |k| = p and |F | = p6 ? Here, the
fn is the subfield
goal is to count all α ∈ F so that k(α) = F . We have the following lattice of subfields, where C
corresponding to the cyclic group of order n:
f1 = F
C
f2
C
f3
C
f6 = k
C
f2 |+|C
f3 |)+|C
f6 | = p6 −p2 −p3 +p.
f2 ∪ C
f3 ). Then we want to count the size of S.63 So, |S| = |F |−(|C
Let S = F \(C
5 December 2011
Transcendence Degree & Transcendence Bases
B Definition: Let k ⊆ K be fields, and S a subset of K. We say that the elements of S are algebraically
independent over k if for any r and any subset {s1 , . . . , sr } ⊆ S of r distinct elements, the k−homomorphism
ϕ : k[x1 , . . . , xr ] → K given by xi 7→ si is injective.
B Note: If {s1 , . . . , sr } ⊆ K is algebraically independent over k, then k(x1 , . . . , xr ) ∼
= k(s1 , . . . , sr ) and the isomorphism
can be given by xi 7→ si , which is a k−homomoprhism.
B Definition: If S ⊆ K is algebraically independent over k for some subfield k ⊆ K, and if K = k(S), then we say
that k ⊆ K is a purely transcendental extension.
B Example: Let k = Q, and K = Q(x). Then k ⊆ K is a purely transcendental extension (this is easily seen by using
the set S = {x}). However, x2 ∈ K, and so Q ( Q(x2 ) ( Q(x) so with respect to the set T = {x2 } (which is also an
algebraically independent set), it appears as if the extension is not purely transcendental since we get that k(T ) ( K.
B Definition: Let k ⊆ K be fields. Partially order the set of subsets S ⊆ K which are algebraically independent over
k by inclusion. We say a subset B ⊆ K which is algebraically independent over k is a transcendence base if B is
not properly contained in any other subset S ⊆ K which is algebraically independent over k.
B Example: Let k = Q, K = Q(x). Then {x} is a transcendence base for k ⊆ K. Also, {x2 } is a transcendence base
for the extension even though k(x2 ) ( K.
B Note: If S is a transcendence base for k ⊆ K, then k(S) ⊆ K is an algebraic extension.
B Comment: Sometimes it is impossible to pick a transcendence base S for a field extension k ⊆ K such that k(S) = K.
√
B Example: For example, Q ⊆ Q( 2) is an algebraic√extension and so its transcendence base is the empty-set. Also,
Q ⊆ Q(π) is purely transcendental, but Q ⊆ Q(π, 2) is probably not purely transcendental.
63 This
will be done using inclusion exclusion.
Math 901 Notes
3 January 2012
32 of 33
B Theorem: Let k ⊆ K be fields. Then there is a transcendence base for this extension.64
B Key Lemma: Let k ⊆ K be fields. Let S ⊆ K be algebraically independent over k. Let u ∈ K. Then T = S ∪ {u}
is algebraically independent over k if and only if u is transcendental over k(S).
Proof. We’ll prove this next time, but first state some corollaries now.
B Corollary 1: Let k ⊆ K be fields, with S ⊆ K an algebraically independent set over k. Then S is a transcendence
base for k ⊆ K if and only if k(S) ⊆ K is algebraic.
B Corollary 2: Let S1 and S2 be transcendence bases for k ⊆ K. If |S1 | < ∞, then |S2 | < ∞ and |S1 | = |S2 |.65
B Definition: We define the transcendence degree of k ⊆ K, denoted trdegk (K) to be |S| where S is a
transcendence basis for k ⊆ K.
7 December 2011
Transcendence Bases
B Corollary 3: Let k ⊆ K be fields, and let S ⊆ K be a subset. If K is algebraic over k(S), then S contains a
transcendence base for k ⊆ K.
B Corollary 4: Let k ⊆ K be fields. Let S ⊆ K be algebraically independent over k. Then there is a transcendence
base S 0 for k ⊆ K so that S ⊆ S 0 .
B Corollary 5: Let k ⊆ K ⊆ L be fields. Then trdegk (K) + trdegK (L) = trdegk (L).
B Note: We’re now ready to prove things:
– Proof. of Key Lemma: (⇒) We’ll prove this direction by contrapositive. So we’re assuming that u is
algebraic (i.e. not transcendental) over k(S) and we need to show that T = S ∪ {u} is not algebraically
d
X
gi i
independent over k. So let f = Irr(u, k(S), x) and write it as f (x) =
x where gi = gei (s1 , . . . , sr ) and
h
i=0 i
hi = e
hi (s1 , . . . , sr ) 6= 0 where gei , e
hi ∈ k[x1 , . . . , xr ] for some r ∈ N and s1 , . . . , sr ∈ S. Then, let fe =
d
X
gei i
x
e
hi
i=0
B
B
B
B
B
e =e
e fe is a polynomial, which we’ll call Fe.
and note that fe ∈ k(x1 , . . . , xr , x). Let H
h0 · e
h1 · . . . · e
hd . Then H
e 1 , . . . , sr )fe(s1 , . . . , sr , u) = H(s
e 1 , . . . , sr )f (u) = 0 as f (u) = 0. So Fe ∈ ker ϕ where
Then Fe(s1 , . . . , sr , u) = H(s
ϕ : k[x1 , . . . , xr , x] → K is given by xi 7→ si and x 7→ u. But Fe is not the zero polynomial since f 6= 0, so that
e 6= 0, and so Fe 6= 0. This means that ker ϕ 6= 0 and so T is not
fe 6= 0, and each hi 6= 0 so each e
hi 6= 0. Thus, H
algebraically independent.
(⇐) Now, we’ll assume that u is transcendental over k(S) and we need to show that T = S ∪ {u} is algebraically
independent over k. Let t1 , . . . , tr ∈ T . If ti 6= u for any i, then ti ∈ S for all i and hence t1 , . . . , tr are
algebraically independent, so that k[x1 , . . . , xr ] → K given by xi 7→ ti is injective. So instead suppose that ti = u
for some i. Without loss of generality, we may assume u = tr . Then ϕ : k[x1 , . . . , xr ] → K given by xi 7→ ti for
i < r and xr 7→ u = tr can be written as the composition of two maps. Namely, ϕ1 : k[x1 , . . . , xr ] → k(S)[xr ]
given by xi 7→ ti for i < r and xr 7→ xr and ϕ2 : k(S)[xr ] → K given by xr 7→ u. Then ϕ1 is injective as
t1 , . . . , tr−1 are algebraically independent over k and ϕ2 is injective as u is transcendental over k(S). Thus,
ϕ = ϕ2 ◦ ϕ1 is injective, so that T is algebraically independent over k.
– Proof. of Corollary 1: S is not a transcendence base if and only if there is some u ∈ K such that S ∪ {u} is
algebraically independent over k. By the Key Lemma, this is true if and only if u is transcendental over k(S)
which is true if and only if K is not algebraic over k(S).
Aside: If t1 , . . . , tr ∈ S and S is algebraically independent over k, then k[x1 , . . . , xr−1 ] → k(S) is injective where
xi 7→ ti . So in the proof of the Key Lemma we use the easy fact that k[x1 , . . . , xr−1 ][xr ] → k(S)[xr ] is also injective.
Comment: The empty-set is always algebraically independent.
Comment: Any subset of an algebraically independent set is also algebraically independent.
Comment: If α ∈ S with S a subset of K and k ⊆ K a subfield, then if S is algebraically independent over k, we
must also have that α is transcendental over k.
Note: We didn’t prove the remaining Corollaries in class, but they’re not terribly difficult so I’ll include them here.
– Proof. of Corollary 2: First suppose S1 = ∅. Then k(S1 ) = k so that K/k is an algebraic extension. Thus,
K has no elements that are transcendental over k and hence S2 = ∅. In particular, this gives |S2 | < ∞ and
|S1 | = |S2 |.
So now suppose that S1 = {s1 , . . . , sr } with si 6= sj if i 6= j, for some r ≥ 1. Then K/k(s2 , . . . , sr ) is not
algebraic since s1 is transcendental over k(s2 , . . . , sr ). Thus, there is a transcendental element in S2 , since
64 The
65 It
proof is just a simple application of Zorn’s lemma.
is true in general that they have the same cardinality, but that’s significantly harder to prove, and requires transfinite induction...ick!
Math 901 Notes
3 January 2012
33 of 33
k(s2 , . . . , sr )(S2 ) = K is not algebraic over k(s2 , . . . , sr ). So let σ1 ∈ S2 be transcendental over k(s2 , . . . , sr ). We
also get that s1 is algebraic over k(σ1 , s2 , . . . , sr ) since otherwise S1 would not be a transcendence base. Thus,
k(σ1 , s1 , . . . , sr ) is algebraic over k(σ1 , s2 , . . . , sr ), and since K is algebraic over k(σ1 , s1 , . . . , sr ), then K is also
algebraic over k(σ1 , s2 , . . . , sr ), then we have that K is algebraic over k(σ1 , s2 , . . . , sr ), so that {σ1 , s2 , . . . , sr }
is a transcendence base for the extension K/k. By induction, we can replace all of the elements of S1 with
elements of S2 . If we run out of elements of S1 to replace, or if we run out of elements of S2 to replace elements
of S1 with, then we get a contradiction to one of the sets being a transcendence base. Thus, we must have that
|S1 | = |S2 |.
– Proof. of Corollary 3: Let Σ be a maximal algebraically independent subset of S. Then K/k(S) is algebraic
and k(S)/k(Σ) is algebraic, so that K/k(Σ) is algebraic. Thus, Σ is a transcendence base for K/k by the first
corollary.
– Proof. of Corollary 4: Let Σ be a maximal algebraically independent subset of K which contains S. Then
K/k(Σ) is algebraic by the maximality, so that Σ is a transcendence base for K/k by corollary 1.
– Proof. of Corollary 5: Let S be a transcendence base for K/k and let T be a transcendence base for L/K.
Noe that T is automatically algebraically independent over k. If |S| = ∞, then L/k has a transcendence base
that contains S. If |T | = ∞, then L/k has a transcendence base that contains T , and so if either S or T is an
infinite set, then trdegk (L) = ∞ and the statement holds.
So now we assume that both S and T are finite sets. Let S = {s1 , . . . , sm } and T = {t1 , . . . , tn }. If S ∪ T is
not algebraically independent over k, then there is a nonzero polynomial f ∈ k[x1 , . . . , xm , y1 , . . . , yn ] such that
f (s1 , . . . , sm , t1 , . . . , tn ) = 0. Since f ∈ k[x1 , . . . , xm , y1 , . . . , yn ] ⊆ k(S)[y1 , . . . , yn ], and since T is algebraically
independent over K (and hence over k(S)), then we have that f (s1 , . . . , sm , y1 , . . . , yn ) is the zero element of
k(S)[y1 , . . . , yn ]. However, since the original polynomial f was not the zero polynomial, then some monomial in
the variables y1 , . . . , yn appears as a term in f with coefficient a nonzero element of k[x1 , . . . , xm ], which when
evaluated at s1 , . . . , sm gives zero. But this contradicts the fact that the set S is algebraically independent over k,
and hence S ∪ T is algebraically independent over k.
It remains to show that L is algebraic over k(S ∪ T ). However, we know that K is algebraic over k(S), so that
K(T ) is algebraic over k(S ∪ T ), and we also have that L is algebraic over K(T ). Thus, L is algebraic over
k(S ∪ T ), so that S ∪ T is a transcendence base for L/k and hence we get the result:
trdegk (K) + trdegK (L) = trdegk (L).
9 December 2011
topic here
B Hilbert & Rings of Invariants: Let G be a group, acting as k−automorphism on R = k[x1 , . . . , xn ] where k
is a field. The big question here is whether the ring of invariants, RG is always finitely generated over k. Hilbert
proved that the answer is yes for certain groups, called reductive groups. The key idea in the proof was that
R is Noetherian! Hilbert’s 14th problem was asking if the ring of invariants is always finitely generated. Zariski
reformulated the problem in terms of an intermediate field, and Nagata found a counterexample. Apparently Rees
was also instrumental in solving the problem, but we weren’t told how in class.
B Fact: Let k ⊆ E ⊆ K be fields such that K = k(S) for some finite subset S ⊆ K. Then E is finitely generated over
k.66
B Special Case: If k ⊆ E ⊆ K are fields and K is a purely transcendental extension of k with trdegk K < ∞, and if
E/k is algebraic, then it can be shown that E is finitely generated over k.
66 “It
is a challenging, but doable problem to prove this fact.” - B.H.