Math 221 - Algebra - GitHub Pages · 2018-12-12 · Math 221 - Algebra Taught by H ector Past en...

Math 221 - Algebra

Taught by Hector PastenNotes by Dongryul Kim

Fall 2016

The course was taught by Hector Pasten this semester, and we met on Mon-days, Wednesdays, and Fridays from 11:00am to 12:00pm. The textbooks weused were Introduction to Commutative Algebra by Atiyah and MacDonald, andRepresentation theory: A first course by Fulton and Harris. There were techni-cally 12 students enrolled according to the Registrar. There was one take-homefinal exam and no course assistance.

Contents

1 August 31, 2016 51.1 Rings . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 51.2 Ideals . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6

2 September 2, 2016 72.1 Operation on rings and ideals . . . . . . . . . . . . . . . . . . . . 72.2 Prime and maximal ideals . . . . . . . . . . . . . . . . . . . . . . 82.3 Radicals . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 8

3 September 7, 2016 103.1 Modules . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 103.2 Submodules and quotients . . . . . . . . . . . . . . . . . . . . . . 103.3 Direct sums and Direct products . . . . . . . . . . . . . . . . . . 113.4 Other stuff about modules . . . . . . . . . . . . . . . . . . . . . . 12

4 September 9, 2016 134.1 Nakayama’s lemma . . . . . . . . . . . . . . . . . . . . . . . . . . 134.2 Tensor products . . . . . . . . . . . . . . . . . . . . . . . . . . . . 14

5 September 12, 2016 165.1 Tensor products of algebras . . . . . . . . . . . . . . . . . . . . . 165.2 Free modules . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 17

1 Last Update: August 27, 2018

6 September 14, 2016 186.1 Rank of a free module . . . . . . . . . . . . . . . . . . . . . . . . 186.2 Direct limits . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 18

7 September 16, 2016 217.1 Exactness and flatness . . . . . . . . . . . . . . . . . . . . . . . . 217.2 Localization . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 22

8 September 19, 2016 248.1 Local properties . . . . . . . . . . . . . . . . . . . . . . . . . . . . 24

9 September 21, 2016 279.1 Primary ideals . . . . . . . . . . . . . . . . . . . . . . . . . . . . 279.2 Primary decomposition . . . . . . . . . . . . . . . . . . . . . . . . 28

10 September 23, 2016 3010.1 Associated primes . . . . . . . . . . . . . . . . . . . . . . . . . . 3010.2 Second uniqueness of primary decomposition . . . . . . . . . . . 31

11 September 26, 2016 3211.1 Integral algebras . . . . . . . . . . . . . . . . . . . . . . . . . . . 3211.2 Notion of a scheme . . . . . . . . . . . . . . . . . . . . . . . . . . 33

12 September 28, 2016 3512.1 Going-up theorem . . . . . . . . . . . . . . . . . . . . . . . . . . 3512.2 Geometric interlude: Morphisms . . . . . . . . . . . . . . . . . . 36

13 September 30, 2016 3713.1 Going down theorem . . . . . . . . . . . . . . . . . . . . . . . . . 37

14 October 3, 2016 3914.1 Noether normalization and Nullstellensatz . . . . . . . . . . . . . 39

15 October 5, 2016 4115.1 Chain conditions . . . . . . . . . . . . . . . . . . . . . . . . . . . 41

16 October 7, 2016 4316.1 Hilbert basis theorem . . . . . . . . . . . . . . . . . . . . . . . . 4316.2 Irreducible ideals and primary decomposition . . . . . . . . . . . 43

17 October 12, 2016 4517.1 Discrete valuation rings . . . . . . . . . . . . . . . . . . . . . . . 45

18 October 14, 2016 4718.1 Local Noetherian domain of dimension 1 . . . . . . . . . . . . . . 47

19 October 17, 2016 4919.1 Fractional ideals . . . . . . . . . . . . . . . . . . . . . . . . . . . 49

2

20 October 19, 2016 5120.1 Dedekind domains and fractional ideals . . . . . . . . . . . . . . 51

21 October 21, 2016 5321.1 Length of a module . . . . . . . . . . . . . . . . . . . . . . . . . . 53

22 October 24, 2016 5522.1 Field extensions . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5522.2 Splitting fields . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5522.3 Normal field extensions . . . . . . . . . . . . . . . . . . . . . . . 56

23 October 26, 2016 5723.1 Separable field extensions . . . . . . . . . . . . . . . . . . . . . . 57

24 October 28, 2016 5924.1 Galois extensions . . . . . . . . . . . . . . . . . . . . . . . . . . . 59

25 October 31, 2016 6225.1 Examples of field extensions . . . . . . . . . . . . . . . . . . . . . 62

26 November 2, 2016 6426.1 Solvability by radicals . . . . . . . . . . . . . . . . . . . . . . . . 64

27 November 4, 2016 6627.1 Representations of finite groups . . . . . . . . . . . . . . . . . . . 66

28 November 7, 2016 6828.1 Structure of finite representations . . . . . . . . . . . . . . . . . . 68

29 November 9, 2016 7029.1 Characters . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 70

30 November 11, 2016 7330.1 Character tables . . . . . . . . . . . . . . . . . . . . . . . . . . . 73

31 November 14, 2016 7631.1 Constructing irreducible representations of Sm . . . . . . . . . . 7631.2 Irreducibility of the Specht module . . . . . . . . . . . . . . . . . 77

32 November 16, 2016 7932.1 Specht modules are all the irreducible representations . . . . . . 8032.2 Restricted representation . . . . . . . . . . . . . . . . . . . . . . 80

33 November 18, 2016 8133.1 Tensor products for non-commutative rings . . . . . . . . . . . . 8133.2 Induced representation . . . . . . . . . . . . . . . . . . . . . . . . 8233.3 Characters of restricted and induced representation . . . . . . . . 82

3

34 November 21, 2016 8434.1 Artin induction theorem . . . . . . . . . . . . . . . . . . . . . . . 8434.2 Connections to analytic number theory . . . . . . . . . . . . . . . 85

35 November 28, 2016 8735.1 Lie groups . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 8735.2 The Haar measure and averaging . . . . . . . . . . . . . . . . . . 88

36 November 30, 2016 8936.1 Irreducible representations of S1 . . . . . . . . . . . . . . . . . . 8936.2 The tangent space . . . . . . . . . . . . . . . . . . . . . . . . . . 89

37 December 2, 2016 9137.1 Left-invariant vector fields of Lie groups . . . . . . . . . . . . . . 9137.2 The exponential map . . . . . . . . . . . . . . . . . . . . . . . . . 92

4

Math 221 Notes 5

1 August 31, 2016

This course will have four themes. The first one is commutative algebras: rings,modules, finitely generated modules, Noetherian modules, local rings, etc. Thesecond one is Galois theory, and we will do some review to make sure everyoneis on the same page. The third one is representations of finite groups. The wayto know about a group is to see how it acts one stuff, and we only know linearalgebra. So we deal with how the group acts on vector spaces. The fourth themeis Lie groups, which is a very large group. You would also want to look at therepresentation of the group, but this is too big. So you look at the tangent vectorgroups and get something called a Lie algebra, which is somewhere between thealgebra and the geometry.

We won’t have an in-class midterm, and there will be a take-home finalexam. You can collaborate on assignments, although you would have to putsome effort before asking. But you are not allowed to collaborate on the finalexam.

1.1 Rings

Definition 1.1. A ring is a set A with two binary operations +, ·, and anelement 0 ∈ A such that

(1) (A; 0,+) is an abelian group,

(2) · is associative,

(3) · is distributive over + (both from the right and the left).

We further say that A is commutative if · is commutative, and A is unitaryif there is an element 1 ∈ A such that 1 · x = x · 1 = x for each x ∈ A.

For some time, we will use “ring” to mean a “commutative unitary ring”.Also, most of the time, A 6= 0.

Definition 1.2. For rings A and B, a function f : A → B is called a ringmorphism (map) if f preserves +, ·, and 1.

Example 1.3. The map f : Z→ Z×Z with n 7→ (n, 0) is not a ring morphism.

One remark is that rings with there morphisms form a category. Still it isnot the nicest category because there is no obvious struction on Mor(A,B). Aring morphism f : A → B is an isomorphism if and only if f is bijective. Thisis a property, not a definition. For instance, in the category of smooth maps, amap might be bijective but not be an isomorphism.

Definition 1.4. For a ring A, we define

A× = x ∈ A : xy = 1 for some y ∈ A ⊆ A∗ = A \ 0.

The elements of A× are called units of A.

Math 221 Notes 6

When A× = A∗, we say that A is a field.

Definition 1.5. An element x ∈ A such that xy = 0 for some y ∈ A∗ is calleda zero devisor. If 0 is the only zero divisor, we say that A is an (integral)domain.

1.2 Ideals

Definition 1.6. Let A be a ring. An ideal of A is a subset a ⊆ A such that

(1) a is a subgroup of (A; 0,+),

(2) aA ⊆ a (which is equivalent to aA = a).

You can take quotients with ideals.

Proposition 1.7. Let A be a ring and a ⊆ A be an ideal. There is a unique ringstructure in the quotient group A/a such that the quotient map πa : A→ A/a isa ring morphism.

Let S ⊆ A be a subset. The ideal generated by S is

(S) = the smallest ideal of A containing S.

To show the existence, you take all the ideals containing S and intersect them.You can easily check that arbitrary intersections of ideals is an ideal. On theother hand, arbitrary unions of ideals is not necessarily an ideal. The rightstatement is that nested union of ideals is an ideal.

But this construction of (S) is not what you learn first. In kindergarten, youtake the span. In fact (S) can also be constructed as

(S) = ∑

finite sjαj : sj ∈ S, αj ∈ A.

Let f : A→ B be a ring morphism. The image im(f) is naturally a subring,and the kernel ker(f) is naturally an ideal. The universal property of the kernelstill holds for rings.

Proposition 1.8. Let A be a ring. For any ring morphism f : A → B, thereexists a unique map f : A/ ker f → B such that f · πker f = f .

A B

A/ ker f

f

πker f

f

Proposition 1.9. Let a be an ideal in A. There is a monotone bijection betweenthe ideals containing a and the ideals of A/a.

b ⊆ A :a ⊆ b

←→

ideals cof A/a

Math 221 Notes 7

2 September 2, 2016

The last thing we discussed the bijection between the ideals of A/a and theideals of A containing a.

ideals A/a

ideals in Acontaining a

b π−1

a (b)

πa(c) c

In this case, we got lucky, because an image of an ideal under f : A→ B is notnecessarily an ideal. At least when f is surjective this is true because it is aquotient. To this to be true, f has to bijective because f(A) needs to be botha subring and an ideal of B. So this works only in this case.

2.1 Operation on rings and ideals

If a, b ⊆ A are ideals, we define the following operations.

a + b = α+ β : α ∈ a, β ∈ b,a · b = (α · β : α ∈ a, β ∈ b).

When we take only the elements that are products, you don’t get an ideal.Two ideals a and b are called coprime if a+ b = (1). This does not require

unique factorization or anything.For rings Ai for i ∈ I 6= ∅, we define the product ring as∏

i∈IAi = (σ : I →

⋃Ai) : σ(i) ∈ Ai for each i ∈ I.

We are not assuming any cofinality of any sort. When I is finite the elementswill look like tuples. Addition and multiplication will be defined element-wise.

There is a natural projection map

pi :∏i∈I

Ai → Ai

σ 7→ σ(i).

This satisfies the universal property of the product. For any ring B and mapsfi : B → Ai, there is a unique map f : B →

∏i∈I Ai such that fi = pi f for

all i.

B∏i∈I Ai

Ai

f

fipi

Math 221 Notes 8

What about the direct sum of rings? One cautionary remark is that theobvious thing to do, which is what you do for groups, doesn’t work. The mapsAi → “

⊕A′′i maps 1 to something like (. . . , 0, 1, 0, . . .) and therefore is not a

morphism. Furthermore, if you take the direct sum of infinitely many rings,then you can’t define 1. The right notion is the tensor product, which we willdo soon.

Theorem 2.1 (Chinese remainder theorem). Let A be a ring and a1, . . . , an ⊆ Abe ideals. Define the map

φ : A→n∏i=1

A/ai.

in the obvious way. Then kerφ =⋃ni=1 ai and φ is surjective if and only if

ai + aj = (1) for every i 6= j.

The proof is left as an exercise.

2.2 Prime and maximal ideals

Definition 2.2. Let A be a ring and a ( A is an ideal.

(1) The ideal a is prime if for any x, y ∈ A, xy ∈ a implies x ∈ a or y ∈ a.

(2) The ideal a is maximal if for any ideal b ⊆ A, a ⊆ b implies a = b orb ⊆ A.

Proposition 2.3. The ideal a is prime if and only if A/a is a domain. Theideal a is maximal if and only if A/a is a field.

Proof. You can check this using the fact that the ideals of A/a corresponds toideals of A containing a.

Now we get to the first theorem.

Theorem 2.4 (Krull, 1929). Let a ( A be an ideal. There is a msximal idealm ⊆ A such that a ⊆ m.

Proof. Consider the set

Xa = b ( A ideal : a ⊆ b

ordered by ⊆. This is nonempty since a is in it, and chains are bounded.Therefore by Zorn’s lemma there exists a maximal element.

2.3 Radicals

Definition 2.5. Let A 6= (0). We define the Jacobson radical J(A) and thenilradical N(A) as

J(A) =⋂

m maximal

m, N(A) =⋂

p prime

p.

Math 221 Notes 9

Proposition 2.6. An element x is in J(A) if and only if 1−xy ∈ A× for everyy ∈ A.

Proof. You can do this.

Proposition 2.7. An element x is in N(A) if and only if x is nilpotent, i.e.,xn = 0 for some n ≥ 1.

Proof. For the ⇐ direction, you note that 0 ∈ p for any prime p. Then by thedefinition of a prime, you see that xm ∈ p implies x ∈ p.

For the ⇒ direction, suppose that x is not nilpotent. Consider the set

Σx = a ⊆ A : xn /∈ a for all n ≥ 1.

We check that (0) ∈ Σx so it is nonempty. Also, chains are bounded becausethe union of the chain will also be in Σx. Therefore Zorn’s lemma tells thatthere is a maximal element p.

We now claim that p is prime. Note that p 6= A. Let s, t ∈ A \ p. It sufficesto show that st ∈ A \ p. Observe that p + (s), p + (t) ) p. By the maximalityof p, there exist i, j ≥ 1 such that

xi ∈ p + (s) and xj ∈ p + (t).

Then the product is xi+j ∈ p+(st) and therefore we get st /∈ p by the definitionof Σx.

One corollary is that the set of nilpotent elements in A is an ideal. You canto it directly, but you will do to do something like binomial expansion. Anothercorollary is that A/N (A) is reduced, i.e., the only nilpotent element is 0. In asimilar vein, define the radical as

r(a) = x ∈ A : xn ∈ a for some n ≥ 1.

Then r(a) =⋂

a⊆p prime p.

Math 221 Notes 10

3 September 7, 2016

3.1 Modules

Let A be a ring.

Definition 3.1. An A-module is an abelian group (M, 0,+) together with amap · : A×M →M satisfying

(1) 1 acts as IdM , i.e., 1 · v = v for any v ∈M ,

(2) for every α, β ∈ A and v ∈M , (αβ)v = α(βv),

(3) for every α, β ∈ A and v ∈M , (α+ β)v = αv + βv,

(4) for every α ∈ A and v, w ∈M , α(v + w) = αv + αw.

This is a generalization of the notion of vector spaces. If A is a field, A-modules are the same as A-vector spaces. If A = Z, then A-modules are justabelian groups. To be precise, there is a forgetful functor from the category ofZ-modules to the category of abelian groups that is an equivalence. We alsonote that if a ⊆ A is an ideal, then it is also an A-module.

We now define morphisms between A-rings. The definition requires the twomodules to be over the same ring.

Definition 3.2. A morphism of A-modules is a morphism f : M → N ofabelian groups which is A-linear, i.e., f(αv) = αf(v).

One obvious remark is that for a fixed A, the A-modules with their mor-phisms form a category. In this category, isomorphisms are the same as bijectivemorphisms.

Unlike the case of rings, the set of morphisms HomA(M,N) from M to Ninherits a structure of an A-module.

Example 3.3. Let M be an A-module. We define its dual module as

M∨ = HomA(M,A).

There is no reason for M∨ to be isomorphic to M . For instance, take A = Zand M = Z/2Z. Likewise, M and M∨∨ are not isomorphic.

Example 3.4. Fix f ∈ EndA(M). Then we can make M into an A[x]-moduleby letting x · v = f(v). You need to be a bit careful here, because if I give youtwo endomorphisms f, g ∈ EndA(M), the A-module M inherits the structure ofan A[x, y]-module only when f and g commutes.

3.2 Submodules and quotients

Although subrings are not what you really want to consider, submodules workquite well.

Math 221 Notes 11

Definition 3.5. Let M be an A-module. A submodule N is a subgroupN ≤M wich is an A-module with the A-action on M .

If f ∈ HomA(M,N), then ker f ≤M and im f ≤ N . You can take quotients,intersections, nested unions. You can also take the sums for a collection Ni ofsubmodules of M as ∑

i∈INi =

∑finite

vi : vi ∈ Ni.

If S ⊆M is a subset, then we define the submodule generated by S as

〈S〉 =⋂

S⊆N, N≤M

N.

Also if M is and A-module and a ⊆ A is an ideal, then we can define

aM = ∑

finite αivi : αi ∈ a, vi ∈M ≤M.

The isomorphism theorems from group theory works the same for modules.You would need to check that the abelian group structure is compatible withscalar multiplication, but this is kind of obvious.

Proposition 3.6. (1) Let f ∈ HomA(M,N). Then there exists a unique f ∈HomA(M/ ker f,N) such that f qker f = f .(2) Let P ≤ N ≤M be A-modules. Then

(M/P )/(N/P ) ∼= M/N

(3) Let N1, N2 ≤M . Then

N2/(N1 ∩N2) ∼= (N1 +N2)/N1

3.3 Direct sums and Direct products

Let Mi for i ∈ I be A-modules. The direct product is defined as∏i∈I

Mi = σ : I →⋃i∈IMI : σ(i) ∈Mi for all i.

The direct sum is defined as⊕i∈I

Mi = σ : I →⋃i∈IMi : σ(i) ∈Mi for all i and has finite support.

There are natural maps pi :∏Mj →Mi which is the projection and ιi : Mi →⊕

Mj which is extension by zero.There are universal properties.

Proposition 3.7 (Universal property of the product). Let N be an A-module.For every collection of fi ∈ HomA(N,Mi) there is a unique f ∈ HomA(N,

∏Mi)

such that fi = pi f for all i ∈ I.

Math 221 Notes 12

Proposition 3.8 (Universal property of the direct sum). Let N be an A-module. For every collection of gi ∈ HomA(Mi, N) there is a unique g ∈HomA(

⊕Mi, N) such that gi = g ιi for all i ∈ I.

Here is a good exercise to do. Suppose you have another object that satisfiesthe same property. Using only this property, you can prove that what we haveconstructed and that other object must be isomorphic.

3.4 Other stuff about modules

Let M be an A-module. We define the annihilator of M as

Ann(M) = α ∈ A : α ·M = 0.

This is an ideal. We remark that M is an A/Ann(M)-module in a natural way,and that it is faithful:

AnnA/Ann(M)(M) = (0) ⊆ A/Ann(M).

Definition 3.9. We say that M is finitely generated if there exists a finitesubset S ⊆M such that M = 〈S〉.

Proposition 3.10. M is finitely M ∼= An/K for n ≥ 0 and K ≤ An.

Proof. Because M is finitely generated, there is some S ⊆ M with S finite.Then we take n = |S| and then there is a surjective map An → M . Then wecan quotient it out by the kernel of this map.

We are going to need this lemma next class.

Lemma 3.11. Let R be a ring and T ∈Mn×n(R) be a matrix with coefficientsin R. There is another matrix T ′ ∈Mn×n(R) such that T ′T = det(T ) · Id.

We note that detT is well-defined as an element of R. As long as you don’tdivide, you can do linear algebra over a ring.

Math 221 Notes 13

4 September 9, 2016

We were discussing finitely generated modules.

4.1 Nakayama’s lemma

Proposition 4.1 (Cayley-Hamilton). Let M be a finitely generated A-moduleand a ⊆ A be an ideal and f ∈ EndA(M) such that f(M) ⊆ aM . Then fsatisfies an equation of the form

fn + α1fn−1 + · · ·+ αn = 0

with αi ∈ ai, as a endomorphism of M .

Proof. Consider M as an A[x]-module with x acting as f . Let v1, . . . , vn ∈ Mbe the generators and let ~v = (vi)i ∈ M = Mn. Write f(vi) =

∑j βijvj with

βij ∈ a with some choice of βij .Let T = [δijx− βij ]i,j ∈ Mn×n(R) with R = A[x]. In more familiar terms,

T = Ix− [βij ].This M is an R-module with coordinate-wise action. Also T acts on M by

matrix multiplication. The matrix T is chosen so that T · ~v = 0.As in the previous lemma, there exists a matrix T ′ ∈ Mn×n(R) such that

T ′T = (detT )I. Then

(detT )~v = (detT )I~v = T ′T~v = T ′0 = 0.

Because vi generate M , detT ∈ R annihilates M , i.e., acts as zero. ExpandingdetT gives the result.

Many problems, like injectivity and surjectivity, reduces to knowing whethera module is the zero module.

Lemma 4.2 (Nakayama 1). Let M be a finitely generated A-module and leta ⊆ A be an ideal. If aM = M then there exists an x ≡ 1 (mod a) such thatxM = 0.

Proof. Take f = id. This is a hard choice because you don’t expect this to givea nontrivial result. Apply Cayley-Hamilton. Then

x = 1 + α1 + α2 + · · ·+ αn

acts as 0.

Lemma 4.3 (Nakayama 2). Let M be a finitely generated A-module and leta ⊆ A be an ideal. If aM = M then there exists an y ∈ a acting as id on M .

Proof. Take y = 1− x.

Lemma 4.4 (Nakayama 3). Let M be a finitely generated A-module and a ⊆J(A). If aM = M then M = 0.

Math 221 Notes 14

Proof. By Nakayama 1, there is an x such that xM = 0 and 1− x ∈ a ⊆ J(A).Then x must be a unit. Then 0 = x−1xM = M .

Lemma 4.5 (Nakayama 4). Let M be a finitely generated A-module and a ⊆J(A) be an ideal. If N ≤M and N + aM = M then M = N .

Proof. Take the quotient module.

Corollary 4.6. Consider a local ring (A,m), i.e., m is only the maximal ideal.Let M be a finitely generated module and k = A/m be the residue field. Thequotient V = M/mM is a finite dimensional k-vector space. Let v1, . . . , vn ∈Mbe lifts of the basis of V . Then v1, . . . , vn actually generate M .

Proof. Let N = 〈v1, . . . , vn〉. The composition φ of the inclusion N → M andthe projection M → V must be surjective.

N M

V = M/mM

φ

Because φ is surjective, N + mM = M . Since m = J(A), we can applyNakayama 4 and get the result.

4.2 Tensor products

A bilinear map is a map b : M ×N → P such that for each v ∈M , b(v,−) ∈HomA(N,P ) and for each w ∈ N , b(−, w) ∈ HomA(M,P ).

Proposition 4.7. Let M and N be A-modules. There exists an A-moduleM ⊗AN with a bilinear map τM,N : M ×N →M ⊗AN satisfying that for eachbilinear b : M × N → P , there exists a unique map b′ ∈ HomA(M ⊗A N,P )such that b = b′ τM,N .

M ×N P

M ⊗A N

b

τM,N ∃!

This M⊗AN is called the tensor product. We also denote τ(v, w) = v⊗w.These pure tensors do not give all the elements of M ⊗A N , but generate it.

These pure tensors satisfy many relations, so you always need to checkwhether a formula involving pure tensors actually gives a map from the ten-sor product. For instance in Z/2Z⊗Z Z/3Z,

1⊗ 1 = 3⊗ 1 = 1⊗ 3 = 1⊗ 0 = 0(1⊗ 1) = 0.

In fact, you can show that Z/2Z⊗Z Z/3Z = 0.

Proposition 4.8. Tensor products have the following properties:

Math 221 Notes 15

(1) A⊗AM ∼= M .

(2) M ⊗A N ∼= N ⊗AM .

(3) (M ⊗A N)⊗A P ∼= M ⊗A (N ⊗A P ).

(4) Tensor products commute with arbitrary direct sums.

You can prove all these by using arrows, not pure tensors.For maps f ∈ Hom(M,M ′) and g ∈ Hom(N,N ′), you can define a map

f ⊗ g ∈ Hom(M ⊗N,M ′ ⊗N ′). We can do this by using arrows.

M ×N M ′ ×N ′

M ⊗N M ′ ⊗N ′

f×g

bil.τ τ ′

∃!

For a ring A, an A-algebra is a ring B with a ring map A → B. This issimply a ring that is also a A-module.

If M is an M -module and B is an A-algebra, then MB = B ⊗A M is aB-module. This is because for any x ∈ B, multiplication by µx gives a mapµx ⊗ id ∈ EndA(B ⊗AM). This gives B ⊗AM an B-module structure.

Math 221 Notes 16

5 September 12, 2016

Last time we were talking about tensor products. For A-modules M and N ,there is another module M ⊗A N with a universal bilinear map τ : M ×N →M ⊗A N . The pure tensors are v ⊗ w = τ(v, w). By bilinearity of τ , thesepure tensors satisfies some relations. To construct M⊗AN , you can write downall the pure tensors, construct the free A-module generated by them, and thenquotient by the submodule generated the relations. This way, you kill exactlywhat you need to be zero, and so it satisfies the universal property.

Given two maps f : M → N and g : M ′ → N ′, there is a map

f ⊗ g : M ⊗AM ′ → N ⊗A N ′

v ⊗ w 7→ f(v)⊗ g(w).

This is actually well-defined, because you can describe it through the universalproperty.

M ×M ′ N ×N ′

M ⊗AM ′ N ⊗A N ′τ

f×g

τ

Suppose you have an A-module M and an A-algebra B. An A-algebra is justa ring B with a map ψ : A→ B. This map allows B to be seen as an A-module,because for α ∈ A and β ∈ B, we can define α · β = ψ(α) ·B B. Now you wantto make M into a B-module. Why would you want to do that? Suppose thatyou have a linear algebra problem but are working in the reals. You want tolook at the eigenvalues but there might exist no eigenvalues in the reals. So youwork in the complex numbers for a while. This is extending the scalar.

Define MB = B ⊗A M . We can give a B-module structure on B ⊗A M asβ(γ ⊗ v) = (βγ) ⊗ v. Why is this well-defined? That map can be described asthe tensor product of the multiplication map µβ : B → B and id : M → M .Then

(µβ ⊗ id)(γ ⊗ v) = (µβ(γ)⊗ v) = (βγ)⊗ v.

These actions of B on MB gives a map B ×MB →MB given by

(β, x) 7→ (µβ ⊗ id)(x).

5.1 Tensor products of algebras

There is a notion of tensor products of A-algebras. Let B and C be A-algebras.Let us look at B ⊗A C first as an A-module. Then this is CB ∼= BC . But it isalso a ring, because you can define

(β ⊗ γ) · (β′ ⊗ γ′) = (ββ′)⊗ (γγ′).

Math 221 Notes 17

The magic of this is that we have canonical ring maps tB : B → B ⊗A C andtC : C → B ⊗A C given by

tB : β 7→ β ⊗ 1 = β(1⊗ 1), tC : γ 7→ 1⊗ γ = γ(1⊗ 1).

These maps are good because they also respects multiplication. The data (B⊗AC, τB,C , tB , tC) has a universal property.

Proposition 5.1. Let A be a ring and B, C be A-algebras. Let D be an A-algebra with A-algebra maps vB : B → D and vC : C → D. Then there exists aunique A-algebra map h : B ⊗A C → D such that

D

B B ⊗A C CtB

uBh

tC

vC

commutes.

Proof. We first build a A-linear map B ⊗A C → D. To get this, we need aA-bilinear map m : B × C → D. We use the map given by

(β, γ)→ uB(β) ·D uC(γ).

The map h is going to be unique map that factors m. This map h actuallysatisfies the diagram and so this shows existence of h. This does not quite proveuniqueness because this uniqueness of h coming from the universal property oftensor products is not quite the uniqueness we want. We first see that the imageof pure tensors are determined. For any pure tensor β ⊗ γ = (β ⊗ 1)(1 ⊗ γ).Then h(β ⊗ 1) = uB(β) and h(γ ⊗ 1) = uC(γ). Then if you want h to be a ringmap, then h(β ⊗ γ) = uB(β)uC(γ). This shows uniqueness.

5.2 Free modules

We assume A 6= (0) for the discussion of free modules. Let M be an A-moduleand S ⊆M be a subset. Then S is

(1) a generating set if M = 〈S〉,(2) a free set if the elements of S only satisfy the trivial A-linear relation,

and

(3) a basis if it is a generating set and a free set.

The module M is a free A-module if it has a basis.

Example 5.2. Take A = Z. Then M = Z is a free module with basis 1. Theset S = 2, 3 is a generating set but not a free set. Although this is a minimalgenerating set, it is not a basis. The module M = Z/2 is not free because it hasno nonempty free set.

Proposition 5.3. A module M is free with basis xii∈I if and only if M ∼= A(I)

where A(I) denotes A direct summed I times.

Math 221 Notes 18


Last time we were discussing free modules. Every free module is nothing butA(I) up to isomorphism. Still this does not trivializes the theory of free modules.

Proposition 6.1. Let M be a free A-module and let S be a basis for M . Givenany A-module N and a set map σ : S → N , there is a unique A-module mapf ∈ HomA(M,N) with f |S = σ.

This is just linear algebra.

6.1 Rank of a free module

Proposition 6.2. Let B be an A-algebra. If M is a free A-module with basisxii∈I , then MB is a free B-module with basis 1B ⊗ xii∈I .

Proof. Consider the case of A(I) with xi = ei. The result follows from thattensor product commutes with arbitrary direct sum.

Proposition 6.3. There is an isomorphism A(I) ∼= A(J) if and only if |I| = |J |.

Proof. Let m ⊆ A be a maximal ideal. Tensor with A/m. Then (A/m)(I) ∼=(A/m)(J) and this is now just linear algebra.

Definition 6.4. Let M be a free module. The rank of M is defined as thecardinal rkAM = |I| for I satisfying M ∼= A(I).

Proposition 6.5. Let M be a free A-module of finite rank r = rkAM .

(1) Every set of r generators is a basis.

(2) Every set of generators has at least r elements.

Proof. (a) Let xiri=1 be a generating set. Let v1ri=1 be a basis. Considerσ : vi →M mapping vi 7→ xi. By the universal property, there uniquely existsf ∈ EndAM such that f : vi 7→ xi. Note that f is surjective. By Nakayama, fis an isomorphism.

For M and N of finite rank, clearly rkA(M ⊕ N) = rkAM + rkAN . AlsorkA(M ⊗N) = rkAM · rkAN .

6.2 Direct limits

Consider a poset (I,≤) such that for every i, j ∈ I there exists k ∈ I withi, j ≤ k. (This is usually known as a filtered.) We are going to call this adirected set.

Given a direct set I, a direct system ofA-modules is a pair (Mii∈I , µiji,j∈I,i≤j)with

• µij : Mi →Mj (or µij ∈ HomA(Mi,Mj))

Math 221 Notes 19

• µij = µkj µij for any i ≤ k ≤ j.

We construct the direct limit of this system in the following way. Take

lim−→Mi = ⊕Mi/D where D = 〈ιi(xi)− ιj(µij(xi)) : i ≤ j, xi ∈Mi〉

that comes with the natural maps

µj = pDιj : Mj → lim−→Mj .

We call (lim−→Mi, µii∈I) the direct limit of the system. We have compressedthe information of the system into this.

Proposition 6.6 (Universal property of the direct limit). Let (Mi, µij) bea direct system. For any A-module N with maps fi ∈ HomA(Mi, N) satisfyingfi = fjµij (for all i ≤ j) there exists a unique f ∈ HomA(lim−→Mi, N) such thatfi = fµi for all i.

This is built out of two universal property: that of the direct sum, and thatof the quotient.

Let M be an A-module. Consider

I = N ≤M : N is finitely generated

which is a poset by inclusion and is also filtered because the sum of finitelygenerated modules is finitely generated.

Proposition 6.7. With M and I as above, we have M ∼= lim−→Mi where Mi = i.

This is a powerful tool because if you see that some property is preservedunder direct limits, and you know it is true for finitely generated modules, thenyou can conclude that it is true for all modules.

Proof. Check that (M, Mi →M) satisfies the universal property.

Proposition 6.8. Let (Mi, µij) be a direct system and let N be an A-module.Then

(lim−→Mi)⊗N ∼= lim−→(Mi ⊗N).

Proof. For eveyr i, there is a map τ : Mi × N → Mi ⊗ N . Apply lim−→ and weget a map τ : (lim−→Mi)×N → lim−→(Mi⊗N) that is bilinear. So by the universalproperty of the tensor product, we get a map

τ ′ : (lim−→Mi)⊗N → lim−→(Mi ⊗N).

Now let us get the other direction. For every i, we consider the map µi⊗ id :Mi ⊗ N → (lim−→Mi) ⊗ N . Choose fi = µi ⊗ id. By the universal property, weget a map

lim−→µij⊗id

(Mi ⊗N)→ (lim−→Mi)⊗N.

Check that they are inverses.

Math 221 Notes 20

Here is a useful construction of direct limit of morphisms. Consider twodirect systems Mi and Nj on the same index set I, and maps φi : Mi → Ni thatare compatible with the direct system, i.e.,

Mi Ni

Nj Nj

φi

µij µ′ij

φj

commutes for all i ≤ j. How do we compute lim−→φi? We first compoes φj withthe natural maps Ni → lim−→Ni to get the maps fi. Then we can use the universalproperty to get lim−→φi.

Mi Ni

lim−→Mi lim−→Ni

φi

fiµi

This suggest that there is a functor from the category of directed systems tomodules.

Math 221 Notes 21


Today we are going to introduce two notions: flatness and localization.

7.1 Exactness and flatness

We say that a sequence

M N Pf g

is exact if ker g = im f . A short exact sequence is a sequence

0 M N P 0

that is exact at M , N , and P .

Proposition 7.1. Let S be a sequence

M N P 0f g

of A-modules. Then the following are equivalent:

(i) S is exact.

(ii) For any L, the following sequence is exact.

0 Hom(P,L) Hom(N,L) Hom(M,L)

The functor HomA(−, L) is contravariant and HomA(L,−) is covariant. Wethen say that the functor HomA(−, L) is left exact, because the 0 is on theleft after applying the functor. This is weaker than every exact sequence beingsent to an exact sequence.

Proposition 7.2. Let N be a A-module. If S : M → P → Q → 0 is exact,then M ⊗N → P ⊗N → Q⊗N → 0 is exact.

Proof. We apply the previous proposition. For every L, the sequence

0 Hom(Q,Hom(N,L)) Hom(P,Hom(N,L)) Hom(M,Hom(N,L)).

is exact. There is a canonical isomorphism Hom(Q,Hom(N,L)) ∼= BilA(Q ×N,L) ∼= Hom(Q⊗N,L). Now use the previous proposition again.

We note that the functor − ⊗ N is not always left exact. Consider theinjection Z → Z given by multiplication by 2. If we tensor by Z/2Z, then weget Z/2→ Z/2 that is also multiplication by two. This is not injective.

Proposition 7.3. The functor N ⊗A − preserves injective maps if and only ifit is exact.

Math 221 Notes 22

Definition 7.4. N is a flat A-module if N ⊗− is exact.

For instance, Z/2 is not a flat Z-module.

Proposition 7.5 (See assignment). (1) The direct sum⊕Mi is flat if and

only if each Mi is flat.

(2) If M is a flat A-module and B is an A-algebra, then MB is a flat B-module.

Example 7.6. A is a flat A-module. Free modules are flat, because it is adirect sum of flat modules. So A[x] is a flat A-module, and vector spaces areflat k-modules.

Theorem 7.7. For an A-module N , the following are equivalent:

(1) N is flat.

(2) For every ideal a ⊆ A, the map a⊗A N → N is injective.

Proof. For (1)⇒ (2), we have an exact sequence 0→ a→ A. Then 0→ a⊗A→A⊗N ∼= N is exact.

This is terribly complicated. See [Liu] Theorem 2.4. The other short proofuses Tor functors.

7.2 Localization

Definition 7.8. A set S ⊆ A is a multiplicative set if 1 ∈ S and S is closedunder multiplication.

For example, S = 1, x, x2, . . . is a multiplicative set for any s ∈ A. Theset S = A \ p is also a multiplicative set for any prime ideal p.

Now let S ⊆ A be a multiplicative set. On A× S define the relation

(α, s) ∼ (βt) ⇔ ∃r ∈ S such that r(αt− βs) = 0.

Then we define the localization S−1A = A × S/ ∼. It can be verified thatS−1A is a ring with the obvious formulas for + and · (of fractions). We denoteα/s = [(α, s)].

It also comes with a natural map

λS : A→ S−1A, α 7→ α/1.

that is a ring map. Thus S−1A is an A-algebra in a natural way. This λS iscalled the localization map. The invertible elements are mapped to λS(A×) ⊆(S−1A)×, and also S is mapped into λS(S) ⊆ (S−1A)×.

Proposition 7.9. Let A be a ring and S ⊆ A a multiplicative set. For allA-algebras f : A → B satisfying f(S) ⊆ B×, there exists a unique ring mapf ′ : S−1A→ B with f ′ λS = f .

Proof. If you want a reference, this is [AM] Proposition 3.1.

Math 221 Notes 23

We denote Ax = 1, x, x2, . . .−1A and Ap = (A \ p)−1A for a prime idealp ⊆ A.

There is also a module version for localization. Let A be a ring and S ⊆ Abe a multiplicative set. Let M be an A-module. We define S−1M = M × S/ ∼where the relation is defined by

(v, s) ∼ (w, t) ⇔ ∃r ∈ S, r(tv − sw) = 0.

The resulting S−1M is an S−1A-module, with the action given by (α/s)·(v/t) =(αv/st).

Again, this has a universal property. As an exercise, write down the universalproperty for λM,S : M → S−1M .

From a categorical point of view, the next thing we have to do is localiza-tion of morphisms. Let A be a ring and S ⊆ A be a multiplicative set. Forf ∈ HomA(M,N), we want to make S−1f ∈ HomS−1A(S−1M,S−1N). If wesucceed, we realize S−1 as a functor from the category of A-modules to thecategory of S−1A-modules. This is easy once we have the universal propertyfor S−1A-modules.

M N

S−1M S−1N.

f

λM,S λN,S

∃!=S−1f

So S−1 defines a functor.

Theorem 7.10. This functor is exact.

Proof. Take any exact sequence

M N Pf g

of A-modules. First we need to check that im fS ⊆ ker gS , but this is clearbecause g f = 0 and then by localizing by S, we get gS fS = 0.

To show im fS ⊇ ker gS , let v/s ∈ ker gS . Then 0 = gS(v/s) = g(v)/s, and sothere exists and r ∈ S such that rg(v) = 0. So g(rv) = 0 and rv ∈ ker g = im f .Let rv = f(w) with w ∈M . Then

fs

( wrs

)=f(w)

rs=rv

rs=r

s.

Math 221 Notes 24


Let me talk a little bit more about localization.

Proposition 8.1. Let S ⊆ A be a multiplicative set and let M be an A-modulethe map MS−1A = S−1A⊗AM → S−1M by (1/s)⊗v 7→ v/s is an isomorphism.

Proof. There is a bilinear map f : S−1A ×M → S−1M given by (α/s, r) 7→αv/s. This gives a map f ′ : S−1A ⊗A M → S−1M that is clearly surjective.You can check injectivity on elements.

Corollary 8.2. S−1A is a flat A-algebra.

Corollary 8.3. S−1(M ⊗A N) ∼= S−1M ⊗S−1A S−1N .

Proof. We have S−1M ⊗S−1A S−1N ∼= (M ⊗A S−1A) ⊗S−1A S

−1N . By thebimodule property, the tensor product commutes and this is isomorphic to

M ⊗A (S−1A⊗S−1A S−1M) ∼= M ⊗A S−1A ∼= M ⊗A (N ⊗A S−1A)

∼= (M ⊗A N)⊗A S−1A ∼= S−1(M ⊗A N).

Likewise, tensoring by a quotient has a natural interpretation. For any ideala ⊆ A, there is an exact sequence

0 a A A/a 0.

Then we can tensor with M to get

a⊗M ∼= aM A⊗M ∼= M A/a⊗M 0.

This implies that (A/a)⊗M ∼= M/aM naturally.In a similar way, you can also get S−1(M/N) ∼= S−1M/S−1N .

8.1 Local properties

Proposition 8.4. Let M be an A-module. The following are equivalent:

(1) M = 0.

(2) Mp = 0 for all prime p ⊆ A.

(3) Mm = 0 for all maximal m ⊆ A.

Proof. It is obvious that (1)⇒ (2)⇒ (3). Let us now prove (3)⇒ (1). Assume(3) with M 6= 0. There is an 0 6= v ∈ M , so 1 /∈ AnnA(v). So there exists amaximal m such that AnnA v ⊆ m.

Let us now look at v/1 = 0 in Mm = 0. There should be an r ∈ A \ msuch that rv = 0 in M . Then r is both in AnnA(v) and A \ m and so we get acontradiction.

Math 221 Notes 25

Proposition 8.5. Let f ∈ HomA(M,N). The following are equivalent:

(1) f is injective.

(2) fp : Mp → Np in injective for all prime p ⊆ A.

(3) fm : Mm → Nm is injective for all maximal m ⊆ A.

The same holds for “surjection” and “isomorphism”.

Proof. (2) ⇒ (3) is obvious. (1) ⇒ (2) directly follows from that localizationis exact. (3) ⇒ (1) follows from the exactness of localization and the previousproposition.

Proposition 8.6. Let N,P ≤M be A-modules. The following are equivalent:

(1) N ⊆ P .

(2) Np ⊆ Pp for all prime p ⊆ A.

(3) Nm ⊆ Pm for all maximal m ⊆ A.

The proof is an exercise.

Proposition 8.7. Exactness of sequences is local.

Proposition 8.8. Flatness is local.

Proof. (2) ⇒ (3) is trivial. Let us prove (1) ⇒ (2). Let M be a flat A-module.Then Ap ⊗M is a flat Ap-module. But this is Mp and so we are done.

Now we check (3) ⇒ (1). Let N → P be an injective map of A-modules.We need to show that if (3) holds for M then N ⊗AM → P ⊗AM is injective.Because N → P is injective, Nm → Pm is also injective. Because Mm is flat,Nm ⊗Am

Mm → Pm ⊗AmMm is injective for every m. Because the diagram

Nm ⊗AmMm Pm ⊗Am

Mm

(N ⊗AM)m (P ⊗AM)m

∼= ∼=

commutes. Then N ⊗AM → P ⊗AM is injective.

An important cautionary remark is that freeness is not local. Locally free isnot the same as free.

Theorem 8.9. Let M be a finitely generated A-module. Then M is flat if andonly if M is locally free, i.e., the following are equivalent:

(1) M is a flat A-module.

(2) Mp is a free Ap-module for each prime p ⊆ A.

(3) Mm is a free Am-module for each maximal m ⊆ A.

Math 221 Notes 26

Proof. Use that flatness is local and problem 1 of assignment 4, which statesthat if A is a local ring then a finitely generated module is flat if and only if itis free.

Proposition 8.10. Let S ⊆ A be a multiplicative set.

(1) Every ideal of S−1A is of the form (λS(a)) for an ideal a ⊆ A.

(2) We have a monotone bijection:p ⊆ A prime :S ∩ p 6= ∅

←→ SpecS−1A = q ⊆ S−1A prime.

(3) S−1 respects the following operations on ideals: finite (not direct) sums,products, finite intersection, and radicals.

In the last three minutes, let me say a word on Spec.For a ring A, we define

X = SpecA = p ⊆ A : p prime.

We want to do geometry here (on X) and we want to see elements of A as“functions” on X. We also want to see A-modules as sheaves of X. Later weare going to put a topology on X.

Math 221 Notes 27


Today we are going to discuss primary decomposition.

9.1 Primary ideals

Definition 9.1. An ideal q ⊆ A is primary if xy ∈ q implies x ∈ q or yn ∈ qfor some n ≥ 1. In other words, q is primary if and only if A/q has the propertythat all its zero divisors are nilpotent.

So maximal implies prime implies primary. For example, (x2) ⊆ k[x] isprimary. If f : A→ B is any ring morphism and q ⊆ B is an ideal, then q primeimplies f−1q is prime. Likewise q primary implies f−1(q) is primary. This isnot true for maximal ideals.

Proposition 9.2. Let q ⊆ A be an ideal.

(1) If q is primary, then r(q) is the smallest prime containing q.

(2) If r(q) is maximal, then q is primary.

Proof. (1) It is enough to show that r(q) is prime, because r(q) is the intersectionof all primes containing q. You can check it on elements.

(2) Say r(q) = m is maximal. Then q ⊂ m, and so m is maximal in A/q,with

m = rA(q) = rA/q(0) = N(A/q) =⋂

p⊆A/q

p.

From this we conclude that m is the only maximal ideal.Now let x ∈ A/q be a zero divisor. Then (x) is contained in some maximal

ideal, but m is the only maximal ideal. So x ∈ m = N(A/q). So x is nilpotent.

It is not hard to check that if there exists a maximal ideal m such thatmj = (0) for some j, then m must be the only prime ideal.

Corollary 9.3. Powers of maximal ideals are primary.

Definition 9.4. Let q ⊂ A be primary. We say q is p-primary if p = r(f).

The way you want to think about this is that p is a shadow of q.

Proposition 9.5. Finite intersections of p-primary ideals is p-primary.

This will be an exercise. On the other hand, it is not true (in general) thatfinite intersection of primary is primary. Here is an example. Let A = k[x, y]and q1 = (x), q2 = (y). They are both primary, but their intersection is notprimary.

Math 221 Notes 28

Definition 9.6. Take a ⊆ A be an ideal, and let x ∈ A. The transporter ofx to a is defined as

(a : x) = y ∈ A : yx ∈ a.

This is clearly an ideal. Using the quotient ring, we can also write (a : x) =AnnA(x mod a) ⊆ A. Also obviously a ⊆ (a : x), and if x ∈ a then (a : x) = A.

Proposition 9.7. Let q ⊆ A be p-primary.

(1) If x /∈ q, then (q : x) is p-primary.

(2) If x /∈ p, then (q : x) = q.

Proof. (1) Let y ∈ (q : x). Then xy ∈ q, and so y ∈ r(q) = p. This shows thatq ⊆ (q : x) ⊆ p, and when we apply r, we get p = r(q : x). The only thing weneed to check is that (q : x) is primary.

Let yz ∈ (q : x) with y /∈ (q : x). Then xyz ∈ q and xy /∈ q, and so zk ∈ qfor some k. Then zk ∈ (q : x). This finishes the proof.

(2) Take an element x /∈ p. Then x is not nilpotent in A/q. Since q isp-primary, this implies that x is not a zero divisor in A/q. Therefore xy ∈ qimplies y ∈ q.

9.2 Primary decomposition

Definition 9.8. Let a ⊆ A be an ideal. A primary decomposition of afinite set q1, . . . , qn of (distinct) primary ideals such that a =

⋂ni=1 qi. It is

minimal if

[Min 1] the radicals are distinct,

[Min 2] qi 6⊇⋂j 6=i qj for all i.

If you have several that has the same radical, then you can collect them andreplace by their intersection, which is again a primary ideal. This is the firstminimality condition, and the second one only says to not use ideals that youdon’t need. We say that a is decomposable if it has a primary ideal.

Proposition 9.9 (First uniqnuess theorem). Let a ⊆ A be decomposable. Letq1, . . . , qn be a primary decomposition. Suppose it is minimal, and let pi =r(qi). Then the primes in the set r(a : x) : x ∈ A are exactly the pi. Thus nand pini=1 are unique for a.

Proof. We remark that for all x ∈ A,

(a : x) =( n⋂i=1

qi : x)

=

n⋂i=1

(qi : x) =⋂x/∈qi

(qi : x).

So r(a : x) =⋃x/∈qi pi.

Let x ∈ A such that (a : x) is prime. Then (a : x) =⋂x/∈qi pi. Then there

exists an i0 such that (a : x) = pi because (a : x) is prime.

Math 221 Notes 29

Now consider any i0 ∈ 1, . . . , n. By [Min 2], there exists an x0 /∈ qi0 suchthat x0 ∈

⋂i 6=i0 qi. Then

r(a : x0) =⋂x0 /∈qi

pi = pi0 .

Math 221 Notes 30


10.1 Associated primes

Definition 10.1. The associated primes of a is

AP(a) = all primes of the form (a : x) for x ∈ A.

Even if a is not decomposable, associated primes always exists. What wehave proved is that if a is decomposition, this is exactly those appearing in thedecomposition. In the non-noetherian case, this can easily be infinite.

Definition 10.2. Let p be an associate prime. We say that p is isolated if itis minimal in AP(a). The prime p is embedded otherwise.

Proposition 10.3. Let a ⊆ A be decomposable. Every p ⊇ a prime also con-tains some element of AP(a).

Proof. Let p ⊇ a =⋂ni=1 qi be a minimal primary decomposition. Taking the

radical, we get

p ⊇n⋂i=1

pi.

Because p is prime, p ⊇ pi0 for some i0.

Corollary 10.4. The isolated primes of a decomposable ideal a are exactly theminimal p containing a.

Localization is useful because you can throw away some primes and selectprimes you want. Taking the quotient throws away things that are below, buttaking the localization throws way primes that are above. Because minimalprimes are the good primes, localization is what you want to do.

Proposition 10.5. Let S ⊆ A be a multiplicative set, and let q be an p-primaryideal.

(1) S ∩ p 6= ∅ implies S−1q = S−1A.

(2) S ∩ p = ∅ implies that S−1q is S−1p-primary and q = λ−1S ((λS(q)).

Proof. (1) If t ∈ S ∩ p then there exists n with tn ∈ S ∩ q.(2) Assume S ∩ p = ∅. We remark that if t ∈ S and xt ∈ q then x ∈ q or

tn ∈ q. But the second is not possible because S ∩ p = ∅. So x ∈ q.Clearly λ−1(S−1q) ⊂ q set-theoretically. Let x ∈ λ−1(S−1q). Then x/1 =

y/t for some y ∈ q and t ∈ S. Then there exists r ∈ S such that rtx = ry ∈ q.Then by the remark, x ∈ q. This shows that λ−1

S (S−1q).Finally

r(S−1q) = S−1(r(f)) = S−1p.

You can check that S−1q is primary using the isomorphism S−1A/S−1q ∼=S−1(A/q).

Math 221 Notes 31

Definition 10.6. The saturation by S of an ideal a ⊆ A is

S(a) = (ae)c = λ−1S ((λS(a)).

Proposition 10.7. Let S ⊆ A be a multiplicative set and suppose the ideal a ⊆A is decomposable. Let q1, . . . , qn be a minimal decomposition with pi = r(qi).Index the qi’s su that S ∩ pi = ∅ for i ≤ m and S ∩ pi 6= ∅ for i > m. Then

S−1a =

m⋂i=1

S−1qi, S(a) =

m⋂i=1

qi,

and the terms appearing are minimal primary decompositions.

Proof. This follows from the previous propositions.

10.2 Second uniqueness of primary decomposition

Definition 10.8. A subset Σ ⊆ AP(a) is isolated if given any p′ ∈ AP(a),p ∈ Σ, if p ⊇ p′ then p′ ∈ Σ.

For example, Σ = AP(a) is isolated, Σ = p is isolated for isolated p.

Proposition 10.9 (Second uniqueness theorem). Let a ⊆ A be decomposableand q1, . . . , qn be a minimal primary decomposition. Let Σ = pi1 , . . . , pik is

an isolated set for pi = r(qi). Then⋂kj=1 qij only depends on a and Σ, not on

the primary decomposition.

Proof. Let SΣ = A \⋃

p∈Σ p, which is multiplicatively closed. Let p ∈ AP(a).There are two cases. If p ∈ Σ, then p ∩ SΣ = ∅. If p /∈ Σ, then p ∩ SΣ 6=.This is because the prime avoidance lemma tells us that p ⊆

⋃ki=1 pij since Σ is

isolated. Proposition 10.7 tells us that

k⋂j=1

qij = SΣ(a),

which is independent of q.

Corollary 10.10. Let a be a decomposition. The isolated components of aminimal decomposition of a are unique.

Example 10.11. Let A = k[x, y] and let a = (x2, xy). You can write it asa = (x) ∩ (x, y)2 = (x) ∩ (x2, y). In both cases, the associated primes are (x)and (x, y), with the first one isolated and the second one embedded. The isolatedprime is determined as (x).

Math 221 Notes 32


Today we will discuss integral algebras.

11.1 Integral algebras

Definition 11.1. Let A → B be an algebra. We say that β ∈ B is integralover A if there is f ∈ A[x] monic such that f(β) = 0. This f is called aintegrality relation.

For example im(A→ B) has only integral elements. Also√

2 ∈ C is integralover Z, with integrality relation x2 − 2.

Proposition 11.2. The following are equivalent:

(1) β ∈ B is integral over A.

(2) A[β] ≤ B is finitely generated as an A-module.

(3) There exists a subring C ≤ B containing A[β] that is finitely generated asan A-module.

(4) There exists a A[β]-module M that is faithful (i.e. AnnA[β]M = 0) andfinitely generated as an A-module.

Proof. For (1) ⇒ (2), let f be an integral relation for β, with deg f = n ≥ 1.Then

βn = −αn−1βn−1 − · · · − α1β − α0.

Then A[β] = 〈1, β, . . . , βn−1〉.For (2) ⇒ (3), take C = A[β].For (3) ⇒ (4), take M = C. This is faithful because 1 ∈ C.For (4) ⇒ (1), use Cayley-Hamilton with a = A and endomorphism µβ :

M →M , v 7→ βv. Then the relation is monic.

Corollary 11.3. Let β1, . . . , βr ∈ B be integral over A. Then A[β1, . . . , βr] isa finitely generated A-module.

Proof. Induct on r.

For C an A-algebra, we say that

(1) C is a finite A-algebra if C is finitely generated as an A-module,

(2) C is an A-algebra of finite type if C is finitely generated as an A-algebra,and

(3) C is integral over A if for any γ ∈ C, γ is integral over A.

Clearly (1) implies (2), and (2) does not imply (1). By the previous corollary,(2) and (3) imply (1).

Math 221 Notes 33

Proposition 11.4. Let B be an A-algebra, and let A ⊆ B be the set of β ∈ Bthat is integral over A. Then A is a sub A-algebra of B.

Proof. The only think we need to check is that A is a ring (with operations and 1of B). Let α, β ∈ A. The rings A[α+β] and A[α−β] and A[αβ] are all subringsof A[α, β]. Since A[α, β] is finitely generated as an A-module, Proposition 11.2(3) ⇒ (1) shows that α+ β, α− β, and αβ are all integral over A.

Definition 11.5. We call A the integral closure of A in B. (Note that A 6⊆ Ain general.)

Proposition 11.6. Let ϕ : A → B and ψ : B → C be morphisms. If ϕ and ψare integral, then ψ ϕ is integral.

Proof. Exercise.

Corollary 11.7. If B is an A-algebra, then Ã = A.

Proof. By definition of the integral closure, both maps

A A Ã

are integral, and so the composition is also integral. Then Ã ⊆ A, but we clearlyhave the other inclusion.

Proposition 11.8. Let ϕ : A→ B be integral.

(1) Given b ⊆ B an ideal, let a = ϕ−1(b). Then ϕ : A/a→ B/b is integral.

(2) Given S ⊆ A a multiplicative set, S−1ϕ : S−1A→ S−1B is integral.

Proof. (1) Use the same integrality relations and reduce modulo the ideals.(2) Let β/t ∈ S−1B. Let f(x) = xn +αn−1x

n−1 + · · ·+α0 be an integralityrelation for β over A. Then

g(x) = xn +αn−1

txn−1 + · · ·+ α0

tn∈ (S−1A)[x]

is an integrality relation for β/t.

11.2 Notion of a scheme

Last time we started with a ring A and set X = SpecA = p ⊆ A prime.Our goal was to do geometry on X, think of A as functions on X, and think ofA-modules sheaves on X.

In the assignment, you will define the Zariski topology on X. For α ⊆ Aan ideal, we define

V (a) = p ∈ X : p ⊇ a.

As a varies, V (a) are the closed sets for this topology.The functions are not really functions in the sense that each point has a

value in different target spaces. Given f ∈ A, given f ∈ A and p ∈ X, define:

Math 221 Notes 34

(1) the residue of field at p which is kp = Ap/p,

(2) the value of f at p is f(p) = f ∈ kp.

In particular, you have the notion of a zero set of f . Let

Z(f) = p ∈ X : f(p) = 0 = p ∈ X : f ∈ p.

This is just V ((f)) so it is closed.

Math 221 Notes 35


Today we will discuss the Going-up theorem. We are going to consider A ⊆ Bthat are integral, and study how prime ideals in A control primes in B.

12.1 Going-up theorem

Proposition 12.1. Let A ⊆ B be integral with A and B integral domains.Then A is a field if and only if B is a field (i.e., 0 is maximal in A if and onlyif 0 is maximal in B).

Proof. We show the forward direction first. Assume that A is a field. Let0 6= β ∈ B, and we want to show that β is invertible. Let βm + αn−1β

n−1 +· · · + α0 = 0 be an integrality relation of minimal degree. Note that α0 6= 0.Otherwise β(βn−1 +αn−1β

n−2 + ·+α1) = 0 and B is an integral domain, so weget an integrality relation of degree n− 1. Then we can rewrite the integralityrelation with α0 = −1, because A is a field. Then you end up with

αnβn + αn−1β

n−1 + · · ·+ α1β = 1.

Factoring β, we see β(αnβn−1 + · · ·+ α1) = 1, that is, β ∈ B×.

Now let us assume that B is a field and let 0 6= α ∈ A. Consider α−1, whichexists in B. By integrality, there exist γi ∈ A such that

α−n + γn−1α1−n + · · ·+ γ1α

−1 + γ0 = 0.

Then α−1 = −γn−1 − γn−2α− · · · − γ0αn−1 ∈ A.

Corollary 12.2. Let A ⊆ B be integral, and let q ⊆ B be prime with p = qc =q ∩ A. (Then p is prime.) We have that q is maximal in B if and only if p ismaximal in A.

Proof. We made a remark last time that integrality is preserved under takingquotients.

Proposition 12.3 (“Uniqueness”). Let A ⊆ B be integral. Let q ⊆ q′ ⊆ B beprimes. Suppose that qc = q′c in A. Then q = q′.

Proof. Let p = qc = (q′)c ⊆ A, which is a prime. Let S = A \ p. ThenAp = S−1A ⊆ S−1B = Bp is integral. Let m = pe ⊆ Ap be the only maximalideal in Ap, as Ap is a local ring. Let n, n′ ⊆ Bp be the extensions of q and q′.Clearly n ⊆ n′, and they are primes because S does not intersect q and q′. Notethat both n and n′ contract to m in Ap. So both n and n′ are maximal in Bp,with n ⊆ n′. Therefore n = n′. Since localization gives a bijection on (suitable)primes, we get q = q′.

Proposition 12.4 (Existence). Let A ⊆ B be integral, and let p ⊆ A be prime.Then there is a prime q ⊆ B such that p = qc (= q ∩A).

Math 221 Notes 36

Proof. We have a diagram

A B

Ap Bp = S−1B

i

integral

λ λ′

ip=S−1i

integral

The prime p extends to the maximal ideal m = pe, and p = mc. Now take anyn ⊆ Bp that is maximal. By the corollary, nc is maximal in Ap, and becauseip is integral and Ap is maximal, we have nc = m. Now take q = λ′−1(n) ⊆ B.Clearly q is prime, and

i−1(q) = i−1(λ′−1(n)) = λ−1(i−1p (n)) = λ−1(m) = p.

Theorem 12.5 (Going-up theorem). Let A ⊆ B be integral. Let p1 ( · · · ( pnbe primes in A. Let k < n and let q1 ( · · · ( qk be primes in B such thatpi = qci for i ≤ k. Then there exist qj primes in B with k < j ≤ n such that

q1 ( q2 ( · · · ( qk ( · · · ( qn

and pi = qci for all i = 1, . . . , n.

Proof. We turn ⊆ to ( by Proposition 12.3. For k = 0, we get q1 by Proposi-tion 12.4. Then quotient by q1 and get q2, and so on.

12.2 Geometric interlude: Morphisms

There is a topology on schemes, as I have said. Let X = SpecA and Y = SpecB.For any ring morphism φ : A→ B a morphism of rings, there is a set-theoreticalmap

ψ : Y → X; p 7→ ψ(p) = φ−1(p).

An exercise is that ψ is continuous.This gives a notion of pullback of functions. If f ∈ A is a function on X,

then we can define the pullback as ψ∗f = φ(f) ∈ B as a function on Y . To gofurther, we need the notion of sheaves. I leave it as an assignment to study thedefinition of a sheaf of abelian groups on a topological space.

Math 221 Notes 37


Today we are going to discuss the going down theorem. Recall that the going uptheorem told us the existence of an extension of a chain upstairs q1 ⊆ · · · ⊆ qk.The going down theorem will tell us the existence of an extension in the otherdirection q1 ⊇ · · · ⊇ qk.

The reason we call A downstairs and B upstairs in A→ B is psychological.In the corresponding map of Specs, the direction of the arrows are reversed andwe are lifting points.

13.1 Going down theorem

Lemma 13.1. Let f : A1 → A2 be a morphism of rings. Let p ⊆ A1 be prime.Then there exists a prime q ⊆ A2 with p = qc if and only if p = pec.

Note that pe has no reason to be prime.

Lemma 13.2. Let A ⊆ B be integral domains and let S ⊆ A be a multiplicative

set. Then S−1A = S−1A in S−1B.

Definition 13.3. Let A be an integral domain. We say that A is integrallyclosed if A = A in K = Quot(A) = (A \ 0)−1A.

For example, Z, Z[√−1], k[x] are all integrally closed. In fact, if A is a UFD

then A is integrally closed.

Proposition 13.4. Let A be an integral domain. Then the following are equiv-alent:

(1) A is integrally closed.

(2) Ap is integrally closed for all prime p ⊆ A.

(3) Am is integrally closed for all maximal m ⊆ A.

Proof. (1) ⇒ (2) follows from Lemma 13.2. (2) ⇒ (3) is trivial. For (3) ⇒ (1),consider A ⊆ K = QuotA. Let f : A → A. This map in injective becauseA ⊆ K. Also f is surjective because surjectivity is local. Thus A = A.

Let A ⊆ B be rings and a ⊆ A be an ideal, and β ∈ B. We say that β isintegral over a if there exist αj ∈ a such that

βn + αn−1βn−1 + · · ·+ α0 = 0.

Lemma 13.5. If A ⊆ B are rings and a ⊆ B are ideals, then a = rA(aA) ⊆ B(which has no reason to be an ideal of B). So it is closed under addition andmultiplication.

Proof. Let us first prove ⊆. Let β ∈ a ⊆ A. Now

βn = −αn−1βn−1 − · · · − α0 ∈ aA.

Math 221 Notes 38

This shows that β ∈ rA(aA).

For ⊇, let β ∈ rA(aA). Then there exists some n with βn ∈ aA. Then

βn = α1γ1 + · · ·+ αrγr

with αj ∈ a and γj ∈ A. Now consider M = A[γ1, . . . , γr], which is finitelygenerated over A because γj are integral. Because βnM ⊆ aM , we can applyNakayama and see that βn ∈ a. It follows that β ∈ a.

Proposition 13.6. Let A ⊆ B be integral domains with A integrally closed(inside its fractional field). Let a ⊆ A be an ideal and β ∈ a (⊆ B) andK = Quot(A). Then β is algebraic over K and its minimal polynomial over Khas the form

f(x) = xn + αn−1xn−1 + · · ·+ α0 ∈ K[x]

with αj ∈ rA(a).

Proof. Let L/K be the splitting field of f over K. Then all roots of f in Lare integral over a, because all roots of f are the also roots of any polynomialhaving β as a root. Since a is closed under addition and multiplication, αj ∈ afor all j. Because A is integrally closed, a = rA(aA) = rA(a).

For example, take A = Z and B = Z[i]. Let a = (4) ⊆ A. Then A isintegrally closed and K = Q. Take β = 1 + i ∈ a because β4 + 4 = 0. Theminimal polynomial is f(x)− 2x+ 2 whose coefficients are in (2) = rA((4)).

Theorem 13.7 (Going down theorem). Let A ⊆ B be integral domains withA ⊆ B integral and A integrally closed. Let p1 ⊇ · · · ⊇ pn be primes in A. Letq1 ⊇ · · · ⊃ qk (k < n) be primes in B with pi = qci (= q∩A). Then there existsqj ⊆ B primes such that q1 ⊇ · · · ⊇ qn and pi = qci .

Proof. The case k = 0 and n = 1 is the same as the going up theorem. Thegeneral case follows from k = 1 and n = 2. It is enough to show p2Bq1

∩A = p2,by Lemma 13.1. We need to show p1Bq1

⊆ p2 because the other direction istrivial. Suppose x ∈ p2Bq1

∩ A \ p2. Then x ∈ p2Bq1implies x = y/s for some

x ∈ p2B and s ∈ B \ q1. In this case, y ∈ p2B ⊆ rB(p2B) = p2 by Lemma 13.5.Then the minimal equation of y over K is

yr + ur−1yr−1 + · · ·+ u0 = 0

with ui ∈ rA(p2) = p2. Because s = y · x−1, with x−1 ∈ K since x ∈ A, theminimal equation for s over K is

sr + vr−1sr−1 + · · ·+ v0 = 0

for vi = ui/xr−i ∈ K. Hence vi ·xr−i = ui ∈ p2. On the other hand, since s ∈ B¡

it is integral over A. This shows that by Proposition 13.6, vi ∈ rA(1 · A) = A.Since vi · xr−i = ui ∈ p2 and x /∈ p2, vi ∈ p. But then sr ∈ p2B ⊆ p1B ⊆ q1.This contradicts B \ q1.

Math 221 Notes 39

14 October 3, 2016

Today we will study some connection between classical algebraic geometry andcommutative algebra. Today, k is a field.

14.1 Noether normalization and Nullstellensatz

Lemma 14.1. Let x1, . . . , xr, z be variables, and let f ∈ k[x][z] be non-constantin z. There are α ∈ k× and N ≥ 1 such that

F = α · f(x1 + zN1

, . . . , xr + zNr

, z)

is monic in z.

Proof. N > deg f works.

Theorem 14.2 (Noether normalization lemma). Let A 6= 0 be a finitely gener-ated k-algebra. There exist x1, . . . , xr ∈ A algebraically independent over k suchthat k[x1, . . . , xr] ⊆ A is integral.

Proof. Let x1, . . . , xn be generators for A as a k-algebra, indexed such thatx1, . . . , xr are algebraically independent over k and all xi are algebraic overk[x1, . . . , xr].

If n = r then we are done. Suppose that r < n and the result holds for anyk-algebra with at most n − 1 generators, by using induction on the number ofgenerators.

Now because xn is algebraic over k[x1, . . . , xr], there exists a polynomialf ∈ k[x1, . . . , xr][z] such that f(x1, . . . , xr;xn) = 0 in A. By the previous

lemma, F = f(x1 + zN1

, . . . , xr + zNr

; z) monic in z. Define yi = xi − xNi

n for

i ≤ r. Then yi ∈ A, and F = f(y1 + zN1

, . . . , yr + zNr

; z) is monic in z. AlsoF (xn) = 0. This shows that xn is integral over k[y1, . . . , yr, xr+1, . . . , xn−1] ⊆ Abut k[y1, . . . , yr] is an k-algebra with at most n−1 generators, and hence integralover some polynomial ring. Because the composition integral extensions is anintegral extension, we are done.

The geometric interpretation is that any variety is integral over some affinespace.

Lemma 14.3 (Zariski’s lemma). Let L be a finitely generated k-algebra andsuppose that L is a field. Then L/k is a finite algebraic extension. j

Proof. Let L/k be a finitely generated k-algebra with L a field. Suppose L/kis not finite algebraic. Then Noether normalization implies that k[x] ⊆ L isintegral for some algebraically independent x. Because it is not finitely algebraic,x 6= ∅. The ring k[x] has at least the ideals (0) and (x1) with (0) ( (x1). Thengoing-up implies that L has at least 2 distinct prime ideals, which contradictsthat L is a field.

Math 221 Notes 40

Theorem 14.4 (Weak Nullstellensatz). Let k be algebraically closed. Let n ≥ 1.Then the maximal ideals of k[x1, . . . , xn] are exactly of the form

(x1 − α1, . . . , xn − αn)

with αj ∈ k. Thus we have bijectionsmaximal ideals ofk[x1, . . . , xn]

←→ Mork-Alg(k[x1, . . . , xr], k) ←→ kn.

Proof. Those ideals of that form are kernels of the maps φ(αi) : k[x]→ k givenby xi 7→ αi.

Let m ⊆ k[x] be maximal. Then L = k[x]/m is a field, and Zariski’s lemmatells us that L/k is a finite extension. Because k is algebraically closed, k → Lis an isomorphism.

k[x] L

k

φm ∼=

Let φm(xi) = αi. Then φm(xi−αi) = 0 so m ⊇ (x1−α1, . . . , xn−αn). Because(x1 − α1, . . . , xn − αn) is maximal, it follows that m is exactly that ideal.

Math 221 Notes 41

15 October 5, 2016

Theorem 15.1 (Weak Nullstellenstaz, High School version). Let k be an alge-braically closed field with n ≥ 1. Let f1, . . . , fr ∈ k[x1, . . . , xn]. The followingare equivalent:

(1) f1, . . . , fr have no common zeros in kn, i.e., the system f1 = 0, . . . , fr = 0has no solution in kn.

(2) (f1, . . . , fr) = (1), i.e., there exist g1, . . . , gr ∈ k[x] such that g1f1 + · · ·+grfr = 1.

Proof. (2)⇒ (1) is clear. Now we show (1)⇒ (2). Say (2) fails. Then there is amaximal m ⊆ k[x] such that (f1, . . . , fr) ⊆ m. Then Nullstellensatz tell us thatm = (x1 − α1, . . . , xn − αn). So fi =

∑nj=1 hij(xj − αj) for hij ∈ k[x]. Then

(α1, . . . , αn) is a common zero of all the fi.

Lefschetz’s principle states that if you have a problem over an alge-braically closed field of characteristic zero, then we may assume that we areworking over C. Let f1, . . . , fr ∈ Q[x], and let K be an algebraically closed fieldof characteristic 0. Suppose that they have a common zero in Kn. Then theyhave a common zero in Q.

This is because if f1, . . . , fr has no common zero in Qn, then there are

polynomials g1, . . . , gr ∈ Q[x] → K[x] such that g1f1 + · · ·+ grfr = 1, becauseK is algebraically closed. Hence also in K[x], g1f1 + · · · + grfr = 1. Thenf1, . . . , fr has no common zero over K.

Here is another way of thinking about this. Say that the solution (αj)nj−1

is in αj ∈ Q[π] ⊆ C for simplicity. Then αj ∈ Q[π] ∼= Q[z] ⊆ C. Then we canspecialize to z = 0 to get a solution in Q.

15.1 Chain conditions

Definition 15.2. Let A be a ring and M an A-module. We say that M satisfies:

(1) the ascending chain condition if given any chain N1 ≤ N2 ≤ · · · ≤M ,there exist and n such that Nj = Nn for all j ≥ n.

(2) the descending chain codition if given any M ≥ N1 ≥ · · · , there existsn such that Nj = Nn for all j ≥ n.

Definition 15.3. M is Noetherian if it satisfies the ascending chain condition.M is Artinian if it satisfies the descending chain condition. A ring A is P if itis P as an A-module, where P is “Noetherian” or “Artinian”.

Proposition 15.4. Let M be an A-module. Then the following are equivalent:

(1) M is Noetherian.

(2) Every collection Σ 6= ∅ of submodules of M has a maximal element (inΣ).

Math 221 Notes 42

(3) For all N ≤M , N is finitely generated.

Proof. (1)⇒ (2) is easy. For (2)⇒ (3), take Σ = Q ≤ N : Q is finitely generated.Let Q0 ∈ Σ be maximal. Then Q0 = N , for otherwise Q1 = Q0 + 〈x〉 forx ∈ N \Q0 is finitely generated.

For (3) ⇒ (1), take N1 ≤ N2 ≤ · · · ⊆ M . Let N =⋃j=1Nj . This is a

submodule because it is a nested union. Then N = 〈x1, . . . , xr〉 and once Nncontains x1, . . . , xr we will have Nn = N .

Proposition 15.5. Let X be “Noetherian” or “Artinian”. We have:

(1) If 0 → M → N → P → 0 is exact, then N is X if and only if M and Pare X.

(2) X is preserved by finite direct sums.

(3) X for A-modules is preserved by arbitrary quotients.

(4) If A is a X ring and M is a finitely generated A-module, then M is X.

Proof. First check (1). Then everything follows.

Proposition 15.6. Let A be a Noetherian ring and S ⊆ A be a multiplicativeset. This implies that S−1A is Noetherian.

Proof. Every ideal of S−1A is of the form ae with a ⊆ A an ideal. Then checkfinite generation.

Next time we will do the Hilbert basis theorem.

Math 221 Notes 43

16 October 7, 2016

Last time we introduced the notion of Noetherian modules and Noetherian rings.

16.1 Hilbert basis theorem

Theorem 16.1 (Hilbert basis theorem). If A is Noetherian and x is a variablethen A[x] is Noetherian.

Corollary 16.2. If k is a field, then k[x1, . . . , xn] is Noetherian. Likewise,Z[x1, . . . , xn] is Noetherian.

Corollary 16.3. If k is a field, then k-algebras of finite type are Noetherian.Likewise finitely generated rings are Noetherian.

Proof of Theorem 16.1. Let a ⊆ A[x] be an ideal. Consider the set

lc(a) = leading coefficients of elements in a ⊆ A.

This is clearly an ideal of a. Because A is Noetherian, lc(a) = (α1, . . . , αn).Let fj ∈ a be such that

fj = αjxdj + lower degree terms,

and let a′ = (f1, . . . , fn) ⊆ A[x]. Take any f = αxm + (lower degree terms).Then

α =

n∑j=1

βjαj , βj ∈ A

because α ∈ lc(a).Now we claim that there exists an g ∈ a such that deg g < d = max dj

with f = g + h and h ∈ a′. If m < d, then we are done. Otherwise look atf −

∑mj=1 βjfjx

m−dj , which has smaller degree. Repeating this step, we get adecomposition of f into something in a and small degree.

Finally let M = 〈1, x, . . . , xd−1〉 ⊆ A[x] as an A-module. Then a = a′+(M ∩a). Because A is Noetherian and M is finitely generated, M is a Noetherianmodule. Then a ∩M is finitely generated. Thus a is finitely generated.

As an exercise, let k be a field and let x be a variable, and prove that k[[x]]is Noetherian. There is an elementary proof, but a more conceptual way is touse that the property of being Noetherian is preserved under completion.

16.2 Irreducible ideals and primary decomposition

Definition 16.4. Let A be a ring. We say that an ideal a ⊆ A is irreducible ifa = b ∩ c implies a = b or a = c.

Lemma 16.5. If A is Noetherian, then every ideal is a finite intersection ofirreducible ideals.

Math 221 Notes 44

Proof. Otherwise there exists a ⊆ A maximal among the ideals that are notfinite intersections of irreducible ideals. Because a is not maximal, there exista ( b and a ( c such that a = b ∩ c. Because a is maximal, there exist waysto represent b and c as finite intersections of irreducible ideals. Then a is anintersection of irreducible ideals.

Lemma 16.6. Let A be Noetherian and let a ( A be an ideal. If a is irreducible,then it is primary.

Proof. We can assume that a = (0) by looking at the quotient. Assume that(0) is irreducible. Let x, y ∈ A with xy = 0 and y 6= 0. We want to show thatx is nilpotent. By the ascending chain condition, there exists an n such that

Ann(xn) = Ann(xn+1) = · · · .

Now we claim that (xn) ∩ (y) = (0). Let z ∈ (xn) ∩ (y). Then zx = 0as z ∈ (y). Also z = αxn as z ∈ (xn). Then 0 = zx = αxn+1, and thenα ∈ Ann(xn+1) = Ann(xn). This implies that 0 = αxn = z.

Now (xn) ∩ (y) = 0, which is irreducible. This implies that xn = 0 ory = 0.

Proposition 16.7. Let A be Noetherian. Then every proper ideal of A has aprimary decomposition.

Lemma 16.8. Let A be Noetherian, and let a ⊆ A be an ideal. Then thereexists an integer n such that r(a)n ⊆ a.

Proof. r(a) is finitely generated, so r(a) = (α1, . . . , αk). Then x ∈ r(a) can bewritten as

x =

k∑j=1

βjαj .

Take n > m1 + · · · + mk where αmjj ∈ a. Then every term in the expansion of

xn has at least one of αmjj in it, and so xn ∈ a. Likewise any x1 · · ·xn ∈ a for

x1, . . . , xn ∈ r(a).

Math 221 Notes 45

17 October 12, 2016

Today we are going to discuss Noetherian domains of dimension 1. Recall thatif A is a ring, then the Krull dimension of A is maximal number properinclusions of prime ideals in A. For example, dimZ = 1 because SpecZ =(0) ∪ (p) : prime. Likewise SpecC[x] = (0) ∪ (x − α) : α ∈ C. In twovariables,

SpecC[x, y] = (0) ∪ (f) : f irreducible ∪ (x− α, y − β) : α, β ∈ C.

As an exercise, prove that there are no other primes. So dimC[x, y] = 2. Thiseasy description does not extend easily to more variables. Also as an exercise,describe SpecZ[x]. It also has dimension dimZ[x] = 2.

17.1 Discrete valuation rings

This is the local case for Noetherian domain of dimension 1.

Proposition 17.1. Let A be a Noetherian domain with dimA = 1. Everya ⊆ A is the product of finitely many primary ideals with distinct radicals.

Proof. Assume a 6= (0). Because A is Noetherian, there exists a minimal pri-mary decomposition

a =

n⋃i=1

qi.

Let pi = r(qi). Because dimA = 1, all pi are maximal. That is, pi are pairwisecoprime. Then

(1) = r((1)) = r(pi + pj) = r(r(qi) + r(qj)) = r(qi + qj).

So qi + qj = (1). So the intersection is the same as the product.

Definition 17.2. Let K be a field. A discrete valuation is a surjective mapv : K× → Z such that

(1) v(xy) = v(x) + v(y),

(2) v(x+ y) ≥ minv(x), v(y).

So define v(0) = +∞ and this is consistent with the previous conditions, givingv : K → Z ∪ +∞.

Example 17.3. For K = Q and a prime p, we have a valuation v = vp whichis called the p-adic valuation. This is defined as

vp

(ab

)= n if

a

b= pn · a

′

b′

with (a′, b′) = 1 and p - a′, b′.

Math 221 Notes 46

Similarly, there is a valuation on K = C(x) for α ∈ C[x], which is vf , andanother valuation v = −deg.

Now define

Ov = x ∈ K : v(x) ≥ 0, mv = x ∈ K : v(x) > 0.

The Ov is called the valuation ring of v. It turns out that (Ov,mv) is a localring, and there is a residue field kv = Ov/mv. For example, for K = Q, we havekvp = Fp.

Definition 17.4. An integral domain A is a discrete valuation ring if thereexists a valuation v on K = Quot(A) such that A = Ov.

From the previous examples, we see that Z(p) and C[x](x−α) are DVRs.

Proposition 17.5. If A is a DVR, then

(1) A is local,

(2) A is integrally closed,

(3) A is a PID,

(4) every nonzero ideal a ⊆ A is a power of the maximal ideal,

(5) dimA = 1,

(6) A is Noetherian,

(7) there exists x ∈ A such that all nonzero ideals are (xr) for r ≥ 0.

Proof. (1) and (2) are clear. Say v : K× → Z with A = Ov and m = mv. Definethe ideals

mr = x ∈ K : v(x) ≥ r.

Then m0 = A, m1 = m, and mi ) mi+1 because v is surjective. Let a ⊆ A be anonzero ideal. Take r = minv(x) : x ∈ a and take xr ∈ a with v(xr) = r. Fory ∈ a, y/xr ∈ Ov, and so a ⊆ (xr) ⊆ a. That is, A is PID and a = mr. For (7),we take x ∈ m1 \m2.

Math 221 Notes 47

18 October 14, 2016

Last time we introduced the notion of a discrete valuation. Now I will explainwhy I call these Noetherian domains of dimension 1.

18.1 Local Noetherian domain of dimension 1

Proposition 18.1. Let A be a local Noetherian domain of dimension 1, andlet m be the maximal ideal with k = A/m. The following are equivalent:

(i) A is a DVR.

(ii) A is integrally closed.

(iii) m is principal.

(iv) dimk m/m2 = 1.

(v) Every ideal 0 6= a ( A is a power of m.

(vi) There is an element x ∈ A such that all ideals a 6= 0 are of the forma = (xr) for r = 0, 1, . . ..

The element x in (iv) is called the uniformizer. Heuristically, local Noethe-rian domains corresponds to neighborhoods of a singular point of a curve, andDVRs correspond to neighborhoods of smooth points of a curve. The process istaking the normalization A of a local Noetherian domain A. For instance, takeA = Z[

√5] and A = Z[(1 +

√5)/2]. If we have a prime p ⊆ A such that Ap is

local Noetherian domain of dimension 1, but not a DVR. Then you can look atp ⊆ A where pc = p.

Proof. I’ll just do the not-so-easy implications. Let us do (ii) ⇒ (iii). Take0 6= x ∈ m. We know that (x) ⊆ m, and so r((x)) = m. The Noetherianhypothesis tells us that there exists an t such that mt ⊆ (x) but mt−1 6⊆ (x).Let t ∈ mt−1 \ (x). Consider z = x/y ∈ K = Quot(A). Note that z−1 /∈ Abecause y /∈ (x). Then z−1 /∈ A = A because A is integrally closed, and itfollows that z−1m 6⊆ m.

On the other hand, we claim that z−1m ⊆ A. For any α ∈ m,(yxα)t

=yt

xt−1

αt

x= β ∈ A

because αt/x ∈ A since αt ∈ mt ⊆ (x) and yt ∈ mt(t−1) ⊆ (x)t−1. Thereforez−1α is a root of Xt − β, and thus z−1α ∈ A = A.

Now z−1m is an ideal of A, but it is not in m. Thus z−1m = A. It followsthat m = zA = (z).

Another tricky implication is (iv) ⇒ (v). Nakayama will imply that m isprincipal, because you can lift the generator. Let t be the maximal exponentsuch that mt ⊇ a, where 0 6= a ( A. Note that such a t exists because mn ⊆ afor some n by the Noetherian hypothesis, and then mn+j ( a for j > 0.

Math 221 Notes 48

We want to show that there exists some α ∈ a such that α = xtu, whereu ∈ A×. Otherwise, for all α ∈ a there exists an u ∈ m such that α = xtu.Then α = xt+1v for some v ∈ A. This shows that a ⊆ mt+1, and we get acontradiction. Then mt ⊇ a ⊇ mt, i.e., a = mt.

Finally let us prove (vi)⇒ (i). We need to construct a valuation v : K× → Z.For β ∈ A, define v(β) = rβ where (β) = (xrβ ). Then you define

v(βγ

)= rβ − rγ .

This is not a priori well-defined, but you can check that it is well-defined.

Theorem 18.2. Let A be a Noetherian domain of dimension 1. The followingare equivalent:

(1) A is integrally closed.

(2) Every primary ideal is a prime power.

(3) Every local ring Ap (for p maximal) is DVR.

Proof. Just localize at p and use the previous theory. (Note that integrality islocal.)

Definition 18.3. Such a ring is called a Dedekind domain.

Math 221 Notes 49

19 October 17, 2016

Last time we introduced Dedekind domains. A is a Dedekind domain if (1) A isNoetherian, (2) dimA = 1, (3) any of the following equivalent condition holds:

(i) A is integrally closed.

(ii) Every primary ideal in A is a prime power.

(iii) Every local ring Ap (with p maximal) is a DVR.

Corollary 19.1. If A is a Dedekind domain, then every nonzero ideal has aunique factorization as a product of powers of nonzero primes.

There was an open problem of whether there is a solution to

xn + yn = zn, xyz 6= 0

for n ≥ 3. Someone had this brilliant idea of writing it as

yn =∏εn=1

(z − εx).

If we have unique factorization or something similar, then it would follow thateach term in the right hand side is a power because they are coprime. But thisdoesn’t work if the ring is not a UFD. So Kronecker came up with the idea ofworking instead with the ideals. In this case, we have unique factorization. Butsome principal ideal being a nth power of some other ideal does not mean thatit is a power of a principal ideal. If the ring is a PID, then we are good, butthen there is no point in looking at the ideals. However, we only need someideals being principal. If we allow some “inverse” of ideals, then the ideals forma group, and we can take the quotient by the subgroup of principal ideals. Ifthis is a finite group, and has order coprime to n, then we are in business.

19.1 Fractional ideals

Let A be an integral domain, and let K = Quot(A). We say that a A-submodulea ⊆ K is fractional ideal if there exists and x ∈ A \ 0 such that x · a ⊆ A.In this context, we call the ideals of A the integral ideals

Proposition 19.2. Let A be a domain with a ⊆ K a A-submodule. Then

(1) If a is finitely generated then it is a fractional ideal.

(2) If A is Noetherian and a is a fractional ideal, then a is finitely generated.

Proof. (1) Just multiply all the denominators of the generators of a.(2) Because xa ⊆ A is an ideal of A and so is finitely generated, a is also

finitely generated.

For S ⊆ K finite, we denote (S) = 〈S〉A-Mod.

Math 221 Notes 50

Definition 19.3. Let A be an integral domain and let a ⊆ K an A-module. Wesay that a is invertible if there exists a A-module b ⊆ K such that ab = A = (1).

Proposition 19.4. Let a ⊆ K be an invertible ideal. Then there exists a uniqueb ⊆ K such that a · b = A, and if fact, b = (A : a). Furthermore, a is finitelygenerated.

Proof. Let b ⊆ K be a sub A-module with a · b = A. Then

b ⊆ (A : a) = (A : a) ·A == (A : a) · a · b ⊆ A · b = b.

Now, since a · b = (1), we get 1 =∑nj=1 αjβj . Then for each x ∈ a, we have

x =n∑j=1

(xβj)αj =∑j=1

cjαj

for cj = xβj ∈ A. This shows that a is generated by α1, . . . , αn.

Invertible ideals of A form an abelian group with identity element A = (1).This is a general fact for an integral domain A.

Proposition 19.5. Let A be an integral domain and a ⊆ K be a fractionalideal. The following are equivalent:

(i) a is invertible.

(ii) a is finitely generated and ap is (Ap-)invertible for every p prime.

(iii) a is finitely generated and am is (Am-)invertible for every m maximal.

Proof. Exercise.

Proposition 19.6. Let (A,m) be a local integral domain. Then A is a DVR ifand only if every nonzero fractional ideal is invertible.

The global version of this proposition is that A is a Dedekind domain if andonly if every nonzero fractional ideals are invertible.

Math 221 Notes 51

20 October 19, 2016

20.1 Dedekind domains and fractional ideals

Proposition 20.1. Let (A,m) be a local integral domain. Then A is a DVR ifand only if every nonzero fractional ideals are invertible.

Proof. For the forward direction, let m = (x). Let a ⊆ K = Quot(A) be anon-zero fractional ideal. Then there exists y ∈ A such that y · a ⊆ A. Thismeans that y · a = (xr) for some r. Then a = (xr/y) and a−1 = (y/xr).

Now let us prove the converse. First A is Noetherian because every nonzeroinvertible integral ideal a ⊆ A is finitely generated. We now show that every(0) 6= a ( A is a power of m. Assume this fails, and let a be a maximalcounterexample, by the Noetherian hypothesis. Note that a ⊆ m. Also m−1 =m−1 ·A ⊇ m−1 ·m = A. Then m−1a ⊇ a and m−1a ⊆ m−1m = A.

If m−1a ) a, then it is an integral ideal that is not a power of m. Thiscontradicts the maximality of a. If m−1a = a, then a = ma so by Nakayama,a = (0). This shows that every (0) 6= a ⊆ A is a power of m, and this furtherimplies dimA = 1.

Theorem 20.2. Let A be an integral domain. Then A is Dedekind if and onlyif every nonzero fractional ideal is invertible.

Proof. Just use Proposition 20.1 and localization techniques.

Corollary 20.3. In a Dedekind domain, nonzero fractional ideals form a groupwith respect to multiplication. This group is free abelian generated by primeideals.

Let A be a Dedekind domain. We write

Div(A) = group of fractional ideals,

PDiv(A) = group of principal fractional ideals = (x) : x ∈ K×,Cl(A) = Div(A)/PDiv(A).

The group Cl(A) is called the class group. This measures the “failure” ofunique factorization of A.

Proposition 20.4. If A is a Dedekind domain, the following are equivalent:

(1) Cl(A) = (1).

(2) A is a UFD.

(3) A is a PID.

Proof. Clearly (3) implies (2) because you can go to ideals and factor and goback to elements. To show (2) ⇒ (3), it is enough to show that primes areprincipal. Let (0) 6= p ⊆ A be prime. Take any 0 6= x ∈ p. Then x factors intoirreducibles x = y1 · · · yn, with n ≥ 1 because x /∈ A×. Then there exists a jsuch that yj ∈ p. Then 0 6= (yj) ⊆ p. Because dimA = 1, (yj) = p. Finally(1) ⇔ (3) is easy.

Math 221 Notes 52

Suppose you have a Dedekind domain A ⊆ K. Consider L a finite extensionof K. Then we can look at the integral closure AL ⊆ L.

AL L

A K

Then AL has dimension 1, and it is integrally closed in L. But we can’t concludethat AL is Dedekind because don’t know whether AL is Noetherian. This is true,but is not obvious. In a future assignment, you will prove that this is true ifL/K is separable.

Theorem 20.5. Let K be a number field (i.e., a finite extension of Q). LetOK be its ring of integers (= ZK). We have:

(1) OK is a Dedekind domain.

(2) O×K is a finitely generated abelian group.

(3) Cl(OK) is finite.

Proof. (1) See [AM] Proposition 9.5.(2) This is not so bad: just use basic properties of heights of algebraic

numbers.(3) This is more delicate: it uses Minkowski’s geometry of numbers.

Math 221 Notes 53

21 October 21, 2016

Today we are going to sketch the theory of modules of finite length. There aretwo key notions of “basic” modules.

• M is simple if the only submodules of M are 0 and M .

• M is indecomposable if M = P ⊕Q implies P = (0) or Q = (0).

For example, Z/4 as a Z-module is indecomposable but not simple.

21.1 Length of a module

We are back to the general commutative algebra setting, where A is just a com-mutative ring. For M an A-module, a chain in M is a sequence (M0, . . . ,Mr)of submodules such that

M = M0 )M1 ) · · · )Mr = (0).

The length of this chain is r. A decomposition series for M is a maximalchain. This is equivalent to each Mi/Mi+1 being simple. There is no reason fora decomposition to exist.

Proposition 21.1. If M has a composition series of length n, then every com-position series of M has length n. Moreover, every chain in M can be extendedto a composition series.

Proof. Define l(M) as the least length of a composition series in M .We first claim that N M implies l(N) < l(M). Given a minimal compo-

sition series (M0, . . . ,Mr) in M , consider Nj = N ∩Mj . Then at least

N0 = N ⊇ N1 ⊇ · · · ⊇ Nr = (0). (∗)

We also have

NjNj+1

=N ∩Mj

N ∩Mj+1→ Mj

Mj+1.

So Nj/Nj+1 are either simple or (0). Deleting all the redundant chains, (∗)can be reduced to a composition series for N . But the embeddings are not allisomorphisms because N 6= M .

We now claim that all chains in M has length at most l(M). Take any chain

M0 )M ′1 ) · · · )M ′r−1 )Mr = (0).

By the first claim, this implies

l(M0) > l(M ′1) > · · · > l(M ′r−1) > 0.

Therefore l(M) = l(M0) ≥ r. This shows that every composition series of Mhas length l(M).

Note also that if a chain has length l(M) then it is a composition series,because if not, new terms can be inserted. This shows that each chain can beextend to a composition series, which has length l(M).

Math 221 Notes 54

Definition 21.2. The maximal length l(M) of a module M is called the lengthof M .

If l(M) <∞, then it is the same as in the previous proof.

Proposition 21.3. If l(M) <∞, then M is Noetherian and Artinian.

Proposition 21.4 (Jordan-Holder theorem). Let M be a finite A-module offinite length, and let (M0,M1, . . . ,Mr) and (N0, N1, . . . , Nk) be decompositionseries for M . Then r = k and the quotients Mj/Mj+1 and Mi/Mi+1 are iso-morphic up to order.

Proof. Induct on length. (We already know that r = k.) I’m not doing becauseit is too long.

Proposition 21.5. If l(M) < ∞ then M is a direct sum of indecomposablesubmodules.

Proof. Induct on length.

Lemma 21.6 (Splitting lemma). Let M,N be indecomposable, and let α : M →N and β : N → M . Let γ = αβ ∈ EndA(N). If γ is an isomorphism, then αand β are both isomorphisms.

Math 221 Notes 55

22 October 24, 2016

22.1 Field extensions

A field extension is an inclusion of fields k ⊆ L. It is called finite if L is afinite dimensional k-vector space. We denote this degree by [L : k]. Finite fieldextensions are algebraic.

Proposition 22.1. If M/L is a finite extension and L/k is a finite extension,then M/k is finite and [M : k] = [M : L][L : k].

Given any x1, . . . , xl ∈ L, we will write k[x1, . . . , xl] the sub k-algebra of Lgenerated by x1, . . . , xl and write k(x1, . . . , xl) the subfield of L generated byx1, . . . , xl and k.

Lemma 22.2. Let L/k be a field extension. For any x ∈ L, we have an evalua-tion map evx : k[t]→ L given by f(t) 7→ f(x). This is a ring homomorphism andif x is transcendental over k then this gives an isomorphism k[t] ∼= k[x]. If x isalgebraic, then ker evx = (hx) for a unique monic irreducible hx ∈ k[x]. This hxis calledthe minimal polynomial of x over k and k(x) = k[x] ∼= k[t]/(ker evx).The degree of hx is deg hx = [k(x) : k].

If L = k(x) then we say that x is a primitive element for L/k. If such anx exists, L/k is called primitive.

Lemma 22.3 (Lifting lemma). Let k(x)/k be a finite primitive extension, andlet hx be the minimal polynomial of x. Let σ : k → L be an isomorphism, withM/L a field extension. Note that σ extension to an isomorphism σ′ : k[x] ∼= L[t].Then the set of roots of σ′(hx) in M is in bijection with the lifts σ : k(x)→Mextending σ : k → L.

Proof. Without loss of generality, we may assume that L = k and σ = idk.Then k(x) ∼= k[x]/(hx) and so we have a bijection

Mork(k(x),M)←→ Mork(k[x]/(hx),M)←→ zeros of hx in M.

22.2 Splitting fields

Let k be a field, and f ∈ k[t].

Definition 22.4. Say that L/k is a splitting field for f

• if f ∈ L[t] splits into linear factors, and

• L is minimal with respect to the above properties. (L is generated by theroots of f in L.)

Theorem 22.5. Let f ∈ k[t]. Then there exists a splitting field Lf/k for f ,and it is unique up to k-isomorphism. Moreover, [Lf : k] ≤ (deg f)!.

Math 221 Notes 56

Proof. Start by picking a irreducible factor g of f in k[t]. Then k[s]/(g(s)) isa field. Call this k1 = k(s). In this field, we have f = (t − s)f1(t). Nowrepeat this construction for f1, and this gives a field k2/k1. Repeating thisdeg f times, we get kn = Lf . In kn, the polynomial f splits into linear factors,and L is generated by the roots of f .

We know show uniqueness. Let L/k be any splitting field for f . We constructan isomorphism Lf → L inductively. We have a map k → L. We can extend itto k1 → L by sending s to a root of g in L. In this way, we get a map Lf → L.By construction, the image contains all roots of f , and this shows surjectivity.

The degree of Lf/k is

[Lf : k] = [kn : kn−1] · · · [k2 : k1][k1 : k] ≤ 1 · · · · · (deg f) ≤ (deg f)!.

22.3 Normal field extensions

Definition 22.6. A field extension L/k is normal if it is algebraic and forevery x ∈ L the minimal polynomial hx ∈ k[t] factors into linear factors in L[t].

Theorem 22.7. If L/k is finite, then the following are equivalent:

(1) L/k is normal.

(2) There exists an f ∈ k[t] such that L/k is the splitting field of f .

(3) Every field extension M/L and k-map σ : L→M satisfies σ(L) = L.

Math 221 Notes 57

23 October 26, 2016

Proposition 23.1. Let L/k be a finite extension. Then the following are equiv-alent:

(i) L/k is normal, i.e., for every x ∈ L, hx splits into linear factors in L[t].

(ii) L is the splitting field of some polynomial f ∈ k[t].

(iii) For any M/L and k-algebra map σ : L→M , the image is σ(L) = L.

(iv) For every x1, . . . , xn ∈ L, there exists an extension M/L such that hxi allsplit in M and any k-algebra map σ : L→M has σ(L) = L.

Proof. For (i) ⇒ (ii), take generators x1, . . . , xn for L as a k-algebra. Letf = hx1

, . . . , hxn . Then this splits in L[t], and the roots generate L.For (ii) ⇒ (iii), suppose L is a splitting field for f . Consider a map σ : L→

M . Then σ(f(x)) = f(σ(x)) because σ fixes K. Then σ sends roots of f toroots of f . Because L is generated by the roots of f , it follows that σ(L) is alsogenerated by roots of f . Also because σ : L→M is injective, σ(L) = L.

For (iii)⇒ (iv), let M be the splitting field of hx1· · ·hxn over L. Then both

conditions are obviously satisfied.For (iv) ⇒ (i), consider any x ∈ L. We extend this to a set of generators

x, x2, . . . , xn of L/k. Take M such that hx, hx2 , . . . , hxn split in M and anyσ : L → M has σ(L) = L. Let y ∈ M be any root of hx. Then there is a mapσ : k(x) ∼= k(y) in M , and we can extend this to σ′ : k(x, x2, . . . , xn) → M .Then σ′(L) = L and y ∈ σ′(L), so y ∈ L. This shows that hx splits in L.

Proposition 23.2. Let L/k and M/L be field extensions. If M/k is normal,then M/L is normal (but L/k might not be normal).

Even if L/k and M/L are both normal, M/k need not be normal.

23.1 Separable field extensions

This is a property that is satisfied by most of the field extensions we care about.Let f ∈ k[t] be a polynomial.

Definition 23.3. f is separable if f has exactly deg f distinct roots in thesplitting field Lf for f/k.

f is separable if and only if gcd(f, f ′) = 1, where f ′ is the formal derivative.If f is irreducible, then f is separable if and only if f ′ 6= 0 ∈ k[t]. This is mostlytrue, but for instance, tp − x ∈ k[t] has zero derivative if t has characteristic p.

Definition 23.4. An algebraic extension L/k, say that L/k is separable if forall x ∈ L, the minimal polynomial hx is separable.

Theorem 23.5 (Primitive element theorem). If L/k is finite and separable,then there exists an x ∈ L such that L = k(x).

Math 221 Notes 58

Proof. If k is a finite field, then L× is a cyclic group, and so we can take x tobe the generator.

Now we assume that k is infinite. It is enough to show when L = k(y, z),because we can induct on the number of generators. Take x = y − λz wherey − y′ 6= λ(z − z′) for every hy(y′) = hz(z

′) = 0 with y′ 6= y and z′ 6= z (in thesplitting field of hyhz).

The polynomials hz ∈ k[t] ⊆ k(x)[t] has z as a root, and H(t) = hy(x+λt) ∈k(x)[t] also has z as a root. Then gcd(H,hz) ∈ k(x)[t] also has z as a root. Butby the genericity condition, it they have no other common roots. This hsowsthat gcd(H,hz) has degree 1, and thus z ∈ k(x). Then y ∈ k(x).

Proposition 23.6. For field extensions, L/k and M/L are both separable ifand only if M/k are separable.

Example 23.7. Let k = Fp. Then f(t) = tpr − t has derivative −1 so it

is separable. Consider the splitting field Lf , which is the finite extension ofFp. Then |Lf | = pn for some n. Consider the map σ : x 7→ xp that is anautomorphisms of Lf . Notice that σr(x) = x for any root x of f , and so theroots of f form a field. This has to be Lf , and thus |Lf | = pr. We call thisfinite field Lf = Fpr . If L is any other finite field with |L| = pr, then L× isa finite cyclic group of order pr − 1. Then every x ∈ L satisfies xp

r − x and soL ∼= Fpr . That is, the field with pr elements is unique up to isomorphism.

Math 221 Notes 59

24 October 28, 2016

24.1 Galois extensions

Definition 24.1. If L/k is normal and separable, then we say that L/k isGalois.

Lemma 24.2. If L/M and M/k are finite extensions and L/k is Galois, thenL/M is Galois. (M/k is not necessarily Galois.)

Lemma 24.3. If L/k is finite and Galois, then we can write L = k(x) wherethe minimal polynomial hx of x splits into distinct linear factors in L[t].

Proof. By the primitive element theorem, we can write L = k(x) for some x.Then hx splits because L/k is normal, and there are no repeated factors becauseL/k is separable.

Definition 24.4. If L/k is a field extension, then we denote

Gal(L/k) = k-automorphisms of L = field automorphism of L that fix k.

This naturally forms an group.1

Lemma 24.5 (Counting automorphisms). If L/k is finite and [L : k] = n, then|Gal(L/k)| ≤ n. Moreover, we have equality if L/k is Galois.

Proof. We use the lifting lemma. Let L = k(x1, . . . , xr). Let kj = k(x1, . . . , xj).Then we have a tower

k0 ⊆ k1 ⊆ k2 ⊆ · · · ⊆ kr = L

with each step being a extension generated by a single element. By inductiveapplication of the lifting lemma, the number of k-maps L→ L is at most

[k1 : k][k2 : k1] · · · [kr : kr−1] = [kr : k] = n.

If L/k is Galois, then L = k(x) for x in the previous lemma. Then the liftinglemma says that

#k-maps k(x) → L = ## of roots of hx in L = deg hx = n.

Our goals are:

• showing that # Gal(L/k) = [L : k] if and only if L/k is Galois.

• for a finite Galois extension L/k, showing that there is a one-to-one cor-respondence (

intermediate fieldsk ⊆M ⊆ L

) (subgroups Hof Gal(L/k)

)M Gal(L/M)

LH H1Some people define this only when L/k is Galois.

Math 221 Notes 60

Definition 24.6. If H ⊆ Gal(L/k) then define the fixed field LH = x ∈ L :σ(x) = x for every σ ∈ H.

It is clear that H ≤ Gal(L/LH) and LGal(L/M) ⊇M .

Lemma 24.7 (Artin). Let M be a field, and H be a finite group of field automor-phisms of M . Then M/MH is finite Galois of degree #H and Gal(M/MH) =H.

Proof. Let k = MH . We first prove that M/MH is Galois. Pick a x ∈ M ,and we want to show that hx splits into distinct factors in M [t]. Let Hx be thestabilizer of x in H. Then ∏

[σ]∈H/Hx

(t− σ(x)) ∈MH [t]

splits into distinct linear factors. This shows that M/MH is Galois, and alsoshows that for any x ∈ M , k(x)/k has degree at most #H = n. But we can’tpick a primitive element yet because we don’t know that M/MH is finite.

Choose x ∈ M with [k(x) : k] maximal. This is possible because [k(x) :k] ≤ n always. We claim that M = k(x). If y ∈ M \ k(x) then k(x, y) ) k(x)and k(x, y) = k(z) for some z ∈ M by the primitive element theorem. Thenk(z) ) k(x) contradicts the maximality of x. This shows that M = k(x), andso [M : k] ≤ n.

Now we have

[M : MH ] ≤ n = #H ≤ # Gal(M/MH) ≤ [M : MH ].

Therefore everything must be equalities, and in particular, [M : MH ] = n andGal(M/MH) = H.

Corollary 24.8. Let L/k be a finite Galois extension and let H = Gal(L/k).Then LH = k.

Proof. We have LH ⊇ k. By the previous theorem, we have

[L : LH ] = #H = # Gal(L/k) = [L : k].

So [LH : k] = 1, i.e., LH = k.

Corollary 24.9. Let L/k be finite with [L : k] = # Gal(L/k). Then L/k isGalois.

Proof. H = Gal(L/k). Then we have [L : LH ] = #H = [L : k] by Artin’slemma. So LH = k, and Artin’s lemma implies that L/LH = L/k is Galois.

I am now going to try and state the fundamental theorem of Galois theory.

Theorem 24.10 (Fundamental theorem of Galois theory). Let L/k be a finiteGalois extension. We have a correspondence between an intermediate field Mand a subgroup H of Gal(L/k), with the following properties:

Math 221 Notes 61

(i) It is order-reversing, i.e., M ⊆M ′ if and only if H ⊇ H ′.(ii) [L : M ] = # Gal(L/M) for each M .

(iii) [LH : k] = [Gal(L/k) : H] for each H.

Why do we have a bijection? We only need to show that LGal(L/M) = M .By Corollary 24.8, L/M is Galois. Then by Artin’s lemma, Gal(L/LGal(L/M)) =Gal(L/M).

Math 221 Notes 62

25 October 31, 2016

If you have a finite Galois extension L/k and an intermediate field extension M ,then M/k need not be normal. So an automorphism σ ∈ Gal /k might move Maround.

L

M σ(M)

k

H

σHσ−1

fin. Gal.

Then the Galois group Gal(L/σ(M)) is given by the conjugation of Gal(L/M).So M/k is Galois if and only if H = Gal(L/M) is normal in G = Gal(L/k). Inthis case, we have an isomorphism

Gal(L/k)/Gal(L/M)→ Gal(M/k)

that is given by restriction.

25.1 Examples of field extensions

Lemma 25.1. Let f ∈ K[t] be irreducible and separable. Then Gf/K =Gal(Kf/K) acts transitively on the roots of f in Kf .

Proof. It follows from the lifting lemma.

Lemma 25.2. For f ∈ K[t] irreducible and separable, by action on the roots off , we can see Gf/K as a transitive subgroup of Sd for d = deg f .

Example 25.3. Let charK 6= 2 and d = 3. Let f ∈ K[t] be irreducible andseparable. Then Gf/K is a transitive subgroup of S3. The subgroups of S3 upto conjugation are:

Order 1 2 3 6Example e 〈(12)〉〈(123)〉 S3

# of conjugations 1 3 1 1Transitive? No No Yes Yes

Table 1: Subgroups of S3 up to conjugation

But in practice, how do you know which one it is? Consider the discriminant,which is defined as

∆f = (α1 − α2)(α1 − α3)(α2 − α3)2 = (−1)d(d−1)/2 Res(f, f ′),

where α1, α2, α3 are the roots of f . Let

δf = (α1 − α2)(α1 − α3)(α2 − α3),

Math 221 Notes 63

for this arbitrary ordering. If σ ∈ Gf/K is a 3-cycle, then σ(δf ) = δf . Soin the case that Gf/K = 〈(123)〉, the Galois group fixes δf , and this meansthat δf ∈ K. On the other hand, if Gf/K = S3 then τ = (12) acts on δf asτ(δf ) = −δf . Then δf /∈ K. Therefore

Gf/K ∼=

S3 if ∆f is not a square in K

A3 if ∆f is a square in K.

For instance, if K = Q then f = t3 + t2 − 3, which is irreducible by rationalroot test, has discriminant ∆f = −231 = −3 · 7 · 11. So Gal(f/Q) ∼= S3.

Lemma 25.4. Sn is generated by any of the following:

(i) transpositions.

(ii) transpositions of the form (1r) for 2 ≤ r ≤ n.

(iii) transpositions of the form (r(r + 1)) for 1 ≤ r ≤ n− 1.

(iv) (12) and (123 · · ·n).

Lemma 25.5. Let f ∈ K[t] be irreducible and separable. Suppose p = deg f isprime. Then Gf/K contains a p-cycle of Sp.

Proof. deg f = p implies that p | [Kf : K] = #Gf/K . Then by Cauchy, thereexists a σ ∈ Gf/K of order p. Since p is prime, then σ is a p-cycle.

Lemma 25.6. Let f ∈ Q[t] be irreducible (hence separable). Suppose deg f = pis prime and f has exactly 2 non-real roots. Then Gf/Q ∼= Sp.

Proof. Complex conjugation gives a transposition and there is a p-cycle. Thesegenerate the whole Sp.

Example 25.7. Let K = Q and f = t5 − 6t+ 3. By Eisenstein at 3, this poly-nomial is irreducible. Also f has three real roots exactly after doing calculus.Therefore Galf/Q ∼= S5.

Math 221 Notes 64

26 November 2, 2016

Definition 26.1. A field K is perfect if every irreducible f ∈ K[t] is separable.

For example, K = Fp(s) is not perfect because f(t) = tp − s is irreduciblebut it factors as (t− p

√s)p in Kf . Fields of characteristic zero are perfect, and

finite fields are perfect. Also algebraically closed fields are clearly separable.

26.1 Solvability by radicals

All fields in this section has characteristic zero.

Definition 26.2. A field extension L/K is pure radical of exponent n if thereexists γ1, . . . , γr ∈ L such that for each i, γni ∈ K and L = K(γ1, . . . , γr). L/Kis radical if there exists a tower

K = K0 ⊆ K1 ⊆ · · · ⊆ Km = L

such that Ki/Ki−1 is pure radical for each i.

Definition 26.3. We say that f ∈ K[t] is solvable by radicals if there existsa radical extension L/K such that Kf ⊆ L.

For example, consider the case K = Q, d = deg f with f irreducible. If d = 1then it is always solvable. If d = 2, then the roots are (−b±

√b2 − 4ac)/2a and

so f is always solvable. If d = 3 then it is also always solvable, and it was foundby Ferro and first published by Cardano. The case d = 4 was solved by Ferrari,and for d = 5 it was fist proved that there are non-solvable polynomials by Abel.

Let us write

µn = ε ∈ Ktn−1 : εn = 1,

and this under × is isomorphic to (Z/nZ,+).

Lemma 26.4. The extension K(µn)/K is Galois and Gal(K(µn)/K) is a sub-group of Aut(µn) ∼= (Z/nZ)×.

Proof. It is Galois because K(µn) = Ktn−1 and charK = 0. Let σ ∈ G =Gal(K(µn)/K). Then σ acts on µn and σ(ε · ε′) = σ(ε)σ(ε′). So we get a map

G→ Aut(µn); σ 7→ σ|µn .

This is injective because µn generates the extension.

Definition 26.5. A pure radical extension L/K of exponent n is Kummer ifµn ⊆ K. (This is for some admissible n.)

Lemma 26.6. If L/K is Kummer of exponent n, then it is Galois, abelian,and Gal(L/K) has exponent dividing n.

Math 221 Notes 65

Proof. Write L = K(γ1, . . . , γr) with αi = γni ∈ K. (Assume without loss ofgenerality that γi 6= 0.) Conjugates of γi are among γiµn, which is the roots oftn − αi. So L/K is Galois.

Let G = Gal(L/K). Define

δ : G→ µn × · · · × µn = (µn)r; σ 7→(σ(γi)

γi

).

Note that δ is a group morphism, and it is injective. Therefore Gal(L/K) isabelian.

Every time you have a Galois extension with abelian Galois group, it isactually Kummer. This is the whole point the Kummer theory, but we are notgoing to do this.

Definition 26.7. A radical extension L/K is prepared if it has a tower

K ⊆ L0 ⊆ · · · ⊆ Lr = L

and an integer n ≥ 1 satisfying

(1) L/K is normal (and hence, Galois),

(2) L0 = K(µn),

(3) for each 1 ≤ i ≤ r, the extension Li/Li−1 is Kummer of exponent n.

Lemma 26.8. Let L/K be radical. Then there is an extension L′/K preparedradical extension with L′ ⊇ L.

I will prove this next time.

Theorem 26.9. Let K be of characteristic 0, and let f ∈ K[t] be irreducible.Suppose that f is solvable by radicals. Then Gf/K is a solvable group.

Proof. Let Kf ⊆ L with L/K radical. Lemma 26.8 tells us that we can as-sume L/K is prepared. Then Lemma 26.4 and Lemma 26.6 gives us a chain ofsubgroups

1 ⊆ H1 ⊆ H2 ⊆ · · · ⊆ Hr = Gal(L/K)

with Hi+1/Hi abelian. Thus Gal(L/K) is solvable. Then Gal(Kf/K) is solvablebecause it is a quotient of Gal(L/K).

Corollary 26.10. The general quintic equation over Q is not solvable by radi-cals.

Math 221 Notes 66

27 November 4, 2016

27.1 Representations of finite groups

Let G be a finite group, and let k be a field, both fixed.

Definition 27.1. A (k-linear finite dimensional) linear representation of Gis a pair (V, ρ) with V a finite dimensional k-vector space and ρ : G→ GL(V ) =Autk(V ) a group morphism.

There are two different (equivalent) points of view:

(i) Linear action: V is a k-vector space and G acts on V in a k-linear way,i.e., g · (x+ λy) = g · x+ λg · y.

(ii) Modules over group rings: If we define the group ring R = k[G], then arepresentation of G is a k-vector space V with a compatible structure ofa left R-module.

Clearly (k, ρtriv) is a representation. If S is a finite G-set, then G has anatural representation (VS , ρS) given by eg · εs = εg·s.

Example 27.2. The vector space Vreg = R is a left R-module and so this givesa representation. This is called the regular representation.

If V and V ′ are representations, then Homk(V, V ′) is a k-vector space witha G-action given by

ρH(g)(f) = ρ′(g) f ρ(g−1).

If (V, ρ) is a representation and W ⊆ V is a subspace such that W is ρ-stable,then (W,ρ( )|W ) is a representation. Also, if V and V ′ are representations, thenboth V ⊕V ′ and V ⊗V ′ have natural G-action. If (V, ρ) is a representation, then(V ∨, ρ∨) defined as ρ∨(g) = ρ(g−1)t is a representation. Note that now we haveduals, homs, and tensor product. It turns out that the induced representationson Homk(W,V ) and V ⊗W∨ are compatible.

Definition 27.3. A representation (V, ρ) 6= 0 is called simple if its only sub-representations are 0 and V . A representation (V, ρ) 6= 0 is called irreducibleif it has no nontrivial direct sum decomposition.

Simple implies irreducible, but the converse is not true. For instance, letG = Z/pZ and k = Fp with V = k2. Then GL(V ) = GL2(k) and let ρ be givenby

ρ : G→ GL(V ); n 7→[1 n0 1

].

The trivial representation ρtriv is a subrepresentation because ρ stabilizes the xaxis. But it is irreducible.

Math 221 Notes 67

Definition 27.4. Let (V, ρ) and (V ′, ρ′) be representations. A morphism is ak-linear map f : V → V ′ such that for all g ∈ G,

f ρ(g) = ρ′(g) f.

Equivalently, one can consider it as a morphism of left R-modules. We are goingto denote this by HomG(V, V ′). This is naturally a subset of Homk(V, V ′), andusing the G-action on Homk(V, V ′), we have HomG(V, V ′) = Homk(V, V ′)G.

Lemma 27.5 (Schur’s lemma). Assume k is algebraically closed, and let V,Wbe simple representations. Then

HomG(V,W ) =

0 if V 6∼= W

k if V ∼= W.

Proof. Note that for every f ∈ HomG(V,W ), ker f = 0 or V and im f = 0 or Wbecause V and W are simple. Now if V 6∼= W , then it is automatic that ker f 6= Vand im f 6= 0. Then ker f = 0 and im f 6= W . So V ∼= W . Suppose thatV = W , and let f ∈ HomG(V, V ). Because k is algebraically closed, there existsa λ ∈ k that is an eigenvalue for f . Then ker(f − λI) 6= 0. Observe thatλI ∈ HomG(V, V ) and so ker(f − λI) = V . So f = λI.

Suppose char - n = #G. For an irreducible representation (V, ρ), defineAvgv ∈ Hom(V, V G) as

Avgv(x) =1

n

∑g∈G

ρ(g)(x).

Note that Augv(x) ∈ V G and Avgv |V G = IV G . Also it is a projection, i.e.,idempotent.

Math 221 Notes 68

28 November 7, 2016

Last time we discussed representations and this averaging operator. Assumingthat char(k) - n = #G, from a representation (V, ρ) we define the k-linear map

AvgV : V → V G ⊆ V ; x 7→ 1

n

∑g∈G

ρ(g)(x).

Then AvgV |V G = IdV G . Also (AvgV )2 = AvgV . So Avg is a projection ontoV G. In the case H = Homk(W,V ), the fixed space is HG = HomG(W,V ). Then

AvgH : Homk(W,V )→ HomG(W,V ).

Theorem 28.1 (Maschke). Suppose char(k) - n. Then every finite dimen-sional k-linear representation of G can be decomposed as a direct sum of simplerepresentations (i.e., “irreducible = simple”).

Proof. Let (V, ρ) be a representation, and let W ⊆ V be a subrepresentation.The goal is to find a subrepresentation U ⊆ V such that V = W ⊕U . Note thatif you have an idempotent operator, then there is a direct decomposition intoits image and kernel.

Let f : V → W ⊆ V be any k-linear projection, i.e., f |W = idW . Letπ = AvgH(f) where H = Homk(V,W ). Then π ∈ HomG(V,W ) and π : V →W ⊆ V . We claim that π|W = IdW . For x ∈W ,

π(x) =1

n

∑g

(ρH(g)(f))(x) =1

n

∑g

(ρ(g) f ρ(g−1))(x)

=1

n

∑g

ρ(g) f(ρ(g−1)(x)) =1

n

∑g

ρ(g)(ρ(g−1)(x)) =1

n

∑g

x = x.

So take U = ker(π). This works.

28.1 Structure of finite representations

Now let’s assume:

• k is algebraically closed (for Schur’s lemma),

• char(k) - n (for Maschke’s theorem).

Corollary 28.2. Let V be a representation, and W be an irreducible represen-tation. Then dimk HomG(V,W ) is equal to the number of copies of W in thedirect sum decomposition of V . In particular, this number is independent of thechoice of a direct sum decomposition.

Proof. Write V = W a00 ⊕W

a11 ⊕ · · · ⊕W ar

r with W = W0,Wi non-isomorphicirreducible. Then

HomG(V,W ) = HomG(W a00 ⊕ · · · ⊕W ar

r ,W0) =

r⊕i=0

HomG(Wi,W0)ai

= HomG(W0,W0)a0 ∼= ka0

Math 221 Notes 69

by Schur.

Corollary 28.3. The direct sum decomposition into irreducibles is unique upto isomorphism.

Corollary 28.4. Up to isomorphism, there are only finitely many irreduciblerepresentations of G over k. Furthermore, if W is irreducible then W appearsdimkW times in Vreg. In particular,

n = dimVreg =∑W irr.

(dimW )2.

Proof. By definition,

HomG(Vreg,W ) ∼= HomR-mod(R,W ) ∼= W.

Example 28.5. Take G = S3, and char(k) 6= 2, 3. Let’s find representations.

• Vtriv = (k, ρtriv) with dimVtriv = 1.

• Vsgn = (k, ρsign) with dimVsign = 1 given by ρsign(σ)(x) = sign(σ) · x.

Now we are missing either four 1-dimensional representations or one 2-dimensionalrepresentation. We have the permutation on on X = 1, 2, 3. This representa-tion (VX , ρX) has dimension 3. Consider the averaging operator AvgVX : VX →V GX ⊆ VX . We have

V GX = span(AvgVX (ei)) = 〈(1, 1, 1)〉.

Let W = ker(AvgVx) = 〈e1 − e2, e1 − e3〉. Is W irreducible? Using bases, it ispossible to check that W does not split into two 1-dimensional representations.So we have all the irreducible representations, because 12 + 12 + 22 = 6.

Math 221 Notes 70

29 November 9, 2016

Today we continue with examples. Assume k = C for convenience. In general,for a (abelian) group, its dual is defined as

G∨ = MorGrp(G,C×) = 1-dimensional representations of G.

Note that if ρ1 6= ρ2 ∈ G∨ then ρ1 6∼= ρ2 because the eigenvalue of ρi(1) is simplyρi(1). Therefore we get an inclusion G∨ → irr. rep. of G.

Example 29.1. Consider G = Z/nZ. Here are some elements of G∨: Leta ∈ Z/nZ and define ψa : Z/nZ → C× as map r 7→ e2πira/n. Thus we get aninjection

ψ : Z/nZ = G → G∨; a 7→ ψa.

So we get n 1-dimensional irreducible representations of G = Z/nZ. Becausethe square of the dimensions add up to n, this is all.

Example 29.2. Consider an arbitrary abelian group G = Z/n1Z×· · ·×Z/ntZ.Let us look at G∨ again. It takes a minute to see G∨ ∼= (Z/n1Z)∨ × · · · ×(Z/ntZ)∨. So G∨ gives n = n1 · · ·nt representations, and there are no more.

Corollary 29.3. For k = C, if G is finite abelian, then all irreducible repre-sentations of G are 1-dimensional, and there are n = #G of them.

29.1 Characters

Now we work in C, because we want complex conjugation. Let G be a finitegroup and n = #G.

Definition 29.4. A map ξ : G → C is a class function if it is constant onconjugacy classes of G.

Let r be the number of conjugacy classes of G, and let Ω ∼= Cr be the spaceof class functions. We define a inner product on Ω as

(ψ, ξ) =1

n

∑g∈G

ψ(g)ξ(g).

For (V, ρ) a representation, define

χV = χρ = tr(ρ(−)) : G→ C.

Because trace is invariant under conjugation, χV ∈ Ω. We call χV the charac-ter of the representation. Note that V ∼= W implies χV = χW . There are somebasic properties:

• χV⊕W = χV + χW .

Math 221 Notes 71

• χV⊗W = χV · χW .

• χV ∨ = χV .

• χV (e) = dimV .

• χtriv = 1.

Theorem 29.5. (χV , χW ) = dimC HomG(W,V ).

Proof. Note that if U is a representation, then AvgU : U → UG ⊆ U is aprojection onto UG. So

dimUG = Tr(AvgU ) = Tr

(1

n

∑g

ρU (g)

)=

1

n

∑g

Tr(ρU (g)) =1

n

∑g

χU (g).

Now take U = V ⊗W∨ = Hom(W,V ). Then

(χV , χW ) =1

n

∑g

χV (g) · χW (g) = dim Hom(W,V )G = dimC HomG(W,V ).

This is incredibly useful because this Hom detects how many irreduciblerepresentations they have.

Corollary 29.6. The set χV for V irreducible are orthonormal. In otherwords, (χV , χW ) = δV∼=W for V and W irreducible.

Proof. Schur.

Corollary 29.7. If V and W are representations with χV = χW , then V ∼= Was representations.

Lemma 29.8. Let ψ ∈ Ω and (V, ρ) be a representation. Define

fψ,V =∑g

ψ(g)ρ(g) ∈ EndC(V ).

Then fψ,V ∈ EndG(V ).

Proof. Let h ∈ G. Then

fψ,V ρ(h) =∑g

ψ(g)ρ(gh) =∑g

ψ(hgh−1)ρ(hg)

=∑g

ψ(g)ρ(hg) = ρ(h) fψ,V .

Theorem 29.9. The set χV : V is irr. is an orthonormal basis for Ω. Inparticular, the number of irreducible representations is r.

Math 221 Notes 72

Proof. Let ψ ∈ Ω and suppose (ψ, χV ) = 0 for every V representation. We needto show that ψ = 0. Take any irreducible V . The previous lemma and Schurimplies that fψ,V = λV · IdV . Then

λV · dimV = Tr(fψ,V ) = n · (ψ, χV ) = 0

by the hypothesis. Thus if V is irreducible, then λV = 0 and so fψ,V = 0. Bylinearity, for every representation W , fψ,W = 0. Take W = Vreg. Then we getthe equation ∑

g

ψ(g)eg = 0.

Therefore ψ ≡ 0.

Math 221 Notes 73

30 November 11, 2016

Recall that so far we know that χV : V irred. is an orthonormal basis for Ω.Moreover, given any representation U and V irreducible, we have that (χU , χV )is the number of copies of V in U . This is called “orthogonality I”.

Corollary 30.1. If (χV , χV ) = 1, then V is irreducible.

Corollary 30.2. Let G be a finite group. Then G is abelian if and only if allirreducible representations of G are 1-dimensional.

Proof. We know that G is abelian if and only if n = r, where n is the order ofG and r is the number of conjugacy. Now use the formula n =

∑dim2.

Before, with some work, we proved the forward direction of this corollary.

Definition 30.3. Let V1, . . . , Vr be the irreducible representations of G. Letg1, . . . , gr be representatives for the conjugacy classes in G. In particular, G =⋃Cgi where Cgi denotes the conjugacy classes. We define the character table

of G as the matrix

T = [χVi(gj)]ij ∈Mr×r(C).

We remark that T is unique up to indices.

30.1 Character tables

Define

D =

#Cg1 0. . .

0 #Cgr

.Then the previous theorem gives the identity

1

nTDT ∗ = I.

Corollary 30.4 (Orthogonality II). For every g, h ∈ G,

r∑i=1

χVi(g) · χVi(h) =

0 if g not conjugate to hn

#Cgif g, h are conjugate to each other.

Proof. Because n−1TDT ∗ = I, we get DT ∗ = nT−1. Then DT ∗T = nI andT ∗T = nD−1. Taking the complex conjugation, we get T tT = nD−1.

Let us work out some examples.

Math 221 Notes 74

e (12) (123)Vtriv 1 1 1Vsign 1 −1 1V? 2 0 −1

Table 2: Character table of S3

Example 30.5. Take G = S3 again. We have Vtriv and Vsign that are 1-dimensional. A set of representatives for the conjugacy classes is e, (12), (123).Using orthogonality, we can fill in the last row using orthogonality II with thefirst column.

After we have the character table, given any representation like permutationrepresentation VX , we know how many each representation is in it. χVX (σ) isthe number of fixed points of σ.

e (12) (123)VX 3 1 0

Then we have

(χX , χtriv) =1

6(3 + 3 + 0) = 1, (χX , χsign) =

1

6(3− 3 + 0) = 0,

(χX , χ?) =1

6(6 + 0 + 0) = 0.

Therefore VX ∼= Vtriv ⊕ V?.

Example 30.6. Take G = S4. We again have Vtriv and Vsign, and also Vstd,which is defined as the kernel of the averaging operator of VX , which has dimen-sion n−1 in general and is irreducible. There are two more representations, andwe can compute the dimensions, because the sum of their squares are 13, and sothey have dimensions 2 and 3. But we still are not there to use orthogonality.

e (12) (12)(34) (123) (1234)Vtriv 1 1 1 1 1Vsign 1 −1 1 1 −1Vstd 3 1 −1 0 −1V3 3 −1 −1 0 1V2 2 0 2 −1 0

Table 3: Character table of S4

Let us now try to make new characters from the ones we have. Try Vsign ⊗Vstd = W . Then

e (12) (12)(34) (123) (1234)W 3 −1 −1 0 1

Math 221 Notes 75

We can easily check whether W is irreducible, because

(χW , χW ) =1

24(9 + 6 + 3 + 0 + 6) = 1.

So W is irreducible and W = V3. Using orthogonality, we can get the last thing.

Math 221 Notes 76


31.1 Constructing irreducible representations of Sm

Let K = C and G = Sm, and n = #G = m!. In this case, r = p(m) wherep(m) is the number of number of partitions. A partition of m is a tupleλ = (λ1, . . . , λk) such that

∑λi = m and λ1 ≥ λ2 ≥ · · · ≥ λk. For λ, the

diagram of λ is

δλ =

where the ith row has λi boxes. A more formal definition is, δλ is a suitablesubset of 1, . . . ,m2.

Definition 31.1. A Young tableau is a function t : δλ → 1, . . . ,m whichis bijective.

For example, the tableau

6 4 21 573

is a tableau of shape λ = (3, 2, 1, 1).

Definition 31.2. Let Xλ = tableaux of shape λ for t, s ∈ Xλ, and for t, s ∈Xλ, define t ∼ s if they have the same elements in each row. An equivalenceclass [t] is called a tabloid. We write Xλ = λ− tabloids.

There is a natural action of Sm on Xλ. And so Sm acts on Xλ. LetMλ = (VXλ , ρXλ) be the permutation representation induced by this action.

For example, if λ = (m), then Xλ has one element, and so Mλ is the trivialrepresentation. If λ = (m − 1, 1), then we get the permutation representation,which is Vtriv ⊕ Vstd. Now here are our goals:

1. Construct the Specht module Sλ ⊆Mλ.

2. Show that Sλ is irreducible.

3. Prove that Sλ ∼= Sµ implies λ = µ.

If we do this then we will be able to conclude that Specht modules are allirreducible representations of Sm.

Definition 31.3. For t ∈ Xλ, define Ct ⊆ S as the column stabilizer of t.In other words, these are the permutations that move elements only withincolumns.

Math 221 Notes 77

For instance,

t =1 23 has Ct = id, (13) ≤ S3.

Definition 31.4. For t ∈ Xλ and s ∈ Xµ, define the vectors:

Et,s =∑σ∈Ct

ε(σ)σ · e[s] ∈Mµ.

This depends on t as a tableaux and s as a tabloid. We also write Et = Et,t.

Note that σEt = Eσt, because Cσt = σCtσ−1.

Definition 31.5. We define the Specht module as

Sλ = C-spanEt : t ∈ Xλ ⊆Mλ.

Note that Sλ = R · Et0 for any t0 ∈ Xλ, because σEt = Eσt and G actson the set of λ-tableaux transitively. So Sλ is a subrepresentation of Mλ. Wecannot conclude that Sλ is irreducible only from the fact that it is generated byone vector.

31.2 Irreducibility of the Specht module

Definition 31.6. We write λ ≥ µ if∑i≤j λi ≥

∑i≤j µi for every j for which

this makes sense.

For example, (3, 3, 1, 1) ≥ (3, 2, 2, 1). This is not a total order, but λ ≥ µand µ ≥ λ implies λ = µ.

Lemma 31.7. Let λ, µ be partitions (of m). Suppose t ∈ Xλ and s ∈ Xµ

satisfying the condition that the entries on each row of s are in different columnsof t. Then λ ≥ µ.

This is an easy combinatorial exercise. For instance, take

t =1 2 345

, s =1 23 45

.

This implies that λ = (3, 1, 1) ≥ µ = (2, 2, 1). This is going to be used inshowing that Sλ ∼= Sµ implies λ = µ.

Lemma 31.8. Let t ∈ Xλ and s ∈ Xµ. Suppose Et,s 6= 0. Then λ ≥ µ. Ifmoreover λ = µ, then Et,s = ±Et.

Math 221 Notes 78

Proof. We assume Et,s 6= 0. If there are x 6= y in the same row of s and samecolumn of t, then (Id−(xy))e[s] = e[s] − e[s] = 0. But 〈Id, (xy)〉 ≤ Ct because xand y lie in the same column. Let σ1, . . . , σl be left coset representatives. Then

Et,s =

l∑j=1

ε(σj)σj(Id−(xy))e[s] = 0.

This is a contradiction. Therefore each row of s, its elements are in differentcolumns of t. Then Lemma 31.7 implies that λ ≥ µ.

Suppose that λ = µ. Then from the condition that elements of each row ofs are in different columns of t, we see that there exists a unique σ0 ∈ Ct suchthat σ0[t] = [s]. So

Et,s =∑σ∈Ct

ε(σ)σe[s] =∑σ∈Ct

ε(σ)σσ0e[t]

=∑σ∈Ct

ε(σσ−10 )σe[t] = ε(σ0)Et,t = ±Et.

Math 221 Notes 79


For a tableau t, we defined a subgroup Ct ≤ Sm, and we defined Mλ has thepermutation representation of tabloids of shape λ. We also define the Spechtmodule Sλ ⊆ Mλ, as generated by Et for t ∈ Xλ. We had two lemmas: oneabout a combinatorial criterion for λ ≥ µ, and one about Et,s 6= 0 implyingλ ≥ µ.

Lemma 32.1. Let u ∈Mλ and t ∈ Xλ. Then(∑σ∈Ct

ε(σ)σ

)u ∈ CEt ∈Mλ.

Proof. Write u =∑li=1 αie[si] for αi ∈ C. Then(∑

σ∈Ct

ε(σ)σ

)u =

∑i

αiEt,si ∈ CEt

by Lemma 31.8.

This operator seems interesting, so define

πt =∑σ∈Ct

ε(σ)σ ∈ R.

Definition 32.2. Define the inner product 〈•, •〉λ on Mλ by

〈e[s], e[t]〉λ =

1 if [s] = [t]

0 otherwise.

Proposition 32.3. πt is self-adjoint with respect to the inner product 〈•, •〉λ.

Proof. We have

〈πtu, v〉 =∑σ∈Ct

ε(σ)〈σu, v〉 =∑σ∈Ct

ε(σ)〈u, σ−1v〉

=∑τ∈Ct

ε(τ−1)〈u, τv〉 = 〈u, πtv〉.

Lemma 32.4 (Submodule theorem). Let W ⊆ Mλ be a subrepresentation.Then W contains Sλ or W is orthogonal to Sλ. In particular, Sλ is irreducible.

Proof. Let u ∈ W and t ∈ Xλ. Lemma 32.1 implies that πtv = αu,tEt forαu,t ∈ C.

Case 1: for some u0 ∈W and t0 ∈ Xλ, we have α0 = αu0,t0 6= 0.Because u0 ∈ W , we also have πtu0 ∈ W and so Et0 = α−1

0 πtu ∈ W . BecauseSλ is generated by any Et0 as an R-module, it follows that Sλ = R · Et0 = W .

Math 221 Notes 80

Case 2: for every u ∈W and every t ∈ Xλ, we have α0,t = 0.In other words, πtu = 0. Let [s] be any tabloid. Then

0 = 〈0, e[s]〉 = 〈πtu, e[s]〉 = 〈u, πte[s]〉 = 〈u,Et,s〉.

Thus letting s = t, Sλ and W are orthogonal.

32.1 Specht modules are all the irreducible representa-tions

Lemma 32.5. Let f ∈ HomG(Mλ,Mµ). Suppose Sλ 6⊆ ker f . Then λ ≥ µ.

Proof. By Lemma 32.4, we see that Sλ is orthogonal to ker f . For every t ∈ Xλ,Et is orthogonal to ker f . So

0 6= f(Et) = fπte[t] = πtfe[t]

because Et 6= 0 and fe[t] =∑αie[si] for si ∈ Xµ. So 0 6= πt

∑αie[si] =∑

αiEt,si . This implies that there exists a t ∈ Xλ and s ∈ Xµ with Et,s 6= 0.Then by Lemma 31.8, λ ≥ µ.

Lemma 32.6. Let f ∈ HomG(Sλ, Sµ). If f 6= 0, then λ ≥ µ.

Proof. Extend f to a C-linear map f : Mλ → Sµ ⊆ Mµ by 0 on (Sλ)⊥. Thenf is R-linear. (You can directly check this.) Then Lemma 32.5 shows thatλ ≥ µ.

Theorem 32.7. The Specht modules Sλ for λ a partition of m give the full list(with no repetitions) of irreducible representations of Sm.

32.2 Restricted representation

I am going to take H ≤ G and RG = C[G] and RH = C[H]. Then we have anatural inclusion RH ⊆ RG. Therefore the restriction of scalars give a functor

ResGH : GRep→ HRep; (V, ρ) 7→ (V, ρ|H).

This is functorial, because there is a map

HomG(V,W )→ HomH(ResV,ResW ); f 7→ f.

Recall that if A → B is an commutative algebra and M an A-module andN a B-module, then what we are doing is restriction of scalars Res : N 7→ AN .But this has an adjoint, which is extension by scalars:

T : M 7→MB = B ⊗AM.

That is, there is an A-linear isomorphism

HomB(B ⊗AM,N) ∼= HomA(M,AN).

The analogue for this is going to be the induced representation.

Math 221 Notes 81


Last time we saw that if H ≤ G, then there is a canonical inclusion RH → RGand so there is a restriction functor ResGH given by restriction of scalars. Todefine the adjoint operation, we need some notion of tensor product.

33.1 Tensor products for non-commutative rings

Let R be a ring, which is not necessarily commutative. Let M be a right R-module and N be a left R-module. A balanced map is a map b : M ×N → Xwith:

(1) X is an abelian group,

(2) b is bi-additive,

(3) b(mr, n) = b(m, rn).

We then define the tensor product M ⊗R N as the universal balanced mapτ : M ×N →M ⊗R N . This exists, because we can construct it as

M ⊗R N =〈m⊗ n : m ∈M,n ∈ N〉Z-Mod

〈relations〉.

If f : HomModR(M,M ′) and g ∈ HomRMod(N,N

′), we get a map f ⊗ g :M ⊗R N →M ′ ⊗R N ′.

M ×N M ′ ×N ′

M ⊗R N M ′ ⊗R N ′

f×g

bal.

f⊗g

Let M be a (R′, R)-bimodule, i.e.,

(1) M is a left R′-module,

(2) M is a right R-module,

(3) (r′m)r = r′(mr).

Then for every r′ ∈ R′, the multiplication map µr′ is in HomModR(M,M ′). Sothere is a natural left R′-module structure on M ⊗R N .

Consider the special case when R′ is an R-algebra, i.e., R→ R′. Then R′ isa (R′, R)-bimodule. So we get a functor

RMod→ R′Mod; N 7→ R′ ⊗R N.

Further note that there is a natural isomorphism

HomR′(R′ ⊗RM,N) ∼= HomR(M,RN)

where M is a left R-module and N is a left R′-module.

Math 221 Notes 82

33.2 Induced representation

Definition 33.1. We define the induced representation functor

IndGH : HRep→ GRep; V 7→ IndGH(V ) = RG ⊗RH V.

There is a more concrete definition of the induced representation, which isgood for computation. Define IndGH V =

⊕li=1 giV where:

(1) g1, . . . , gl are left coset representatives for G/H so that G =⋃giH,

(2) giV are C-isomorphic copies of V with the formal isomorphisms V ∼= giVwith g 7→ gix,

(3) g · gix = gjhx where j, h depends on g, i as follows: ggi ∈ G so we letggi = gjh.

Example 33.2. Take H = 〈1〉 ≤ G. Then IndVtriv,H = Vreg,G. We cansee this from RG ⊗RH C ∼= RG. Alternatively, you can look at the concreteconstruction. Because H has one element, g1, . . . , gl has to have all elementsG and the action must be the same thing as the action on the representationbecause g · gi = (ggi) · 1H .

Proposition 33.3 (Frobenius reciprocity). For V a H-representation and Wa G-representation,

HomG(IndGH V,W ) ∼= HomH(V,ResGHW ).

Thus, (χIndV , χW )G = (χV , χResW )H .

These stuff are used in number theory. In the proof of the infinitude theprimes in an arithmetic progression, Dirichlet used the twisted Riemann zetafunction, with characters 1-dimensional representations in the denominator.Then other people started to consider more general representation and Heckedeveloped a full theory of this and Artin proved some theorem about represen-tations coming from induced representation.

33.3 Characters of restricted and induced representation

Clearly χResV = Tr(ρV |H) = χV |H . For the induced representation, we can usethe concrete definition of Ind and realize g as a huge block matrix. Then

χIndV (g) =∑

g:giV→giVTr(g|giV ) =

∑g−1i ggi∈H

Tr(g−1i ggi|V )

=1

#H

∑g0∈G

g−10 gg0∈H

Tr(g−10 gg0|V ) =

1

#H

∑g0∈G

g−10 gg0∈H

χV (g−10 gg0).

Denote

ΩH = C-class functions on H, ΩG = C-class functions on G,

Math 221 Notes 83

and also let

ΛH = Z-linear combinations of G-representations

and likewise ΛG. Then ΛH ⊆ ΩH and ΛG ⊆ ΩG. Also the definition of inducedand restricted representations gives maps Res : ΛG → ΛH and Ind : ΛH → ΛG.

Math 221 Notes 84


We have a map Ind : ΩH → ΩG given by

ψ(−) 7→ 1

#H

∑g0∈G

g−10 (−)g0∈H

ψ(g−10 (−)g0).

By definition, Ind(ΛH) ⊆ ΛG. Likewise, we have restriction maps Res : ΩG →ΩH and Res(ΛG) ⊆ ΛH .

34.1 Artin induction theorem

Lemma 34.1. ΛG and ΛH are commutative unitary rings.

Proof. We only need to check that ΛG and ΛH are closed under multiplication.To do this, use the tensor product.

Lemma 34.2. Res is a ring morphism and Ind is a group morphism.

Proof. This is clear.

Lemma 34.3. Ind(ΛH) ⊆ ΛG is an ideal.

Proof. The image is clearly closed under addition. Let ψ ∈ ΛH and ξ ∈ ΛG.We want to show that Ind(ψ) · ξ is in Ind(ΛH). Let g ∈ G. Then

((Indψ) · ξ)(g) =1

#H

∑g0∈G

g−10 gg0∈H

ψ(g−10 gg0)ξ(g)

=1

#H

∑g0∈G

g−10 gg0∈H

ψ(g−10 gg0)ξ(g−1

0 gg0) = Ind(ψ · Res(ξ))(g).

So we have (Indψ) · ξ = Ind(ψ · Res(ξ)).

The image has no reason to be the whole ring. But we will soon see that itis close to being the whole ring. In fact, a suitable multiple of 1 is in the image,so it is a full sub-lattice.

Definition 34.4. If K is a cyclic group, define the “special” character ξK ∈ ΩKby:

ξK(k) =

#K if 〈k〉 = K

0 otherwise.

Math 221 Notes 85

Let us denote by Cyc(G) the set of cyclic groups of G. We remark that∑H∈Cyc(G)

IndGH(ξH)(g) =∑

H∈Cyc(G)

1

#H

∑g0∈G

g−10 gg0∈H

χH(g−10 gg0)

=∑

H∈Cyc(G)

∑〈g−1

0 gg0〉=H

1 = #G.

This looks very promising.

Lemma 34.5. It H is cyclic,then ξH ∈ ΛH .

Proof. We proceed by induction on #H. It is clear if #H = 1. For the inductivestep, we use the remark. We have∑

K∈Cyc(H)

IndHK(ξK) = #H.

So ξH = # · ξtriv − (some representations). This shows that ξH ∈ ΛH .

Theorem 34.6 (Artin). Let V be a G-representation. Then ξV is a Q-linearcombination of characters induced from cyclic subgroups of G.

Proof. We have

IG =∑

H∈Cyc(G)

IndGH(ΛH) ⊆ ΛG

as an ideal. By the remark, the constant function #G · ξtriv is in IG. Therefore(#G) · χV ∈ IG.

Brauer showed a more stronger theorem, by allowing groups that are calledelementary subgroups, but restricting to integer linear combinations.

34.2 Connections to analytic number theory

In 1737, Euler defined

ζ(s) =

∞∑n=1

1

ns

for s > 1, and noted that this diverges as s→ 1+. He figured out that you canfactor it as

ζ(s) =∏p

(1 +

1

ps+

1

p2s+ · · ·

)=∏p

(1− 1

ps

)−1

.

Math 221 Notes 86

So there has to infinitely many primes, for analytic reasons. Even now, this isthe best way to count primes.

Around 1837–1838, Dirichlet proved the fact that there are infinitely manyprimes in any congruence class p ≡ a (mod N) for (a,N) = 1. Here is Dirichlet’samazing approach. Take G = (Z/NZ)×, which is abelian. For χ : G→ C×, hedefined

L(s, χ) =∑

(n,N)=1

χ(n)

ns=∏p-n

(1− χ(p)

ps

)−1

again by unique factorization, for s > 1. Then using the fact that χ(p)/ps issmall, we can write

logL(s, χ) =∑p-N

− log(

1− χ(p)

ps

)=∑p-N

∞∑r=1

1

r

χ(pr)

prs=∑p-N

χ(p)

ps+ gχ(s),

where |gχ(s)| < C independent of χ and N . Using orthogonality, we can say

FN,a(s) =1

ϕ(N)

∑χ

χ(a) logL(s, χ) =1

φ(N)

∑χ

∑p-N

χ(p)χ(a)

ps+ bN (s)

=∑

p≡a (mod N)

1

ps+ bN (s)

where bN is uniformly bounded. So the goal is to show that FN,a(s) → ∞ ass→ 1+.

There are basically two cases. If χ = χtriv then L(s, χtrvi) is∑

(n,N)=1 1/ns

and this diverges. If χ 6= χtriv, you can show that L(1, χ) 6= 0,∞.

Math 221 Notes 87


35.1 Lie groups

Definition 35.1. A Lie group is a group (G, 1, ·) satisfying:

(1) G is a C∞ R-manifold,

(2) the multiplication map · : G×G→ G is C∞,

(3) the inversion map i : G→ G;x 7→ x−1 is C∞.

A morphism between Lie groups is a group morphism which is C∞.

Clearly R and C with addition are Lie groups. Also R× and C× with · areLie groups. The circle S1 = z ∈ C : |z| = 1 with multiplication is also a Liegroup. This is isomorphic to R/Z with addition. More generally, Rn/Λ withaddition is a Lie group, where Λ is a lattice.

More interesting examples are GLn(R) and GLn(C) with multiplication.Likewise SLn(R) and SLn(C) are Lie groups. The group B(n) of upper tri-angular matrices over R and Bn of upper triangular matrices over C with mul-tiplication are Lie groups. There is a confusion convention that real Lie groupshave n in the parentheses and complex Lie groups have n in the subscript. Wealso define

O(n) = A ∈ GLn(R) : AtA = I,U(n) = A ∈ GLn(C) : A∗A = I.

Note that U(n) is not a complex manifold, because the condition is not holo-morphic. We have SO(n) and SU(n), which are O(n) and U(n) that havedeterminant 1.

We have an isomorphism R ∼= B(2). Note that this is precisely the coun-terexample we saw when talking about irreducible representations being simple.

Definition 35.2. A representation of a Lie group G is a pair (V, ρ) with

(1) V a finite dimensional C-vector space,

(2) ρ : G→ GLC(V ) is a morphism of Lie groups.

For example,

ρ : R→ GL2(C); x 7→[1 x0 1

]is not semi-simple.

So we start to panic. But we can make this through. The key feature of thetheory of Lie groups are linearization of two kinds: of representations and of Liealgebras.

Math 221 Notes 88

35.2 The Haar measure and averaging

What went wrong with the map R → GL2(C)? The whole point of Maschke’stheorem was the averaging operator. But in this case, groups are infinite. Sowe would like integrate instead of adding elements.

Theorem 35.3 (In Knaap, Ch. VIII). Let G be a Lie group. There is a Borelmeasure µ on G which is left-invariant, i.e.,∫

G

ψ(g)dµ(g) =

∫G

ψ(hg)dµ(g).

Such a µ is unique up to scalar. If G is compact then µ is unique if we requirethat it is a probability measure (i.e.,

∫Gdµ(g) = 1). Furthermore, µ is bi-

invariant. This µ is called the Haar measure.

Example 35.4. Take R/Z with the Lebesgue measure. This is the Haar mea-sure. If G is finite, then the measure

µ =1

#G

∑g∈G

fg

is the Haar measure.

Assume G is compact. Let (V, ρ) be a representation. Define the averagingoperator

AvgV : V → V G; x 7→∫G

ρ(g)(x)dµ(g).

Lemma 35.5. If (V, ρ) is a representation, then ρ is unitary for a certain 〈 , 〉ρon V .

Proof. Let 〈 , 〉 be any inner product in V . Define

〈x, y〉ρ =

∫G

〈ρ(g)x, ρ(g)y〉dµ(g).

Check that it works.

Theorem 35.6 (Weyl). Let G be a compact Lie group.

(1) [easy] All finite dimensional C-linear representations of the Lie group Gare semi-simple.

(2) [requires some analysis] The set χV : V irred. is an orthonormal Hilbertbasis for Ω. In particular, they are countable. Here,

Ω = L2(G,µ)G-conj.C , (ψ, ξ)Ω =

∫G

ψ(g)ξ(g)dµ(g).

Math 221 Notes 89


Let G be a compact Lie group. We have the space Ω = L2(G,µ)conjC that is a

Hilbert space under the inner product∫Gψ(g)ξ(g)dµ(g). Then our theorem will

say that χV : V irreducible is a Hilbert basis of Ω.

36.1 Irreducible representations of S1

Let us look at an example first. The simplest compact Lie group is U(1) = S1 ∼=R/Z. Then µ is just the Lebesgue measure. Because the group is abelian, wehave Ω = L2([0, 1],Leb). We claim that if V is an irreducible representation,then dimV = 1. If (V, ρ) is irreducible, then ρ|〈1/n〉 = ρn : Cn → GL(V ) isa representation of a cyclic group. Note that the sets 〈1/n〉 equidistributes in(R/Z,Leb). This is saying that

1

n

∑j=1

δj/nw∗−−→ Leb

as n→∞. Then we get

limn→∞

∫〈1/n〉|χρn |2dDirac =

∫G

|χρ|2dLeb = 1

because χ is irreducible. But the sum∫〈1/n〉|χρn |

2dDirac is an integer. This

shows that for n 1, ρn is irreducible.

Proposition 36.1. The only irreducible representations of G = R/Z are: forn ∈ Z, en : θ 7→ e2πinθ ∈ GL1(C). This can also be written as en = e⊗n1 withe⊗−1

1 = e∨1 .

Proof. First note that the en’s are distinct irreducible representations. Let(V, ρ) be an irreducible representation. Then dimV = 1 and ρ is unitary. Soρ : G→ S1.

Take ρ(θ) = exp(2πif(θ)) where f : R → R is the lifting to the universalcovering. Then f(t) = αt for some α ∈ R. Going back to S1, we figure out thatα has to be an integer.

Corollary 36.2. The functions en(θ) = e2πinθ for n ∈ Z form an orthonormalHilbert basis for L2([0, 1],Leb).

This recovers the Fourier theory for L2 spaces.

36.2 The tangent space

Let M be a C∞ manifold in general, with n = dimM ≥ 1.

Definition 36.3. For a point p ∈ M , a curve at p is a map γ : (−ε, ε)→ M ,smooth, with γ(0) = p.

Math 221 Notes 90

We say that γ1 ∼ γ2 if for every f ∈ C∞U with small enough U 3 p,(f γ1)′(0) = (f γ2)′(0). Define the tangent space as

TpM = [γ] : γ a curve at p.

So far this is a pointed set with the point being [constp].Let ϕ : U → Rn be a chart at p. Define

Ψϕ : TpM → Rn; [γ] 7→ (((πi ϕ) γ)′(0))ni=1.

It is an exercise that Φϕ is bijective. Furthermore, given two different chartsϕ1, ϕ2, we have Ψϕ1

Ψ−1ϕ2∈ GLn(R). Also [constp] 7→ 0 by Ψϕ.

Corollary 36.4. TpM has a unique R-vector space structure such that [constp] =0 and for every chart ϕ at p, Ψϕ is a linear isomorphism.

Definition 36.5. The differential is a functor

d : pointedManifolds→ R−vsp; d(M,p) = TpM,

with morphisms given by

F ∈ Mor((M,p), (N, q)) 7→ (dpF : TpM → Tq; [γ] 7→ [F γ]).

Definition 36.6. A vector field is a function X : C∞M → C∞M satisfying

(D1) X is R-linear,

(D2) for every f, g ∈ C∞M , X(f · g) = X(f) · g + f ·X(g).

For example, if M = U ⊆ Rn, then all the vector fields are of the form

X =

n∑i=1

hi∂

∂xi

for hi ∈ C∞U .

Definition 36.7. The tangent bundle TM is the C∞M -module of vectorfields on M .

Given a point p ∈ M and a vector field X ∈ TM , we can get a Xp ∈ TM .There are two ways of getting this. We can argue that there exists a curve γ atp such that for all f ∈ C∞M , X(f)(p) = (f γ)′(0). This γ is not unique, but[γ] is uniquely defined. So Xp = [γ] is well-defined. Another constructive wayof doing this (in the local case) is to first write X =

∑hi(∂/∂xi) and define

Xp = Ψ−1IdU

((hi(p))mi=1). These two give the same answer.

Lemma 36.8. The function τX : M →∐p∈M TpM defined by p 7→ Xp deter-

mines X uniquely.

Proof. X 7→ τX is R-linear. If τX(p) ≡ 0 then Xp = 0 for every p, and thenX(f)(p) = (f const)′(0) = 0. This means that X(f) = 0 for every f ∈ C∞M .That is, X = 0.

Define for F : M → N the map dF : TM → TN ; X 7→ X F ∗. Then

((dF )(X))F (p) = (dpF )(Xp).

Math 221 Notes 91

37 December 2, 2016

Last time, for a n-manifold M , we defined the tangent space TpM at p. Wealso define TM , which is the C∞M -module of vector fields. Here, Xpp∈Muniquely determines X.

37.1 Left-invariant vector fields of Lie groups

Let (G; e, ·) be a Lie group and let n = dimG ≥ 1.

Definition 37.1. Given g ∈ G, define Lg : G→ G as p 7→ g · p. This induces amap L∗g : C∞G→ C∞G given by L∗g(f)(p) = f(g · p). We say that X ∈ TM isleft-invariant if for every g ∈ G, L∗g X = X L∗g.

Equivalently, we can say this as for every group element g ∈ G and a pointp ∈ G,

(dpLg)Xp = Xg·p.

For example, if G = R, then the invariant vector fields will be X = d/dt upto scalar multiplication. Then f ′(t + α) and (f(t + α))′ are the same. Let uslook at a more interesting example. Let G = R>0 with multiplication. ThenX = t(d/dt) is an invariant vector field. The commutation of L∗g and X can bechecked as

(L∗g X)(f) = L∗g(tf′(t)) = (tg)f ′(tg),

(X L∗g)(f) = X(f(g · t)) = td

dtf(gt) = tgf ′(gt).

Definition 37.2. Define LTG the R-vector space of left-invariant vector fieldsof G.

Proposition 37.3. The map LTG→ TeG given by X 7→ Xe is an isomorphismof R-vector spaces. The inverse is determined by

v ∈ TeG 7→ Xv the only X such that for all p, (deLp)(v) = Xp.

The only thing you need to check is that Xp defined as in the propositionvaries in a C∞ way. This can be done locally.

Definition 37.4. We define Lie(G) = g = TeG ∼= LTG.

Definition 37.5. Let X ∈ TM . An integral curve for X is a C∞ functionγ : I0 →M such that for all t0 ∈ I0, [γ(t+ t0)] = Xγ(t0).

Proposition 37.6. Let X ∈ LTG and let X 6= 0. Then there is a unique globalintegral curve γ : R→ G for X satisfying γ(0) = e.

Math 221 Notes 92

It turns out that γ(s0 + t0) = γ(s0) · γ(t0). That is, the group is abelianalong this curve. The idea of the proof is to define another curve γ(t) = γ(s0)−1 ·γ(s0 + t). The by uniqueness, γ = γ.

In our previous example G = R>0 and X = t(d/dt), the integral curve forX is γ(t) = et. This is because γ(0) = 1 and (f γ)′(t0) = f(et)′|t0 = et0f ′(et0)and X(f)(γ(t0)) = tf ′(t)|et0 = et0f ′(et0).

Definition 37.7. A 1-parameter subgroup is a morphism γ : R→ G of Liegroups.

So far, from a vector v ∈ g, we have obtained a left-invariant vector fieldXγ and then a 1-parameter subgroup γv : R → G. Given γv, it is not hard torecover v. It is just [γv] = v ∈ g.

37.2 The exponential map

Definition 37.8. The exponential map exp : g→ G is defined by v 7→ γv(1).

In the example G = R>0, a vector λ ∈ R = T1G gives a left-invariant vectorfield Xλ = λt(d/dt). Then we get the curve γλ(t) = eλt. So expG(λ) = eλ.

The reason we care about this map is because it allows us to go back frommorphisms of the tangent space to the morphisms of Lie groups.

Theorem 37.9. Let G,H be Lie groups and suppose that G is connected. Thenthe map

d : MorLieGrp(G,H)→ MorR−vsp(g, h)

is injective.

The reason this is true is because the diagram

g h

G H

dF

expG expH

F

that commutes. Because exp is a diffeomorphism near 0 7→ e, dF determines Flocally, and then determines the whole map F locally.

If you have two X,Y ∈ TM , then they are maps C∞M → C∞M . In general,X Y doesn’t satisfy the Leibniz rule, but X Y −Y X does. So we can define

[X,Y ] = X Y − Y X ∈ TM.

It is also true that if X and Y are left-invariant, then [X,Y ] is also left-invariant.Thus g also have this [ , ]. This bracket satisfies some relations.

(1) [ , ] is R-bilinear.

(2) It is alternating, i.e., [X,X] = 0.

Math 221 Notes 93

(3) The Jacobi formula

[X, [Y,Z]] + [Y, [Z,X]] + [Z, [X,Y ]] = 0

holds.

Definition 37.10. A R-vector space with such an operation [ , ] is called a Liealgebra over R.

Theorem 37.11. Let G,H be Lie groups. Suppose G is simply connected. Then

d : MorLieGrp(G,H)→ MorR−LieAlg(g, h)

is bijective.

Index

algebra, 15annihilator, 12Artin induction theorem, 85Artinian, 41ascending chain condition, 41associated primes, 30

balanced map, 81bilinear map, 14bimodule, 81

character, 70character table, 73class function, 70class group, 51commutative ring, 5coprime ideals, 7curve, 89

decomposition series, 53Dedekind domain, 48descending chain condition, 41differential, 90direct limit, 19direct sum, 11discrete valuation, 45discrete valuation ring, 46domain, 6dual group, 70

exact sequence, 21exponential map, 92

field, 6field extension, 55filtered, 18finite algebra, 32finite fields, 58finite type algebra, 32flat module, 22fractional ideal, 49free module, 17Frobenius reciprocity, 82

Galois extension, 59going-up theorem, 36

Haar measure, 88Hilbert basis theorem, 43

indecomposable, 53induced representation, 82integral, 32, 37integral algebra, 32integral closure, 33integral curve, 91integrally closed, 37irreducible representation, 66isolated prime, 30isolated set, 31

Jacobson radical, 8Jordan-Holder theorem, 54

Krull dimension, 45Kummer extension, 64

Lefschetz’s principle, 41left-invariance, 91length of module, 54Lie algebra, 93Lie group, 87localization, 22

for modules, 23locally free module, 25

Maschke’s theorem, 68maximal ideal, 8minimal polynomial, 55module, 10

finitely generated, 12submodule, 11

nilradical, 8Noether normalization, 39Noetherian, 41normal extension, 56

partition, 76

94

Math 221 Notes 95

perfect field, 64prepared extension, 65primary, 27primary decomposition, 28prime ideal, 8primitive element, 55Primitive element theorem, 57product, 11product ring, 7

radical extension, 64radical of an ideal, 9rank, 18regular representation, 66representation, 66, 87

morphism, 67restricted representation, 80ring, 5ring morphism, 5

Schur’s lemma, 67second uniqueness theorem, 31separable, 57simple module, 53

simple representation, 66solvable, 64Specht module, 77spectrum, 26splitting field, 55

tabloid, 76tangent bundle, 90tangent space, 90tensor product, 14, 81

of algebras, 16

uniformizer, 47unit, 5unitary ring, 5

valuation ring, 46vector field, 90

Young diagram, 76Young tableau, 76

Zariski topology, 33zero divisor, 6

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