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Prelims > Galois Theory > Flashcards

Galois Theory Flashcards

1
Q

What is the main idea of Galois theory

A

Turn problems about polynomials into problems about groups. Look at field generated by roots of polynomial. Galois group - all permutations of roots preserving algebraic relations between roots. Not the modern approach to construction Lecture 1

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2
Q

Historical examples of problems motivating Galois theory?

A
  1. Can a polynomial be solved by radicals?
  2. Ruler and compass constructions

Lecture 1

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3
Q

What is the Galois group?

A

Look at Galois group of field extension K/L - symmetries of K fixing L.

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4
Q

Main theorem of Galois theory?

A

Lecture 1: Subfields of field extension Subgroups of Galois Group

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5
Q

Applications of Galois theory?

A
  1. Langlands program: Representations of Galois groups of fields have something to do with modular forms
  2. Galois Group Fundamental group in alg topology

Lecture 1

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6
Q

What is a field extension? Examples?

A

Two fields K < L sometimes denoted L/K. The field K is called the base field of the extension

Every F can be considered a vector space over its prime field

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7
Q

Define: degree of field extension

Examples?

A

Denoted [L:K] is the dimensionm of L as a vector space over K

[C : R] = 2
[R : Q] = inf

Lecture 2

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8
Q

Define: finite extension

A

Finite extension if [L : K] finite

Lecture 2

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9
Q

When is an element alpha in L algebraic over K? transcendental?

algebraic extensions?

Examples?

A

If alpha is the root of polynomial p(x) in k{x}

Algebraic extension if every element in L is algebraic over K

Transcendental otherwise

fifth root of 2 algebraic, root of x^5-2
pi, e transcendental - but difficult proof
Q(x) field of rational numbers over Q clearly has x transcendental over Q

Lecture 2

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10
Q

Is alpha = cos( 2pi/7) algebraic or transcendental? Proof?

A

Lecture 2

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11
Q

Prove: alpha in M is Algebraic over K iff alpha is in a finite extension of K

A

Lecture 2

DF 521 - alpha algebraic over K <=> the simple extension F(alpha)/F is finite

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12
Q

Prove: [M:K] = [M:L]{L:K]

A

Lecture 2

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13
Q

Prove: If alpha, beta are algebraic over K, then so are sum difference product division

A

Lecture 2

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14
Q

Prove: If alpha is a root of polynomial with algebraic coefficients over K then alpha is algebraic

A

Lecture 2

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15
Q

Define: splitting field

examples?

A

Let p(x) in K[x] and L be an extension of K s.t.

  1. p factors into linear factors in L[x}
  2. L is the smallest possible field in which 1 holds - i.e. L is generated by roots of p over K

then L is called a splitting field of p(x)

D&F: The extension K/F is called a splitting field for the polynomial f(x) in F[x] if f(x) factors completely into linear factors (splits completely) in K[x] and f(x) does not factor completely into linear factors over any proper subfield of K containing F pg 536

Examples

  1. x - a — linear —- splitting field is just K
  2. x^2+bx +c — irreducible quadratic —- splitting field K[x}/(p(x))
  3. x^3 -2 — cubic — careful need [M:K] = 6
  4. cos(2pi/7) example: 8x^3+4x^2-4x-1
  5. x^4+1 over Q
  6. (x^2-2)(x^2-3)

D&F 537

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16
Q

Prove: Existence/Uniqueness of Splitting Fields

A

pg 20-22

D&F 536, 541 - 542

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17
Q

Define: algebraic closure

A

A field extension K bar of K s.t.

  1. Any polynomial in K[x] factors into linear factors over K bar
  2. K bar is generate by roots of polynomials in K[x] – minimal

D&F: F bar is algebraic over F and every polynomial f(x) in F[x] splits completely over F bar.

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18
Q

Construct algebraic closure of a countable field

A

pg 24

if uncountable, just well-order the polynomials by axiom of choice

D&F 544

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19
Q

Define: algebraically closed field

A

A field L is algebraically closed if all polynomials in L[x] have roots in L

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20
Q

Prove: the algebraic closure of a field is algebraically closed

A

pg 25, D&F 543

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21
Q

Discuss closing a field under square root. Motivation?

A

pg 26

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22
Q

What is Fundamental Thm of Algebra? Proofs?

A

Louiville’s Thm - Stein pg 50 - 51

Topological proof - Borcherds pg 27

Topological proof - degree Hruska pg 164 - homotopy of loops

Algebraic proof - still uses intermediate value thm – and Galois Borcherds pg 73

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23
Q

Examples of algebraically closed fields

A

C, Q bar - algebraic closure of Q (focus of algebraic number theory), Puiseuz series

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24
Q

Is algebraic closure unique? functor? Discuss analogy to topology.

A

Any 2 are isomorphic, but no canonical isomorphism.

Uniqueness of algebraic closure = Uniqueness of fundamental group

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25
Q

Define: characteristic of field

A

ker map Z —> field, 0, 2,3,5,7, …

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26
Q

Show that if F is a finite field, F must have order p^n for some prime p

A

pg 32

Comes down to vector space structure over F_p

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27
Q

Classify all finite fields - existence and uniqueness

A

Exists a unique field of order p^n for any prime p, positive integer n

use magic polynomial x^q -x where q = p^n

Lagrange’s Thm - prove this too
pg 33-35

D&F 549 - 550

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28
Q

Explicitly construct finite fields of order 2^n for n = 2,3

Show two different constructions for n=3 are isomorphic

A

pgs 36-37

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29
Q

Is there a canonical finite field of a given order?

A

No

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30
Q

Give an example of how to find the number of irreducible polynomials of a given degree over a finite field

A

pg 39-40

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31
Q

Define: normal extension

3 equivalent conditions

Example, non example

A

D&F 537: If K is an algebraic extension of F which is the splitting field for a collection of polynomials f(x) in F[x], then K is called a normal extension of F.

Borcherds: TFAE - if any holds called a normal extension for an algebraic extension K < L

  1. Any polynomial p in K{x] that is irreducible and has a root in L factors into linear factors in L[x]
  2. L is the splitting field over K of some set of polynomials
  3. K < L < K closure. Any automorphism of K closure over K maps L to L

Examples

  1. x^3 -2. Q adjoin cube root 2 not normal. Adding root of unity yields normal - splitting field
  2. 8x^3 + 4x^2+4x -1 is
  3. Quadratic always normal

Very closely related to normal subgroups - provides origin of term

pg 41 - 43

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32
Q

Is the normal extension of a normal extension normal?

A

No - Q < Q[root 2] < Q[4th root 2] - both degree 2 so normal but x^4 has imaginary roots not in this extension

44-45

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33
Q

Define: separable polynomial, element, extension

A

A polynomial f in K[x] is called separable if no multiple roots in K bar <=> (f, f’) = in K[x}

an element alpha in L is separable over K if it is the root of a separable polynomial in K[x].

L/K is called separable if all alpha in L are separable

Otherwise inseparable

All we are trying to do is avoid dealing with multiple roots

D&F 551

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34
Q

Prove: A polynomial f in K[x] is separable <=> (f, f’) = in K[x}

A

46

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35
Q

Compare separable and normal

A

L/K, f irreducible poly of alpha in L. deg f = n

L normal => all roots of f are in L
L separable => all roots of f are distinct

Normal + Separable => f has n distinct roots in L

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36
Q

Examples of separable extensions?

A
  1. K < L char K = 0, then L is separable over K. Say f irreducible, deg f’ = def f -1 => (f’, f) = 1 since f irreducible can’t have a common factor with something of lower degree
  2. This can fail for char K != 0 pg 48
  3. All extensions of finite fields are separable - roots of x^p^n - x = f(x). Note: f’(x) = -1 so (f, f’) = 1

All extensions you come across tend to be separable unless you look quite hard

D&F 546 -547
x^2 -2 separable, (x^2-2)^n inseparable for n>1
D&F 546

x^2 - t over the function field F_2(t) of rational functions in t with cofficients from F_2 is not separable x^2 -t = (x- root t)^2

x^p^n -x over F_p and x^n -1 over field of char not dividing n are both separable - n distinct roots of unity

37
Q

Opposite of separable

A

Purely inseparable extensions - only one distinct root. x^p^n = a for some a in K where char K = p. Factors. pg 49

38
Q

Define: Galois Extension

5 equivalent definitions

prove 1-4

A

Let K < M be a finite field extension. G = Aut(M/K) automorphisms of M fixing all elements of K. TFAE

  1. M is Normal and Separable (over K)
  2. [M : K] = |G|
  3. K = M^G
  4. M is splitting field of separable polynomial
  5. Galois Correspondence
39
Q

Show for any finite group H we have a finite extension with H as Galois group

A

Let M = K(x1, x2, x3, x4, x5) be the field of fractions in 5 indeterminants acted on by S_5 by permutations of x1, …, x5. M^G = K(e1, … , e5) the elementary symmetric functions

Gal(M: M^G) = S_5

More generally, let M be the field of fractions of the regular rep of any finite group. HARD over Q - inverse Galois problem pg 51

40
Q

Prove: Aut(M/K) <= [M : K]

A

pg 52. Prove something a little more general. If K < M and f: K –> X is a field hom, then there are at most [M : K] homs extending f to M. Taking M = X gives desired result.

Create a chain of fields k < k{a1] < k[a1, a2] < … < M

Any extension of k(a1) must map a1 to a root of p1…

41
Q

State and prove the Fundamental Thm of Galois Theory

A

If M/K is a finite Galois extension, then there is a 1-to-1 correspondence between intermediate fields K < L < M and subgroups 1 < H < Gal(M/K)

L —-> Gal(M/L)
M^H

42
Q

Discuss the subfields of Q(4th root 2, i) using Galois theory.

A

pg 64

43
Q

When is a number constructible with ruler and compass?

A

+, - , x, / and sqrt - pg 66. a constructible iff a in normal extension of degree 2^n

44
Q

When can you construct a p-sided polygon?

A

<=> [Q(zeta) : Z] has degree power of 2 pg 70 - 72

45
Q

Define: primitive element

A

An element that generates the entire field extension

46
Q

Prove: If M/K is separable, then M/K has a primitive element

A

Extend M to normal extension over K - so M normal + separable = Galois.

Idea: Choose x not in any extension L with K < L < M, L != M. Then K(x) = M (since it can’t be in any L).

We just need to prove we can make this choice - i.e. that M != union L_i.

For char 0, this comes down to the linear algebra proposition that if K is an infinite field, and M a f,d, vector space over K, then M is not the union of finite number of proper subspaces.

For char p, multiplicative subgroup is cyclic - has a primitive element

pg 77 - 79

47
Q

What is Abel’s theorem?

A

The general quintic cannot be solved by radicals

48
Q

What is the approach to proving Abel’s theorem?

A

Show if alpha can be expressed by radicals (over k) then alpha is in a finite SOLVABLE Galois extension of k –> i.e. the Galois group is solvable - we can split the extension into abelian or even cyclic extensions

Thus, to show 5th degree polynomial cannot be solved by radicals, we want to find a degree 5 polynomial whose Galois group is not solvable - S_5

49
Q

Examples of polynomials with non-solvable Galois groups?

A
  1. Q(e1, … , e5) where ei are the elementary symmetric functions
  2. f(x) in Z[x] irreducible, degree 5, 3 real roots, 2 non-real roots — x^5 - 4x +2

pg 84 - 86

50
Q

Prove: Solvable by radicals ==> solvable Galois extension

A

pg 88 - 90

51
Q

Does solvable Galois extension ==> solvable by radicals

A

Yes in char 0, no in general pg 90

52
Q

Define: prime subfield

A

The subfield of F generated by the multiplicative identity of F. It is isomorphic to either Q (if char(F) = 0) or F_p (if ch(F) = p)

53
Q

Let F be a fields and p(x) in F(x). Does there exist an extension K of F containing a solution to the equation p(x) = 0 (i.e. containing a root)?

A

Yes, DF 512

54
Q

Find a basis for the field K = F[x] / ( p(x) ) as a vector space over F, where p irreducible.

Discuss addition and multiplication of these elements

A

If p degree n, then K is an n-dimensional vector space over F with basis 1, theta, theta^2, … theta^n-1

where theta = image of x.

Just prove they span - Euclidean domain and are independent - assume relation

DF 513-514

55
Q

Examples of the K = F[x] / ( p(x) ) construction?

A
  1. C/R
  2. Q(i)
  3. Q(root(2))
  4. x^3 -2 - discuss how inverses can be computed using Euclidean algorithm and fact that any lower degree polynomial is relatively prime to the irreducible x^3 -2
56
Q

Let K/F and a, b, c, … in K.

Define: subfield generated by a,b,c,… over F, simple extension, primitive element

A

The subfield generated by a,b,c, … denoted F(a, b, …) is the smallest subfield of K containing both F and the elements a, b, … intersect all subfields

If K is generated by a single element F(a) over F, then K s.t.b. a simple extension and a called a primitive element for the extension (every finite extension of a field of characteristic 0 is simple - has primitive)

57
Q

Let p(x) be irreducible in F[x] and alpha be a root of p(x) in some extension K/F. What is the relationship between F(alpha) and F[x] / ( p(x) ) ? proof?

A

They are isomorphic

DF 517

This implies the roots of an irreducible polynomial are algebraically indistinguishable - the fields obtained by adjoining any root of an irreducible polynomial are isomorphic

58
Q

Let phi : F –> F’ be an isomorphism of fields and p(x) be irreducible in F[x]. Discuss the relationship between F[x]/ ( p(x) ) and F’[x] / (p’(x) )

A

Isomorphic

DF pg 519

59
Q

Define: minimal polynomial, degree of alpha

prove existence/uniqueness

A

The unique monic irreducible polynomial m(x) which has alpha as a root.

60
Q

Prove: If K/F is finite <=> K is generated by a finite number of algebraic elements over F. What is degree?

A

DF 522 & 526

61
Q

What do quadratic extensions look like?

A

DF 522 F(root(D)) where D is not a square in F

62
Q

Say L/F is finite extension and F < K < L. Why does [K : F] divide [L :F]?

A

Because [L:F] = [L:K] [K:F] pg 524

Compare to Lagrange for finite groups

63
Q

Show sixth root of 2 is irreducible over Q(root 2)

A

DF 524 Example 2

64
Q

How can you construct a finitely generated field F(alpha, beta, …) ?

What is its degree over F? Basis?

A

Recursively by a series of simple extensions. F(alpha, beta) = (F(alpha))(beta). Simple minimality argument pg 525

65
Q

Discuss the extensions Q(6th root 2, root 2) and Q(root 2, root 3)

A

DF 526

66
Q

If K is algebraic over F and L is algebraic over K, need L be algebraic over F? Proof

A

Yes 527 - 528

67
Q

Define: composite field

Degree over F?

A

If K1 and K2 are two subfields of K/F, then the composite field K1K2 is the smallest subfield of K containing both K1 and K2. -intersection of all subfields containing K1 and K2

DF 528-529

68
Q

How can we bound the degree of a splitting field of a polynomial of degree n over F?

A

At most n! - if f(x) irreducible over F, then adjoining a root is a degree n extension. Now f1(x) has a linear factor, adding another root is at most a degree n-1 extension …

The general polynomial of degree n over Q has a splitting field of degree n! - generic situation

D&F 538

69
Q

Discuss: cyclotomic fields, primative nth root of unity, degree of extension

A

Consider splitting field of x^n -1 over Q - roots are the nth roots of unity - e^2piki/n

Euler phi-function gives number of nth roots of unity

Splitting field of x^n -1 is generated by one primitive nth root of unity - called the cyclotomic field

Degree of extension:
prime = p-1
in general = phi(n) - Euler phi function
pg 539-540

70
Q

Discuss the splitting field of x^p -2, p a prime

A

extension of degree p(p-1)

pg 541

71
Q

Define: Frobenius endomorphism

Consequences for finite fields?

A

Let F be a field of characteristic p. Then for any a, b in F,

(a + b)^p = a^p + b^p and (ab)^p = a^p b^p

Put another way, the pth-power map defined by phi(a) = a^p is an injective field homomorphism from F to F.

Called the Frobenius endomorphism

If F is a finite field with char F = p, then every element of F is a pth power of F.

Injectivity implies surjectivity in case of finite field

D&F 548 -549

72
Q

Define: perfect field

Prove: Every irreducible polynomial over a field F is separable <=> F is perfect.

A

If char F = p, then perfect if every element of F is a pth power in F i.e. F = F^p. Also any field of char 0 is perfect

D&F 549

73
Q

Define: nth cyclotomic polynomial phi_n(x)

What is degree?

A

Polynomial whose roots are the primitive nth roots of unity

Degree is phi(n) = number of integers less than n relatively prime to n

D&F: 552

74
Q

Factor x^n-1 using cyclotomic polynomials

Show how to compute on examples

A

product over d dividing n phi_d(x)

compute recursively

D&F 553

75
Q

Prove: The cyclotomic polynomial phi_n(x) is an irreducible monic polynomial in Z[x] of degree phi(n)

Corollary?

A

Monic and degree are immediate from def o f phi_n as product over linear factors, one for each primative root.

In Z[x] follows by induction and Gauss’ Lemma

Finally, irreducible takes some work…

The corollary is that the degree over Q of the cyclotomic field of nth roots of unity is phi(n) : [ Q( zeta_n ) : Q ] = n
D&F 554 - 555

This follows since the thm essentially says that phi_n(x) is the minimal polynomial for any primative nth root of unity

76
Q

Define: automorphism, “fixing” alpha, Aut(K/F)

A

An automorphism is an isomorphism of a field with itself. Denote the collection of automorphisms Aut(K)

Fixes alpha if sigma(alpha) = alpha. Can also talk about fixing a subset with obvious definition

Aut(K/F) = the collection of automorphisms of K which fix F

77
Q

Why must automorphism of K fix prime subfield?

A

Since prime field is generated by 1 and any automorphism must take 1 to 1

78
Q

How can we understand what an automorphism does to K? Proof?

A

It permutes roots of irreducible polynomials.

Let K/F be a field extension and let alpha in K be algebraic over F. Then for any sigma in Aut(K/F), sigma alpha is a root of the minimal polynomial for alpha over F i.e. Aut(K/F) permutes the roots of irreducible polynomials. Equivalently, any polynomial with coefficients in F having alpha as a root also has sigma(alpha) as a root.

Proof. The idea is that polynomial equations just describe relations in K. Any automorphism must preserve all relations.

pg 559

79
Q

Examples of Aut(K/L)?

A
  1. Aut(Q( root(2) ) ) = Z_2

2. Aut(Q(cube root 2)) = trivial

80
Q

Define: fixed field of H

A

If H is a subgroup of the group of automorphisms of K, the subfield of K fixed by all the elements of H is called the fixed field of H

81
Q

In what sense is the association of groups to fields and fields to groups inclusion reversing? Proof?

A
  1. If F1 < F2 < K are two subfields, the Aut(K/F2) < Aut(K/F1)
  2. If H1 < H2 < Aut(K) are two subgroups with associated fixed fields F1 and F2, then F2 < F1

pg 560

82
Q

Discuss what can go wrong in passing from field to group back to fixed field

A

Consider two examples

  1. Aut(Q( root(2) ) ) = Z_2
  2. Aut(Q(cube root 2)) = trivial

For (1) everything works well. For (2) it doesn’t: not enough automorphisms to force the fixed field to be Q - this happens because not enough roots of 3rd root 2 in field

pg 560 -561

83
Q

Discuss the size of the automorphism group for a splitting field

A

Let F be a field and E the splitting field over F of f(x) in F[x]. The size of Aut(E/F) <= [E : F] with equality if f(x) is separable over F.

Proof. By induction on [E : F].

DF 561 - 562

84
Q

Define: K Galois over F, Galois extension, Galois group of K/F, Galois group of f(x)

Equivalent conditions? What about Q?

A

Let K/F be a finite extension. Then K is stb Galois over F and K/F is a Galois Extension if |Aut(K/F)| = [K:F]. Then call the group Aut(K/F) the Galois group of K/F, Gal(K/F).

The Galois group of f(x) over F is the Galois group of the splitting field of f(x) over F

  1. K is Normal and Separable (over F)
  2. [K : F] = |Aut(K/F)|
  3. F = K^Aut(K/F)
  4. K is splitting field of separable polynomial
  5. Galois Correspondence

Note: For Q, the splitting field of any polynomial over Q is Galois since the splitting field of f(x) is clearly the same as the splitting field of the product of irreducible factors of f(x)

DF 562 - 563 and Borcherds

85
Q

Examples of Galois Correspondence?

A

Borcherds

  1. C/R {1, complex conjugation}
  2. Splitting field of x^3 - 2 = S_3 DF 564-565
  3. FInite fields - cyclic, generated by a –> a^p Frobenius DF 566
  4. z^7 -1

DF 563 - 564

  1. Q( root 2, root 3) = Z_2 x Z_2
86
Q

Is the Galois Extension of a Galois Extension Galois?

A

No. As with Normal extensions, the case to consider is
Q < Q(root 2) < Q(4th root 2).

D&F 566

87
Q

Define: character of a group, linearly independent, embedding of K into L

A

A character of a group G with values in a field L is a homomorphism from G to the multiplicative group of L

Characters x1, … , xn of G are linearly independent if they are linearly independent as functions on G i.e. there is no nontrivial relation a1x1 + … + anxn =0.

An embedding of K into L is an injective homomorphism of field K into L

88
Q

Prove: If x1, … , xn are distinct characters of G with values in L, then they are linearly independent over L

A

Suppose not independent and choose a minimal dependence relation….

pg 569

89
Q

Why are characters important in field theory?

A

We can consider restriction of an embedding K into L to the multiplicative group of K. This defines a character of K^x –> L^x. The character caries all the information of the original field homomorphism since the only thing left out is 0 and we know that goes to 0.

By the result about distinct characters being independent, we see also that distinct embeddings of K into L are independent as functions on K. In particular distinct automorphisms are linearly independent

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