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
2
Q
Historical examples of problems motivating Galois theory?
A
- Can a polynomial be solved by radicals?
- Ruler and compass constructions
Lecture 1
3
Q
What is the Galois group?
A
Look at Galois group of field extension K/L - symmetries of K fixing L.
4
Q
Main theorem of Galois theory?
A
Lecture 1: Subfields of field extension Subgroups of Galois Group
5
Q
Applications of Galois theory?
A
- Langlands program: Representations of Galois groups of fields have something to do with modular forms
- Galois Group Fundamental group in alg topology
Lecture 1
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
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
8
Q
Define: finite extension
A
Finite extension if [L : K] finite
Lecture 2
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
10
Q
Is alpha = cos( 2pi/7) algebraic or transcendental? Proof?
A
Lecture 2
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
12
Q
Prove: [M:K] = [M:L]{L:K]
A
Lecture 2
13
Q
Prove: If alpha, beta are algebraic over K, then so are sum difference product division
A
Lecture 2
14
Q
Prove: If alpha is a root of polynomial with algebraic coefficients over K then alpha is algebraic
A
Lecture 2
15
Q
Define: splitting field
examples?
A
Let p(x) in K[x] and L be an extension of K s.t.
- p factors into linear factors in L[x}
- 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
- x - a — linear —- splitting field is just K
- x^2+bx +c — irreducible quadratic —- splitting field K[x}/(p(x))
- x^3 -2 — cubic — careful need [M:K] = 6
- cos(2pi/7) example: 8x^3+4x^2-4x-1
- x^4+1 over Q
- (x^2-2)(x^2-3)
D&F 537
16
Q
Prove: Existence/Uniqueness of Splitting Fields
A
pg 20-22
D&F 536, 541 - 542
17
Q
Define: algebraic closure
A
A field extension K bar of K s.t.
- Any polynomial in K[x] factors into linear factors over K bar
- 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.
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
19
Q
Define: algebraically closed field
A
A field L is algebraically closed if all polynomials in L[x] have roots in L
20
Q
Prove: the algebraic closure of a field is algebraically closed
A
pg 25, D&F 543
21
Q
Discuss closing a field under square root. Motivation?
A
pg 26
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
23
Q
Examples of algebraically closed fields
A
C, Q bar - algebraic closure of Q (focus of algebraic number theory), Puiseuz series
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
25
Define: characteristic of field
ker map Z ---> field, 0, 2,3,5,7, ...
26
Show that if F is a finite field, F must have order p^n for some prime p
pg 32
Comes down to vector space structure over F_p
27
Classify all finite fields - existence and uniqueness
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
28
Explicitly construct finite fields of order 2^n for n = 2,3
Show two different constructions for n=3 are isomorphic
pgs 36-37
29
Is there a canonical finite field of a given order?
No
30
Give an example of how to find the number of irreducible polynomials of a given degree over a finite field
pg 39-40
31
# Define: normal extension
3 equivalent conditions
| Example, non example
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
32
Is the normal extension of a normal extension normal?
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
33
Define: separable polynomial, element, extension
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
34
Prove: A polynomial f in K[x] is separable <=> (f, f') = in K[x}
46
35
Compare separable and normal
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
36
Examples of separable extensions?
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
Opposite of separable
Purely inseparable extensions - only one distinct root. x^p^n = a for some a in K where char K = p. Factors. pg 49
38
# Define: Galois Extension
5 equivalent definitions
prove 1-4
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
Show for any finite group H we have a finite extension with H as Galois group
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
Prove: Aut(M/K) <= [M : K]
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
State and prove the Fundamental Thm of Galois Theory
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
Discuss the subfields of Q(4th root 2, i) using Galois theory.
pg 64
43
When is a number constructible with ruler and compass?
+, - , x, / and sqrt - pg 66. a constructible iff a in normal extension of degree 2^n
44
When can you construct a p-sided polygon?
<=> [Q(zeta) : Z] has degree power of 2 pg 70 - 72
45
Define: primitive element
An element that generates the entire field extension
46
Prove: If M/K is separable, then M/K has a primitive element
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
What is Abel's theorem?
The general quintic cannot be solved by radicals
48
What is the approach to proving Abel's theorem?
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
Examples of polynomials with non-solvable Galois groups?
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
Prove: Solvable by radicals ==> solvable Galois extension
pg 88 - 90
51
Does solvable Galois extension ==> solvable by radicals
Yes in char 0, no in general pg 90
52
Define: prime subfield
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
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)?
Yes, DF 512
54
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
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
Examples of the K = F[x] / ( p(x) ) construction?
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
Let K/F and a, b, c, ... in K.
Define: subfield generated by a,b,c,... over F, simple extension, primitive element
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
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?
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
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) )
Isomorphic
DF pg 519
59
# Define: minimal polynomial, degree of alpha
prove existence/uniqueness
The unique monic irreducible polynomial m(x) which has alpha as a root.
60
Prove: If K/F is finite <=> K is generated by a finite number of algebraic elements over F. What is degree?
DF 522 & 526
61
What do quadratic extensions look like?
DF 522 F(root(D)) where D is not a square in F
62
Say L/F is finite extension and F < K < L. Why does [K : F] divide [L :F]?
Because [L:F] = [L:K] [K:F] pg 524
Compare to Lagrange for finite groups
63
Show sixth root of 2 is irreducible over Q(root 2)
DF 524 Example 2
64
How can you construct a finitely generated field F(alpha, beta, ...) ?
What is its degree over F? Basis?
Recursively by a series of simple extensions. F(alpha, beta) = (F(alpha))(beta). Simple minimality argument pg 525
65
Discuss the extensions Q(6th root 2, root 2) and Q(root 2, root 3)
DF 526
66
If K is algebraic over F and L is algebraic over K, need L be algebraic over F? Proof
Yes 527 - 528
67
# Define: composite field
Degree over F?
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
How can we bound the degree of a splitting field of a polynomial of degree n over F?
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
Discuss: cyclotomic fields, primative nth root of unity, degree of extension
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
Discuss the splitting field of x^p -2, p a prime
extension of degree p(p-1)
pg 541
71
# Define: Frobenius endomorphism
Consequences for finite fields?
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
# Define: perfect field
Prove: Every irreducible polynomial over a field F is separable <=> F is perfect.
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
# Define: nth cyclotomic polynomial phi_n(x)
What is degree?
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
Factor x^n-1 using cyclotomic polynomials
Show how to compute on examples
product over d dividing n phi_d(x)
compute recursively
D&F 553
75
Prove: The cyclotomic polynomial phi_n(x) is an irreducible monic polynomial in Z[x] of degree phi(n)
Corollary?
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
Define: automorphism, "fixing" alpha, Aut(K/F)
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
Why must automorphism of K fix prime subfield?
Since prime field is generated by 1 and any automorphism must take 1 to 1
78
How can we understand what an automorphism does to K? Proof?
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
Examples of Aut(K/L)?
1. Aut(Q( root(2) ) ) = Z_2
| 2. Aut(Q(cube root 2)) = trivial
80
Define: fixed field of H
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
In what sense is the association of groups to fields and fields to groups inclusion reversing? Proof?
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
Discuss what can go wrong in passing from field to group back to fixed field
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
Discuss the size of the automorphism group for a splitting field
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
# Define: K Galois over F, Galois extension, Galois group of K/F, Galois group of f(x)
Equivalent conditions? What about Q?
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
Examples of Galois Correspondence?
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
Is the Galois Extension of a Galois Extension Galois?
No. As with Normal extensions, the case to consider is
Q < Q(root 2) < Q(4th root 2).
D&F 566
87
Define: character of a group, linearly independent, embedding of K into L
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
Prove: If x1, ... , xn are distinct characters of G with values in L, then they are linearly independent over L
Suppose not independent and choose a minimal dependence relation....
pg 569
89
Why are characters important in field theory?
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
