galois theory: main results - kennesaw state...

Galois Theory: Main Results

Philippe B. Laval

KSU

Current Semester

Philippe B. Laval (KSU) Galois Theory: Main Results Current Semester 1 / 27

Introduction

Recall the summary of what we are aiming for, given in the previousset of slides.Galois wanted to answer the question: when is a polynomial a (x)over a field F solvable by radicals? That is, given a (x), is there aformula for its roots like the quadratic formula which gives us theroots of polynomials of degree 2?Galois found that if a (x) is solvable by radicals then its Galoisgroup is a solvable group.We will see that a group G is said to be solvable if there exists asequence of subgroups {e} ⊆ G1 ⊆ G2 ⊆ ... ⊆ Gn = G if for each k,Gk C Gk+1 and Gk+1/Gk is Abelian.We will see that the Galois group of the root field K of somepolynomial ∈ F [x ], denoted Gal (K : F ), is the group ofautomorphisms of K fixing F .First, we will need to study automorphisms of K fixing F . This willcome from field extensions.


Introduction

Of course, we now have covered several topics. In this set of slides,we will focus on Galois groups.Let us begin by summarizing some important results of the previouschapter.

Let K be a root field of some polynomial over F and let a ∈ F .1 Any isomorphism with domain K which fixes F is an automorphism ofK . (theorem on slide 26 of Galois theory: preamble).

2 If a and b are roots of an irreducible polynomial p (x) ∈ F [x ], there isan automorphism of K fixing F and sending a to b. (theorems onslides 21 and 27 of Galois theory: preamble).


Introduction

Let K be the root field of some polynomial a (x) ∈ F [x ]. Ifc1, c2, ..., cn are the roots of a (x) then K = F (c1, c2, ..., cn).Recall we noted in the previous chapter that any automorphism h ofK which fixes F permutes c1, c2, ..., cn.We also noted that every element a of F (c1, c2, ..., cn) is a sum ofterms kc i11 c

i22 ...c

inn . where k ∈ F .

Since h fixes F , it does not change k hence h is completelydetermined once we know h (c1) , h (c2) , ..., h (cn).Thus, every automorphism of K fixing F is completelydetermined by a permutation of the roots of a (x).This is very important. It means that we may identify theautomorphisms of K which fix F with permutations of the roots ofa (x).We will now focus on the automorphisms of K which fix F . Inparticular, we want to know how many such automorphisms there are.The answer is given in the next theorem.


Galois Groups

TheoremLet K be the root field of some polynomial over F . The number ofautomorphisms of K fixing F is equal to the degree of K over F that is[K : F ].

Sketch of a proof:

Suppose that [K : F ] = n. We show that K has exactly nautomorphisms fixing F .

Explain why there exists a ∈ K , a /∈ F , such that K = F (a).Let p (x) be the minimum polynomial of a over F . What is deg p (x)?

If b is any other root of p (x), by a previous theorem, there is anautomorphism of K fixing F and sending a to b.

How many choices are there for b?

hence there are exactly n automorphisms of K that fix F .


Galois Groups

Example

How many automorphisms of Q(√2)fixing Q are there? Describe them.

ExampleHow many automorphisms of C fixing R are there? Describe them.

Example

How many automorphisms of Q(√2,√3)are there? Describe them.


Galois Groups

Example

How many automorphisms of Q(√2)fixing Q are there? Describe them.

We know that Q(√2)is the root field of p (x) = x2 − 2 which is also

the minimum polynomial of√2 over Q. Thus,

[Q(√2): Q]= 2 (we

already knew this), hence there are exactly two automorphisms ofQ(√2)fixing Q.

Since the roots of p (x) are ±√2, one automorphism sends

√2 to

√2

and the other one sends√2 to −

√2.

To describe them more completely, we need to know how they mapelements of Q

(√2).

Recall that the elements of Q(√2)are of the form a+ b

√2 with

a, b ∈ Q.Hence the first automorphism is the identity functiona+ b

√2→ a+ b

√2 and the second is the function

a+ b√2→ a− b

√2.


Galois Groups

ExampleHow many automorphisms of C fixing R are there? Describe them.

We know C = R (i) is the root field of p (x) = x2 + 1 which is alsothe minimum polynomial of i over R. Thus, [C : R] = 2.Following the same steps as above, we see that there are twoautomorphisms of C fixing R.They are the identity function as the function a+ bi → a− bi .


Galois Groups

Example

How many automorphisms of Q(√2,√3)are there? Describe them.

We know that[Q(√2,√3): Q]= 4. Hence, there are exactly 4

automorphisms of Q(√2,√3)fixing Q.

We also know that Q(√2,√3)is the root field of

p (x) =(x2 − 2

) (x2 − 3

). Since any automorphism of Q

(√2,√3)

fixing Q sends the roots of any polynomial to the roots of the samepolynomial, it must send the roots of

(x2 − 2

)to roots of

(x2 − 2

)and roots of

(x2 − 3

)to roots of

(x2 − 3

).

Thus, the 4 possibilities are

ε :

√2→

√2√

3→√3

α :

√2→ −

√2√

3→√3

β :

√2→

√2√

3→ −√3

γ :

√2→ −

√2√

3→ −√3


Galois Groups

Since every element of Q(√2,√3)is of the form a+ b

√2+ c

√3+ d

√6,

we see that

ε(a+ b

√2+ c

√3+ d

√6)

= a+ b√2+ c

√3+ d

√6

α(a+ b

√2+ c

√3+ d

√6)

= a− b√2+ c

√3− d

√6

β(a+ b

√2+ c

√3+ d

√6)

= a+ b√2− c

√3− d

√6

γ(a+ b

√2+ c

√3+ d

√6)

= a− b√2− c

√3+ d

√6


Galois Groups

TheoremLet K be an extension of F . Then, the automorphisms of K which fix Fform a group under composition.

Sketch of a proof:

For now, let G be the set of automorphisms of K which fix F (we’llchange its name shortly).

Suppose that [K : F ] = n. Clearly, G is a subset of Sn, the set ofpermutations on n elements.

It is enough to show that G is a subgroup of Sn.


Galois Groups

Though this result is fairly easy to prove, it is extremely important. Itallows us to use all our knowledge on groups to study the solutions ofpolynomial equations. That is what Galois theory is about.

DefinitionIf K is the root field of a polynomial a (x) in F [x ], the group of all theautomorphisms of K which fix F is called the Galois group of a (x). It isalso called the Galois group of K over F . It is designated by the symbol

Gal (K : F )


Galois Groups

Example

We saw above that Gal(Q(√2,√3): Q)= {ε, α, β, γ}. It is easy to

check that its operation table is

◦ ε α β γ

ε ε α β γ

α α ε γ β

β β γ ε α

γ γ β α ε

Note that since γ = αβ, we could have used αβ instead of γ as the fourthelement of this group. We did not need to introduce a new letter.


Galois Groups

Example

Let K be the root field of a (x) = (x − 1)2(x2 − 2

) (x2 + 2

)∈ Q [x ]. Find

Gal (K : Q), describe its elements, and find its operation table.


Galois Groups

Example

Let K be the root field of a (x) = (x − 1)2(x2 − 2

) (x2 + 2

)∈ Q [x ]. Find

Gal (K : Q), describe its elements, and find its operation table.

a (x) is a polynomial of degree 6, its roots are 1, 1,±√2,±i

√2.

It is clear that the root field of a (x) is K = Q(√2, i)and [K : Q] = 4

(why?). Hence, there are 4 distinct automorphisms of K that fix Q.Since these automorphisms must map the roots of any polynomial toroots of the same polynomial, the toots of x2 − 2 must be mapped tothe roots of x2 − 2 and the roots of x2 + 2 must be mapped to rootsof x2 + 2.


Galois Groups

Thus, the 4 possibilities are

ε :

√2→

√2

i√2→ i

√2

α :

√2→ −

√2

i√2→ i

√2

β :

√2→

√2

i√2→ −i

√2

γ :

√2→ −

√2

i√2→ −i

√2

Since a basis for Q(√2, i)over Q is

{1,√2, i , i

√2}, every element of

Q(√2, i)is of the form a+ b

√2+ ci + di

√2. We need to figure out

what each automorphism will map i to.


Galois Groups

ε (i) = ε

(12i√2√2)

= ε

(12

)ε(i√2)ε(√2)

=12i√2√2

= i

α (i) = α

(12i√2√2)

= α

(12

)α(i√2)α(√2)

=−12i√2√2

= −i

β (i) = β

(12i√2√2)

= β

(12

)β(i√2)β(√2)

=−12i√2√2

= −i

γ (i) = γ

(12i√2√2)

= γ

(12

)γ(i√2)γ(√2)

=12i√2√2

= iPhilippe B. Laval (KSU) Galois Theory: Main Results Current Semester 17 / 27

Galois Groups

Therefore

ε(a+ b

√2+ ci + di

√2)

= a+ b√2+ ci + di

√2

α(a+ b

√2+ ci + di

√2)

= a− b√2− ci + di

√2

β(a+ b

√2+ ci + di

√2)

= a+ b√2− ci − di

√2

γ(a+ b

√2+ ci + di

√2)

= a− b√2+ ci − di

√2


Galois Groups

The operation table is

◦ ε α β γ

ε ε α β γ

α α ε γ β

β β γ ε α

γ γ β α ε


Fixer and Fixfields

For the remaining of this chapter, F and K will remain fixed. F willbe an arbitrary field, K will be the root field of some polynomiala (x) ∈ F [x ].In our discussion, we will consider fields I such that F ⊆ I ⊆ K that isfields "between" F and K .

We will refer to these fields I as intermediate fields.Since K is the root field of a (x) over F , it is also the root field ofa (x) over I for each intermediate field I .

We will call G = Gal (K : F ).

DefinitionFor each intermediate field I , let I ∗ = Gal (K : I ). I ∗ is a subgroup of G .We call I ∗ the fixer of I .

Remark: I ∗ is the group of all automorphisms of K which fix I . It isobviously a subgroup of G .


Fixer and Fixfields

DefinitionFor each subgroup H of G , We consider the setH◦ = {a ∈ K : π (a) = a for every π ∈ H}. It is easy to see that this set isa subfield of K . We call it the fixfield of H.

Remark: The fixfield of H is a subfield of K . It also contains F sinceevery element of H ⊆ G fixes F . Hence, it is one of the intermediate fields.Before we state an important result, let us recapitulate what we have.

Every subgroup H of G fixes an intermediate field I , called thefixfield of H.Every intermediate field I is fixed by a subgroup H of G , called thefixer of I .


Fixer and Fixfields

The above suggests that there is a very tight relationship between thesubgroups of G and the intermediate fields between F and K . In fact,there is a one-to-one correspondence. This is the heart of Galois theory. Ingeneral, it is much easier to determine the subgroups of a group than thesubfields of a field. It is why Galois theory is so important. We give,without proof, an important result which captures this.

TheoremSuppose that F is a field of characteristic 0, K is the root field of somepolynomial a (x) ∈ F [x ], G = Gal (K : F ), I is an intermediate field, andH is a subgroup of G.

1 H is the fixer of I ⇐⇒ I is the fixfield of H.2 If I is the fixfield of H then |H| = [K : I ].3 If I is a root field over F then Gal (K : I ) is normal in Gal (K : F ) andGal (I : F ) ∼= Gal (K : F ) /Gal (K : I ).


Fixer and Fixfields

In the next chapter, we will relate this to a polynomial being solvable byradicals. We finish this chapter with an example.

Example

Consider Gal(Q(√2,√3): Q). Draw a lattice of the subgroups of G

between {ε} and G . Draw a similar lattice of the intermediate fieldsbetween Q and Q

(√2,√3). Find the fixfield of each subgroup of G and

find the correspondence between the subgroups of G and the intermediatefields between Q and Q

(√2,√3).


Fixer and Fixfields

Example

Consider Gal(Q(√2,√3): Q). Draw a lattice of the subgroups of G

between {ε} and G . Draw a similar lattice of the intermediate fieldsbetween Q and Q

(√2,√3). Find the fixfield of each subgroup of G and

find the correspondence between the subgroups of G and the intermediatefields between Q and Q

(√2,√3).

Recall that G = {ε, α, β, γ}. See previous slides for the definition ofthese symbols and the operation table of the group.

From Lagrange’s theorem, we know that G can have subgroups oforder 1, 2, 4. They are {ε}, {ε, α}, {ε, β}, {ε, γ}, {ε, α, β, γ} = G .The corresponding 5 intermediate fields areF = Q ⊆ Q

(√2),Q(√3),Q(√6)⊆ Q

(√2,√3)= K .


Fixer and Fixfields

For any subgroup H of G , let H◦ denote the fixfield of H. Recall thatH◦ = {a ∈ K : π (a) = a for every π ∈ H}

{ε}◦ = Q(√2,√3).

{ε, α}◦ = Q(√3).

{ε, β}◦ = Q(√2).

{ε, γ}◦ = Q(√6).

{ε, α, β, γ}◦ = Q.


Fixer and Fixfields

For any subgroup H of G , let H◦ denote the fixfield of H. Recall thatH◦ = {a ∈ K : π (a) = a for every π ∈ H}{ε}◦ = Q

(√2,√3).

{ε, α}◦ = Q(√3).

{ε, β}◦ = Q(√2).

{ε, γ}◦ = Q(√6).

{ε, α, β, γ}◦ = Q.


Fixer and Fixfields

Conversely, for any intermediate field I , the fixer of I isI ∗ = Gal (K : I )

Q∗ = {ε, α, β, γ} = G .Q(√2)∗= {ε, β}.

Q(√3)∗= {ε, α}.

Q(√6)∗= {ε, γ}.

Q(√2,√3)∗= {ε}.


Fixer and Fixfields


Q∗ = {ε, α, β, γ} = G .

Q(√2)∗= {ε, β}.

Q(√3)∗= {ε, α}.

Q(√6)∗= {ε, γ}.

Q(√2,√3)∗= {ε}.


Fixer and Fixfields


Q∗ = {ε, α, β, γ} = G .Q(√2)∗= {ε, β}.

Q(√3)∗= {ε, α}.

Q(√6)∗= {ε, γ}.

Q(√2,√3)∗= {ε}.


Exercises

1 Do the following problems at the end of chapter 32 of your book:B1-B6, D1-D7, G1.


galois theory: main results - kennesaw state...

Documents