8.1. Extensions solvable by radicals. Solvable groups. Example.

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來自 National Research University Higher School of Economics 的課程

Introduction to Galois Theory

98 個評分

National Research University Higher School of Economics

Introduction to Galois Theory

98 個評分

A very beautiful classical theory on field extensions of a certain type (Galois extensions) initiated by Galois in the 19th century. Explains, in particular, why it is not possible to solve an equation of degree 5 or more in the same way as we solve quadratic or cubic equations. You will learn to compute Galois groups and (before that) study the properties of various field extensions. We first shall survey the basic notions and properties of field extensions: algebraic, transcendental, finite field extensions, degree of an extension, algebraic closure, decomposition field of a polynomial. Then we shall do a bit of commutative algebra (finite algebras over a field, base change via tensor product) and apply this to study the notion of separability in some detail. After that we shall discuss Galois extensions and Galois correspondence and give many examples (cyclotomic extensions, finite fields, Kummer extensions, Artin-Schreier extensions, etc.). We shall address the question of solvability of equations by radicals (Abel theorem). We shall also try to explain the relation to representations and to topological coverings. Finally, we shall briefly discuss extensions of rings (integral elemets, norms, traces, etc.) and explain how to use the reduction modulo primes to compute Galois groups. PREREQUISITES A first course in general algebra — groups, rings, fields, modules, ideals. Some knowledge of commutative algebra (prime and maximal ideals — first few pages of any book in commutative algebra) is welcome. For exercises we also shall need some elementary facts about groups and their actions on sets, groups of permutations and, marginally, the statement of Sylow's theorems. ASSESSMENTS A weekly test and two more serious exams in the middle and in the end of the course. For the final result, tests count approximately 30%, first (shorter) exam 30%, final exam 40%. There will be two non-graded exercise lists (in replacement of the non-existent exercise classes...)

從本節課中

Week 8

We finally arrive to the source of Galois theory, the question which motivated Galois himself: which equation are solvable by radicals and which are not? We explain Galois' result: an equation is solvable by radicals if and only if its Galois group is solvable in the sense of group theory. In particular we see that the "general" equation of degree at least 5 is not solvable by radicals. We briefly discuss the relations to representation theory and to topological coverings.

8.1. Extensions solvable by radicals. Solvable groups. Example.15:07

8.2. Properties of solvable groups. Symmetric group.13:20

8.3.Galois theorem on solvability by radicals.11:33

8.4.Examples of equations not solvable by radicals."General equation".13:59

8.5. Galois action as a representation. Normal base theorem.14:35

8.6. Normal base theorem (cont'd). Relation with coverings.12:13

與講師見面

Ekaterina Amerik
Professor
Department of Mathematics

[SOUND] Our next subject

is the solvability

of equations by radicals.

So we consider K of characteristics 0 and

embed it into some algebraic closure, K bar.

A definition, a finite extension.

E of K is solvable by radicals.

If there exists the elements alpha 1 and

so on are generating.

E and such that alpha

i to the power ni belongs

to K of alpha 1 and so

on, alpha i- 1 for

some natural numbers and i.

In other words E is

obtained by adjourning

nis roots of some ais.

For example one can take q of the cubic root

of 2 + 3 of square root of 7,

something like this.

But one can also take several expressions of this form.

Cubic root of 2 + 3 square root of 7.

And then the 5th root of 4

+ 5 square root of 11, okay.

Then we will have the square root of 7 and

the square root of 11 as alpha 1 and alpha 2.

And then alpha 3 will be under this cubic root.

And alpha 4 will be the 5th root.

So another definition.

Let P be a polynomial over K.

We say P is solvable by radicals.

If this [INAUDIBLE] field of P is contained in some extension.

Solvable radicals if there exists an E solvable by radicals.

And containing all roots of P.

Well maybe it's better to say in this case that the equation be equal

to 0 is solvable by radicals.

So more precisely, It would say that the equation,

P equal to 0 is solvable, by radicals.

But to make the definitions shorter,

we shall say simply that the polynomial P is solvable by radicals.

So let me list a few elementary properties.

Clearly a composite extension of solvable extensions is solvable.

Secondly, if we have,

L an extension of K solvable by radicals.

By definition, such an extension is finite.

Then there exists a Galois.

Finite Galois extension.

Containing.

Well, let's give it a name.

E containing L, and also solvable by radicals.

This is something we have already seen.

It suffices to check for E.

The composite extension of all images of L in K bar.

Indeed, take

the composite of

all images of L in K bar.

Now, you can also view it as the images

of l by the Galois group of K bar over K.

Or this is the same

as images of L by

the Galois group

of K bar over K.

Well, I should better use the plural of

those are the images of L and K bar are the same.

As the images of L by the Gal group of K bar over K.

Let me now talk about solvable groups.

This shall be a brief reminder since this is not a course on group theory,

you are supposed to know some group theory already.

So I somehow I presume that you are familiar with this definition but

I will recall the definition of basic properties.

So let G be a group.

Which is called solvable,

If it has a filtration.

A filtration is the following thing.

We have G = to G0.

Contains G one contains.

Gr- 1 contains Gr, which is trivial.

So a Gir subgroups and

we shall require that those

are normal subgroups.

Gi is a normal subgroup.

Of Gi- 1.

So we can form a quotient.

And we require the quotient to be abelian.

So Gi- 1 / Gi is abelian.

So for example, the group of permutations on three elements is solvable.

Since S3 contains A3,

contains the trivial group.

[COUGH] A3 is a normal sub group.

The quotient A3 by A3 has only

two elements in particular it

must be cyclic of order 2.

And A3 is cyclical for order of 3.

S3 itself is not abelian, but it is solvable.

It has a filtration which is just written here with a abelian factors.

Abelian quotients, okay?

S4 is also solvable.

Indeed S4 contains A4.

Even permutations on four elements contains a group which I shall

denote by K, contains e.

I will explain what K is.

Here K is the subgroup of double transpositions.

So it consists of, Identity,

then (12)(34),

(13)(24), and

(14)(23).

So subgroup, Of double transpositions.

A double transposition is

a product of two transpositions

with distinct support, right,

which permute the distinct elements, okay.

So now, let me rewrite this sequence.

I have S4 contains A4 contains K contains e.

So A4, the subgroup of even permutations

is a normal subgroup of, of S4.

The quotient is

cyclical for the two.

Then K is a normal sub group in A4.

Of course, a conjugate of a double transposition

is always a double transposition and the quotient,

Is of order 3, so it is also cyclic.

K is not cyclic, but it is abelian.

K is abelian.

Isomorphic to

The product of two cyclic groups of order 2.

So it follows that S4 is solvable.

[MUSIC]

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