3.5. Separable degree, separable extensions.

Loading...

Do curso por National Research University Higher School of Economics

Introduction to Galois Theory

98 classificações

National Research University Higher School of Economics

Introduction to Galois Theory

98 classificações

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...)

Na lição

Week 3

We recall the construction and basic properties of finite fields. We prove that the multiplicative group of a finite field is cyclic, and that the automorphism group of a finite field is cyclic generated by the Frobenius map. We introduce the notions of separable (resp. purely inseparable) elements, extensions, degree. We briefly discuss perfect fields. This week, the first ungraded assignment (in order to practice the subject a little bit) is given.

3.1 An example (of extension). Finite fields.12:37

3.2 Properties of finite fields.14:03

3.3 Multiplicative group and automorphism group of a finite field.15:57

3.4 Separable elements.15:07

3.5. Separable degree, separable extensions.15:07

3.6 Perfect fields.9:06

Conheça os instrutores

Ekaterina Amerik
Professor
Department of Mathematics

Let me generalize this property to the case of

fields which are not necessarily given as K(alpha).

So now let me take an arbitrary finite extension

of K.

And let me define the separable degree

a separable degree as a number of possible

homomorphisms of L into algebraic closure of K.

Again, if L is generated

by one element alpha, then this degree is just the number

of distinct roots of its minimal polynomial,

if L is K(alpha) then this is just

the number of distant root

of it's minimal polynomial.

And let me say that

the extension L

is separable over K if

the separable degree is equal to the degree.

the separable degree is equal to the degree.

And I can also define the degree of inseparability

of L over K, as the quotient

the degree divided by L over K separable.

But this won't be much very important on the sequel.

So let me formulate a theorem.

First part sais that the separable degree is multiplicative.

If I have L an extension of K, and M an extension of L,

then M over K separable is equal to

M over L separable times L over K separable.

And M is separable over K if and only if M is separable

over L and L is separable over K.

So this is part one, part two says that

the following things are equivalent:

1) L is separable over K

2) any element of L is separable over K

3) L is generated over K by a finite number of separable elements.

3) L is generated over K by a finite number of separable elements.

And 4) L is generated

over K by a finite number of elements

and each alpha is separable over

the preceding stuff K (alpha_1, ..., alpha_(i-1) ).

the preceding stuff K (alpha_1, ..., alpha_(i-1) ).

Remark: the same holds

when we replace

separability by pure inseparability.

Okay, so let me give a proof of this theorem.

Well, part one.

We know that any phi from L to K bar

extends to phi tilde from M to K bar,

this is the extension theorem.

And in fact, there are exactly

M over L separable ways to do this.

Since K bar is also L bar,

since given phi one

considers K bar as L bar.

An algebraic closure of K is also an algebraic

closure of L once an embedding of L into K bar is given.

Then the number of ways to extend is computed just by definition,

by definition it's the separable degree of M over L.

This implies the first part,

thus we have M over K separable

thus we have M over K separable

equal to a product of L over K separable M over L separable.

Now the equivalence

of separability of M to the separability

of M over L and of L over K, so

equivalence of separability.

This is just the inequality, in fact.

Just the fact that

Just the fact that

the separable degree of an extension

over K does not exceed the true degree, okay?

So if the separable degree and the degree are both multiplicative,

and if one of them does not exceed another than we can conclude everything.

And this fact, now this is proved the last fact

is proved by induction

using the fact of that this is true

for E generated by one element.

Well, part two, 1) implies 2).

So 1) is L is separable over K and 2) is

any element of L is separable over K.

This is the consequence of the first part,

the first part implies that

any subextension of a separable extension is separable.

K(alpha) over separable extension L

is itself separable.

2) implies 3).

So if any element is separable over K,

then L is generated by separable elements. It's clear.

If any element is separable then of course, the generators

are also separable. 3) implies 4).

Well, if ai is separable over K then a_i is also separable over an extension of K.

This is clear because the minimal polynomial

of a_i over K of a_1, ..., a_(i-1) divides

of a_i over K of a_1, ..., a_(i-1) divides

the minimal polynomial of a_i over K

and we have seen these in the last lecture.

And so if the big one is separable,

that is to say has distinct roots so then the small one is also separable.

If P_min of a_i over K is separable,

which means that it has distinct roots,

then so is it's divisor,

P_min of a_i over

K(a_1, ..., a_(i-1) )

And finally 4) implies 1), well, this can be proved by

induction as above, I shall not give more details.

So now one might ask, is the notion

of separability defined for extensions which are not necessarily finite?

Yes, in this case

it is best to define a separable extension as

such extension that all its elements are separable.

In particular,

if L over K is not necessarily finite

and algebraic extension, we can define L separable,

the separable closure of K in L,

as a set of all x such that x is separable over K.

A separable element is by definition algebraic,

of course it has minimal polynomial.

So the preceding theorem implies that this L separable is a subfield,

L separable is a subextension

called separable closure of K in L.

called separable closure of K in L.

Well, I think normally one presumes that L is algebraic

over K and of course,

L is purely inseparable,

over L separable.

Well and finally maybe I will repeat

the important remark: if

the characteristic of K is 0, then any extension

is separable.

And if characteristic of K is p, then

a purely inseparable

extension has degree

p^r, and always

this degree of inseparability is p^r.

Explore nosso catálogo

Registre-se gratuitamente e obtenha recomendações, atualizações e ofertas personalizadas.

O Coursera proporciona acesso universal à melhor educação do mundo fazendo parcerias com as melhores universidades e organizações para oferecer cursos on-line.

© 2019 Coursera Inc. Todos os direitos reservados.

Baixar na App Store

Baixar no Google Play