7.2. Kummer extensions.

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来自 National Research University Higher School of Economics 的课程

Introduction to Galois Theory

90 个评分

National Research University Higher School of Economics

Introduction to Galois Theory

90 个评分

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...)

从本节课中

Week 7

We continue to study the examples: cyclotomic extensions (roots of unity), cyclic extensions (Kummer and Artin-Schreier extensions). We introduce the notion of the composite extension and make remarks on its Galois group (when it is Galois), in the case when the composed extensions are in some sense independent and one or both of them is Galois. The notion of independence is also given a precise sense ("linearly disjoint extensions"). This week, an ungraded assignment is given.

7.1 Cyclotomic extensions (cont'd). Examples over Q.12:18

7.2. Kummer extensions.14:28

7.3. Artin-Schreier extensions.11:14

7.4. Composite extensions. Properties.13:02

7.5. Linearly disjoint extensions. Examples.15:14

7.6. Linearly disjoint extensions in the Galois case.12:41

7.7 On the Galois group of the composite.7:39

与讲师见面

Ekaterina Amerik
Professor
Department of Mathematics

[SOUND] The next example I

would like to give,

is that of Kummer extensions.

So consider K, a field such that

the characteristics of K is prime to

a certain number n, and such that.

X to the power n- 1 is split in K.

So K contains all ns roots of unit and

we suppose that n is prime to characteristics.

So we consider a and

k, so let alpha

the nth root of a.

Well an nth root of a of course, a root

of x to the power n- a.

The same in some algebraic equations, it does not have to lie in.

And take d which is minimal i aside

from that a to the power of i is k.

Preposition d divides n,

the minimal polynomial

Of alpha is X to the power d

minus alpha to the power of d,

and K of alpha is Galois

with cyclic Galois

group of order d.

Over k, of course.

So let's prove this proposition.

It's clear that K of alpha is Galois

because all the roots of unity,

all nth roots of unity are in K.

So K of alpha contains.

All roots of X to the power n minus a.

So this is.

A split in field of X to

the power n minus a.

So this is normal and also separable, because this is a separable polynomial.

N is prime to the characteristics.

So normal.

And separable.

Since n is.

Prime to the characteristics.

Okay, so this is Galois.

Now, let's define a homomorphism

from the Galois group to

the group of nth roots of unity.

We will send g to g of alpha over alpha.

This is correct because g sends alpha to another root of the same polynomial.

So g sends.

Alpha to a root of X power n minus a,

so the quotient is a route of unity.

Okay, so this is well defined of course, this is injected.

So let's give it a name,

should we call it just f?

But f is injective.

Since g of alpha determines g.

Alpha generates our field.

So g of alpha determines g, and then what is it's image?

In some sub group of un, un is cyclic a sub group of a cyclic group is cyclic.

Both un cyclic, let delta be it's order.

Of the image.

So we want to show.

This is the only thing which remains to show, that delta is d.

Okay, so let's write g of

alpha to the power of delta.

This is f of g to the power delta times alpha to the power delta.

F of g is a number, it's a root of unity.

And it's exactly the root of unity by which the action of g multiplies alpha.

And this must be equal to alpha to the power of delta, because

delta is exactly the order of f of the Galois group.

So f of g to the power of delta is that entity.

Then it follows that alpha to the power of delta isn't K and.

Alpha to the power i is not in K form for

i less than delta.

Since otherwise.

[COUGH] The minimal polynomial of alpha will be of degree.

So inferior to delta, but this is impossible, but it was.

The order of the Galois group so

the degree of the feud extension

Is exactly delta.

[COUGH] So this proves the proposition delta is equal to d and

the minimal polynomial of alpha is

x to the power d minus alpha to the power d.

So proposition two and

conversely.

Any cyclic extension.

Of degree n.

Prime to the characteristics of k.

Is obtained in this way.

Generated by an nth root.

Generated by.

The nth root of a for sum a and k.

Or let us prove this proposition.

So take consider L an extension of K such that the Galois group

is cyclic generated by sigma, then we have sigma to the power n is identity.

Then by linear algebra.

I shall not prove it, but you should know it from the linear algebra course,

such a thing is diagonalisable.

Sigma is diagonalisable.

Now, let us show that all eigenspaces have dimension 1.

All eigenspaces.

Have dimension 1.

Indeed if you have x and

y in the same, eigenspace

then sigma of x over y.

Is x over y, because x and y are multiplied by the same number.

So x over y is in K and this is exactly

what means that the dimension of

this eigenspace is equal to 1.

So x and y proportional.

Over K.

So it follows that all roots of unity are eigenvalues.

Of sigma.

Now it suffices to take an eigenvector

take alpha such that sigma alpha is zeta alpha for

zeta some primitive root of unity.

Then the orbit of alpha.

Has.

N elements.

And also it follows.

That the degree of

k of alpha over k is n and

alpha to the power n is nK.

Since sigma of alpha to the power of n is

zeta n times alpha to the power of n.

So this is just alpha to the power of n,

and we see that alpha is a root

of X to the power of n minus

alpha to the power of n.

That well, it's better to write some here.

And this is irreducible by degree reasons.

So this is what we're going to prove.

[SOUND]

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