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

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Del curso dictado por National Research University Higher School of Economics

Introduction to Galois Theory

90 calificaciones

National Research University Higher School of Economics

Introduction to Galois Theory

90 calificaciones

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...)

De la lección

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

Conoce a los instructores

Ekaterina Amerik
Professor
Department of Mathematics

[MUSIC]

Last time, we discussed synchrotronic extensions,

that is the extensions generated by the nth root of unity.

So cyclotonic.

Extensions.

These were split in fields.

Of cyclotomic polynomial.

So this was generated

by a nth root of unit.

And, of course, it is sufficient to consider only primitive roots,

which are exactly the roots of this Phi-n.

And we got a very precise description of those

extensions in the case when phi N was irreducible.

For instance, over Q.

So we have seen

that Q1 of zeta n

where zeta n is e to the power two pi i over n.

So this is a primitive nth root of unity.

Was Galois.

Of Galois group.

Z over nZ star.

So it's the multiplicative group over numbers module n,

and so it adds, right?

Zeta n goes to zeta n to the power a where

a is an element of this.

Zed over n zed there, that is,

we can view it as an integer relatively prime to n.

Okay, so let us consider a few of, more concrete examples.

Let's take n equal to 8,

then, of course, zed over 8,

zed star, has four elements.

Namely, this is,

these are the classes of one of

course three, five and seven.

So our Galois group also has four elements.

The identity.

And then the thing which sends zeta eight to zeta eight to the over three.

Then it can send zeta eight to zeta eight to the power five.

And then it can send it to the power of seven.

So let us denote them.

Sigma three, sigma five and sigma seven under

notice of that sigma seven is something very easy.

This is just complex conjugation.

Sigma 7 sends to these zeta 8 to zeta 8 bar.

So this is complex conjugation.

And it's fixed field is of course the intersection of

our cyclotomic field with r.

So I'll have Q of zeta 8 fixed by sigma 7,

this is just Q(zeta 8) intersected with R,

this is the real part.

And this is generated over Q by zeta

8 + zeta 8 bar which is the square root of two.

And of course, it has degree two over Q,

so that's quadratic extension

Of Q.

And there other two quadratic extensions.

Our Galois group has three sub-groups of order two.

That is to say the one generated by sigma three,

the one generated by sigma five, and the one generated by sigma seven.

Our Galois group has three subgroups of order two.

So we have three quadratic.

Sub-extensions.

So one of them we have already found.

This is Q of square root of two.

So let's find the two others.

If we have, consider the fixed field

of sigma 3, then what could it be in?

Sigma 3 sends zeta 8 to zeta 8 to the power 3.

And it is easy to see that the fixed

field here can be seen, for instance, as Q over their sum.

Zeta eight plus zeta eight to the power of three, right?

And the sum is i square root of two, all right.

Let me draw the circle.

So here I have this zeta eight and here I have a zeta eight Q.

So the sum will be some where

on the imaginary axis if you look at the precise formula.

This will be i square root of two.

Okay, now so this is Q one of i square root

of two, and finally, Q of zeta eight

fixed field with respect to sigma 5.

These can be viewed as a, well,

we cannot take now, Zeta plus zeta to

the power 5 because this is going to be 0.

Okay?

But we can take the product, of course, of

zeta 8 times zeta 8, with a power

of 5, so it's Q of zeta 8 to the power of 6,

and this is just Q(i), this is minus i,

so this is Q(i), okay?

So here we described the three

subextensions, okay?

Let's consider another example,

let's consider the fifth cyclotomic extension.

Let's take now the fifth root of unit here.

Then the Galois group is

zed over five zed star, so

this is cyclic of order four.

And it is generated by the transformation

of these squares zeta five,

for instance.

And it has only one subgroup.

Only one proper subgroup of where there are two of those,

so the Galois group has only one proper subgroup,

which is cyclical for which is cyclical for.

And so our field has only one proper subfield different from Q of course.

It has only one proper subfield.

Different from Q.

And this is going to be the real part.

All of the complex conjugation.

Which is of course an element of Galois group.

Now this is the same as the real part, and

if we look, we would generate,

we see that this is the same as Q of sigma 5 plus sigma 5 bar,

and this is going to be the same thing as Q of,

coincidence of 2pi over 5.

Okay?

So these were the examples of cyclotomic extensions of Q and

of course the picture is exactly the same as long as

the cyclotomic polynomial is irreducible.

If it is not reducible, which can happen as we have seen,

the Galois group becomes smaller.

[MUSIC]

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