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1.1 Field extensions: examples

We introduce the basic notions such as a field extension, algebraic element, minimal polynomial, finite extension, and study their very basic properties such as the multiplicativity of degree in towers.

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...)

Impartido por:

Ekaterina Amerik
Professor

Transcripción

My name is Ekaterina Amerik.

I am working at the mathematical department of Higher School of Economics,

and I'm going to read a course on Galois Theory.

This is a course about field extensions.

We assume familiar the basic notions of abstract algebra,

like groups, rings, fields,

modules, ideals, and their basic properties.

We also assume a certain knowledge of linear algebra.

All rings we consider will be commutative, associative, and with unity.

Let me give you the first definition.

K and L will be fields.

L is said to be an extension of K,

if L contains K as a subfield.

An equivalent definition is as follows:

L is an extension of K if L is a K-algebra.

Let me remind you, what is a K-algebra.

A K-algebra is a ring,

with a structure of a K-module

such that these two structures are compatible,

such that the multiplication

which goes from A times A to A is bilinear.

K-bilinear, of course.

Well, if you wish, you can reformulate it, saying

something like

k_1 a_1 times k_2 a_2

is k_1 k_2 acting on a_1 a_2,

where k_i are elements of K,

and a_i are elements of A.

Why is a field containing K as a subfield the same thing as a K-algebra?

Well, in fact, given a K-algebra structure

on A is the same as giving a homomorphism of rings

from K to A.

Well indeed, if I have a K-algebra,

then I can define a homomorphism by sending k

to k as it acts on 1.

And conversely, if I have a homomorphism of rings

from K to A, I can define a K-algebra structure by setting

ka equal to f(k) times a.

So now, if A is a field L,

then any homomorphism from K to L is injective.

Well, there are several ways to see this.

One of them is saying that f(k) is always invertible.

Indeed f(kk^(-1))

is the same as f(k) times f(k^(-1))

this is f(1)

which is 1.

So we see that f(k) is invertible, so it cannot be 0.

If, of course,

k is not 0 itself.

So f(k) is never 0 when k is nonzero itself.

In particular, f is injective.

Or a more sophisticated way to say the same thing, is to say that

the kernel of f is always an ideal.

So alternatively,

the kernel of f is always an ideal.

And a field does not have nontrivial ideals,

the only ideals are 0 and K itself.

If you don't know this, you are strongly encouraged to do it as an exercise.

Now let me give some examples.

Of course, C is an extension of R.

And R is an extension of Q.

Second example.

Any field has what is called a characteristics,

so if L is a field, there are two possibilities:

either, if you take the unit element and

start adding it to itself, you never obtain 0.

In this case, we say that the characteristics of K is 0.

And we see that in this case, of course, L contains Z.

Well, but L is a field

so if it contains Z, it must contain Q.

So L is an extension of Q.

Or, if you take the unit element and start adding it to itself,

you obtain 0 at certain point.

Then it is easy to see that the first time it happens,

it happens for some prime number m,

so minimal m with this property

is prime.

And then one says that this

prime is the characteristic of this field.

Then L does not contain Z anymore.

What it contains is Z/pZ.

For p - prime, Z/pZ is a field.

And when we want to emphasize that we consider Z/pZ

with its field structure, we denote it by F_p.

denoted by F_p.

And our L is an extension of F_p.

So we have two possibilities:

if the characteristic of L is 0, then L is an extension of Q.

If it is p, then it is an extension of F_p.

One calls Q and F_p the prime fields.

Q and F_p are prime fields.

Any field is an extension of one of those,

and they don't contain any proper subfields.

is an extension of a prime field.

Third example is very important.

So let's take the ring of polynomials in one variable over K.

And let's consider the quotient ring

by an ideal generated by an irreducible polynomial P.

Then I affirm that this is a field.

There are two ways to see these.

Again, in the elementary language, you can see the following thing.

So if Q is a polynomial which is not

a multiple of P, then Q is prime to P.

The greatest common divisor of Q and P is 1.

And then you have the Bézout identity.

There exist A and B

polynomials, such that

AP + BQ = 1,

so what you see is that

BQ = 1 modulo P,

and therefore

B is an inverse of Q in K[x] modulo P.

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