3.2 Properties of finite fields.

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Do curso por National Research University Higher School of Economics

Introduction to Galois Theory

89 classificações

National Research University Higher School of Economics

Introduction to Galois Theory

89 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 now prove this Theorem 1.

So recall that we have seen that the map F_(p^n)

which takes x to x^(p^n)

is a homomorphism.

Then it follows easily that the set of x

such that F_(p^n) of x is equal to x is a subfield

such that F_(p^n) of x is equal to x is a subfield.

And it contains of course F_p

And it contains of course F_p

But these are exactly roots of x^(p^n) - x.

Let me call this polynomial in what follows Qn of x.

Let me call this polynomial in what follows Qn of x.

Of course this subfield is exactly the splitting field of Qn,

Of course this subfield is exactly the splitting field of Qn.

And since Qn does not have multiple roots

this can be seen for instance by

verifying that Qn is relatively prime with its derivative.

verifying that Qn is relatively prime with its derivative.

Indeed, this derivative is just equal to 1.

So they're exactly p^n of the roots.

So we have the subfield of p^n elements and

this is a splitting field of Qn

of Qn, is exactly the field of roots of Qn and

this is a field of p^n elements.

So conversely, we want to show that any

field of p^n elements is a splitting field of Qn.

Let the cardinality of K be p^n,

and alpha an element of K.

Then alpha^(p^n - 1)

is equal to 1 provided

that alpha is nonzero.

Indeed, the multiplicative group of K

has cardinality p^n - 1.

So any element of K*

to the power of (p^n - 1) must be

equal to the unit element.

So alpha is a root

of x^(p^n) - x

and 0 is of course also a root.

So, K consist of roots

of Qn,

and this also answers

the question why there is a unique subextension of F_p bar,

which has p^n elements.

This is just because these subextension

consists exactly of the roots of Qn.

The unicity of the

subextension, of the embedding,

well, not exactly of the embedding, but

of the image of the embedding of K into F_p bar also follows.

Let me now formulate and prove a few other properties of finite field.

These are very much in the spirit of Theorem 1,

so let me probably call them Theorem 1.2,

1.3, and so on.

So Theorem 1.2 claims

that F_(p^n) contains

F_(p^d) if and only if

d divides n.

One direction is the multiplicativity of degrees in towers.

Indeed, in this case we have that the degree of

F_(p^n) over F_p

is equal to the degree of F_(p^n)

over F_(p^d) times

the degree of F_(p^d) over F_p.

So this is n, and this is d.

So we have d divides n.

Conversely, suppose that d divides n,

then if x^(p^d) is equal to x,

then if x^(p^d) is equal to x,

the same is true for x^(p^n),

since taking x to the (p to the power n)'th power

is just iterating taking x to the power (p^d).

is just iterating taking x to the power (p^d).

So F_(p^d) is

So F_(p^d) is

contained in F_(p^n).

So these are very small theorems.

Theorem 1.3, it's better maybe to consider all of them as parts of the Theorem 1.

So theorem 1.3 says that F_(p^n)

is a stem field

and a splitting field

of any irreducible

polynomial P from

F_p[x] of degree n.

So the part about stem field is clear,

so indeed a stem field of P has degree n over F_p.

So this is F_(p^n).

Then the part about the splitting field can be shown

for instant as follows: let

alpha be a root of P.

So alpha is an element of F_(p^n),

then Qn of alpha is 0 since Qn has as roots

exactly all elements of F_(p^n).

So P divides

Qn and so

P splits in F_(p^n).

P splits in F_(p^n).

This has a simple corollary, so

Qn is the product over all d dividing n

of all P, where P is

irreducible monic of degree d.

Now here's the proof.

Indeed, we have already seen

that all such P divide Qn.

Since, the stem field is

Since, the stem field is

F_(p^d) and this is contained in F_(p^n),

so Qn of alpha is 0 if alpha is a root.

So they are relatively prime of course, and their product divides Qn.

Also, Qn has no multiple roots, so there are no multiple factors either

and then what remains to prove is that

there are no other irreducible factors of Qn.

So let R be

an irreducible

factor of Qn.

Then if alpha is a root of R

Qn of alpha is

0, of course.

So this means that F_p of alpha

is contained in F_(p^n),

and this means that F_p of alpha

is F_(p^d), where d divides n.

And so this means that the degree of R divides n.

So there are no other irreducible factors.

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