3.3 Multiplicative group and automorphism group of a finite field.

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Do curso por National Research University Higher School of Economics

Introduction to Galois Theory

89 ratings

National Research University Higher School of Economics

Introduction to Galois Theory

89 ratings

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

Our next goal is a theorem you probably already know.

The theorem saying that the multiplicative group of a finite field is cyclic.

To make it instructive I will prove it in a slightly more general version.

So let K be a field.

And G, a finite subgroup

of the multiplicative group of K, then G cyclic.

And let us prove this.

The idea is to compare G and

the cyclic group of order N

where N is the cardinality of G.

So, let me denote psi(d) the number of elements

of order d in G.

So we need to prove that

psi(N) is not 0, and

we know that

N is equal to the sum of those psi(d),

and let me denote phi(d)

the number of elements

of order d in Z/NZ.

But Z/NZ is a cyclic group,

and it contains a single

subgroup of order d, which is itself cyclic. So, as

subgroup of order d, which is itself cyclic. So, as

Z/NZ contains

a single (cyclic) subgroup

of order d

for each d dividing N.

Namely this is the subgroup generated by N over d, so

Namely this is the subgroup generated by N over d, so

Namely this is the subgroup generated by N over d, so

the one generated by N over d.

phi(d) is also

equal to the number of generators

of Z/dZ.

This is well known to be the number of numbers

This is well known to be the number of numbers

between 1 and

d - 1,

which are prime to d.

So we know,

of course that phi(n)

is nonzero, so I claim

that either

psi(d) is 0, or

psi(d) is equal to phi(d).

This is of course sufficient

since the sum of the phis and of the psis

must be equal.

So, proof of the claim.

If there is no element of order d,

then of course psi(d) is 0.

If there is no

element of order d in G,

then psi(d) is zero.

If there is one element of order d in G

then this x is a root

of the polynomial x^d -1.

of the polynomial x^d -1.

And if you look at all the rules of such a polynomial

you see that they form a group

of this polynomial

form a cyclic subgroup of G.

So G as well as Z/NZ has

So G as well as Z/NZ has

a single subgroup

of order d, or cyclic

(which is cyclic)

Or no such subgroup at all.

Or no such subgroup at all.

So this is a comparison between G and Z/NZ,

but the difference is of course that Z/NZ

has such a subgroup for any d dividing N.

And G not necessarily

a priori. Okay, so

this implies that if, this in fact implies the claim.

If phi(d) is nonzero,

then there is such a subgroup.

And the number of elements of order d,

phi(d), is the number of generators of that subgroup.

Well, obviously this is the same thing as phi(d).

So whenever psi is nonzero it is equal to phi, okay?

In particular psi(d) is always less or equal then phi(d),

but in fact there must be equality, because the sum of those are equal.

So the sum of psis is equal to the sum of phis,

and then the psis all must be equal to phis.

In particular, psi(n) is nonzero.

Well, and we have proved the theorem.

There is an element of order N in G, so G is cyclic.

Okay, this has a couple of corollaries

about field extensions.

Corollary 1 is that if K

is an extension of F_p of degree N,

so K is F_(p^N), then is generated by one element.

There exists an alpha such that K is equal to F_p(alpha).

Well, you might

object and say this has already been shown.

We have seen that F_(p^N) is a

stem field of any irreducible polynomial of degree N,

but this corollary is actually stronger because when we've been discussing

those stem fields we did not say that such polynomials existed.

So, in particular, there exists

an irreducible polynomial of degree

an irreducible polynomial of degree

N over F_p, okay?

This is the thing which we don't know yet,

but the proof is of course very easy, since

we know that the multiplicative group is cyclic

it suffices to take a generator of this group.

So for the proof it suffices to take for alpha

a generator of K*, okay.

Corollary 2, the group of automorphisms

Corollary 2, the group of automorphisms

of K, well, F_(p^n) over F_p

of K, well, F_(p^n) over F_p

is cyclic generated by the Frobenius map.

where Frobenius map F takes x to x^p.

As we have already remarked this is a field automorphism.

Well, let me prove this.

Of course, X^(p^n) is X for any x in

Of course, X^(p^n) is X for any x in

F_(p^n), this we have already seen several times.

So, Frobenius to the power n is the identity.

On the other hand the order of this Frobenius is exactly n.

Since if m is strictly less than n,

Since if m is strictly less than n,

then F^m is not the identity because

x^(p^m) - x = 0

has only p^m roots.

And p^m is strictly less than p^n.

So it cannot take all elements of our field to itself.

Okay. Well, and

finally F_(p^n) is F_p of alpha

finally F_(p^n) is F_p of alpha

where alpha is a root of irreducible

polynomial of degree n,

so it cannot have more than n automorphisms.

Well, this alpha goes to another root of P

Well, this alpha goes to another root of P

Well, this alpha goes to another root of P

under an automorphism.

So there can not be more than n of them.

So the cardinality of this group is at most n.

So the cardinality of this group is at most n.

Well, then it is n and the group

is cyclic generated by F.

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