Loading...

7.5. Linearly disjoint extensions. Examples.

Course video 41 of 55

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.

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

[MUSIC]

So, these were the genuine properties of composite extensions.

Now, let me give you a definition of linearly disjoint extensions.

Maybe it's best to state it as a definition,

which is a theorem at the same time.

Definition, theorem,

the following properties are equivalent.

A, The tensor product is a field.

Of course, equivalent for algebraic extensions.

Secondly, j is injective.

C, if we have x1 and so on, xn, in l1,

which are linearly independent over k,

then they are also linearly independent over l2.

D, you have two families,

x1 and so on xn and l1 and

y1 and so on, ym.

In l2, if the two families

are linearly independent over k,

then the products xi yj

are linearly independent over k.

And in the case of finite extensions,

When L1 is finite over k,

this is also equivalent,

all this is equivalent to

the fact that L1, L2,

degree over L2 is equal to

the degree of L1 over k.

And so on.

The degree of L1, L2 over k is just the product of the degree of L1 and L2 over k.

In other words.

L1, L2 degree over k,

should be just for the product.

Okay, in this case,

We say that L1 and L2 are linearly disjoint.

Linearly disjoint over k,

so let us prove the theorem.

A equivalent to B is clear.

Since we have L1, L2 which is just the image of j.

And so, a and b both say that

L1 turns L2 over k is equal to L1, L2.

Then b implies c.

We have x1 times 1, and so on.

Xn times 1.

These are linearly independent over L2.

By the base change property.

Now, if j is injective.

Their images, X1 and

so on, xn are also linearly independent over L2.

An injective map sense,

a linearly independent family into a linearly independent family.

C implies d.

Now, this is very easy.

If we have some relation.

Aij, xi, yj over k.

Then, we have the sum

of aij yj over j,

which should be equal to 0.

Since the xi are linearly independent over k.

But now, aij also must be 0 since

yi are linearly independent.

Yj are linearly independent.

And now, finally, we have to show that d implies b.

Remember that b was the injectivity of j.

So, let's take that in the tensor product,

and suppose that j of z is 0.

Remember, that z by definition is the sum

of aij, the xy tensor yj, okay?

And j of z is just

the sum of aij xi yj.

So, if j of z is 0 then all aij are 0.

And so, z is 0.

[COUGH] So, this theorem is proved,

well the part about finite extension and

degrees is pretty obvious.

Now, the part involving degrees.

Easy to follow from our four properties.

So, let me give you a few examples.

First of all, the extensions which have relatively

prime degrees are always linearly disjoint.

If L1 over k is of degree m, and

L2 over k is of degree n,

then obviously, well, m,

n are relatively prime, then obviously.

L1, L2 are linearly, Disjoint.

Indeed, m and n must divide the degree of their composite.

And the degree of the composite is, At most mn.

So, this degree is equal to mn.

And this is exactly one of the definitions of being

linearly disjoint that you multiply the degrees.

Okay, so in particular.

If you have say Q of the fifth root of 2 and

Q of the fifth root of unity,

they are linearly disjoint.

The degrees are 5 and 4.

The degrees are 5 and 4.

This is logically 5 and a Q of the fifth root of unity

is a cyclotomic extension it is of degree 4 is what we have.

Already central d.

On the other hand, Q of fifth root of 2 and

Q of the, how to denote it?

Well, let's see.

Let's denote by the fifth root of one

exactly the exponential of 2 pi i over 5,

so if we take this e to the power of 2 pi i

over 5 times the fifth root of 2,

these guys are not linearly disjoint.

Not linearly disjoint.

Since both degrees are equal to 5, but

the composite of the spliton field of x to the power of 5- 2.

And this is of degree 20.

Indeed, in both cases, L1,

L2 is at the spliton field

of x to the power of 5- 2.

Since in the first case, we are linearly disjoint,

the degree of L1, L2 over Q is equal to 20.

In the second case,

Both L1 and L2 are of

degree 5 over Q.

And 5 times 5 is not equal to 20.

Let's say that L1 and L2 are not quite independent over Q,

in the second example, whereas they are in the first.

So, you see that the difference is rather subtle.

Well, the obvious reason in the second case

is that those extensions are generated

by rules of the same polynomial, but,

Still some effort is needed to formalize

why this is not linearly disjoint case.

In particular, we see,

That it is not sufficient, in order to be linearly

disjoint to have a very small intersection.

So, L1 intersected L2 is

Q and does not imply that L1 and

L2 are linearly disjoint over Q.

This is exactly what happens in the second example.

So, this is what happens in the second case.

The degree is our prime,

so we cannot have any sub extensions.

[SOUND]

Acerca de Coursera

Cursos, programas especializados y títulos en línea impartidos por los principales instructores de las mejores universidades e instituciones educativas del mundo.

Join a community of 40 million learners from around the world

Earn a skill-based course certificate to apply your knowledge

Gain confidence in your skills and further your career

Coursera brinda acceso universal a la mejor educación del mundo, al asociarse con las mejores universidades y organizaciones, para ofrecer cursos en línea.

© 2019 Coursera Inc. Todos los derechos reservados.

Descargar en la App Store

Consíguelo en Google Play