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2.3 An example. Algebraic closure.

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Introduction to Galois Theory

National Research University Higher School of Economics

4.3 (100 ratings) | 19K Students Enrolled

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...) Do you have technical problems? Write to us: coursera@hse.ru

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From the lesson

Week 2

We introduce the notion of a stem field and a splitting field (of a polynomial). Using Zorn's lemma, we construct the algebraic closure of a field and deduce its unicity (up to an isomorphism) from the theorem on extension of homomorphisms.

2.1 Stem field. Some irreducibility criteria.13:20

2.2 Splitting field.11:45

2.3 An example. Algebraic closure.14:50

2.4 Algebraic closure (continued).15:07

2.5 Extension of homomorphisms. Uniqueness of algebraic closure.11:47

Taught By

Ekaterina Amerik
Professor

Transcript

Okay let me give you an example. Let's take the same polynomial x^3 - 2 over Q. Its roots are cubic root of 2, , and j times cubic root of 2 and j squared times cubic root of 2, where j is the exponential of 2 pi i by 3 so its a primitive third root of unity. Okay, the splitting field L is then Q of cubic root of 2, and j, right? All our roots are in this Q cubic root of 2, j and also j must be contained in a splitting field as it is just the quotient of two roots, so this is easy, and now let us find the automorphisms of L. Well, for these let me write two towers. So I have this Q and I have two extensions Q of j and Q of cubic root of 2. Here the degree is 3, here the degree is 2, and I have the composite extension Q of j and cubic root of 2. Well, the degree of j over Q of cubic root of 2 is of course 2. The minimal polynomial of j is x^2 + x + 1. Well, both over Q, but also over Q of cubit root of 2. It's obvious that such a thing must be irrreducable also over Q cubic root of 2 because well, for instance j is not a real number, which cannot be included in this field. So this degree is 2. This degree by the same token is 3. The cause of course is the total degree is 6, so here it cannot be anything but 3 because the degrees of finite extensions multiply in towers. So, and we see that there is a Q(j)-automorphism a Q(j)-automorphism of L taking cubic root of 2 to j cubic root of 2, right? Because L is a stem field of x^3 - 2 over Q(j), so there are automorphisms interchanging roots. And there is also a Q(cubic root of 2)-automorphism of L a Q(cubic root of 2)-automorphism of L taking j to j^2, right? taking j to j^2, right? Since these are two roots of the same minimal polynomial. So L is a stem field of the minimal polynomial over Q of cubic root of 2, and there is an automorphism exchanging those roots, right? So we have a whole group of automorphisms, which is in fact equal to the group of permutations on three elements, so let's call this sigma, let's call this tau Well, the group of automorphisms of L over K is embedded into S_3. So permutation groups of 3 elements. And in fact, well, how it's embedded is because of those automorphisms permute the roots of x^3 - 2. In fact, Aut L over K is equal to S3 since those sigma and tau they already generate S3. If you remember what kind of permutations of roots are they so sigma was taking cubic root of 2 to the other thing. So sigma is a cyclic permutation of roots. Yes, and tau is also, tau is a transposition of roots, so if you tau takes j to j squared and keeps cubic root of 2 where it was. So this tau fixes cubic root of 2 and it exchanges j cubic root of 2 with j squared cubic root of 2. So this is a transposition. This is a 3-cycle. Together they generate S_3. Okay, let me now talk about algebraic closure. So, a definition: A field K is algebraically closed, if any nonconstant polynomial has a root in K. This is the same as to say that any nonconstant polynomial splits as a product of linear factors. For example the field of complex numbers has this property. In fact, we will give a proof of this in the future. Shall be proved by almost pure algebraic means. Well, definition. An algebraic closure of K is a field L, which is algebraically closed and algebraic over K. So Theorem 2, any field K has an algebraic closure. Note that I don't say at this point anything about uniquness. Up to isomorphism or whatever. We will treat this later. Now, let's concentrate on existence. Proof is a bit weird. How are we going to proceed? So we first construct K_1 such that any polynomial with coefficients in K has a root in K_1. Well, this is not yet a victory because we don't know whether any polynomial with coefficients in K_1 has a root in K_1. Maybe we have introduced some new rootless polynomials. Then construct K_2 such that Then construct K_2 such that any polynomial in K_1 of x has a root in K_2. Well, I'm not explaining you yet how to construct them. I've just explained the general strategy. So and so forth, so let's say this is the stratagy, and so forth. So what you will have is K in K_1 in K_2, and so on, in K_n, and so on. Take K bar the union of all K_n. I claim that K bar is algebraically closed. K bar is algebraically closed. Indeed, any polynomials with coefficients in K bar really has its coefficient on some floor of this tower. P is in fact in K_n of x. So there exists an n such that P is in K_n of x. So it has root in K_(n+1), so in K bar, okay? So, if we learn to construct such K_1, K_2 and so on this will solve the problem. So construction of K_1. But because of the time reasons I have to do it in the next part of this lecture.

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