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6.2 The Galois correspondence

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Introduction to Galois Theory

National Research University Higher School of Economics

4.3 (101 calificaciones) | 19K estudiantes inscritos

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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4.3 (101 calificaciones)

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De la lección

Week 6

We state and prove the main theorem of these lectures: the Galois correspondence. Then we start doing examples (low degree, discriminant, finite fields, roots of unity).

6.1 Some further remarks on normal extension. Fixed field.13:14

6.2 The Galois correspondence14:19

6.3 Galois correspondence (cont'd). First examples (polynomials of degree 2 and 3.14:53

6.4 Discriminant. Degree 3 (cont'd). Finite fields.15:04

6.5 An infinite degree example. Roots of unity: cyclotomic polynomials14:00

6.6 Irreducibility of cyclotomic polynomial.The Galois group.14:20

Impartido por:

Ekaterina Amerik
Professor

Transcripción

[SOUND] So the Galois correspondents. Let's take a Galois extension. Let L over K be a Galois extension. The group of automorphisms of L over K is code by definition the Galois group. And is denoted like Galois of L over K. Some theorem, By Galois. So if L is finite, Over K. Then there is a bijection between the subfields of L, sub-extensions of L and the subgroups in the Galois group. Or sub extensions Correspond objectively to the subgroups. H in Galois. So how do they correspond? If we have a subfield F, then we just send it to the Galois group. Of l over f. L over f is Galois, because it's separable and normal. It's f, which is not necessarily Galois over K, okay? And then, in the other direction, you will have H, then we send it to its fixed field, LH. The second part, as I just said, F is not necessarily Galois but one can see when it's Galois. So the second part, f is Galois over k. If I don't leave for any g in Galois in l over k. G preserves F globally. G of F is F, so if G fixes F more than twice than this means of course that F is equal to K, but here we ask for something with her we ask that g fix f as we said little bunny. Okay and this is also equivalent to seen the Galois group of l over f. Is a normal subgroup in Galois L over K. In this case G goes to g restricted to f. Is this rejection? From Galois of l over k to Galois of f over k. And the kernel is just Galois of l over f. Okay so let's prove it. In fact the first part has been more or less done right? So l, fixed fueled by Galois of l over f is f, this has been done. So n is operable extension. [SOUND] For separable extensions. Then also H is a part of the Galois group of L over LH. This is obvious by definition. But also by Artin's theorem. They have the same cardinality. The degree of L over LH is equal to cardinality of H. But this degree, as we have seen, For normal extensions they have as many automorphisms as to the degree of the extension is, right? So this is Galois the cardinality of Galois group of l over L h so by normality. And separability of course. We have seen that normal and separable extensions have many automorphisms. Okay but in general the cardinality of the group of automorphism is less or equal than the degree of the extension and the Galois case they are equal. So, we have a subgroup which has the same cardinality as the group. And so they must be equal. So one must have H is equal to the Galois group of l over l h. So this means that the maps which we had in the theorem, the map which sends f. To the Galois group of L over F which sends H to the fixed field LH. Well, this means these are mutually inverse. Inverse. Other than that, it's invertible, it's a bijection. So, in particular These are bijections. So the first part was more or less older median observed. Before we were discussing urgency and other things, now the second part, something has to be done here. So we had. Three equivalences. FIrst statement was that f was Galois. The second statement was that a and g. G of F was F for a of g and Galois and the third statement was, of course that the Galois group of L over F was a normal sub group. This is a notation for normal subgroup in Galos, L over K. So we must prove equivalence. Well let's show that 1 implies 2. Let's take an X in F. We have the minimal polynomial of X/K. It splits in L, but it has a root in F. And has a root. In F. So it must have all rules in F by normality. So, F is normal since it's the one. This means that this P minimal Splits in F. This means, of course, that any map from Galois group preserves F since it premutes the roots. Any g in the Galois group, permutes the root of the minimal of f over k right? Since this is true for any x in F, this means that g of F must be contained in F. Right since these f is generated by all sectional. F is generated for the conceive of e. Right. Now it's also very easy to remark that 2 implies 1. Now if g(f) is an F, then, All roots Of the minimal x k for x and f are in f. Since again, g permits those rules. Well, since the Galois group, Acts transitively on the roots of an irreducible polynomial. This is something which is very very important. We have seen this last time, it results from this theorem about extension of homomorphisms and it is very important to memorize it, that the Galois group is transitively on roots of any irreducible polynomial. Okay. So this means that any polynomial which has a root in f is split over f and this is the definition of normality. This is the definition. I will continue after break. [SOUND] [MUSIC]

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