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3.4 Separable elements.

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Introduction to Galois Theory

National Research University Higher School of Economics

4.3 (103 classificações) | 20K alunos inscritos

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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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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

Ministrado por

Ekaterina Amerik
Professor

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Transcrição

Our next topic is separable extensions. We would like to say that a spliting polynomial spliting field of an irreducible polynomial has many automorphisms. So E a spliting field of an irreducible polynomial. We would like to say that it has many automorphisms. Well, what do I mean by this? I mean the following things. So if alpha and beta are roots of P, then E contains K(alpha) and K(beta). And I have also an isomorphism of K(alpha) with K(beta) over K. No, wait, sorry, over K. Better to say like this. And by extension theorem, I can extend it to an automorphism of E to an automorphism of E. So there is one problem. There is one problem about this. One problem. Is it true, that an irreducible polynomial has many roots? Of degree n has many, Of degree n has many, that is to say n, it cannot have more than n, roots? And the answer is yes in characteristic 0. But, of course, not always if characteristic of K is a prime number p. An irreducible polynomial in characteristic p can have multiple roots. So P has multiple roots. So P has multiple roots. This means that the greatest common divisor of P with its derivative is not 1. Well in characteristic 0, this cannot happen. In characteristic 0 this is never the case when P is irreducible. Well, just because if you are not prime to an irreducible element, you should be divisible by it, but this is, of course, impossible since the degree of the derivative is strictly less than the degree of P. And P prime is nonzero when P is nonconstant. So P does not divide P prime. In characteristic p, P prime can vanish. And then this GCD can be equal to P. Well, to the big P, right? Not to the characteristic, of course, but to the polynomial P. And the GCD is a polynomial. Well, how can P prime vanish? P prime vanishes exactly when P is a polynomial when P is a polynomial in x^p. That is to say, P is a sum of ai x^i, with a_i nonzero only if i is a power of p. No, not a power of p, if i is divisible by p. So let me then take r = maximal h such that P is a polynomial in x to the power p to the power h. in x to the power p to the power h. That is ai is 0 whenever p^h does not divide i. p^h does not divide i. Okay, then I can write P(x) as a polynomial Q(x^p^r). P(x) as a polynomial Q(x^p^r). And Q is not a polynomial in x to the power p anymore. So Q prime is non-zero. In particular Q is prime with Q derivative and Q does not have multiple roots. And in addition, all roots of P have multiplicity p^r. Well, let me formulate it as a proposition. Proposition, I think I did not have propositions today yet so this is Proposition 1. Of course the proof is completely trivial. So all roots of P have multiplicity p^r. Well, in fact, it's easy to see if lambda is a root of P, then P is (x - lambda) times, I don't know, R. And then mu, which is lambda^p^r is a root of Q. And then mu, which is lambda^p^r is a root of Q. So Q(y) is (y - lambda^p^r) times I don't know, times S, So Q(y) is (y - lambda^p^r) times I don't know, times S, So Q(y) is (y - lambda^p^r) times I don't know, times S, where lambda is not a root of S. And then we'll just set y equal to x^(p^r). And then we'll just set y equal to x^(p^r). So we know that P(x) is what? (x^p^r - lambda^p^r) times S(x^p^r) (x^p^r - lambda^p^r) times S(x^p^r) (x^p^r - lambda^p^r) times S(x^p^r) And this is exactly (x - lambda)^p^r and lambda is not a root of this. and lambda is not a root of this. Okay, so we know that multiplicity of lambda is p^r. So, let me give a definition. So, let me give a definition. Let P, a polynomial over K, be irreducible. Then it is called separable if it is prime to its derivative so it does not have multiple roots. If it is prime to its derivative so it does not have multiple roots. The separable degree of P is the degree of Q is above. The separable degree of P is the degree of Q is above. And the degree of inseparability of P is, of course, the degree of P over the degree of Q. So this is this famous p^r. And P is called purely inseparable if the degree of P is equal to the degree of inseparability of P. Well then, of course, P is just x^p^r - a. Well then, of course, P is just x^p^r - a. And an element of a field extension, of an algebraic field extension, is called separable or purely inseparable if its minimal polynomial has this property. Definition: if L is an algebraic extension of K, then alpha in L is called separable then alpha in L is called separable or purely inseparable over K if its minimal polynomial has this property. Well, let me formulate a proposition, Proposition 2, which is obvious in fact. If alpha is separable over K, then the cardinality of the set of homomorphisms over K of K(alpha) into a fixed algebraic closure K bar is equal to the degree of its minimal polynomial. And if alpha is not necessarily separable in general this number, is equal to the separable degree of this minimal polynomial. Well, the proof is obvious because the separable degree is just the number of distinct roots of this minimal polynomial. And we can send alpha into any of those roots.

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