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1.2 Algebraic elements. Minimal polynomial.

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Introduction to Galois Theory

Université nationale de recherche, École des hautes études en sciences économiques

4.3 (100 notes) | 19K étudiants inscrits

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 (100 notes)

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À partir de la leçon

Week 1

We introduce the basic notions such as a field extension, algebraic element, minimal polynomial, finite extension, and study their very basic properties such as the multiplicativity of degree in towers.

1.1 Field extensions: examples14:24

1.2 Algebraic elements. Minimal polynomial.12:27

1.3 Algebraic elements. Algebraic extensions.14:34

1.4 Finite extensions. Algebraicity and finiteness.14:34

1.5 Algebraicity in towers. An example.14:09

1.6. A digression: Gauss lemma, Eisenstein criterion.13:40

Enseigné par

Ekaterina Amerik
Professor

Transcription

So let me continue with my third example. We have been considering the quotient of the polynomial ring over K by the ideal generated by an irreducible polynomial P. And we have seen that this is a field. I would like to see the same thing in a slightly more sophisticated language, but it amounts to the same. Instead of writing down the Bézout equality, one can also say that an ideal generated by an irreducible element of K[x] is a maximal ideal. And the quotient by a maximal ideal is always a field, so alternatively one can say that (P) is a maximal ideal of K[x] and the quotient of a ring by a maximal ideal is a field. But of course the proof of this amounts to the same Bézout equality. How do you prove (P) is a maximum ideal? You just consider some potentially bigger ideal, that is to say containing P and some element Q not belonging to the ideal generated by P. This Q is going to be prime with P, you can write the Bézout equality and it is in this way that you see that such an ideal will be necessarily equal to the polynomial ring K[x]. Okay, so this is an extension of K in an obvious way. Maybe here it's even better to say that it's an extension of K because it's a K-algebra, hence an extension of K. And let me give you a more concrete example. Let's take K equal to the field of two elements. So Z by 2Z, it consists of 0 and 1 and 1 + 1 is 0. Let us take P = x^2 + x + 1, so this is an irreducible polynomial over F_2. Then K[x] modulo (P) is a field of four elements. It contains 0, 1, the class of x modulo P, let me denote it by x bar, and the class of x plus 1 modulo P, let me denote it by x plus 1 bar. And you see that x bar squared is equal to -x-1 since you know that x^2+x+1 is 0 in our field. Well, the characteristics is 2, so -1 is the same thing as 1, so it's just x+1 bar, in the same way x+1 squared is x bar, and they are inverse of each other. x bar times x+1 bar is one. x bar times x+1 bar is one. So this is the structure of a field of four elements. The cardinality of K is 4, one writes then K=F_4. Well, this might be strange at the first sight, because we only know that K has four elements. And if you write F_4 you somehow mean that there is only one field of four elements. Well, it is true, there is only one field of four elements. In fact, all finite fields of the same cardinality are isomorphic, and we will see it very shortly. Okay, now let me talk about algebraic elements. So I have a field extension K and L, and I take alpha in L. I say that alpha is algebraic, if I can find some polynomial: if there exists P in K[x] such that P of alpha is 0. Otherwise I say that alpha is transcendental, alpha is transcendental. So this is a definition of course. Well, let me give you a lemma. So, if alpha is algebraic then there exists unique unitary polynomial of minimal degree with this property: such a polynomial is irreducible. Any other polynomial with this property is divisible by P. Any Q... so let me call this unitary polynomial P. Any Q such that Q(alpha)=0 is divisible by P. One calls P the minimal polynomial of alpha over K. Such a P is called the minimal polynomial of alpha over K. So the proof of the Lemma is very simple of course. It is a direct consequence of definitions, let me say it in sophisticated terms. We know that K[x], the polynomial ring in one variable, is a principle idea domain. And the polynomials vanishing in alpha certainly form an ideal. So this ideal is generated by one element. Let me denote this set by I, so I has a generator P. And this is certainly, a unique up to a constant element of minimal degree in I. P is unique up to a constant element of minimal degree in I, furthermore, if P was not irreducible, if P is Q times R, if P was not irreducible, if P is Q times R, then P(alpha) is Q(alpha) times R(alpha). So Q(alpha) or R(alpha) must be 0, which is a contradiction with the minimality to the degree. Then Q(alpha) is 0 or R(alpha) is 0. And this contradicts the minimality of the degree. If you are not familiar with the fact that the polynomial ring in one variable over a field is a principle ideal domain you are strongly encouraged to learn this. This is a very simple effect, it is a consequence of Euclidian division, and this time I will not give any more details on this.

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