My name is Ekaterina Amerik. I am working at the mathematical department of Higher School of Economics, and I'm going to read a course on Galois Theory. This is a course about field extensions. We assume familiar the basic notions of abstract algebra, like groups, rings, fields, modules, ideals, and their basic properties. We also assume a certain knowledge of linear algebra. All rings we consider will be commutative, associative, and with unity. Let me give you the first definition. K and L will be fields. L is said to be an extension of K, if L contains K as a subfield. An equivalent definition is as follows: L is an extension of K if L is a K-algebra. Let me remind you, what is a K-algebra. A K-algebra is a ring, with a structure of a K-module such that these two structures are compatible, such that the multiplication which goes from A times A to A is bilinear. K-bilinear, of course. Well, if you wish, you can reformulate it, saying something like k_1 a_1 times k_2 a_2 is k_1 k_2 acting on a_1 a_2, where k_i are elements of K, and a_i are elements of A. Why is a field containing K as a subfield the same thing as a K-algebra? Well, in fact, given a K-algebra structure on A is the same as giving a homomorphism of rings from K to A. Well indeed, if I have a K-algebra, then I can define a homomorphism by sending k to k as it acts on 1. And conversely, if I have a homomorphism of rings from K to A, I can define a K-algebra structure by setting ka equal to f(k) times a. So now, if A is a field L, then any homomorphism from K to L is injective. Well, there are several ways to see this. One of them is saying that f(k) is always invertible. Indeed f(kk^(-1)) is the same as f(k) times f(k^(-1)) this is f(1) which is 1. So we see that f(k) is invertible, so it cannot be 0. If, of course, k is not 0 itself. So f(k) is never 0 when k is nonzero itself. In particular, f is injective. Or a more sophisticated way to say the same thing, is to say that the kernel of f is always an ideal. So alternatively, the kernel of f is always an ideal. And a field does not have nontrivial ideals, the only ideals are 0 and K itself. If you don't know this, you are strongly encouraged to do it as an exercise. Now let me give some examples. Of course, C is an extension of R. And R is an extension of Q. Second example. Any field has what is called a characteristics, so if L is a field, there are two possibilities: either, if you take the unit element and start adding it to itself, you never obtain 0. In this case, we say that the characteristics of K is 0. And we see that in this case, of course, L contains Z. Well, but L is a field so if it contains Z, it must contain Q. So L is an extension of Q. Or, if you take the unit element and start adding it to itself, you obtain 0 at certain point. Then it is easy to see that the first time it happens, it happens for some prime number m, so minimal m with this property is prime. And then one says that this prime is the characteristic of this field. Then L does not contain Z anymore. What it contains is Z/pZ. For p - prime, Z/pZ is a field. And when we want to emphasize that we consider Z/pZ with its field structure, we denote it by F_p. denoted by F_p. And our L is an extension of F_p. So we have two possibilities: if the characteristic of L is 0, then L is an extension of Q. If it is p, then it is an extension of F_p. One calls Q and F_p the prime fields. Q and F_p are prime fields. Any field is an extension of one of those, and they don't contain any proper subfields. is an extension of a prime field. Third example is very important. So let's take the ring of polynomials in one variable over K. And let's consider the quotient ring by an ideal generated by an irreducible polynomial P. Then I affirm that this is a field. There are two ways to see these. Again, in the elementary language, you can see the following thing. So if Q is a polynomial which is not a multiple of P, then Q is prime to P. The greatest common divisor of Q and P is 1. And then you have the Bézout identity. There exist A and B polynomials, such that AP + BQ = 1, so what you see is that BQ = 1 modulo P, and therefore B is an inverse of Q in K[x] modulo P.