Now, let me introduce an important notation. So, I have L an extension of K, and alpha an element of L. I call K of alpha the smallest subfield of L containing K and alpha. And I call K of alpha is the square brackets the smallest subring or a K-algebra, if you wish, containing K and alpha. containing K and alpha. Well, let me give you an example. Well, first of all let me say that K of alpha in the square brackets is generated as a vector space over K by 1, alpha, alpha squared and so on, alpha to the power of n, and so on: by the constants and all the powers of alpha. So, let me give you an example. Of course, C is equal to R of i. as a field. But it is also equal to R[i] as a ring. Because any element of C is x + iy where x is a real part and y is the imaginary part. So, of course this is a vector subspace generated by 1 and i. And this is a general phenomenon. So, Proposition 1: the following conditions are equivalent: First, alpha is algebraic over K. Second, that K[alpha] as a ring is a finite dimensional vector space over K and third, K of alpha as a ring is equal to K of alpha as a field. So, let me prove this proposition for you. Let me prove that 1 implies 2. If alpha is algebraic, then I can write alpha to the power of d plus a_(d-1) alpha to the power of d-1 plus and so on, plus a_1 alpha plus a_0 is equal to 0 where a_i are elements of K. This is just P of alpha, where P is a minimal polynomial. So, we see that alpha to the power of d is minus a linear combination of the sum of a_i alpha to the power i where i is from 0 to d-1. So, you see that alpha to the power of d, alpha to the power of d+1 and so on are going to be linear combinations of the lower powers of alpha. This implies of course that K of alpha is generated by 1, alpha and so on, alpha to the power of d-1 over K. And in particular it is finite dimension. Let me prove that two implies three. Well, it is enough to prove that K of alpha is a field. Since, of course, K of alpha in the square brackets is contained in K of alpha with the round brackets. Well, how do you prove this? It is very simple. Well, I'll let x be an element of K of alpha. I want to show that this is invertible. Consider the multiplication by x. So, this is the multiplication by x. This is a homomorphism of vector spaces, this is an injection of vector spaces over K. But, we know from linear algebra that if you have an injection, of finite dimensional vector spaces of the same dimension, then it is also a surjection. Since K of alpha is final dimensional this is a surjection. So there exist there exists a y in K of alpha, such that y times x is equal to 1. So, x is invertible. X is invertible and K of alpha is a field. Of course, I have forgotten to say that x was supposed to be nonzero in order to have the multiplication by x injective. But, I guess everybody has understood. So, we were assuming x non zero from the beginning. Okay. So, let me prove that 3 implies 1. So if K of alpha in the square brackets is a field, then alpha is algebraic. This is maybe the easiest part. So, if alpha is not algebraic, then there exists no polynomial P such that P of alpha is zero. But what does this mean? This means that a natural homeomorphism, let me call it i, from K[x] to L which sends P to the P of alpha is injective But, K of x is not a field. K of x is not a field. And the image of this homeomorphism is K of alpha which is a field. And the image of i is a field. Because this was just K of alpha. So, this is a contradiction. Definition: L an extension of K is called algebraic over K, if every element of L is algebraic. Well, let me give you some properties of those algebraic extensions. Proposition 2: if L is algebraic over K, any K-subalgebra of L is a field. any K-subalgebra of L is a field. Well, proof. Let's take L prime in L a subalgebra. Fix an alpha in L prime. I have to show that it's invertible. Well, I know it's algebraic. And then I know that K of alpha which is also sub algebra of L is a field, and then I know that alpha is invertible, and since I can do it for any alpha different from zero, it follows from this, that L prime is a field. Well, another proposition, which will be important is as follows: Let's have K in L in M. If alpha in M is algebraic over K, then it's algebraic over L, and it's minimal polynomial over L divides and it's minimal polynomial over L divides it's minimal polynomial over K. Well, this is of course clear since I can just consider this minimal polynomial of alpha over K as an element of L of x.