So, now let me tell you about the structure of a finite algebra over a field. So, K will be a field, and A will be a finite K-algebra. "finite" means a finite-dimensional vector space. Well, before telling you this, I would like to recall the Chinese remainder theorem. Let A be a ring. And I and J be ideals. One says that these are relatively prime If they generate A. Lemma: If I, J are relatively prime then IJ is equal to the intersection of I and J. If I_1 ... I_k are relatively prime with J, then so is their product. So, this is the product. These are many ideals, each of them is relatively prime to J, and so the product then is also relatively prime to J. And then 3), which is a direct consequence of the 2) if I J are relatively prime, so are I^k and J^l. if I J are relatively prime, so are I^k and J^l. So, the proof of the lemma is very easy. So, the inclusion IJ sits in the intersection is clear. This is just by definition. Now, if we are relatively prime, then we have that 1 is i + j for some i in I, and j in J. Then for any x from the intersection Then for any x from the intersection we have x = x_i + x_j. And this sits into IJ, and this also sits in IJ, so x is in IJ. So this was 1), 2) is also not difficult, let's suppose for simplicity of notations that k = 2. The general case is similar. So, we have 1 = i_1 + j_1 = i_2 + j_2, where i_1 is in I_1, i_2 is in I_2, and j_1, j_2 are in J without indices this time. Well, now let's just write 1 = (i_1 + j_1)(i_2 + j_2) and this is i_1 i_2 + j_1 i_2 + + j_2 i_1 + j_1 j_2. i_1 i_2 + j_1 i_2 + + j_2 i_1 + j_1 j_2. so this is in J, and this is in I_1 I_2. So, this is just what we want to proof. And the 3) is obvious from 2 by induction. So, now the Chinese remainder theorem, Theorem 2, I guess, or Theorem 3, I don't remember anymore. Chinese remainder is as follows: let me have I_1 ... I_n ideals is as follows: let me have I_1 ... I_n ideals Consider the map pi from A to the product of the quotients by those ideals. Well, of course, the kernel... How the map is made? It sends a to a mod I_1, ... a mod I_n, so the kernel of pi is the intersection. Now, I affirm that pi is surjective, if and only if our I_1, ... I_n are pairwise relatively prime. I_n are pairwise relatively prime. And then, of course, A modulo the intersection of I_k is equal to A modulo the product of I_k is equal to the product of the quotients. Okay, so, let me prove it. If pi is surjective, this means that there exists an a such that pi(a) is equal to the following lines, 0, ..., 0, 1 on the i-th place, and then again 0. This means, that well, let me call it a_i, right? This means that a_i belongs to an I_j for j different from i, and 1-a_i belongs to I_i. And this, of course, implies that I_i is relatively prime to any I_j, since 1 is (1-a_i) + a_i. So, conversely. Suppose, that all our ideals are relatively prime. This implies that I_i is relatively prime to the product of I_j's, where j is different from i. So, there exists x_i in I_i, and y_i in the product of the others, such that x_i + y_i = 1. And such an element y_i maps to 0, ..., 0, 1, at the i-th place and then again 0s. Of course, if we take some linear combination sum of, say, b_i y_i shall map to b_1, ... ..., b_n, for any b's. So, our map is surjective. Now let us consider A a finite algebra over K. Before proving a general theorem on the structure of A, I would like to state a proposition. First of all, if A is an integral domain so, it does not have 0 divisors, then A is a field, and 2) which is a rephrasing of 1), says that any prime ideal of A is maximal. Well, I shall prove only the first part, the second part is just a consequence of definitions. In fact, a factor over a prime ideal, a quotient over a prime ideal is an integral domain, and a quotient over a maximal ideal is a field. If you don't know this, please look it up in any book. For instance, in the very beginning of Atiyah & Macdonald introduction to commutative algebra. Let me prove the first part, What means, that is an integral domain? So, integral domain, that is to say, no zero divisors, means that for any element a, the multiplication made by a is injective. But now, A is a finite-dimensional K-vector space. So, it is surjective. So this implies, that the multiplication by a is an isomorphism. In particular, it is surjective. Surjective, and so, there exists b, such that b times a = 1. So, 1 here has a preimage b. Well, which means of course, that A is a field.