Theorem on the structure of finite K-algebras. Let A be a finite K-algebra This means, that A is a finite dimensional K-vector space. Then, first of all, there are only finitely many, maximal ideals m_1, ..., m_r in A. Secondly, let J be the intersection of those maximal ideals. Since they are relatively prime, this is the same as the product. Then J^n is 0 for some n. And the third point is the structure theorem: A is isomorphic to the product of A/(m_1^n_1) times ... is isomorphic to the product of A/(m_1^n_1) times ... times A/(m_r^n_r) for some n_1 and so on n_r. Proof. So, why there are only finitely maximal ideals? Pick a number of maximal ideals. Let m_1, ... m_i be maximal ideals. By Chinese remainder theorem, we have A over the product of m_j's isomorphic too A/m_1 times ... ... times A/m_i. Now all those persons are finite dimensional K-vector spaces. A/m_j and A over the product, are finite dimensional K-vector spaces. And the dimension of A over K is greater or equal than the dimension over K of A over the product. This is the same as the sum of the dimensions over K of A/m_j 's. And each of those dimensions of A/m_j is at least 1, so this is greater or equal than i. So, the number of maximal ideas is at most the dimension of A over K. In particular, there are only finitely many. The second part. J is also a finite dimensional vector space. A finite dimensional vector space over K. And so are it's powers. Now consider the following sequence: J contains J^2 contains J^3 contains ... ... contains J^k and so on. This is a decreasing sequence of finite dimensional vector spaces. The dimension decreases as k increases. So this sequence must stabilize. There exists an n, such that J^n is equal to J^(n+1). I claim, that in this case J^n is 0. Well indeed, if not let e_1, ... ..., e_s be a basis of J^n. ..., e_s be a basis of J^n. As J^n is equal to J times J^n, we can write e_i equal to the sum of lambda_ij e_j, for some lambda_ij in J. If you consider the matrix M, which is the identity minus lambda_ij, we have that M times the vector of e_i's is 0. This is just the identity I have written with those lambda_ij's. Since M is a matrix over a ring, not over a field, this does not immediately mean that the the i's are 0 but, we can always find the matrix M-tilde such that M-tilde M is equal to the determinant of M times identity. This is possible over a ring. So the corollary is, of course, that the determinant of M annihilates my vector of e_i's. But, if you look at this matrix M, you see that the determinant of M is of the form 1 + lambda, where lambda is in J. Since J, remember, is the intersection of all maximal ideals, lambda is in every maximal ideal. So, 1 + lambda is in no maximal ideal. But this means, that 1+lambda is invertible. And this already means that the e_i's are 0. So this is a contradiction. Part 3, so by part 2, We can find n_1 and so on, n_r such that m_1^n_1 times ... ... times m_r^n_r is 0. We can, for instance, take all n_r's equal to n. All n_i's equal to n. Then by Chinese remainder theorem, A is isomorphic to A/(m_1^n_1) times ... A is isomorphic to A/(m_1^n_1) times ... ... A/(m_r^n_r). Since this A is just equal to A/(m_1^n_1 ... m_r^n_r). And all those m_i^n_i are relatively prime. m_i^n_i are pairwise relatively prime. So this proves the theorem. Well, remark: The n_i's are not uniquely determined. I could have taken all n_i's equal to n, but mostly you also can write this identity with n_i's, with some n_i's at least different from that n. The n_i's are not uniquely determined. And for instance, if we take A, the quotient of the polynomial ring, maybe with a big K, K[x] over the ideal generated by x^2 times (x+1)^3. Then we have two maximal ideals: m_1 is generated by x, m_2 is generated by x+1. And then A is equal to A over m_1^2 A over m_2^3. But also, n of the part 2 of the theorem is equal to 3, so, A is isomorphic to A/m_1^3 times A/m_2^3. In fact, the reason is very simple. Then m_1 squared is in fact equal to m_1 cubed. One inclusion is obvious. In A the ideal, generated by x^2 contains the ideal generated by x^3, but also, the converse is true. (x)^3 contains the ideal generated by x squared. Of course, this is not true in the polynomial ring but it is true in A. And I leave the verification as an exercise. So, let me give you a couple of examples of finite dimensional K-algebras. One of them we have already seen, if we take the tensor product of C with itself over R, then you obtain C times C. And you can also consider other examples of this kind. Q(square root of 2) x Q(square root of 3) is, in fact, a field Q(square root of 2, square root of 3). And you see that those algebras are products of fields. So all n_i's may be taken equal to 1. In other words, we don't have nilpotents in our algebra. So, it is a reduced algebras. Reduced, by definition, is without nilpotents. And in the next lecture, you will see that this is a general phenomenon. That the presence of nilpotents is due to the inseparability of extensions come from inseparable extensions. And I should stop here today.