So, let me summarize briefly what we have done in the last lecture. We have been considering L over K, a finite field extension. And we have defind separability. We have seen that, if L is generated over K by a finite number of separable elements, then the number of homomorphisms over K from L to K bar, to an algebraic closure of K was equal to the degree of L over K. And this is "if and only if" in fact. Well, in general this number of homomorphisms is less or equal than the degree. These number of homomorphisms, we have called it "the separable degree" of L over K. Well, if L was generated by only one element alpha, this was clear since those homomorphisms were taking alpha to other roots of the minimal polynomial. And so, the number of homomorphisms was equal to the number of roots of the minimal polynomial. And in general, one can use induction and the multiplicativity over the degree. This was just linear algebra. And, the number of homomorphisms this was a theorem about extension of homomorphisms. Extension theorem. A separable extension was exactly an extension which had the right number of homomorphisms into the algebraic closure. Today, well, not exactly today, maybe next time we will characterize separability in terms of tensor products. So, the subject of today's lecture is tensor product. And structure of finite K-algebras. This will be a general digression which does not have much to do with field extensions. We consider a ring A. and M and N - two A-modules. The tensor product M tensor M over A is another A module M tensor M over A is another A module together with an A-bilinear map together with an A-bilinear map A-bilinear map, say phi from the usual Cartesian product to this M tensor N over A. Which has the following universal property: if P is any A-module and f from M times N to P A-bilinear. and f from M times N to P A-bilinear. What means "A-bilinear"? This is very simple. So, for any m, the map f_m, from N to P, sending n to f(m,n) and f for any n, f_n from M to P, sending m to f(m,n). are A-module homomorphisms. One also sais that these are A-linear homomorphisms of A-modules. homomorphisms of A-modules. Then there exists a unique map, a unique homomorphism of A-modules f tilda, from M tensor N to P, so, there exist a unique f tilda homomorphism of A-modules, such that f is equal to f tilda times phi. This property characterizes this pair consisting of phi and tensor product this pair consisting of phi and tensor product because if there is another pair like this, if say, phi bar M otimes N bar is another pair with this property, then, by the very definition, we have mutually inverse homomorphisms of A-modules between them. Ieverse homomorphisms of A-modules between our tensor products. Now, so the unity of such a thing follows directly from the definition. But of course, the problem is existence. Why does such a thing exist? And then one has to give a construction. The construction can be as follows: so, consider E consisting of maps from M times N to A, as sets. So, without any structure. the only thing I want from these maps is that they shall be almost everywhere 0. Which are 0 almost everywhere. What means almost everywhere? Almost everywhere is outside of a finite set. Outside of a finite set. For instance, I can take some kind of delta functions. This delta_m,n takes the value 1 on the pare (m,n) and 0 otherwise. Then, this E is a free A-module with base delta_m,n. Now, I have a map from M times N to E. Map of sets, which is not a bilinear, of course. It just sends (m,n) to delta_m,n. But, I can make it bilinear by changing E. So, this is not bilinear. There is no reason for it to be bilinear, but I can take a quotient. So, take F in E a submodule generated by delta_m+m',n minus delta_m,n minus delta_m',n. Same for n's: m,n+n' minus delta_m,n minus delta_m,n'. Then same for multiplication by an element of A: delta_am,n minus a delta_m,n and same from the other side. delta_m,an minus a delta m,n. delta_m,an minus a delta m,n. And of course, the map from M times n to the quotient is bilinear. And it is very easy to see that this has the desired universal property.