One also has other more or less elementary properties of the tensor products. For instance, one has a kind of associativity M1 times M2 over A times M3 is the same thing as M1 times over A M2 tensor M3. How to prove such a thing, the easiest is to introduce such a triple tensor product as a universal object for trilinear maps. Introduce M1 times M2 times M3 as a universal object for trilinear maps. And then show that both parts are isomorphic to this pbject And, of course, one can devise many other things, but let me talk now about the base change. So, I have a ring A. Another ring B, which is an A-algebra. Also an extension of A as a ring. And let's have also M an A-module. And N and B-module. I can make N into A-module. N into an A-module just forgetting the multiplication by b's, by forgetting the B-module structure. So, I forget about the action of b's and only remember the action of a's, right? And the tensor product is a way to turn M into a B-module, well, not exactly M, we can tensor M with B over A, and this will be a B-module. You can make a B-module of M by replacing it by considering B tensor M over A. Indeed we can introduce the B-module structure on B tensor M over A by setting b times (b-prime tensor m) equal to b b-prime tensor m. This might look sophisticated, but you have certainly encountered some examples before. For instance, the complexification of a real vector space. If you have a complex vector space C^n, then you can make a real vector space R^2n of this just by forgetting the complex structure. If here you had basis e_1 ... e_n, then you just forget that you can multiply it by imaginary numbers and so, you obtain the basis e_1 ... e_n ie_1 ... ie_n. But then you forget all together about this i and you redefine them as v_1 ... v_n. Now, if you complexify, if you want to make a complex vector space out of R^2n. Now, can take C times R^2n. This will be C to the power 2n with basis of e_1 ... e_n, v_1 ... v_n. You just don't remember that those v_i's were ie_i's and you consider them as vectors independent over c. Well, more precisely, one should write, of course, 1 tensor e_1 ... ... 1 tensor e_n, 1 tensor v_1 ... ... 1 tensor v_n. Of course, you can also go in the other way. You can take R^n with basis e_1 ... e_n, then make it into a complex vector space by tensoring with C. So, now you have a C-basis 1 tensor e_i, and then you obtain R^2n by forgetting the complex structure. So, here we shall have our basis, which consists of 1 tensor e_i's and i tensor e_i's. So, in general, if M is a free module free A-module. with base e_1 ... e_n and I tensor it up with B, this will be a free B-module with base 1 tensor e_1 ... 1 tensor e_n. We also have a couple of important maps. We have maps of A-modules from M to B tensor M, which sends m to 1 tensor m and also I have a map in the other direction. So, this is of A-modules and equally of A-modules, but from N tensored with B over A to N itself. Sending b tensor n to bn. Sending b tensor n to bn. Well, recall N is a B-module, but the map is, of course, of A-modules. Of B-modules too, of course. The proof of this story about the basis is, of course, the same we just have seen before. So, the proof here is the same as that of Proposition 1. Again we construct certain bilinear maps and say that those factor over the tensor product and this implies that certain families are linearly independent. Now let me formulate a very important theorem. So, Theorem 1, Base change theorem. Notations are as before. A-homomorphisms between M and N are the same, well, in bijection but, of course you can also say that the corresponding groups of homomorphisms are isomorphic or whatever, so A-homomorphisms between M and N are the same as B-homomorphisms between our B-modules, so, B times M over A and N. How to prove such a thing? this is very easy. So, if I have a homomorphism from B times M to N, I can, of course, composite with my embedding map from M to B times M. Let me call it, say, alpha. So, in one direction, we have f goes to f times alpha. In the other direction, if I have... The other direction. If we have a map from M to N, g, then we can tensor it up with B. And obtain B times M to B times N. This is just identity times g. And then we compose it with the map we have defined on the last page. I don't know if we have given it a name. Let's call it mu. This is the map to B, which was sending b tensor n to bn. So, g goes to mu times identity tensor g. And then we check that those maps are mutually inverse.