So let me give you an example of such a base change which deserves the name of a proposition. I think it is Proposition 2. Let's take I, an ideal in A, so my ring B, the A-algebra, will be A/I. And I am going to base change M to an A/I-module, then I affirm that this is just M over IM. So IM is a submodule of M. I take the quotient module, and I affirm this is the same as the base change of M to A/I. Well, how it's done? I can define in one direction the map from M to A/I times M. This is my old map alpha, right? m goes to 1 times m. Then I remark that this sends IM to 0. Because if I have im, where i is an element of I, this will be 1 times im, but now my tensor product is over A. Everything is A-linear, so I can put i into the other side. So this is i times m, and this is 0 times m, because now I am in A/I so this is 0. So this factors alpha induces a map from the quotient, say alpha-bar to A/I tensor M. Now in the other direction we apply the "Base change theorem". Apply base change. We have the projection, consider the projection from M to M/IM. This is a map of A-modules. And this is B-module. Well, this gives then B = A/I times M to M/IM, a map of B-modules. And one checks again that this is the inverse of alpha-bar. Okay, so let me give you some examples. Let's take Z/2Z and tensor it over Z with Z/3Z. What do we obtain? So we may consider it as a base change of Z/3Z to Z/2Z. So this is Z/3Z quotient over the ideal generated by 2 times Z/3Z, but 2 is invertible modulo 3. This is just -1 in fact, but 2 times Z/3Z is just equal to Z/3Z. (2) is not a proper ideal in Z/3Z, it's equal to the whole ring, so the quotient is 0. Another obvious example is as follows: if I base change a polynomial ring, I obtain again a polynomial ring only over B. Is B[x]. If I base change a quotient, this is a very interesting example, if I base change A[x]/(P), what do I obtain? I obtain B[x] modulo again the ideal generated by P, but in B[x]. So this is now ideal generated by P in B[x]. Okay, so finally, let me tell you a few words about tensor product of A-algebras. So, let me fix two A-algebras B and C are A-algebras. and alpha from A to B and beta from A to C and alpha from A to B and beta from A to C our structure maps, which define the A-algebra structure on B and C, then I can introduce a new A-algebra B tensor C as follows: this is ring, this tensor product is a ring with respect to the following operation. Is the ring with respect to b tensor c times b-prime tensor c-prime equal to b b-prime tensor c c-prime. In fact, this has the following universal property. I shall be very brief about this, so I have this A, B, C. And B tensor C over A. So I have alpha, beta, these maps, let me denote them by phi and psi. So phi sends b to b times 1, and psi sends c to 1 times c, then, for any A-algebra D and psi sends c to 1 times c, then, for any A-algebra D A-algebra D, one has Hom from A, Hom over A, sorry, from B times C to D is in bijection with Hom over A from B to D times Hom over A from C to D. If I have some homomorphism, say h from B times C to D, this is the same as giving two homomorphisms, let's say f and g. So giving h is the same thing as giving f and g such that all maps in this diagonal commute. The diagram commutes. So if we have h, of course, we can define f and g by what? by h times phi and h times psi, and conversely, given f and g, can define h(b times c) as f(b) times g(c). Okay, but I shall not say any more on this, this is just for the general culture. The main point for us is that the tensor product of the A-algebras is itself an A-algebra by this very simple rule, componentwise multiplication, and let me give you an example. So, consider C tensor C over R. How to describe the ring structure? Well, what kind of ring is this? I mean formally you have the formula, of course, but what does it mean? Let's say, this is C times R[x]/(x^2+1), right? This is the same as C[x]/(x^2+1), as I have told you a few minutes ago. This is the same as C[x]/(x^2+1), as I have told you a few minutes ago. And now this is isomorphic by Chinese remainder theorem, which we will recall in a couple of minutes, this is isomorphic to C[x]/(x+i) times C[x]/(x-i) this is isomorphic to C[x]/(x+i) times C[x]/(x-i) this is isomorphic to C[x]/(x+i) times C[x]/(x-i) which is, of course, isomorphic to C x C. So you see for instance, that this ring is not a field. C is a field but this tensor square of C over R is not a field, has zero divisors, and so not a field. So how to identify a zero divisor? This can be done, for instance, as follows: Here, one sees that the class of x+i is a zero divisor. And then, of course, this x+i under this identification is represented as 1 times the class of x + i times the class of 1. So, in C times C over R, this will be just 1 times i + i times 1.