[SOUND] [MUSIC] Let me also give you, A couple of other properties of norms and traces. First of all, consider, The map from the square of E to K, which sends (x, y) to the trace of the product. Then this is a non-degenerate, K bilinear form. Okay, bilinear form is also what we have to prove is that this is non-degenerate. So, indeed. If, x is in the kernel of this, well, let me call it T. This means that the trace, Of xy is 0 for any y in E, but this cannot be the case, When xy is a non-zero element of K. By definition, the trace is the sum of the quandary gate. So of the, Trace of xy, Is then, The degree of the fluid extension times xy. In this situation, when xy is in k. And finally, if alpha is integral over Zed, And K = Q. Then the norm of the trace with respect to this extension. E / Q. That one does a normal, so with the respect to the extension, E / Q are integers. Well, how do we see this? Now we know that the trace, for instance, of E / Q of alpha is equal to the trace of Q of alpha over Q1 of the trace with respect to K / Q of alpha, of alpha. Now this is just some integer times alpha. So the degree of the extension times alpha. So this is this integer times the trace of alpha with respect to this extension of Q alpha / Q. And about this last trace, we know that this is a coefficient of minimal polynomial which is integral. So this E and Zed, Since this last trace, Is a coefficient up to sine of the minimal polynomial. Whose coefficients are integers as we have seen. Why, Are they called norms and traces? Why such names? The source is as follows. So consider fa from E to E which is the multiplication by a. Then, The trace of a is exactly straight. And the norm of a, is as determinate. Now this fa is a linear map, a k-linear map. It's an endomorphism of a vector space you are working, and the trace of a is the trace of this endomorphism, and the norm of a is the determinant of this endomorphism. Of course, I should always write with respect to which field extension. Do I take the trace and the norm of ly and leave the proof as an exercise. So now, let us finish the proof of our theorem. So end of proof of theorem one. So we have to prove, we have to show, That Ok, the ring of integers of a number field is contained in a free zed submodule of rank N where N is the degree of the extension. We have already seen that there is a basis, e1, and so on, eN, so let e1 and so on, eN be a Q-basis of K contained in Ok. Then use this trace which gives you a non-degenerate bilinear form. So consider (x,y) goes to the trace of, with respect to Q / K of the product. This an non-degenerate bilinear form, And on K. It follows that the residual basis, V1 and so on, vN dual basis. So these are elements of K. And we have the property that the trace, With respect to our extension of (ei, vj) = delta ij. So this is 0 when i is different from j, and 1 when i = j. So I claim that the Zed submodule, Generated by those VI contains, Ok. Indeed take some alpha in Ok. Write alpha equal to the sum of alpha i, vi where alpha i are Q, you can always do that because the vi is a Q-basis of k. But in fact, one can easily see that the ai, r and zed because the ai are exactly the traces with respect to our field extension of alpha ei. By the very definition of the vj's. And alpha ei is an element of Ok. Since alpha and ei are elements of Ok, so is alpha ei, then it follows that the trace of alpha ei, with respect to our extension, is an integer, and since this is exactly alpha i, this is what we had to prove. We have expressed any element of Ok as a combination of vi with integral coefficients. So Ok is contained in a submodule, Zed submodule, generated by vi. [SOUND] [MUSIC]