[MUSIC] So, these were the genuine properties of composite extensions. Now, let me give you a definition of linearly disjoint extensions. Maybe it's best to state it as a definition, which is a theorem at the same time. Definition, theorem, the following properties are equivalent. A, The tensor product is a field. Of course, equivalent for algebraic extensions. Secondly, j is injective. C, if we have x1 and so on, xn, in l1, which are linearly independent over k, then they are also linearly independent over l2. D, you have two families, x1 and so on xn and l1 and y1 and so on, ym. In l2, if the two families are linearly independent over k, then the products xi yj are linearly independent over k. And in the case of finite extensions, When L1 is finite over k, this is also equivalent, all this is equivalent to the fact that L1, L2, degree over L2 is equal to the degree of L1 over k. And so on. The degree of L1, L2 over k is just the product of the degree of L1 and L2 over k. In other words. L1, L2 degree over k, should be just for the product. Okay, in this case, We say that L1 and L2 are linearly disjoint. Linearly disjoint over k, so let us prove the theorem. A equivalent to B is clear. Since we have L1, L2 which is just the image of j. And so, a and b both say that L1 turns L2 over k is equal to L1, L2. Then b implies c. We have x1 times 1, and so on. Xn times 1. These are linearly independent over L2. By the base change property. Now, if j is injective. Their images, X1 and so on, xn are also linearly independent over L2. An injective map sense, a linearly independent family into a linearly independent family. C implies d. Now, this is very easy. If we have some relation. Aij, xi, yj over k. Then, we have the sum of aij yj over j, which should be equal to 0. Since the xi are linearly independent over k. But now, aij also must be 0 since yi are linearly independent. Yj are linearly independent. And now, finally, we have to show that d implies b. Remember that b was the injectivity of j. So, let's take that in the tensor product, and suppose that j of z is 0. Remember, that z by definition is the sum of aij, the xy tensor yj, okay? And j of z is just the sum of aij xi yj. So, if j of z is 0 then all aij are 0. And so, z is 0. [COUGH] So, this theorem is proved, well the part about finite extension and degrees is pretty obvious. Now, the part involving degrees. Easy to follow from our four properties. So, let me give you a few examples. First of all, the extensions which have relatively prime degrees are always linearly disjoint. If L1 over k is of degree m, and L2 over k is of degree n, then obviously, well, m, n are relatively prime, then obviously. L1, L2 are linearly, Disjoint. Indeed, m and n must divide the degree of their composite. And the degree of the composite is, At most mn. So, this degree is equal to mn. And this is exactly one of the definitions of being linearly disjoint that you multiply the degrees. Okay, so in particular. If you have say Q of the fifth root of 2 and Q of the fifth root of unity, they are linearly disjoint. The degrees are 5 and 4. The degrees are 5 and 4. This is logically 5 and a Q of the fifth root of unity is a cyclotomic extension it is of degree 4 is what we have. Already central d. On the other hand, Q of fifth root of 2 and Q of the, how to denote it? Well, let's see. Let's denote by the fifth root of one exactly the exponential of 2 pi i over 5, so if we take this e to the power of 2 pi i over 5 times the fifth root of 2, these guys are not linearly disjoint. Not linearly disjoint. Since both degrees are equal to 5, but the composite of the spliton field of x to the power of 5- 2. And this is of degree 20. Indeed, in both cases, L1, L2 is at the spliton field of x to the power of 5- 2. Since in the first case, we are linearly disjoint, the degree of L1, L2 over Q is equal to 20. In the second case, Both L1 and L2 are of degree 5 over Q. And 5 times 5 is not equal to 20. Let's say that L1 and L2 are not quite independent over Q, in the second example, whereas they are in the first. So, you see that the difference is rather subtle. Well, the obvious reason in the second case is that those extensions are generated by rules of the same polynomial, but, Still some effort is needed to formalize why this is not linearly disjoint case. In particular, we see, That it is not sufficient, in order to be linearly disjoint to have a very small intersection. So, L1 intersected L2 is Q and does not imply that L1 and L2 are linearly disjoint over Q. This is exactly what happens in the second example. So, this is what happens in the second case. The degree is our prime, so we cannot have any sub extensions. [SOUND]