Let me now define what is a finite extension. L is said to be a finite extension of K, if it is a finite dimensional K-vector space. Well, the dimension of L over K is called the degree of the extension. Is called degree of L over K. And it's denoted as follows: [L:K]. Let me prove a theorem. Suppose L is an extension of K and M is an extension of L, then M is finite over K if and only if M is finite over L and L is finite over K. Moreover, in this case the degrees multiply. And the degree of M over K is a product of the degree of M over L and the degree of L over K. Let me prove this theorem for you. One direction: suppose M is finite over K. Well, any family of elements of M which are linearly independent over L of course is linearly independent over K. Any m_1, and so on, m_n which are independent over L are also linearly independent over K. This is obvious because what is linear independence of this is non-existence of linear combinations, of linear combinations with coefficients from L or from K, which is zero. Of course, if there is no such combination with L coefficients a fortiori, there is no such combination with K coefficients because K is a subfield of L, it's a part of L. So, thus, the dimension of M over L is finite because you cannot have a linear independent family of more than dimension of M over K elements. Well, now, on the other hand, L is a K-vector subspace of M. So if M is finite-dimensional over K, then L is also finite dimensional over K since the subspace of a finite dimensional vector space is also finite dimension. So one direction is easy. The other direction is also easy, but one must make a little computation. So let e_1, ..., e_n be an L-basis of M and let epsilon_1, ..., epsilon_d be a K-basis of L. of M and let epsilon_1, ..., epsilon_d be a K-basis of L. So, let us prove that e_i epsilon_j form a K-basis of M. Well indeed, any x from M is a linear combination of e_i with L coefficients. x is a sum of a_i e_i with a_i in L. Well, each a_i is also a linear combination. Let's say, sum of b_ij of epsilon_j, Where the b_ij are from K this time. So we can write x as a sum by over i and j over b_ij epsilon_j e_i, and so on. That is to say that epsilon_j e_i, which of course is the same thing as e_i epsilon_j, generate generate M over K. And we only have to check that these are linearly independent over K. So, on the other hand, if we have some linear combination, some linear combination over i and j: c_ij e_i epsilon_j, if this should be 0, then of course we can recompose the terms. Then we have the sum over i the sum over j c_ij epsilon_j e_i is 0. But e_i, so these coefficients are in L, right, but e_i form a basis, so for any i the sum of с_ij epsilon_j is 0. But this means, as epsilon_j form a basis that с_ij is 0 for all i and j. So Theorem 1 is proved. Now I would like to prove a second theorem. Let me introduce one more notation which is quite obvious. I will call K(alpha_1, ..., alpha_n) in L K(alpha_1, ..., alpha_n) in L the smallest subfield of L containing K and the alpha_i. I will also often say that this is generated by alpha_1, ..., alpha_n. Is geberated by alpha_1, ..., alpha_n over K. This is the smallest subfield, containing K and alpha_i. containing K and alpha_i. So theorem two. L is finite over K if and only if L is generated by a finite number of algebraic elements over K. Well again, one direction is obvious. If L is a finite dimensional K-vector space, then we can take a basis. Let alpha_1, ..., alpha_d be a K-basis of L, then L is certainly equal to the smallest subring of L containing alpha_1,..., alpha_d, which is a field, so it is also a smallest subfield of L containing alpha_1, ..., alpha_d. Now, all alpha_i are algebraic, this is just because, well moreover, each K[alpha_i] is a finite dimensional K-algebra, since it is a subring of L which is already finite-dimensional. So then by Proposition 1, alpha_i is algebraic. Well, in the other direction it is also not difficult. We have K[alpha_1] is finite dimensional over K, K[alpha_1, alpha_2] is finite dimensional over K[alpha_1], and so on. K[alpha_1, ..., alpha_d] is finite dimensional over K[alpha_1, ..., alpha_(d-1)], and so on. All of those elements are algebraic so all of those guys are fields K[alpha_1, ..., alpha_i] is a field so K[alpha_1, ..., alpha_i] is a field so it's equal to K(alpha_1, ..., alpha_i). And now, just use Theorem 1 to conclude that L which is K(alpha_1, ..., alpha_d) is finite over K.