Algebraic extensions, satisfy a similar property to that of finite extensions. If you have a tower of algebraic extensions, then it is algebraic if and only if each floor of this tower is algebraic. So, let me prove Theorem 3. Well, again let L be an extension K, and M an extension of L. Then M is algebraic over K if and only if M is algebraic over L and L is algebraic over K. So, one direction: let alpha be in M. Of course, if it satisfies a polynomial relation with coefficients from K, if P(alpha) is 0, for some P in K[x] then also, this P is in L[x]. So alpha is algebraic over L. And if now alpha is an L, and we want to prove that it's algebraic over K, then it's even easier. Then also alpha is in M, and so alpha is algebraic over K. And so L is algebraic over K. Another direction. So, I have L, which is algebraic over K and M algebraic over L. I must prove that M is algebraic over K. So, take alpha in M and consider the minimal polynomial of alpha over L. It's coefficients are elements of L, so they are algebraic over K. By the previous theorem, they generate an extension, E which is finite over K. Now, E(alpha) is also finite over K. Now, E(alpha) is also finite over K. E(alpha) is also finite over K. Since E(alpha) is finite Is finite over E, alpha is algebraic over E. And this means that alpha is algebraic over K because since E(alpha) is finite over K, then there will be some linear dependence between the powers of alpha. There is a relation between powers of alpha. Okay, let me give you an example. So consider the extension obtained from Q by adjoining, let's say, the cubic root of 2 and or the square root of 3. This is clearly algebraic and finite over Q. And what is it's degree? The degree of this extension is 6, indeed, we have Q inside Q of the cubic root of 2 inside Q of the cubic root of 2 and square root of 3. The minimal polynomial of the cubic root of 2 over Q is x cubed minus 2. This is an irreducible polynomial over Q, which has the cubic root of 2 as a root. And so, Q(cubic root of 2) is generated over Q by linearly independent elements. 1, cubic root of 2, and cubic root of 2 squared. So, the degree of Q with cubic root of 2 adjoined over Q is equal to three. Now, the square root of 3 does not belong to Q with cubic root of 2 adjoined. Well, maybe the easiest way to see it is as follows: because otherwise, one would have Q embedded in Q of square root of three embedded in this Q with the cubic root of two adjoined. And the degree of this extension is clearly two. And the degree of this extension is three, well 2, the degree of Q with square root of three over Q would divide 3, which is Q of cubic root of 2 degree over Q. We have seen that the degrees multiply in towers of finite extensions. So this is impossible. Therefore x square minus 3 is irreducible over our extension, and is the minimal polynomial of cubic root of 3 over this extension. And the degree of our big extension over Q cubic root of 2 is equal to 2. Therefore the degree of Q cubic root of 2 square root of 3 over Q is 2 by 3 is 6. Well, I should have said this earlier. This fact is completely general that the degree of an extension generated by an algebraic element alpha over K is equal to the degree of the minimal polynomial of alpha. So in general, Proposition: the degree of K(alpha) over K the degree of the minimal polynomial of alpha over K, if alpha is algebraic. I should have said this earlier, that the proof is obvious. Since we have already remarked several times that K(alpha) is generated by 1, alpha, ..., alpha^(d-1) if this is dim and in fact this is a basis, independent. Well in any case this is a good practical tool to compute the degree of finite extension. And I would like to finish with the little remark which might be helpful to understand the next lecture. So a proposition: let K and L be a field extension. So consider L prime, which is a subset of L, formed by all elements algebraic over K. Then L prime is a subfield of L. One calls it the algebraic closure of K in L. Well, the proof is, of course, easy. If alpha and beta are algebraic over K, we have to prove that alpha plus beta and alpha beta are algebraic. But this is trivial by theorem, I guess, 2. Follows from theorem 2, since alpha plus beta and alpha beta belong to K[alpha, beta] which is a finite, by Theorem 2, extension Of K.