Let me generalize this property to the case of fields which are not necessarily given as K(alpha). So now let me take an arbitrary finite extension of K. And let me define the separable degree a separable degree as a number of possible homomorphisms of L into algebraic closure of K. Again, if L is generated by one element alpha, then this degree is just the number of distinct roots of its minimal polynomial, if L is K(alpha) then this is just the number of distant root of it's minimal polynomial. And let me say that the extension L is separable over K if the separable degree is equal to the degree. the separable degree is equal to the degree. And I can also define the degree of inseparability of L over K, as the quotient the degree divided by L over K separable. But this won't be much very important on the sequel. So let me formulate a theorem. First part sais that the separable degree is multiplicative. If I have L an extension of K, and M an extension of L, then M over K separable is equal to M over L separable times L over K separable. And M is separable over K if and only if M is separable over L and L is separable over K. So this is part one, part two says that the following things are equivalent: 1) L is separable over K 2) any element of L is separable over K 3) L is generated over K by a finite number of separable elements. 3) L is generated over K by a finite number of separable elements. And 4) L is generated over K by a finite number of elements and each alpha is separable over the preceding stuff K (alpha_1, ..., alpha_(i-1) ). the preceding stuff K (alpha_1, ..., alpha_(i-1) ). Remark: the same holds when we replace separability by pure inseparability. Okay, so let me give a proof of this theorem. Well, part one. We know that any phi from L to K bar extends to phi tilde from M to K bar, this is the extension theorem. And in fact, there are exactly M over L separable ways to do this. Since K bar is also L bar, since given phi one considers K bar as L bar. An algebraic closure of K is also an algebraic closure of L once an embedding of L into K bar is given. Then the number of ways to extend is computed just by definition, by definition it's the separable degree of M over L. This implies the first part, thus we have M over K separable thus we have M over K separable equal to a product of L over K separable M over L separable. Now the equivalence of separability of M to the separability of M over L and of L over K, so equivalence of separability. This is just the inequality, in fact. Just the fact that Just the fact that the separable degree of an extension over K does not exceed the true degree, okay? So if the separable degree and the degree are both multiplicative, and if one of them does not exceed another than we can conclude everything. And this fact, now this is proved the last fact is proved by induction using the fact of that this is true for E generated by one element. Well, part two, 1) implies 2). So 1) is L is separable over K and 2) is any element of L is separable over K. This is the consequence of the first part, the first part implies that any subextension of a separable extension is separable. K(alpha) over separable extension L is itself separable. 2) implies 3). So if any element is separable over K, then L is generated by separable elements. It's clear. If any element is separable then of course, the generators are also separable. 3) implies 4). Well, if ai is separable over K then a_i is also separable over an extension of K. This is clear because the minimal polynomial of a_i over K of a_1, ..., a_(i-1) divides of a_i over K of a_1, ..., a_(i-1) divides the minimal polynomial of a_i over K and we have seen these in the last lecture. And so if the big one is separable, that is to say has distinct roots so then the small one is also separable. If P_min of a_i over K is separable, which means that it has distinct roots, then so is it's divisor, P_min of a_i over K(a_1, ..., a_(i-1) ) And finally 4) implies 1), well, this can be proved by induction as above, I shall not give more details. So now one might ask, is the notion of separability defined for extensions which are not necessarily finite? Yes, in this case it is best to define a separable extension as such extension that all its elements are separable. In particular, if L over K is not necessarily finite and algebraic extension, we can define L separable, the separable closure of K in L, as a set of all x such that x is separable over K. A separable element is by definition algebraic, of course it has minimal polynomial. So the preceding theorem implies that this L separable is a subfield, L separable is a subextension called separable closure of K in L. called separable closure of K in L. Well, I think normally one presumes that L is algebraic over K and of course, L is purely inseparable, over L separable. Well and finally maybe I will repeat the important remark: if the characteristic of K is 0, then any extension is separable. And if characteristic of K is p, then a purely inseparable extension has degree p^r, and always this degree of inseparability is p^r.