Last time we have defined Galois extensions. So we have L over K. It is Galois if and only if it is separable and normal. This is the definition. Of course, it's the same thing as to say that L is a splitting field, of a family of separable irreducible polynomials over K. And we have seen this is not a triviality, this needs a proof. But still it is equivalent to, well, in the case when L is a finite extension of K this is equivalent to saying that the number of automorphisms of L over K is equal to the degree of L over K. I forgot to make a few remarks on normal extensions. These actually don't satisfy exactly the same properties as the other types of extensions we have seen before. So, remarks on normal extensions. So normal extension is just a splitting field. And we have seen for instance that an extension L over M over K was finite or algebraic or separable or purely inseparable if, and only if, it was true for L over M and M over K. So, for normal extensions this is not the case anymore. So let's have a tower of extensions K in L in M. Then if M is normal over K, then of course M is normal over L. This is clear. Since, if M is a splitting field of a family of polynomials over K, one can just consider them as being polynomials over L and say that M is a splitting field of a family of polynomials over L. But, of course, this is not true anymore that L is normal over K. L does not have to be normal over K, right? Because L can be just a stem field of Because L can be just a stem field of a polynomial of which M is a splitting field. And when they are not equal, of course, L is not normal by definition. For instance, if K is Q, then we can embed it into Q... with the fourth root of two adjoined, and this is embedded into Q of this and i adjoined. This is a splitting field of x^4-2, of course. And this is just a stem field of x^4-2, which is not a splitting field, so it is not normal. Not normal over Q. The polynomial x^4-2 has two roots in this extension, but it also has two roots outside of this extension. One more trivial remark is that quadratic extensions are always normal. In fact, a quadratic extension is a stem field of a quadratic polynomial. But you know how to solve quadratic equations and you'll see that in this case, the other root of this quadratic polynomial is also in this field. So this is just by the formula for roots of a quadratic equation. Or else in fact it's even easier if you have a quadratic polynomial, if P is quadratic over K and has one root in L, of course it also has another root in L because P shall be the product in L of x minus this root and something else and something else is up to a scalar multiple just x minus the other root. Okay? So maybe it is just useless to involve this formula, it's much simpler than that. And this makes it easy to construct a normal extension over normal extension which is not a normal extension. One often has K in L in M where L is normal over K and M is normal over L, but M is not normal over K. Well, if you take for instance, quadratic extensions, then in most of the cases, their composition will not be normal over the base field. For instance, we'll just take Q in Q of square root of 2 in Q of this fourth power root of 2. Both of those extensions are normal. Since those are just quadratic extensions. But Q of the fourth root of two is of course not normal over Q. So these were the remarks on normal extensions I forgot to make last time. What we have also seen last time... so L is a field If G is a group of automorphisms of L over something, over some K, over prime field, something like this. Then we can consider L^G, which is the fixed field. So, these are x from L such that gx is equal to x for any g in G. Okay? gx is equal to x for any g in G. Okay? On the other hand, if we have a subfield, Then we can consider the group of automorphisms Then we can consider the group of automorphisms automorphisms of L over K in the case when L is normal. because otherwise the group will be too small to give information about L. But in the normal case it makes sense to consider the group of automorphisms of L over K. So what we have seen we have seen that if L is separable over K then the fixed field by those automorphisms was just K. Because the group of automorphisms was permuting the roots of the minimal polynomial of x over K so, if it was fixing x, then it was meaning that x was the only root over its minimal polynomial, so this was meaning that x was purely inseparable over K. We also have seen, that if G was finite, then L was Galois over L^G and the degree of L over L^G was equal to the cardinality of G. And now we are going to summarize all these in a theorem which is in fact the main subject of this lecture course and this theorem is called the Galois correspondence. But let us interrupt just before stating the theorem.