 This lecture is part of an online graduate course on Galois theory and will be about normal extensions of fields. So what we have is an extension of fields, which we will assume to be algebraic throughout. And we want to consider the following properties. Suppose a polynomial p with coefficients in k is is irreducible and has a root in l. And then we can ask are all the roots of p in l? Whatever that means. I mean, it's not quite clear what we mean by root of p in general. But informally, the question is if an irreducible polynomial has one root in l, are all its roots in l? So let's start by looking at a couple of examples. We will use the example we used several times before of the rationals contained in the rationals with the cube root of 2 adjoined. And this extension does not have the property above. So if we take p to be x cubed minus 2, cube root of 2 contains one root of p, but not the others. Because the others of the cube root of 2 times a complex cube root of unity, which certainly doesn't lie inside this field. On the other hand, we had the example earlier of 8 alpha cubed plus 4 alpha squared minus 4 alpha minus 1. This is the polynomial of p of alpha. And if this is equal to zero, then if we look at the extension q contains q of alpha, you remember that alpha turned out to be cosine of 2 pi over 7. And the other roots were 2 alpha squared minus 1 and 2 times 2 alpha squared minus 1 squared minus 1. And so all three roots of this polynomial lie over this field here. So this is the question we want to discuss with normal extensions. If a polynomial has one root, are its other roots in there? So there are actually three equivalent conditions for an extension to be normal. So the sort of theorem. The following conditions are equivalent for an algebraic extension k contained in l. So the first property is that any polynomial p in k of x that is irreducible and has a root in l factors into linear factors in l of x. So this is a sort of precise way of saying that all roots of p lie in l. I mean, it's better to say factors into linear factors because it's not quite clear what we mean by the roots of p if it doesn't actually have roots in l. The second condition says that l is the splitting field over k of some set of polynomials. Well, in the electron splitting fields, we only really define splitting field for one polynomial. But later on, for algebraically closed fields, we pointed out that you could actually define splitting fields for other sets of polynomials. So we're going to allow any set of polynomials. If the extension is finite, then you can just use one polynomial here. It doesn't really matter. And the third condition says that let's extend l to an algebraic closure of k containing l. So the third condition says that any automorphism of k fixing all elements of the original field k maps l to l. So it sort of says l is fixed under automorphisms of the algebraic closure of k, whereby an automorphism of the algebraic closure, we mean it must actually fix all elements of k. So the automorphisms of the algebraic closure are more or less the absolute Galois group of the field k, as we will see later. And all of these three conditions are fairly easy to prove. So let's first prove that one implies two. So condition one says that p has a root implies all roots in l. And condition two says that it's a splitting field k contains in l. And one implies two is easy, because all we do is we take an irreducible polynomial p alpha for each vector element alpha in l. And then it's easy to check that l is the splitting field of this set of polynomials. The condition, it must be a splitting field because each irreducible polynomial by assumption splits into linear factors in l. That's what condition one says. So it is automatically a splitting field for this. Now we show condition two implies three. So condition two says that l is a splitting field. And condition three says that any automorphism sigma of the algebraic closure of k over k maps l to l. And this is easy because the automorphism sigma maps the set of polynomials p alpha. So l is the splitting field of these polynomials p alpha to itself. And this is because sigma acts trivially on k and so just fixes all these polynomials. And l is the splitting field. So l is the set of roots of all the polynomials p alpha in k bar. So since the polynomials p alpha are fixed by sigma, l is fixed by sigma. So condition two implies condition three is also very easy. Finally we have condition three implies condition one. And this is the condition that's a little bit tricky to prove. So three says that any automorphism sigma, let's say it's an automorphism of k bar over k. So this says that any automorphism sigma fixes l and condition one says that if p has a root alpha in l, then all roots of p are in l. And here we were taking k contained in l contained in an algebraic closure of k. So let's see why this is true. Well, we observe for any root beta of p in k bar, there is an automorphism sigma of k bar taking alpha to beta. And we can see that quite easily because we've got k, which is k, and this is contained in k of alpha, which is contained in k bar, and this is contained in k of beta, which is contained in k bar. And since alpha is a root of an irreducible polynomial, k of alpha is actually isomorphic to k of beta. These two fields here are both just isomorphic to k of x over p of x, whatever where p is the irreducible polynomial. So we have an automorphism of k of alpha to k of beta. And now since k bar is the splitting field of this field, it's the splitting field of all polynomials over this field here. So by uniqueness of splitting field, we can extend to an automorphism from k bar to k bar, mapping that one to that one. So we've got an automorphism of the algebraic closure mapping alpha to beta. So we have sigma alpha equals beta. On the other hand, by assumption, we know that sigma of L is equal to L. And we have alpha is in L by assumption. So beta, which is sigma of alpha, is in sigma of L, which is equal to L. So beta, the other root, is also in the field L. So this finishes verifying that the three properties we had earlier are equivalent. And any of these three conditions can be used as the definition of a normal extension. So we say that an extension with one of these properties is a normal extension, and different authors have different views on which of these three properties is the best definition of normal, but it really doesn't matter because they're all equivalent. So we had a couple of examples earlier. We saw that q contained in q of the cube root of two was not normal. On the other hand, if we extend q, the cube root of two, and then extend this further to the cube root of two together with omega, where omega is a cube root of one, this extension here is a normal extension. And that's because it's the splitting field of x cubed minus two. And in fact, the usual way of producing normal fields is just to write down a random polynomial and take the splitting field. So normal extensions are easy, easy to construct. Also, we can see that if k contains an L as degree two, it is normal. And this is pretty easy to see because if L is k of alpha, then alpha satisfies alpha squared plus b alpha plus c equals naught for some b and c. But then the roots alpha and beta of this polynomial satisfy alpha plus beta is minus b. So the other root beta is equal to minus b minus alpha, and therefore it's in L. So L is automatically a splitting field and therefore, and therefore a normal extension. So finish by discussing the following questions. Suppose k is contained in L is contained in M. And suppose this is normal. And suppose this is normal. And then we can ask, is k contained in M normal? And the answer is no in general. And this may be a bit surprising because there seems to be a rather obvious proof that this is normal. So first give us sort of fake proof that k contains in M is normal. And then I'll explain what's wrong with it. So we want to show that sigma M contained in M, where sigma is an automorphism of the algebraic closure over k. So this is all contained over k bar. So let's look at what sigma does. Well, first of all, sigma maps L to itself because L over k is normal. But then we look at the action of sigma on M over L, and we see that sigma must then map M into itself because L contained in M is normal. Therefore, sigma maps M to itself. Therefore, M is normal. Well, this proof is wrong. And let's see a counter example. A counter example is just given by q contains in the square root of sorry, q with the square root of two adjoined. And then we have q with the fourth root of two adjoined. And as we saw earlier, this extension here has degree two and this extension here has degree two. And all degree two extensions are normal. So these are both normal. So this extension is normal. And this extension here is also normal. But the extension q contained in the fourth root of two is not normal. And the reason for this is that an x to the four minus two has four roots, which are the fourth root of two minus the fourth root of two. But then it's also got the fourth root of two times i and the fourth root of two times minus i. And these two elements here are none real. And so they obviously not in the field generated by the fourth root of two. So what's gone wrong? Well, let's look at the element sigma. So sigma, suppose sigma is an automorphism of the algebraic closure. Well, sigma certainly maps q root two to q root two. And then we said that anything that then since this extension is normal, we said that any automorphism fixing this field must also map this field to itself. Well, now there's a sort of ambiguity about what we mean by the automorphism fixing this field. So when we say sigma fixes q root two, do we mean, well, we could just mean sigma of q root two is contained in q root two, or we could mean that sigma alpha is equal to alpha for all alpha in the q time with the square root of two adjoined. So when you say an automorphism fixes a field, it's a little ambiguous. And this ambiguity is actually what the error was in the supposed proof I gave, because we have an automorphism, we can have an automorphism of the algebraic closure of k bar that acts non-trivially on this field here. I can map the square root of two to minus the square root of two. And therefore we can't deduce that it maps this field to itself, because this is a normal extension. So we can only deduce that an automorphism that fixes every element of this field acts true, maps this field to itself. An automorphism that fixes this field in the sense of just mapping the field to itself doesn't necessarily map this field here to itself. So a normal extension of a normal extension need not be normal. By the way, as we will see fairly soon, the term normal for normal extension is very closely related to the term normal for normal subgroup in group theory. In fact, the term normal for normal subgroup comes from the term normal in field theory. And we will see that when we do Galois theory, we show that intermediate extensions of a field M turn out to be normal, if and only if they correspond to normal subgroups of the Galois group. Okay, that's covered normal extensions. As I said, that there are two properties of Galois extensions we want to discuss which are normal extensions and separable extensions. And in the next lecture, we will discuss separable extensions. And after that, we will put them together and discuss Galois extensions.