 This lecture is part of an online course on Galois theory and will be about Galois extensions of fields. So what we're going to do is define Galois extensions, or rather we're going to define them in about five different ways and show that these are equivalents. So I suppose we've got an extension of fields, which we will assume as finite. Now we can define the Galois group, which we'll denote by g. So it's the Galois group of the extension m over k and this is equal to the automorphisms of m fixing all elements of k. You may think it might be more natural just to define the group to be the group of automorphisms of m and we sometimes do this. For instance, m might be the field of all algebraic numbers and then you're interested in its full automorphism group. However, as we've mentioned before, it's very very useful to be able to have a sort of relative notion where we just define a Galois group of an extension of fields instead of of a single field because then we can sort of build up from one field to another in a series of steps. So anyway, what we're going to do now is to write down five different conditions that the extension might satisfy and these conditions are as follows. First of all, we can say m is normal and separable over k, of course. So this is probably the most common definition of Galois. I'm not really so keen on it because it looks kind of rather artificial. I mean, why normal and separable rather than some other two random properties of field extensions? So this definition is sort of convenient but not very it's not very intuitive. The next equivalent condition says that the degree of the field extension is equal to the order of the Galois group G. So what's going on here? Well, we're going to show in a few minutes that for any field extension the order of G is always less than or equal to m over k. So what this says is that the symmetry group of m is as large as possible. So we can think of this as saying that Galois extensions are the ones where m is as symmetric as possible. So Galois extensions are the symmetric ones. The third equivalent condition says that k is equal to mg. Well, what on earth is mg? Well, this means the fixed points of m under the action of the group G. So in any case k is certainly contained in mg because we define g to be a group of things fixing m. This condition is very useful for examples. So in a few minutes we will see how to use this condition here to construct Galois extensions with any given finite group. The fourth condition says that m is a splitting field of a separable polynomial over k. And again, this condition is really useful because it can be used to give many examples. You can just take your favorite polynomial, which is separable and just take its splitting field and that will give you a Galois extension. So the fifth condition is in some sense the central theorem of Galois theory. What we can do is we can look at extensions of fields that lie between k and m and we can also look at groups that lie between g and 1. And the main theorem of Galois theory says that there's a one-to-one correspondence between these sub extensions and these subgroups. And the correspondence is given as follows. If we've got a subfield l, then we can look at the Galois group m over l, not l over k by the way, an easy mistake to make. So we look at all the symmetries of m that fix all elements of l. And similarly if we've got a subgroup h then we can just look at the set of elements of the field m that are fixed by h. Sorry, that's an h not a g. And the key point is that for Galois extensions these maps are inverses of each other. So we get a one-to-one correspondence between l and h. So any of these conditions here can be used as the definition of a Galois extension. That's for finite extensions. If you want to work with infinite Galois extensions, you need to be a little bit more careful. For instance, you can't use condition 2 because the degree will be infinite. And you can't use a single polynomial here because you usually need an infinite number of polynomials. Um So what we're going to do sooner or later is prove these conditions are equivalent. What we're going to do this lecture is prove the following implications And we're going to prove that 4 implies 1. Let's just notice that 4 implies 1 is completely trivial because we've shown that any splitting field is normal. So this case is trivial. Um The equivalents of 4 and 5 we will do later. Proving 5, it takes a little bit more work than the other implications. So we'll postpone it to a later lecture. Um, I should point out the biggest difficulty in Galois theory is that this correspondence here is actually order reversing. So if you change l to a bigger field, the corresponding group becomes smaller because there are elements that that fix all elements of the bigger fields. And similarly if you make the subgroup bigger, then um the the the corresponding fields of things fixed by it gets smaller because there are there are more elements. So it's it's harder for something to be fixed by it. And this is somehow a big psychological difficulty in dealing with Galois theory. Human brains are not very good with the fact that if coping with the fact that if something gets bigger, then the corresponding thing over there gets smaller. So that's a common source of errors. In particular one common error is to absolutely think that the field l corresponds to the Galois group of l over k rather than m over l. Um, so um before proving that let's first give the examples that we promised. So I said here that this condition here was was good for generating lots of examples. So let's see some examples of that. So if we've got a group g, find that group g actually on a field m, then um the extension of fields mg contained in m is always Galois. So let's use this to give some explicit examples of Galois extensions. So first of all, let's take m to be some field and we want it to be a field with lots of obvious automorphisms. Well, it's an easy way to do that. Let's just take it to be field of rational functions in five independent variables. And then this is acted on by the group s5 which just acts as permutations of x1 up to x5. And then we can take the fixed field mg which is if you've been to a standard algebra course which covers symmetric functions, you know that this fixed field is generated by the usual symmetric functions. So e1 is x1 plus x2 all the way up to x5. And e2 is x1, x2 plus all Proletes of pairs. And so on up to e5 which is x1, x2, x3, x4, x5. So this is a fairly explicit field and we know the Galois group of this field extension m over mg is now just the symmetric group s5. So we found a Galois extension with with group s5. And more generally let's just take any subgroup of s5. Then the Galois group of m over mh will be the group h. So any subgroup of the symmetric group of five elements is the Galois group of something. And of course there's actually nothing special about five. You can do this with any symmetric group. And we know by Cayley's theorem that any finite group is contained in some symmetric group. So we found for any finite group we found some Galois extension with that as its Galois group. It's a much much harder problem to find a Galois extension where you fix the base fields. So we can ask can we find an extension of the rational numbers with Galois group h. And this is the inverse problem in Galois theory and it's still unsolved in general. We can solve it for quite a lot of groups h but for instance Shafarevich did it for all solvable groups h but we don't know how to do it for all of them. So it's easy to find Galois extensions with a given Galois group as long as you're not too fussy about what your base field is. So next we're going to prove a useful lemma before going on to prove the equivalence as I stated in the in the first sheet. So here's the lemma. Suppose we've got a field k contained in m and we let g be the Galois group of m over k. Then the order of g is at most m over the degree of m over k. More generally suppose we've got a map from k to some field x and we don't really care what it is and we've got k contained in the field m and we can ask how many extensions are there of this map. So we want how many homomorphisms from m to x are that make this diagram commute and there are at most m over k homomorphisms no matter what this field x is. So if we take m equal to x then we get the result that the Galois group of m over k is at most this. So this result is a bit more general and it's sometimes useful and this is easy to prove because m is finite over k. So let's suppose it's generated by elements alpha 1, alpha 2, up to alpha n and what we're going to do is we're going to look at the chain of fields and so I'm going all the way up to m. So this is a typical example of why it's useful to have relative versions of results because we can prove things by hopping up one extension at a time and we want to ask how many homomorphisms are there from these fields to this field x and let's first of all look at this one. Well alpha 1 is going to be a root of p1 which is some irreducible polynomial of degree k alpha 1 over k. Now this must map alpha 1 to a root of p1 in x and once you've mapped alpha 1 to a root that determines the homomorphism. So we want to know how many roots of p1 are there in x and the number of these in x is at most the degree of p1. So the number of extensions here is at most this number here and now we can do the same thing for the next step. We can ask how many extensions from this of this map are there to this field here and by exactly the same argument the number of extensions is going to be at most the degree of these fields. So we take k alpha 1 alpha 2 over k alpha 1. So there are most this number of ways to extend the map to this field and for each of these extensions there are most this number of ways to extend the map to this field. So the total number of ways of mapping this field to here is at most the product of these two numbers which is the degree of k alpha 1 alpha 2 over k. Here we recall that m over k is equal to m over l times l over k for any field extensions k contained in l contained in m. So by continuing this argument we see the number of maps from m to x is at most the degree of m over k. So that proves the lemma. Next we can start proving the various implications at the beginning. So I said I was going to prove the implications 1 implies 2 implies 3 implies 4 and we'll start with 1 implies 2 and you have doubtless totally forgotten what conditions 1 and 2 were so I'll stop by reminding you. So condition 1 said normal plus separable and condition 2 said that m over k is equal to the order of g. So recall we've got k contained in m and g is the Galois group of m over k. And what we do is we've got k contained in m and we now embed this in some algebraic closure of k containing m. And now we're going to use the fact that m over k is separable. So this implies that there are exactly m over k extensions from well what we want we've got a map from k to the algebraic closure of k and we've got a map from k to m and we want to ask how many extensions are there from m to the algebraic closure. So m is contained in the algebraic closure but we're kind of forgetting this that the map from m to the algebraic closure doesn't have to be the identity on m. And the proof of this is almost the same as what we had on the previous sheet. What we do is we look at k contained in k alpha 1 contained in k alpha 2 and so on and we're mapping this to the field k bar the algebraic closure and we want to know how many extensions are there from k alpha 1 to k bar. Well we saw that are at most k alpha 1 over k extensions in general. But now we can see that there are actually exactly this number. So instead of having an inequality there we actually have equality and the reason for this is as follows. So we get equality because first of all alpha 1 is separable so p1 has k alpha 1 over k roots. If it wasn't separable it might have less than that number of roots. And secondly we observe that this equation here is algebraically closed so contains all these roots. So there are this number of roots of p1 and the algebraic closure contains that number of roots so we can map alpha 1 to any one of those roots so the number of homomorphisms is actually equal to this number here. And now of course we can just repeat this in all the next steps and we find that are exactly this number of extensions provided this is separable and this is algebraically closed. Now we use the fact that m over k is normal and this implies that any map any homomorphism from m to k bar extending the map from k to k bar maps m into m and this is because m is a splitting field of some polynomial p in k of x and m is generated by the roots of this polynomial. On the other hand all the roots of this polynomial must also be mapped into m because m contains all the roots so the homomorphism must actually map m to itself. So on the one hand we've got this number of homomorphisms from m to the algebraic closure on the other hand they must all have image in m so this means there are this number of homomorphisms from m to itself which is what we wanted to prove the the symmetries of m is equal to this number here. So that proves the implication one implies two. Now let's show that two implies three and as usually we better stop by remembering what two and three are so two was the condition that m over k is equal to g and three was the condition that k is the fixed points of m and of course we've got k can take is a subfield of m and g is the general group of m over k and now what we do is we look at k is contained in k so k is contained in mg which is contained in m and we notice that the extension m over mg has this number of homomorphisms. So if we look we now have the following inequalities so we've got the degree m over k is equal to g and this follows by assumption the assumption is up here this was our condition too and this is less than or equal to m over mg and this follows by the previous lemma the order of the Galois group is of an extension is at most the degree of the extension and finally this is less than or equal to m over k and this follows because k is contained in mg so we know that m over k is equal to m over mg times mg over k so it seems we're running out of room is that actually visible yes now you notice that these two terms on each side are the same so these inequalities must actually be equalities so we see that this is an equality so these two are the same so this is equal to one and let me write this out again because it seems to be a little bit illegible so this just says that mg over k is equal to one which is the same as saying mg equals k which is of course what we wanted to prove up here so that shows that implication that that that condition two implies condition three finally we're going to show that condition three implies condition four so condition three was the one that said mg equals k and condition four says that m is splitting field of a separable polynomial and as usual we've got k is contained in m and g is the Galois group of m over k and what we're going to do is let's pick any alpha in m and we look at all the conjugates of alpha under g so we've got alpha beta gamma and so on so beta might be g1 of alpha and gamma might be some other g2 of alpha and so on and now that there are two things we could do we could either take these with multiplicities or we could take just one copy of each and we're going to take just one copy of each so we're going to make sure these are all distinct and this isn't one of the cases where you where you have to repeat things with a multiplicity you know if there are several elements of g fixing alpha we still take only one copy of alpha and now let's look at the polynomial px which is x minus alpha x minus beta x minus gamma and so on and we observe first of all that p is separable because we said that alpha is not equal to beta gamma and so on these are all distinct so it's a separable polynomial secondly we notice that it's fixed by g because g just permutes its roots so it maps the polynomial to itself so the coefficients are in mg that's what being fixed by g means and this is equal to k because of this condition here so p is actually a polynomial in k of x so any alpha in m is a root of a separable polynomial in kx with all roots in m and this just means that m is the splitting field of a separable polynomial because we just take a polynomial p for a finite set of generators of the extension m and multiply all those polynomials together and that gives m as a splitting field okay so that has completed the proof of all the implications of the first four conditions we had at the beginning of the lecture what we're going to do in the next lecture is have a pause and give several examples to illustrate condition five and then in the next lecture after that we will prove that these four conditions are in imply condition five