 This lecture is part of an online course on Galois theory, and we will be proving the fundamental theorem of Galois theory. So let's just recall what this says. It says that if k contained in m is a Galois extension, we're going to assume as usual that these extensions are finite, then there's a one-to-one correspondence between intermediate fields, l, and subgroups of the Galois group. So g is just the Galois group of m over k. And we recall this correspondence is given as follows. So a subfield corresponds to the Galois group of m over l, so that's automorphisms of m fixing l, and not as you might guess, automorphisms of l. And on the other hand, a subgroup corresponds to the fixed field. That's the set of elements of m fixed by h. And what we want to do, the fundamental theorem says that these maps are inverses of each other. So in other words, if we take a field l, and we go to the corresponding group, and then we go back to the fixed field of this, so we take the fixed field of m of Galois of m over l, this should be equal to l. Well, first of all, it's obvious that l is contained in that. On the other hand, if we start with a subgroup h, then we go to the fixed field mh, and we now want to go to the Galois group of m over mh. And h is obviously contained in this, and what we want to show is that these two are actually equal. So we want to say, are these equal? If we can show that, we will have proved the fundamental theorem. Well, in order to show they're equal, all we need to do is to show they're the same size. Because if we've got a group contained in another group, and they're the same size, then they have to be equal. Well, what do we mean by the size? Well, the size of h, we just mean the order of h. And by the size of l, you might think we're going to mean the index of k in l, but we're actually going to take it to be the index of l in m. And this is a little bit funny because it means if l becomes bigger, then its size becomes smaller. And the reason we've done it is so the size of l will turn out to be the same as the size of the corresponding group. I mean, it seems more natural to say the size of l is the index of k in l, but then we would find that the size of a group didn't correspond to the size of the field. Now, if we can show that the size of a field is the same as the size of the group, and the size of the group is the same as the size of the field there, then this will show that these are the same size and these are the same size so they're equal. So to summarize, what we have to show to prove the fundamental theorem is we need to show two equalities. First of all, if we go from h to mh, we need to show the order of h is equal to the index of mh in m. And the second thing we need to do is to show that if we go from l to the Galois group of m over l, we need to show that the size of l is equal to the order of the Galois group of m over l. So let's just summarize in green, these are the two things we have to prove. And if we can prove these two inequalities, we'll prove the fundamental theorem. Now, this one is easy to prove because we know m over mh is Galois, because we showed earlier that if you take the fixed points of a field m under a group h, then that's a Galois extension. And since it's Galois, this implies that h must actually be equal to the order. So this is automatically true. So this is true. So the problem is to prove this. Incidentally, you notice so far we have not used the fact that the extension is Galois. So we haven't used the fact that m over k is Galois. So this is not used yet. However, we need to use the fact that m over k is Galois in order to prove this inequality here. So in order to show this, what we do is we look at k contained l contained m, and we look at the Galois group of m over k. So this is going to be all maps from m to itself that fix all elements of k. So the order of this is going to be equal to the sum over maps from l to m extending the map from k. And for each of these, we have the number of extensions from the image of l to m. In other words, to map m to itself, we first map l to m and then we have to extend that to a map from m. And let's count and see how many we've got. Well, the number of these is at most the index of k in l. And each of these terms here is at most the index of l in m. Now we know that since m over k is Galois, this is equal to m over k. So this is the key point at which we're using the fact that m over k is Galois. Now you see that this sum here, we have at most l over k things, each of which is at most m over l. So this side here is at most m over l. Well, it's actually equal to m over k, which means we must have equality here and equality here. So the fact that m over k is Galois implies we actually get an equality there and an equality there. We don't really care about the equality here, but we care about the equality here because this says that the number of maps from m to itself extending the identity map from l is equal to m over l. So this says the Galois group of m over l must have order equal to the index. But that's rather nice because it's exactly what we were trying to prove here. So this proves the fundamental theorem of Galois theory. It says for Galois extensions, we get a one-to-one correspondence between subgroups and subextensions. You may wonder what happens if k contained in m is not Galois. Well, it's not too difficult to check what happens there. What we do is we look at k and we look at the field of fixed points of g and we look at m. So g is going to be the Galois group of m over k, which is all automorphisms of m fixing elements of k. And we get a one-to-one correspondence between subgroups of g and subextensions. k contained l contained m with mg contained in l. It's very easy to check this because this bit here is in fact a Galois extension. So the Galois correspondence just tells us that subgroups of the group g correspond to fields between mg and m. So all that happens in general is we can't see what's going on between k and the fixed subfield of m. So let's just finish by giving a slightly more complicated example of a Galois extension and the intermediate fields. What we're going to do is look at q and we're going to look at a fourth root of 2 rather than a cube root of 2 as we were looking in the previous example. And this isn't Galois because we can multiply this by i so we'd better extend this by having the fourth root of 2 together with i. And let's write down the obvious subextensions. Well there's q with the fourth root of 2, there's q with i times the fourth root of 2. And there are some obvious quadratic extensions because we can have the square root of 2 and we can have i and then we can multiply these together and we get the square root of 2 times i. And then we can put these two together and we can get 2 with i and the square root of 2. And of course there's q and q with the fourth root of 2 and i. And it's not completely obvious that there are any other extensions. I mean these are the ones that are easy to write down. In fact there are a couple more. And what we're going to do is to find these other two extensions by using Galois theory. So we need to know what the Galois group is and the Galois group isn't too difficult to work out. Now what we do is we draw the fourth roots of 2 like here. So they form a nice square. So here's the fourth root of 2 times i and here's the fourth root of 2 times minus i and here's minus the fourth root of 2. Now the Galois group must act as permutations of these and it must in fact act as a group of symmetries of this square because this element is minus this element and the Galois group must always preserve negation. So the Galois group is a subgroup of the group of automorphisms of a square. On the other hand this extension is degree 4 and this extension is degree 2. So the Galois group is order 8 and the group of symmetries of a square has order 8. So these are all the elements of the Galois group. So let's write down all subgroups of the group of symmetries of a square. So this is just the dihedral group of order 8 which is easier to write down. So we've got the trivial subgroup and then we've got subgroups of order 2 and we can get most of these by reflections. So I'm going to draw a green line to say we're looking at reflections. So here's an element of order 2 which just switches these two elements and fixes these two. So in fact this one is complex conjugation but whatever. And then we have another element of order 2 which is just reflection in that diagonal. And then we've got an element of order 2 that is just rotation by 180 degrees. And then we've got two more reflections because we can reflect in this line or we can reflect in that line. So these are the subgroups of order 2 which are just correspond to the five elements of order 2. Now we look at the subgroups of order 4. Well what we can do is we can sort of look at the group of symmetries of a rectangle which is obviously a subgroup of the group of symmetries of this square. So this gives us one subgroup of order 4. Then we can look at the cyclic group of all rotations by a quarter of a revolution. So this generates a group of rotations. And finally we can look at the group of symmetries that preserve this rectangle. So that gives us another the third group of order 4. And finally we've got the whole group D8. And now we should draw which of these subgroups are included in others and the inclusions look like this. So these two groups here are z modulo 2z times z modulo 2z. So they're three subgroups whereas this is a cyclic group of order 4. So it is only one subgroup of order 2. So this is a complete list of all the subgroups. Now let's write in what are the fields corresponding to each subgroup. Well this is going to be the field qi times the fourth root of 2 rather obviously. And this is going to be the field q. So those are pretty easy. The cyclic one, well if you think about it you'll see that this rotation actually fixes i. So we get qi here. And here we get q with root 2. You see all the symmetries that fix the square just fix root 2. And here we get q of root minus 2 or q times root 2i. And this one obviously fixes the fourth root of 2. And this one obviously fixes i times the fourth root of 2. And rotation by 180 degrees, well it obviously contains all these fields here. So it's just q root 2i. And now we see that there are two extra fields that we didn't notice when we were writing down the obvious fields. So let's figure out what these other two fields are. Well the elements of this field are fixed by reflection in this line here. So it's sending the fourth root of 2 to the fourth root of 2 times i. So if we add up these two elements that will be fixed by this reflection. So this is the field q where you take the fourth root of 2 and you multiply it by 1 plus i. And this is kind of similar except you take the fourth root of 2 and multiply it by 1 minus i. So the fundamental theorem of Galois theory is found these extra two sort of slightly hidden subfields of this field here. We can also ask which of these extensions are normal. Well you notice that these two elements of order 2 are conjugate. So let's put a ring around them to show they're conjugate and these two are also conjugate. And the subgroups are conjugate and obviously the corresponding fields are also going to be conjugate under the same element because subfields just correspond to subgroups. So we see that almost everything is a normal extension except that these four fields here are not normal extensions of q but pretty much every other sub extension is a normal extension of fields. So next lecture we'll be giving a rather famous example of a Galois extension which is how to use Galois theory to construct the regular 17 sided polygon. As I mentioned earlier this was an amazing discovery Gauss did as a teenager.