 This lecture is part of an online course on Galois theory, and it will be about examples of the Galois correspondence. So we recall what this is. So we have a Galois extension of fields, which we will take to be finite. And do you remember there's a one-to-one correspondence between subfields of M containing L and subgroups of the Galois group of M over K. And the correspondence went as follows. For a subfield L we associate the group of automorphisms of M that fix all elements of L, which is the Galois group of M over L. On the other hand, if we've got a subgroup of the Galois group H, the field corresponding to it is the set of elements of M fixed by H. And in the next lecture we will show that these correspondences are bijections, but in this lecture we're just going to work out several examples of them to see what's going on. Let's start off with a trivial one just to warm up. Here we're just going to take the reals contained in the complex numbers. And in this case the Galois group just has two elements, which are one and complex conjugation. And there are two subgroups, and there are obviously two subfields of C containing C and R, and the subfields are the complex numbers itself and the real numbers. And these correspond to the following subgroups. The complex numbers correspond to the trivial subgroup, and the real numbers correspond to the subgroup containing one and complex conjugation. And this is rather trivial. The only thing to watch out for is that the larger field corresponds to the smaller subgroup. And as I mentioned, this is one of the biggest source of errors in Galois theory of thinking the bigger field should correspond to a bigger subgroup. So let's do a slightly more interesting example. So let's take the field Q contained in Q with the cube root of 2 added. Well, as we saw earlier, this isn't actually a Galois extension because the other cube roots of 2 aren't in it. So we should extend it a bit and take Q with the cube root of 2 and also add an omega where omega is a cube root of 1. Omega squared plus omega plus 1 equals 0. So this is now a Galois extension, and its Galois group G is isomorphic to S3, which is just the group of permutations of the three cube roots of unity. So we have the cube root of 2, cube root of 2 times omega, and the cube root of 2 times omega squared. So what I'm going to do is I'm going to write out all the subfields we can think of and all the subgroups we can think of. So let's do the fields. So we've got the first one, we've got the full field up here, Q with the cube root of 2 and omega added, and down at the bottom we've got the rational numbers. And in between them, well, there's a field extension of order 2 generated by omega and there's a field extension of order 3 generated by the cube root of 2. But then there are two other cube roots of 2 generating other fields. So we have two more fields here. So we've got the cube root of 2 times omega and the cube root of 2 times omega squared. So these fields are embedded in each other like this. And we can put down the relative degrees. So the relative degrees here are 3, 3, 3, 2, 3, 2, 2, 2, 2. Now let's write down the subgroups of the symmetric group S3. And this time we should remember to put the symmetric group S3 on the bottom, not the top. And we have a trivial subgroup there. And the symmetric group S3 has six elements which are 1, 1, 2, 1, 3, 2, 3, 1, 2, 3, 1, 3, 2. So it's got some elements of order 2 and some elements of order 3. And you remember from group theory the subgroups of this aren't terribly difficult to figure out. First of all there's a subgroup of order 3 which consists of 1, 1, 2, 3, 1, 3, 2. So that's generated by the elements of order 3. And then it's got 3 subgroups of order 2 generated by these elements. So we've got the subgroup generated by 2, 3, subgroup generated by 1, 2, and the subgroup generated by 1, 3. So here we've got subgroups and we can write down the index of each subgroup in the other subgroups. So the indices here are 2, 2, 2, 3, 2, 3, 3, 3. And you notice this lattice of subfields looks just like this lattice of subgroups except of course the inclusion gets reversed. And we can also do things like write down which fields are normal in other subfields. So which extensions are normal? Well this extension is normal and so are these two. And this extension is normal and this extension is normal because it has a degree 2. But these extensions are not normal. You remember this extension is the standard example of a non-normal extension. Similarly we can write down which subgroups are normal. Well any subgroup next to is normal so these are normal and that's normal and this is normal. And these three are not normal subgroups of S3. In fact you notice that the subgroup S3 permutes these three subgroups under conjugation. So this is a conjugacy class of three subgroups. And the Galois group similarly permutes these three subfields under conjugation. And again you can see normal subfields correspond exactly to normal subgroups. I didn't mention by the way that this is a normal subgroup of the full group and this is a normal extension of Q. But I don't have room to draw in absolutely everything. So this is an example of the Galois correspondence. If you've got a Galois extension then the lattice of subfields looks just like the lattice of subgroups up to duality or something. So now let's look at the example of the field of order 2 contained in... Well it's either the field of order 4 or it's order 16 or a field of order 16 depending on whatever. And the first thing we want to do is to work out what is the Galois group. So can we find any automorphisms of the field of order 16? Well there's one obvious automorphism which just takes A to A squared. You remember that the map taping A to A to the P in characteristic P is a homomorphism because A plus B to the P is equal to A to the P plus B to the P and AB to the P is equal to A to the P, B to the P. So this is the famous Frobenius endomorphism and it acts on here. And we can ask what its order is. Well you notice that phi squared takes A to A to the 4 and phi cubed takes A to A to the 8 and phi to the 4 maps A to A to the 16 which is equal to A. So we see that phi to the 4 is equal to 1. So phi generates a cyclic group of order 4. On the other hand we know that the order of the Galois group has to be the degree of this extension. The degree of F16 over F2 is just 4. So we found the full Galois group. The Galois group is just generated by phi and is order 4. And you notice there's nothing special about the field of order 16. You find that similarly if you take Fp containing Fp to the N, the Galois group is cyclic generated by A goes to A to the P. Incidentally this is one of the reasons why finite fields are really easy to deal with. It's because not only are all extensions of finite fields Galois extensions, but their Galois groups are cyclic and cyclic groups are of course particularly easy to deal with. So finite fields are easy because cyclic groups are easy. Anyway, let's draw a picture of the subfields and the corresponding subgroups. So subfields aren't very difficult to figure out. There's only one subfield other than the obvious maximal and minimal ones which look like this. And the full Galois group contains 1, phi, phi squared and phi cubed. And the trivial subgroup corresponds to the full field. And this is one subgroup of order 2 which is just the elements 1 and phi squared. And here all extensions are normal and all subgroups are normal. So there's nothing terribly exciting to say about this. By the way in here that the chain of subgroups and the chain of subfields is linear, of course that's not always true. For instance if we took the Galois field of order 2 to the 6 for example, then we would have a Galois field of order 2 cubed and a Galois field of order 2 squared. And these would contain the Galois field of order 2. Similarly we would have a cyclic group of order 6 and here we would have a group of order 1 and here we would have a subgroup of order 3 and here a subgroup of order 2. My notation isn't very good but anyway. So the subfield lattice of a finite field isn't always linear but it's always very easy to work out. So for the next example I'm going to look at the field over the rationals generated by a seventh root of unity. So zeta to the 7 is equal to 1 and zeta is equal to e to the 2 pi i over 7 or some power of it. Well of course the polynomial x to the 7 minus 1 is not irreducible. It's equal to x minus 1 times x to the 6 plus x to the 5 plus x to the 4 plus x cubed plus x squared plus x plus 1. And you remember there's a sort of trick of showing this is irreducible using Eisenstein's criteria and if you change x to x plus 1 which I won't bother repeating. So this is an irreducible polynomial. And now if zeta is one root of this the other roots, well if zeta is a seventh root of unity so zeta squared, zeta cubed, zeta to the 4, zeta to the 5 and zeta to the 6. So in particular this extension is separable because it's character 6-0 and it's normal because it's a splitting field of a polynomial so it's a Galois extension. And there's nothing special about seventh roots of unity here. We could do something fairly similar for at least any other root of unity of prime order for composite order we need to be a little bit more careful. So what's its Galois group? Well any automorphism must take zeta to one of the other roots so it must take zeta to zeta to the i for i equals 1, 2, up to 6. And what's the group operation? Well if we take zeta to the i and then apply the automorphism raising things to the jth power this is just zeta to the ij. So as you see from this the Galois group is just the non-zero elements of the integers, sorry, the integers mod 7 under the operation of multiplication because here the composition of these automorphisms just corresponds to multiplication. So the Galois group has six elements which can be denoted by 1, 2, 3, 4, 5, 6 and the group operation is multiplication mod 7. And we remember that this is cyclic of order 6 and of course 1 is not a generator of this group, it's the identity element. So a generator might be 3 because if you look at the powers of 3 we have 3 to the north equals 1, 3 to the 1 equals 3, 3 squared equals 2, 3 cubed equals 6, 3 to the 4 equals 4 and 3 to the 5 is 5, 0, 1, 2, 3, 4, 5. So it's a cyclic group of order 6 and a cyclic group of order 6 has just four subgroups. So let's write out the subgroups and the subfields. So the subgroups look like this, there's a subgroup 1 and there's a subgroup 1, 2, 3, 4, 5, 6, the whole group and then there's a subgroup of order 2 which is generated by the cube of the generator and there's also a subgroup of order 3 generated by, consisting of the elements 1, 2 and 4. So it looks like this and we can write down the indexes of all these groups so that is index 2, that is index 3, that is index 3 and that is index 2. And these correspond to fields so let's write down the fields. First of all up here we have the fields Q of zeta and down here we have the field Q of course and here we have a field generated by the elements 1 and 6. Well what does the element 6 correspond to? Well this takes zeta to zeta to minus 1 which is the complex conjugate of zeta. So this element here is really just complex conjugation looking at the elements of this field that are fixed under complex conjugation in other words the real elements and it's sort of pretty obvious what that's going to be. It's going to be the element Q zeta plus zeta to minus 1 and you remember this is 2 cosine 2 pi over 7 and we studied this field quite a bit in several examples earlier except we used the element cosine of 2 pi over 7 instead of 2 cosine 2 pi over 7 but that doesn't really make any difference. So anyway the indices of these fields are like this and then we've got one other field so we've got a sort of mystery field here and if you haven't seen this example before it's probably not at all obvious what this field is so what the Galois theory is telling us is that there's a subfield of this which is a degree 2 extension of the rationals well what on earth is it? Well how do we find it? Well it consists of the elements that are fixed under this group here so how can we find an element of this field? Well let's take an element zeta and we're going to have zeta plus zeta squared plus zeta to the 4 and you notice this is obtained by you start with zeta and then you act on zeta by all three elements of this group and add them up and that's obviously going to be invariant under this group so this is a good candidate for a generator of this field okay so this should be q of alpha but what is alpha? I mean we know that any quadratic extension of q should be generated by the square root of something so what's the something that this is generated by a square root of? Well let's play around with alpha a bit so you can find that alpha squared is equal to zeta squared plus zeta to the 4 plus zeta to the 8 which is just zeta plus 2 zeta q plus 2 zeta to the 6 plus 2 zeta to the 5 and then you notice that here we've got zeta squared plus zeta to the 4 plus zeta which is just alpha again and here we've got 2 times the things that aren't in alpha and if you sort of fit around with it a bit you notice that alpha squared plus alpha is now going to be 2 times every power of zeta except for zeta to the nought so if we now add on a 2 we get this is 2 zeta to the nought plus zeta to the 1 plus zeta squared plus all the way up plus zeta to the 6 and now the sum of all these powers of zeta is just 0 so this is just 0 so now we've found out what alpha is it's a root of this equation here so alpha, well we can solve this using the formula for quadratics it's just minus 1 plus minus the square root of 1 minus 8 which is minus 7 over 2 so we can now see what this field is it's q of root minus 7 and again it's index 2 over q and index 3 in this field here so just as before we can see the subgroup the lattice of subfields is just like the lattice of subgroups and of course all groups here are abelian so everything is normal and there's no need to mark the normal ones so what we're going to do next lecture is actually prove the fundamental theorem of Galois theory showing that we always get such a correspondence between lattices and sorry between subfields and subgroups and then we'll give a few more examples