 This lecture is part of an online course on Galois theory and will be about splitting fields. So suppose we have a field k and we have a polynomial with coefficients in k. So we might have a polynomial p of x in the ring k of x. And the problem we want to solve is find an extension of k. So we want to find an extension field l so that all roots of p are in l. Well, saying all roots of p are in l doesn't quite mean anything yet because, you know, what is a root of p? So we shouldn't really say that. What we should say slightly more precisely is that p factors into linear factors in the ring l of x. In other words, l contains all roots or at least l contains n roots of p where p has degree n. Well, if we've got an extension of this property, then any bigger extension will also have the same property. So we really want the smallest extension with this property. So this is property one that p should factor into linear factors in l of x. And property two, we want l to be as small as possible. So we could say l is generated by the roots of p over k. That means it has to be a minimal field containing all the roots of p. Or in other words, everything in l can be written as some polynomial in the roots of p with coefficients in k. So if a field has these two properties, l is called a splitting field of k. Actually calling it a splitting field is slightly sloppy terminology because what we want is not just the field l, but the field l together with an embedding of k into it. But as usual in mathematics, the terminology is a little bit sloppy because if you are completely precise about everything, it becomes completely impossible to understand what's going on. So let's have some examples of splitting fields. So let's take a polynomial p of x to be linear x minus a naught. Well, then the splitting field is just k itself. So the splitting field is just k. It's completely trivial. So let's do a slightly more complicated example. Let's take p to be quadratic x squared plus a1x plus a naught equals naught, for example. Then if this is reducible and splits into linear factors, the splitting field is k again. So suppose it's irreducible. Well, then we can construct a new field l by just doing the usual construction. We take the ring of polynomials over k and quotient out by the ideal p. And this is now a maximal ideal because p is irreducible, so this is a field. And it contains a root of p. I mean, the image of x in this field l will be a root of p by construction. In fact, it contains all roots because if p factorizes as x minus alpha times x minus beta, then alpha plus beta is minus a1. And a1 is in the field k. So if an extension of k contains one of the roots of this polynomial, it contains both of them. So this is now a splitting field. So polynomials of degree one or two, splitting fields aren't really terribly interesting. You just get them by joining a root. Things become a little bit more complicated for polynomials of degree greater than two. So let's look at the next example. So let's just look at x cubed minus two and let's work over the field of rational numbers. And then we can form a field l by adjoining the cube root of two. Or if you like, you can think of this as being q of x modulo x cubed minus two since you can easily check this is irreducible. So these two fields are the same. So here we're thinking of this field as being a subfield of the real numbers. And we can think of this as just being some sort of abstract field. It doesn't really matter what you do. And now we can ask, so l contains one root of this polynomial. What about the other roots? Well, it doesn't contain the other roots because if you think of r as being contained in the complex numbers, you can draw where the other roots are. So one of the roots is going to be the cube root of two. And the other two roots are going to be the cube root of two times cube root of unity. So this is the cube root of two times omega. And this is the cube root of two times omega squared where omega equals minus one plus root minus three all over two. So two is a cube root of unity. And now what you notice is that this field here, l, is contained in the real. So it can't possibly contain these two other roots. So l does not contain all roots of p. And as usual, this is a sloppy way of saying that p doesn't factorize into linear factors over l. So p, the polynomial p which is x cubed minus two now factors as x minus the cube root of two times x plus, x squared plus the cube root of two, x plus the cube root of two all squared. And this is now irreducible over l. So what do we do? Well, we can now just join a root of this to l. So we form a new field m, which is just l. And now we join some polynomial x. Maybe you'd better call it a different variable so we don't get confused. And we're now quotient out by y squared plus the cube root of two y plus the cube root of two squared. And now this is again a field. So what we've got is we've started with k, which was the rationals, and we extend it to l, which is the rationals with the cube root of two adjoined. And we extend this to a new field m, which is q with the cube root of two and also this number omega adjoined. And now m is now a splitting field. And let's work out its degree. Well, this is degree three because we've adjoined a root of a cubic polynomial. And this is degree two because we've adjoined a root of a quadratic polynomial. All together, m over k has degree six. So we found a splitting field of this polynomial, which has degree six. Well, let's look at another example of a degree three polynomial where something a little bit different happens. So now we're going to use the polynomial 8x cubed plus 4x squared minus 4x minus one. And this is the polynomial we had in the previous lecture. And the roots were cosine of 2 pi over 7 and cosine of 4 pi over 7 and cosine of 6 pi over 7. And you can easily check it's irreducible. So we can start with q and we can form a field q of x modulo. Let's call this polynomial p of x, so I don't have to write it out every time. And this extension has degree three. And now you may think we're going to do the same thing in the previous example. We're now going to take, we've found a degree three extension which contains this root. And now we're going to find another degree two of extension of that which contains this root and this root. Well, we don't need to because cosine of 4 pi over 7 is equal to cosine of 2 pi over 7 all squared times 2 minus one. This follows from the fact that cosine of 2 theta is 2 cos squared theta minus one. So we see that if we are joined one of the roots of this polynomial to q, we've automatically joined the second root because it's a polynomial in the first root. So this is now the splitting field and its degree over q is 3 and not 6 as in the previous example. So when you construct a splitting field, you sort of start off by joining a root and then maybe you have to join another root, maybe you've already got the roots already there. As you can imagine for polynomials of degree greater than three things can get more and more complicated. So let's just give a very simple example of a degree four polynomial. So as we look at the polynomial x the 4 plus 1 which you can easily check is irreducible. And let's work over q although we don't really have to. And if alpha is a root, well so is alpha cubed alpha to the 5 and alpha to the 7 because if alpha to the 4 is equal to 1 then alpha cubed to the 4, so if alpha to the 4 is equal to minus 1 then so is alpha cubed to the 4. In fact you can draw these roots in the complex plane. They're all roots of unity so they sort of live here. This might be alpha, alpha squared, alpha cubed and alpha to the 4. So again if we had joined one root of this polynomial to the rational numbers we automatically had joined the other three. So the splitting field is q of alpha which is just isomorphic to q of x over x to the 4 plus 1 and it has degree 4 and not 4 factorial or something as you might expect in general. So now that we've seen some examples let's just show that splitting fields always exist and we're going to show they're sort of unique although as we will see there's a little bit of a problem about uniqueness. So let's look at the existence of a splitting field. Well here we're given a field k and we're given a polynomial p and it's not necessarily irreducible. So let's write p equals p1, p2 and so on where the pi are irreducible. This means irreducible in k of x because of course they might become reducible in some bigger field. And let's set k naught equal to k and let's put k1 equal k naught and let's join a root of one of these polynomials p1. So we might quotient out by p1 of x. So this is now going to be a field because p1 is irreducible. And we're going to assume the degree of p1 is greater than 1 because if it's equal to 1 then it's kind of stupid of joining a root of it. And now we say if p splits into linear factors then we can stop. But if p has a factor of degree greater than 1 we do this construction. And now we repeat with k1 and p. So again if p splits into linear factors over k1 we stop and if it's got a nonlinear factor we repeat and get k2 equals k naught of x over something else. So over k1 p will split more than over k0 because you see we've joined a root of p1 so p1 is going to split as a linear factor times something else. So we can continue reducing the degrees of the factors of p until they're all of degree 1. So we just continue until p splits into linear factors. And it's pretty obvious the field we construct like this will be a splitting field because by construction the polynomial splits into linear factors. On the other hand it's also pretty obvious that the field extension we've got is generated by all the roots by roots of p. So this gives us a splitting field. However it's not entirely obvious that this splitting field is unique up to isomorphism. You see there was some choice. Instead of first joining a root of p1 we could have joined a root of p2 and maybe only joined the roots of p1 later. And does the field we get depend on what order we add these in? Sounds plausible it's independent but we do actually need to check this. So let's talk about uniqueness. What we're going to show is that any two splitting fields of p over k are isomorphic. And you've got to be a bit careful because I don't just want to say they're isomorphic as fields. I want to say they're isomorphic as extensions. So what this means is if we've got two splitting fields L and L prime over k what this means is there's an isomorphism from L to L prime which sort of commutes with the embedding of k into L and L prime. So this is a little bit stronger than just saying L is isomorphic to L prime. But as usual we're quite often a bit sloppy and just say that the splitting fields are isomorphic and it's kind of understood that we should remember to include the inclusion of k as part of the structure. Well in fact it's slightly easier to restate this theorem slightly. So what we're going to do instead is suppose you've got a field k and an isomorphism to a field k prime. And suppose we've got some sort of splitting field of k and some sort of splitting field of k prime. So this is going to be a splitting field of the polynomial p and the polynomial p is going to correspond to a polynomial p prime and we're going to take a splitting field of p prime. What we want to do is to construct an isomorphism from L to L prime. So this is the map we've got to construct. In fact we can do it a little bit more general than this. Instead of assuming L is a splitting field let's just assume that L is generated by some roots of p. So it might not actually contain all roots of p. And here we're going to assume that L prime contains all roots of p prime. That means p prime splits into linear factors. So this might actually be bigger than a splitting field and this might be smaller than a splitting field in general and we want to show there's a map from there to there. Well we do this as follows. What we do is we factor p over k. So let's write p equals p1 p2 and so on with p i irreducible. And then we're going to extend k to a field k1. We're just going to look at the field generated by a root p1. And now over k prime this p prime splits as p1 prime p2 prime and by assumption L prime contains all the roots of p1 prime. So we can have a root of p1 prime here. Sorry that shouldn't be k of p1 that should be k of alpha. So here this is going to be a root of p1 prime. And now we can construct a map from this field here into L prime just by mapping the root of p1 that we selected to the root of p1 here. And now we can continue like this. So now we factor p over this bigger field and we factor k2 and k3 and so on. And we just sort of keep going and we build it up and we can construct a map of fields from L to L prime. In particular we see that the degree of L must be less than or equal to the degree of L prime. That's the degree of L over k, the degree of L prime over k prime. Because we've embedded L into L prime. And now suppose that L and L prime are splitting fields. Well then we can reverse the argument and have a map the other way so we find the degree of L prime over k prime would be less than or equal to the degree of L over k. And therefore these degrees are equal to the degree of L over k is equal to the degree of L prime over k prime. And since these have the same degrees in other words they're the same dimensions as vector spaces this map reconstruction must be an isomorphism. So we've shown that any two splitting fields are isomorphic. Well you might think this shows that there's a splitting field for any polynomial and in some sense you're right but the problem we have is it a splitting field or the splitting field. In other words we would talk about a splitting field if there are lots of splitting fields and there's no particular reason for choosing one of them and we would call it the splitting field if there's a sort of canonical way of choosing a splitting field. So yeah I apologize to anyone who's Russian I'm trying to understand this because Russian doesn't actually distinguish between the definite and indefinite article or whatever. The problem is suppose you've got a splitting field of k. So I've constructed a splitting field and you've constructed a splitting field and fine we've just shown there's an isomorphism between them but the problem is this isomorphism is not unique. So you could say they're the same splitting field but in some sense they're not really because if I pick an element of my splitting field which elements of your splitting field does it correspond to? Well it's not really clear because there's a choice of isomorphisms. There's an actual example of this we can give so if we take the real numbers and let's take the splitting field of the polynomial x squared plus one. Well we all know what the splitting field of this is it's the complex numbers which is r a joined i times the square root of minus one. Well then you go over to the engineering department and you discover the electrical engineers have also constructed a splitting field of the reals for this polynomial but their splitting field is different it's rj. They claim they can't use i because i is already used for current or something so they have to use j. And we know these two splitting fields are isomorphic but you know what's the isomorphism? What does i correspond to? Because there are two possibilities. We could map i to j or we could map i to minus j and it's not clear which we should use. You might think we should map i to j because that's the most sensible thing to do. But you have to remember these are electrical engineers we're talking about. We've got the sign of the electrical charge wrong and you know maybe they got the sign of the square root of minus one wrong as well so we should map i to minus j. So what's the right thing to do? Well it's not really a meaningful question. There's just this sort of ambiguity that turns up all the time when if you've got a splitting field that you... it's a little bit hard to pin it down because there's a lot of ambiguity. We will see this problem turning up a little bit later on in the course. It's not really a major problem. It's just a sort of minor bookkeeping problem but it occasionally trips you up because it causes minor problems if you're not watching out for it. So what we're going to do in the next two lectures is give a couple of applications of splitting fields. We're first going to define algebraic closures of fields and then construct some finite fields.