 This lecture is part of an online Galois theory course, and will be about the Heptadecogon, or the regular Heptadecogon. So Heptadecogon is 7, and deca means 10, so this is a 17-sided polygon. And one of Gauss' most famous early results was showing how to construct a 17-sided polygon using a ruler and compass. Of course, constructing a 17-sided polygon using a ruler and compass is completely nuttily pointless, but it still gives a nice application of Galois theory. Incidentally, when I was checking Heptadecogon, I tried looking at it on Google, and it claims there's a word called Heptadecophobia, which, if you believe this, is apparently a morbid fear of Heptadecogons. There's a Wikipedia article that claims 17 as unlucky in southern European countries, in the same way that 13 as unlucky in northern European countries. I'm not quite sure whether there's some sort of weird Wikipedia hoax. I mean, they do have a picture of an Italian plane where row 17 of seats has been omitted, but for all I know, this is some sort of weird scan by the aeroplane manufacturer to make their planes look bigger or something. Anyway, let's get back to Gauss. First of all, we recall that a number is constructable by ruler and compass is equivalent to saying you can construct it from the rational numbers using the field operations plus, minus, times, divide, and the square root operation. So we take the usual field operations and also allow taking square roots. I'm not going to prove this. It's not too difficult if you know some Euclidean geometry. You first use analytic geometry to show that any number you construct by ruler and compass can be defined using these operations, and conversely you show that using a ruler and compass you can carry out these operations on lengths of lines. So we say a number is, we're going to use an algebraic definition of constructable just to say it's constructable if you can write it using these. So 2 plus root 3, and then you take the square root of that minus 7 and plus the square root of 3 plus root 5, and so on would be a typical constructable number. So at the moment this isn't terribly useful. So let's rephrase it a bit. So alpha is constructable if and only if alpha is contained in a normal extension of the rationals, of course, of degree 2 to the n for some n greater than or equal to 0. And this actually allows us to tell where the numbers are constructable because now we've reduced it to a problem about finite extensions which we can go around applying Galois theory to. So let's just quickly sketch the proof of this. So first of all, suppose alpha is constructable. What this means is we can start with the rationals and then we have, let's call this k0, and then we construct from that a field k1 which is the rationals and then we join the square root of some number alpha 0 in k0. And then we can have a field k2 which is k root of alpha 1 for some alpha 1 in here. All these extensions are degree 2 and eventually we'll get up to kn and our alpha is an element of kn. So alpha is in an extension of degree 2 to the n. But this is not good enough because kn need not be normal. And it's quite easy to see an example of this. In fact, I gave one earlier. All we do is take q containing q the square root of 2, containing q with the square root of the square root of 2. And this is not a normal extension because it doesn't contain the complex square roots of 2. So this construction doesn't prove that alpha is in a normal extension of degree of power of 2. So it's not too difficult to modify this. So we start off with q which is equal to k0 and then we take k1 which is k0 with alpha 0 added. So we take the square root of something in alpha 0. And then we take k2. We don't just take k1 the square root of alpha 1. We also join the square root of all conjugates of alpha 1. And so this is degree 2 and this is degree of power of 2. And similarly, when we take k3 to be an extension of 2, we take this as k2 and then we have the square root of alpha 2 and the square root of alpha 2 prime, square root of alpha 2 double prime and somewhere. These are all the conjugates of alpha 2. And this ensures that each of these extensions is indeed normal. And we obtain k2 from k1 by adding in a lot of square roots and each of that doubles the degree. So this extension still has degree of power of 2. So we find that alpha is in a normal extension of degree of power of 2. It's important that we put normal here because there are lots of degree 4 extensions whose elements are not constructible because you can't solve an arbitrary fourth degree polynomial just by using square roots, for example. So that's shown that if alpha is constructible, it's in a normal extension. Now let's go back the other way. So let's suppose alpha is in a normal extension of degree 2 to the n for some n. Then we look at the Galois group G and this is also as order 2 to the n. So that's a normal extension and we're characteristic zero. So it's a Galois extension and therefore it's Galois group that you can control everything in as order 2 to the n. Now you recall any group of order 2 to the n is nilpotent. So we can find this chain of subgroups g n, g n, g n minus 1, g n minus 2 such that each is normal in the one before and each has index 2 in the previous one. And now we just look at the corresponding field. So we have q contained in k1 contained in k2 and so on contained in kn and alpha is in here. And now each of these field extensions is degree 2. So k2 for example over k1 is of order 2 and this means that k2 can be generated over k1 by taking the square root of something because if we take some element alpha in k2 and it satisfies some, it's a degree 2 extension so it satisfies alpha squared plus a alpha plus b equals naught for a b in k1 and then we can just apply the quadratic formula alpha is minus a plus or minus the square root of a squared minus 4b over 2. I should have used b and c but never mind. And we're not working characteristic 2 so we don't need to worry about dividing by 2 and we can see that k2 is generated by the square root of something over k1 and so on for all the others. So alpha in here is a constructable number. So that shows that a number can be constructed using square roots and field operations if and only if it's in a normal extension of degree 2 to the n. So for example we cannot construct a regular 7 sided polygon. Well constructing a regular 7 sided polygon is more or less the same as constructing the number 2 pi over 7 or the cosine of 2 pi over 7 because if you take a regular 7 sided polygon on 2, 3, 4, 5, 6, 7 of a circle of radius 1 then this distance here is going to be cosine of 2 pi over 7 and constructing this distance is the same as constructing the polygon. And we've seen that this number here is a root of an irreducible degree 3 polynomial whatever it was, 8x cubed plus 4x squared minus 4x minus 1 or something like that. So any extension containing cosine of 2 pi over 7 has degree divisible by 3 by the multiplicative formula for degrees so is not a power of 2. So this number here can't possibly be a constructable number and we can't construct a 7 sided polygon. So now let's look at 17 sided polygons or more generally let's look at p sided polygons. So we recall that constructing a p sided polygon is more or less the same as constructing a p-th root of unity so it's a complex number when we say a complex number is constructable we just mean it's real and imaginary parts are constructable rather obviously. So we recall that x of p minus 1 is not irreducible divided by x minus 1 we get x to the nought plus x to the 1 plus all the way up to x to the p minus 1 and we recall that this is irreducible and the standard proof of this is you just change x to x plus 1 and invoke the magic word Eisenstein and Eisenstein's criterion then implies that the polynomial is irreducible. So suppose zeta is a p-th root of unity not equal to 1 so it might be cosine of 2 pi over 7 cosine of 2 pi over p plus i sine 2 pi over p for example then we know the other roots of x to the nought plus x to the 1 plus x to the p minus 1 are the other p-th roots of unity so they are zeta, zeta squared, zeta cubed all the way up to zeta to the p minus 1 so the extension q of zeta is normal because it contains all the other roots of this polynomial it's just the splitting field of this and it's obviously separable because we're working in characteristic nought so it's a Galois extension this is sometimes called a cyclotomic cyclotomic extension cyclotomic extensions are just extensions generated by roots of unity and they're incredibly important and they're the easiest sort of extension and we'll be talking about them a bit more later so anyway now we can work at its Galois group we sort of did this for p equals 7 earlier and the general case is rather similar so we have an automorphism it must take zeta to zeta to the i for i in z modulo pz a non-zero element of that and if we compose these if we take zeta to the i to the j this is just zeta to the ij so the group operation is just multiplication so this is the Galois group and we can see it's the whole Galois group because we know the Galois group is order p-1 and here we found p-1 automorphism so that's everything so we can now say when can you construct a p-sided polygon? well this is equivalent to saying that q of zeta over q has a degree a power of 2 by what we just said we know it's a normal extension so all we need for it to be constructed is a power of 2 well the degree is just p-1 so p is equal to 1 plus 2 to the n for some n and these are the famous Fermat primes Fermat showed that n itself must be a power of 2 so the Fermat primes are 2,3,5,17 257,6,5,5,3,7 so the Greeks knew how to construct regular polygons with 2,3 and 5 sides you can find a construction of a regular pentagon in Euclid's elements and 2,000 years, more than 2,000 years later Gauss really amazed everybody by showing you could also construct regular polygons of this number of sides well so far we've got a we've got a rather abstract existence proof of a construction but if you stop and actually try and use this you'll discover that it's not very explicit I mean we've sort of claimed that we can write zeta in terms of square roots and rational numbers so how do you actually write it down explicitly? and so what I'm going to do now is show how we can give an explicit description of zeta in terms of square roots and for this we're going to use Galois theory so let's just do it for p equals 17 so we know the Galois group is z over 17z times which is 1, 2, 3, up to 16 and we recall from number theory that this is cyclic because the non-zero elements mod p always form a cyclic group and a generator is called a primitive root in fact this is generated by the number 3 it's not too difficult to check and since it's cyclic of order 16 it's got subgroups of order 1, 2, 4, 8 and 16 and let's write out these subgroups explicitly we've got the subgroup 1 and then it's contained in the subgroup with 1 and 16 and then it's contained in the group with 1, 4, 13, 16 which is contained in the group with 1, 2, 4, 8, 9, 13, 15, 16 which is contained in the whole group so here are the subgroups written out explicitly and corresponding to these we get fields we get a field q of zeta which contains another field which contains another field which contains another field which contains the rational numbers and what we want to do is to figure out generators for these three fields explicitly and write each of them in terms of square roots in terms of the other field so first of all what we want is an element in this field this means it has to be fixed by this group here so that's obvious we just take zeta to the 1 plus zeta to the 16 and this is 2 cosine 2 pi over 17 which is the number we want to construct so there's not much to do there but what you should do is you note that this group 1, 16 has various cosets so let's write out all its cosets so it's got cosets 1, 16, 2, 15, 3, 14 and so on and when I said let's write them out all of them I didn't mean I was actually going to do that because I'm kind of lazy and from that we see there are various other corresponding elements you can write down as zeta to the 1 plus zeta to the 16 zeta squared plus zeta to the 15 and so on and these are all conjugates of zeta to the 1 plus zeta to the 16 under the Galois group let's call this element alpha and we can do the same thing here we can write out the cosets as 1, 4, 13, 16 and then another coset would be 2, 8, 9, 15 and another coset would be 3, 12, 5, 14 and the other one is 6, 7, 10, 11 if I've got them right so there are 4 cosets and here there are just 2 cosets there's 1, 2, 4, 8 etc and the other coset is 3, 5, 6, 7 etc and again for every coset you can write down a corresponding sum over zeta so here we have zeta to the 1 plus zeta to the 4 plus zeta to the 13 plus zeta to the 16 and here we have zeta squared plus zeta to the 8 plus zeta to the 9 plus zeta to the 15 and so on I'm not going to write out the others you can imagine what those are and let's call these elements things so here's alpha and here's beta might be that one there and these are all these things here are the 4 conjugates of beta under the full Galois group and what we can do now is we can write this number in terms of alpha rather easily because so we can write zeta in terms of alpha because you see that zeta squared minus zeta alpha plus 1 is equal to 0 so here we have a quadratic equation in zeta with coefficients in alpha so we can write zeta explicitly in terms of alpha so zeta is alpha minus the square root of alpha squared minus 4 over 2 and so on and now we want to express alpha in terms of something well we have a conjugate of alpha over this field here we can take the conjugate of alpha to be alpha prime which is zeta to the 4 plus zeta to the 13 and the reason for choosing 4 and 13 is because we then have alpha plus alpha prime is equal to beta and you can work out that alpha times alpha prime is equal to one of the conjugates of beta zeta to the 5 plus zeta to the 14 plus zeta cubed plus zeta to the 12 which is this one here so we can express alpha plus alpha prime and alpha times alpha prime in terms of explicit elements of this field so we have this quadratic equation for alpha alpha squared plus something times alpha plus something is equal to 0 where the something and the something are the explicit elements of this field and now we can do exactly the same for these elements we can take this element here and we will find that this element plus this element is equal to this element here and this element times this element will equal something here so we can get an explicit quadratic equation for beta and similarly if gamma is this element here we can get an explicit expression for gamma we find gamma is equal to the square root of 17 minus 1 over 2 if you want to do it explicitly and check your answer so we can get a whole series of quadratic equations each expressing an element of this field in an element of the next field and if you unravel that all you find an explicit expression for cosine of 2 pi over 17 in terms of square roots writing it out explicitly is a little bit of a mess I'm not going to do so because Gauss has already done it so this is Gauss's famous book on number theory Disquisitione's Arithmetica and if you go almost to the end so it's about two or three pages from the end let me magnify this so you can see oops I've overshot a bit right here we have Gauss's expression for the cosine of the angle p over 17 and p is Gauss's notation for 2 pi so here's cosine of 2 pi over 17 expressed in terms of repeated square roots of numbers you see there's the square root of 17 turning up all the time and so on so I guess that's enough about Hector Deckergun's next lecture we'll be doing some other applications of Gauss's theory in particular I'll be giving an algebraic proof that the complex numbers are algebraically closed