 This lecture is part of an online course on Galois theory and will be a sort of review of field extensions. So we just recall that a field extension just means two fields, one of which is contained in the other, so k and l are going to be fields. And this is sometimes denoted by l over k for some reason. And Galois theory is mainly about the study of fields, but it turns out to be really useful to study pairs of fields or field extensions. And the reason for this is that if we want to study a field l, it's quite often useful to sort of build up to it by starting with a smaller field l0 and building a chain of bigger and bigger fields until you reach l. And then we sort of prove things about l by starting with l0 and kind of hopping up one field at a time. So we need to study field extensions in order to be able to do this. So here are some basic examples. We can have the rational numbers contained in the reals. So this is rationals and this is the reals. And the real numbers are contained in the complex numbers. So these are two standard examples of extensions. If we've got an extension, we define the degree of the extension l over k. This degree is denoted by lk in square brackets. I have no idea why. And this is just the dimension of l as a vector space over k. For example, the degree of the complex numbers over the reals is just 2 because the complex numbers form a two-dimensional real vector space. And the degree of the real numbers over the rationals is infinite because the real numbers are an infinite dimensional vector space over the rationals. If you're a sort of set theory type person, you know that this symbol infinity is actually kind of a bit sloppy because there are lots of different infinities in set theory. You can ask which, you know, is this a countable infinity or an uncountable infinity and so on. We don't really care. Galois theory is mostly about finite extensions. And if an extension is infinite, it's pretty much out of reach no matter what sort of infinity it is. So we won't worry about that. So the extension is called finite. So the extension l over k is called finite if the degree is finite. So, for instance, the complex numbers are a finite extension of the reals. Now, suppose we've got an extension k contained in l and we've got some element alpha in l. Alpha is called algebraic over k if alpha is the root of a polynomial p of x with coefficients in the field k. Alpha is called transcendental over k if it's not. Transcendental has nothing to do with any religious meaning of being transcendental. And if it's algebraic, there's a smallest irreducible polynomial that it's a root of. So the degree of alpha is the smallest degree of a polynomial p. And if alpha is transcendental, I suppose you could say it as degree infinity if you like but I don't think people use that much. So let's see some examples. Let's take k to be the rational numbers. If you talk about a number being algebraic or transcendental without specifying the field, then you usually mean you're working over the field of rational numbers. So if we take k to be the rational, you can look at say the number alpha which is a fifth root of 2. And this is obviously algebraic because it's a root of x to the 5 minus 2 equals 0. And it has degree 5 because this polynomial has degree 5. If you look at the numbers pi or e in the reels, then these are transcendental. This is sort of difficult to prove. So Hermite proved e was transcendental back in the 19th century and Lindemann proved that pi was transcendental a few years afterwards and started off a large huge theory of transcendental numbers where you try and prove various numbers are transcendental. Well, these are rather difficult examples of transcendental numbers. So let me give you an easy example of a transcendental element of a field. If you look at the field q, it's contained in the field of rational functions over q. So you remember that rational functions are going to be any polynomial divided by any other non-zero polynomial. And now you notice that x is transcendental over the field of rational numbers. And here's an example of a number where it's not perhaps immediately obvious whether it's transcendental algebraic. What about the number cosine of 2 pi over 7? Is this algebraic or transcendental? And if you didn't know about it, you might sort of guess it's transcendental because cosine is a very transcendental function and pi is transcendental and so on. But it's actually algebraic. And one easy way to see that it's algebraic is to notice let's call this number alpha. Alpha is equal to zeta plus zeta to the minus 1 over 2 where zeta is a fifth root of unity e to the 2 pi i over 7. So let's just draw a picture to see what's going on. If you look at the unit circle, then you can look at the seventh roots of unity and you know that they all lie on the unit circle in the complex plane and form a nice regular seven-sided polygon. And zeta is going to be this one here. It's traditional to use zeta for roots of unity. So we know zeta to the 7 is equal to 1. And alpha is going to be just this point here. So you recall from Euler's form that this is cosine of 2 pi over 7 plus i times sine of 2 pi over 7. Well, this isn't actually irreducible because it's got a factor of zeta minus 1. So we can find an irreducible polynomial with the root of zeta by questioning out by that and we find 1 plus zeta plus zeta squared plus zeta cubed plus zeta to the 4 plus zeta to the 5 plus zeta to the 6 equals 1. And now from this we want to find an explicit irreducible polynomial satisfied by alpha. Well, let's divide this thing by zeta cubed so we get zeta to the minus 3 plus zeta to the minus 2 plus zeta to the minus 1 plus zeta plus zeta squared plus zeta cubed equals 0. And now we notice that 2 alpha cubed can be written as zeta to the minus 3 plus 3 zeta to the minus 1 plus 3 zeta plus zeta cubed. And what we're trying to do is to express this in terms of alpha. So if we look at 2 alpha squared then it's 1 times that plus 2 times 1 plus 1 times zeta squared and if we look at 2 alpha it's equal to 1 times zeta to the minus 1 plus 1 times zeta and 2 alpha to the 0 is just equal to 1. So if we take the linear combination 2 alpha cubed plus 2 alpha squared that will give us 1, 1, 3, 2, 3, 1 and then we can, if we subtract 2 times 2 alpha and then subtract 1 you see this is equal to zeta to the minus 3 plus zeta to the minus 2 and so on all the way up to plus zeta cubed which is equal to 0. So here we found a polynomial satisfied by alpha so it's 8 alpha cubed plus 4 alpha squared minus 4 alpha minus 1 equals 0 so here's a polynomial with alpha as a root and it's got rational coefficients. By the way you might wonder what the other roots of this polynomial are well that's not very difficult to see because you can do the same thing with zeta squared or with zeta cubed so you find the other roots are going to be cosine of 2 pi over 7 times 2 and cosine of 2 pi over 7 times 3 and in case you're wondering why you don't get cosine of 2 pi over 7 times 4 the answer is you do it's just that cosine of 2 pi over 7 times 4 is equal to cosine of 2 pi over 7 times 3 So next we observe that there's this simple criterion Suppose we've got a number to be algebraic Suppose we've got two fields k and m then alpha in m is algebraic over k if and only if alpha is contained in a finite extension of k and this is very easy to prove let's first suppose that alpha is in a finite extension l well then we have l over k we'll have degree n less than infinity for some n and now all we do is look at 1 alpha alpha squared up to alpha to the n and what we notice is that these are n plus 1 elements of an n-dimensional vector space over k so there must be a linear relation a non-trivial linear relation so we have a nought plus a 1 times alpha plus a n alpha to the n equals 0 and this just is a polynomial satisfied by alpha so it implies alpha is algebraic over k of course on the other hand we want to show that if alpha is algebraic then alpha is contained in a finite extension and for this we first recall a way of constructing finite extensions which is suppose p of x is an irreducible polynomial in k of x then we can form an algebraic extension we take the ring of all polynomials over k and we quotient out by the ideal of all multiples of p so then this is a field well it's automatically a ring because if you take any ring and quotient out by an ideal then that's always a ring I guess I should say commutative ring but whatever so the problem is existence of inverses in other words if we've got an element of this ring does it have an inverse well suppose we have some element qx in kx over p with q not equal 0 in this this means that q and p are co-prime in k of x because p is irreducible and we're assuming q is not equal to 0 in here that means q is not a multiple of p you should recall that the ring of polynomials over a field is a Euclidean domain and a unique factorization domain now since it's a Euclidean domain since these two elements are co-prime we can find polynomials Ax times q of x plus B of x times p of x such that this is equal to 1 and for do this we use the Euclidean algorithm as in a basic ring theory which is supposed to be a prerequisite for this course and now if you look at this it just means that A of x is the inverse of q of x if p is equal to 0 so in k of x over p of x so elements of this ring have inverses and you can actually find the inverse using the Euclidean algorithm for polynomials if you like so let's get back to what we are trying to show we're trying to show that if alpha is algebraic this implies alpha is in a finite extension of k so well alpha is algebraic means that alpha is a root of p x in k of x where p can be assumed to be irreducible so all we do is we look at the field k of x modulo p of x and this contains k and it also contains l and now we can have a map from this field to l which just takes x to alpha and it's well defined because alpha is a root of this polynomial p which makes it easy to extend which means that the map from k of x to l vanishes on p of x the image of k of x over p of x in l is a field containing alpha and furthermore we see this field has dimension is equal to the degree of the polynomial p over l so it's a finite extension so this is a finite extension of k so this shows that not only are elements of finite extensions algebraic but that algebraic elements are always contained in finite extensions so next we ask how does the degree behave if we've got a tower of extensions so as I mentioned we're going to spend a lot of time working with long towers of extensions and we want to know what's the degree of m over k and it's equal to the degree of m over l times the degree of l over k so the degree is multiplicative in extensions and you can ask why? well this is very easy what we do is we pick a basis x1 up to xm of l over k where m is equal to l over k and we pick a basis y1 to yn for m over l where n is of course just the degree of m over l and then we check that the numbers xi, yj for one less than or equal to i less than or equal to m one less than or equal to j less than or equal to n form a basis for the vector space m over the field k and I'm not going to carry out this check because it's very easy you can do it as an exercise or something if you want the main thing we're going to use in the next few minutes is that in particular if this extension is finite and this extension is finite then this extension here is finite so we really need a very special case of this although we'll be using the full result later so now we're going to show that if alpha and beta in some extension l are algebraic over k and so are alpha plus beta alpha times beta alpha divided by beta and alpha minus beta probably beta is non-zero of course this isn't completely trivial so let's look at the following example suppose we look at the square root of 2 well that's algebraic and the cube root of 2 is also algebraic and the fifth root of 2 is also obviously algebraic but if you add them together and try and find a polynomial of these as a root you probably won't be able to unless you're extremely persistent or have a big computer that's because the smallest polynomial over the rationals with these three numbers as a root has degree 30 so it's not immediately obvious that the sum of two algebraic numbers is algebraic it's not so easy to write down an explicit polynomial with this as a root so we can say this is algebraic but maybe not immediately obvious that it's algebraic well here's one way to see it, this is algebraic what we do is we look at the field k and it's contained in the field k of alpha and this is going to be a finite extension so this means the smallest subfield containing k and alpha as alpha is algebraic this is some finite extension which degrees the degree of alpha and then this is containing the extension k alpha beta because beta is algebraic over k so it's certainly algebraic over k of alpha so this is also a finite extension and the degree is less than or equal to the degree of k of beta over k it might actually be less than this because beta will satisfy an irreducible polynomial over k because it's algebraic but this polynomial might become reducible over k of alpha so this degree here might actually be a little bit less than you expect but this doesn't really matter the point is these two extensions are finite so this whole big extension is finite and what this means is that anything in this extension is finite over k and therefore algebraic so this contains alpha plus beta alpha beta and so on so alpha plus beta and alpha beta and so on all algebraic I'm just reminding that although this proof looks almost trivial if you actually try and unravel it and find the polynomial explicitly you'll suddenly discover you're doing some tedious and messy linear algebra there's a similar result where we can show that if alpha is a root of a polynomial with algebraic coefficients this means again we're working over a field k and alpha is in some field l so then alpha is algebraic and what this means is so if alpha to the n plus a n minus 1 alpha to the n minus 1 and so on plus a 0 equals 0 and if all the a i are algebraic this implies alpha is algebraic well we can prove this in a very similar way to our proof of the sum of two algebraic numbers as algebraic so all we do is we take our field k and then we join a 0 and then we join a 0 and a 1 and we go all the way up to k a 0 up to a n minus 1 and then we join k a 0 up to a n minus 1 and at this point we do something a little bit different we join alpha so I'll write alpha in fluorescent pink to remind you that's not a n and all these extensions here are finite extensions that's because we assumed all the a i's are algebraic and this extension here is finite and the reason this is finite is because we've got a polynomial satisfied by alpha with roots in this field so just as before we see this big extension here is finite and since alpha is now in a finite extension of k we see that alpha is algebraic so let's see an example of this we showed that e and pi well we mentioned that e and pi are transcendental it's a very hard open problem say is e plus pi transcendental I mean everybody believes it is it would be really astonishing if it wasn't but proving it's transcendental seems to be beyond what transcendental number theory can do at the moment we can also ask is e times pi transcendental and again this is an incredibly difficult open question the answer to both of them is probably yes but nobody knows how to prove it but what I'm going to do is I'm going to solve one of these so let's have a theorem e plus pi or e times pi is transcendental and the proof of this is almost trivial what we do is we look at the polynomial x squared minus e plus pi x plus e times pi and the roots are obviously e and pi now if e plus pi and e times pi are both algebraic this implies the roots this implies the coefficients to this polynomial algebraic so this would imply the roots e and pi are algebraic well we know they're not by Hermit and Lindemann's theorem so e plus pi and e plus pi can't both be algebraic so here I've got two impossibly difficult questions almost impossibly difficult but I can answer one of them I just don't know which of them I can answer actually there's a sort of slight logical problem here that there's a philosophy of mathematics called intuitionism which says that if you assert you've proved either a or b this means you've either proved a or you've proved b and an intuitionist would deny that I've proved that one of these two numbers is transcendental because I've no idea how to prove e plus pi is transcendental and I've no idea how to prove e times pi is transcendental so an intuitionist would say I'm not allowed to assert that I've proved that one of these numbers is transcendental but whatever in classical mathematics I've proved that at least one of them is transcendental okay that's enough for the review of extensions and algebraic numbers the next lecture will be on splitting fields of polynomials