 This lecture is part of an online Ganawar theory course and will be about the norm and the trace of an extension of fields. So we will take k contained m to be a finite extension of fields. Now, if alpha is an element of m, then multiplication by alpha. So we might take any element v in m to alpha times v is a linear transformation. So here we're thinking of m as being just a vector space over k and we're thinking of the linear transformation of m as a vector space. So we don't really worry about its field structure. And this linear transformation has a trace and a determinant, like all linear transformations, which are just going to be elements of k. And in the case of the special linear transformation and multiplication by alpha, these are called the trace of alpha and not the determinant of alpha, the norm of alpha. So as you see, terminology has, as usual, got slightly muddled up. For example, let's just work out the trace and norm when the extension is r contained in the complex numbers. So it's degree two. And let's take alpha to be a complex number x plus i y. And we want to compute what this linear transformation is. So let's pick a basis, one and i for the complex numbers. And then multiplication by x plus i y is given by two by two matrix in this basis. And you can easily check it's x y minus y x. So the trace is two x, which is two times the real part of alpha. And the norm is x squared plus y squared, which is the absolute value of alpha squared. So you see the trace and the norm are a sort of generalization of the real part and the absolute value of a complex number if you ignore these factors of two here. So let's try and calculate the trace and norm for an element of an extension. So first suppose that m is equal to k of alpha. In other words, alpha generates m as an extension of k. Well, then look at the minimal polynomial of alpha. So it's going to be alpha to the n plus a n minus one alpha to the n minus one and so on plus a naught. And now this is the same as the characteristic polynomial considered as a linear transformation. And there are two ways of seeing this. You can either use the Cayley Hamilton theorem, which implies it immediately. And if you forgot what the Cayley Hamilton theorem is, I'm not going to tell you what it is, but you'll then just have to use the other way, which is we can calculate the characteristic polynomial using the basis one alpha, alpha squared, up to alpha to the n minus one. So you can either do it by explicit calculation or by quoting the Cayley Hamilton theorem. So the trace of alpha is just minus this coefficient of the characteristic polynomial or minimal polynomial. So it's minus a n minus one and the norm of alpha is plus or minus a zero, depending on whether n is even or odd. And in particular, we see the trace is just the sum of the roots of the minimal polynomial. So if I call this p of x, this is just the sum of the roots of p. And this is equal to the product of the roots of p. Notice that the roots are not necessarily in the extension m because this might not be Galois, but the roots are in K contained and K alpha contained in the splitting field. So the roots are contained in the splitting field and you can then add up all the roots and that's an element of K. And you can also think of this as being the sum of all conjugates of alpha under some large Galois group. You've got to be a little bit more careful if alpha does not generate m. It's a bit annoying that things behave differently if alpha does not generate m, because otherwise you could define the norm and the trace just by looking at the minimum polynomial, but this just doesn't work in general. So in general what happens is K is contained in K of alpha which is contained in m and you can then check that in general if alpha has minimal polynomial, alpha to the n plus a n minus 1 alpha to the n minus 1 and so on equals 0, then you can see the trace of alpha is now a n minus 1 times the index of K alpha in m. That's not very difficult to check. I'll just leave it as an exercise. And similarly the norm of alpha, which I'll denote by n, is equal to plus or minus a 0 to the power of this degree. In particular we notice that the trace of alpha and the norm of alpha depend on which field you are using. So the trace of alpha in the field K of alpha is not usually the same as the trace of alpha in the bigger field of m. For this reason people sometimes write trace of m over K of alpha and norm of m over K of alpha in order to be completely unambiguous. But the problem with being unambiguous is the notation gets so cluttered up it becomes hard to read. So quite often it's better just to write trace or norm and cope with the ambiguity. By the way in some older books the trace is sometimes called the spur of alpha and written sp. If you get annoyed about the fact that the trace depends on which field you're working on you can work with the absolute trace. So the absolute trace is you take the trace of m over K of alpha for any extension m containing alpha and you then divide it by the index and you can see that this then doesn't vary if you replace m by bigger field. I don't really recommend doing this because the problem is this might be 0 if you're working in a field of characteristic p greater than 0. So you sometimes find this breaks down. You can try and do the same and define an absolute norm. You take norm of m over K and you raise it to the power of 1 over the index of m over K and this is even worse than the trace because you can't always take these roots and even if you do take roots it's kind of ambiguous. It occasionally works. One case when it does work is if you take the extension r contained in C and then the absolute trace is just the real part because you're dividing the trace we worked out earlier by 2 and the absolute norm is then just the absolute value. So apart from this special case the absolute trace and the absolute norm seem to be more trouble than they are worth. Next we'll have a few examples of traces and norms turning up in number theory. There's a sort of problem you get quite a lot in algebraic number theory, especially in class field theory where you have some extension of fields or possibly rings of a dels or whatever and you want to know what is the quotient of the non-zero elements of K by the norm of the non-zero elements of m. So this is going to be a group and it turns out quite a lot. It turns out to be a sort of zero-taked cohomology group if you're into taked cohomology which partly explains why it's so common. So let's just give some examples of this. First of all let's just take r contained in C. Well then you've got the non-zero real numbers divided by norms of non-zero complex numbers which is just the real numbers over positive real numbers which is a group of order two. So it's a group of order two whose elements we might write as plus or minus one. Let's look at another example. Let's just take q contained in the Gaussian numbers q of i and then you know q star is sort of generated by elements minus one, two, three, five and so on and q of i is similarly generated by units and then we have to take the primes of q of i which are one plus i and three two plus or minus i, seven, eleven, three plus or minus two i and so on and if you take the norms of these you get one, two, three squared, five, seven squared, eleven squared, thirteen and so on. So what's happening is primes that are one mod four factor into the product of two Gaussian primes and primes that are three mod four don't. So you can now see what q modulo norms is. It's basically a sum of a lot of groups of order two corresponding to minus one, three, seven, eleven and so on. So we see that q star modulo the norm of q i star is sort of encoding a lot of information about which primes are sums of two squares and things like that. So this just indicates that this group can be both interesting and quite complicated. So the final example of k star over norm of m star we're going to look at is the case of finite fields. So let's take k and m to be finite. So k has order q for some prime power q and m has order that's number of elements q to the n. So k star has order q minus one and m star has order q to the n minus one. And let's calculate what is the norm of some element alpha for alpha in m. Well, if we've got a Galois extension you can see the trace of any element alpha is just the sum over g in the Galois group of g of alpha. And similarly the norm of alpha is just the sum of g over the Galois group of, so it's just the product of g in the Galois group of g of alpha. If alpha generates this Galois extension then this follows almost immediately because these expressions here are the two coefficients of the minimal polynomial giving the trace and the norm. If alpha doesn't generate m then you have to think about it slightly longer which I'm not going to bother to do. So anyway, using this we can now calculate the norm of the number alpha because we know what the Galois group is, g is generated by the Frobenius and in this case the Frobenius element takes alpha to alpha to the q which is the order of k. So this is not quite the same as taking alpha to alpha to the p which is the Frobenius over the finite field of order p. So the elements of g consist of one phi squared up to phi to the n minus one. So the norm of alpha is just equal to alpha times alpha to the q times alpha to the q squared up to times alpha to the q to the n minus one which is alpha to the one plus q plus q squared plus q to the n minus one. So the norm in a finite field is rather easy to write down and obviously you can write down a similar expression for the trace which we won't need. Now let's look at the map from k star rather the map from m star to k star and this has a kernel consisting of the things with norm alpha equal one and this is order q to the n minus one and this is order q minus one and let's think about how many elements alpha have norm one. Well this says alpha to the one plus q plus q to the n minus one minus one equals naught and you notice this is a polynomial equation so it is at most one plus q plus q to the n minus one roots. So this group here has order at most that. Well now the image has order greater than or equal to q to the n minus one over the order of the kernel so the order of the kernel is at most that so the image has order at most this sorry that's the one in the exponent by action which is equal to q minus one so the image has the same order as k star so the map from m star to k star given by taking the norm is onto. So in this case k star over norm of m star is equal to the trivial group. So that's one place the norm turns up it turns up because you want to take k star module of the norm of an extension field quite often. The other application we're going to look at is the following problem. Find the algebraic integers in some algebraic number fields. So just recall what an algebraic integer is. An algebraic integer alpha is something that is a root of x to the n plus a n minus one x to the n minus one plus a naught equals naught with all a i in z. And what distinguishes this from a general algebraic number is the coefficient of the leading term is just one. So you can see that ordinary integers are just the rational numbers with this property. And we can recall that alpha is an algebraic integer is equivalent to saying that z of alpha is a finitely generated module over z. And from this it easily follows that if alpha and beta algebraic integers, so are alpha plus or minus beta and alpha times beta. So the algebraic integers in any given algebraic number field form a ring. And the problem is to find out what this ring is and this can be quite tricky. For example, let's look at z root minus, sorry, let's look at q root minus three and ask what are the algebraic integers in it. And the obvious guess is z root minus three. And this is wrong because as we've seen there's a cube root of unity omega which is minus one plus root minus three over two. And this doesn't have integer coefficients when you write it as a polynomial in root minus three but omega squared plus omega plus one equals zero, so it's still an algebraic integer. And finding the algebraic integers in a given algebraic number field is actually a really rather tricky problem. But the norm and the tracing give some help in finding this and I'm just going to illustrate this for quadratic number fields. So we can ask what are the algebraic integers of q root n where I'm going to take n to be square free because you would just miss out any square factors and I'm going to take n to be an integer. So let's take an element alpha equals a plus b root n in q root n. Here alpha a and b are going to be rational numbers and we want alpha to be an algebraic integer and find out what a and b is. Actually it's really easy to figure out when alpha is an algebraic integer just by kind of looking at the formula for the root of a quadratic equation but we're going to do it using norm and trace just to illustrate that norm and trace is sometimes useful. Well the key point is that if alpha is an algebraic integer this implies the norm of alpha and the trace of alpha are integers because they are algebraic integers and they're in the rational numbers they must be ordinary integers. So the trace of alpha which is 2 alpha, 2a is an integer and the norm of alpha which is a plus b root n times a minus b root n is a squared minus b squared n is also an integer. And let's see what we can deduce about this. Well obviously a and b can be any integers and that will give alpha an algebraic integer so the question is when can we have an algebraic number alpha with a and b not integers and we see from this that a must either be an integer or it could be an integer plus a half and we can then just subtract the integer and ask can a be a half? And if a is a half this implies that norm of alpha equals a half squared minus b squared n is an integer. So this means b squared n is in the integers plus a quarter so 4b squared n is an integer and since b is square free this implies 2b is also an integer. And if b is an integer then you can easily check that a can't be half an integer so the only possibility is a equals a half b equals a half so we can ask is a half plus a half root n an algebraic integer and if you work it out from the norm of this is a quarter minus a quarter n so we must have n congruent to 1 mod 4 so we found out what the algebraic integers of q root n are here n is in z and is square free so if n is congruent to 1 mod 4 it's generated by z and we can have 1 plus root n over 2 on the other hand if n is congruent to 2 or 3 mod 4 then all we get is the obvious ones where we where we include the square root of n over z so next lecture we'll be looking at another application of the trace for defining a bilinear form on m in fact we define a bilinear form on m by saying alpha beta is the trace of alpha beta and this bilinear form turns out to be really useful