 So this lecture is part of a Galois theory course and will be about the discriminant of a field extension k contains an m, which we're going to assume as finite. So we first recall that if phi is a vector space over k, again, finite dimensional, suppose it's got a symmetric bilinear form, then we can define the discriminant of this form as follows. What we do is we pick a basis v1 up to vn of v and we look at the matrix whose entries are just all the inner products, v1, v1, v1, v2, and so on, v2, v1, whatever. And the determinant of this matrix is called the discriminant of the bilinear form. Well, there's a bit of a problem because it depends on the choice of basis. Suppose you choose a new basis, w1 up to wn, how does the discriminant change? Well, the new basis w, let's just write w for the collection of all w i's, is given by a times v for some non-singular matrix a. And you find the discriminant with respect to v is, so with respect to w, is then equal to a squared times the discriminant with respect to the basis v. It's a rather easy calculation. So the discriminant is a well-defined element of k up to multiplication by elements of, the squares of non-zero elements of k. So if the discriminant is non-zero, then it's a well-defined element of the group of elements of k modulo, the squares of elements of k. So it gives you an invariant of a symmetric bilinear form which doesn't quite take values in non-zero elements of k but in non-zero elements of k modulo squares. So obviously if we've got a field extension, then we have a bilinear form on m defined by, a, b is the trace of a, b. You remember we defined the trace of an element of m in the previous lecture just to be its trace as a linear transformation on m. So we now have the following problem. Calculate the discriminant of m. We should really say the discriminant of the extension m over k but let's be sloppy and not bother mentioning k all the time. Well first of all the discriminant can be zero and this is a slightly unpleasant thing if the discriminant is zero and if something unpleasant is going on you've probably got an inseparable extension and indeed if you look at the standard inseparable extension so if we take the field of all rational functions in t over some field k prime characteristic it's got a subfield generated by t to the p and you can check the trace for this is always zero so in particular the bilinear form is identically zero and the discriminant is zero and that's not so nice. But what we're going to do is to prove that it's always non-zero for separable extensions. So suppose m is separable over k and it's finite of course so by the primitive element theorem m is equal to k of alpha for some element alpha and alpha has a minimal polynomial so alpha to the n plus a n minus 1 alpha to the n minus 1 and so on plus a zero is equal to zero so this is going to be the minimal polynomial and we remember a polynomial also has a discriminant. There's a discriminant of polynomial given by the product of alpha i minus alpha j all squared where this is a sum over i less than j and the alpha i are the roots of the polynomial and what we're going to do is to show that the discriminant of this extension is more or less the same as the discriminant of a polynomial. Well the discriminant of a polynomial is an element of the k and the discriminant of the extension is only an element of k modulo squares so they're not quite the same but we mean the obvious thing that the discriminant of the extension is the image of the discriminant of the polynomial under the obvious map. So let's calculate the discriminant of k of alpha. Well we have to pick a basis for m as a vector space over over k and there's a really obvious basis we're just going to take one alpha alpha squared up to alpha to the n minus 1 doing anything else would be kind of perverse. The n is of course the degree of m over k. So the discriminant is then going to be the determinant of the following matrix. We take the trace of 1, the trace of alpha, trace of alpha squared, trace of alpha, trace of alpha squared and so on. So we just take a matrix whose elements are traces of various powers of alpha because here the basis is given by V i equals alpha to the i minus 1 so what we want is the trace of V i, V j which is just the trace of alpha to the i plus j minus 2 so we're getting this matrix here. And the trace of alpha to the n is just the sum of alpha i to the n where alpha i are the roots of our polynomial p. The roots need not be in m so we should really take k contained in m contained in some splitting field and the roots will lie in the splitting field not in m but that doesn't really matter. So if we write the trace of alpha as the sum of alpha to the n our discriminant is the determinant of the following polynomial. We take sum of alpha i to the 0, sum of alpha i to the 1, sum of alpha i to the 1, sum of alpha i squared and so on so we just have a big matrix whose entries are sums of powers of the roots so these sums are just over all roots of the polynomial and we can factor this as follows it's just a product of this matrix and so by this matrix so here the columns are just powers of the roots and here the rows are just powers of the roots and now you recognize these matrices these are just the van der Mond matrices and their values are plus or minus the product of i less than j of alpha i minus alpha j there's some sort of sign in the van der Mond matrices that I can never actually remember but it doesn't really matter what the sign is because we're squaring the van der Mond matrix so each of these is given by this expression so this matrix here is just the product of i less than j of alpha i minus alpha j or squared which is the discriminant of the polynomial that's the minimal polynomial of alpha so we've shown that the discriminant of a field extension is almost the same as the discriminant of the minimal polynomial of something generating a field extension in particular if m over k is separable this implies that alpha i is not equal to alpha j for i not equal to j of course so the discriminant which is the product of alpha i minus alpha j or squared is none zero so the bilinear form trace of a given by a b was trace of a b is non singular and it's not difficult to check that conversely if the trace is non singular then the extension is separable so that's the discriminant of a field extension now let's give a few applications of it so the first example says are the fields let's take q of alpha and q of beta isomorphic where let's take alpha cubed plus alpha plus 1 equals 0 and beta cubed plus beta minus 1 equals 0 and in general it can be quite difficult to tell whether or not two fields are isomorphic I mean is there some polynomial in beta satisfying the equation for alpha well you would guess there probably isn't seems rather unlikely but can you actually prove it well we can just calculate the discriminants so you recall the discriminant of x cubed plus bx plus c is minus 4b cubed minus 27c squared so we can work out the discriminant of these these this the discriminant is equal to minus 4 minus 27 which is minus 31 here the discriminant is minus 4 plus 27 which is equal to 23 so these two fields are not isomorphic because the discriminants are different and you notice these have to be different in q star modulo q star squared you can't just say that they're different as integers you have to say they're not the ratio isn't the square of a rational number in fact the discriminant turns out to be a fairly powerful means of telling whether finite extensions of the rationals are isomorphic or not there's a theorem in algebraic number theory that says that for a given discriminant there are only a finite number of algebraic number fields with that discriminant actually in that theorem the discriminant is a slightly finer invariant because you can take the ring of integers of an algebraic number field and look at the discriminant of that and that will actually be a well-defined integer not just to not just something modulo squares by the way it's not actually a complete invariant for example suppose we take the three algebraic number fields q alpha q beta and q gamma where alpha cubed minus 18 alpha minus 6 equals 0 and beta cubed minus 36 beta minus 78 equals 0 and gamma cubed minus 54 gamma minus 150 equals 0 these all have discriminant 2 2 3 5 6 but they're all different as fields but this seems to be a fairly rare phenomenon as far as anyone has calculated so for the next question let's look at this field q of alpha where alpha cubed plus alpha plus 1 is equal to 0 and ask the following problem what are the algebraic integers in the field well the algebraic integers contain z of alpha because alpha is obviously an algebraic integer and the question is are there any other algebraic integers not not not in this field and this isn't you remember this is in general a rather subtle question for instance if you if you take beta squared plus 3 equals 0 then the algebraic integers are strictly bigger than z of beta because the algebraic integers as we saw last time are actually contained in z of minus 1 plus root minus 3 over 2 so we sometimes need to use fractions to get all algebraic integers well we can use the discriminant in order to in order to identify the algebraic integers here because let's look at the discriminant with the basis 1 alpha alpha squared is minus 31 and suppose that the actual ring of algebraic integers is bigger so we've got z of alpha is contained in the algebraic integers and this might have a basis say 1 beta and gamma and in this basis we will get some we can work out the discriminant of this basis well we can write 1 alpha and beta in terms of the basis 1 beta and gamma using some matrix a so so we would and then we find the discriminant using the basis 1 alpha alpha squared the discriminant using the basis 1 alpha and alpha squared is equal to the determinant of a squared times the discriminant using the basis 1 beta and gamma and so the problem is that this discriminant is minus 31 which is square free so the determinant of a has to be plus or minus 1 so in other words you can't find a strictly bigger ring of algebraic integers containing this ring here because the discriminant is square free so z alpha is all algebraic integers in q of alpha well you may wonder what happens if the discriminant isn't square free well if the discriminant contains a square factor then it can be really quite difficult to work out the full what the full ring of algebraic integers is and for that you can see a course on algebraic number theory okay so the next lecture we will be looking at a famous theorem called Hilbert's theorem 90 and I explain where this rather unmemorable name comes from