 So this lecture is part of an online commutative algebra course and will be about fitting ideals. So first of all where does the word fitting come from? Well it's actually the name of a guy who invented it, it's not some funny adjective or anything like that. That was back in the 1930s. So the fitting ideal is sort of invariance for a finitely generated module M over any ring R. You might think if the module is finitely generated the ring should be notary, but in fact we don't need to use that. And the fitting ideals are going to be a sequence of ideals. There's the zeroth fitting ideal of M, which is contained in the first fitting ideal of M, which is contained in the second and so on, you get the idea. And the ith fitting ideal, let's call it the nth fitting ideal of M, is a sort of obstruction to generating M by n elements. Well it's not really because people got things a bit muddled up. Instead of working with the fitting ideals it would really be better to work with the quotient by the fitting ideal. So you take the quotient of R by the nth fitting ideal and this is now a module, a cyclic module. And it's really this module which is the obstruction because if M is generated by n elements then this quotient, this module vanishes. So when you say this is an obstruction, it's a sort of funny obstruction, that if M is generated by n elements this isn't zero, it's actually equal to the whole ring R. So as usually in mathematics definitions and terminology have got a little bit muddled up. So it's for historical reasons we use the fitting ideal rather than the quotient of R by the fitting ideal, which should probably be more sensible. So let's start by defining the zeroth fitting ideal. So we're going to define the zeroth fitting ideal of a module M. Now M is finitely generated, so it has a presentation. Well which means first of all there's a map from R to the n onto M. So this has a basis corresponding to the generators, let's call them G1 up to Gn of the module M. And then we need some relations and there may be an infinite number of relations. Remember we're working over a ring that might not be fine, might not be notherian, so there's no particular reason why the kernel of this map should be finitely generated. So we get some possibly infinite module, infinitely generated module of relations. And let's write down what the relations are. Well the first relation is going to be given by elements R11 up to R1n and we'll give you the relation R11G1 plus R1nGn equals zero. And then we get R21 up to R2n and we get a relation R12G1 plus all the way up to R, it should be 21, sorry about that, up to R2nGn is equal to zero. And so on. So we get R31 something and what we get here is a matrix with n columns corresponding to the generators and possibly an infinite number of rows corresponding to the relation, so it might be an infinite matrix. And now we can define the i-fitting ideal is the ideal generated by the determinants of all n by n minors of this matrix. What that means is you pick n rows and the elements of those n rows will give you an n by n matrix and you take the determinant of that and you do that for every possible subset of n rows. So that's really rather a lot of them and that gives you loads and loads of determinants and the ideal of R generated by all these determinants is a zero fitting ideal. So it's a rather complicated looking definition. Okay fine, well there's one obvious problem. This seems to depend on the presentation because a module m will have many different presentations and if someone takes a completely different presentation they will have a completely different matrix here and get a completely different collection of determinants and are they really going to end up with the same fitting ideal if you do that? Well the answer is yes you do end up with the same fitting ideal. So let's explain why. So how can you change a presentation? Well there are some obvious ways of doing it. First of all we can add a new generator g to m but then we also have to add a new relation g equals R1 g1 plus plus Rn gn. If we've got n generators we can add an n plus 1th generator to provide we also add a relation saying it's a linear combination of the generators we first thought of. Secondly we can add a new relation as a linear combination of known relations and we can add an infinite number of new relations. If our ring was finitely presented we'd only have to add a finite number of new relations but in general you might need to add an infinite number all at once as a single step. And we can also do the inverses of these. We can remove a generator if one of our relations says it's a linear combination of previous relations and we can remove a relation if it's a linear combination of known relations. And any two presentations are related by these operations and their inverses. And this is quite easy to see because if we've got two presentations so we've got R to the m goes to m goes to 0 R to the something else goes to R to the n goes to m goes to 0 and what we do is we take these generators and then we add the generators of this other presentation one by one so we apply step one a few times so we now get a presentation of m with m plus n generators. And then we can add all the relations of this second presentation to that which we can do using step two. So without modifying m we've added all the generators of the second presentation to this and we've added all the relations of the second presentation to this. And we can now undo these steps we can now throw out all the relations of the first presentation since they are now redundant and then we can throw out all the generators of the first presentation since they're now redundant using the inverse of step one. So all we need to do is to show that the fitting ideal doesn't change if we do these two steps. So the fitting ideal of m does not change under steps one and two. And let's see why this is so. Well let's have a look at what step one does to the matrix of generators. So we've got this original matrix of relations like this and then we're adding a new generator which means we add a new column and obviously this new generator doesn't affect the old relations but we have this new relation saying it's equal to r1, it's equal to some linear combination of the old generators. So our new matrix adds one extra row and one extra column to our old matrix. And now we're looking at the old definition of the fitting matrix is the determinant of n by n matrices of the old matrix. The new definition means we're looking at determinants of n plus one by n plus one matrices, sub matrices of this new one. But you can see that these two are actually the same because if we've got n by n matrix in here then we can just add this row and this column to it and that will have the same determinant up to sign because if you add a new column with only one one in it then the only way of getting a non-zero determinant is then the determinant of this n plus one by n plus one matrix will be the same as the determinant of the smaller n by n matrix. And conversely, if you choose n plus one rows here, you're generally better have the first row as one of these rows otherwise your last column will be zero so you're getting the determinant of n plus one rows here including this first row and then if you just cut out the first row and the first column that's not going to change the determinant. So the determinants of this matrix, the n by n determinants of this matrix are more or less the same as the n plus one by n plus one determinants of this bigger matrix. So step one doesn't change the fitting ideal. Step two on the other hand, what we have is we might have our old matrix and then we're adding in some new relations which are linear combinations of these old ones and again this doesn't change the determinants of an n by n matrix because if you take n rows of this new matrix including this row then that's going to be a linear combination of the matrices you get by taking one of the rows in this linear combination. So it would be a linear combination of determinants you get by taking n rows from your original matrix. So both step one and step two don't change the ideal you get by taking all the determinants. So the zero fitting ideal of n actually doesn't depend on the presentation. So let's see an example for this. Let's take m to be a finite abelian group which is just a z module of course. And try and work out what a zero fitting ideal is. Well, m can be written as a sum of cyclic groups so it's going to be z over n one z plus z over n two z and so on. And this gives an obvious presentation and the corresponding matrix of generators is just going to look like n one n two up to whatever the last one is. And there's only going to be one way of taking an n by n determinant of this so the determinant is going to be n one n two up to n whatever it is. And the fitting ideal will be generated by this so the fitting ideal zero fitting ideal of m is just the ideal generated by the order of the abelian group m. So in other words z modulo fitting ideal of m is just a cyclic abelian group with the same order as m. So in particular we see that this vanishes if and only if m is generated by zero elements. So this is an example that the zero fitting ideal is the obstruction of generating m by zero elements in this case. Now that's done the zero fitting ideal. What about the higher fitting ideals? So let's have a look at the ith fitting ideal of m. Well this is going to be defined in the same way. So you remember we've got a presentation r to the something goes to r to the n goes to m goes to zero. And we've got this big matrix r one one r one n r two one and so on. And we take the ideal generated by the determinants of the n minus i times n minus i minus. So we take n minus i rows and n minus i columns and take all the determinants of those. Well as before this does not depend on the presentation. The proof is similar but maybe a slightly more tricky because instead of the n by n minus you've now also got the possibility of choosing only some of the columns. But otherwise the proof is fairly similar. So let's try and work out the fitting ideals of finitely generated Abelian group. So I suppose m is a finitely generated Abelian group. So we can write m is equal to z to the j. So we might take j copies of z plus z over n one z. So on plus z over n k z. So now what we're going to do is we know we can do this so that n one divides n two and two divides n three and so on. And if we do that then n one n two and so on are uniquely determined. So we have the following presentation where we can write down the presentation as a sort of matrix. Write it as a big one. We can write some zeros at the beginning and then we get n one n two n three and so on up to n k down here. So we can easily write the presentation as a diagonal matrix which makes it really easy to work out what the determinants of sub matrices are. And now you can see from this that the zero fitting ideal of m is going to be zero as long as j is greater than one. And this will be, you go up to the jth fitting ideal is zero. And then if you go up to the j plus one fitting ideal of m, this will be n one n two up to n k because it will be the determinant of this sub matrix. And then the jth plus two ideal of m, well you've got a bit of choice here because you can pick all but one of these. But you want to take the ideal generated by all possibilities. You take the smallest possible determinant you can get by throwing out one of these. So that turns out to be n one n two up to n k minus one because k is the biggest number. And this would go all the way up to the fitting j plus k is now equal to the ideal one. So here we see that m can be generated by j plus k elements. So this j plus kth fitting ideal becomes the whole ring of integers. So in other words, the obstruction to generating m by j plus k elements vanishes. You'll also notice that the finitely generated Abelian group is actually completely determined by its sequence of fitting ideals. This isn't actually true for more general rings. You've kind of different modules with the same sequence of fitting ideals. You can also think of the fitting ideal in this way. If you take the z-modulo, the i-fitting ideal of m, you can think of this as being the size of what is left over after killing i elements of m, or rather this is the minimum size of what is left over. So you try and choose i elements of m to make the quotient by those i elements as small as possible and then z-modulo, the i-fitting ideal, that that's its size. For example, if you just... never mind. So for any module m, m being generated by i elements implies the i-fitting ideal of m is equal to zero. Sorry, it's equal to the whole of r. It's the quotient by this that's equal to zero. Converse is true for finitely generated modules over the integers or over local rings as well. We won't actually prove this, although they mention it next time when we're looking at local rings that are local complete intersections. Okay, so that's the end of the summary of fitting ideals. We're going to use fitting ideals next lecture when we talk about local complete intersection rings.