 This lecture is part of an online course on commutative algebra and will be about notarian modules. So we're going to define them and then give an application of them to a theorem of Nerter, showing that invariance of finite group are finitely generated, even over fields of positive characteristic. So let's first have the definition. So module M, so this is a module over some ring R is called notarian, if all sub-modules are finitely generated as modules. So in particular, if R is considered as a module over itself, we see that R is notarian as a ring, it's just the same as saying that R is notarian as an R module, because in this case, the sub-modules of R considered as an R module are just exactly the same as ideals. Now for notarian rings, we had two other equivalent conditions and in exactly the same way for modules, we have two other conditions. So this condition that all sub-modules are finitely generated is equivalent to the condition that every non-empty set of modules, or sub-modules has a maximal element. And this is also equivalent to the condition that every strictly increasing chain of sub-modules M1 contained in M2 contained in and so on. So these are sub-modules of M is finite. And the proof that these three conditions are equivalent is almost exactly the same as the proof we gave for ideals. So I'm not going to bother doing it. Next, we would like to know that the sum of two notarian modules is notarian. And in fact, we can prove a slightly more general condition. Suppose we've got an exact sequence of modules. So here A, B and C are R modules. Then B is notarian is equivalent to A and C both being notarian. So the implication in this direction is very easy and I'll just omit it. The implication in this direction is also kind of easy but you need to think for about three seconds to see it. So what we do is let's take a sub-module M of B. Then we want to find a finite set of generators for it. And we can form a finite set of generators for M as follows. First of all, we take a finite set of elements of M such that the images in C generate the image of M in C. And we can do that because the image of M in C is finitely generated because C is notarian. Then we take a finite set of elements of M generating M intersection A. And we can do that because A is notarian. So the intersection of M with A is finitely generated. And then you can combine these two sets and you see that fairly easily these generate the module M. They've checked that they generate M for anyone who wants an exercise. So a special case, if A and B are notarian this implies A, the direct sum of A and B is notarian because there's just an exact sequence. Norc goes to A, goes to A plus B, goes to B, goes to zero. And we can see from this that any finitely generated module over notarian ring is also notarian. And this follows because if the module M is finitely generated it means there's a map from R to the N on to M. And we know that R to the N is notarian because R is notarian and the sum of N copies of a notarian module is notarian. So M is a quotient of a notarian module so M is also notarian. So this is very widely used in commutative algebra and awful lot of commutative algebra spends its time studying finitely generated modules over notarian rings because these modules are much easier to handle than arbitrary modules. So now we'll have a couple of applications of notarian modules. So the first application which we've mentioned several times is that scissorges are finitely generated if invariants are. So remember invariance when you say they're finitely generated it means finitely generated as an algebra whereas scissorges you're talking about being finitely generated as an ideal. And we've mentioned this before so I'll just go through it fairly quickly. So suppose we've got a ring of invariants and suppose you've managed to show it's finitely generated that means we've got a ring R which is a polynomial ring in N generators mapping on to the ring of invariants. And this means that the first order scissorges are given by the kernel of this map and they're the relations between these basic invariants and it's finitely generated so we can have a map R to the N on to R where N is finite. So these are the first order scissorges and they're finitely generated because the ring R is notarian. So I should have said here we're really doing higher order scissorges of finitely generated. And now the key point is that since R is notarian and this is a finite module over R we see that this is a notarian R module. And this means the kernel of the map from this to R which are the second order scissorges is also finitely generated. So these are the second order scissorges and this is finite because this module here is notarian. And now we can just repeat this process. This module is also notarian so the kernel which is third order scissorges is also finitely generated and so on. So all higher order scissorges are finitely generated because finitely generated modules over notarian rings are notarian. We can actually do the same thing here for any module M that's finitely generated over a notarian ring. Since it's finitely generated we can have a map from R to the N1 onto M and this module here is notarian so the kernel is finitely generated. So we can have a map from R to the N2 onto that and the kernel of this is finitely generated and so on. So we can take N1, N2, N3 or to be finite. For general modules over arbitrary rings we can do something like this except the numbers here might all be infinite which means we kind of lose control over them. So in particular M has a free resolution by finite free modules, finitely generated free modules. And you can try and extract information about M from this and there are two slight problems. The first problem is that this chain might be infinite and the second problem is that this resolution is not unique. In some cases, however, the chain can be taken to be finite it is almost unique up to trivial variations. So if the ring is R is KX1 up to XN for example Hilbert showed the chain can be taken to be finite and it's almost unique. It's almost unique in the sense that if it's not that there's a sort of minimal resolution you can do for graded resolutions of M and any other resolution has got from this minimal resolution by adding something stupid. So you could just add an R here and an R there and isomorphism from that one to that one and that'd be a sort of stupid way of making the resolution bigger. So apart from things like that for graded modules over polynomial rings we really do get a finite chain that is almost unique and you can use that to find lots of invariance of M for instance one obvious invariance is just these experiments here and there are some other more subtle invariance related to fitting ideals and things like that that we will discuss later. So the next application is a theorem due to Emmy Nerter which says that if G is a finite group acting on a vector space V so this is finite dimensional over the field K then the invariance of G are finitely generated. Okay, well you may think we Hilbert already proved this or Hilbert proved this for K having characteristic zero. So key point in Nerter's theorem is that K can have any characteristic and this gives an extra problem because if we're allowing characteristic greater than zero there's no Reynolds operator row in general. So you remember the Reynolds operator was the key ingredient of Hilbert's theorem that allowed us to go from finite generation of ideals to finite generation of algebras. And to see that the Reynolds operator might not exist we can look at the very following very simple example. This is the sort of key counter examples and awful lot of things. You just look at the group G consisting of all elements like this. So this is contained in say SL2 over the field K so we're allowing any entry there. So G is isomorphic to K under addition and it acts on two dimensional space over K. Now suppose we take K to be the finite field of order P so that G is finite. Then we see that we've got an exact sequence and nought goes to K, goes to K squared, goes to K, goes to zero with G acting on K squared and you can see that there's a one dimensional subspace of K squared that is acted on a trivial by G and the quotient G also acts trivially on. So G acts trivially on the subspace and the quotient space. However, the action of G on this space is not trivial. So here the action of G is non-trivial. So trivial just means it fixes all elements of these. So we've got this slightly bizarre case where G can act trivially on a subspace and on the quotient space but doesn't act trivially on the whole space. And what this means is suppose you want to find a Reynolds operator. What we would like to do is to find a map from here to here which splits it. Well, it can't split because if this representation was the sum of this representation and this representation, then G would act trivially on K squared. So this does not exist and a Reynolds operator is a sort of slightly more complicated and general version of this. If we've got a, here we've got an action of G on a space and we want to find a Reynolds operator mapping it onto the fixed vectors and it just doesn't exist here. So we seem to be stuck. So what do we do? Well, Nert has found an alternative argument using the fact that sub-modules of notarian modules are always finitely generated. So here's the proof of Nert's theorem. So suppose G acts on a ring of polynomials K X1 up to Xn. So G is acting on the space and n dimensional vector space over K and then we're going to look at the ring of polynomials over K. And now we notice that each XI is a root of the following polynomial. So we take X minus XI, X minus G1, XI, X minus G2, XI and so on, where the GIs and the identity element are all elements of G. So this is just a product of X minus G, XI product of all elements G. So if this ring is R, this is just a polynomial with coefficients in R. And now we notice the coefficients are invariant under G. In fact, they're just the elementary symmetric functions in the elements G1, X1, G2, X1, G3, X1 and so on for GI running through G. And since this set is permuted by G, all the elementary symmetric functions in this set are invariant under G. So we take the ring, so let's take the ring S to be the algebra generated by all these coefficients for all the XIs. So we've got S. Now this is contained in the algebra of invariants, which is contained in the ring R, which is K X1 to XN. And we know this is a finitely generated algebra because there are only a finite number of these invariants. The number of generators is at most N times the order of G because for each X that there are N possible XIs and for each XI, it's a root of a polynomial of degree G. So we're taking these G coefficients. And now we want to show that R is finitely generated S module. Okay, we're switching between being finitely generated as an algebra and finitely generated as a module. So you've got to be really careful to pay attention to this word here. It's trivial that it's finitely generated as an S algebra because it's just generated by these. What we're saying is that it's also finitely generated as a module over S. And the point is that each XI is integral over S. This means it's a root of a polynomial with leading coefficient one and the other coefficients in S. And so we can write down an explicit generating set for the ring R, which is K, X1 and so on as an S module. It just consists of the elements X1 to the N1, X2 to the N2 and so on, XN to the... I guess I shouldn't have used N as an exponent with all the NIs less than the order of G. That's because X1 to the power of the order of G is a linear combination of one X1 to X1 to the G minus one and so on because it's a root of a polynomial with leading coefficient equal to that. And similarly X1 to the G plus one is also linear combination of smaller powers and so on. So these monomials with all values of exponents are certainly basis for the mod, since they span the module R over S. And using the polynomial satisfied by the XIs, we can reduce all these exponents to being at most G. So we have at most G to the N generators for R as S module. We don't need the exact number of generators. I'm just giving this number because it might help see what's going on. So let's go back to this. We've got S, which is contained in the invariance, which is contained in R, which is K, X1 and so on. And let's see what we know. This is a finitely generated K algebra. So first of all, we notice that that's a finitely generated K algebra. Second, we see that this is a finitely generated S module, point three, so it is a notarian S module. So since this is a finitely generated algebra, so it is notarian ring. And we know that any finitely generated module over a notarian ring is a notarian module. That's what we've just proved before. So step four says that this is a finitely generated S module. Because this follows from step three, that any sub-module of a notarian module is finitely generated. So now we have step five. So it is a finitely generated K algebra. Again, we're switching from being finitely generated as a module over S to being finitely generated as an algebra over K. And we can see this because the generators, are given as follows. First of all, we take the generators of S as a K algebra. And second, we add the generators of the invariance as an S module. And it's so easy to see that if we take all these, then they generate the invariance as a K algebra. So that gives the proof of Nertis theorem. You might say, well, Nertis theorem works in characteristic P and in characteristic zero. So why don't we just use that instead of Hilbert's result, which only works in characteristic zero? Well, the reason is that Hilbert's result, so Hilbert's proof works for some infinite groups, such as special linear groups. Nertis proof only works for finite groups, because you remember, we had to take a product over all elements of the group. And if the group is infinite, that doesn't give you, that doesn't make much sense. So you can ask, if G is an infinite group in characteristic P greater than zero, is the ring of invariance a finitely generated algebra? So in characteristic zero, Hilbert's work gives a satisfactory solution for finite groups, Nertis group gives a satisfactory solution, so it leaves this case here. And this turns out to be much harder. The answer is usually yes. The exact result is as follows. So the following the equivalent, here G is an algebraic group, that just means essentially a group defined by some polynomials, such as a typical example of the special linear group over field K. So the following equivalent, first of all, if G acts on a finitely generated algebra over a field K, the ring of invariance is always finitely generated algebra. This is equivalent to the condition that G is reductive. Well, what does reductive mean? This means there's no normal subgroups that form isomorphic to the sum of N copies of K. Let's take K to be algebraically closed because I can't quite remember if this is necessary. And the third condition is that if G acts on a vector space, K to the N with a fixed vector, V not equal to naught, then there is some polynomial fixed by G, some polynomial F fixed by G has F of naught does not equal F of V. So this was proved by Nagata and Habouche. So Nagata's role was to show that condition three implies condition one, and Habouche showed that condition as if G is reductive, then condition three holds. So condition three sort of says there's a non-linear Reynolds operator. So in characteristic zero, we can find a polynomial F that's of degree one. It's just a linear polynomial. And this polynomial F is very close to relation to the Reynolds operator we wrote down. So what happens in characteristic P is you can't take the Reynolds operator to be linear. It has to be some sort of funny non-linear function. This makes everything rather more difficult. So it's also a little bit paradoxal is you might have thought that the easiest algebraic group is just the additive group of the field K that things like the special linear group look much more difficult to deal with. But it turns out that it's the additive group of the field that's causing all the problems with things being not finitely generated. So roughly speaking, G, invariance of G of finitely generated provided the additive group of the field isn't involved in G. I guess I should have said that the condition that one implies two is also essentially due to Nagata who found an example where the ring of invariance isn't finitely generated if G is not productive. Okay, so that's all I want to say about the two in modules for the moment.