 Vijay Lakshmi Trivedi, who is from the Tata Institute of Fundamental Research in Mumbai, India, and she brought numerous contributions to combative algebra and algebraic geometry on topics like Hilbert-Samwell functions, Hilbert-Koz multiplicities, Frobenius splitting, and more recently introduced the concept of Hilbert-Koz basic function. With that being said, Professor Trivedi, please go ahead. Yeah, thanks, Florian, for a kind introduction. And I would like also to thank the organizers for giving me opportunity to give this talk. So this is the first time actually I'm giving a talk on a pen tip, so please bear with me. I'll be a little slow. So now today's talk, I'll be talking about the existence of HK multiplicity. And in fact, the proof is pretty self-contained and I'll be following the proof given by Monsky. So before that, let me start with the hypothesis which I'm going to presume throughout the lecture. So here, R is a local ring with a maximal ideal M and the residue field K. And the emission of R, actually you have seen in various lectures and of course the characteristic of R is prime P, which is positive. So M is a finitely generated R module. Now, this you have seen many times before, but just to be very clear, I'll repeat the notation. So E-theterated Frobenius denoted by F, this is no, sorry, there is no star here right now. It's the map, which is endomorphism given by R goes to R to the power P. And this makes, gives the another modular structure to the R, which is denoted as flora star R. And the model, I recall the elements of this particular ring F flora star E. I'll denote always this for just following the previous talks like this R that means it's an element R in R. But it's denoted F flora star E R. And as you know, the multiplication as a R module is given by, suppose R1 is an element of ring R, then F flora star E R is nothing but F flora star E R1 to the power P to the power R into R. So this you have seen many times before, but I'll be repeating it. Similarly, there is a model structure on the model denoted as F flora star M. And everybody's familiar with that. Now, another notation which I'm going to use, this is for a given ideal I. Okay, as defined as the set of ideal generated by set of elements X in X to the power Q. Yeah, this is, I'm sorry, this is total messed up. So here it is the set of ideal generated by elements of the type X to the power Q, where X belongs to I. So Q for me, it always denotes like a power of a prime. So let's say P to the power E or P to the power N. So it's very easy to check if I have a generator given by say elements X1 up to XM, then I to the power I Frobenius Q is given by ideal generated by the elements X1 Q XM Q. So this also you have seen earlier. So I'll recall the following theorem of which has been actually recall several times in the past lectures. So, but it's the fundamental theorem which begins this, I suppose the study of characteristic P methods for commutative algebra. So here, your R is a Noetherian ring. So note that I'm not saying it's a local ring, it's a Noetherian ring of characteristic P positive. And then R is regular if and only if F or lower star E of R is a flat R module. So that is for some E you can say for some E greater than one or for all E greater as equivalent. So note that this is so very funny the regularity assumption has nothing to do with characteristic, but it's translated in terms of a Frobenius map. And he's a one is the first person, I suppose, one of the leading ones who realize the importance of. So in this proof, be none of us approved this theorem but he, the ingredient one of the ingredient in this proof is the importance of looking this length function where M is a maximal ideal and relate to the regularity of the ring. So I'm going to state that part of the thing regarding the length function in the next lemma soon. So, before that, let me say make few important remarks. So, one of the thing you have noticed in many of the lectures, you require the assumption that R is F finite, because you'd like F lower star R which is a model here to be a finite R model you would like to basically would like to stick in a category where the all the models are finitely generated or R. So, that assumption was made, of course, many of most of the general things to satisfy that, but in this today's setup actually we can get by without assuming the F fine net assumption, and I'll explain why. So, this is one remark. So the thing is what you do, you take R and replace it by replace by another ring s. So what is the property of this ring is Sir s is complete local. So, advantage of R is like we correct the statement. So of course it is contains a field. So if I take a, for example, completion of our the it will, it will contain the residue field. That's one thing. And of course the maximum I might ideal of R if I extend to s is going to be the maximal ideal for s. That's the hypothesis and you can further you can assume the residue field is algebraically closed. So that you have seen in mass talk also so for example if I take a polynomial ring I completed I take a power series ring and I take the algebraic closure. And of course, taking replacing power series ring over K replacing it by K bar over power series ring over K bar is not exactly you are tensoring K bar over K, but you are further taking completion. But nevertheless, that operation is a faithfully flat extension so R to s is a faithfully flat extension. So this ring R by another ring s which is complete local, and it has a maximum ideal is generated by the same elements and the residue field is all directly closed. So that way you have assume your ring is actually a finite. Okay, wait a minute. So, so, but this doesn't. The English remark are you assuming R is local. My ring is the local unless I say that is. Okay, okay, okay. Okay, because previously in the country is all you do not assume that R is local for the reason I am asking. Yeah, so the, I'll assume my ring to be local. This exactly this hypothesis unless I mentioned something. And this faithfully flat extension maybe regarded as some have a particular case of what is calling burbaki conflemon or something like this right. Can I say, can you repeat it? Yes, in burbaki there is a construction called conflemon in French, which more or less is what you are doing here, which you construct a faithfully flat just in case someone knows what this means. Okay. All right. So here. So, why this is sufficient. Why can you do this so because I'm looking at the length function so here. So let me be careful here, because I can't talk of. So, suppose I take the ideal I, which is M primary so it makes sense to talk of length function. So, now what happens if I take a length of model. M, though I'm writing this elaborate structure for model M I'll reduce everything to the ring thing soon, but look at this length. Now, suppose I have replaced R by S so why so it doesn't length wise it doesn't make a difference. So I'm just writing it here. M mod you take I provenance Q M tensor S or R is a faithfully flat extension so I can take this as M tensor S or R modulo IQ M tensor R or S. I'm writing this is S. So, so basically, instead of looking at this length I'll be competing this length or the S. Okay, here I already have written it. So, so this is actually just now so already in my file. So, suppose if I replace R by S. So what I also have is my field is algebraically closed. So, look at this length function this is so I'm claiming the second one I mean this one. So, this one. So, so I'm saying that this length function F lower star EM and I extended ideally same as length of M mod IQ M. So, why is it so this is this is very important equality for us because this we are going to repeatedly use and So, how do I put this so look at this one so you have a F E lower star M. I dot F lower star EM F lower star ER as an R module and this is a length of as an R mod I can take it R which is same as taking FK module now F lower star EK. case the recipe for remember and this is came. Now, this one. Now, once you have gone into Froben yes so basically this one more step I should write it here. So, this is. Okay, perhaps here I should have changed it. So, this I can. So, here, this one I can write as F lower star E of I Froben is Q M. Right. This one always. So now everything is like a module applied Froben is over it so it is same thing as length or R M IQ M. And this one this is algebraic closed so nothing is happening here the length is one so this so this equality is this what so board work is not very nice here. So, this is. This is now. So, this is going to be my star two. This I'll be looking at this length function certainly reminds all of us the ordinary length ordinary power and corresponding length which is Hilbert Samuel polynomial. So today the aim is to study this length function. This one. Now, so before that we'll assume this property of this length function so which is suppose if you have a local ring and I'm primary ideal finally generated module. So here, D one is a dimension of the model M. That's important. So length of M mod I and M is a multiplicity plus and plus D one minus one choose D one and so on. So it's a polynomial of degree D one and all the coefficients are rational. So when when you look at looking at this length function very much would like to have something like this some property. So, so aim of the talk is to study this. So this one. Hilbert Samuel polynomial is also telling you something that if I look at just the limit of I and M upon and to the power D and take the limit as n tends to infinity. This does exist with EMI upon the factorial. So, so we would like to study the length function this one. But, or at the most can I say something about the limit. So before that we come to that why we should be interested. Let me give a recall the lemma of the quiz. Since he's the one who started this so it's say suppose if I have a local ring now or dimension of the R is D. Length of R of M Q Frobenius Q is greater than Q to the power D. That's the first one. And the second one is says equality holds if and only if your R is regular. So this this can be for all for some e greater than equal to one or for all greater greater than equal to one it's the same. So, for a greater than equal to one. So this is this is the significance of length function key you know the length of length of the arm or Frobenius Q is the least. It implies the ring has a nice property. Okay, so, so I this is of course the second part is harder so we'll just look at the easier part just to get a feeling of the. This Frobenius powers. So, let us see the first one, which is this one will put the second one, the proof of the one. So here, we're looking at the length function. So you assume, again, R is complete. And K is K bar without loss of generality. And further, I can replace R by R mod P, a piece of prime ideal which gives a dimension. If I replace R by R mod P, if at all, if, if I prove that the length function of this for R mod P is greater than equal to QT, then of course it is going to be sold for R also so we can assume R is a domain. So is this reduction clear that I can assume my R is complete local domain. Now, you have this, then what you do, you choose elements xt inside the maximal ideal giving a minimal reduction. So, now your R is complete local, so it contains a field. So, I just call this ring to be a, this should be x, I suppose, xt. This ring is contained inside your ring R. Now, I'm claiming rank of F lower star E R as an R module is same as rank of A F lower star E A module. So why is that? So let us see. So you look at the map F lower star E A. So this is injective. Everything is a domain. So Frobenus exists inside the Frobenus. So an A itself is sitting inside R and R is sitting inside F lower star E R. It has two Frobenus. Now, so this is, everything is, this is finite over this, this is finite over this. So everything is finite over A actually. Okay. So, if I look at this map, and if I look at module like this. So rank of this module as a module via this map is same as this one. Okay. But rank of F lower star E or R or this is same as R over E. Nothing is changing. So that means rank of this should be equal to rank of this. So that's what it is. And this is equal to Q to the power D. That is easy to check because it's a poly power series ring. It has a basis. So F lower star E A has a basis. I think we have seen it before is a F lower star E X1 I1. F lower star E E XT ID where IJ 0 less than equal to IJ is less than Q. So we got rank of this module is equal to QD. However, rank of F lower star E as R is a certainly. I want this mu of F lower star E. So this, the minimal generator of this module as a R module is going to be greater or equal to the rank of F lower star E and R. This is just the minimal number of generator of F lower star ER localizing further at the quotient field. So obviously it has to be dropped further and which we have just now seen is equal to Q to the T. However, now this one is nothing but by NACA, am I using or this thing is F lower star ER. This is R module and which we have put by our star condition is length of R of MQ. So what we have proved that this is greater or equal to Q to the power T. So this is always true. So that was one thing. So and this part that when equality when actually exactly it is equal to QT implies R is regular. It's a harder and I'm not going to prove, but it tells you something just that looking at the length function tells you about the nature of the ring R. So, in fact, you could have taken a little cruder invariant and and this is what actually we are going to do is the length of R of in fact slightly more general form of this modulo Q to the power D where Q is changing to infinity. So he is saying for one Q the this equality is equal to one, then your ring is regular. But what about the limit? Maybe it does have some it tells you something about the ring. And this in fact turns out to be our candidate for the day. I mean, we are going to study this and this limit function in fact was considered by Kunz, but he thought that this limit doesn't exist. So he dropped drop it there. But later Ramon ski again rediscovered and in fact he put this limit exist. And this is what today we are going to follow his proof. And this. So this is our aim of the day is exist. So in fact, we'll prove it for any model and all. And why by the way, why after so many years, once you looked at this, that's another story will come to that later. So let me go to the sums. Let me set up some notation. So this setting used more in number theory, this particular kind of notation. If F and G are two functions. F and G, I hope you can read this one. We say FN is of order G of N. If there is a positive constant says that the norm of F is less than equal to C into norm of GN for all N. So for all N or for all positive, it doesn't matter because we are free to choose our C. So, okay. So we begin towards the theorem of Monsky. So first thing, what can you see first of a length of the function? So this is what we are going to do. So we'll do it full generality like Monsky did. He looked at the module and looked at this length. So recall my M is finitely generated and ring is a Noetheran local ring and I is M primary to make sense of the length. So this function is of order Q to the power dimension of M. This is important. So this is actually not very difficult. So what you do? Suppose mu of I is given by M elements. Then I to the power M Q contains I for Venus Q for all Q. I mean, it's easy to see. Like if I have I given by generator X1 up to XM, then I M Q will be given by Monomials summation X1 I1 X. I'm sorry, my X keeps turning into capital and a small both actually I M where summation of I J is equal to M Q. So all the I J's cannot be less than Q. So one of them has to be greater than or equal to Q. And so in particular it belongs to the Q. So, so of course then length of M upon I Q M is less than equal to length of M upon I M Q which is equal to P of M of I of M Q. Where P M Q I have defined as a Hilbert-Sammel polynomial for the maximum for the year. So here this is. So it's a polynomial of degree equal to the dimension of the module. We all know that. So here. So suppose so this is equal to suppose your PM PM I of X is given by something like C naught extra the power D1 plus C1 extra the power D1 minus one. I'll just write in the simplest form C of D1 remember D1 is a dimension of M. Then choose your C to be maximum of CI where I zero less than I less than D1 multiplied by M is a fixed number. So I can do that to the power D1. So this implies that PM I M Q is less than or equal to C into Q to the power D1. So that proves the lemma. So are there any questions on that? So at least we have some bound on the length of this function in terms of the dimension of the module and that's going to be very important. So all the arguments actually involved here are pretty basic but you had to just arrange them carefully. So this is my notation and in fact I think it's a pretty common. So lemma to be the set of all elements all primes of the spec of R says that dimension of R mod P is dimension of R. Now in the whole game you'll see the only components of R matters are which are which are the maximal components giving the dimension of the ring. Others one don't matter in the competition in this limit. So let me see. So this is the first instance of that. So here suppose you have M and N. So here is a comment that the inclusion should be in the reverse manner. What what what? I think in the previous one. But actually the first line in the proof. One minute. Yeah, tell me. The first line in the is the inclusion. I mean, I think the inclusion should be. Yeah, this one. I don't know why I'm getting this. One minute. This one. Just I don't know why is I'm sorry. Hello. And I apologize for this. I'm in my office. I don't know why I'm getting phone call here. I this is correct. I am Q continue. Yeah. That's okay, right? Yes. Otherwise the second line inequality. We will not get. And the next. This one. This one. Second line in the proof. The inequality we will get if the inclusion is I mean the contentment. I don't get. So IQ is a bigger ideal, right? So when I go model a bigger ideal length will decrease. Is that did I answer your question? Okay. Yeah, thanks for the correction. Yeah. So. MNN a finitely generated model. So such that. MP is isomorphous isomorphic to NP. So this is for all P in a spec of R. So I'm sorry. This is for all P in this. So I'm not asserting that MNN have any map between them. But suppose there is a map when I localize at this minimal, this minimal prime of special type MP isomorphic to NP. Then. I can compare their Rubenius length. M minus length of N of IQ M. This is of order. Q to the power D minus one. So this is saying that if on the maximal dimensional component. Modules are same. Then their length differs only by a smaller number which is Q to the power smaller order which is Q to the power D minus one. Okay. So. In fact, this way of reasoning you have seen it before also, but let me repeat it again. So, you take S to be R minus union of P where P belongs to lambda. Then S inverse R has only the primes in its spec of S inverse not only the minimal primes of this. So that means all this ring is an Artean ring S inverse R. And where maximal ideas are S inverse P. So it's a product of RP by Chinese remainder theorem. Okay. Now what is happening? S inverse M I look at. So of course this is going to be product of MP where P is in lambda, which is same as there is an isomorphism for each MP. In fact, into NP P in lambda and which is by definition is same as S inverse of N. So, using this piece of maximal dimensional component, you have you have a isomorphism at least that from S inverse M to S inverse N. And it's an S inverse R linear morphism. But we have seen several times this. That M comma N because it's a because M and and both are finally generated. We can it will come in with localization and this is what is happening. S inverse M S inverse. So you have a map here so you can there exists a map here corresponding map and call that map to be free belonging to home. I don't know why this says that S inverse fee. Now, you look at the map. Now, which is given by this map fee. So this map is telling you that at each prime coming from maximal component, they are isomorphic. So they have no co-carnal on those primes. So, so in particular, the mention of secret dimension of the module C is less than or equal to D minus one. This is okay. This this part. So now, if I look at this one, this function M upon IQ M. So M you go again go from M to M of IQ M by tensoring with R mod IQ. So, so tenser isn't right exact functor. So I have the sequence C upon IQ and C to zero. So this implies N of IQ and this length is less than equal to length of M upon IQ M. Plus now C is of dimension less than or equal to C is a module of dimension less than or equal to D minus one. So it's this length should grow to the order Q to the power dimension C, which is less than or equal to D minus one is for this. Similarly, I can there is nothing special about ordering M and N so I can reverse the argument replace interchanging N and M and I'll get the same thing. So it will give the theorem. Is this okay. I'm presuming that I'm not going very fast because I'm writing it very slowly so it is not possible for me. I think it's just perfect. What did you say. I think it's just perfect. Okay, thanks. Very kind. So here, so this is the one thing now we are approaching the proof into small small steps. So, but each of them are actually some of the very nice arguments. So you have a short exact sequence so. Okay. So here, the length of M of IQ M is actually equal to length of M prime IQ M prime. Plus length of IQ M prime. So actually, I forgot to mention one thing when I mentioned the Hilbert Hilbert Samuel function. So there the one thing was very important in the proof if you remember is he you can consider graded ring the length function because let me just write it here. So, you want to look at the R mod I n for example, for each end so you had a ring structure on this like R mod I direct some I mod I square like this. And this graded drink structure has been used to prove the existence of Hilbert Samuel function if you look at the proof. If I replace n by Frobenius Q Frobenius n, then it is, there is no ring structure because simply here like I n into I m there's a multiplication is actually I n plus M, but I can't say like that IQ into IQ prime is equal to something it is that is not correct. So this is this is a problem. So we are going through this laborious argument. So, okay, so let's come to the proof. So, so what you do assume you are just needed a length so I can assume again R is complete case k bar. Now suppose R is reduced this is begin with this one. So suppose R is reduced. Then, what is happening. Then S involved, do I need to serve then RP is a field for all P belonging to this thing to for any pre minimal prime. So you have. So when I localize the sequence. This one at P such a P then it is just exact sequence of vector spaces over RP so of course they will split and there will be. So, this kind of isomorphism let me write. Direct sum and be double prime is equal to MP. This happening for all P in lemma. So by previous lemma we have we have the assertion so let me not write it here so we have this thing. Okay, now suppose this is this is very nice argument actually are is not reduced here you are using that. And there is an endomorphism given by Frobenius. This is this is beautiful. So, suppose R is not reduced, then it has a nil radical and not call it to be a nil radical of R. So, so there should exist you can always choose a q naught large enough say P to the power e not such that and not to the power. I mean q naught is zero. On the other hand, if I have any module, and I look at its e not F e not M this one and multiply this by this. So it is going to be F lower star e of n naught of q naught into M. Right. That's the multiplication but and not q naught is zero. So this is zero. So this is telling me if I take e not. Maybe you need the q zero between brackets right. And when when you write the axis q zero such that and zero raised to the q zero a zero maybe is q zero between brackets, in order to use afterwards. No, no, no, no, no, no, no. No ordinary power. Okay, thank you. But, but maybe you're right. I can do this also Yeah. Yeah, sure. Yeah. Thanks. Yeah, I can choose this right. So, you agree with that right. So, yeah sure. So this is like, yeah, thanks. Yeah. This is telling you if I take E0 iterated Frobenius then your module F lower star E0 annihilated by N0. So it's an R mod N0 module. I mean F E0 lower star M is now you had a short exact sequence between modules. So remember your Frobenius is exact functor actually. I mean you're not doing anything here. So it's a F lower star E0 R module. It's in particular it is R module in particular since it's annihilated by N0 it's R mod N0 module. So it's an exact sequence of R modules. Hence R mod N0 module because N0 is annihilating all of them. I thought I read. Here I have written actually already this I'm going very slow in fact. So here actually I have already written to save the time. So here so you have this exact sequence. I'm writing it as N prime N and N double prime. So it's a sequence of R mod N0 module. So that means R mod N0 is a reduced ring. So by the previous assertion when we dealt with the case R is reduced you can always write length of this thing as kind of additive for the radius case. So now but this module this module is same as if I open it. So perhaps I should write it here. So here because so this module is a F lower star E of M iq dot F lower star E M. So when it goes here F lower star E M into F lower star E of iq into q0. So you got the length function here. So this is same as the length of this here. Now so this is this M mod q q0 M is equal to this into this plus or qd minus one. But remember q q0 d minus one and then this same as order of qd minus one. So in fact you're getting for all q large enough. So for all q greater than equal to q0. So you have the additivity of the length function. So okay is it okay this one. So now we more or less prove the result about the existence of hk multiplicity for the ring. So here I'm assuming RMK no ith and local but I'm also adding the assumption it's a complete domain. I'll just explain in a moment why am I suddenly also complete domain and dimension of r is d that's important. Then there exists a constant which I so my i is always m primary if I don't say anything. So half r i I said it's a greater than one. This actually already I have proved it. If you look carefully I'll just explain again such that here it is. So first time you see that something like a Hilbert-Samuel function you have obtained not quite a polynomial but up to the lower order. So that's the first thing you assume. Okay so we'll give a proof here. So I say assume k equal to k bar. So here why didn't I assume that it's a complete domain. Of course without loss of generality you could have assumed your r is complete and k is k bar but I cannot I am not assured of the being a domain property. So I start with case when r is a complete domain. It's important for me. So now we have already seen and that's the first lemma we have seen. F lower star of r as a rank of this thing is actually p to the power d. In fact using the power series argument what we had proved was the thing if you remember is actually p to the power e to the power d actually is q to the power d in fact we had assumed. So this we know. Now r is a domain so what we have is that means so many copies of r of pd are sitting inside F lower star of r. I suppose it's okay right because I am saying if I have S inverse localized the quotient will I just say it properly is isomorphic to F lower star r tensor S inverse r or r. Now this is a domain this one so localization doesn't kill anything so it is sitting inside here right. So then you can choose a last S and some element in the S such that there is an injective map from here to F lower star a we are choosing appropriate S here. So you have a map like this. So here you are using as a domain. Okay so this is isomorphic at the generic at the minimal prime. So if I look at its a co kernel so dimension of C is less than equal to d minus one. So that means now additivity of the length function says p to the power b length of r mod iq minus length of F lower star of r iq my floor is star. This is important is of order qt d minus one. Okay so now multiply the whole thing with the p to the power d q to the power d. So what you get is a length of r of iq into q to the power d minus so this length these are all over r of course when I don't say anything it means over r. So this is this length was actually if I can change it right here it's actually r of iqp because you have taken one probinus it will get multiplied. So here you have one upon qp d length of r mod iqp so one upon q. So if you look at the survey article of you know he said that there are four basic facts about why the probinus works. So one of the thing he said was summation one upon p and converges that is and this was actually very useful for me several times I have used it and other thing he said the map r to F lower star of r essentially is the map you take e of r to F lower star e plus one r. So the way one looks at it you want to compute some comparison with this or this or so you better see the at this level and see what amplification it has when you take the morphobinus. So so here we have c actually that was the obstruction and this is the amplification we have so so we have this one now. So what you do now q just denotes pn actually. Yeah so this I'll denote a sequence of numbers cn or to be maybe I should not talk like that it's very confusing so cn where n belongs to n I just define cn p to the power nd equal to length of r mod iq which is iq is pn actually. So n I'm not saying as cn is a constant it very much depends on q I'm right now I'm saying so if I look at this way so this this star one is telling me just that there is a set of numbers real numbers of course in fact rational number here cn plus one is less than equal to some constant which is here coming c0 into one upon pn d something pn. So this is a convergent sequence so this defines limit if I look at length of r mod iq upon qd q tends to infinity is same as limit and tends to infinity cn which converges so this limit is called e hk of ri. So rewriting everything from the equation you just telling me that length of r iq is equal to e of hk of for this new constant q to the power d plus of order q to the power d minus one. So for a complete local domain we we have already found the thing the limit exist assertion. It might be a good place to stop here because I think we're really over time. Oh I'm so sorry. So what is it? It's over the time. My understanding is that the lectures are 50 minutes but I'm okay okay then I'll stop here actually. Yeah so maybe next time I'll take up this result lemma. Apologize I completely forgot. Oh no it was a good place to start anyway. So let's see if there are any questions for Professor Trivedi. All right I hear no questions I don't see any questions in the chat either so thank you very much Professor Trivedi. Okay yeah thanks so and we'll meet again next Wednesday on the regular schedule for the remaining two lectures. Yeah yeah okay yeah have a good day everybody. Bye.