 So, okay, so, so last, so if, so RMK as you remember is a Noetherian local ring with residue field K. If this is a complete domain, then there is a constant EHK of RI. So remember I is always a ideal of finite co-length such that length of R mod IQ and here, okay, let me I'll just say it clearly. So this I'm just recalling the last time's theorem. So D is a dimension of the ring. So we are still quite have not proved the existence of the limit for arbitrary module. So this is what I'm bringing. So here it is. This we have proved as a complete domain. So remember, I'm sorry, I think you forgot to share the screen or do you plan to share the screen? We don't see a screen. Oh, okay. All right, perfect. Oh, yeah. So last time we were on the way to prove the existence of Hilbert-Kunj-Madeep city. So we had proved the result when R is a complete domain. And so now this, in fact, this is a, we can prove it for arbitrary module. So, so this theorem, this is terrible. So what is happening? So let M be a finitely generated module, R module. So here the D was a dimension of the ring. So again, there is a constant depending on M and I. Now I'm saying it is positive. And here, in fact, it was greater than one such that length of the length function M upon IQ M is equal to EHK of MI, Q to the power D. Here D is again the dimension of the ring actually. So, so we have all ingredients to prove the result now. So what we'll do again, you can assume R is complete local ring and K is a residue field, which is algebraically closed. Then consider the prime filtration. So let 0 equal to M0 containing M1 be a prime filtration of M. So I'll quickly recall what is the, how do you, so that means the definition is that means MI plus one of MI is isomorphic to R mod PI for some prime ideal. So how do you get it? It's very easy. Just you are saying that, let me quickly say M has an associate prime. So there is a ideal P1 container. This is, sorry, this is ring R inside M. So now you can take this to be your M1, this one. Now go modulo M upon M1, repeat the same process. So there will be R mod P2 sitting inside this. Now pull back P2 all the way to M. And this is the way you get the primary prime filtration. Okay. Now we have already proved the additivity of the length function. So that implies that your length of M upon IQ M should be equal to summation length of MI plus one upon MI into IQ MI plus one upon MI plus some element of order D minus one. Okay. But MI plus one mod MI is a, is a, is just a complete local domain. So applying the previous result to what you get is it is equal to E summation I running EHK of MI plus one upon MI at IQ to the power D plus order Q to the power D minus one. Here if you have MI plus this R mod PI is of dimension less than D, then this will get absorbed here. So I can, without losing anything, I can just take the elements where MI plus one upon MI is equal to D. If they're less than D, they'll get absorbed here. So it's okay. So such for MI plus one upon MI for such thing, which is isomorph to RPI, it means that PI belongs to lambda. So if I look at R local as a PI, then it's a field. So the prime filtration gives me just M zero PI and PI. So many of them will actually become zero. So it means the number of times the models which are isomorphic to this one will occur in this filtration is exactly equal to the length of the model MPI. So I just look at this again. And it tells me this is summation of EHK R mod PI length of MP where P is in lambda into Q to the power D plus order Q to the power D minus one. So you have this length formula. Now in this formula, if you go modulo Q to the power D, so it just tells you EHK of M with respect to I is same as summation EHK R mod P into length of MP. So this is the final thing you have proved the length formula as well as or the second assertion. I didn't write it anyway. So this associativity formula, this is called so associativity formula. Okay. So this associativity formula has a nice corollary. So what it says is a following. Suppose you are is a domain. So you're again, the condition is RMK where R is a domain, then your multiplicity of M for I is nothing but multiplicity of the ring R at I into the length of or rank of this thing. So this is rank of M as R. So that means basically studying the multiplicity for module boys don't study the multiplicity for the ring. So you can always assume your ring to be a module. You can just study for rings and another thing if I take any other ring, which is a finite module over R, then I can recover the multiplicity back and forth. So let me say it. So R to S finite map of rings where M is local, then EHK of S of extended ideal that is the problem is same as EHK of the ring R at I into rank of R at S. So if I have this ring, which I understand well, or this ring, which I understand well in the sense, I know what is the multiple HK multiplicity of this or this, then I can recover the another one merely by knowing the rank. So I have given you exercise. So this one example was a Verones subring inside a polynomial ring. So remember polynomial ring or power series ring, we understand very well, because the length of the each length function is actually equal to Q to the power D. So is equal to one the quotient. And then then the ring of invariance if I take so G is a fine, suppose G is a group of order finite order acting on a polynomial ring on a polynomial ring ring R, then RG to R is a finite extension. So I give you another exercise, but that is before proving the existence of HK multiplicity. So it says, suppose you are in the situation RMK, where characteristic is positive, then multiplicity of the ring R with respect to I, mod D factorial, D is the dimension of the ring, is less than or equal to EHK of I, is less than or equal to E of I. So is everything is okay so far? So so immediately tells you something, it says if dimension of your ring R is one, then your EHK of I is same as E of I. So it's an integer always, it's nothing makes, there's no changes. So now we are looking at the some properties of the HK multiplicity. So what, so you have this number. So what does it mean? I mean, what kind of properties of the ring it reflects? So the first thing first, so if you know regular multiplicity, it is one, if and only if your ring is regular, of course, provide the ring is unmixed. So here also the same thing, it's a formally unmixed, suppose this is an important assumption, formally unmixed means if I take a completion, all the all the prime, all the irreducible components are of same dimension. So EHK of R is one, if and only if your R is regular. So we are not going to prove anything. So of course, if R is regular, then EHK of it's, you can assume it's a power series ring for completion and it's easy to check that this is one for the converse that takes bit of a, that's a lot of work and it's given by Watanabe and Yoshida, I think. So remember that if you recall the first result we stated of Kunz was saying that length of length of R more, this is with respect to maximal ideal, so I should say maximal ideal. And Q is equal to QD, which is that means I'm saying that length of QD is one, if and only if R is regular. So this, so this, this, this assertion is certainly stronger. So this is saying I don't have to look at one length if I take a symptotic length, which is happens to be one, then it implies that R is regular. So here I would like to mention that this, in fact, this length, this limit function of this length, where Q tends to infinity was actually thought by Kunz and he somehow he thought this limit doesn't exist, but he did think that this limiting function will be something very important. So many years later in another contest in the context of number theory, Monsky considered this limit and he proved it existed when he came to know that Kunz also has thought about it. So in his honor, instead of putting say Hilbert Samuel multiplicity, he called it Hilbert Kunz multiplicity. That's one thing. Another thing here, I would, you may have wondered why we have a formally unmixed condition. It's important because lower dimensional component don't contribute anything to the multiplicity. So let me give an example. Suppose if I take a power series ring in three variables and I look at x, y comma x, z, this ring, suppose this is R, then EHK of R, or maximal ideal I mean, is same as EHK of the ring K. That's what in associativity formula has proved is x, which of course we know one, but certainly your ring R is not regular. So are there any questions so far? Okay, so now perhaps this following lemma might be one of the reasons it really caught on with community algebraic community as the following. If you remember, if I have two ideals I and J, so this is let me recall here now. He's containing J, I mean their M primary ideal. Again, I'm in a local setup such that multiplicity of I, so this is your hypothesis says that E of I is same as E of J. This implies that J is contained in the I closure. So here perhaps you had to assume your ring is again formally unmixed just to avoid the unnecessarily lower dimensional component. And in fact, it's a converse is obvious if so. So this is not a thing. So the integral closure versus usual Hilbert-Samuel multiplicity, here the tight closure versus HK multiplicity run exactly parallel. So it says if so consider I containing J and are formally unmixed, then J is contained in the tight closure of I star if and only if E of HK of I is same as EHK of J. So most rest of my talks perhaps I won't give much of a proof. So let us see the proof of at least easier side. So suppose your J is containing I star. So I want to prove that in that case, HK multiplicity is same as of I same as HK multiplicity of J. So what you do for this, the very J is contained I star and so for any element of J I would find X belong to J. For example, there exists some C X as that C X MC, I mean, this is not a multiplication, just a subscript X to the power Q belongs to I Q for all Q. Since J is finitely generated for each X we have. So we just collect the generator C X for each generator and pick the product. So which of course will belong outside the minimal prime. So belong to R is zero. I forgot up or down for such that C into C into J of I Q is contained in I Q. So this implies that I look at this module J Q mod I Q. It is a R mod I Q colon C module as a finitely generated module. So in the sense generator of J Q will suffice. But remember if J is generated by M elements, J frobenius Q will also be generated by M elements. So in fact, there exist M copies you can have or such that direct sum of R R I Q colon C subjects to J Q upon I Q. So this implies length of this one is less than or equal to length of this. But what is the length of this? So it is length of R I Q minus length R upon J Q. This this one this length which is equal to M into which is less than equal to M into length of R of I Q C. So this this module remember this is over R mod C where C doesn't belong to any minimal prime. So dimension of R mod C is less than dimension of is strictly less than D. So this is a more of order Q to the party. So here in fact I can write it. So here I don't have to write it. So I go more low Q to the party and take the limit. So which implies your EHK of I minus EHK of J is zero. Okay, so this is one thing what how about the converse converse? So suppose I'm not going to prove but I would like to mention something. Suppose your EHK of I colon one more element say XR is same as EHK of I. So you would like to say that X belongs to the tight closure of I. So this is what you want that will prove the converse. But what happens here? In fact, the result, this is a theorem of a Hawks-Renek. It says if suppose X doesn't belong to the tight closure of I star. You remember if the X belongs to tight closure, there should exist C. So in particular C should exist some Mk minus Mk plus one independent of Q says that C into XQ, I'm sorry here I'm writing it belongs to IQ, etc. That means a letter of IQ colon XQ is intersects this number. And here I'm just going a bit fast. So this one says it's even better if X doesn't belong to I star. So the colon shrinks like anything. So that's the main point here. So it says the set of C says that C XQ belongs to IQ. These kind of elements in fact belong to MQ over KL for some constant L. So that means the letter of this IQ colon XQ, this thing shrinks like anything. It means it is proportional to the power of Q. So that will contribute the length function to be some epsilon to the Q to the power D something like that. So this is the hardest part to prove actually. And there they use like an element of a small order and the valuation. So they have kind of quote a weaker criteria for tight closure, but they say these are equivalent. So this is a tricky argument, but it's very intricate. So you can look up. So this is very important point that your annulator is actually keeps shrinking rapidly at the power of Q. So that is the property of HK multiplicity. Now I would I am actually not covering all or I cannot cover all the important results, but you would like to know what else kind of what in the okay tight closure is one thing, but there are other characteristic P singularities of the ring, which EHK tells. So let me quote a few of them. So one is, I think EHK of M first of all. So the philosophies as follows is smaller the multiplicity, EHK multiplicity, better the singularity will be. So already you have seen one sample. If smallest is one here for M. So if it is one, the ring is regular. Nevertheless, suppose your ring is not quite one, but this is still smaller than one plus one upon D factorial. D is the dimension of the ring again. Then R is when we call a F rational. So I think this result was proved in several steps. First some bound involving characteristic P was the Blickeler Inescu and then this one, this one I am not sure maybe Inescu can tell me. So, so the here the point is not here, but generally, unlike usual multiplicity, EHK multiplicity is not a discrete set. I mean, so it can take many arbitrary value. So one question is suppose your ring is not regular, then how big the EHK multiplicity can be. So the answer is this is greater than equal to one plus one upon D factorial into D to the power D. This is just these are fun facts. So, sorry, D to the power D. That's what you said? Yeah. Okay. So I have not shown you any non-trivial examples of EHK multiplicity. So that so far cases I have listed like where on is ring or the ring of invariance, etc. Since they come from the polynomial ring, which is doesn't involve any characteristic turn on in the EHK multiplicity, you won't see it there. So, so let me the famous example of Hansen. So, when Monski had this introduced later on, he computed the multiplicity for the next non-trivial class of ring, which is say plane curve modulo, you can go a cubic equation. Quadric is nothing not so spatial here right now. The cubic equation he did compute for, I don't know which one, one set of people have proved for characteristic two and other people for not characteristic not equal to book wise and Chen. But their EHK multiplicity is not interesting for us because it doesn't reflect any characteristic. And that I will perhaps tell you later the reason for that. So, let R be this x4 plus in the node sided x4 comma y4 that's that's nonsense. So, this is x. So, you are k x y z modulo, this equation x4 plus y4 plus z4. Okay. And characteristic of k is p. So, what they have shown is EHK of R is equal to three plus one upon p square. If p is congruent to plus or minus three mod eight and equal to three if p is congruent to plus or minus one mod eight. So, this example is actually interesting for several reasons. So, one reason is it tells you something. If I look at EHK multiplicity, the deformation things are not going to work for the computation of EHK multiplicity because suppose there is exist x in R. Remember, the usual multiplicity we always come one way of computing Hilbert-Samuel multiplicities, you go modulo general hyper plane section or general element. In fact, that's the one definition of your usual multiplicity. You keep on going D number of elements. The ring becomes quotient ring becomes zero dimensional. So, just count the length of the ring. That's one thing. So, or what we call reduction ideal. But here, this kind of induction is not possible. Because suppose there is some element x in R or maximal ideal such that EHK of R is same as EH, I don't know why this is doing this of R mod XR. Then this now your ring is of, ring is two dimensional ring, cool dimension two. And this is cool dimension one. And we have just now seen that for one dimensional ring, this is same as R mod XR. Now, for Hilbert-Samuel multiplicity, there is no characteristic involved, right? So, this is a characteristic free thing. So, obviously, this cannot be equal to this. So, this is not possible. Now, I would also now, I'll try to give some, basically, many I'll talk here, is the following. What happens? You look at this, this, this is called in geometry, we'll call this as a plane curve, because if I look at proge of R, it will become a one dimensional projective variety, and which is sitting inside the plane, which is P2. So, for these kind of curves, we have a notion of a locally free shape called vector bundles. So, I'll, I'll very briefly touch on that. So, so slightly generality, not the example of Monsky. So, you take k x y z modulo h, h is a homogenous irreversible primate polynomial. Suppose you have this, then the theory is the following. Suppose and degree of h is d. So, sorry, this is h, d we have been using for the dimension of the ring. So, here I'm using for the degree of the polynomial. So, the general theory tells me the following. So, this, this is my S. So, call it SS. This is 3d mod 4 plus L square. I'll just explain all these terms 4d P2S, where P is the characteristic of the ring. So, P is characteristic of S. So, what are these L? So, L is an integer. It's important. It's bounded by d minus d3. That's important. And S is also an integer. I'll explain the significance of this one. So, here do stop me and ask the thing. So, what happens? You project if I S. So, it becomes take approach of S. So, now X is a curve here. So, X is a curve. So, you have a locally free sheaves on that. So, one is a structure sheave. And then line bundle non-trivial one, OX1. So, what do you do? OX1. So, this is OX. So, this is locally the ring sheave on X. So, there are three things. This is just given by the elements X, Y and Z. This three copies of OX. One goes to X. One goes to Y. One goes to Z. So, degree 0 goes to degree 1. That's why there is a shift here. So, this is a locally free sheave on X. This is a locally free sheave on X. So, the kernel has to be locally free sheave on X. So, this is called vector bundle here. So, coming back to the... So, wait a minute. So, here, if you look at the people, when the people who work on vector bundle theory, they often like to work on something called semi-stable vector bundle. So, that means that by definition, if a W is a sub-sheaf of V, here it is a curve. So, everything is a locally free. Otherwise, you have to stick to our torsion free. So, mu of W, which is degree of W mod rank of W should be less than or equal to, I'm sorry, mu of V. So, this is some definition. So, this definition tells you that lot of homology of V will disappear if V were semi-stable. So, that's the one reason people in vector bundle, we would like to work on semi-stable because it also forms a bounded family if you restrict some parameters of the vector bundle. So, and on the other hand, any vector bundle have a filtration by say, say something like a V0 and continue V, Vn equal to V says that VI plus 1 upon VI is semi-stable. So, you can break it into this kind of factors. Otherwise, so these are called hardener cement filtration. So, that's one thing. So, that's useful for the formulation of this one. But one other thing is a problem. If characteristic of k were 0, then if I take pi to be any finite map, y to x, then pullback of, so pullback here will be a tensoring with a ring. So, pi upper star is semi-stable. So, V semi-stable implies pi upper star V semi-stable. However, that's not the case when you come to the characteristic P and the major obstruction is the Frobenius map, which is the finite map. And you start with the semi-stable bundle, pi upper star V may not be semi-stable. So, you may start, okay, I'll start with the hardener cement filtration of V where every subcocean is semi-stable. But when I pull up that hardener cement, still I don't know it remains hardener cement for the pull up. So, these problems are there. So, this is the theory of semi-stable vector bundle. So, here what is happening? My rank is of V is 2. Then, perhaps I'm not saying it right. So, if Fs upper star V is not semi-stable for some S, then by hardener cement theory, there is a line bundle sitting inside Fs upper star V. And this is the hardener cement filtration. Okay. So, okay, having talked about all these things. So, what is the coming back to the HK multiplicity? So, this one is telling me if I have this L and S. So, you forget about this generality. So, and consider the Suzuki bundle V for that. So, V F upper star V, S minus 1 iterated of V, they're all semi-stable and Fs upper star V is not semi-stable. So, that's N, the hardener cement filtration which is here, Fs upper star V. So, that means mu of L should be greater than mu of F, this one. So, this is equal to L mod 2. So, I come back to looking at this number. So, this is telling me at which stage your bundle will destabilize. Till then, it will remain semi-stable. If S is this one is 0, that means your bundle is going to remain semi-stable for all S, that means. So, this S is telling me in the place where it's a provenance iteration where the bundle is not going to remain semi-stable and this number L is telling me how much it has destabilized. So, these are so HK. So, if you have these two information, you know HK multiplicity and compute it right away. So, generally this is that's not easy. So, on the other hand, suppose if I know the HK multiplicity of SM by chance, this is there some number then and P is greater than D into D minus 3 then recover L and S. So, for P bigger than this number HK multiplicity will give me the information about the hardener cement filtration of your vector bundle. So, this is the crucial point here. So, coming back to our example of Monsky. So, D is actually 4. So, it is 3 plus this thing. So, this is telling you something. So, HM example. So, it is telling you if P is congruent to plus or minus 1 mod 8 then FS upper star V is semi-stable for all S for all SM and if P is congruent to plus or 1 minus 3 mod 8 then F upper star V is not semi-stable. So, this example is a bit of a surprise for vector bundle people who work in vector bundle because the thing is generally if suppose V is semi-stable in characteristic zero then V reduction mod P is semi-stable for all P positive. So, the semi-stability condition is an open condition that means but this one is saying however if I look at the Frobenius pullback condition this is not an open condition. I may start in characteristic zero the bundle which is semi-stable. So, of course it is going to be semi-stable for all finite pullbacks but this is not an open condition. So, just a computation of HK multiplicity gave you this example. So, so, are there any questions now? Okay. So, talking of coming back to this example of Monsky. So, it tells you that for infinitely many P things can it doesn't achieve minimal and for infinitely infinitely many P it can differ. This is no open criteria for this HK multiplicity but there is something nice. I said that this L is bounded. So, as P tends to infinity this number is going to be zero. So, this leads to the open question. So, you have a RMK. So, R is the noether and local ring and residue field here and characteristic of K is zero and now RP MP and suppose your ideal is I this is let the question is does EHK RP IP limit P times to infinity exist. So, see the answer to this question will give you a notion of Hilbert-Drucking-Multipstick characteristic zero or well-defined notion and also for at least for M primary ideal suppose we know the equivalence of tight closure of an ideal in terms of the HK multiplicity perhaps it will give you a notion of tight closure in characteristic zero but of course for the ideal which are M primary. So, I really don't know the development there. So, talking of these notions there is notion of epsilon multiplicity. I think people can look up. So, this is Katkowski and all and then there is a generalized I think that's by Neil Epstein or something. I don't know HK multiplicity. So, these are another different notion of this one minute. I'm just I don't know what happened to my screen. Can you see the screen? We cannot see the screen but it's 950. So, thanks. So, should I stop? Yeah, I mean we can have a 10 minute break and come back. Thanks. Yeah, sure. And then hopefully we can resolve the screen issue. Yeah, thanks. Thanks a lot. Let's see if are there any questions for Vijaya for this lecture. I don't see any questions. So, listening to the speaker. Thank you. And we'll take a