 So in this case, here's a push forward of Rabinius. So the target of this map becomes a finitely generated flat model over local ring. So it is a finitely generated free model. So we can get measures of singularities now by trying to measure how far is in general this model from being free. Measures of, let's say, freeness. So you can invent many measures for non-freeness. The two that we consider today are going to be the minimal number of generators divided by generic rank. So if my model M is free, right, I know that the number of generators is equal to generic rank, so this number is one. And you see that the converse also holds, right? I have this fraction is equal to one if and only if M is free. The other one, slightly more exotic, is the free rank. So I can take maximal integer N so that my model M surjects onto free model of rank N and I can still divide it by the generic rank. And we see, of course, if M is free, the maximal free model that you can surject on is M itself, so again, we are going to get that this is equal to one if and only if M is free. So this is the two measures of freeness that I can easily define, right? You can define more. And the next natural idea is that, well, if I don't have just one model, right? I have several of them. Well, I have a sequence of them. As E varies, all of them are going to be finally generated free models. So a natural question is to look, right? What happens with those invariants as I vary E? And this is what give us the definitions of Hilbert-Kunz multiplicity and F signature. So Hilbert-Kunz multiplicity of R will be the first one. So limit as E approaches infinity of number of generators. So let me abbreviate it, mean gen. So this one was already defined. And this one is, I think, was not mentioned today. So F signature is going to be defined by the second measure. Does it make sense? Well, now I make these definitions and the natural questions are basically, well, whether my limits exist and whether those invariants still detect singularity. So the main purpose of my lectures is to explain to you why this is true. So limits exist. So goals, the special value one, which is the smallest possible value for a Hilbert-Kunz multiplicity and the largest possible value for F signature. So equals to one if and only if R is red. And then if you will have time left, we will talk a little bit more of different properties of them, but the minimal program for the four lectures is to discuss those results. Any questions? So as a starting point, I want to basically look a little bit. How do we get from part two to part three? Or equally, how do we get from this definition of Hilbert-Kunz multiplicity that I've given here to the definition that, for example, Holger stated today. So let's try to take this definition and simplify it a little bit. So first, I have the denominator, the rank of F lower star ER, right? And I want to know what it is a bit better. So for that reason, I want to quote another wonderful result of Kunz, who says that if we take local ring and K will be its residue field, and it's still a finite, and it's a domain. So still the same assumption that I started with. Then we can say that the generic rank is going to be given by P to the power E dimension R, and here we'll get the residue field extension. So K over Kp. Basically, it's a formula that tells me how to compute the generic rank. All right, maybe I should ask, have you seen this formula before? Raise your hand if you've seen it. That's good, yeah. All right, so let's try to prove it. So let's start with the simplest possible case, right? The one which comes there with the theorem of Kunz. So let's take our ring R to be a power series ring over something. In this case, I can understand Frobenius very visually. That's what's already mentioned today that I can just say. I mean, you just look at how Frobenius act. That I'm going to, first of all, take p roots of my scalars, and then I want to take, of course, p roots of my variables. So one over p, xd one over p. And then what do I see? I have, well, first I have this extension, let's say from R to adjoining, for example, variables. I mean, just extending scalars, and then I can extend, well, to the full thing. Both of these extensions are free, and I can clearly say what are the basis for each of them. So the basis here is given by what? It's given by monomials of appropriate degree. So the basis here will be given by monomials x1 to the power alpha one to xd alpha d, such that each alpha is from, well, from zero to p. So I see that this is a free extension of the rank p to the power edu, right? Well, this extension, it's just extension of the field, right? And the degree of its extension is this thing. So here, the degree is k one over p. So the basis is just given by the coefficients, right? Over there. Does it make sense? So that was the basic case, right? If you take power series rank, so now let's go a little bit further than that. Let's assume that now R is a complete domain, not necessarily regular. And then I know that if I take a system of parameters, well, I know by construction theorem that R contains its coefficient field, right? So there is a field as a morphic to k inside R. And then we can easily see that if you take a system of parameters of R, then the ring generated by it is going to be as a morphic to power series. So I can find a power series ring of the same dimension as R inside R, and this is going to be a finite extension. And then basically I just refer to the computations that I made. So I can use the transitivity of the rank, right? I can say that the rank of, so maybe I should give a name to this. So this is A, right? And we have this diagram A to R, and here is F lower star E, A, F lower star E, R. And I know that basically ranks here are multiplicative, right? So I can compute the rank of this over A in two ways, right? So the rank over A of F lower star E, R, it is both equal to the rank of R over A times rank of, well, the ones that we're interested in. And it's also equal to the other way, right? So it is rank of, let's write A and the rank. Well, that one we already know, right? So if I go this way, my theorem tells me that this is PDD as an extension of fields. But what I want to mention is that because, well, if you look carefully, right, what we have here, I mean those two sides are the same, right? The rank of R over A and the rank of F lower star E, R over F lower star E, A is the same because it's the exact factor. So those two things are equal, so I can actually cancel them and I have a quality over here between the stop term and the bottom term, right? So this deals with the second case. It's all just consequence of existence of a power series subring such that it's a finite extension and then using the transitivity of rank. So that leave us the third case where R is not complete. All right, let's move also way back. So of course I want to complete R and reduce my work to the previous case as it is natural to do, but there is one little subtleties. The problem is that if you complete a domain, it may not be a domain anymore. But I can still say that the completion is going to be reduced. It's part of the package of being a finite. I get it for free. So as it was said in one of the short talks, I know that R is reduced if and only if the Frobenius map, so yeah, let's write Frobenius map is injective. So it's injection. But because I am finitely generated, I know that the completion is just given by the tensor product. So being injective is not going to change because the tensor in, because completion is flat, right? And from here, this is equivalent to R hat being injective inside of R tensor R hat. And this is isomorphic to the completion of this. So this is very fine. It comes right as the completion is the same as tensoring with the completion. Okay, so what I know is that my domain, if I completed, it may not be the domain, but it's still reduced. Consequence is that the completion is reduced. So let me take P a minimal prime. And then because it's reduced, I know that localization at P is a field. Well, in fact, I can take a minimal prime. So in addition, I can add that dimension of R mod P is equal to dimension of R, not R, R completion. I know that the dimension of the completion is the same as dimension of the original ring and the reforzer is a minimal prime such that the dimension is the same. So now, what do I know? I know that, well, let me call this L, this field L, and then by looking at what I have in my second case, I know that the field L, if I look at the degree of the extension with Frobenius, it's going to be the same as before, P to the power D, and then extension coming from the residue field, right? Because the residue field of completion is the same as the original R. So what is left is to go back, right? So I need to infer now information about R. So what do I know? I know that if I look at the Frobenius push forward of L, it's going to be the same just by definition of L as localization and localization commutes with Frobenius. So I'm going to get just R and here is R, right? So this is, but what I also know, I know that my P, by the choice of it, right? Because it is a minimal prime of the maximum dimension. If I contract it to R, it's going to be zero because R is a domain, there is only one minimal prime of the maximum, well, there is only one minimal prime. So it follows that over here, right? When I localize things inside L, I'm also going to invert all nonzero elements of R, right? So this is actually going to be L, K, where K is a fraction field, just because here I invert elements, right, already, the ones that are in the fraction field. And from this formula, I can compute the, I mean, it is, those are fields, right? So the degree of the field extension is going to be the same, right? If I do Frobenius in L or do Frobenius in R, A and K is going to be the same because it's just tensor. And this is the invariance that I'm interested in because this is what? This is a field of fractions of R, right? And here what I get, this is by definition the generic rank, so this is generic rank. And this one we already computed. All right, so what do we get from this? Is that in this definition, if you like, I can remove the generic rank and I can write clearly a formula for it, P to the power E, and then the extension of the residue fields. Okay, we still have the top on the both formulas. Well, on the formula one, right, as I told you, I want to get, to confirm that the definition of Hilbert-Kunz multiplicity that I get this way as a kind of singularity measure Hilbert-Kunz multiplicity, it coincides with the definition of Hilbert-Kunz multiplicity that Holger gave us in the morning. So let's try to look at the number of minimal generators over here. So how do we compute the number or the minimal number of generators of a model over local rank? You know that we can just compute it using the Kayam limit just as a length. So I'm going to use LG as Holger introduced of m modulo mm. So if I use it here, right, I'm going to get length of f lower star E r modulo m f lower star E r. But let's look what we have here, right? As Holger explained to us, right, we get the bracket powers by looking at tracing Frobenius, right? So if you look carefully at the definition of this object, what we are going to see is this is equal to just push forward of the actual bracket power because what are the elements, right? It is a linear combinations of elements in m, right, with elements in f lower star E r, but I know that scholars act on f lower star E r by p powers, right? So if you just write an expression mi f lower star E r i, you can bring this back, right, and you're going to get something with q root, right, a q power. So what I get is the length of f lower star E r, and here I get this quotient that we already seen, the q as the Frobenius power of f. Why did I write that? So it remains to understand how does Frobenius push forward affect the length. But this is easy to do because we can just look at the composition series. So we have, I mean, it just comes from the exactness, right? So by definition, how do you compute the length? Length of m is going to be equal to what? It's going to be equal to the maximum L so that I have a saturated chain of some model m1 through m2, so on, so on, until m, right? So mL equals to m. And because it's a saturated chain or my quotients over here, mi plus one over mi, are going to be isomorphic to the residue field, right? And if I apply Frobenius to this chain, I still will get by exactness a sequence of sub-modules. But the quotients will not be just case, they're going to be p roots of case, right? So what I'm going to get over here, I'll get that, well, let's move this aside and write that this is going to be the length of just model r mod mq and then I need to multiply by the residue field extension because length is additive, right? So now each of these factors contributes to me that much. Okay, so now if I plug this to my formula, I will retrieve precisely the definition of Hilbert-Kunz multiplicity that Holger told us this morning, right? So it's going to be, so the field extension and then it's going to be the length of the Frobenius power. And in the same way, right, we will see how if you remove the residue field, right? It's the same thing over here, right? You'll get this q to the power d. Okay, does it make sense? All right, very good. So a long story short, right? I gave you a different point of view on Hilbert-Kunz multiplicity and we see that it's really just a different point of view. It gives you the same environment. Now what I want to do is to start working towards the proven existence of the limit. So what I want to observe is now I want to observe a couple of important corollaries from the theorem of Kunz. So let me get the eraser. So the first corollary is that I can apply this result just to a pair of prime ideals in a given a finite ring. So if R is a finite, P and Q are two different primes, containing each other, then I can go module of P, I can localize at Q and I will get to the situation. So I will get the formula which allows me to compare this as a ranks, I mean the pyramid vacation here between the two fields. So I have residue field at Q and I have residue field at P. I do the P roots on both sides. So this becomes, if I localize at Q, it will be the residue field. If I go module of P, it will be the generic rank over here. So I should have switched them. And what we get is that here we get P to the power E and what will be the dimension. So I localize at Q and I go module of P so it is going to be the dimension of RQ module of PRQ. So for example, from here, you can see immediately another observation of Coons is that if you take a finite ring, it's going to have a finite dimension. So R is a finite. Because there are no theorem rings constructed by Nagata which have infinite dimensions. So the height, the raw maximal ideals of increasing height. But this is not possible here because my right-hand side is bounded. I have finitely many prime ideal, minimal prime ideals. I have finitely many just take E equals to one. I have just one maximum for them, right? And then I see that the height of maximal ideal has an upper bound to it. Okay, so that was corollary one or I guess corollary two. And the corollary three is the ones that I'm going to work with for the proven existence of Hilbert-Cons multiplicity. So from this formula for the rank, we can do now exact sequences. So what we're going to do is to say that if there are, so still local ring killed, then I have exact sequences F star E R2 G1. Basically because I know what is a generic rank, right? I can find exact sequences that basically confirms the rank, right? So here this is the same number. So let me call it P gamma is equal to K one over P, K P to the power D, right? And this is P gamma and this is P gamma. And what I want to say is this modules T182 are torsion, so a dimension of T1 and dimension of T2 are both strictly where the dimension of. So this just works by applying the second corollary. So what do we get? We take that if we take PI, let's say P is a prime such that dimension of RP is equal to dimension of R. Then by the Kuhn's theorem, this free module and my Frobenius are going to be isomorphic, right? Because they're vector spaces of the right of the same dimension. So this P gamma RP is isomorphic to F lower star E R. They're vector spaces of the same dimension. So they're isomorphic. And then the trick that we do is that we say that that's U to be the complement of all such primes. So it is R minus union of P, let's call it mean H. By definition, this is a set of all primes of the maximum dimension. And then what happens? If I localize, maybe I should have, sorry, let's say that R is reduced, so I don't have to think about it. Yeah, so R is reduced. Then if I localize R at this multiplicatively closed set, it's going to be what? It's going to be product of localizations at HPI, right? So it's going to be product of field. And I see that those two guys are still isomorphic, right? So this is product of fields. So I still get that this free module and Frobenius are going to be isomorphic. And then I know that because we're finally generated, right? We'll finally present it. We know that HOME localizes, vice versa. So basically just that HOME localizes. And then I take a map that gives me this isomorphism in one way or another way. I lift it to a map. I mean, it comes from localization of a map in R and we see that the Cochurnal, because it is killed by localizing at U, it must be something of small dimension. And this is what gives me this exact sequence. So what we are going to do with this theorem of Coons. We want to use this theorem of Coons to show existence of, well, the corollary to the theorem of Coons to eventually show the existence of this limit because we can look at those guys over here, right? And we can compare the two exponents, right? We can compare Q, well, E and E plus one or we can compare E and not E. So it allows me to work with different terms of this sequence. So I think the actual existence of the limit we'll go to the next time, but we can do one more step, right? Before we say that this limit actually exists, right? Let's do, so here we get automatically, right, that this. So what I want to say is that, sorry, let me organize my thoughts. So what I want to say, so before proving the existence of the limit, let's confirm that this thing actually is appropriately normalized. So if you take this length, right, and we divide it by this denominator, we're really going to get something bounded, right? Because if I claim that the limit exists, then it should be bounded. So the lemma, if you take local ring, if I take I m primary, if m finally generated, then I can find a constant c, of course, going to be bigger than zero such that the length of the quotient of m mod I bracket q, m is going to be less than c, and then here I'll get q to the power d. So this in particular means that that sequence over there is bounded. Let me actually see. So there are multiple ways to prove this lemma, and yeah, I want to double check what exactly, okay, I just want to, so let's go very combinatorial. So let us say that we can generate this ideal I by bunch of elements, F1 through Fm. So I want to deduce this bound from the existence of Hilbert-Samuel polynomial. So therefore I want to connect regular powers to Frobenius powers. And there is a trick to do so just from combinatorial analysis, right? So I have that elements in, if I take power m to the power pe, right? So what do I get? I get that this ideal, this power is generated by monomials on my fi, right? Of the total degree m to the power pe. So you just see that if I write down such monomials, the power of one of these variables has to be actually pe, right? Because it cannot put something less than pe for all of them and get something of the total degree m pe. So you will see that this power is now going to be contained in the Frobenius bracket power and maybe not peq. I'm trying to follow Holger who insists that I use q. But unfortunately I've been working too long with Kevin who insists that I use p to the power e, right? So I get conflict. Is this good? All right, so from here, what do I get? I get the bound, right? I get that the length of m mod bracket q is going to be less or equal than the length of actual normal power i to the power mq. And now I refer to the classic Hilbert theorem. I know that this is polynomial as m varies, right? So this is polynomial large enough. This becomes polynomial and it has degree exactly d. The leading coefficient multiplicity, there is d factorial and then there will be mq to the power d, right? Leading term, if you like. So because it's a polynomial starting from some point, I can take a constant that bounds this polynomial and then increase it if I need to incorporate the starting pieces. So from here, it follows that there is c to the power q to the power d. Well, m, if you like, it was a power d which bounds this length for all m, for all q. q, yeah, it's q. All right, so let's, I still have five minutes, so okay, let's actually try to largely finish the existence of the limit, right, for Hilbert-Kons multiplicity. So what do we do? We have this limit over here, we have that corollary and then we can just combine them two and see that things fallen together. So existence of Hilbert-Kons multiplicity. So I have these two exact sequences and I want to get Hilbert-Kons multiplicity from them, right, so what I want to do is to tensor them with R mod i bracket q. So I'll get exact sequence. So let's, yeah, let's take first this guy, big gamma, R mod i bracket q and then I'm going to get f lower star, R mod i bracket qp and then I'm going to get which one was it, t1, t2, that was t2, not i bracket q and I get the same sequence with those two guys switched. So I know that length is additive in short exact sequences so I can bound the length of the middle guy using the things on the sides and I also know that this mod was t1 and t2, they were not a full dimension, so therefore I will get a bound coming from the lemma showing that they're actually small, right? So if I erase this, I will get a simple bound. So I get that the, let's, what I get, I get absolute value of the difference of length because I have two exact sequences. I'll get a length of R mod i bracket q and we already did computation so I know what is the length of this. So this is minus k p, so k over kp, the residue field extension length of R mod i bracket qp. So the difference of these two lengths is going to be less than a maximum of this length, right? So maximum. So their dimensions are at most d minus one, so if I apply a lemma, I can bound to the length of either of them by some constant c, right? I can choose the maximum of these two constants, so I'll get a bound like this. And then what do we get? We just divide everything by q to the power, oh sorry, so I'm missing here a term, right? So no, I'm not missing the term. No, I'm not missing the term. So what do we get? We, yeah, what I'm missing is that I'm missing to write P gamma over here, that's what I miss, right? So then I divide both sides by q to the power d, right? I note that this P gamma is nothing else as P to the power d times as the same field extension, k to the power kp, so if you divide things together, you're going to see that the length of R mod iq minus over q to the power d minus the length of R mod iqp, pq together to the power d is going to be less than some new constant, c prime one over, ah, this one is one over, so let's see, yeah, I will divide by q to the power d, thank you, thank you, of you so much today. So we have q to the power d minus one, right? I divide everything by one to the power qd, so of course it's going to be one over q. All right, good. So does it make sense? All right, so we get a bound like this and then next time I will continue from this inequality and we'll finish proof of the existence. It's only a couple of lines left.