 to resume from where we left last time. So last time we were trying to prove this theorem. So if R is finite, reduced, then Hilbert-Kunz multiplicity exists. And moreover, what we want to get from this theorem is even more, right, that we want to get a convergence estimate. So let me try to recap where we left and what we were doing. So the starting point was, well, the convergence estimate will come from this basic lemma that we can actually bound the length of quotients by bracket powers in terms of just some constant, which depends on the model M, right, and Q to the power dimension of the model. So how do we apply it? We apply it to the sequences, and the sequences will come from the following observation. So let me choose a magical constant gamma to be given by the field extension and then P to the power dimension. And then if I take any minimal prime of R such as the dimension of R mod Q is equal to the dimension of R, so this is what is the set, then we see that if I localize M at this prime, well, R is reduced, right, so localization at any minimal prime is a field, right? So MP is a vector space, and then if I look, so it's going to be some number of copies of KP, right? So it's a vector space of some rank. So then if I apply Frobenius to MP, it's going to be what? It's going to be same number of copies and then it's going to be just KPE one over P. Why did I say P? Should be Q, right? Sorry, Q is my prime, right? P is my characteristic. So I see what? I see that if I compare MQ and I compare the Frobenius of it, right? What do I see? I see that M, so if I take gamma copies of M and localize them at Q, those are going to be isomorphic to localization of Frobenius plus Frobenius plus Frobenius at Q, just because those are going to be vector spaces of the same dimension, right? And by the Kuhn's rank theorem that we discussed last time, the degree of this field extension of P roots of KQ over KQ is precisely this constant gamma, right? So this is why we have this isomorphism. Because those two guys are isomorphic at all minimal primes of maximal dimension, we can get by lifting this isomorphism to R, we can get these two exact sequences where the models, the Kuhn's have dimension, less than dimension of R. And then we just tensor them with I bracket Q. So because of the Frobenius, we are going to get in two places I bracket QP and then we know how to compute the length with Frobenius. This is going to spawn another field extension. So this is why I get field extension over here in the denominator because I divided both sides and this is how my gamma turned into P to the power D. So this is what I get just by tensoring those two exact sequences with I bracket Q, R mod I bracket Q, good? All right. So what remains to say is that those two models are torsion. They have small dimension so I can use the lemma over there to say that this is less than the constant C, let's say of T1, right? Q to the power D minus one and this is less than constant C of let's say T2, Q to the power D minus one, good? So what I then get is that if I call them, I think Holger calls them Phi E, right? The normalized length. So I hope he corrects me. So this is length of M mod I bracket Q, M divided by Q to the power D, or maybe it was Q, but anyway. Phi E, then if you look at this, right? What I get here by dividing by QP to the power D, I'll get the absolute value of Phi E plus one minus Phi E is going to be less than, well, maximum of C of T1, C of T2. And well, I will divide it by QP to the power D, right? So here I will get over QP to the power D. Here I'm going to get Q to the power D. So this is how I get my sequence. So those guys definitely will be less than just divide them by, so I'll divide here by QP to the power D. Here QP to the power D, P to the power D, and this P to the power D I can see. All right, so this is almost what we need. So it remains to note that using this estimate, I can get actually the Kashi property for this sequence. So if I take now, let's say E and E prime, I can use this inequality on the absolute power, a little repeatedly, right? I also say that this is less or equal than the sum of differences between Phi E plus I minus Phi E plus I minus one, let's say, right? So consecutively, and this is all bounded using this constant as, so let me just say C over Q, right, less than C, and then it's going to be the sum of one over QP to the power I, right? Maybe I should just call them P to the power E plus I. So just use this inequality repeatedly, but my E increases, right? So the power here also increases. And then I know that this is a convergent series, right? So I can write even better using the formula for geometric series. I can write that this is going to be less than C to the power here. It's a finite geometric series, but it is less than the infinite one. So I'm going to get C so over P to the power E and here's the sum of infinite geometric series one over one over P. So this is going to be definitely less than C, let's see, P E minus one. So from here, using the convergence of geometric series, I get a Kashi estimate, right? So I now know that for any E prime, right? And for all E large enough, I can make it as small as you want, right? And it follows from here that the sequence converges. So this gives convergence. Once I know the convergence, I can also set E prime to go to infinity. And then what do I get? I get that the first thing will become its limit, Hilbert Kuhn's multiplicity of I on the model M. And I can write the definition of IE to get the theorems that we want. So minus length of M I bracket Q M divided by Q to the power D is less than what they have here. C to the power P E minus one. Does it make sense? Okay. So we proved our theorem and from here I have a few things more to say. So first of all, if we look at this proof, right? We can see what is the origin of this convergence constant C. So what exactly happens? I have these two kernels, T1, T2. I have these constants coming from the lemma. So the constant C bound in the coland as this I bracket Q for the model T1 and for the model T2. And after that, all I do to them is just proportional things, right? Here, well, I don't need to do anything, right? I mean, I get smaller number here. So here I just choose the constant 41, constant 42, and then I just need to scale it a little bit, right? So the convergence constant that I get at the end, so maybe to get the theorem, I should say this is C prime over P E. This constant at the end that I get C prime is just proportional to the constants C, T1, and C, T2 that I get from the lemma. So this is really a key observation for the theory. So I'm going to write it as key observation is that C prime is just proportional to this constant C of T1 and C of T2. Well, it's something. It's the maximum of these two times some constant. Why is this crucial for the theory is because this is a gateway to uniform convergence. So next time we will see an example or maybe also on Friday, we will see an example of how to use it, but basically this is what enables it, right? You replace this lemma with something finer and you are going to refine this constant. So instead of I bracket Q, I can take a different sequence, a different filtration for which I, for example, can refine this bound, right? If it will depend, for example, on the parameter. And then this dependence on parameter is going to pop up in these two places. And then these operations do not really depend on anything, right? And they're just going to directly appear over there in the constant C prime. So this is the gate to all uniform convergence techniques in Hilbert-Kunst theory. It's just this convergence proof pretty much outputs you constant solely dependent on the bound on the length of the quotient of T1 and T2 that you get from that lemma. Good. So we won't see any uniform convergence today because I need to still prove to you that a signature exists, but we will get to it later. So let's also record a couple of properties of Hilbert-Kunst multiplicities that we'll use later. Basically, just one of them. So by looking at this proof, right, I can show that Hilbert-Kunst multiplicity is additive in short exact sequences. So if I have an exact sequence of lm to m to zero. Actually, let me say one more thing before it comes there because let me make one comment, right? So this theorem assumes that r is a finite and reduced, but it holds without this assumption, so let me just make a few comments. We don't have too much time to elaborate on it, but let's explain how to reduce the general statement to the situation where the ring is a finite and reduced. So first, how to get reduced? So the recipe for this is simple because we can just take our module m and replace it by f over star e, m, let's say e naught. And we can choose e naught so that any nilpotent element multiplied to the power p e naught is going to be zero, right? So we can just assume that the nilradical of zero to the power p e zero is going to be zero. And then this module over here is naturally a model over the reduced ring of r. So this is going to be model over the reduced ring of r. So we can prove existence of Hilbert-Kuhn's multiplicity here, but because this is just Frobenius, we know that Hilbert-Kuhn's functions are the same, right? I mean, just the scaling if you like, right? If I divide this by Frobenius powers, they're just going to, if I want to quotient this by Frobenius powers, I'm just going to get another factor of q naught and this is not really going to affect anything. It just at the end of the statement, I will need to assume that my power e in the convergence estimate should be greater or equal than this e naught, right? I just need to move my sequence a little bit further and then I'm automatically over the reduced ring. The second thing is how to get a finite. So to get a finite, we need to extend the residue field because if I take r and then I complete r and then I do completed tensor product with perfection of k, this is going to be what? So my original completion of r, right? We know by Cohen's structure theorem this is going to be quotient of a power series ring by some ideal, right? Then what is the completed tensor product? They're just going to be replacing the coefficient field by perfect field, right? So I leave the same defining ideal and it's still power series here. I just replace my coefficient field with its perfection. So this is all it does. And then this is a nice extension. It's faithfully just change, faithfully flat to just change the residue field. And then the only problem is that it will generally introduce new potents, right? Because we add roots of elements, but we know how to get rid of new potents using this recipe. Because of faithful flatness, you see that I mean this is, and because the maximal ideal extends to the maximal ideal, you can check easily that this is not going to change your lens, right? And therefore you can just do your convergence proof over there. Okay, but if you're not happy with this, I mean we can just assume everywhere over here that our field is, well, perfect and that our ink is reduced. Okay, so this was the only comment about affinitness that I'm going to make. Otherwise I will pretend that things are going to be affinit and they will work for not affinit things too. Now let me do the property of Hilbert-Kunz multiplicity that I wanted to do. So if you have an exact sequence, let's say L to M to N, then we get equality on Hilbert-Kunz multiplicities. So in other words, Hilbert-Kunz multiplicity is additive in short exact sequence. So the proof of this is essentially the same proof as what we did on that board where our exact sequence came from here, right? We just need to get isomorphism at minimal primes and we get what we need to get. So first, if my R wouldn't be reduced, I could put large enough Rabinus push forward to make this model to be over reduced ring. So let me just pretend that R is already reduced and after that, if R is reduced, then localizations at minimal primes are fields for any minimal prime and then when I localize this, I get exact sequence of vector spaces. So of course it's additive, right? So I'll get that Mp is going to be isomorphic to direct sum of NP plus LP. And because this isomorphism holds for any minimal prime, I can do the same trick as I did over there, right? I can look at a model M and I can compare it to the direct sum of N plus L and they'll get the two exact sequences as we got last time, right? We get exact sequence of M mapping to direct sum of N and L with the cokernel T1 and we get exact sequence of N direct plus L mapping to M with the cokernel T2. And the properties that I get from this isomorphism is that this cokernel T1 and T2 are going to have small dimension. And then I just do the same thing as we did here, right? I will see that these cokernels are not going to affect Hilbert-Kunz multiplicity and the Hilbert-Kunz multiplicity of direct sum is clearly a sum of Hilbert-Kunz multiplicities and the theorem is proved. Does it make sense? Okay, so it's the same trick. And from here we deduce a very useful formula that we are going to need in future is that if we apply this formula for any prime filtration of N, we are going to get that Hilbert-Kunz multiplicity of M is going to be equal to a sum as P varies through this minimal primes of the maximal dimension of Hilbert-Kunz multiplicity of I r mod P and then the length of M localized at P. So that's a useful formula because it's essentially for free often just reduce you to the domain case, right? And the proof of this is simple. So you take prime filtration, so it's submodules M0 through Mi inside M. So there's equations isomorphic to r mod sum prime PI and then we apply that Hilbert-Kunz multiplicity is additive in short exact sequences and we get from here that Hilbert-Kunz multiplicity of I and with respect to M is going to be the sum of all those Hilbert-Kunz multiplicities of I and r mod PIs. But this is Hilbert-Kunz multiplicity here as an R module. This is not Hilbert-Kunz multiplicity of R mod PIs ring, right? So this thing by that lemma over here in the corner is going to be zero unless R mod PI has a full dimension, right? Because I have the estimate. So this is going to be each of them is going to be zero unless my PI is inside this minimal primes of the maximal dimension. So you get the only PI that survive here and then you count the number of them by localizing prime filtration. It will be composition series of M. So you're going to get this number of primes as the length in the localization. So this is often called associativity or dTvT form. All right. So now let's switch our attention to a signature. So we've worked a bit and we show that Hilbert-Kunz multiplicity exists but I also promise to you to show that signature exists. So let's try to come here pieces. So the statement that we get for a signature is exactly the same. So then a signature of R exists and I can even get the same convergence estimate if I want. But before we can actually write the statement we need to do the same preliminary work that we did for Hilbert-Kunz multiplicity. So last time, right? I explained to you how to get from this more natural model theoretic if you like definition or singularity theoretic definition of Hilbert-Kunz multiplicity as the number of generators, right? Limit whatever number of generators for being used over generic rank. I can show that this fraction number, minimal number of generators and generic rank. Well, we computed it and we showed that this is equal just to the length of R mod M bracket Q over Q to the power D. So to enable me to use the same machinery I need to do the same thing for a signature. So if you like this as a lemma, so I defined for you a signature as coming from the sequence of maximal surjection on a free model, so maximum of M so that we have a surjection from Frobenius push forward to direct sum of N copy so far and divided by generic rank. So I want to say that we are going to get similar type of formula. So this is going to be the length of a quotient by an ideal IE divided by still Q to the power D. And the only difference is that I need to explain to you what is IE. This ideal IE, the Frobenius splitting ideal has a description in terms of maps from Frobenius to R itself, right? So it's going to be, so IE by definition consists of element X such that for any map coming from, well, let's say home over R, Frobenius push forward to R, I'm going to get that the image of X is going to be inside the maximal ideal. So this is elements that do not give me a splitting, right? If the image is not contained in the maximal ideal, right, this is a unit and therefore, right, I get a surjection onto R so I get a splitting, right? So this ideal IE consists of elements that never, never are going to give me a splitting. So, yeah, here, yeah, IE, no, IE depends on R, right? This is elements on R, thank you, yeah, this is elements on R so that for any map writers they're going to never be surjective. So is the lemma clear? So the benefit of this lemma is that we'll get to ideals, right? Again, what was the benefit of this statement is that we got to ideals to Frobenius powers and then we can do all this, then the ring can set all the machinery, right? So this is going to be the same benefit to us. So let's prove this. This is really simple once you just know what to prove. So what we do is that we write now this direct sum decomposition. So I have this maximal free model, so n maximal. As in the definition of a signature and then I have some leftover part, let's call it ME. But because I have surjection, right, I split, so I get decomposition like this. So let's try to apply any map from here. So if I apply a map and I take an element inside ME then what can happen? Well, there cannot be, it is not possible that the image of ME is not going to be contained in M because if I get something outside of M, I'll get one more splitting but my n was chosen to be maximum, right? I cannot extend it to n plus one. So it must happen that this image is contained in M because n is maximum. If not, I could get one more splitting right by taking a pre-image of element mapping to a unit. So what does it tell me? It tells me that if I take ME, right, and I add to it M times, well, let's write it M or plus M, this is going to be contained in a flow over star E, IE, right? If I take element in R, right, which corresponds to something laying in ME, right, which is probably news, I cannot have a map, right, sending it to a unit. And so this is the first containment, right? And the second containment, I just, well, if I multiply, I mean, it is clearly from these definitions that M times F over star E R, right, is going to be contained, you know, any map will send this by linearity into M, right? Just because my maps are maps of our models, right? So this is how I get direct sum of M's here, just by multiplying everything by M. So we have one containment, but then if you look at it a little bit closer, right, we will get that it's actually not going to be containment, this is going to be inequality. Why is this going to be inequality? Because any minimal generator over here, any minimal generator of this free model, naturally gives you splitting, right? Because it's a minimal generator, you organize them and this is a splitting. So we just get the equality like this, and then you compute the length and you're going to get the statement, right? So the length of F over star E R mod IE is just going to be equal to this number M, right? And then it's the same machinery, we plug what is the rank and this cancels the field extension and we're only left with Q to the power D. So once we have this observation, the rest of the proof is same as what we had before. So let me remind you the key part, right? So we had exact sequence F over star R to some direct sum of Rs to T1 to zero, and we have here opposite. So direct sum, I think I called it gamma to T2. So I can do the same thing, I can tensor them, so let me just write IE, here is IE T1, and here I'll get IE bracket P, here I get IE, here I get IE bracket P. So to continue my proof, I need to do two things, right? I need to say that this thing is small, right? I mean, because T1 and T2 have small dimensions, right? And then I need to also somehow get rid of the bracket power. To do so, I need to observe two properties. So this description that I have here, it immediately tells me that those ideals IE they have good behavior with respect to Frobenius. So the properties that I get is that if I apply bracket power to IE, it's going to be contained in IE plus one. And the second property is that if I take any map that sends, well, from Frobenius to R, then it's going to do opposite. It's going to send IE plus one inside IE. So why does this happen? Well, the first one is just from this, right? And the second one, it comes just by looking an exact sequence, right? We have F lower star R to R, right? So I take my map phi and then I could take F lower star E of phi. And then by composing it with arbitrary map psi to R, I'll get map from E plus first push forward to R, right? And I know by definition that IE plus one should be sent by this composition inside M, right? So if you look at it, then the image over here also needs to be sent by M, right? To M, so by definition it means that it ends up in IE, right? So this is just from the definition and the fact that we can push Frobenius, right? First, does it make sense? Okay, very good. So what do we see then from this exact sequences? So let me say A, B, right? From A, I'm going to use the property two to see that A actually can be refined to a map from IE plus one. I mean this submodule IE plus one, right? Is going to be still contained in the kernel, right? Because those maps over here, they are what's called P minus E linear maps, right? Because if you look at the projection on each of the individuals a month, right? This is one of these maps from the home, right? So the map over here is just natural, gives me exact sequence of this form. And from here I will get the length of R mod IE plus one. No, other way around. So gamma times length of R mod IE minus length of R mod IE plus one is going to be length as the length of the quotient of the kernel. But if you look at this statement over here, or perhaps if you just look at this containment, I know that, well, if you like this zeroes property, that bracket power of the maximal ideal is going to be contained in IE. So I can bound the length of the kernel still by the lemma that I had over there, right? So it's going to be still C over, well, times Q to the power D minus one, right? Well, maybe I should say P to the power E D minus one. So the first sequence just easily patched using property two and it gives me the estimate that I need. And the second sequence is even easier because I have this bracket power in the middle, right? And I have this containment, the length of this quotient is going to be greater or equal than the length of the quotient by IE plus one, right? So I'll get that the length of R mod IE bracket P is going to be less, so greater or equal than length of R mod IE plus one, which is bounded by the sum of those two things. So gamma times length of R mod IE plus again this constant C Q D minus one. The rest of the proof is absolutely the same, right? I have a sequence now and I can get the Kashi estimate for the sequence and then it converges. I switched right and E is on tops. This should be reverence about. So it's the same. Once you have the lemma and once you observe that this lemma implies this two useful properties, this exact sequence is again do the same job, right? And we have the same bound on the car kernels. So they will give me the same convergence estimate. So this is a bit, okay. Any questions? Any questions? Okay, good. So we proved the existence of Hilbert-Kunz multiplicity. We proved the existence of a signature and now I can start working toward approving their approach. So maybe just so that you have it, right? What we get as a result of my proof is the same statement as we had before. Colant of R mod IE over Q to the power D. It will be uniformly controlled, right? We get this convergence estimate C over Q. Okay, so I now know that the signature exists. I know that Hilbert-Kunz exists and now it's then the right time to start studying their properties and try to actually understand what does it tell about singularities. So next time, I think I will talk about this key property of detection of singularities and today in the remaining minutes, I want to actually explain the connection between a signature and stronger for clarity. So the theorem that we will prove is that, well, I'll take my local finite ring and then it is equivalent for R to be strongly of regular and to have positive F signature. So F signature, as we discussed, it has two special values, right? Zero and one. Next time, we will say that the value one generalizes Kunz theorem and it shows that R is regular and today we will do the other special value, zero, the minimal value. It precisely tells me whether the ring is F regular or not. So let's do the proof, right? Let's start by assuming that R is not strongly of regular and trying to prove that the signature is equal to zero. So suppose R is not strongly of regular. So what does it mean? What is the definition of a regularity? So definition, let me remind it to you, say that R is strongly of regular if for any C not contained in minimal primes, in any of the minimal primes, then there will be E and the map, Frobenius push forward, so P minus E linear map that will send C to one. So this is the definition of strongly of regularity. Now I need to negate it, right? So I assume that R is not strongly of regular. So what does it mean? So not for any C, right? There exists a special C inside R zero, right? So R zero again is the complement of the set of minimal primes. So that what? For all E and for all phi, what do I have? I have that it will never go to one, right? So it will never go to unit. I'll get that phi of F lower star E C is always going to be inside the maximal ideal, right? So this is the negation of stronger regularity, right? There is an element C which will never, never going to give me a splitting. So now what does it mean? It means by definition over there, right? That this element C is contained in all my ideals IE, right? Because what is IE? IE consists of all elements, right? That for a given E, right? Will never give me a splitting, right? And this element C will never give me a splitting for all E's, right? So then I look at the statement over there, the conversion statement, and I get the estimate that the length of R mode IE is going to be less or equal than the length of R mode CR plus M bracket PE. Let's say, why? Because I know that my IE contains both the Frobenius bracket power, right? And it contains element C. So I give this ideal contained in IE. But what do I know about element C? It's not inside minimal primes. So it follows that dimension of R mode C is going to be less than dimension of R, right? It's a parameter element. So by the lemma on the Frobenius length, we are going to get that this is less than C to the power PE dimension of R minus one. And therefore, if I take the limit, I'm going to get zero, right? Because I'm dividing by E to the power dimension of R. So this direction is really easy from the definition of our Frobenius splitting ideals IE. Good? The second direction is more complicated. So we will have to work for it. Let's start. Well, yeah, I guess what I can do in the remaining time is to prove that the strongly F-regular ring is a domain. So basically, for the other direction, I will need to do some reductions to end up in a better case. So let me just start doing this and then we continue next time. So let's say that if R is strongly F-regular, is a domain. So first, R is going to be reduced. Why is it going to be reduced? Because if I look at the definition of stronger F-regularity and I apply it to C equals to one. So if you take C equals to one in the definition, as this is definite element, not inside minimal primes, then I will get that for large E, I'm going to get a splitting, right? Which will send one to one, right? So I'm going to get a surjection and it's a splitting. So it will give me an injective map, right? From F to F lower star ER, but we know that existence of such injections implies that R is reduced, right? Now let's prove that it's a domain. So let me just take minimal primes. So say P one through P whatever M, I'm minimal primes, and let's take magically elements Fi to be contained in the intersection of all PJs which are not equal to I minus PI. So I have those elements Fi and then let me take F to be their sum. Actually not F, let's call it C. So I call it C because I'm going to apply the definition to it. As you see from the definition, I can write this element C cannot belong to any minimal prime. So I can apply, is not contained in any PI. So I can apply the statement of the theorem. So I'll get some of Fi. Well, I can write, I can take the sum out. So I get the sum of Fi F lower star E Fi's is going to be equal to one, right? So this is just the definition of stronger figure. I will find E large enough so this works. So I have a sum of elements in a local ring which add up to one. So it follows that one of them is a unit, right? I cannot have elements all inside maximal ideal to add up to something not inside maximal ideal. So one of them needs to be outside of the maximal ideal. So without loss of generality, I assume that this is let's say F1, right? So Fi of F lower star E F1 is a unit, good? So then what happens? If I look at F2 times U on one side, this is a nonzero element, right? So this is not equal to zero because U is a unit and F2 was a nonzero element. But on the other side, what do I get here? I'll get Fi of F lower star E of F2 to the bracket PE times F1, right? So it just be, just our linearity, right? So I move F2 here, then I move F2 inside the Frobenius and I get this. But what do I know? I know that F2 times F1 is contained in the intersection of all minimal primes, right? Just by the definition of them. But my ring is reduced so the intersection of all minimal primes is equal to zero. So therefore F2 times F1 is going to be zero. Therefore this is also going to be equal to zero and what I get is a contradiction with the fact that I have several minimal primes, right? I must have one minimal prime. So this contradiction is M not being equal to one, right? If I have only one minimal prime, there is no F2. Okay, very good. We are right on time, 2.50. So let me finish today and then tomorrow we are going to finish this proof. Sorry, I didn't hear you. Sorry, Hilbert-Kunz multiplicity and F-signature. Well, in general it's not so much related, right? But in the Gorinstein case you can say a bit more. So, well basically, yeah, let me say that if R is Gorinstein, then the original paper of Heuneky and Lushkin that introduced F-signatures, they proved that F-signature of R is going to be equal to Hilbert-Kunz multiplicity of a parameter ideal X, right? Minus a circle. So, Hilbert-Kunz multiplicity of X at U, so U, circle element, right? So this is the relation that we have as an equality. Otherwise, we have a conjecture of Watanabe and Yoshida, a theorem of Paul's Tren Tucker that you can write F-signature outside of this Gorinstein case. Generally, you can write that F-signature of R is going to be in phenome of all relative Hilbert-Kunz multiplicities of that sort. So I'm going to write Hilbert-Kunz multiplicity of I minus Hilbert-Kunz multiplicity of J over the coland. Well, I just say that I, U, right? U-socal element. But here I is not a parameter ideal if you are not in the Gorinstein case. And from the first formula, if you'll get something of minimal multiplicity in Gorinstein, so multiplicity is going to be equal to two. If you take a parameter reduction, so you're going to get a direct formula relating F-signature number two and Hilbert-Kunz multiplicity, right?