 Put immediately the goal that I was announcing last time, right? Today our goal is to prove this theorem original due to Watanabe and Yoshida which generalizes celebrated Nagata's theorem to Hilberg-Kunz multiplicity. We will show that a local ring is regular if and only if it is formula unmixed and Hilberg-Kunz multiplicity is equal to one. So again formula unmixed means that it's equidimensional and there are no embedded components. For the completion. So one direction is not easy, but it's already done due to Kunz. We discussed that if it's regular rings and Hilberg-Kunz function is identically one and we will fight to get the other direction. So our strategy will be similar to what we did yesterday for F-signature. We will do induction and we'll use equimult multiplicity. But things are going to be a bit more challenging because F-signature leaves in a strongly irregular ring, but Hilberg-Kunz multiplicity not. So we will need to work a bit with tight closure. But I think it's quite appropriate for this conference to talk a bit more about tight closure and honor Melon Craig in this way. But first let me say what is the key tool? So the key tool for the theorem of what the Naby-Ushida is not surprisingly lm of what the Naby-Ushida is. So if Hilberg-Kunz multiplicity of R is equal to one and I is an primary ideal such that it's Hilberg-Kunz multiplicity is greater or equal than its co-land, then R is regular. And I need to put one more assumption so we also need I to leave inside M bracket P. Just simply, if we know that Hilberg-Kunz multiplicity is one and we can find a special and primary ideal then that's enough to deduce that R is regular. So let me prove this lemma. For that I will need two simple ingredients. The first one is a slight version of the result we discussed last time. So last time I proved this for J equals to R by a filtration result. We took just the definition of co-land, well in this case we'll take the length of J mod i, we'll take the composition series and we apply the Frobenius and then we can't think and we'll get this inequality. Last time we did it for J equals to R so there was no Hilberg-Kunz multiplicity of J but here we filtered not all the way to R but to J. The second result is also very simple, we say that the Hilberg-Kunz multiplicity of a bracket power is going to be P to the D original Hilberg-Kunz multiplicity. So this is an easy result coming just from changing index in the convergent sequence. So what do we have by definition here? We'll get on top IQP but in the denominator we have something slightly scaled. So we multiply both sides by P to the power D and we'll get the original Hilberg-Kunz multiplicity. It's just a formal consequence for any convergent sequence. If we, with such normalization, if we change the index it's going to pop out. Same way how it works for the usual multiplicity. So two simple lemmas to prove the lemma for the NAPI and your shape. So let me give you the proof. Well, we need to use that this ideal i has all these wonderful properties. So I am going to use this lemma for i contained an m bracket P. I have Hilberg-Kunz multiplicity of i and it's going to be less or equal than Hilberg-Kunz multiplicity of r times length of m bracket P over i plus Hilberg-Kunz multiplicity of m bracket P. And I can compute many things here. So I know that Hilberg-Kunz multiplicity of r by the assumption is one. Well, the length, I don't know, it still stays. What do I have here? By my lemma I get this P to the d goes in front. Then I get Hilberg-Kunz multiplicity of m left, but that's by definition. Hilberg-Kunz multiplicity of r again equals to one. So the other part is P to the power d by combining with lemma. Good? All right, now what else we know? We know that this is going to be greater or equal than the co-length. So length of r mod i. How can I write the length of r mod i? Well, I can still use it. I have a bigger ideal. So this is length of m bracket P modulo i plus co-length of m bracket P. I just made an extra step in the computation of co-length. So what do we see from this chain of inequalities? This is the same length as here, right? So the inequality between those two quantities will imply the inequality between P to the d and the co-length. So I get that the length of r mod bracket m bracket P is less or equal than P to the power d. And now we look at this. We remember the Kuhn's theorem, the computations that we did before on Tuesday in the first day of classes. And we see that this immediately will imply by the Kuhn's theorem that r is regular. So we know that actually the inequality goes the other way around. And if its equality holds, then r is regular. This is the equivalent characterization of flatness that we discussed. So as you see, it's a very simple lemma, but somehow it has a strong punch. I don't know any good way to prove the Watenabe-Ushida theorem without going through the Watenabe-Ushida lemma. Somehow it is really, really crucial for all the proofs that I know. Because you could ask, we have a very easy proof for a signature. We can get as easy proof for Hilbert-Kuhn's multiplicity if we assume that r is regular, r is stronger than regular. But then, if you could prove from Hilbert-Kuhn's multiplicity equals one that r is stronger than regular, you could, may as well, just prove that it's coin Macaulay. And then, this is what Watenabe-Ushida did in their original proof. Once you know that the ring is coin Macaulay, you can just take i to be a parameter ideal deep enough, right? Because we know that for parameter ideals in coin Macaulay ring, we have the equality. Because Hilbert-Kuhn's multiplicity equals multiplicity for parameter ideals, and that's the characterization of coin Macaulay rings. But proving that it's just coin Macaulay from this condition is not an easy task. Other ways to do it, well, we could use integrate closed ideals. And the other way to prove this, and basically some of the ideas that I will present, they come back to the paper of Craig Hewneke and Yanwei Yao who did induction, the same way how we do induction, and then they choose a special i to be symbolic power of a smaller prime ideal plus parameter element. So we will do something similar, but our construction of i will be based on induction and equimultiplicity. Okay, so let's proceed. So I want to start building equimultiplicity machinery. So last time we built, we understood when f signatures are equal, right? That was the equimultiplicity. Today we need to understand what happens when Hilbert-Kuhn's multiplicity doesn't change when you localize. And to do so, I need a little bit more about tight closure, so tight closure. So I'll mention some of these things that we already said, let me remind you the definition that x is inside tight closure if and only if there exists c, which is not contained in any minimal prime, and c sends bracket powers of x into Frobenius powers of i for all q large enough. So that's the definition of tight closure. I also need definition of test element. So c is a test element. If it works for all x and all i over here in this definition, it does not exist c, it's this c will always work. So it works for all x, all i. So this is the element you can always use to test the tight closure. That's why it's called test element. I will need the result of Hoxter and Heunicke, which give us existence of the test elements. If r is complete, well, for me it suffices complete and reduced, then test element exists. And I want to remind you one more thing which was already mentioned by Holger, well all those things more or less we've mentioned by Holger, is that if under these assumptions of the Watenabe you should have theorem, so r had unmixed, we have the tight connection between tight closure and Hilbert-Kunz multiplisit, then if i is contained in j and Hilbert-Kunz multiplisit of i is equals to Hilbert-Kunz of multiplisit of j, if and only if j is contained in tight closure of i. Okay, so I will need these four facts about tight closure. And I need also a couple of statements about it. So first I want a simple lemma, so if we have a test element, test element c, so i, any ideal, x is not contained in i, and I think, yes, that's all I need, x inside m, then the intersection of tight closures of i plus xnr star intersection of all m greater or equal than 1, greater or equal than 1 is going to be equal to the tight closure. So intersect tight closure by adding deeper and deeper powers of x, and it's still going to intersect just to the tight closure of i. So I'll leave this as an exercise on the definition of test elements, it's very simple, you really need to use that c is uniform for all those tight closures, and then the assertion will just follow. The second lemma that I will need is about filtration of ideals, lemma. So under the same assumptions, we have a local ring which is unmixed. So now we are going to use the connection with Hilbert-Kunz multiplicity, and suppose that, so we have local ring such that this completion is unmixed, and then le, so i is m primary, and le is a sequence of m primary ideals such that I have three conditions, maybe a. So this sequence contains bracket powers of i, this sequence behaves well with respect to bracket powers, and c, it converges to the Hilbert-Kunz multiplicity of i. So limit as e equals to infinity, the length of r mod le over p to the power ed is equal to Hilbert-Kunz multiplicity of i, then I want to say that my ideal le is contained in tight closure of bracket power for all e, so lemma, pause here for a second. So basically what I'm saying is that if we have Frobenius well behaved sequence of ideals which still converge to Hilbert-Kunz multiplicity of i, then all of them needs to be still within the tight closure, right, they cannot be too big. Is assertion clear? Okay, so let's prove it. So the proof actually is very easy, let's fix le, let's say, well let's just fix le, right, and what do we know about le? If I take e plus e prime, I'm going to have what, inside of it I'm going to, no, I want to say, so I want to compute Hilbert-Kunz multiplicity of le, so I need to take it to the bracket power, right, that's by definition how I compute Hilbert-Kunz multiplicity, and now I want to sandwich it between things and what I can use, I can use property a and say that this is going to be containing p to the power e plus e prime, right, and on the other side I can use property b and say that it's going to be contained in le plus e prime. So I sandwich the bracket power of le between those two things, and then I need to compute the limits, so if I take the column of those guys and divide by p to the power e prime d, right, so I divide it by p to the power e prime d because I want to compute Hilbert-Kunz multiplicity of this ideal, right, so e prime is available, so what I'm going to get by the lemma over there, here I'll get Hilbert-Kunz multiplicity of i pe, right, which is equal to, well pe something, pe d Hilbert-Kunz multiplicity of i, so this is the lemma, right, so this is the definition of Hilbert-Kunz multiplicity of pe i to the power pe and this is the lemma. Then what do I have on this side? By my assumption, my sequence still converges to Hilbert-Kunz multiplicity of i and I'm changing the index, so it is the same proof as what I have here, right, I changed index in this sequence, I got that the limit is multiplied by p to the power d, I change index in that sequence and I will get that the limit is going to be multiplied by p to the power e d and times the original limit, which was by assumption that I can get by Hilbert-Kunz multiplicity of i, right, so what happens? It follows then Hilbert-Kunz multiplicity of le is equal to Hilbert-Kunz multiplicity of i bracket pe and therefore I can use, well one of these properties, the last one, right, to say that le must be contained in the tight closure of p, yeah, here I used property c, does it make sense? Are there any questions? Okay, good, so as long as my sequence is respect in Frobenius, I can deduce statements about tight closure, not just from Hilbert-Kunz multiplicity of one ideal, but from the entire sequence. All right, so that was the tools from the tight closure theory that I will need in order to show, in order to derive the equimultiplisty machine, so let me start raising things and now we can talk a little bit about Hilbert-Kunz equimultiplisty, so our goal is to understand better what does it mean for Hilbert-Kunz multiplicity of localization prime to be same as Hilbert-Kunz multiplicity at the closed point, right, that's the maximal ideal. And the tool that we will use is the same tool that we used for the localization inequality, so let me remind you that we proved that if I take Hilbert-Kunz multiplicity of p plus x and r, so if my, so in this setup we had the dimension of r mod p was just equal to one, right, and maybe let's call them q. So we proved last time, right, that this sequence, Hilbert-Kunz multiplicity of my prime q, and then I add element not inside q, any element, so x inside m minus q, so it becomes q m prime ideal, x plus q, and then if I take limit of the sequence, it's going to be Hilbert-Kunz multiplicity of x r mod q times Hilbert-Kunz multiplicity of r localized, or this is a parameter ideal so I can write multiplicity, maybe I prefer to write multiplicity. So we had this lemma, and that's the lemma that connects m primary things to something q primary, right, that's the one that allows me to localize if you like. So I want to build more understanding from this lemma, and if you remember how did we then show the localization inequality, I bounded this Hilbert-Kunz multiplicity using the filtration lemma by Hilbert-Kunz multiplicity of m times the co-land, and it's also converged to multiplicity. So we want to build more understanding from this result, and I will do this by showing more about the sequence, so the sequence does not just converge, it's actually a monotone sequence. So the lemma, crucial for me, is that under this assumption, so dimension of r mod q is equal to 1, x is inside maximal ideal but outside of q, I can show that Hilbert-Kunz, I will show to you that Hilbert-Kunz multiplicity of x plus xn plus q over n is going to be greater equal than the same thing with xn plus 1. So it's a monotone sequence. Last time we just proved this convergence, but I actually explained better why does this sequence actually converge. And the important part for me, because I will try, I'm interested in equi-multiplicity, moreover, if equality holds over here, then we are going to get, well, rather technical look and condition on limits. So limit as, no, not limit, yeah, so limit as e approaches infinity, length of r mod, so let's see, so xp e, r plus r is equal to 1, xn minus p, no, not q bracket, p e colon xn, so this is my ideal, I take bracket power of p e of x, right, and then I take colon of bracket power of q, so this is different thing. So I take colon, ah, n should be p e, and then if I divide it by p to the power ed, this is going to be the Hilbert-Cons multiplicity, so this is going to be equal to Hilbert-Cons multiplicity, well, 1 over n Hilbert-Cons multiplicity of q plus xn. So we will see where this technical look and limit comes from. So the proof of the lemma is actually just by following the proof of this statement from the last time, but with paying attention to the details. So the key part from the last time was that if we had a one-dimensional, then we had exact sequence of, sorry, it was a, a mod xn, a, no, n plus 1, right, it surjects clearly on a mod xn, a, and we can compute the kernel as a mod x plus an annihilator of xn. So we have this exact sequence. Well, what I can get from this exact sequence? I am going to plug, well, I am going to plug into this exact sequence for a b-news powers, right? My a is going to be what? It's going to be r bracket q to the power pe, right? And my x will be appropriate powers of x, right? But maybe, let's see, yeah, let's maybe write carefully what do I get? What do I want to remind you? So I want to remind you that, so how did the proof go? So we have proven that length of a mod xn, right, a is equal to the length, also length xn minus 1 a plus this colon over there, so length of ax, ax minus 1 a plus this colon with xn minus 1, right? So this was the first step of utilizing this exact sequence and then we kept going, right? Now we can go from xn minus 1 to xn minus 2 and so on. So we would get what? We will get that this is equal to the sum of the length with different colon ideals, x a plus 0 colon xk, right, where k varies from 0 to n minus 1, right? It just utilizes the sequence over and over to decrease this power eventually just to power 1. And what else we observed in the just original proof? So I'm repeating the argument that we used over there. So we observed that this is going to be greater or equal, so each of these colon ideals is going to be less or equal than what I have here, right? So this is length of a modulo x a plus 0 colon xn, because my colon ideals, they increase, right, with them. So I have the containment 0 colon x i into 0 colon x i plus 1, right? The annihilators get bigger and bigger, so I get this inequality just because each of them individually is smaller exponent, right? Okay. So now I can look and apply my observations to what I had over there, right? So I can write to use at the length of, so I want to prove to use this inequality, right? So I need to get something with xn plus 1 and something with xn and I could multiply n plus 1 to that side, right? So the length of a modulo xn plus 1 is equal to the length of a modulo xn a. And then I have this colon piece, right? But I can, okay, let me write it. So I'm actually not worth to write it. So now I'm going to use this inequality to remove this colon ideal, right? So this is going to be less or equal than 1 over n length of what they have there, right? Length of, so what I have, so x, so the length of a modulo xn is greater or equal than n times this length of the colon ideals. So I replace this guy, right? By 1 over n length of just a modulo xn a, right? So I have the same thing here and the same thing there, so they add up and they'll give me this n plus 1 over n length of a modulo xn, makes sense? So if this was statement star, then here I use statement star together with this bound over there. So this is where this inequality comes. Into this equality I now replace, I give you a bound on this. Make sense? Take a look. Any questions? Okay, so now plug instead of a, r mod q to the bracket power. And the statement just follows, right? So take a to be, a to be what? a to be r modulo q to the bracket power pe. Then what do I get from this inequality? I get that the length of r modulo my q bracket power pe. And let's replace x to be xn to the power pe. What, so I get here x to the power n plus 1 pe is less or equal than n plus 1 over n, right? Length of r mod pe plus xn pe, right? So this is what I get by plugging a to be r modulo q bracket power. And then what do you see? You take the limit, right? And you precisely get the statement over here, right? I just left to take the limit to derive the first statement and the second statement just comes from looking at this defect, right? If you look at this inequality and I have the quality between this and that, then I should have the quality over here, right? Well, I'm taking limit, so I won't get exact equality, but I get equality in the limit, right? This column term over there, so this one over here, the ones that I replaced should converge in the limit to this additional term that I have here, right? Because this is the inequality that I used. So if the limits are equal, this thing and this thing should have the same limit. Does it make sense? So here, take limit and you're going to get Hilbert Kuhn's q plus xn plus 1 less or equal than n plus 1 over n Hilbert Kuhn's q plus xn. Okay. Good. Any questions? All right. So that's a crucial level because what does it provide to us? It provides to us an interesting filtration inside my, I mean, it gives me something fitting into this level, right? Because I get now some filtration which has the same limit as Hilbert Kuhn's multiplicity, which has the limit equals to Hilbert Kuhn's multiplicity of some ideal. So from here, I can deduce something interest. So what I can actually deduce? So let me raise the proof. Let me leave the lemma. So same assumption. So let me just say dimension of r mod q is equal to 1, x not inside q. And suppose that Hilbert Kuhn's multiplicity of, Hilbert Kuhn's multiplicity of r localized at q is equal to Hilbert Kuhn's multiplicity of r. And then what I claim to you is that, well, let me just try that. If I take q to the power pe and I take type closure of them, then this is going to be a modically saturated, right? So this is going to be equals to itself. So the type closure in other ways because r mod q has dimension 1, m is the only potential associated prime. You can restate it as the type closures of Rabinu's powers are q primary. Equivalently. Equivalently, this type closures, yeah, maybe this is how I should reward, is q primary. So let's prove this. Well, I want to use the lemma. I have this family of ideals. So let's look at what I get. So if I take le to be q bracket pe colon xn pe plus xper. Then over there, I prove to you that equimultiplicity is going to force maybe add, let me add to you the actual proof, right? Let me actually remind what we deduced last time, right? So here we wrote this is less or equal than Hilbert Kuhn's multiplicity of r times the same multiplicity. This was how we proved the localization in equality, right? We filtered this thing and we got upper bound and this lemma that we had give us equality on the limit, right? And from here, we deduced that Hilbert Kuhn's multiplicity of r is greater equal than Hilbert Kuhn's multiplicity of r localized at q, right? So if I assume these two things are equal, right, then I am going to get equality over here, right? But then I have a descending sequence, right? So the equality must hold all the way through, right? From starting point q plus xn x, then going down and down, they all must be equal to this thing because I sandwiched it between two equal things, right? So I have a descending sequence which starts and adds at the same point, right? This is what happens. Okay, does it make sense? If Hilbert Kuhn's multiplicity of r is equal to Hilbert Kuhn's multiplicity of r localized at q, this monotonicity forces that Hilbert Kuhn's multiplicity of q plus xn, right? r is going to be equal to n times Hilbert Kuhn's multiplicity of r localized at q times this thing, right, e of xrk. In other words, the sequence over here will become a constant sequence. And this is why I can get this limit condition. Does it make sense? Okay. Good. So then I look at this sequence of ideals. It is really easy to see using the properties of colon ideals that the sequence satisfies the assumption of the lemma. So lemma, I don't know, I should give it a name. So lemma a. So by this lemma a, I will get that le star is going to be contained in what? In q bracket pe plus x bracket pe, right? Together tight close. I look at the definition and I see that this part is already contained in the right hand side. So from here, it follows that this saturation, this colon ideal and x and pe is contained in this tight close q pe plus x pe. This holds for all n, right? Because nothing depends on n on the right hand side. This is increasing chain. So eventually I'm going to get here the entire saturation. And x is a parameter, q has dimension one. So this is nothing else as just q bracket pe, saturated with respect to the maximal ideal. Because modulo q, x is parameter, right? It is n primary, right? Modulo q. All right. And then what happens then is that I have this containment, right? So what I have is that q bracket pe saturation is contained in q bracket. We have this thing. But this statement now holds for all x. In particular, I could take one element x and take its powers, right? I could take x and pe here, right? And greater or equal than one, right? Because on the left hand side, nothing depends on x. On the right hand side, I was working with arbitrary x. So I can take x to be anything in particular powers. And then I can use the tight closure lemma to say that this is going to be contained in actually tight closure of q pe star. So the saturation is contained in the tight closure, okay? Okay. And then an easy exercise. So I won't explain it. It actually forces our tight closure to be saturated. You just write down imagines that it's not saturated. There is an element sent by power of maximal ideal into the tight closure. And you apply this condition that the tight closure still converge to the same limit as the regular powers. And you are going to get that this condition implies that the tight closures are already saturated. So this is a tight closure. So you see, I mean, using all these properties of tight closure, we get at the end from this rather technical lemma over there, a rather pretty condition, right? We just get that the tight closures of Rabinius powers, they're going to be q primaries, they're not going to have m primary components. Okay, so let me then erase this lemma and write one immediate consequence of this corollary, corollary of a corollary. So if we know that an ideal is q primary, so the quotient by it, because it's one dimension that's going to be coin Macaulay, so we have the consequence between length and multiplicity. So under these assumptions, assumptions from here, what I'm going to get is that the multiplicity of x r mod q bracket p star is going to be equal to the length of r bracket q p plus x r, one dimensional coin Macaulay. So just the condition of x being a regular element in this quotient, this is all it is. All right, so I can erase now everything and I can finish the proof of the, what do I mean by the theorem? I have now everything that I need. So we do induction on d, d equals to zero, we have Hilbert-Kunz multiplicity of r is the same as the length of r. So being equal to one implies that r is a field and field is regular, so we are done. So if d is bigger than zero, then I have a prime q such that Hilbert-Kunz multiplicity of r q is equal to Hilbert-Kunz multiplicity of r, right, proper prime contained but not equal to n just by taking a chain of prime ideals. So this is equimultiplist condition and this being equals to one implies that Hilbert-Kunz multiplicity of r, so because this is equal to one and this is always greater or equal than one, so from this chain of inequalities right, the localization we proved last time, the equality must follow through, right? So we get that Hilbert-Kunz multiplicity of r q is equal to one and in particular is equal to Hilbert-Kunz multiplicity of r and by induction, we assume that r localized at q is written, what do we need to do? We need to construct the sequence satisfying the what an ideal satisfies the Watenabio-Schiddler conditions and I'm saying that now let's just look at ideals that we can construct over there. So I'll take q bracket P star. I take x inside m bracket P, right, without q. So I take this ideal. And I need to verify that it will satisfy the assumptions of the Vatanabian Schroeder theorem. So it is easy to see that if my E is large enough, this is going to be all contained in the bracket power of mP, right? So this condition is OK, right? So this IE, well, IE is bad. Let's call it LE. LE is contained in m bracket P for IE large. You can use that it's going to be, I mean, you can just use that the risk algebra of integral closures is finally generated. And this is contained in integral closure. So I need just to compute length and multiplicity, right? So the Hilbert-Kunz multiplicity of q, q bracket P, right, star plus xr, I want to compare it to actually to the length, right? But there, I mean, right, so I need to compute this length, right? But there, I can just use the multiplicity condition, right? And let me quickly speed it up. So what we are going to see over there is that, so what do we see? So we see that if I take, so basically what I see is that if I take, basically, look at this thing, right? So let me take P bracket U, right? So this is contained, this ideal over here, it is contained, of course, in tight closure of qx, right, together to the bracket PE, right, star. So this Hilbert-Kunz multiplicity is going to be equal to the Hilbert-Kunz multiplicity of that ideal, so it is Hilbert-Kunz multiplicity of qx together, bracket PE. So this is going to be, yeah, I should say, PE here. So this is just going to be equal to PE to the power d, right, and then Hilbert-Kunz multiplicity of qx, right? But then this is one of these things in the filtration that I used, right? So I know that this is equal to PE to the power d, Hilbert-Kunz multiplicity of xr mod q times Hilbert-Kunz multiplicity of rq. So I get this on one side. And then on the other side, I just use the consequence of this corollary, right, the length. So I could put P bracket E here. This is going to be E bracket here. So this is going to be PE. And then I use the associativity formula. This is multiplicity of xpe r bracket q. So PE are removed times length of rq, q bracket PE star. And because I assume that the localization is regular, this is just equal to PE d minus 1, right? So at the end, I get PE d times the multiplicity. And over here, I assume that, I mean, it is consequence that Hilbert-Kunz multiplicity of rq is equal to 1. So I see that those two things are equal, right? And from this, I can apply the lemma of Atanabe and Yashida. And it tells me that my ring is regular. OK. It's a place to stop. Thank you. Thank you, Greg.