 So, always the Italian and today still has a key, yeah. So, recall that a, what's the question now, for every system parameter, the tight closure, so system parameter is tightly closed, okay. And so, so first remark that if it's a fractional, so it's, namely, so the fractionality localizes. So this is proved by Nakamura and also by Hoxton-Hunek. So, so you, so how about a frigate, a regular localized, so this is open, I believe, still. And back to, so there is notion of strongly a frigate, that's localized. So, second thing is that if a is a fractional, sorry, f is a fractional, and a is a fractional, and Gorenstein, so it's a frigate. It's, so let me prove this, so idea of the proof, so this proof, so it suffices for every, and probably I start going, I, but we can take parameter idea. So, since a is Gorenstein, this is Gorenstein too, and we have here, so W dual come back to the original one by Gorenstein property, and then so there is some easy lemma. So if I is tightly closed, then for, any idea, so is, so of course, this is parameter idea, by assumption, tightly closed, and so this is column of this, yeah, so this is tightly closed too. Okay, and then we come to the new section, that closure of modules, and FPR. So, of course, we have Frobenius map, and in this case, we write, so E here, so this is a module, the operation is obtained by Frobenius map, and okay, so for a module M, we put FE of M, M tensor. The easy way to recognize is, so if M is finitely generated, we have presentation of N in this way, and this map is described by some M by N matrix, and then tensor in E, we have the same, and only thing is, so this is changing to the power, choose power of each element, and for sub module, so this is a functor, and this is a functor, was like the exact functor of the category of M modules. Yeah, so if M is finitely generated, ah, so this is for general, so this is explanation in the case of finitely generated. So if we take a sub module, sub module, so we have induced by inclusion, but not necessary injective anymore, and so we note q equal the image of this map, this is okay for you, yes. So note that if, so i is an ideal in i, so the definition, bracket q is the same, as before. Okay, and yes, so the tight closure, so N tight closure in M is defined as x is in closure of N in M, if I don't leave for some note for every q, c, x to the q, ah, sorry, one more thing. So N, x, for x here, so the image of yeah, x here is x to the q, so xq is the image, c, okay, equal, so x tends to one, cx to the q is included in q, and note that, so it is easy to show. So we can, yes, consider the factor module and zero is the same as tight closure and tight closure of N in M, module of N. So in some sense it is suffices, suffices to consider the case equal to zero. So if M is not finite regenerative, so we take, so the regenerative union of zero colon M prime, where M prime runs every sub-module of M, which is finite regenerative, and yes, okay. So the most important, there are two very important modules in this dimension is D, one is HD, HD, so this local homology, another is the injective envelope, injective envelope. So perhaps everyone is familiar with injective envelope, so is there somebody not familiar with injective envelope? Okay, so, yeah, okay, everybody knows injective envelope very well, okay, let's go on. So, okay, so then the next M dimension D, so A is weakly irregular, the tight closure of zero is zero. Second A is a rational tight closure of zero, local homology is zero. In this case, so of course, this is not a finite regenerative module, but since this is a natural union of a factor module parameter ideal, so the local homology A is common, computed by FSC, the parameter ideal. So in this sense, so this part is the same as saying that every parameter ideal is tightly closed. Ah, my resume is not available yet, yeah. But anyway, I want to skip this. So I'd like to say one thing, so for E, yeah, so we consider the circle of this, so it is well known that this is clean file, and this element plays very important role, so in this, so I hope the resume will be ready soon, but sorry, it's regular in this sense. So not to zero is equivalent to say this element is, this is a very trivial fact, still very important, and yeah, so this is the same to say for every C, yes, x is some, it's not zero. Okay, so we come to the definition of purine. Ah, so sorry, yes. Ah, yeah, sorry, sure. Thank you, thank you. Okay, and then define a pure, so we defined as pure, and yeah, this is, this notion is the older than tight closure and the oldest of, so, and this is another way, yeah. So this is another way, yeah, yeah. We need some condition, maybe we need some condition. At least, so A is finite, finite model, so this appeared already, so, lectures of, yeah, you, and so, yeah, this is at least okay, so if we assume it's finite, and so to repeat K is a perfect field, yeah. So I prefer the notation, yes, and yes, this is, yeah, also used by Kevin, and then x1 to xd, one over q, of course, this is, and perfect, of course, as since we can take the same case here, so, we come to the feta square criterion, so assume that B, so B is a regular, then A is F pure, if and only if I bracket P column, I not include it, maximal ideal, bracket P. I wrote something in my resume, but not a complete proof, and so it is, the easiest case is hypersurface, then, so, of course, Ip, I means this is, of course, generated by Fp, and this is F, so generated by Fp minus one, so, and A is F pure, if and only if Fp minus one is not in bracket and qp, it's me complete proof, is not in bracket and qp, it's me compute one example, twice the cube, and this is the expression, so we can allow B equal 3, 2, and yes, and of course, when is A, so, of course, F equal x cube plus, and so if P, it is very easy to see that, so this breaks down for P equal 2 or 3, and so if P equal 3n plus one, then so P minus one equal 3n, we take expansion of, so up to 3n, we are okay, so we distribute this to, yeah, so expanding the monomials, so here this, so this expansion, contains x2n plus and, yeah, 2n, sorry, so we can put n here and n here, and still this is not contained in bracket P, so this means that, so if P is 3n plus one, then the A is F pure, so instead if P is 3n plus two, so again, we distribute, so this, yeah, we distribute x cube plus y plus z, yz, because in this case, 3n plus one, so the most, so we take at most n, so and we have still 2n plus one, and then it is very easy to, so this expansion as somewhere, either y or power z is bigger than P, okay, so this gets in the bracket P and then A is not F pure, so it is proved by electric for that light cube, so F is a homogeneous cube, yeah, isolated singularity, in this case it is proved this A is F pure for infinite many P, also not F pure for infinite many P, and this is the most important is number theory, this is related to, for this, okay, so the guy who showed, yes, so this elliptic curves is super regular for infinite many P, and no, no, yeah, ordinary, and super singular for infinite many P, for the, okay, yes, yes, so this is proved by, kiss, yeah, a kiss, sorry, yes, it was good to check, yeah, thank you, so this part is, so it, the same thing becomes in higher dimension, and so this is settled up to dimension three, so in some sense we, we have two world, world of character P, world of character zero, so in some sense we can come to here, model of P direction, so I will explain that tomorrow, and so for F pure, so usually, so if we have good things here, and this direction is, so usually okay, and so you can support that, if you support, if this is character zero, then we have something like that in character P, and in this case so, but we would assume that, that property in character zero is called love canonical, so if F pure for sufficiently, or yeah, maybe in future for single P, then we have log canonical, for single P, so I say, if we are fortunate, perhaps not yet, but if sufficiently many pay P, so it is okay, but we would like to say the converse, so something good, given something good, then we want to ask, so if this is so F pure for infinitely many P, and I think that part was okay in dimension three, but at least from dimension four it is widely open, okay, so, and sorry, so in some sense working on, so not this problem that about, so relationship between relation F pure, rational singular, rational F regular, so we just erased, but so F regular and F pure are both characterized by injective envelope, so we have this implication here, if the ring is F regular, then it is F pure, so of course if tight closure is zero, then of course it should be injective, probably it should be injective, and so in the relation between these two, in some sense I'm working since 1980, so there was a meeting in Japan in 1981, and that time, so Hockster informed me that that's criteria, that's criteria, but still we didn't know, also for greater rings, so R is a rational singularity, we have negative invariant, and F pure, non-positive invariant, so from this we would like to suppose rational singularity imply F pure, but actually that's not the case, so let me, so example, so T inverse T to the P, C and T, so this is a two dimensional greater ring, normal greater ring, and so if of course this is generated by, so it's obvious to see that this times this is X, Y, minus T to the A plus B, and so what you'll see, one more thing, sorry, some mistakes, anyway, so this is YZ minus, so you can have this by easy computation, because so if you know that X minus one X equal, yeah, one, so by this relation you can find the relation, okay, so the important thing is this R of T, R, X, Y, Y, Z, Z, Y, Z, G, X, now this is famous standard least learning, in this case delta is just three points, and so it is for any, yes, some idea generated by square free monomials, and so this is called standard least learning, so in this case, yeah, since there is no square free, it's very easy to show that injectively, yeah, so this is a direct sum out, so this means that standardized learnings are FPR, and so when this R is FPR, so actually, so I, yeah, I'm out of time, so what to, yes, so if, yeah, this is FPR, this is always F rational, always F rational, for every F rational, FPR is utmost one and F regular, so you can find plenty of ABC which does not satisfy, so there are so many rational things which are not FPR, and one more thing is that if we divide R by, so singular element T, where it is, so this is FPR, but there are many ABC, choice of ABC, which does not give FPR, so that means FPR doesn't deform, so sorry for, to be late, ah, yes, yes, yes, yes, F regular, yeah, of course, yes.