 OK, to je veliko prav, da je tukaj, in da sem tukaj oportunit, da se pravite na vrštju. Vzlušam, da se pravite na vrštju, da se pravite na vrštju, da se pravite na vrštju, da se pravite na vrštju. So, let's start with introducing some basic definitions and results. R would be a standard gradient algebra over a phi k, here. So, the Hilbert function and the Hilbert series of R are defined like this. So, the Hilbert function is a function from the natural numbers to themselves, and counts the dimension of R in degree i, and the Hilbert series encodes the Hilbert function in a formal power series. So, if D is the cruel dimension of R, the Hilbert probabilist that the Hilbert function is actually a rational function, and the denominator is one minus t minus t to the d. Why? The numerator is some polynomial, hr of t. And this polynomial is called the H polynomial of R, and we are interested in these coefficients. Here h0, of course, is one. H1 is the dimension of R1 minus d, and then the other numbers are less difficult. I mean, they have a not so clear description. So, this vector is the h vector of R. Notice that some of these hi can be negative, can be negative. While, of course, these dimensions are all natural numbers, some of these coefficients can be negative. So, this is Britten here. They were found in the Hilbert series, of course, and the coefficients are natural numbers. Instead, the hr of t may not be here, a polynomial with coefficients in n. And, of course, if the dimension of R is zero, then the h polynomial is equal to the Hilbert series. Also, this number counts dimension, count dimension of vector spaces, so they should be negative. So, also, notice that if we take a regular element of the one l, if you go module l, the h polynomial doesn't change. It's an easy computation to do this. So, if I raise one call, then all the coefficients in the h polynomial are negative. However, without the calling assumption, things change and here is an example. If you take this polynomial ring, which is in 2 times r plus 1 variables, you take this ideal here. One can track that r as dimension r plus 1 and that r and the h vector is this one. So, already the h2 is negative and such an r is even spawned. One can track that this is spawned. So, in particular, it's generalized on the punctual spectrum. So, we have, I think, enough depth is not enough to infer some non-negative that some coefficients of the h polynomial are non-negative. Of course, the sum of all these coefficients is always positive because it's the multiplicity of r. OK, now let's take also. If any is the dimension of r1 as a k vector space, r can be seen as a quotient of a polynomial ring as in n variables over a 4k, over the 4k module in ideal, homogeneous ideal i. And this ideal i does not contain linear forms because n is really the dimension of r1. So, OK, in this case, if you denote by c the height of i, one can see that really, because there are no linear forms in i, each one is exactly seen, the co-dimension, the height of i. And also, one is this, that the fact that h2 is non-negative means that the quadrics in i2 are at most the sum, and so, OK, since there are no linear forms, these quadrics are all minimal generators, so, the minimal generators, which are quadratic, are at most the sum, if h2 is non-negative. And also, if more generally, if one has that ij is 0, I mean, i is 0 in degrees less than k, in number k, then the fact that hk is bigger than or equal to 0 is equivalent to this, to have in this inequality on the polynomials of degree k in i, which are also minimal generators in this k. And, OK, this bound is, especially this, we will not study, we will not see today this, but this is the bound predicted, I mean, the bound that gives Bertini for a prime ideal when k is an algebraically closed field, here the assumption that k is algebraically closed is crucial, because if you extend the ideal, maybe that even before it was prime, it will not be prime again. OK. So the main motivation for this work was this theorem, by Moray and Terai, that says that if r is a standardizing ring, that means that i, the idealized generator is calculated by square 3 on polynomials of s, of the polynomial ring s, then if it satisfies this air condition sr, then all these hi are non-negative for any i smaller than or equal to r. And, moreover, the multiplicity of r is bigger than or equal to the sum of the h0 plus up to hr minus 1. Pardon, Moray, you can also say that if sum of the hi is 0, when i is smaller than or equal to r, then the ring r has to be quamacoli. OK. Ser condition sr is this, that means that the depth of r is bigger than or equal to r, but not only this, it also means this, that... What did I do? OK. And that... I mean, when you localize the prime ideal p, the depth of the localization is bigger than or equal to r, and if the dimension of the localization is smaller than r, localize that p should be quamacoli. This is Ser condition. OK. Now, notice that if r is generalized quamacoli, then of course, in this case, r satisfies Ser condition sr. If no lift depth of r is at least r, because... Whenever you localize p, well, p is not the maximum ideal, the ring is already quamacoli, so it's enough to look at this one, p is the maximum homogeneous idea. So in particular this says, but the previous example that sr alone is not enough to infer this, because in the previous example that we saw, if you remember, I go one moment back by one slide, even before this one. This ring as mod i was generalized quamacoli of depth r, so in particular it is it satisfies sr, but already h2 is negative. So one... The point is that mod i and per i at the assumption, the strong assumption, that r is a standard ring, and this means that somehow you need something more on the ring, and we... I mean this paper, this talk studies which conditions do you need on the ring? And... OK, let's say, let's introduce this notion. We say that the ring r satisfies this condition mtr, mt comes from moraiter i. mtr, if you have this upper bound on the regularity of the efficiency modules of r. For any i from zero to dimension of r minus one, so dimension of r is not the last... I mean the first x module which is not zero doesn't come. After this condition you can also introduce for modules, but for this talk I just speak of the ring. OK, this notion is good for several reasons. For example, mtr, first of all, does not depend on us in the sense that if you present her by another polynomial ring as maybe with more variables, the condition doesn't change because these efficiency modules are the at least dual of local commodity modules. OK, this. Then mtr one can check that it is preserved by general hyperplane section and moreover mtr is preserved by saturating. So these three especially the last two things are allowed to do some inductions. So we have the same that I mean is already shown by moraiter i we gave a different proof also in the case of modules but using essentially the induction that I is allowed by these properties that I have. So if r satisfies mtr, so if there is this upper bound or the regularity of these efficiency modules then h i should be negative for any i smaller than or equal to r and the sum of the last from r plus s hs was the last possibly non-zero is non-negative and this is equivalent of course to the fact that the multiplicity of r and also you have that if the regularity of r is smaller than r or h i zero for some i smaller than or equal to r then r should become mtr. So we have this lim that is for modules in this case you have in this case you have to assume that the module is generated in the query non-negative so the main theorem that we could show with long and length is this if r is dimension d and satisfies sr conditional sr and one of the following holds or k is characteristic zero r is no boa incodimation d minus one so this means that for example if it is the project of a smooth project scheme is ok so in this case or ok this other case we can say r to boa incodimation d minus two but this time we need that the dimension of r is exactly r plus one where so the depth of r satisfies sr conditional sr where r is one less than dimension always characteristic zero instead of characteristic p r is fewer the sense that the Frobenius morphism is pure it's a pure morphism then in this case in all these cases r should satisfy this conditional ntr so those satisfies those bounds on the regularity and of the deficiency module so in particular we have all the consequences on the h factors so there should be no negative for all r smaller than or equal to r and the multiplicity should be at least this sum and moreover you have that r is called if the regularity is smaller or sum h is zero for i smaller than or equal to r ok ok so we have this theorem and what we really prove is these bounds upper bounds on the regularity of these deficiency modules and in fact we prove something more we prove that some kodara vanishing holes in these cases and some type of kodara vanishing here ok now ok so in standard asnarins arduboa if the characteristic is zero and they are pure positive characteristic we we recover the demorator higher result but however later later demorator higher result kumini amorator kumini amorator prove that mtr this property holds true for any monomial ideal not just square free monomial ideas such that the quotient satisfies sr condition and so it makes sense to ask the following questions true that r satisfies mtr and so all the beautiful positive properties of each vector if r satisfies the sr condition as r so is enough depth and for example k is characteristic zero and s mod the radical of i isduboa in codimension R minus 1 so just look at the radical is characteristic zero and r isduboa in codimension R minus 2 so we add the codimension D minus 1 and then we add another result that says that it is true if duboa in codimension R minus 1 but the ring has dimension R plus 1 and I mean we believe to this because this and this is would be somehow sharp because if r is 2 this is true because if r is 2 r satisfies s2 so the project of r is connected in codimension 1 satisfies s2 you can go to the algebraic closed or it does not depend on the you can extend the field so you can assume k is algebraic closed satisfies s2 so connected in codimension 1 duboa in codimension 0 plus mtr 0 and you have also s2 so reduced so if connecting codimension 1 reduced by bertini you have that the true should be non-negative so it makes sense to expect this would be quite sharp in particular for example this is the first case that we don't know if r is normal in codimension 4 for example and satisfies s3 is h3 might be greater than or equal to 0 so normal r1 so in particular duboa in codimension 1 I mean this we don't know and then also if k is positive characteristic and the reduced ring is pure k is positive characteristic and r is f-injective f-injective is something which is weaker than f pure and actually f-injective is the counterpart is the characteristic key counterpart of duboa because there is a result of space that says that one thing is f-injective is f-injective for infinite prime speed then it should be duboa in the conference is conjective to be true so I mean f-injective is the right on the other hand codimension is not true in characteristic so here we assume that that is f-injective and r is really f-injective not the last proger is f-injective that would be much weaker I want to finish just discussing the bit this last point this fourth point ok, suppose that k has a positive characteristic then we give this definition we say that r is deformation equivalent to ring so we abbreviate by missing that r is depth pure depth pure if essentially r is the title so there exist domain A which by technicality we assume to be essentially finite type of k and a a algebra rA which is finite rA is an algebra and such that all fibres are targeted we want that one fibre is r and one is a pure so in particular this means that there are two prime and a, p and q such that the fibre rp is as more fit to r and the fibre rq is a pure so ok, first point notice this that if r is a pure it is for short depth pure because in this case is enough to take as a the field as r and p equal to q equal to z but there are many rings which are not a pure but r depth pure and ok as well we will see so proposition always in the same paper is that if r is depth pure and satisfies start condition as r then also in this case you have empty r and so all this bounce on the the inspector ok, recall that just last slide to see some examples of depth pure, recall that r is as mod i where as is a polynomial ring and i is a homogeneous idea ok, one example is this if the initial of i is square free for some monomial order you take some monomial order compute the initial of i and it is a square free monomedia then in this case one can check that r is a pure and we will see the process by the formation at the generic fiber is r the special fiber is r as mod initial of i and as mod initial of i is a pure because the initial is square free but in this case there are many examples which are is not the size of the fure for example if i is an idea generated by is a binomial idea r is not a fure in general but the but the initial idea is square free here is another example if you take this idea here so in trip variables then in this case r is a fure the as mod i in fact as a k t as j this idea ok this idea and r a s t mod j mod this idea j then in this case you look at the fiber over t-1 this is just r a mod mod out by t-1 this is r but if you look the fiber at t this is r a mod t which is just this which is as mod square free monomial and this is a pure so ok in particular note is that the character is spagai r is not even f-injective there are also some non-f-injective rings which may be that pure so one question that I mean I don't know I don't expect that the answer is positive but I also don't know any obstructions for the moment so one question is this that can you find an example of a f-injective standard idea which is not that pure ok finish here thank you are there any questions? again yes I mean all the I mean the ideal there in the which is ideal which is square free is that pure so there are many non-quamacoli not that pure and also not that I mean no these are all f-injective they are also f-injective and that full because they because of purity the forms to to f-injective and that but in this example here this deformation is not a grobner deformation for this kind of deformation is different you had proved some results on the on coder vanishing for symbolic powers if you are sufficiently smooth ok this gives some two results here on the age coefficient for powers or not I mean for symbolic powers but some results I mean if you assume that you have vanishing yes I mean it solves for symbolic powers but you have this really this upper bonds on the regularity to start with i-manusar I mean this sharp I mean you had a question where the regularity of the deficiency module is smaller than or equal to i-manusar and so you are asking if I mean you think that there is a way to to have this upper bound for symbolic powers for smooth things for instance if you have good singularity ok I don't know how do you think we can talk later maybe because of another talk but ok maybe we have a break of about 10 minutes we will start 5 minutes later I think it's ok 45 please