 Vzostajte vse in na način, da se je nekaj droži,ideri, Kreg, Haere in Mel. Zame se, da je ovo boj, v vse, da je Alessandro, Steve. So, ... ... vzostajte o, kako se je tega izgleda za vse. Vama človek, zato srečnem vse, izno tem, ki smo izgledati površenje na uniformne obladi. Aspečno, ki se je tudi na uniformne vršenje, je priber, da je v teženje vsega vsega, u vsega vsega vsega, na Kreg. Zato vsega izgleda, da vsega je vsega z vsega komputacijna, zato nekaj vsega. ... is going to be a polynomere ring over a field. K, so k field. I, for all the talk, is going to be a great homogeneous ideal. And what I want to do, I want to fix some data. And once I fix the data, what I would like to do is, well, let's say, I want to... So I... Skakov iznovno. Widnem skasviti dokranje ogdang nimi preveni s pajamaser, dar poš tabs evenly. zagvaž presidentseli ngrazar in premiesti, bahe oven vine kaj je izgleda ideja, ki je sodavljena, sodavljena tezunja, ki je sodavljena, je spremljena in ideja, je jezna za njegovatve obziv. In je to še poslednje list. Ne, časno list. Zato, včest so včetno bom zvoniti halje na otvorili, da je z njo, da je znašljena, da je z vse zelo, da je zelo na vse. Selo, mi je vse na vse. to bez stvari. As in вот tukaj prejda. Kaj da jim je nič, ko držičen z F1, F r, da je v degrai delje, V 1 in so on. Pči drža in s deljev s deljev, da je zarazla v vkorej vsebran, v d. Tako zelo, da je data, je zelo druga variabila delenja. Tako zelo, da je data, zelo, da je T-R-M. Zelo, nekaj versišlje, in v Bajers tesis, v 1982, in in izvah, da je to vseh. Molo, molo mora. Du be. Menej vseh počuče, da je to taj počutek. Mamo vseh Marka, Masha Den, Hoa, menej vseh, tudi na mjim vseh, na mjim vseh vseh, Leanga. Vseh je to, na vseh, termorder. The site of a Greenberg basis is under control. So basically the largest degree of a minimal generator. In the initial idea with respect to this monomial order, termorder of I, is bounded above by a basic double exponential function of D and N. Let's see, I have to write it two times. I am going to write to the Dubet version, this square plus over two, plus D to the two, to the N minus one. So it's a double exponential on D and N. So that's the first result of this finish of fixed data and invariant is sites of the Greenberg basis and here is the boundary. Example number two. Kastronom for regularity. Tami? The what? Ah, capital D, you're right. So little D, let's say smaller than D, and I have to make my life easier. So it's the largest literally, perfect. Thank you. So just take a little by a month further. Several different versions of similar bounds. Improvement on the type of bounds. Myself and Enrico, Bara, Machsadena, Ova, many other people perhaps even more recently. Myself and Stefan in a different collaboration. One get the regularity of the idea for the fixed data is bounded above by D to the two, to the N minus two. I mean, to get rid of some trivial case, let's say that N is at least three. If you're just into variable, everything is trivial. Another type of this flavor perhaps, actually I'm going to call this a corollary. A corollary of this is basically a bound on all the, but in number. I and J, the I and J, but in number of I, for instance, can be bounded above by, I'm going to take this number, D to the two, to the N minus two, choose I, for instance. Actually, let me write this slightly differently. I'm going to write it as D to the two, to the N minus one. And the reason of this is basically because you have a, but in number, A, J, which is bounded above by total betty number of I. Total betty number of I. Once you know the sites of the group net basis, by a person may continue it, you can bound this with the betty numbers of the initial. And the initial, the resolution is easy from, I mean, some control, the resolution is easy from Taylor. So basically this, thanks to the Taylor resolution, plus perhaps the proof of two, something like this. So you get D to the two, to the N minus two, choose I, the choose I comes from Taylor, which is smaller easily than D to the two, to the N minus one. Anyway, so three examples, once again, sites of the group net base, irregularity and betty numbers. Now, the problem becomes way much harder if the data, instead of fixing the, if instead of fixing the number of variables, we work in an unknown number of variables. And that we go into still bound conjecture and conjecture type, type of problem. Now the theorem, due to mal and theorem. So, what's the setup here? Still I, in state polynomial ring. Now I have x1, xb again. Let's say number of variable begin unknown. So the data here is just the degree D. So with this type of data, the solution of still bound conjecture is basically theorem by an onion. This says that if you look at, if you care for the projected dimension, among other invariant, they didn't place the projected dimension first because I had N as my number of variables, so that N would be an easy bound for the projected dimension. But if you look at every possible, every possible ideal, so let's say for every I and for every N, and let's say we're looking at the projected dimension of I, this number is actually finite. Not clear what the number actually of value is, but the point is that there exists a bound that only depends on the data, the degree of the generator. It's independent of the number of variable. Similarly, by product of the proof also from the paper, one get bounds in basically of the data that you can get from the resolution. So for instance, regularity of I is finite. And the supremum, let's say of every, for every I, for every N, the I and J of the value of the betting number B, I, J of I is also finite. Again, not clear what the value, what the bound can be, but it's, I mean, there's a horrible, horrible recourse from this proof and it's unthinkable what the value actually are. But if we care for betting numbers, let's say I'm gonna change sides. If you care for a betting number, especially for a single betting number, let's say not the class of all the possible betting number in the resolution, then the task become easier. Let's say two bound is single B, I, J of I. In this case, let's say, fixed the pair I, I and J. It's easier. For instance, let's say a start with an idea generated by three quintics, F1, F2, F3, whole degree 5, so 5, 5, and 5, that's my degree sequence. Then I can ask, for instance, what's a good bound for the betting number in normal logical position, 6, shift, 12, of I. So, can we bond that? Again, no variables. I mean, still malend in some sense, but I fixed a single betting number, so task is easier, as I was trying to say, but if you just follow the bet, if you try to control stuff with the proof, it will be absolutely crazy. But yeah, the things are not too bad and what you can do, you can compute an initial of I, and you start with three quintics, so the initial is going to start with three quintic monomials, and those are, let's say, let's give a name, or Q1, quintic, Q2, Q3, so these have degree 5, and I'm going to compute this using, basically, I'm going to compute a Grobner basis. If you compute a Grobner basis, you have initials, and you have to do S-pairs, leading, leading, S-pairs, you're going to look for a list of multiple, maybe something's going to reduce to zero, but between them there are only three S-pairs that you can consider. At most S-pairs correspond to at most S1, S2, S3, perhaps they're zero, but these are at most the number of things of degree 6 that they can expect in this in, plus larger degrees. So if the in contains those monomials, three quintics, and three things of degree 6, we can estimate the Betty numbers, because basically we have a, you know, Taylor resolution, a hypothetical Taylor resolution using this data. So the Betty number Bijj of I is bounded above by the Betty number Bijj of the in, which is bounded above by essentially the table that you expect from Taylor. Zero, one, two, three, four, five, and six. I'm going to stop at six. There's nothing here, I start in degree five, then there is degree six. Degree five, three, in degree six I have three generators, and from the, let's say, Taylor, or if you want the LCM lattice, you can get those numbers under control. So let's say three, one, after that zero, zero, zero, zero, in this trend, three, 15, 20, 15, six, one, and finally I get a zero here, and that's the position that I care for. So the boundary was zero. Mistake. So what's next? So I want to change the data a little bit, and I want to focus on primes, prime ideals. So focusing on prime, my data is going to be the fact that the idea i is actually a prime, and another part of the data that I want to fix is that the multiplicity of S mod i of the prime is fixed, it's fixed, and let's say some number here. Fix multiplicity, and it simply fixes the fact that we're working with prime. So here there's a Heisenberg-Gotho, it was Heisenberg-Gotho conjecture, now false, because of the counter example of McCulloch-Jason, which basically says if I look at the supreme of all the possible p satisfying my data and I care for the regularity of p, this number cannot be bound, cannot be bounded by a polynomial function of E. By the way, I wrote Heisenberg-Gotho conjecture, so the prediction of the conjecture was that the regularity of prime was bounded by multiplicity minus height of the prime plus one, but in particular this is less than the multiplicity, so the Heisenberg-Gotho predict multiplicity as a bound for this type of things. So they prove, they have a counter example, great, but after, there was a counter example, you start realizing that the focus for many, many years of people working on this conjecture was actually to prove the conjecture, which is perfectly fine, was the optimal statement, but once you realize that, the next question is, okay, well fine, let's settle for something else. What's a good bound? Anything, I'll take anything. And the point is that in their paper, Irina in Heisen, ask, well, is there a bound at all? Because, so, can the regularity be bounded, not even like that, but by anything as a function of the multiplicity? Is there a bound at all? So, this bound cannot be bounded, so question, is it at least finite? Is there a bound? Or perhaps, I can look at generator. E is the supremum over the prime of the number of generator, b0 of p. Yes, so, to both question is basically positive. So, as a consequence of the, of the theorem of Ananian and Mehl, so of, as a consequence of the solution of the statement conjecture, one get, is a result, we have it in a joint paper, myself, Mark, Michel Den, Jason and Matteo, yes, it can be bounded, the bound depends on the solution of the statement and is of the same magnitude. So, basically, this problem and the solutions to the statement are, they are connected to each other, a good bound for one, a reasonable, anything, for one corresponding to a good bound for another one. If I just follow this proof of the, yes, through the proof of stillman conjecture by Mehl and you start with something ridiculous, like five. So, I have multiplicity of five. Which bound do I get? Well, think about taking five, five to the five to the five to the five, one, one exponent in top of the other and you go up to hundred exponent. And you say, well, no, is it good? No, shrink. So, they take that, that number and do on top of each other is that our exponent and you do it again hundred times. You do it again itself hundred times and so on. How many times should I repeat this? Well, as many times is the very first exponent. OK, so it's unthinkable. So, it's absolutely unthinkable. So, but, well, we would expect something reasonable, like a double exponential, perhaps. So, in those type of totes, double exponential bound are not bad. So, that's the point I want to, I want to make. So, sometimes I treasure the fact that there are bounds sometimes for some invariants that are independent of n. And so, anytime I see one, it's a little gem. It's a very rare. So, let's say results with good, but I didn't even care about good any bounds for something independent of n, of the knowledge of n are very rare. One of those is this very old theorem attributed to Castenuovo. 1889, which says the following. So, if you start with a prime, so same data as before. So, p prime, I'm going to fix the height of the prime, let's say, let's call it h. And then I want the fact that the field is algebraically close. Then what happened? Then the number of minimal generators that are quadrics in your prime ideal. So, p02 of the prime. So, how many quadrics you have among the generators is bounded above by a nice thing. h plus 1 choose 2. No mention of number of variable whatsoever. And that's remarkable. I mean, it's nice. But, so, once you see that, say, well, what's next? Not so much. So, can give a couple of proofs. So, first proof, using Bertini. Using Bertini, what do you do? You go modular linear general, sorry, general linear forms. Plus, 1 has to saturator. And you bring the prime down, right? Cut, saturate. Cut, saturate. Go down. If the dimension is greater than 2, you go down. You still get primes. Between dimension 2 to dimension 1, you go down to radical. And from dimension 1 to dimension 0, you get a no zero divisor. So, basically, without problem, without loss of generality, you reduce to some ideal. So, you lost prime, you lost everything. But leaves inside k of x1, xh. And so, the second betting number of your prime are bounded above by the second betting number of i, which is bounded above by h plus 1 over 2, how many quadrics you have in that ring. The key point that in this process, the degree one part of i, so, let's say, without loss of generality, prime in degree one was zero, so was non degenerate. And you don't gain any linear form in this process. So, this is thanks to the fact that the field was algebraically close. There are counter-examples otherwise. And, basically, without this, there are counter-examples. Counter-examples without this assumption in the field are very close. So, that's the part where you need it. And because as you saturate, the ideal gets larger. So, as you mod out the general linear form, saturate because of the Bertini, the saturation may add things. If you have quadrics, no big deal. You bound something, it's even better. I mean, you bound something that will be larger than what you care. But if you have the linear form, the linear form can kill anything. Can kill quadrics. How many quadrics can kill? Quite a lot. I mean, a single linear form can kill as many quadrics as the number of variables. So, can kill and not under control data. So, that's crucial. What's surprising is, basically, so, that's a proof one. And since the conference in about, let's say, second proof is what characteristic P method. This is due to Craig. Basically, what you do, you reduce to characteristic positive. Then you take your S mod P. This is inside the T, this is a graded plus closure. And you take a system of parameter here, lift to a regular sequence. And when you mod out to the system of parameter, if in this process you lost some quadrics, then the system of parameter in the plus closure will be more sizzzy than what you would have expected. So, the quadrate that you kill here, if something goes wrong in your estimate, correspond to more sizzzy. So, that's a sketch of this part. So, what do we need? And again, as far as I know, so, from that, I mean, I expected it to be some attempt to extend this thing, but the methods, the reduction of characteristic P of Craig and all the other standard methods that rely on cutting down, they don't work for higher degrees. So, I wonder, can they bound the cubics? Can they bound quartics? But, see, sure, if you are there, so the point is this condition, not there be an extra generator, but if you get even a quadric as you saturate, a quadric can kill cubics. If the saturation give you a cubic, a cubic can kill quartics, a quintic and so on. So, if the saturation add extra stuff, those things kill a unknown number of elements, and you lose total control of that. So, what we did with Alessandro is basically, we found a different route. So, we found a different method to prove this result of this type, and our sound is the following. So, you start with prime, let's say in S doesn't matter. So, prime P of height of pH, and that's our data, then for every J, the number generator in degree J of this prime are bounded above by a double exponential. So, H to the 2, to the J plus 1, minus 3. What's the proof like? I can just spend the last minute discussing the proof. So, basically, what you want to do, you care for, actually maybe, let's just do the first step. So, what you do, you write prime, your prime has a bunch of form, as many as the height, that is your regular sequence in the prime, colon sum G. You assume the degree of G is very high, and so, instead of studying the generator of prime, you basically end up studying, so the B0J of P correspond to studying the B1J plus D of effigy of this almost complete intersection. How do you study the first schizogy? Grobner basis? Try your theorem. So, the first schizogy is controlled by a site of a Grobner basis. Which Grobner basis? You cannot use term order. Term order use the last variables and add the next noise completely out of control. So, instead, we use generic coordinate, we use weights, and somehow we reduce the amount of noise. So, the proof relies on the calculation of a Grobner basis with respect to weight, is that the weight don't get refined to monomial order. If you use McColley or anything, you care for an in respect to weight, that weight get refined to term order and then the calculation is run. We kept the calculation at the weight level instead of a sphere with its schizoges, and basically, you compute this hypothetical Grobner basis and you can check that's under control. Anyway, so my time is up and I think I'll stop here. Thank you. So, any questions? Or comments? But see, the spirit of this is exactly the proof I give you where I bound the beta 612. So, as the algorithm of Grobner goes, you bound what's going on in each step, but it might go forever. And yet the trouble is that this degree are not known. So, you run a Grobner basic calculation here that could start in ridiculously high degree. But it's like a Grobner basis of a principal ideal. Once you get rid of the noise. So, it's like computing a Grobner basis generated by a single element. There's interaction between those two, but it can go forever. Actually, immediately also this historical result of debate, more by, are surprising because the way a Grobner basis terminate, Bookberger, is because of netarianity. Even that thing is not clear. The proof typically rely on the knowledge of number variable and the fact that worst case scenario is basic. Regular sequence and working relax. And you deal with the standard decomposition. But here things cannot terminate. Yes, Serenem. Just a moment. Please use the microphone for the best recording. Sorry, I just missed it probably. Is this bound sharp, you say? Well, so this is sharp, clearly. Same example. The bound is not as you change without this assumption there's no bound. So, what I'm going to say, without the field being closed, there's no bound whatsoever. And actually without this stuff, the bound that we get here is beta 0,2 of p bounded by 2h squared plus h. So, we actually do get a bound without an assumption on the field which absolutely unduable with the standard method. But they are not sharp. But the point is that the spirit of those bounds, they are double exponential, which kind of fit perfectly the technique. Those technique, I mean, again. When you go without knowing the number of variable you end up in this tower, soft tower, soft tower. So, this is a, you know, kindergarten bound compared to the towers. The CG, thank you. So, the CG are completely under control as well. I might not have the specific number written. But, yeah, for every fix i in j, beta ij of p is bound in also by a double exponential. OK. So, this can go because obviously we... Well, it's not obvious because of the proof. Even later it used to a NIN, but it's not monomia. But somehow what happened and to get the CG, well, we do something. It's not obvious right away, but you get a similar bound. So, any CG whatsoever or any premadeal in a single spot is a double exponential of your data. Thank you for the question. Yeah. I forgot to mention it. But here, the important point is to stay connected over K bar. I mean, it's not really to be prime. Not really? To be prime. Yeah, yeah, of course. Yeah. Same here, actually. So, basically the proof that we do even our bound works for radical, a mixed radical, for instance, same bound. And if you don't even care for a mixed, you know, any radical and you get the bound instead of the height so any radical idea, if you know the big height, any spot on any betting numbers is a double exponential of the big height. No matter with the variables. It's quite a step. Some other questions? If not, so let's thank the speaker again.