 And welcome to the last but definitely not least session of the conference. There are a lot of exciting mathematics to come. And I'm very honored to introduce the first speaker of this session. Together with Professor Chung and Professor who's a key advisor. This speaker was part of the first generation of Vietnamese community of Angiobrists. They did their degree mostly during the war. And you know, during the war and after that Vietnam suffered a lot. There were not just a war but the crippling embargo and isolation. Up until the early 90s, there were no meat electricity, very little electricity. And when you talk to the people in the older generation, you will learn that they were not even white paper to write math on or to tie papers on. So but they somehow overcome all of that and produce mathematics and many students and build a very strong group there. I was not, I didn't have much chance to interact with them but having them, the presence in the field make a big difference. Just to know that some Vietnamese name I mentioned in the papers and books about the result really were very inspirational for me. So without further ado, Professor Le Tuan Hoa from the Institute of Mathematics in Hanoi will tell us about Castel Nú Vô Mumford's regularity of powers of ideals. So thank you very much for the introduction. So internationally he's gone down but in Vietnam everybody called him by long. He goes down. Many people have shared the same family name. So I would like to thank your organizer for inviting me here to give a talk and also to ICTP for generous support. So this is my fourth participation in this series of five workshop. And now I talk about the Castel Nú Vô Mumford regularity. And this is a joint work with Nguyen Dang Hoa so in public he uses the family name Nguyen and Ngu Vy Trung. But in Vietnam now we have at least four competitive entrepreneurs, it's named four. That's why we insist to talk on N Vy Trung. Otherwise people can miss you. So I talk, this is the title of my talk and so the content of my talk is the following. First I give a motivation and then I consider the key of the equi-generated ideal in a gridded ring. Here we consider the four functions related to the Castel Nú Vô Mumford regularity. And the third part I talk about the polynomial ideal. In this case, I'm not sure the ideal is equi-generated so it can be related in different degree. So let's start with this one. The setting is a standard gridded algebra of a few. That means I is the direction of the IN so that the zero component is a few. And then the ring is related by the element of the degree one. And M is the maximum gridded ring. M is the finitely generated arm of the one. So in that case, if E is the Arthendian gridded ring, which we know by E is the last component where it is not zero, then the Castel Nú Vô Mumford regularity can be defined as a maximum of G on this A invariant of the local co-homology plus I. And equivalently, we can write down just... Equivalently, we can write down just Castel Nú Vô Mumford regularity. So Castel Nú Vô Mumford regularity of E equivalently is a maximum of T. So Z is I, Z, A, I, R plus E, T minus I is not zero. And if I is equal to the polynomial ring in this case, then we can define the Castel Nú Vô Mumford regularity via minimum free resolution. We have E here. When we have a copy, say B, B0, I, and so on. Then the Castel Nú Vô Mumford regularity is equal to the maximum of BI, BJI minus J, this one for all J from zero to Q. So over the polynomial ring, we have a finite minimum free resolution. There is another way to define the Castel Nú Vô Mumford regularity. So it is why it is important because Castel Nú Vô Mumford regularity controls the complexity of the grid structure of M. And it is this, this one is written in the paper by Bayer and Mumford. So now let I grid it, I did. We suppose that it is not important. And I always assume that I is generated by degree D1, degree D1, D2, etc, Dp. And Dp is the maximum degree. We denote by DM, the max of I. Okay. With this one, we know that usually the Castel Nú Vô Mumford regularity is bound, up above for the maximum degree, but usually it is a huge one. In one talk earlier in this conference, it only mentioned by talk by Kavik Lea, only mentioned that in the worst case, the ball is a double exponentian to compare the number of the variable. But when we take a large ball, we take a sufficiently large ball, it is a very nice work by Kacovsky and Sok Chung at the same time also by Kodala. They say that when N is big enough, the Castel Nú Vô Mumford regularity is just a linear function, not only bound by a linear function, but it is really a linear function for N big enough. I should say, and then this reason extended to any module, and then the co-home logic, the Castel Nú Vô Mumford regularity should define via co-home logic, because in this case we don't have a minimum free resolution. Okay, but the work, the raisins can hold. So I write here, this raisins so that we can N big enough, and then we call the asymptotic. And what is interesting is that D is one of this, one of this degree. And of course, only at most marks. And when I is related, is equi-generated, of course because this, we see that D is exactly delta. So this, the D in some chain is easy to compute. So there is a nice story about this raisin initial, but I have no time to record this story here, okay? So as I said, D is called the asymptotic degree, and that is one can define in some sense explicitly. That is a kind of degree of the reduction number, reduction idea, okay? So, and that D is one of the, but the free coefficient, the intercept E remains mysterious. And there is a problem with, when does this regularity become a linear function or equivalently give an upper bound of the index of stability, we know by a red step. This is the minimum place from which it becomes, a linear function. So we call it the index of stability of the constant of a morphology. And the problem is when does it become a linear function? So this is a very hard problem. And until now, very few raisins on this one, even in the case when I, I am a primary idea. So to my knowledge, we have a three work on this case by Billicam, by David Eisenberg and Ulrich, and by Mark Shadang. And in this case, even they cannot give, they give some bound for the red step, but they cannot, there is no explicit bound, except this case, except the case that I, monomial idea, I'm a primary monomial idea, and asymptotic degree D, then the number, and we know by the number of the generator of this type, so it's easier, Mark, S is less than or equal to R. And then we have this small bound, quite a good bound, okay? And at first I thought that, at first I thought that for the monomial case, we should have a linear bound or a polynomial bound. But in fact not, the case. That is quite unsurprising for me. Namely, in the higher dimensional case, if we consider the integral closed, we also have a similar from the paper by Kotkovsky, Herzog-Krum, and by Kodailam, because of course we assume that the fin-tration, the corresponding fin-tration is a nutrient. For example, we consider the case of the integral closed or in the eye is a polynomial idea, okay? When eye is a polynomial ring, then we have a similar one. And in this case, very recently, I could give a upper bound, which is a huge bound, but still, it's a big bound, okay? But in view of the gravity, so it is still acceptable bound, okay? So in this case, we can give for the integral closed, we can give it a bound, but for the usual bound, still no bound. I try hard, but still no bound. But so, just still no bound. For the integral closed, I'm not sure whether it's about the ultimate, but I think it's not ultimate at all, very far from the ultimate. But for the monomial case, I can give a class, a example of monomial case, where stability at least is, so it must be exponential function. What's the big one, okay? And this is a consequent of my study on the maximum degree of power of polynomial idea. So, the question is, why it is so difficult to study optimal E and the index of stability? And a way to answer this question, in our opinion, that we should study this problem, study the behavior, initial behavior of the whole function. If these functions are quite, behave quite freely, then the bound must be very big one, because it cannot control. That is one reason. And because we have this function, the first part E n, so it is, this problem is equivalent to study the behavior of the difference. Because the first, this is in some sense, when now. So we should study E n, and we call that the defect sequence of the function. And we think that the name defect sequence was first even introduced in paper by Billy John. Not yet, David, sorry. The paper by David, Eisenberg, and Harris, there are some recent passion on this case, so assume that M is a relative degree zero. And when the dimension of M is big, but the ideal is M primary. And assume that I equate it, then this sequence is a weakly decreasing of non-negative integer. So I write here to remember. So, the reality of I n M equal to dn plus E n. So that is the sequence. And in the paper by David, by Eisenberg and Unrich, under this assumption, and assume that the zero local community of M is zero. Then we have a further constraint. So E n minus E n minus one cannot be bigger than d, d is the degree of the generating degree. Because in this case, we consider the equationality. If the dimension is bigger than zero and I equate it, then this sequence need not to be weakly decreasing. And in fact, Stumpfland gave the first count example. And in this, he also, the count is also raised from the other author, but resumed in this paper. And then Konka gave also very interesting and now until now, in my opinion, is a unique example when this sequence can be zero very long. And then as a player and certainly at the place E n plus one can be bigger than zero. So it's quite interesting. And you will see later that this is really very, very, very, very, we need to have a higher dimension in this case. So if not, I is not equating it. And even the model is a polynomial ring. Welcome, soldier. The sequence can be initially increasing and later degree soldier. This sequence is neither is never monotone. It's not equating it. Even when the case, the idea is M primary, okay? So the, about the partial reason, such as the sequence E and could be arbitrary. And our main reason in this talk confirm this yet. We say that it is the sequence could be arbitrary, but we can prove that for the K of the ring, not for the any polynomial ring. And you will see that, okay? For the simplicity, we already consider the K of ring, the model M of the ring. And the algorithm idea, which is not important. And we will study three function, the function of the quotient, the function of the model and this function. Why we study this one? Because already in the paper by Eisenberg-Harris and Eisenberg-Hundrich, they study this function via, via this, this function is studied via this function. And we follow this way, and we will see that some reason we got is follow from this, the study of the regularity of i and minus one over modulo i. So now, I recall, we consider the case of equi-generated identity in the greatest ring. And in this case, i is a standard greater ring, okay? And from this short exact sequence, this short exact sequence, one can easily show, very quickly show yet. This regularity equal to this regularity and equal to dn plus e minus one for n big enough. That means they share the same, this one. They share, so they all modify by i n, by the regularity i n. So I write here, so regularity of i n minus one, i n minus one i n equal to dn plus i n minus one. And regularity of r, r, r, r, r over i n equal to dn plus bn cn minus one. And this proposition says yet, e n equal to a n equal to cn equal to e n for n big enough. I said cn, this one. This definition is equivalent where I wrote here. And as I said yet, also this is equal to e for n large enough, but they are different for small n. At the beginning, they can be very different. And in the paper by Kotkovsky-Hershov-Chung and Kodyram, we can see that e n is non-negative. But it is not clear which a n and cn. We can prove yet under the assumption i is generated by the same degree, then a n always is non-negative. And cn is also negative, non-negative, if the height is positive. We think yet it is always the same, but always k is yet cn is non-negative. And it is under this assumption. But until now, we could not prove this one, okay? But that is not so very important in the study of the behavior, initial behavior. So now we start with the case, we will function this function. And the k are the dimension zero, that means i is m primary iodine. Then there is a constraint on this case. We only know that by previous reason by Eisenberg-Undrich and Eisenberg-Huerke that it is not, this is a, follow this one, we can also adapt this one, we can also prove that this from the defect sequence of this function also quickly decreasing, okay? And it turns out that this is a, this is necessary condition, but it is also the sufficient condition. So we can prove that a sequence of non-negative integer is a defect sequence of this function for an equi-generated iodine i in a standard degree algebra, which is this one, if and only if it is a quickly decreasing sequence. And in order to prove this theorem, we give a explicit constructor, we give a explicit, we give even a class of monomial iodine which satisfies this. So that is the construction. You see here the ring, S, the ring R is just a quotient more ring, model on R monomial iodine. And the iodine is very easy, this is one. Okay, it's an all monomial iodine, in time just because they are monomial iodine, we can compute, I don't think it's not so easy. I don't think it's not so easy. Oh, sorry. For the case dimension is positive, we can prove that similar one, but that here is only the sufficient condition. So the defect sequence of this function can be any conversion sequence. So there is no weekly, nothing. Only conversion sequence of non-negative integer with a property. But this property is not necessary. I can formulate this in the equivalent form. So that means that this function is any numerical asymptotic linear function of slope D, so that FD is bigger or equal to N minus one. And as I said, it is only a sufficient condition because the condition Z, this is, the condition Z is, this is not, so we can have this example, interesting example, given by Nguyen, Danghub, and Bukong Thang. They give an example that rate E is M plus three, but rate E to N is a sixth N, and that is a very big one. Now we consider this function, the regularity of R modulo I to N. And again, in the case as the dimension is zero, that means the M primary I to N, I z and B, and here it shows that this defect sequence is a weekly decreasing. And in this case, even to the under the sum set, that it is related by the same degree, we have further constraint. Namely, this defect sequence, not only weekly decreasing, but not too much. So I and minus I and plus one must be bounded by the degree D, is a really big. And quite interesting yet, we can show that this together with the weekly decreasing is sufficient, this is necessary, but it is also sufficient consequence. Namely, the sequence of non-negative integer is a defect sequence of the idea related by form of degree D in a standard gradient algebra, which dimension zero, if and we leave this sequence is a weekly decreasing and the difference is at most D. So for the case of the dimension, positive dimension, we have a sufficient condition, sufficient one. So that if the sequence is a non-negative, satisfy this property, then the defect sequence, the defect here, it can be realized as a defect sequence. Okay, and we can reformulate this in the form of the function of Casano-Fermont-Pott regulates. So it is, this function could be any American and asymptotically in the function of D, so just this, so this far, and okay. So I mentioned earlier that thing and how they can give, so that means the condition before is not necessary. And again, the proof of this is a explicit construction, which is similar to previous one, but of course it's a modification. Namely, we have this a little bit more complicated. More complicated, and then the formula also more complicated. And as a passion case, then we can get, so all are monomial, I.D. So just we can compute the computation of the locomotive, reduce to the computation of the quotient of certain monomial I.D. So now we consider the third one, the main function, the ability of the power of I.D. And again, in the case of I mod I, so the zero dimension, so the defect sequence is only proof that it is weakly decreasing. And we can use the previous I.D. to compute the Casano-Fermont-Pott regulates for this. I, as I said, yet in each case, we should compute separately. We cannot reduce it from the corresponding region of the quotient ring or I-1 of modulo I.D. So that is the main obstacle in the, why we should study the three functions separately and compute separately. And then we can prove this, the defect sequence of this function can be any weakly decreasing sequence of non-negative integer with this property, where M is a least integer, so that is. And again, like in the case of the quotient ring, this condition is not necessary. It is not necessary. And this condition is opposite to the property in the previous result. So, and if this local community is zero, then this EN has exactly the opposite property. But in our construction, we have this is non-zero, of course. Otherwise, it could be true, could not be true. So, there is a question in the paper by, oh, sorry, by David Eisenberg and Undrich. They asked whether this sequence always weakly decreasing and using our, using our region here, this here, this proposition, and also this theorem. We can give a explicit, oh, sorry. We can give a explicit sequence of EN so that this is D plus N, so it's much bigger than, it's never weakly decreasing. So, they give a counter-example to the question by Eisenberg and Undrich. So, that is about the Casino-Vombo-Forty-Reverty. And then, which, there is a recent paper, recent study on the saturation degree. So, we call that the saturation is this definition, defined by this one, okay? Saturation of iron is exactly. And then, the degree, the saturation degree is defined by the last place where it is not zero. And it is the last component of the zero-global community of this kind. And recently, very recently, I, Tai Ha, a larger fan, they prove that if C, R is a polynomial ring over the complex field, field, field, complex number, and I, the I is generated by this, by form, which here is a reverse one, order. So, that's the procedure scheme cut out by this, is a non-regular. Then, the saturation degree is bounded by a linear function. Extending the method in the paper by Chung and Wang about Casino-Vombo-Forty-Reverty, we can prove that this saturation degree is in fact, you see, in this case, it is a constant. And if this is not zero, it is asymptotically a linear function with a slope, then at most, the asymptotic degree. And it is really the asymptotically degree. Yep, this I did, generated by form of degree D. That means there is no lower genetic monomial, polynomial of the smaller degree. So, okay, they, in this paper, they asked whether the ball is the optimum one, and use our result, we can show that, we can show that, there is an example that the ball for N is a big enough, the ball is far from optimum. And then, we can prove that this is really, as I said, when this is bigger than zero, not zero, then this is asymptotic, is a linear function for N, big enough. And we want to study the initial behavior of this one. Okay? So, similarly, we can write out that saturation degree, iN is equal to delta N plus BN. And in this case, when i is related by the same form, the form degree, it is equal to the D, when i is equally generated by lm of degree D. And we also want to study the initial behavior of this sequence. Then, we can prove that the defects, we can show that the defect sequence of this function can be any conversion sequence of non-negative integer with this property. And again, we use the construction in your previous proposition. So, note that, note that if, so this is a sufficient condition, and we can give an example that this integer could be negative. So, it's really only a sufficient condition. Okay? Now, we go to the third part, the polynomial, iN. So, in the second part, we can get quite a, in some sense, a final reason about the defect sequence. Because all the difficult complexity is included in the ring. In the ring, i mod, i mod, i. So, the ring i is not polynomial ring, and we take a suitable question ring of the polynomial ring, and then we can get, we can construct short kind of monomerizing. But when i is equal to h, the polynomial ring, we don't have this freedom. And the problem become more difficult. Okay? So, now, the third thing is this one. So, the ring is a polynomial ring, and i, all motion iN, we can be created in different degree. So, here, we should go away from the assumption of accuracy. If accuracy is too much constraint, and then we cannot control, we cannot. Okay? In this case, in this case, the regulative i to n is equal to this one. So, we need only to study the function, customary motionality of the power. In small dimension, even generated by different degree, we still have some restriction. So, cannot be free in this case. Namely, when the dimension is most one, then we have this property, iN plus M, bounded by the sum of the previous two. In particular, if this is zero for some n, then this should be zero for all n, for all n. So, it is very clear that it cannot be arbitrary. In the case of the dimension is zero, so M primary, we have even a more strict, more constraint, we have this one. So, if this is zero for some one, or for some n zero, then from that, all must be a zero. In this case, for n a big enough, but in this case, we only know that from, start from there, all should be zero. So, it's more constraint. So, that is not clear yet, not any bounded, non-increasing function can be realized, or the effect sequence. However, we can prove that, any non-increasing function, could be realized as the effect sequence of M primary monomial iN. And the construction in this case, are much more complicated than in the previous part. Namely, the first we construct a monomial iN, that is equal to dN, that means the effect sequence, the effect is zero for all n big enough, but at the beginning, the effect could be quite big. Could be quite big. So, the second step, we truncate this iN by power of M, and there is a reason in the paper by Eisenberg and Ulrich, say Z. So, we know that when n big enough, it is zero. But the first one is a big one, but now it truncated by this one, then it becomes constant, statement. So, that means, the effect sequence has an elementary type, we call it the type two. So, here is the type two. So, the effect sequence is Z at the beginning, statement, and then becomes zero. After step two, we have this one. The effect sequence is constant and then becomes zero. And the third step, we consider the second fiber product of iN. So, the fiber product of iN is in the set of the two ring of the disjoint set of variable. And this one will take all product of the two variable into setting, and for this one, we can prove a technical layman. We can show that this effect sequence, if we consider this fiber product, then the effect sequence is the maximum of the two. The maximum of two. And thanks to this one, we can show that the effect sequence is the maximum of two. And thanks to this one, we can prove the theorem by induction. So, when the number of the value is only one, it's only elementary. And then, for the bigger one, then we extend this z, so from z to z, we only have a by induction. And now we consider this iN with this one. Then use z layman before, then in this part, it should be this. And then become zero, then the maximum. So, we can prove the theorem by combining this iN of elementary type. How about the higher dimension? For the higher dimension, it is still not clear, but we can show that for any sequence of positive number, then there is a monomial gradient show that the effect sequence of the customer moment for quality, we consider as a numerical function, get local maxima exactly as this point. So, any, we cannot prove that the effect sequence is arbitrary, but we can show that it is, it's a behavior quite counter control, can get maximum at any point where pre-described point. And the idea of the proof is, we use, we first construct, we use the construction in a bellicombe where we have the effect sequence is increasing and we call this is a type one. So, this is a type one, oh, sorry. Type is decreasing and then become stable, called this type one. And that it can follow from, can be given in the paper by Bellicombe. And then combining each, which are technically lemma before, that we consider the product of the two ideas in different ring, then the effect sequence, the effect is a sum, not the maximum, but the sum. It was given some year ago in my joint paper with thumb, okay? And using this, we can prove the theorem. So, you see, by the fourth theorem we have any decreasing function. We have, we can realize the function. And then we have this one. And now we use the lemma, we can sum, we can have a sum of these two function. We do this, this, this. So, from this, we go to, we have, go to this construction. Any convergent sequence of non-negative integer can be realized as the effect sequence of the function of Casanova morphology in the polynomial case. And we are working on this connection right now. And we hope that we can solve this very soon. So, that is the end of my talk. Thank you very much for your attention. Are there any questions for Hua? Anything you can say about the SOC or degree? No, we don't, we didn't think about that, but maybe could be. All right, if there's any, if there's no other question, let's take the speaker again.