 Great. Hi, everyone. Welcome back now for the last part of this short lecture series. So in part two, we just finished proving this beautiful theorem by I and Les Feltz-Smith, Hoxson-Hunikey, Mashweed, and Muayama in the setting of character CP. And the main point to prove this theorem was to, well, to prove it in character CP, as Hoxson-Hunikey did, the main point was to cleverly use tight-closure techniques and to use the pigeonhole principle. So we use tight-closure because it's easier to prove that something is in the tight-closure of the power than it is to prove that it is on this power. And then our ring was right there, so ideals were tightly closed. The pigeonhole principle really gave us this containment when I take not an N, but when I take a power. And remember, the idea was that you reduce to the local setting and realize that you really just want to prove something about an ideal generated by each element. And you want to see which Frobenius power does the HQ power live in. And we noted that we could do this sort of slightly better containment by sort of using the full power of the pigeonhole principle. Now, I want to point out that if you follow the exact same proof we did, we sort of applied the pigeonhole principle K times. So I take some K, that's at least zero. So when K is zero, this is exactly the same that I just showed you, right? The first symbolic power is I. So there's a kind of souped up version you can do by kind of repeatedly applying the pigeonhole principle to kind of keep getting things in the Frobenius power. So you're kind of in the local setting, this thing will turn into a power. And, you know, it's essentially the same proof. If you want to do it carefully, I'd recommend first go over what we did today and try to do it on your own first, right? Reduce to the local setting, see what it is they want to prove, try to apply the pigeonhole principle. And then you can look at my notes where there is a complete proof of the statement. But I want to write it out because this is going to be important in this second half. So I want to talk about how do, how do we, one can improve this theorem. And there are many ways to sort of go, right? One could think about non-rightly remains. And there's lots I have planned to comment on about that, that we might not have time for. If we do have time, I'll do it at the end of this lecture. If not, you can take a look at the notes. Or remember, there's this much, much longer version, 200 pages in my website that you can look at. But I want to stay in this setting a regular range instead. And I want to talk about whether we can try to write tighter containment. So remember, my goal was, give an I, I want to find and a power, I want to find the best possible symbol of the power you can put in it. And we prove a positive statement. We prove a certain containment holds. We did not say that was best possible. So let's look at some examples. On Monday, we thought about this square free monomial ideal and three variables. So here k is a few. Well, there's theorem. Well, what is the big height here? Remember, we wrote on a primary decomposition for this black light. All the minimal primes, all this is your primes of height two. So the big height is two. So the theorem says the two n symbolic powers can be in the n power for all n. So for example, it says the fourth symbolic power is contained in the square. Okay, can we do better? Last time? Well, okay, two times to go on Monday and lecture one. We saw that the second symbolic power is different. And that's not contained in the square. So if you could do better, the only other possibility would be considered the third symbolic power, right? I already know four words and two dozen. Three is what's left. And you can check this. This is an easy exercise that the third symbolic power is contained. So sometimes you can do better. You can even do better for primary. So for primes, this big height is just the height. And so for example, if you look at the prime p that defines the curve, t to the fourth, t to the fifth, three, four, five. So I really mean like the kernel of the map. So it's the prime and three variables that defines this curve, right? So if you like, it's the kernel of the map to kt, that's anything to be valuable to one of these powers. That's that's what I mean by this. Well, it's a really cool exercise to prove that the second symbolic power is p is not the square. And I wish I had time to do it. It's one of my favorite examples to do. Never seen it. Think about it or look at look at my my my longer notes. We'll see we'll see this and carefully. But anyway, this is a prime of height two, right? It defines a curve in three dimensional space. And so the theorem again says two n symbolic power is contained in the n power for n. And so in particular, again, the fourth symbolic power is contained in the square. It's a cool exercise to prove the second symbolic power is not contained in the square. It's different. But once again, we can improve this a little bit. And write three instead of four. So I'm going to keep first heard. So before he and Mel Hoxer had done that poof, we just covered when he heard of iron lasso spells and sniff words. This is a very first question he asked. Well, first he was surprised. It was a truly wonderful, surprising result. And then he said, Well, you know what? I cannot find any prime of height two in a regular local ring for which I can't get the third symbolic power. Not only couldn't we find it. No, yes, found such a prime in the 22 years since he asked this question. And despite this what I'm going to say next, spoiler alert. And there are good reasons to think about this. So let's see, this is a situation where our big height is two. And there are some things we have, we have seen that are better than two n contained. Right, we proved in frankly, you can do that. So in fact, Harbourn asked a few years later, a very reasonable question. Now I'm back to the setting of any radically ill. Okay, I should be careful here. His original question was for genius ideas in in in phone on the wings. So I'm kind of restating it slightly. But I think it makes sense to consider this this generality. If I have a rock leg, you know, the guy H in a regular ring, regular ring. Can I take the statement from that nice theorem? And could I possibly always improve it? Like subtracting h minus one. So remember, Oxford and he told us, this is true. If you take a power of p, in characteristic p, right, like the pigeonhole principle told us that it would sort of it's, it's, it's a very fundamental elementary fact. So it seems extremely reasonable that you should be able to do this problem. So in particular, notice that when you're in primary care to see two, you get this containment of the third symbolic harm in the square for Bay high two automatically, right? That's that's what this says when you take people to do an n equal to the thing that is strange. And sometimes mathematics is very bizarre. The thing that is truly shocking is that there are radical ideals that don't satisfy this conjecture problems. So the first kind of example was signed by doing each key number. And so there's three algebraic diameters from full ends. These are experts on on points configuration. So what they did was they found a very explicit configuration of points of 12 points into that field. I'm very bad at drawing. And I forgot to prepare drawing beforehand. So if you want to see a picture of the special configuration of 12 points into two, please look into it. If you're more out to break, maybe I can write down what it is. Because it looks very simple. But I'm actually going to write, not just your original code example, but I'm going to write a generalization of it. My hard one in Switzerland. The first current example was published in 2015. And this is a paper from 2015. And I'm going to be over some fields. Okay, I just did not to. Because it's going to be a kind of example to put some bulk on the square, which is true in characteristics. We're going to fix the number and to be at least three. Oh, maybe shouldn't be called in, but I'm really bad at letters, C, maybe. And it's x y to the C minus x to the C, y to the C minus x to the C, z, x to the C minus z. When it takes C to be three, this gives you 12 points. This is in general some special configuration of points. So this is a radical ideal. It defines, as I just said, some points in P2. So that has the height, too. And it fails the third symbolic value. So somehow the magical thing in here is that somehow, even though we have this containment, so by the way, it is originally a receipt but then however in Switzerland, you can do this over in the field, you've got to be careful about what you pick. So notice that, you know, somehow, even though for, for powers of your characteristic, you're going to get this containment, that's somehow not sufficient to get the containment for all the other ends. It isn't forcing to be true for all the other ends. If you go back, and you will, you look at our proof, you know, our tight closure proof in the Hoxer-Kineke theorem, we didn't use this more powerful version of the vision hope but if you try to put it in there, you'll see that there's a true obstacle. Like you can't just, you know, it shouldn't work, right? Because otherwise, this thing would be kind of example, but it really, it really doesn't work. It's sort of a sad, sad way. But in my opinion, it just kind of makes the story more exciting. Now, I will say that despite the fact of now there are other families of kind of examples, all of them are so very special, you kind of have to take some special configuration of points or, or hyperplanes or whatever in some really, really special way. And in some sense, I want to convince you that actually, many ideals do satisfy this, this context. And we're going to prove that a large class of ideals and characteristics need to satisfy this context for all and not just for powers of people. So maybe let me start by saying that hard ones can get your goals for various things. Let me list some and then we'll, we'll prove we'll think about a large class. It's satisfied by minimal ideal. I added the word square feet that is not necessary. But I added in because for us, all ideals are radical. So I should think quickly. It's satisfied by generic sets of points in P2 and P3. So P2 is a result of boxing and hard one. P3 it's a result of the niche. It's likely to MPI. There's no proof. These proofs sort of rely on geometric things that are especially in these dimensions. It's true for some special configuration, like things like metroid configuration, star configuration. So sort of for certain special nice things is all And today we're going to prove what I think of as a generalization of this monomial case. So this is sort of a philosophy that I learned from the recent aspect of core. Whenever you see a theorem that is true for monomial ideals, ask yourself, hmm, is this true for ideals defining efferents? Because I want to say most of the time, now this is me talking most of the time, it ends up being the version of that theorem that makes sense for all ideas defining efferents as a large class of ideals in character CP. And it contains the class of spur-fee monomials. Now I believe you've seen if your if your range a little bit in these lectures, but again, it's kind of the end. We're all tired. So I'm going to just give us a quick reminder, but I'm going to cheat. I'm going to first tell us what is an F split ring? Well, a radius of character CP is F splits. If for being a splits, right, that's what the name says. So, but for being a map, you can think about it as a map of our models. I don't know what notation he's been using before, but I'm writing this to mean the R module that has underlying billing group R, but our module structure is given by the Frobenius map. So this is Frobenius. And for Frobenius to split, that means that there's a map of our modules data that gives a splitting for this. So beta F should be the identity. So when R is F finite, so I guess maybe let me put it like this. So today, so that I don't accidentally say something lucky, all rings are F finite for the rest of my lectures. So remember that means Frobenius makes R into a finite derivative model that's F will restart R as finite. So that's adding F split is the same thing as F pure. You've probably seen the proper definition of the pure A means is Frobenius map is pure. I'm not going to write it carefully, because I think it's easier if you're not comfortable with these things, it's easier to think about the splitting. But I mentioned our pointings of definition at all, because I'm going to think about this condition in terms of a criterion by Fetter. So theorem I think you've seen several times that I want to remind you because it's so important. Fetter's criterion was proved by Richard Fetter as part of the species. This is Sunamal Hoxter and this is in his thesis in 1983. So if I have a regular local ring, and a prime characteristic, I'm going to be lazy to write P because I'm zero. We take some wrap with you. The quotient is F pure. So F splits. If and only if when you look at your ideal I, and you ask, who sends you into a Frobenius power. You want this colon ideal, so the things that send I into the Frobenius power to be pretty big. So you want it to not be contained in the Frobenius power of M. And you can take this for all or some. This is a typical thing to do. So you can rewrite this to just P if you like. And now you have a very computable statement, like you could put this in Macaulay, for example, compute the colon ideal and decide, oops, whether you're not sorry, whether you are not in the Frobenius power of M. So I think again, I think of this statement as a measure of largeness, right? This is saying this colon is really big. I will say, you know, this also works. If you take a polynomial ring of our fields, you know, you grade it, and you take a homogeneous ideal instead, this also applies, right? So I said, my ring was regular local, but you could also take this graded setting. So in particular, wink, wink, it is okay to use this in the graded setting in the in the problems that you're working on. So in unrelated news, one thing you will prove is that if i is a square free monomial ideal, then our mod i is guess what? So that's, that's an example of such a thing. These guys are contained in ideals that define a few rings by which I mean what I could be quotient by them, I get it. So this is a small somehow square free monomial ideals that are a small subclass inside of the world of after things. So for example, the renaissance rings, well, renaissance rings, the polynomial rings are renaissance rings of other afterings, they're all afterings. If you take the generic, the terminant ring, these are all afterings. More generally, if you take sort of a nice swing of the variance, you also get an aftering. So this is to say, this is a nice, interesting class of singularities. These rings that are pure, they sort of have mild enough singularities that you have some control over them, but it's also a really large class that includes lots of classical rings that you want to be interested in. So what we're going to prove is that we're going to prove, so we'll show is that our mod i is pure. Okay, I might as well write the theorem now and then I'll just move it to the next stage. Theorem. This is the theorem of myself with pregnancy, published officially in 2019. I guess I should say that ours regular, it has characteristic p. It should be a finite. i is radical of big height h. And if our mod i has nice singularities, if it is at pure, then it satisfies carbons convention. Well, let me copy page. But before I write a proof for the theorem, and when I give you an idea of what it is that we're going to do, so I want to motivate that the proof. Because I think this is a sort of technique that is helpful in obviously, what we want to do is we want to prove some containment. So we talked a little bit about proving containment. We said that proving a containment is a local statement. So in particular, that's going to help, right? Because this this condition of authority is something that localizes well, so we can easily reduce the problem. Actually, let me let me add that here already, prove kind of the first step is reduced to the local statement. How do you do that? This localizes proving containment is something that's sufficient to do after you localized. So you can you can start by doing that. So I mean, it's sufficient to prove the setting. But a different idea, if you ever want to prove an ideal is contained above, you can sort of measure the failure of that containment by studying the colon ideal that corresponds to it. So this is the same as showing, is it? So proving that an ideal is contained another is the same as showing that the things that send i into j contain one. So it's the same as showing that this quality ideal, let's always contain it, but it's the same as showing that this colon is R. What does it mean for an ideal to be R? Well, it means it's very, very big. It's as large as an ideal can possibly be. How do you show? So if R is local, so maybe let me say Rn, when the way you would have guessed that's what I meant. I take a local ring and I want to show some ideal does not R. This is equivalent to showing that M is not contained. Sorry, but this guy is not contained in M. You were so big that you're not contained in the largest possible ideal. So to prove a containment, it's sufficient to prove that this colon ideal is not contained in M. And in fact, you know, in character Cp, you can even take for convenience powers. For convenience powers should preserve containment. It's sufficient to prove that the, the forbenious power of your colon ideal is not contained in the forbenious power of the maximum. You know, whatever this aim they are, this is a strategy that would always look to prove a containment. What do I have at my disposal? I want to prove a theorem or I'm assuming that I have an ideal that defines that theorem. This fetish criterion gives me a fantastic ideal to look at that somehow, you know, is too big. That does not contain an M to the forbenious power. So what you could do is, you know, if I guess now this I is not the same eye, but whatever, if arm or eye, they're pure. Okay, maybe I need a different letter. Let me call this I everywhere. A suppose that I know that arm or eye is a pure. All I need to show is that that quote, that special corn, I for being his cute lie is contained in this special colon I study. Why? Because this guy is not contained by Fetter, Fetter tells me the special colon is not in the forbenious power of the maximum ideal. But if I have shown that this guy is contained in here, you know, that forces this bigger thing not to be contained in the forbenious power. Because if this thing was contained in a forbenious power, so would the other thing do. So in general, you want to prove some containment. And you have access to some special ideal that you know, defines an F theory. If you somehow prove that this special colon is contained in the forbenious power of the colon that has that comes from the container you want to prove. Then that's what you do. Then you're done. So in my special case, I want to prove this containment. I really want to prove this implication. This is the one thing we couldn't do if we're in person. So to prove this, you show this, what you're going to show is that kind of independently of his security condition, you're going to show that the special contained colon ideal is always contained in the colon. Well, the forbenious power of the colon of the thing. And then that will say, so we'll say this always for large q. And if I prove this, and by this always, I mean, no conditions on I, right, give me a radical ideal I, big I H, I'm going to prove such a containment always holds. And this is going to force that when our money isn't pure, this guy is too made to be convenient. That's the thing is part of the maximum ideal. So this guy is not in the previous part of the maximum ideal. And then you trace back to this, and you're going to find this colon is all of R. And so the containment you want to prove is true. Does the strategy make sense? Are there questions about the strategy? Okay, now I'm going to do something evil. I'm going to prove not this. The thing I just described is what we prove in our paper. I'm going to give you a better version of the theory, which is what we should have written in our paper, but it's not always written there. Because I'm going to prove something was wrong. But it's the exact same technique with that same idea. So I will show coffee. So So now I have our am local. And I will show the following. Well, in principle, I would show this that I'm going to change the total want to change it. I'm going to prove it slightly different contain. So I'm going to do, okay, whatever this is. This implies the following. So suppose I show this containment that I'm claiming, right? Maybe let me call this claim. This is what I'm going to show. This imply that if our mind that pure, this is going to give me the containment and plus H containing an I am. Remember the strategy when I'm an eyes of pure, this thing is not convenient, the convenience power the maximum ideal. So this thing cannot contain the convenience power the maximum ideal. So the containment corresponding to this point. So when I'm an eyes of pure, this is the containment I will show. And somehow this from this containment, we will obtain the big theorem. So I should make a decision now what we're going to do first. Okay, let me first tell you let's leave the claim. Okay, so the claim gives this containment. Let's prove that theorem from this campaign. Right? So I have. So if I n plus H is contained in an eye, I am. I want to show you that hard ones conjecture is true. So how do I do this? And what does this even mean? This looks strange, right? I think of the statement of saying the problem. Each time you take your symbolic power, and you raise the order a little bit, you raise the order by H. That kind of gains you an extra power. So it's more the more copies of H you have in your the order of your symbolic power, the higher power you're going to win. This is kind of what I love the most made of oxygen, we can actually build a black hole that's right. But this is stronger. Because now if you want to consider this is the the symbolic power involved in harboring, right? The one that somehow should end up in either the end of the end of the story. If I go from here, you know, you can even think of this as kind of like H appears n minus one many times. Each time I subtract an H. So for example, I could write H times n minus two. So I subtract an H, right? I went from this guy to this guy, and I gain a power of up. Okay, so I'm literally applying this statement. It's bad that I'm using the letters and the both right, but I'm applying this statement with like, I don't know this and here is actually what I'm calling here. A gen minus two, plus one. But the idea is each time you lower this guy by H, which is what I did from here to there, I lower the power by H, I gain a power of up. And now I'm going to repeat this. And you keep doing this, you keep subtracting for as long as this makes sense, you can keep subtracting H from it. So notice that throughout the process, the sum of this number, and this number are going to remain constant today n minus one, right? Here I have zero, there's no power of I appearing. And I have an n minus one, then I have n minus two, one, which also sum to n minus one. And then you keep doing this, right? Like the next term would be hn minus three, plus one, I squared, and then you keep doing this. It's kind of an induction. It's an induction that stops. Where does it stop? Well, when you get to the end, you've subtracted H as many times as you could, you get this one left, right? Because you had like a multiple of H plus one. And then, so I guess this is like h times zero. And that plus whatever I write here should add up to n minus one. But now the first symbolic power is I. And so this is this idea, which is where I wanted to learn. That was already written before. So kind of the point is that from here, this sort of gives you a simple induction that goes all the way from this guy, the idea at the end of the one. And this is actually a stronger statement than the one than just proving hard ones. This is actually more powerful. And I may or may not have time to tell you why. But let me actually prove the claim, right? This all only makes sense if I do this claim. So let me prove. So I will prove this awful looking thing, but I promise it's actually not awful. So to prove this, I'm going to do the problem. I'm going to take an element F in here. And then I'll prove that it lives in here. Okay, you know, you are going to prove in your problem sets, the Frobenius powers commute with columns. So this is so I can I can move the Frobenius to inside the phone. So on this side of the phone, and I have I and plus eight. On this side, since Frobenius power commute commutes with products. This is I for being is key. Like to be anything. Oh, and here is the thing. Right. So what I did was I applied for being is on both sides, instead of having the Frobenius outside of the column. This is it. This is only true because firing is right. Right. This is what you're going to prove in your problem set. So hints to the regularities is necessary. I think I added a problem to prove that to prove that the regular is missing. Okay, so somehow I want to start with an answer in this column and land in this God awful, horrible looking thing. Wow. What does it mean to live in this horrible quote? Well, to live in a corner and heal means that you take the thing on the right inside of the thing on the left. Right. So I really want to show that as times I n plus H. Q is contained in that mess on the other side. That's my goal. That's what I'm going to do. And all I know about it is that it sends I into I for being. So let's start. I start off, I'm going to write a string of in a plot of containment that lead to where I want to start here. Along the way, I'm going to make a lot of steps that seem wasteful. This is the most wasteful of all the steps. I'm going to replace the convenience power by the ordinary. So the convenience power is contained in the truth power. And then I'm going to take that ordinary power. Okay, I'll write it out. This is the ordinary power first. So now that means I'm going to break that ordinary power into two pieces. One of them will get q minus one. And the other one will get just one power of the symbolic power. But now I'm going to do another really wasteful thing. I'm going to remember that a symbolic power is contained in I itself. So this seems like the most wasteful step of all times. But somehow I claimed that at the end of the day, even though we were very wasteful, we're going to end up with the best possible statement. So it's kind of kind of magical. Okay, so so far, I just replaced a convenience power by a ordinary power. Took one power out. And then remember the symbolic powers continue the ordinary in the ideal itself. Okay, what did I know about it? Yes. Remember, is something that sends I into the Frobenius town. So when I do as times I, I land inside the Frobenius town. So so far, I started on the left. And I haven't gotten here yet. But this is where I'm at. And then we compare where you want to go with what you have. And I'm halfway there. I want to land in a product of two things, and I got one of them down. So I'm going to keep that put and going to focus on the other. So I actually want to show that what's left here, and plus H, the q minus one, I'm going to show that that is contained in. I want to show that it's convenient. So I'm going to continue and I to the end. That's where I'm going. That's my goal. And remember, I'm really trying to show this horrible looking thing. So sorry, I've only done it very silly steps, right? I took Frobenius power, replace to the power, reorganize stuff, use the definition of where s lives. So now I have nothing to do with that s, I have nothing to do with that complicated stuff. I just want to do this new, still horrible looking thing, but maybe not as horrible looking now, right? It's a little slightly less confusing. So first step. So now again, I want strength containment. I'm going to do something I've done today, which is remember, though, that when you take a product of symbolic powers, it lands in the symbolic power of the sum of the orders, right? So this guy lives inside q minus one, and that's where this lands on. And now I want to somehow go land in here into the symbolic power. In the beginning of the final part, I wrote down a limer that I claim what's going to be helpful. And this is when it's going to be when I get it back. Okay, I'm not going to get all of it. I'm going to get this. So remember, I said this is just the general principle again, and just use in a fancier looking way. So recall, I have this. So let me even put it in a different color or something. Color. This is a recall from earlier. I'm going to apply this thing in gray. What am I going to do? I somehow want to land in here. That's my goal. And this is a previous power of the symbolic power. So it looks like this, right? So let me apply this statement, right on the other side, I just have a symbolic power, which is awesome. So I'm going to apply this apply, apply with k equals what where, well, I want this side to be this side, right? So I want to keep this one to the end. So I want k to be n minus one. So I'm going to copy what's in here and put an m minus one i h q plus n minus one q minus h plus one. My God. So that's theorem from before guarantees, maybe let me put it in a box or something. Okay, the theorem from before somehow that was just a given whole principle guarantees this containment hole. I want to prove this funky thing lives in the symbolic power. I've already shown the left hand side lives in this other symbolic power. If I somehow prove that this symbolic power lives in this symbolic power, then I can complete the dots, right? I can go here, here, here, here, and then I'm done. Like that's why I want. So it's sufficient to show. Somehow I want to show that inside of here lives that other symbolic power I have that lives the q minus one h minus one symbol. If I prove that I connect all the dots and I go from here to there and we've already reduced the problem to this thing. But now look at the amazing thing. I have a containment of two symbolic powers. Note that should go ahead. Yes. There shouldn't be in plus a two n plus gauge. Where should there be? Oh, down here. I see it. I see it. Thank you so much. You mean I miscopied this thing over here. It's q minus one 10 then plus h. Thank you so much. Wonderful. Any more comments or questions? Good, we're almost there. So when is the containment of symbolic powers true? This is when it's easy, right? It's just they're both symbolic power. It's like taking two pounds. This is saying a is bigger equal than b. What's that? What do I want to do? I want to prove the symbolic powers in another symbolic power. So I want to show that h, oh, sorry, q minus one times n plus h is bigger equal than h to plus n minus one q minus eight plus one. It looks awful, but I've reduced the problem to an inequality of integers. It can't get any easier than this. I have an inequality of integers. I just want to prove it. But remember, I don't need this for all q. I made a claim earlier. I actually only need this for q large. Because what I'm going to apply remember is fetish criterion and fetish criterion just needs large enough q. So I'm actually going to show not that this inequality always holds because it doesn't always holds. But that it holds for all q very well. And once you have an inequality involving integers, where all that matters is that a particular integer be really big. It's actually sufficient to focus on the coefficients on that very big thing. So we're going to look on both sides of inequality and read the coefficient of q. On this side, I have n plus h. On the other side, I have eight and n minus one. I need the coefficient on the left to be bigger than the coefficient on the left. Because the point is, I'm going to make q to be really big, right? So when q is really big, the terms involving q are going to win against the rest because the rest is not going to change. It's constant. Or if you like another way to think about this is you can take it in a problem and divide by q. And it's going to hold whenever that, you know, the parts that you didn't have q's on. Now I divided by q, q is going to infinity. So that those pieces become zero. So only the things that had a coefficient of q are going to matter when you send q to infinity. And so you can see here very explicitly that this is extremely high. You can't do this any better. Because this was literally as good as you could have been, right? I mean, you know, the only thing you can do better is would be if they would be the same. And then you would have to worry. So actually, not saying it's right, I really want this to be bigger, because if not, then I actually, you know, if they were the same coefficient, then I'd actually have to worry about all of those. So this is both great timing and terrible timing. It's great because we finished this proof. It's not great because I have one minute I had so many more things I wanted to tell you. But let me in one minute summarize some of the other things that I wish I had time to do. So first, can you do better than this? And I told you where you'd know, there are examples of things that are pure for which you really cannot do any better. This is the harbour to conjecture is the best possible scenario. If you want to prove something for all ideals that defy a theorem. But if you restrict the class of a pure to strongly afraid, you're going to exclude things like square free monominal ideals, but you're still going to keep interesting things like vergenesis and two minantal rings. And you can do better. What the crag and I showed is that when our my strong way of regular, let me at least try that out. And I think you've seen the definition of strongly afraid, so I'm not going to write it. This isn't the same paper. It proves that if our more I is strongly at regular, let me write in a serious statement can replace age by age minus one. So I satisfy this harbourn's conjecture, but it's going to behave as if its big height is one less than what it actually is. So you're going to get I like to the age minus one and minus eight minus one plus one, which looks awful that there are ways of rewriting this that don't look awful. I'm just like literally writing harbourn's conjecture and putting in age minus one with age is three. But no, this is a corollary. What happens when each is two? This is a really great question to ask the audience to end, but it's very dangerous to do this in a zoom call. So tell me, what happens when you do is you plug in one where these ages are. And you end up proving that the age, the n symbolic power is contained in the n power. So you prove equality. And there are plenty of ideal of defining after range of big height two that are not generated by regular sequences. So this equality doesn't follow for many elementary facts about symbolic problems. It's giving you new things, where you have this equality. You could also have talked a little bit about what happens when it brings up. But I really am out of time. So I think I'm going to have to stop there. So thank you for your attention. Alright, thank you so much for the beautiful talks. Are there any questions? Hi, Eluisa, I had a question. Yeah. Um, so most of the containment problem they've all around ideals in a regular look for some ideas. Why? Most of the containment problems revolved around radical ideas in a regular way, which either define FPO or strongly of regular singularities. Do the does the containment problem become completely intractable if we move away from regular rings to security, say, taking ideas in a war in team ring or re-defining a good singularity? Yeah, very good question. So can we study things like carbon's conjecture of the containment problem when a ring's not regular? There are many obstacles that one faces, right? We use regularity in so we can see much of waves in various places. One big question that is open, and I think would be really beautiful to answer, is the problem. And maybe that explains how hard the problem is, more generally. So when I told you this beautiful theorem, I made a comment, the big theorem that we started from. I made a comment that says in particular that you can take this constant instead of eight, you can take the dimension of R, like you lose best possible mess for all ideals, but now you have a statement that works for every ideal, and it does not depend on. So one problem that is open is if you take a ring that is not regular, say a complete local domain. And so you focus just on prime ideals. Is there a constant D for which this happens for all prime ideals and for all N? The constant doesn't depend on the prime you consider. That's an open question in general. It's proved that such a D exists in some special classes, especially, but I work with various people, but especially by what the community cats and about last year and unique in the past, they have various results of that. Now, some of the results, they're not even constructive, right? They're not finding this one's in constant. They're saying that one exists that is independent of five. So a different kind of question is, can you find something constant explicitly? So one thing I didn't have time to discuss, but it is possible to get in the notes, is that it's a different proof of this theorem using test ideals and characteristics P and multiplier ideals and characteristics zero. And if you follow along with that proof, it depends on essential properties of test ideals or multiplier. And there's some work in particular by Daniel Smokin and Javier Tavarra Juárez, where they sort of examine that proof and examine the one fact that really, really needs a regular way. The rest of it works a corresponding way of regularity. That's right. We've defined something called diagonally a regular range, where they can do a sort of a different proof that still works. And they actually end up proving this exact same result. So for primes, they prove you can take this once in constant to be that high over these diagonally So, for example, that includes the hyper surface defined by the maximal minus of a two by two matrix, for example. And that might sound like a simple ring, but that was open for a long time, finding this constant for primes in that ring. So it gets really tricky. And another thing you can do is you can also study things like hardware instances, right? You can say, okay, maybe can I, first of all, can I do better than like, is this sort of hardware and like saving the best thing to do? And at the same time, you can try to find sort of good, you know, a good, good swanson constant that works. So in particular, this theorem that we proved in the second half is theorem of myself and Freikiniki. This required them to be right here. And the main point was that we were using, okay, we were using some of these components that use regularity. But on top of that, we were using federal criteria. Federal criterion gives you a criterion for an ideal to define an F theory, but the ambient ring should be right. So if you want to prove a version of this theorem overrings that are not regular, you need a federal like criterion, right? Overrings are not regular. So with Lin-Schulman and Carl T. We have a paper where we prove a theorem like this, where our rings are born sign, and we prove a sort of a feather like criterion for born sign wings. But there's the cats. We still need our ideals to have finite projected dimensions. And it is not just so the finite projection dimension comes in several ways. But in particular, our feather criterion really requires finite projected dimensions. But in particular, I think that's a criterion that that possibly has other applications, you know, besides the lower symbolic values. So this is a very long answer to your question. This is to say, like, there is something one can do. The problem can still be attacked outside of regular rings, but it's just much more difficult. And it's still exciting. And I think, in particular, there's more than characteristics techniques can do for the problem. But I think we're just starting to distraction. Any other questions? If not, let's send Luisa again for the beautiful talks. And now on Monday, Baiba was going to tell us about the problems and the problems that are already posted. Yes, I have a question. Yeah, so why we are not considering about the something similar questions like symbolic for a brilliant symbolic power of I, there exists the M such that that is content inside I to the power normal power I to the power of Mn. So, yeah. So, let's see if I if I understand your question, is your question about when was there? No, yes. So, so here we are considering about only suppose for every end. So, large in the symbolic power of N, when it can for this end, can you say that there exists a Mn such that the symbolic power of N is content inside the normal part of Mn? Oh, you mean backwards? Backwards, yes. So, when you do this backwards, it becomes really easy because the end power is always containing the end symbolic power. This is always true. And then, of course, you can now make this guy bigger or make this guy smaller and keep the complete. No, I am saying that symbolic power of I that content inside that normal power of Mn. So, for every end, yes, yes, yes, for every end, is it is there is there any question like this? Oh, you mean you want to fix the end instead? Yes. And you are asking does there exist an end? Now I am sorry. So, see, a silly thing we could do is you could take this Mn to be one, right? So, you always have that the end symbolic power is contained in high. So, that's that's always true. And so, the thing that is not clear is when you start changing this power here, can you still find the symbolic power? Does that make sense? Like once you fix this guy, there's always something to put in here, but it's not very exciting because you can always put the one at the end of the day. You can ask, you know, what is the best thing you can put in here? But that's the same thing that we're asking to continue. But asking for existence of something on this side, asking for existence of an Mn, then that's easy. The answer is yes, when always exists. No, but it's I'm asking about the best thing which can. That's right. So, that's exactly the containment problem asks. The containment problem asks for the best possible thing you want to put. So, asking, you know, given and what is the best possible Mn such that this holds, we're asking, given A, what is the best possible, I don't know, FA to put in here? These are equivalent questions, right? So, I'm asking this for, for like, ah, ah, OK, but maybe now, now maybe I understand your question a little bit. Maybe you're saying, OK, but we're putting like some funky functions here, but what if this guy is not a funky function? What am I going to put in here? But in the end, it's sort of still equivalent, right? Because, for example, when I put statements like this, this is kind of saying that if you want something, maybe I should be using a different way. Maybe you want to use this is kind of saying, let's give me this containment. But in fact, this is equivalent of this, where is the naive big theorem we started from? Kind of equivalent of this. And this somehow should make the problem seem even sillier, right? Because it seems like this should be a very short difference between these two, right? Taking a ceiling or taking a floor seems almost the same. But it really, oh, this is not how I want to write it. I'm writing this back today. Sorry about that. Ha, taking a ceiling or taking a floor feels really similar. But the truth is that there is a massive difference between these two when you think about the problem from this perspective. So yes, the containment problem is also asking the same thing that you're asking. But then you have to sort of switch around how you write with convenience. So if you play with these, you'll see what I'm saying. With these two arguments. Good question. I mean, we can also stop the recording. And if somebody wants to ask more questions, I can stay here and answer more questions.