 Alright, great. Well, welcome everyone to parts two and three of this last lecture series. There were some technical problems in the beginning we started a little late so I'm going to try to steal in this first half more or less on time, or maybe cut the break a little bit but try to add the second half, when it's supposed to. So we started talking about symbolic powers on Monday. And we're really focusing on the case of radical ideals. So the end symbolic power of an ideal is what I obtained when I take the ordinary power and a computer primary decomposition for it, and I collect all the minimal components. So you're going to take your power, you localize that all the minimal primes of I which in this case they're also the associated friends of mine, and contract back to our. So these are really the elements that live in your power up to maybe multiplying by something that's not in any of those minimal primes. Now, at the end of last time. I said a little bit about some of the many difficult problems about symbolic powers that people think about. And for most of today we're going to be talking about a problem known as the containment problem is very easy to phrase, but it isn't a difficult question and in particular. It's the question that sort of includes a sub problem, some of the other questions that I mentioned on Monday. So roughly speaking the idea is I'm going to hand you an idea. And I want to know which containments of this kind that can be so with symbolic powers are containing which powers. And many remarks to make maybe the first one is that this contains. contains some other problems. So for starters. This contains the quality problem that I mentioned last time. If you want to decide if a particular power is equal to the symbolic power notice. In fact, it was here in my reminders that the symbolic power always contains the power. And I'm talking about a sort of backwards kinds of containment. So really, if I can make these numbers to be equal. So, if a particular symbolic powers contain the power. That's really equivalent to asking for equality. So, in a way, you know what we really want to do this from a sort of best possible situation. And so, given the be I really want to find the best possible a that I can put in here. And the best case scenario is that I can make this number be equal to this one, a priority, it can be bigger it can't be smaller. So, you might, if you cancel this problem completely for each be finding the best a, you will in particular decide when you have a quality. But even if you don't have a quality, you can think of this question as a way of sort of comparing the, these two families of ideals you have your ordinary powers or smart powers. Compare the two really because you understand the powers really well, but the symbolic powers are much harder to understand. And so you want to sort of extract information about the symbolic powers from the powers. So last time I mentioned very briefly, that when you have a homogeneous ideal and a great ring symbolic powers are also homogeneous. And I asked about what are the degrees of elements in those symbolic powers and we saw examples where those degrees were sort of unexpected, right, the kind of the example we saw this sort of too low to be in the power but you might also, you know, see sort of unexpected things. Maybe let me introduce a little bit of notation that I actually will not use in the rest of the lecture but just to explain the problem. If I have a homogeneous ideal today. The smallest degree of a non zero homogeneous. The smallest element in J out of notice alpha. And notice that, you know, if you happen to know that a symbolic power is contained in a particular power. Then this automatically tells you something about these degrees. So if I want to compare the smallest possible degree of something in each one of these well. Obviously a bigger ideal might have things of smaller degree but whatever is the smallest degree of something in here that's also in here. And so this guy is bigger. So in particular, notice that the degrees of elements and the powers are really easy to compute like this is just the times whatever is the smallest degree of something I. I mentioned last time that getting upper bounds for this smallest degree is really easy, but getting lower bounds is what's hard. So somehow if you if you solve. If you if you get a particular statement of this kind, you might gain information about lower bounds on degrees on symbolic values. And in fact, there's been some some recent developments in terms of finding bounds on these lowest degrees that sort of involve not quite this containment but related related statements. We did useful to consider containers like this. And then there are the reasons you might consider this problem right as I said before it's sort of a comparison question. This is a question that first appeared in a paper of Senegal in 1985. But that really gains more traction. In the late 90s, early 2000s. And this starts, thanks to work of the NS one son. Who first understood a little better what it meant to say. Okay, maybe first let me say like this. So, chencel and others back in the 80s were interested in the question of when it even make sense to ask this. Right. So I was saying given a be I want to find the large, the best possible as the smallest possible a. But I haven't even addressed the question of whether that makes sense right given to be is there. We can make them all precise by saying that these two families of ideals are co final with each other. Maybe for the purposes of time. I won't write things. So precisely. But let me tell you about a theorem of nonsense, which was published in 2000, which essentially says that there exists an answer for the containment problem. So let me put it like this for all a is there's the. Oh, I think I'm switching the letters or no. For all these are existing. So really, there's an answer to the containment problem for all powers for this given I. So maybe let me clarify that I mean given I. So this is really equivalence to actually the existence of a constant see that presumably as I'm writing it might depend on your ideal I such that the CNN symbolic power is going to be the end power for all in. So you can think of this as sort of saying, if you can answer the containment problem. Your answer is no worse than linear, right the function that gives you the best possible thing you can can put in here for each end is worse case scenario linear function. And this sort of opened the door to trying to understand what what is this this constancy. So her proof didn't didn't give us, it gave us the existence of a city right it didn't tell us with what to see is explicitly. But even more surprising. Soon after inspired by this, there came an answer. Over regularings. So, I'm not about to write. I'm going to need a little bit of notation to write it was proved to receive I am not a fellow Smith, using the theory of multiplier ideals. And then soon after. They proved the theorem in characteristic P using type closure techniques, and then they use reduction to characteristic P to get the theorem whenever you're in the field. So it's going to be some theorem about regular range. And then, for a long time, this is all we knew. Very recently, thanks to Andrea solution of the direct summit conjecture using perfect like spaces techniques. We started seeing solutions to all these other questions and it's good to stick with opera. And one of the first things we saw. We sort of established a theory of multiplier slash test ideals and makes it interesting. And I think that's now officially, I want to say 2018. Do I have it here. Ah, yes. Their work require the ring to be excellent. And I'm not going to write a theorem about excellent rings. Thanks to work of the coming what I am. Who just a few months ago, so I think disappeared on the archive in November. So sort of doing similar things to my suite. I'm going to put 2021. This is not published yet. So it's very, very recent, but he was able to remove that excellent assumption so now we have a clean theorem for all regular ins. So this is going to be some answer to this containment problem. It's going to tell us over regular ring. Why is this constant C from Swanson's theorem. I just want some constant. That that is going to appear for a few weeks. For that I need a definition so I can write the theorem. You just tell me what the big height of an ideal is. So it's the maximum of all the heights of the associated rhymes of your ideal. Yes, in our setting our ideals are all radical. So you really just look at all the minimal times and deciding what's their maximum. So this is sort of a more honest version of the height than the usual one right the usual one is going to look at all the heights of your minimal primes and choose the smallest one. And I will choose the largest. So in particular for a prime ideal, right. So if, if I is prime. Notice that the big height is the site. But the fear we're about to write is a film that really requires this this big height invariance, we can't replace it by the height. And what is the statements. Well, let me say that I is radical of big height age. Do I need the assumption that the ideal is radical no I don't. But then we have to talk about, well we have to see a little bit more about what this invariant is the big height works but you can do a little better. I mean you can't do better for radical ideals this this is the constant that that you need to put in there. We also didn't define symbolic powers in full general. So the theorem. I talked to for so long you probably already guessed if you didn't know is that this one's in constant in this case is this big height. So, I'm looking at the time is maybe something I'm going to cut at the end so maybe let me make a comment right away. But as a beautiful consequence of this. We can actually take this one's in constant to be independent of the ideal. Because if you take D to be let's say the dimension of bar. That's the largest possible big height. Actually, we could take dimension of our minus one but let's not worry about that. It tells me in particular. If I take the D and symbolic power. That's presumably bigger than the H and symbolic power equal ideas, each or more. And so this guy is continuing here which is containing the power. But the amazing thing about this is that now we wrote a statement that doesn't depend on you, you have a uniform content to the works for all ideas. And this is sort of a beautiful thing that often happens. We have to sort of uniform behavior. So what we're going to do for the rest of this first half is I want to prove this theorem. So really we're going to cover the orchestra to make you prove in a computer to see P, and how we're going to do that. And, and the sort of really, we really will use very elementary characteristics techniques which I think, maybe to me this is one of the first terms I ever saw I really saw the power of characteristics techniques, because you know this is not, this is not an easy theorem. And, and the I was supposed to prove in a way, it's also sort of beautifully elementary in some sense, in the sense that you, you just have to show a certain string of containment. But it's actually very technical because each of those containments depends on some, you know, difficult properties of multiplier ideals. But I love how, you know, we turn this complicated theorem into something very elementary using type closure. So we're going to need a few, a few tools that you will prove as exercises. So, actually this one there. Well, okay, I'll write as exercise. It's spelled out with a few more details in the in the problem set. So first of all, we're going to need that if you take a regular ring of prime characteristics, then taking for being a specialist does not change if it would be a different. So, this is really a result of this can't be wrong. It doesn't actually need a ring to be regular but it needs the ideal to have funny projected mention. You write this down over a regular ring but the proof the right is essentially the same you're just going to use that you're able to have fun. Well, okay, maybe you'll use a little, a little more than that. But it is true more generally. So, one of the points of this is that remember last time we talked quite a bit about associated crimes and we said how one of the difficulties of dealing with symbolic powers is that it's really hard to study associated problems. And when you do things like taking ordinary powers. You lose control over your associate runs, but the beauty of being in courtesy P and being able to take for being as powers is that for being as powers do preserve associated problems. And this is going to be a key idea in the proof. And one more thing we're going to need. So we're going to prove a containment of two ideas. And we're going to need a very elementary factor the containment are local statements. So to prove a containment. You want to prove that I've contained in J. It is sufficient to prove that containment holes after you localize. Probably then this is an exercise when you're first committed as a book was. Maybe you did it for all crimes, or maybe you did this sort of final version that I'm going to need. I'm going to need that actually it is sufficient to look at the associated crimes of the larger. So if somehow you can control the associated crimes of the thing you're going to land on. You're going to find your containment after all the localizations, then you, you won the game. And now, thanks to this. The first part of the puzzle, I'm going to call it a theorem business call the theorem in my notes to me this should be a letter. This is, I'm about to write my very favorite version of the principle. So what is it. If I have a radical idea. If I have a high age in a regular ring. Then, if I take. So I want to prove something about the HN symbolic power being contained in the end power that's my goal. But now I'm going to prove that that's true for powers P. And I'm going to prove one better. I'm going to prove that I can land in the for being itself, which is smaller than your name. I'm going to prove. So step one, by the last exercise I just went down. It is sufficient to prove this at the associated primes of the potential larger. So I just want to focus on the associate of friends of this guy so sufficient to show after localizing. So the associate of primes. And of the convenience. But then the first exercise I wrote down was that the associate of friends of the previous power are very well understood to be associated friends of I. So we only need to prove this containment holds after we localized at all associate of primes of. Let me write Q instead of P, just because he's the character steak and having two things. Okay, so I need to prove this containment actually localized in the associate prime of I, I was some radical ideal so I'm really talking about the minimal friends I. And this isn't something I said last time what happens when I take a rock like you. I localized an associate prime of it. Well, I just had an intersection of a bunch of primes and I localized that one of those. The remaining crimes are going to become the whole thing so I'll just get that one prime. So after localization, this is a statement about the maximum ideal. In what is now a regular local ring. And also the taking for being is powers. It's easy to show that that's something that commutes with localization. So again in the localization IQ is just the maximum ideal. So what we need to show. I'm going to rewrite a statement now in this sort of easier local setting. So I have a now I have a local ring in fact I have a regular. And I want to prove a statement about the maximum ideal. Well okay maybe there's a little bit more that I still need to do. So, and on this side when I localize, I is going to become the maximum ideal. And I'm going to take it for being it's power. On the other side. I have a symbolic power. The symbolic powers remember are the things I get when I localize at the associated crimes and saying, saying something different that's written here these are the same my guys back. I'm going to look like that each of those primes, and then look like the power can drop back. So if I don't take this and I localize again. So if I take my symbolic power and I localize I just get the ordinary power. Now I'm here. When I take. I hq. And I localize you. That's like taking IQ, which is our maximum ideal now remember, and just taking an ordinary power. Again, because I localize it on one of these. So on this side I have my maximum ideal. And I took it's ordinary. So what I need to show is that in a regular local ring. Something with less to understand here is what is the age have to do with anything. Well, it was the big height of I. So it was the largest height at one of these cues that I might have localized that. So what I know about this regular local ring. Sorry, I'm calling it are now. Well, it's dimension. Maybe let me write up here it's dimension is the height of this prime that I localized that. And that's at most the largest height upon us to see the final high, which is that the height. So I might as well pretend that each is the dimension of my ring because I know the dimension of my ring is that most worse case. And now we want to prove this. So this is what we'll start now. Now we can pretend this is the theorem. And this is what I'm going to do. And now yes I said earlier this is my favorite version of the vision hope principle, and this is a very easy application of the picture. Why, well, my ring is regular. I know about it. So since my ring is regular. I know that this maximal ideal is generated by at most each many things. Like, it might actually be that the dimension is something smaller right that this was a minimal primary smaller height that the big height but it doesn't matter I can always choose each things that are going to generate this and so really now I want to show that when I take an ideal generated by each element, and I take an HQ of power that is contained in the Q for Venus power. So I really want to prove it. Now, it is of course sufficient to think of generators of the power. The power is generated by monomials on these axes. So the generators here are things that look like monomials on the axis. And what you know about them. Well, I guess if I wanted to pick a minimal set of generators I could even be this equal to let me just say bigger equal than HQ. So for all the monomials of this kind I get a generating set for this ideal. And so if I prove that all these guys live in here I am done. But what does it mean to be in this ideal. Well, I just need to check that in this monomial. I need at least one of these powers so I mean at least one of these ai to be at least. Well, this is where the pigeonhole principle comes in. I'm summing a bunch of integers positive non negative integers. They sum up to HQ. If none of them was bigger than Q. Right if they were all up to Q minus one. Then no way they could together sum up to HQ. So this is automatically actually let me raise the need this is automatically going to get me what I want. I have each many things summing up to HQ with least one of them. And so therefore, sorry for the funny writing, this guy, this generator is indeed a different. And so the thing that we proved after all this so we reduced to the local setting. And then we apply the pigeonhole principle and show that the HQ symbolic power is always containing the Q's to be in this power. Any idea a radical idea of high, a big high age. So no, I can do better. When I was applying the pigeonhole principle. I only needed that at least one of these guys was at least. Yeah, it was at least Q. So I wanted to conclude that at least one of them was Q. So let's apply the pigeonhole principle a little more carefully. How do I make this fail. I want to conclude that one of them is Q that the worst or best case scenario depending on how we look at it is that of these age many things that they're all smaller than Q. Right. So, if all of them were Q minus one or less, actually follow them were exactly Q minus one right, they would sum up to eight times Q minus one. So in this situation, it would fail to have one of them to be at least Q. But now if I go and I add one more. Over here. Necessarily, at least one of these must be Q, right, because we took sort of like the extremal case and we added one to that. This is the best thing that the pigeonhole principle can possibly tell us. So we can improve our theorem to say that the HQ minus H plus one symbolic power I'm just distributing the age. It's contained in the Q for being a seller. Okay, so this was the key piece of the puzzle. And it was, like I said, it's just the pigeonhole principle. This is what I think makes it so beautiful. Now we're ready to prove the theorem. The theorem, this is the Hawks are going to be a paper from 2001. You have a regular ring. But now I'm going to say that it has crime characteristic B. I'm going to take a radical ideal. We've been high age. And we're going to show that the H and symbolic power is contained in the end power. We have already proved this in the case one ends of power. Now we have to do all the other things. So how do we do this. Let's fix and then we'll prove the statement for this particular end. So we're going to take a U in the H and symbolic power. And what we're going to show is actually that you. Well, I wanted to show it's in the pattern, but what I'm going to show is that is in the type closure. And, you know, R is regular. You've been talking about type closure for months now. So I'm going to assume, you know, I'll be the same side. Your ring is regular. If you like it, weekly F regular. And so that all ideals are tightly closed. And so in particular, the type closure of the power is just the power itself. And so by proving and living the type closure, you prove you are in. That's right. So I see a question in the chat. The maximum ideal should be generated by as many elements as a dimension of R. That's correct. So if we go back here, this is what we use in this group. We had reduced the problem to the local setting. And we had a regular local ring. So this maximum ideal was generated by as many elements as a dimension of R. Well, but now R is not my original R. So this is my fault for you in the same letter, right? Now this is my localized ring, right? I took the original regular ring and I localized and associate a prime of I. So the dimension of this localized ring is actually the height of the primary localized that. And what you know about that prime is that it was associated to the convenience power. So it is associated to the original ideal. And by saying that H is the big height, I'm saying that H is at most, sorry, H is the largest of all the heights of all these cues. So for whatever cues that you localize that you always have this in the cloud. So, you know, and for the cues that have actually smaller heights, you know, you could maybe take less elements here, but for at least one of these cues, you will need, you know, the full on age elements. So I'm not saying these are minimal generators for any particular reason. Does that answer your question? Okay, thanks. Wonderful. Any more questions anyone want to ask? Please do interrupt me. Okay, feel free to interrupt me anytime now I open the chat. And so I will be, I'll be trying to pay attention. Okay, so coming back here we're trying to prove this containment. So we've remembered our ring is regular so we're going to take advantage of type closure. So we're going to take an element in the symbolic power prove it's in the type closure of the power. So I know you're all experts on type closure now this is the very final lecture series of this workshop. And so I barely have to remind you of this but it's early morning it's been a long few months so let me write down what we want. So, you will show that there exists some C. I guess my ring is a domain so I'm just going to say it's a non zero C such that C times you to the queue is in I to the end to the queue. Well to be in the type closure I only need this for a large queue, but I'm actually going to prove that for all queue this happens. I'm going to find a C independent of you that makes this happen forever. So it's very much in the cycle. Okay, so now you'll see it's just simple calculations. So what will we do. I fixed them to begin with. And I guess I fixed you but for now ignore the view. I fixed and so let's use the division algorithm, first thing you learn in the first out of the class. But division algorithm. Oh, I'm doing this backwards. I'm going to now take half. Okay, now. So for each queue. So for a minute, I'm going to fix you I guess in my calculation. But then in the end you'll see that what I'm going to end up writing we're not involved use at all. Okay so temporarily I'm going to fix the queue. I'm going to take this queue and I'm going to apply the division algorithm. I'm going to use the remainder so it's at least zero and it's smaller than them. So divide q by n is essentially the point. And now, and I guess I can say, also, a at least zero because choose bosses. Okay, so with this. I'm going to start with my you that we have done I to the age at this is where I started right remember, and I want to end something like this. This is my goal. I'm going to put it on the side so we can see it. Okay, so I started here. So if I take you to the a. It lives in the symbolic power to be a. So if you remember our lecture on Monday, we said that when you take a product of two symbolic powers. This lives inside a symbolic power. Well this is a product of multiple one, but when you take about two, it lives inside the symbolic power with order to some of them. So I took a product of a many things that were the same. I guess I'm going to add hna times. So I'm going to live inside of here. I'm not equal, but I have a team, which is all in. Next step, I'm going to now take my you to the a. And I'm going to multiply it by i to the agent. So this is really inside i to the agent times the symbolic power. I'm just using, I'm using the previous slide directly. Now, remember that we said that the ordinary powers are we contained in the symbolic powers. So you can think of this now as a product of symbolic powers, which is contained the symbolic power of the sun. I'm writing a really complicated looking thing. But I am going to do this one better. Actually, sorry, erase this, I'm going to replace it by something slightly better. I'm going to note that this remainder are kind of just by division algorithm is something that's strictly smaller than n. So if I take a smaller power, a smaller order, right, the power is a bigger ideal. So this guy is contained in here. And I guess I just said the power is contained a symbolic power. So here I'm just using and bigger equal actually and strictly bigger than our, but I just needed to be equal. And then I use the power to continue the symbolic power. And so I'm going to do the same thing I was writing down earlier. But now I'll play HR. Now the beauty of this is that this is I to the age times a n plus R. This is Q, right, it's up here. This is Q. So this is I, HQ. So so far this long string of things, I conclude, tells me that I to the age and you to the a is in I symbolic HQ. So this is at the limit that we put we call the theorem that for this one, the pigeonhole principle. What is the pigeonhole principle say he says that when I take the HQ symbolic power. So whenever I take a symbolic power that at each times about P, I land in a for being as power. So, I'm going to use this. This lands in the for being as power. So we're almost there. I'm going to take powers of n on both sides. Right. So, take a power of n here you get h n square. Take a power of n here you get a m. Oh, this is contains the exercise that we're going to use again today that symbolic or ordinary powers and convenience powers commute. That's not true. This is what I almost said online, right, ordinary powers. Community. For being a source community products. Elementary. I'm going to do just one more silly thing. Notice. Q is a n plus R. Let me write that Q is a n plus. So in fact, when I make a very silly statement. Q is at least a n. So if I had written I to the h n square times you to the queue. That would, that would be inside of here. When we write that out. And that would be contained in either the end. So what do we do all this. Because this power now. First of all, you see that there's no q's involved in that power there's just Asian n square, nothing else, no, no q's involved. So this is independent of the, and more over, you know, our is a domain. Our is a domain and I is non-zero. I should have said that in the beginning. Okay, I'm going to put that in my assumptions that I think we all agree that the theorem we're trying to prove is silly. So it's a zero idea. So we're not going to consider the zero ideal. When I take a power of a non-zero ideal in a domain, I get a non-zero idea. I have a non-zero ideal. There's nothing to do with q. I can choose a c that's inside of this power. And now choose that favorite c that's independent of you. And you conclude that this happens. In fact, this was true for all q. And so we conclude that you live in the tight closure of this power. And this is the definition of living the tight closure. I mean, better than that. But our reading was regular. And as we said in the beginning, the type, ideals are tightly put. So this is either the end. So indeed we proved either the hn is contained in either the end. And we started about eight minutes late. So I'm going to, I went three minutes over in this first part, but I think I'm going to, this is a good place to stop. I'm going to stop here. We're going to take a seven minute break. And then we're going to resume at the time we were supposed to at the, I was going to say, 9am, but that's not true for everybody. So that's the hour in six minutes. We will resume. But if you want to ask me questions in the meanwhile, you can. I'll pause here for a minute. See if anyone has questions. I do have one.