 Okay, welcome back to the lecture series of the Summer School, and the speaker is Luisa Grifo from the University of Nebraska, and she will talk about symbolic powers. Great, well, thank you. Welcome everybody to the very last lecture series of this very long workshop. I am thankful to the organizers for inviting me and thankful to all of you for sticking around until the end. So I am going to be talking about symbolic powers, and in particular, about how we can use characteristic P techniques and tight closure in particular to solve problems on symbolic powers. So we won't really see any characteristic P until Wednesday. So today I'm really gonna be talking about background, like what are symbolic powers? What are questions people are asking? And then we'll use primary prime characteristic tools on Wednesday. Now, this is a very short lecture series, and so there's of course a lot of symbolic powers that I will not be able to cover today or on Wednesday. So if you wanna learn more about the subject, there are many resources you can use. So if you look at my notes, there are many references, first of all, in particular, a couple of surveys. And if you go on my website, even from the main page, you should be able to access long 200 page set of notes that I wrote about symbolic powers. So if you wanna learn more, you can read that. Though I wanna warn you, it is not complete. In particular, it definitely still has typos, and there are several chapters I'm still going to write that are not there yet. So if you look later, you might find a longer, better version of the notes. So anyway, so today I wanna start with background, what are symbolic powers? And the very first few things we need to remember are really things that we already all know. And so I'm gonna go very fast in the beginning, but if you're not, when we get to actually talking about symbolic powers, we'll slow down, okay? So these are just, this is a recall moment. First of all, I need you to recall what our primary idea was. So being primary is sort of a generalization of being prime. An ideal is primary if when you have a product in there. If one of the terms in your product is not in your primary ideal, the other one must at least be in the radical. So we must have some power in that idea. So immediately from this definition, we see that the radical of primary ideal is prime. And so we're gonna wanna say, we wanna record what that radical is. So we'll say something is P primary if its radical is the prime P, and it's a primary ideal. Now, big, big warning, just because your radical is prime, this doesn't mean you're primary, right? It's very easy to give examples. So for example, if you take the ideal X square, X, Y, take K to be your favorite fields, then it's easy to see that this is not primary. The problem element of this X, Y over here, but also the radical of I is also easy to compute, it's the prime X. So this is an ideal that's not primary, it's radical is prime, but still. An easier way to see what the problem is with this ideal is that actually being primary is an equivalent to having only one associated prime. And the problem with this guy is it has only one minimal prime, but it has an embedded component. So there's an embedded prime X, Y. So I will say right away, we've seen our first exercise that there are things I'm calling exercises that are suggestive exercises to think about. The ones I intend for you to turn in are actually called problems. So when you look at the notes, you wanna go at the very end and the things that are called problems are the ones that I intend for you to turn in. Things called exercises are just good questions to think about. So anyway, here's another simple exercise that we'll use today. The intersection of primary ideals with the same radical is also primary to that same prime. Okay, so having primary ideals, you can now talk about primary decomposition. So giving a primary decomposition of an ideal is to write your ideal as a finite intersection of primary ideals. Now, if you have, I just told you in this exercise that if you intersect primary ideals with the same radical, you still get something primary. So if you have two components with the same radical here, you might as well intersect them and make just one component with that particular radical. So I wanna think about irredentant decompositions which are ones where I've already done that. I've already collected together all the components from the same prime. And of course, I wanna think about decompositions where I can't delete anything, because if I can delete something, then why even write it? If it's not doing anything in intersection, I don't wanna count them. Now, I'll notice from what I've said so far, it's so easy to see that if I hand you a primary decomposition for something, you can very easily turn it into an irredentant one by deleting things you don't need and then collecting together components with the same radical. Now, one of the, maybe one of the first theorems that we would definitely say this was definitely primitive algebra. If you were definitely doing primitive algebra to do this, it's this beautiful theorem that was proved in full generality by Emmy Noether in the paper where she introduces Nazarian rings, which of course she didn't call Nazarian rings, but this is where they first appear. And it had been proved before by Emmanuel Lasker. And yes, that is the chess champion, Lasker. He had proved that the setting of polynomial rings and power series rings, and then she proved the full generality. And the statement is, of course, give me any ideal in a Nazarian ring, and then it has a primary decomposition. So for my entire lectures, unsurprisingly, all the rings are Nazarian rings. And notes, I could give you examples, but I think you've all seen examples before. I'm just going to remind you that primary decomposition, they're not unique, but there are some parts of it that are unique. So there are two uniqueness theorems. One tells me that whenever I hand you any redemptive composition, the primes that appear as radicals of those components, they are actually the associated primes of your IQ. And so the associated primes of the mongealarmor eye, those are what I call the associated primes of the ideal eye. And those are exactly the primes that appear in any redemptive composition of mine. You might have different decompositions, but what you know for sure is what P's you see a P primary component for. So in particular, the number of components is fixed. It's the number of associated primes. Now, some components are unique. The components that are coming from minimal primes, those are unique, right? Remember, the set of associated primes contains all the minimal primes. And so in particular, the P primary component associated to each minimal primes is actually always the same in any redemptive composition. And it's given by what you get when you look at the ideal eye, you localize at that prime P and then you contrast to active R. So another way to say this is you're asking who are the elements that lived in eye once I've localized P, that locally around P, you want elements that lived in eye. So in other words, the elements up in your ring that are sent into eye by something on P. So these minimal components are important in particular. And that's the beginning of the story of symbolic values. So let's first talk about symbolic powers of prime ideals. So let P be a prime, what is the N symbolic power of P? Well, we write it P to the N, but with parentheses around it to distinguish it from the ordinary powers. And a simple way to describe this is to say, this is that unique P primary components in P to the N. So when we write the power P to the N, fun fact, you'll see examples soon, a power of a prime does not have to be a primary ideal. Again, it's radical P, but that doesn't make it a primary. In fact, if the powers of primes were all primary, there would be no such thing as the theory of symbolic powers. We wouldn't be here talking about this. But even though P to the N is not primary, I know for sure there's a unique minimal prime as P. And so I know exactly what that unique minimal component in any redundant decomposition looks like. So the theorem we just wrote tells me that this unique P primary component is what I do when I look at P to the N, I localize it P and I contract to R, right? So writing this more explicitly, it's the elements that live in P to the N up to maybe multiplying by something not in P. Now, it's an easy exercise to check that in fact, this can also be described equivalently as the smallest P primary ideal. Continuing P to the N. So how would women check this? Well, you would take any P primary ideal that contains P to the N, right? And I noticed that you intersect it with P to the N and you still get the same thing because it contains P to the N. But now that's something you can play with your primary decomposition. And if it's something P primary, you can put it together with your P primary component and you'll find that nothing changed. So this guy must have contained that small, sorry, must have contained this symbolic power, which is that minimal component. Anyway, I know I said that fast, but if you've never thought of this before, try to check it a little slower later. Okay, so here's an example that in particular, I will explain what I just said that powers of primes don't have to be pregnant. So take your favorite fields and take any constant you like see. And now let's take, this is actually the same computation if you do the polynomial ring, but let's do the local setting. So let's take the power series ring, k of x, y, z. And then let me take x, y minus z to the c, just some hyper surface. And in this hyper surface, I wanna consider the prime ideal x comma z, okay? Now, I wanna think about the powers of this prime ideal. I claim that this is going to be an example where the powers are not primary. And so we're gonna say something interesting about the symbolic powers of this guy. Maybe let me say immediately that of course, if this is the p primary component and p to the n, or if you like by looking at this line, one can quickly see this is going to be something containing the power, okay? So if I'm gonna show you, I'm going to show you that this symbolic power of this prime is not the power. Maybe let me in fact write that as a remark first, right? So when are the symbolic powers powers? That's exactly when p to the n is primary. So I'm gonna show you that there's symbolic powers of this guy that are not the powers by showing you that the power is not primary. I wanna think about the power p to the c. What is nice about p to the c is that in here, of course, you have z to the c because z is a generator of p. But in our ring, that's the same as x, y, okay? But now notice, of course, that y is not in this prime p. And it sends x into p to the c. So it's kind of hidden, but from this definition over here, we see right away, this tells me that this x is in the cth symbolic power of this prime p. But on the other hand, notice that x is very much not in p to the c. Well, I guess it's important here that I declared that c was at least two, okay? And so in fact, this tells me that we just found an element in the symbolic power that's not in the power. So it tells me the two things are different. And in particular, it tells me that this cth power is not a p primary p. So maybe another comment to make here is that one thing that's interesting, well, okay, let me save that comment for a little bit. Sorry, you can guess what I was going to say. Okay, so I give you a definition for primes. What is the more general definition? Well, I'm going to want to tell you what is the nth symbolic power of an ideal i. All examples we will consider in these lectures is we're actually going to be radical. And that's already extremely interesting. In fact, the symbolic powers of primes alone are already extremely interesting. And if we only talked about primes, there'll be plenty to say. So you don't need to go think about all ideals to have something interesting to say. Now, for technical reasons, I want to have an i with no embedded primes. And the main point is that when the ideal has embedded primes, there's sort of two choices of definitions for symbolic powers that are different. They're both good for different reasons. And we have such little time to talk in these lectures. So I don't want to get into the weeds of all the technical details, it really isn't, I think it's not the point of the lectures. Again, even if you just think about primes, it's exciting. So in fact, most of the time, even you can take this for the entirety of the lectures, I'll take an ideal that it's radical. So what is the symbolic power? Well, again, we write it like this. And what we do is we take our power, i of the n, and now we're going to localize it a bunch of primes and contract back to R. But now we don't have one minimal prime to look at. We're gonna do this by looking at all the minimal primes of i. Now, notice I said my ideal had no embedded prime. So this is in fact the same as going over all the associated primes, right? So I'm collecting what I'm doing here is I'm collecting the minimal components of i to the n. I'm writing the primary decomposition for that power i to the n, and I am collecting the pieces that come from minimal primes, right? The minimal primes of i or i to the n are the same thing. So I'm just collecting these minimal components. So in a way, computing symbolic powers is like computing a primary decomposition but focusing on the minimal components and ignoring all the embedded components. So if you want to write this in a friendly explicit format, you could say that this is all the f's that are sent into the power by now some s that is not in any of the minimal primes. Which is the same as saying it's not in their union. So like before, notice, this is really a question about associated primes. So the symbolic power is the power if and only if the power has no embedded primes. So essentially the thing, the power only has the minimal primes which are actually the same associated primes as the ones that I have. You didn't gain anything by looking at the power. I do note that if I take a radical ideal, so what's a radical ideal? One way of describing being radical, yeah, you can see you're equal to your radical. But another way to say that is your e-code to the intersection of your minimal primes. So these are the minimal primes. When you have a radical ideal like this, what happens when you take I and you localize? I hope you can't hear her but my dog just decided to start barking very loudly. So actually, let me say for simplicity, let me say I take any of these PIs, any of these minimal primes. What happens when I take I or even I to the end? Actually, let me write I first. What happens when I take I and I localize at a minimal prime? Well, all the other components are certainly not contained in that minimal prime, right? So they're all going to explode and become the whole ring. The only component that gets saved is the PI component. So this is the same as taking that prime and localizing it. And now, of course, same thing happens when you take powers. So notice in particular, let me see if I can show you everything at the same time. So now here, when I compute these things, it's the same thing as taking the power of the prime and localizing that and contracting that. So in other words, if you like, we could have defined the symbolic powers of a radical ideal as just taking the intersection of the symbolic powers of the primes, right? So you really only need to know what symbolic powers of primes are in some sense. Now, before we do some more examples, there's a few more comments to make. So so far we've, I think we've said that computing symbolic powers is something about understanding associated primes. So another way to describe this, completely equivalent. So in fact, let me say this like this first. So to compute the Anismbach power of I, I compute a primary decomposition for I to the N and I want to delete all the embedded components. How do you delete certain primary components of an ideal? That's taking a saturation with respect to something. So let me write that down explicitly. So this is, I guess, a separate definition if you'd like, it's kind of a side note. So if you take I and J ideals, the saturation of I with respect to J. So this is kind of like a column but there's an infinity of here. So what you do, you take the columns of I with the powers of J. These make some ascending change. My ideal is an ovarian so that ascending change stops. So whatever is that largest of these columns, that's your saturation. So really you're asking about elements, the sand like J to the N inside I for some N. But eventually you send all the powers of J inside of I. So you are going to show, this is one of the problems you're gonna turn in. And it's not very hard, but it's I think as good as a science to think about. You've never done it before. You're going to show that what such a reading does, wanna write it in words and it's written in math in the notes. What this does is this deletes primary components, certain primary components of I. And which ones is it delete? It deletes the ones, let me say it like this. So with the primes corresponding to them, so associated to primes that contain this J. So to compute a symbolic power, what you wanna do is you wanna delete certain primary components. So you just have to find an ideal J that is sort of contained in the right primes and not contained in the right primes, right? So that it deletes exactly the correct components that you wanna make disappear. So at the end of the day, any symbolic power is just a saturation. So if you wanna make statements about symbolic powers, you can also just study saturations and you're proving statements about symbolic powers as well. Now, we can think about this as sort of a way of computing symbolic powers would be, go find the correct ideal J to saturate with. And then you're in business, saturations are really easy things to compute. The problem of course is finding the correct thing to saturate with is not such an easy thing to do. But in fact, you're going to, one of the problems you're gonna do, no need to copy this, it's in the notes. One of the problems you're going to do is you're gonna show that there exists an element, okay, maybe I should be more precise. So given an ideal I, and maybe for simplicity, we're gonna say it's radical. And you know, whenever I assume our ideal is radical, it's not actually necessary, it just makes the technicalities much easier, okay? You're gonna show that there exists some elements, I'll call T in your ring R, such that saturating the powers with respect to that T give me back to the symbolic values. So saturating with one element is really easy, the hard part is figuring out what elements you wanna take, okay? So I'm really in a way sort of getting ahead of myself because I'm answering a question that I haven't explicitly said out loud, but I should right now. Why are symbolic powers difficult? Because they have something to do with primary decomposition, and primary decomposition is complicated to calculate. One of the difficulties about studying this topic is that you wanna compute examples. Examples can be very difficult to compute. Even if you have a computer, you have Macaulay, if you wanna go through primary decomposition, good luck, right? If your power is low, okay. If your ideal's gonna ring with, not very many variables, let's say, okay, but things get very difficult very fast. So going through this definition is tricky. If you have such a T, oh my God, that's amazing, right? You just have to compute a power and computing saturation is super easy. It's one of the easiest things to do computationally. So for some types of ideals, one can say what this is, but in general, in general, in general ring, the story is much more complicated. But okay, take this as a side note. This is something you're going to do in the exercises. I do wanna point out though, that one thing that helps, okay, maybe I'm gonna point out one thing that's bad and one thing that's good. So I think that's bad. Associated primes of powers are difficult to find. Things that's good, you have a little bit of control because okay, maybe first, let me say problem, the function that computes the associate primes of power is very difficult. So like for example, when you increase the power, the newest associate primes don't even have to contain the old ones. So crazy things can happen, all sorts of crazy things, even in nice rings. But one thing that is really nice is that actually when you compute, when you look at all the powers together, even though each power might have an insane, crazy set of associate primes, there's actually only finally many among all the powers, which is a relief, I think, right? It makes things work a little better. So the fact that there's finally many associate primes is a theorem of atlas. Actually it's a line and a proof of something else, people by atlas. But as I know, this is the first time this appears in writing, although if you look at those notes that I was mentioning, I think the proof is much longer than one line. But you know, it's sort of written by Eckhardt-Simmons paper. But then Rodman proved something stronger. So Emmy attributed this to Ratliff in 76 and then Rodman in 79. They show that the set of associate of primes. So this first statement is Israeli Ratliff. This set is finite. And then Rodman showed more. So I was just telling you that when you go from one power to the next, you don't have to contain the old associate of primes, right? I mean, you contain the minimal primes for I, but then crazy things can start happening. But Rodman showed that there eventually the associate of primes stabilize. So if you go look high up and up, the set of associate of primes becomes constant at some point. But up until that point, where they become constant things can be crazy. There's not that much in the literature about what this point, like this stability index is. You'll find some results in a paper by Susan Morey. But in general, it's really hard to say where this stability begins. Okay, so I've been talking about the sort of abstract idea of taking symbolic powers, but I really owe you examples and properties. So let me start by writing properties first because it'll help us with computing examples. So properties. I won't prove these properties. They're all elementary. Anything I don't prove, I should say, you can find them in various places in the literature. One place I can promise you has every proof of every theorem I will say in these lectures is those notes, those long notes that you'll find on my website. So here's some easy properties that you can try to do on your own. Very good. So one, I already said this and this follows immediately from the definition, but think about it for a second if you don't see it right away. Symbolic powers at least contain powers. The first symbolic power is I itself. Well, that's, you know, caveat, right? I'm assuming my ideals have no embedded prime. So in fact, let me just write to make this quick that my ideals are optimal. If you take a higher symbolic power, so if A is bigger equal than B, then your symbolic power gets smaller the same way that the ordinary power gets smaller. And in fact, maybe let me write this number four, I'm not quite following the order, this is in my notes, sorry about that, I just switched to, okay, let me say that R is a domain and I is non-zero. So if you happen to know that you have a power inside some symbolic power, well, we already know this works if you take the same number. So if you take a power that's bigger, this is certainly going to work, right? But this is in fact, and if and only if. So A must be bigger equal than B, for this to make sense. And I'll finally, one that we'll talk a little bit more about today, although I'm going a little slower than I intended, so I should pick up the page. This is the greatest family of ideals. If you take a product of something and symbolic power by something, not a symbolic power, land inside a symbolic power corresponding to the sum of the orders of those two. Now, something that I want to write separately, these are all elementary, the thing I'm going to write next is not elementary, although you can prove it in many different ways. It's sort of a fundamental fact about symbolic powers is that sometimes the symbolic powers are indeed the powers. And one situation when that happens is when your ideal is generated by a regular sequence. So for simplicity, let me say I have a radical ideal, complete intersection generated by a regular sequence, then the symbolic powers are found. Let me say something really fast. Let's maybe explain a little better in the notes. The real statement is that if your ideal is generated by a regular sequence in any room, the associated primes of the powers are the same as these sort of primes of the I. So in particular, if you're ring with Conan McCauley, that says that I has nothing better primes. And so it says that it satisfies, we can use our definition of symbolic powers and symbolic powers of powers. So instead of saying that it's from McCauley, I said that my idea was radical, which easily makes this work anywhere. Okay, now with all this, I'm now ready to give you some examples. So let's say that I take this square free monomial ideal. So I'm in a polynomial being in C variable and I have an ideal generated by monomials where the powers that appear are one and zero. So this is square free monomial ideal. Those are exactly the monomial ideals that are radical. Computing primary decompositions of monomial ideals a little easier than computing primary decompositions of some random thing. One way to see that what I'm writing is correct is you can think about this geometrically, if you like to look at things geometrically. This is the ideal of the three coordinate points in P2 and or if you're in a fine space, you have the three coordinate lines, right? That decomposes, why don't I draw the picture? That decomposes as the union of those three lines and each one of them corresponds to one of these three ideals, right? So for example, this is the axis. Okay, so one of the symbolic powers of this guy, well, I have these three minimal primes and I'm going to compute their symbolic powers. So I'll take x, y to the n, x, z to the n, y, z to the n. Now notice these three ideals are generated by a regular sequence. Their symbolic powers are powers. So this is just the intersection of the powers. Now sadly, or maybe not sadly, it's a good thing it makes symbolic powers interesting. Intersections and powers don't commute, right? I can't fool this power out. So for example, if you compute the second symbolic power, now I should have used the copy face but I'm halfway there. So now it's good to do that. In here, it's easy to see that you have the element x, y, z. Like for example, x, y is in here, like z in here, y, z in here. And I don't know, it's either early or late for all of you. So maybe I won't ask the audience for help. But now you've had a second to see that this element x, y, z has degree three. And that's bad because I is generated in degree two and that means that I square is generated. So I square is generated in degree four. So everything in here is degree four or bigger. So there's no way that this element of degree three is in I square, but it is in the second symbolic power. So this is a common phenomenon that you'll see elements in your symbolic powers that kind of live in the wrong degrees to sort of be in the past, right? You'll see, I wanna see unexpected degrees that you weren't expecting to see in these symbolic powers. I have one more example in the notes that maybe I'll skip for now, but it's a cool example to try to do by hand. It can take the two by two minors of a generic three by three matrix. And that's a prime ideal whose symbolic power is not the powers. So in particular the determinant of your matrix is in the second symbolic power of the month as well. And so that very fast, please check out the example in the notes later. I have 10 more minutes, I think. So I'm gonna use those 10 more minutes to tell you two more things I really wanna cover today. So first, I wanna tell you one reason, okay, there's a thousand reasons to take care of our symbolic powers, right? First of all, primary decomposition is one of the most natural fundamental things that can come in rounds with us. So of course we should care about symbolic powers because they come from primary decomposition in a very natural way. We care about associative primes. We wanna understand associative primes as powers. Those are very natural questions to ask. But another reason you might care, you're a denominator in particular, you might care because symbolic powers encode geometric information that the ordinary powers don't. I think of the ordinary powers as sort of the natural algebraic motion of power. And somehow some of the powers that appear naturally algebraically actually contain geometric information. So one way to make that precise is by a classical result of Zewiski and Nagata. So this is something that Zewiski proved half of in 49 and Nagata proved the other half in 62 in his local rings book. So let me say that my ring, let me start over C and then I'm gonna delete C and put any field. So let's say that you take a poem on a ring over C and you take a rack like you. Well, actually let me re-organize this. Okay, this is my fed up for now. What does Nussel and Zat say? Well, if there are various ways of phrasing Nussel and Zat, it tells me that this ideal I is the ideal of all polynomials that vanish along the variety that I divide. It tells me that for each point in my affine space it corresponds exactly to a maximal ideal. So to say that I vanished at all the points in this variety, so to vanish at one point is to be in a certain maximal ideal this means this is a maximal ideal. To be a point on that variety means the corresponding maximal ideal contains the ideal I. That means to be a point on the variety that I defined. And this is to be a part of an I to vanish along that variety means that you vanish at every single point in that variety that I defined. But each one of these is like going over every point in that variety. So in fact, the statement I just wrote, you know, if you're now going, don't interpret it. Like, of course I could put any algebraically closed field and give the same interpretation I just gave. But actually it's true more generally that erotical ideal is equal to the intersection of the maximal ideals that contain it. That's just saying that this is a Jacobson thing, this polynomial, but this is through a variety of fields. So the Zarisky and Nagata theorem is a high order version of this. I'm gonna add, I'm gonna decorate this equality with symbolic powers on the side and powers on the side. So you should read as the saying that to be in the end symbolic power is to be a polynomial that vanishes to high order. Let me say along Viva. So think of this as saying that at that point of course point to F, you vanish to order, right? That's sort of the interpretation. You wanna give this in the algebraically closed field saying. Okay, so one can also write this statement in a sort of a projective looking statement. So if you take a homogeneous ideal, you would then instead of maximal ideals here, you could put the ideals of each point which now are not maximal anymore, right? They have like height one less than the dimension of the ring. I could write that explicitly, it's written explicitly in the notes. But since I only have six minutes, I'm gonna use those six minutes to tell you something else. So if you wanna see a projective version of this theorem, please, please check out the notes and you'll find a proof in those long notes in my website. Okay, so on Wednesday, I wanna talk about efficiency P problems. So today to prepare to get there, I want to talk a little bit about what are some of the big open problems related to symbolic powers. And in all fairness, I'm saying something that the problems I'm going to list here are a little more than just open problem. Like if sort of some of them are too open-ended to ever get a complete, you know, nice solution, but sort of the types of questions that we wish we knew more about, right? So here's some problems. Oh, I should write the word open because otherwise there's a danger that it sounds like these are the questions you need to solve for your homework. You should not, do not attempt these at home. Okay, I mean, do but not this week. Okay, so here's one problem. What to me is maybe the first obvious problem is equality. When is a symbolic power just a power? It's just kind of like the best case scenario, right? If the symbolic power is just a power, you know, everything you could want about it. You can say which polynomial is finished or what are in along this variety, right? You can tell me what degrees they have. You can tell me anything I ever wish to learn about them. We have a sufficient condition already, right? We said this always happens when I was generated by write the sequence. You know, module, technical things. And so is that an if and only if? No, not at all. Very, very, very far from being an if and only if. There are many other conditions that might lead to this equality. So you could take this question as how do I characterize which ideals are I that satisfy this equality for all N or you might ask, you might hand me an I and ask for which ends this is true. Is it true for any N? Is it true for all? Is it true for some? You might have equality for some values and not for others. Again, there are all things you can find examples of in my notes, not the notes for these lectures, but the longer ones. And there's many nice theorems to cite here, but I have limited time. So I would at least cite one theorem that is not a statement in character CP, but it's a very recent result. And it was proved via characteristic P like techniques. So sometimes you can apply, you know, you learn all this characteristic people stuff and then you find out, you can sometimes use it ideas like that in characteristic zero as well. So here's an example of that. It's the recent results of Jonathan Montaigneal and Luis Munez Fethicula. This is a 2021 paper that's the publishing date. So you take a field, you take a polynomial ring over that field and you take a square free monomer. You would think that this is easy dealing with structure monomer like deals and polynomial rings. What is there to know about some of these things that we don't know? Well, lots turns out. So maybe take this also as the kind of things we didn't know until 2021. Okay, so this came in as the following. Suppose you've tested the equality of symbolic powers and powers for small values of N. So you went up to the minimal number of generators over two and you took the feeling that's sufficient to show that you have equality for all N. So this answers the question, can I test equality by testing it up to a point? It's given me a finite number of steps to test that will guarantee that I have equality for all values. We don't have such a result even in a polynomial ring. We don't have such a result for all ideas. We have such a thing in dimension three. Wow. There's no theorem correct me if you the affairs that in dimension three, you're either equal, the second symbolic power kind of determines everything. They're either all equal or they're all different. But that's very special, right? If you take more numbers of variables, we don't have a theorem like this in more variables. But the proof of this theorem sort of involves techniques that look like characteristics. You're inspired by Provenius like things. Okay, so my clock says that it's 9.50. I think I've started two minutes late. So I don't know, is it okay if I speak for two more minutes? Or should I stop here? I can also stop here. It's okay if I go two more minutes? Okay. Yes, you can go in for two more minutes. Okay, great. Well, I think I started at 9.02. So I'll go two more minutes. Okay. Oh, no. This one, let me say something very vague. What is the smallest degree of some elements in i-symbolic again? So if i is homogeneous, it's easy to prove the symbolic powers are also homogeneous. So I have a homogeneous ideal and some graded rain. I wanna ask about the degrees of things in the symbolic powers. Notice that since the symbolic powers contain the ordinary powers, giving upper bounds on this degree, on the degrees of things, it's easy, it's giving lower bounds this far. There's so much to say about this question, but my time is limited. So I'll just mention that's an interesting question that people think about. And let me mention very quickly that another big question is to study what is called the symbolic phase in algebra. And here's what might be a surprising difficult question. When is this a finitely generated or algebra? Which when a ring is excellent, is it equivalent to saying that this is in the beginning? And this is also interesting and difficult. This is sort of asking, can you describe all symbolic powers by describing finitely meaning? This is some graded rain, thanks to the fact, the last property on symbolic powers there, but thanks to this fact, that's a nice graded rain. And this being finitely generated is kind of saying finitely many symbolic powers describe all the symbolic powers. And this can fail. So you might have something where this is not an ordinary rain, even for primes in polynomial. And there's a lot to say about that. I may mention this briefly again next time. So what we're going to do on Wednesday is we're going to talk about one problem in particular and some theorems that have improved using character 60 techniques. And so we'll see in particular why character 60 techniques are so useful to prove things about symbolic powers. And now I'm definitely over time. So thank you. I'll stop here. All right. Let's thank Louisa for the nice talk. Are there any questions? All right. So if not, let's thank Louisa again. Thank you. All right, so let's move on again on Wednesday. Yeah. And you are all welcome also to email me questions. Although I am quite busy and this week in particular and I don't promise that he won't take me like a day to reply to you, but I'll turn on this or you can ask him questions on Wednesday during the lecture. Thanks. I'll see you all then. All right. Bye. I think you can stop there with Quentin.