 Can you talk about blow-up algebras of determinants ideas in prime characteristics? Please. Thank you very much, Bern. And I'd like to start by thanking the organizers for this great opportunity. I'm really grateful to be here speaking at this conference. So the title of my talk is a bit longer than this, but I was running. I was short on time, so I shortened it up. I will talk about some F singularities of some blow-up algebras objects, and I'll be more precise later. And I will say most of this is joint work with Jonathan Montaño and Luis Núñez Petancur. And if I have time in the end, I doubt it. But if I have time, I'll talk about some more recent work with the same authors. And Lisa Seche, I'm Matteo Barbaro, but I don't think I'll get that far. OK, so what I'm going to say can be done much more generally than this, but for the sake of simplicity and for this talk, I'm going to stick to the following setup. So k will be a perfect field of characteristic p. And I'm going to work in the graded setup. So for me, s will be a polynomial ring over k in n variables with some positive grading. I'll just write n graded, by which I mean the degree 0 part should be a field, and the variables have a positive degree. I'm going to work with an ideal i inside s, a proper ideal. I'm not going to say it again, but it's always going to be homogeneous. All objects, maps, elements are going to be homogeneous. So if I don't say it, it's intended. And I will assume, again, for a sake of simplicity, that it's a prime ideal. I know i is not a letter for a prime, but in my main application, in my main example, it will kind of make sense. And also, again, it's a reminder that this holds for generally. It doesn't have to be a prime. But for simplicity, I will stick with the prime case. And r will be s mod i. OK, so I'm the first characteristic p talk kind of the workshop, definitely not of the conference of the program. So I feel like I have to maybe remind some of the definitions, even though, of course, everyone here, by this point, knows them. But I'll do it anyways. So r is a pure, or a split, if the natural inclusion of r inside the ring of its pitfruits, this is just a natural map that views an element as itself inside the ring of pitfruits, if this map splits as a map of our modules. So if there's a map back, an earlier map, basically sending one to one. r is strongly a fragular. And here I don't really need strongly, but I'll probably say it anyways. If, for all non-zero elements, in r, the map from r to r1 over p to the e, so iteration of this pitfruits construction sending one to c to the 1 over p, this map splits for e sufficiently large. OK, so clearly strongly a fragular rings are a pure, the converse doesn't hold. And yeah, so the goal of this talk is to study these two f singularities, singularities related to the Frobenius map, for certain blow-up objects. So in particular, I will focus on blow-up objects, resalgebras, and associated graded rings, associated to symbolic powers, which Dale already talked about. So I'll basically look at the symbolic filtration of i. So I'll take the symbolic powers of my ideal i. And they form a filtration, as Dale pointed out, not necessarily an aetherian. And associated to this, we can construct the symbolic resalgebra. Let me just remind the definition. And the associated graded ring. Symbolic associated grid. That's part of my point in making this assumption. I'm not worrying about definitions of symbolic powers for non-prime ideals, I'd say. But it can be done. OK, so since this question will be coming anyways, probably. So I don't need it here, but for simplicity, let me assume that i is an aetherian filtration, so that these rings are actually in the aetherian. I don't need it for some of the things I'll say, but just in case. I mean, again, in the main application I have in mind for today, these will be in aetherian. So it doesn't, for now, harm to assume that they are. All right, and then at this point, I'm going to make a definition, which is going to be a key definition for this talk. So I'm going to say that i is a symbolic of pure ideal. There exists a map, which is a splitting, so which sends one to one. Polynomere ring is regular, so it's definitely any of those, so we can find this map, which in a sense respects the filtration. In the following sense, can people see back there? Down here, is it a bit too low? OK, maybe I'll just move it up. Thank you, Claudia. Yeah, I forgot about this. But maybe for later, I wanted to stay here. But OK, but you're right. So you're right, you're right, you're right. If this happens for every n greater or equal than 0. So basically, when you take a certain symbolic power, and you raise it to the power 1 over p, so this is an ideal of the ring as 1 over p or a submodule, if you want. So if you restrict the map to that, you will send this symbolic power into an appropriate symbolic power for every n greater or equal than 0. So if there's some compatibility with this filtration. So let me point out the first immediate consequence of the definition, that if you set n to be equal to 0, then you immediately get this. So in particular, this gives a splitting of r. It has a phi induces a map, which I'll still call phi, still sending 1 to 1, because that's what phi does in s. So in particular, the quotient has to be f pure. So this is an immediate consequence of the definition. But more happens, because of course you have that for every n, and you see where I'm going, I think. So let's look at the symbolic result, just to fix ideas. We have s in degree 0, and then you have i in degree 1, and then you have i squared, symbolic 2 in degree 2, and so on. Down here, you have the same to the 1 over p. I should say this is graded with respect to tn graded, and this will be n over p graded. So degree 1 over p is i. Let me skip a little bit. In degree p, there's symbolic p to the 1 over p, degree 1. And then maybe one more. It's all to the 1 over p. So phi induces, I mean phi is defined here. But then phi sends this guy into here. So you get an induced map here. i to the p is contained in i. So i to the p to the 1 over p is contained in i to the 1 over p. So basically, all this chunk is mapped here. The next p elements will be mapped here now by choosing n equals 1. You get that this is in here and all the next. So basically, you get a map going up. So phi is a map from the reese algebra to the 1 over p to the reese algebra, which is, I mean, I guess, s linear in principle, but it's really linear over the reese algebra. And it sends 1 to 1 by construction. So it is a splitting. So the reese algebra is f pure, the symbolic reese algebra. And the same argument with a little more effort, but kind of the same argument works for the associated graded, they are f pure. So this justifies at least the terminology and the definition. It's a symbolic f pure ideal. If it satisfies that, and as a consequence, you get that the symbolic reese algebra and symbolic associated graded ring are f pure. All right, now, I mean, I've given a definition. I told you why it could be interesting. Now, of course, it might never happen, right? I mean, it could be like a not really useful definition. So the next goal is to convince you that this does indeed happen. And to do that, I need to do a review of Federer's criterion, and most actually, it's proof more than the statement in a way, and then see how this translates into this setting, and we move from there. Let's review Federer's criterion. OK, so let me call m maybe the maximal ideal of s. All right, so r is a pure, or a split, let's say. So, well, we want a map from s1 over p to s, which sends i1 over p into i, and which doesn't send, let me phrase it this way. It's a bit odd, but let me phrase it this way. It doesn't, I mean, of course, if this happens, it will not be surjective, and it will not be a splitting, but in this case, they're equivalent. Remember, everything is graded here. Maps, or elements, ideals, it's all graded. And now, a different way of saying this, using the fact that s is a polynomial ring, is that there's an element, homogeneous, which sends i inside i to the bracket p. This is this compatibility condition. And f should not be in m to the bracket p. So this is the loss. So Federer's criteria is telling you that you can find an element in here, not in here. OK, so now let's try to use this idea to rewrite this statement. OK, so i will be symbolic after. Well, if we can find an element which, for every n, sends this ideal to the 1 over p into this, so what I want is an element inside. This is the same requirement as this, but now with an ideal which is not i, with two different ideals, I guess. I guess if you plug in n equals 0, you get exactly this column over here. And we want this to happen for every n. And we want this to be not inside m to the bracket p for the same reason. So symbolic of purity in terms of Federer's criteria, let's say, or revisited is this condition. Now, of course, this is not really an ideal condition to check. You have to check it for every n. And it could be complicated. But OK, this is a conference for Craig Uniki and Mel Ochster. So I'm happy to cite at least one result of theirs. And this is a result I really like. So let's set h to be the height of the prime of our ideal. Then there's a uniform symbolic power that we can stick inside all of these column ideals, which is i to the symbolic h times p minus 1. So it's a use of the flatness of Frobenius, of course, and a pigeonhole principle type of argument. And it's in the Ochster Uniki. I mean, not exactly this statement maybe, but these ideas are from the Ochster Uniki paper on symbolic powers, containment on symbolic powers. OK, so what's the point of this? Well, if we want to show that this is not in here, then all we have to do is show that this ideal, this one, this single ideal is not here. So this will be our goal to show that something is not symbolic or pure. And we can do even better, in a sense. So let me show you what more we can say. Suppose if we can find an element which I'll call g and a monomial order such that g is in i to the h, but the initial form of g is a square free monomial, then I claim we're done, meaning i is symbolic and pure. So we reduce the question to a one element question. Why is this? Well, it's really simple, because if we set f to be equal to g to the p minus 1, then on the one hand it will be in the correct symbolic power. And on the other hand, it will not be in m to the bracket p because of this condition. I mean, this monomial to the p minus 1 will be in the support of it. So all we need to do to get symbolic and purity is to find such an element and such a monomial order. So that will be our goal. Our goal where? Well, as I said, I have a main application in mind for today, which is the case of ideals of minors of a generic matrix. So my ideal will be i sub t, will be the ideal of t minors. OK, so as I said, I mean, maybe i being a prime looks a bit less weird now. So i is a prime, i t is a prime, ideal of s. And in fact, the quotient is known to be strongly fraggler by another result of unic and ox and unic. But I'll get to that later. And OK, so in this setup, our goal is to construct this element g. And rather than doing it generally, I'll do it on one example. And then we should, I mean, from there we induct. So let's say m is 4 and is 6 and t is 3. So I'll be looking at the 3 minors of this 4 by 6 matrix of variables. OK, so I'm going to draw some lines and then explain what they mean. So starting from the first, so moving along from top left to bottom right, I start with the first 3 minors and then I have a 3, length 3, anti-diagonal, then length 4, and so on. So my polynomial g will be 1 times delta 2 times delta 3 times delta 4 times delta 5, where the deltas are the minors associated to this anti-diagonals. And I take the maximal minors, meaning delta 1 is 3 by 3. These 3 are 4 by 4. And delta 5 is again 3 by 3. OK, so this by construction is in i3 squared i4 cubed. Now h here is 8, so I need to be able to say that this element is in i3 to the symbolic h. Now there's a general statement if you want a consequence of the Ritz-Kinagata type of theorems, or just prove it directly, which says that if you trade this sub-index, the size of the minors, and you reduce it, then you gain symbolic powers on top. And if we use it here, we have that this is in i3 squared, and i4 is contained in i3. So each copy of i4 is contained in i3 squared, and I have three copies of them. So in the end, I get into i3 to the 8 symbolic, which is what I want. And the point is, this works in general, not just for this example. So to conclude, I need to show you what the monomial order is. And for this, we can take any anti-diagonal term order, which means any order, monomial order for which if you take a minor in your computer initial form, you'll get the anti-diagonal. For example, you can choose Lex, just Lex order from this box left, and then moving down. So that would work, for example. But there are other choices possible. Well, if you do so, the initial form of G will be, well, the product of these lines. So it's the product of the entries corresponding to the lines, which is a square free monomial. And so we're done. So the symbolic resalgebra and associated graded ring of ideal of minors of a generic matrix are pure. Now I should say that the symbolic resalgebra, first of all, is netherian. So these are all netherian filtrations, which was my point before. And second, the symbolic resalgebra was known to be effrational by work of prunes and coke. But that doesn't imply if pure anyways. And there was no mention of the associated graded. But also, we can prove more. And I want to take the last five minutes or so to show you we can actually prove strong effregularity. So this is, I mentioned at the beginning, but it's with Jonathan Montagno, Luis Núñez Petancor, and myself for ideas of t-minors. We're in this setup. OK, so how does the proof go? So let me go back to this picture. You'll notice that I started at the three minors. But I really could have started earlier, meaning taking this one minor, these two minors, and also the last ones. Because in terms of being inside the symbolic power, that's even better. I mean, the more factors you put, the better. It wouldn't help, but why not? And also, in terms of the initial form being square free, that wouldn't matter. Because I would get this monomial, the product of these two monomials. So it would still be square free. The reason I didn't do it is because I want to use a criterion of Oxford and Unique, which will give strong effregularity of these objects. So what's the criterion? Well, let me call, let me. I don't want the camera to go crazy. So I'll just redraw the matrix. I'm going to call delta this two minor, the top left two minor. I'm going to use maybe like a wiggle for the anti-diagonal. So, well, clearly, delta times f is still in I symbolic, already generally, it symbolic H. It's really in this example, I3 to the 8, but of course this works in general. If f was then to the p minus 1. I'm taking f, which is g to the p minus 1. It's the same notation I used before. So if f was then delta times f clearly is, but also this element is not in m to the bracket p. Why is that? Well, again, because of initial forms. If you compute the initial form of this, you'll have a square free monomial to a p minus 1 times the anti-diagonal. Maybe since I introduced this notation, let me use it. It's pretty clear, but. So you'll get the initial of f and then you'll also get the anti-diagonal, but it's still not in m to the bracket p. So that guy will not be in m to the bracket p. What's the effect of this? Well, using this framework, there are maps for the resalgebra and the associated graded rings, which send delta to the 1 over p, seen in degree 0, depending on which object you look at. It will be maybe a class or something, but it doesn't matter. In appropriately interpreted, this will go to 1. So there is a splitting of this element. And finally, and with this, I will conclude, s mod i t localized in delta is a regular ring. The singular locus of this is the ideal of t minus 1 minus, this is a t minus 1 minus. So this fact, appropriately used, tells us that the symbolic resalgebra localized in delta and the associated graded ring localized in delta are strongly irregular, not necessarily regular. This doesn't have to be, but they are strongly irregular. And since they are theetherian, as I was mentioning, since they are theetherian, we can use the criterion of Oxford and Unicke. They're, in particular, finite in this setup. We can use a criterion of Oxford and Unicke, which says that if you have a localization which is strongly irregular, and the splitting of the element you're localizing at, then the original rings are strongly different. And before quitting, I will just say that we can do this for more general filtrations, for, I mean, doesn't have to be the graded setup. And in the recent work, and for other type of determinant objects, like fuffians, minors of symmetric matrices, ankle matrices. And in the more recent work with Lisa Seicha and Matteo Varvara, we're focusing more on the ladder setup, but still determinant objects, but leathers. So thank you very much for your attention.