 Let's start the recording. Thank you. All right, well hello everybody. Welcome to back to the class. And let me introduce, reintroduce Thomas Polstra who will be talking today. And continuing to talk on applications of big color modules. Awesome. All right, wonderful. Thank you very much. So, yeah, so today's goals is. Right, so the first big goal is, I want to, right, finish off some applications of big color Macaulay modules and algebras to the homological conjecture specifically want to go over a proof that direct summands of regular rings. Get to enjoy the property being called Macaulay. We will do this via the so called banishing maps of tour conjecture. And after that, I'll spend a little bit of time introducing these so called BCM rational singularities, which one Ma and Carl sheet that are studied and defined in terms of they call it Macaulay algebras. And how exactly they mimic the, the, the, the theory and the study of a rational singularities and characteristic P. All right, so remember the, the greatest tool at our advantage is Andre's theorem on the existence of so called weekly functorial big cone Macaulay algebras and I wrote it a little bit different here than what I wrote last time so I wrote it in a bit of a stronger form. So I just wanted to put the strongest form up here so it just says just take a map of complete local domains. And suppose that you're given a big corn Macaulay our algebra, and such algebras always exist for such rings. And for any choice of big corn Macaulay algebra, there's going to be a diagram. This little commutative square that's on your screen. So that the, the B sub S is a big corn Macaulay S algebras so like, not only do you have weekly functorial big corn Macaulay algebras for ring maps R to S, you know, an arbitrary ring map. You, you have weekly functorial big corn Macaulay algebras, based off of any big corn Macaulay algebra on your source ring are the really quite an impressive statement. So, okay, so, so we're going to use this and right the goal is is we want to forget. We want to use this and want to show. So the goal. So, at least the way I'm thinking about the goal here is that if R to S splits. R is C L. Then, or excuse me. That's the conclusion. Passes regular regular. Then R is from the college. And so the idea behind doing this is what we're going to be able to notice is that by standard reductions we can reduce to the case. R is a complete local domain by standard reductions. This splitting out of a regular ring. We'll get that at least R as a product of domains but then you can check the color Macaulay property on each of the products. And the color Macaulay properties of course a local property and the local properties a complete property as well. So it's kind of the heuristic of the reduction there. In which case, we have code structure theorem. It tells us that R is module finite over a regular local ring. So there exists a nother normalization. So a to R finite map of complete local domains for night such that a is a regular local ring. And I suppose that the, you know, say for example the maximum ideal of a is generated by the elements. Right. Then X one through XD is a system of parameters on R on R. And we want to show, we just call this X bar underline X underline X is a regular sequence. And here's somebody's cameras on in the background so that happens to be just turn it off. And we want to show X bars regular sequence on R. Now to do this. So to do this, it is enough to show. It is enough to show or equivalent to show. So the I casual homology of R equals zero for all I greater or equal to one, where H bar equals I casual homology group. All right, just as a reminder, but one way to realize the I casual homology group of a right of us of a system of parameters on a ring R where that system of parameters is the extension of say the maximum ideal. I'm a regular from another normalization or regular suffering. This is these casual homology groups are actually come down to a tour computation. So you could first just take a free resolution of a modern maximal ideal or a mod these X's by the casual complex on a or with respect to X one through XD on a, and then that's a free resolution of our X or the casual complex, you can tensor that with R to obtain the same casual complex on R. And then you take a knowledge. But right in that process, we started off with a free resolution of something with tension with something. And we took homology and so what we're really doing is computing a tour. So this is why this, you know, these tour computations are going to be showing up. So a mod X bar, actually kind of mimic the way I'll write things in terms, let's put R in the first slot X bar, right, but it's just the residue fields of a particular that is a regular ring. I mean, for what it's worth, but and then what toy is isomorphic to. So what we're really trying to do is get to some certain core modules to vanish and that's where this vanishing map of torque and texture comes in. So, so the theorem for texture. And I'm going to present this in not the most general fashion, but let's suppose or let's a to R be a finite extension of say, complete local rings, let's even do domains and a is regular ring. R to S be any ring wrap to a regular ring S particular will apply this later to as a regular ring for which are split out of to a regular regress but this could be any arbitrary graph doesn't matter then for every a module m the induced maps of tour. So the tour I a are to tour I a S is the zero map zero for all I that's this can be a theorem here so take a finite extension of complete local domains take a to be a regular local ring so thinking of like a north or normalization for a complete local ring. We're going to take our map it to any regular ring like and then for any a module m we're going to have a set that is canonical map of tour modules as functora map of tour modules has to go away. It's it's a trivial map. So here's the proof. And you'll see the I think the true power of not only big column of college modules in here but the algebras. So, here's the proof. I'm right there. So of course you go like this so so there's a reduction to ask is to be a complete regular local ring is as well. And so the point is if you look at this tour module right here tour I s so so first and foremost I assume him is finally generated. So assume him is a finally generated a module and this is achieved by direct aren't direct limit argument. Okay. Then. So, so we always assume it was finally generated on top of this. When we look at s we want this map to vanish the vanishing is you know we're looking for the image of the map to vanish and this target here which is a naturally an s module and something being zero is a local condition and right that local condition being zero can be checked after completion. And that's the that's going to allow us to say that we can also assume as well that s is a complete regular local and now we're going to set up where we can use this Andre's theorem a weekly punctorial big column of collage with respect to R and s. All right, so there exists a diagram. All right, so by Andre s s we have a great show. So we have this is a finite her. All right. So, R to BR say BCMR algebra. Okay. So this implies since a finite. So since eight are that a two piece of R is a BCM a algebra. And this is where I think one of the true powers of big column of Holly modules comes in this, you know, I think it's a very important to remember. So why are college modules just so useful. Well, the point is, is that over any north or normalization, that big Colin McCauley module is a big Colin McCauley module over regular and big Colin McCauley modules over regular rings are automatically flat. That's going to be one of your exercises so so a to BR is actually flat since a is regular to this statement that if you're actually if you only assume a is Colin McCauley and you have a flat algebra, it's automatically Colin McCauley. That's a, that's also going to be an exercise and that's easy. That's a bit easier. And so this this statement right here is going to be exercised for and so the point is then is therefore so tour I a of the sea to get this right so B are has to be zero all I very cool all right so so the diagrams above so there exist can be the diagrams so we have diagrams right so we're going to we have our tour I a r m mapping to tour I a assume right this is the map we want to be zero that's what we're trying to get to be zero. We know this maps up to but out of the big Colin McCauley module and the left piece there. Right and we said that this has to be zero. So for all one right this is important here. And then they have this score I a piece of s you know but now we're going to use that our hypothesis that as is regular as well so as to be as this again. I mean by the same reasoning but this is a different regular ring. And so this map so right here is one to one this horizontal arrow is the right horizontal arrow is one to one. I'll call that star so so star is injected. And that's going to be enough to prove what you want so right we want to show that this maps a double star zero. Well the point is right at this double star if you map up through this injective map up to this top tour factors through a zero module and so it has no other option but to be zero. So right so a couple things happen there right so the community diagram of the weekly functorial to your call McCauley algebras but then more just as important. Very important fact if somebody ever asked you for some strange reason why are they calling the college modules nice well, they're automatically flat over any normal music. One of the more. I think my internet's coming out. Okay, I think I should be back. So now let's prove direct summands of regular rings. So theorem. So direct summands regular rings are she. So, okay so let's say R to S splits to assets regular. All right so standard reductions right. R is complete local domain. And we can have this module finite extension so a to R finite. Hey, those regular local ring. So I'm assuming ours complete domain here to make this happen. And so, what we know is is that okay. So we know so by vanishing maps of tour or we're going to have that tour I a of R against say a modest maximal ideal to tour I a S a modest maximal ideal is the zero map. Great. But this thing is actually split splits. It's injected in particular. But this map is injective. And that is just simply because our test splits. And so that's that should prove. Right because remember this is the ice cosmology on a regular sequence. And this is of course for a high degree. That's the proof that direct summands of regular rings, always get to be calling the colleague. Okay, so that's, that's the end of the discussion I wanted for the homological conjectures and so I want to move on now to these so called BCM rational signatories. Thomas can I ask you a question with a quick. Please. I think you just proved you are not assuming anything about the characteristic correct. That's right there's no assumptions on the characters. Okay. Yeah, yeah, yeah, yeah, so, I mean this was so amazing about this theorem I love the students remarkable statement. And it is truly the case that this theorem and characters to P is 100 times easier than the other two cases. 10 times easier than the characteristics zero case and maybe a thousand times easier than the mix characteristics set up but like, but, but there's no assumptions on the characteristics of us and so like I mean this could be maps. So, you know, well yeah, exactly. Exactly just saying there's no assumptions on characters. Okay. So BCM rational signatories are big Colin McCauley rational singularities so all right so let's recall something about local homology and characteristic team. So let's we are going to recall so let's say a local ring, say dimension D. Let's do a prime characteristic thing. So the the Frobenius induces a Frobenius action of local homology. I mean, it does this and all the local homology modules but I want to focus on the top one. All right so in particular I mean how does this action work just as a reminder so right so if we choose that identification of this local homology modules, you know kind of a limit system based upon a system of parameters. X1 through XD. Then the class of something that looks like our plus say X1 to the T up to XD to the team. And what is, what is the Frobenius action do well shouldn't be too much of a surprise but the class represented by this element and the direct limit system is going to be mapped to the class of the direct limit system represented by art of the P plus X1 to the P. And so, and then so right so definition. You know for being a stable sub module, if stable sub module in containment. Is sub module such that the Frobenius on a DMR maps in back to itself. And so the, the, the, the theorem here is that. So the theorem right so some theorems to Karen Smith so I'll list a few really important facts here. So let's let's take arm K to be a fleet local domain characteristic team. All right so. So maybe fact one is that the tight closure of zero. There's a unique. The tight closure of the zero sub module, the top local homology module is a Frobenius stable sub module. And moreover, not only is a Frobenius stable but it's the largest one with respect to inclusion right and this. You know you should think about this for we're talking about an Artinian module. And we're finding a module in a class of sub modules, namely the Frobenius stable sub modules, and that class of sub modules admits a maximum element with respect to inclusion it's a really phenomenal. So zero star hdm r, which, all right, so. Okay, is the largest for the stable sub modules, so which I'm going to say a couple things here. Gonna be the kernel. Okay, so the largest so the tight closure zero. Two important things you already said one out loud and write it down a second but also not only is it the largest for being a stable sub module but it's a kernel of a particular map. It's a kernel from the top local homology module bar to the top local homology of the absolute interval closure of our somehow the union of all finite extensions of our motor bar. Zero star. Largest f stable so. And in conclusion, a couple things come out of this so. So, if our CM, the following are equivalent. So what comes out of this is that a ours a rational and to. And then the other thing that is actually going to come out of this as well there's another equivalence to just being a threshold like, which is going to be the following see that hdm r to hdm b is one to one. For all BCM. And so one of the key points here to prove B imply see so let's see why B imply see. The point is is this is that if, if R to be is a BCM algebra. You're going to have two things happen. One, the kernel of hdm R. R to be the title. Oh, thank you. You're right. Thank you so much. Alright, so far to be is a big home to call algebra than the the kernel of this top local homology modules is an f stable sub module hdm R. And this does not require, require B to be BCM. This works for any algebra is an exercise. And then to where does the BCM part come in is you know that this thing is a this map has to be non zero. And this is an exercise. Right, so let's say you're in this situation of R to R plus induces an injected math and hdm R to hdm R plus, meaning you know that ours f rational and that the kernel of that map. Not only is it trivial but it's the largest f stable sub module. So once you know that, you know, this number two that the map is non zero, and you know that they're the only f stable sub modules are zero and the whole module automatically forces, right, the, this, this map to be injected. So, all right, so thus zero. So our is your is a injective one zero and hdm R the only stable so much as that's that's all the information you need to know to conclude that hdm R the hdm BS injected. It feels to me the right way to to extend the notion of that rationality and characteristic free manner to all complete local domains is the following definition so this is definition to lunch on law. So, a complete complete. So, the Big Column Macaulay local domain is called BCM rational standing for Big Column Macaulay rational if for all BCM algebras to be the induced map of local columnology at the top level. And so, you know, and so a couple things that happen with these senior leaders are only two things will prove for people to talk about so. So one exercise. This will be one of the exercises that's assigned to you. So exercise 10 in the notes says that BCM rational implies normal. So every F rational ring, for example, is known to be a normal domain and so hopefully this type of thing they stand to BCM rational and it does and there's a two part exercise here that outlines what to do. And but the one I want to focus on is actually I it's it's originally in an exercise in the notes but I want to actually work it out. So it's a really nice proof to very beautiful proof so. So it's going to be exercise 11, which is going to be the statement that all BCM rational singularities, the form. So, so Paul. So a singularity class to form so something like. Key of notary rings. So think of, you know, properties such as Colin McCauley, Gorenstein, maybe the property being regular property being a pure. Whatever. Adjective use tried local domains. So that property of nothing rings is set to the form. If the form, let me say, in a ring R or so the form. And a ring R. If our has property P. There exists a non zero divisor X and R. Such that. Armada XR. I saw examples of this. Right, so the properties of saving regular. Gorenstein. Colin McCauley. All the form. If you give me a non zero divisor ring R and armada X happens to be a regular local ring and a R has to be a regular local ring as well. Non examples so famous non example. F regularity does not deform form in general. This is due to honor xing producing some famous counter examples to this. This was an open problem and tight closure theory that was settled in a negative. But it does the form under certain hypothesis. So what are rings which are called Q Gorenstein this property does the format. And if you know what that is great if you don't. And this is a result to have a cat's room in the Kerman. And then of course another very important example is the notion that rationality deforms as well and I believe flooring and asking about this so if you have a local ring of characters. We look at domain of characters and P and armada X happens to be a rational, meaning all system of parameters are tightly closed in that ring. And that same properties enjoyed my arm. And so the proof we'll see that is we'll just perform BCM rational singularities and it's a relatively easy proof. So this is a in the notes this is exercise. Let's R and a be a be a complete. Local domain and X and R a non zero. Suppose that arm at XR is BCM national. Then R is BCM national. There's the other solutions. So, all right, so let's R to be be a BCM algebra. We want to show HDM R to HDM B is one. Now, R and B is a BCM algebra, meaning that B is a balanced big column of collie module over R meaning if I give you any system of parameters on R, then it's necessarily a regular sequence on. In particular, when you look at our mod or excuse me be mod X. That has to automatically be a balanced big column of collie algebra for our mod X. That's what that's going to then fly. So notice that our mod XR to be mod XB is a BCM or our mod XR algebra. And so by assumption, we get to know that this map is injective. And so we'll use this so what we're going to do is we're going to look at a short exact sequence. Let's do the kind of standard short exact sequence involving a non-zero visor. The point is for this short exact sequence we can put on top of it another short exact sequence. All right, so now we have this short exact sequence. This commuter diagram is short exact sequence. And so what we're going to do is we're going to look at the resulting commuter diagram of local collie module. So a little proposition. This will be a little segue and something else for a second. So if M is BCM over, you know, RMK, you know, for all dimension of R is D. So this implies all lower local collie module are zero. But careful, careful, careful here. The converse statement is not necessarily true. You have to be careful. Converse is not necessarily true. But there is at least one very important case where this is true where the converse can be turned around. It's a very important case, which is if. Right, so if you look at the proof, right, that Gennady is going to be talking about here, you know, after me, of R plus being big column colleagues. You can reverse this statement for R plus. So, so this type of statement is reversible obviously for finally generated modules. It's kind of old theory of common collie modules is building up those types of statements. But once you throw in a non finite generation assumption, things get weird, but, but, but I do want to emphasize that this is reversible for M equals R plus. That's a kind of an interesting statement there. Like if you want to prove R plus is big column colleague, it's actually enough to show all the lower local collie modules, but that type of statement isn't true for any module just just our plus. All right, but okay, so let's use this proposition we have lots of damaging a vocal collie. So we're going to apply the proposition to this big old commuter diagram. And the thing that we're going to keep in mind is that everything in site or be our mod X and B mod X, our big column the collie modules over the rings that we're studying them over. So, we apply proposition, proposition to the diagram to get a commutative diagram. The local collie modules. And so, so, so let me just redraw the diagram to begin with zero R times X. Right, so this is the diagram we start with. And so the diagram we end up getting will look like this. So, right, we start going, work our way up the local collie module chain and the first non vanishing one will be the D minus one terms on the R mod X and B mod X. Map into hdm B, this will map into our multiplication by X onto hdm B. All we get to assume is is that this left arrow is one. And we're going to take, let me call this map be right and we're trying to show that he is one to one. All right, so let Ada be an element at the top of the homology mobile bar, and suppose that be a beta equals zero party. So what we can do in this assumption so imagine if Ada is non zero. So let's say we're in a situation where he is not. So if Ada is non zero. There exists a T and the natural numbers such that X to the T Ada is zero. Okay. Right. So, right. So we have this element. Living inside this local collie module. And we know that every element of a local collie module is multiplied, but it is killed by high powers of elements from the maximum equal. And so what we can do is all right, we look at this element Ada, we know that there's a T so that X to the T Ada kills it. And so what we'll do is pick T minimally. So, hey, so pick T minimal replace. Ada by X to the T minus one Ada, and assume that X Ada equals zero. All right. But the point is right at X Ada equals zero. That means this green Ada here is being mapped to X Ada equals zero and so it actually belongs in this kernel. So what we'll call model gene module here. And now you can do a little diagram chase. Right. You have a one to one map here, one to one map here. And so if Ada was non zero inside this module, the image here cannot be zero either. And so that's that's how I'm just include. I can't scroll down anymore, reach into the page. So a simple diagram chase. Excuse me for the poor writing here finishes the argument. I think with that is a perfect place to stop. Thank you, Thomas. Do we have any questions. Go ahead and ask. I don't see any questions so let's go ahead and thank Thomas for several beautiful talks. And put those claps. I never know how to respond to those when I'm the talker, the speaker. I don't know how to respond.