 So in the third lecture, I'm going to show that direct summand of regular rings are comacotic. And we already showed that direct summand of regular rings are, okay, so how are we going to do this? We're going to sort of split the proof in, okay. So we know regular rings are strongly regular, and we know direct summand of strongly regular rings are strongly regular. So to show that direct summand of regular rings are comacotic, so what remains to show that this strongly irregular rings are comacotic. So this is the remaining thing that we wanted to show. And in order to do that, we need some, okay, this is, if you're, okay, so let's just, so the first thing I want to show is that the completion of a strongly irregular ring is still strongly irregular, okay. And to do that, I need to, you know, so far when we track some ring is strongly irregular, we just say we pick any C and then we track like 1 to F e lowers our C for some E, 1 to F e lowers our C splits. So it will be really, really convenient if we can only track this splitting for one C, and indeed that's indeed the case, but you have to pick this C sort of carefully, okay. So here's a theorem that I want to, it's extremely useful. So let R be an affinite ring of characteristic P, so suppose there exists C, not the animal prime such that R local as at C is strongly irregular, so, okay, for example, regular. Okay, so that is, you find some C such that after inverting C, the ring is regular, let's say. So then I claim that to track R is strongly irregular, you only need to split off that C. So then R is strongly irregular if and only if there exists E bigger than 0 such that the map from R to F e lowers our R sending 1 to F e lowers our C splits. Okay, so you see the theorem is, okay, of course, the only if direction is trivial, right, because I mean, so if R is strongly irregular, so then of course, you know, you, so if R is strongly irregular, so then this should be host for every C, for every C you should be able to find an E such that this much splits. So the interesting direction is the converse. So you only to track R is strongly irregular, you only need to track this one single C. You know, you can find this E large such that this is great. Okay, so let's sketch a proof of this. Okay, so let's give him the D. Not in any minimal prime of R. Okay, so then the image of D is not in. So the image of D is not in any minimal prime of RC. And so our assumption is that RC is strongly a regular. So things are local as C is strongly a regular. So we have D zero. And the map fee in the harm set such that feel happy zero lower star D is equal to one inside RC. Okay, so now, I'm going to use this trick again and again so I feel or are always I find it so I feel or sorry is a finite module. So the harm sort of commutes that this is harm over RC. I should just say, right. So, so things are is a finite. So we know that this harm RC, I feel yours are RC RC, but this is just RC tensor. So I'm using harm commutes with this flat base change when the modules inside are finally generated. Okay, and so therefore in particular, I have my fee look my fees inside here. So fees inside left hand side. So we know fee has to become from the right hand side. So, this fee equals to this phi over C to the end was some. Zero. And some fun inside harm. I feel yours are all. Okay. Right. It follows that if you if you apply this fine to this, I feel yours are D. Oh, sorry, I feel sorry. This is zero. Okay, so this is zero. Right, because sorry. Sorry, this is easy rule. Right, because I said they're just easy real such that, you know, fine inside harm easy real RC or see a search that feed easy real fun this field. I feel it was very easy real D equals to one. So, so this should be zero. This is zero. So this is C to the end, because after I, you know, if I divided by C to the end, I got, I got, I got one inside RC. So this is some part. So this is, this is a C to the end. Okay, so as of now. Okay, so now basically the here is clear right so I take any D, not any minimal prime of R. So I want to split like this F you lower star D for for some large, for some large E. Okay. And so what I know by, by this assumption that RC is from a regular is that I have an easy real search that feel this F zero, F zero lower star D is some power of C. Basically, what you left to do is just to left, what you left is to split this C to the end. That's the, that's the thing. But then that sort of, you can imagine that sort of coming from this assumption you can, you can split off C, but you can iterate this search that you can also split up C to the end. That's the, the idea. Okay, so let me just give the details. So, so next kind of technical to ready right on everything but let me just try to do that so so next pick you want bigger than zero such that this and this less than P to the you want minus E. Okay. Sorry, I guess there's one minus zero. Okay, so such that the image of F E zero lower star C in F E one lower star is a multiple F E one lower star C to the end. Okay, so because the image of this inside here is F E one lower star C to the P to the E one minus E zero and it's less than that. So that's a multiple of this element. And so now things are to F E zero lower star R. Sorry, this is, sorry. I mean, this is not easier. This is, I'm just, this is the E. Yeah. Okay, so things by our assumption that our assumption says there exists E such that this map sending one to a field. So, so this E here is this E that I begin with. So there exists E such that R to a few lowers our are sending one to a few lowers our six bits. So I have this. I pick you want very, very large such that so this end is less than this P to the E one minus E. And I know this image is a multiple of F E one lowers our C to the end. And so since this map sending one to a few lowers our six bits. It follows that the map from R to F E one lower star R sending one to F E one lower star C to the end. So this map split. Okay, because you can just apply like after the E one minus E lower star. And you apply to that splitting you have R to F you. Let me just write it down so you have R to F. Okay, so I have from R to F E lower star R. And what I'm going to do, I'm going to, I'm going to apply F F E one minus E lower star. So I have F E one minus E lower star R. So here I have F E one lower star. Okay. And so now, so this, this is one to F E lower star C. And this is one to. Well, so this is F E one lower star C to the P to the E one minus. And I'm assuming this is a, this is a multiple of F E one. So, so, so, so F E one lower star C to the end divides sort of F E one lower star C to the P E one minus E. Okay, so this is a multiple of F E one lower star C to the end. So if, so if this map splits. So then the map from the map goes from F E one minus E lower star. So if this map splits. So then the map sending F E one minus E lower star one to F E one lower star C to the end. So this map also split. Okay, so maybe you should really do all these as an exercise but then I so I'm coming something more I'm saying the map from R to F E one lower star are sending one to F E one lower star C to the end splits. So what you do is that you compose with our here. Right, and I realized that this map always splits, because our is F split because our is, you know, you have some maps ending like one to F you lower star C split. So in particular, the maps from R to F you lower star are sending one to F you lower star one splits. So our is always a split. So you can. So this map splits, and then this map always fits so the composition split so that's how I got to this conclusion is actually a bit of work so I. You can refer to the notes for more details but the point is that you have to increase your E to incorporate this, this secret again. Okay. So then you, so then, so then you pick a splitting theta. That is F E one lower star R to R sending F E one lower star C to the end back to one. So then you check the map theta composed with F E one lower star. And that's going to go from F E one plus zero lower star R to R. Right, and this sending F E one plus zero lower star D to one. So again, there are some details you have to work out it's perhaps the easiest thing is to just draw a diagram. But the idea is that your first, your first, your first, okay, I have to like E one plus zero lower star so I view this at F E zero lower star, and then I apply F one lower star. So you first send this F E zero lower star D to C to the end, and they will raise it to F E one lower star sending that C to the end to one. So they sort of like the composition of two maps, the first map sending, you know, F E zero lower star D, or F E one plus zero lower star D to F E one lower star C to the end, and then the second map sending F E one lower star C to the end to one. Okay, so this is the composition of two maps. This is where you find this E one plus E zero that maps this element to one. So you do this for every D. So you track this splitting. Okay, so this is a technical results but I hope at least the idea is clear. So you, you first use the assumption that our C is running a regular you're sending any D to some part of C. And then you, you have to trickle with the Frobenius power, you send that C to the end at some large Frobenius trace of C to the end back to back to one. Okay, so that's the idea but there are some details you have to check. Any questions. Okay, so I'm using this, we can prove. There's a collared. This is the, this is the case that I'm going to use this theorem that I just said. So, so I find it. Local ring RMK cell pairs P is strongly of regular if and only if the completion is strongly afraid. Okay, so let's just prove this. So we may assume our is a domain. Let me just say a normal domain. Okay, because we already showed strongly a regular local rings are domain and strongly a regular rings are always more so we have, we have this. So, there's a fact. Now that I'm going to, I'm going to sort of skip. I won't prove this. The fact I'm going to use is that there exists a non zero element, so you said are such that our local as a C and the completion of our local as a C are both regular. Well, let's say a finite and regular. Okay, so it is always a finite because ours at finance to the completion is also a finite so localization of final results finite. So, so they are finite automatic. Okay, and the point is that I can find some non zero C such that both are regular. And so this follows from the fact that strongly, sorry, it's followed from the fact that at finite rings are excellent. In particular, the regular locals are open so you can find some C. I mean it's a domain so generically is regular so you can find some non zero elements such that our localism inverting that single element. And this other fact is the completion of our local C is right, but this also follows from the, the excellent, excellent ring hypothesis. So, you know, basically the fibers are all geometrically regular so our C is regular then the, the, our C is also regular. Okay, but this fact, I mean, use it some, some results about some non trivial result about at finite and excellent ring so which I'm not going to prove in this lecture but anyway, let's just grant this fact I'm just telling you for at finite rings. There exists some non zero element before I mean this is actually true for any at finite domain I think so, for any at finite domain, you can find a non zero element such that inverting that element, both are and the completion of our are regular. Okay. And so now the, you see the reason I, I have this is because not now I want to invoke this theorem that we just proved. So we want to show our is strongly a regular if and only the completion of our is running a regular. So now, to check, they are strongly a regular I only need to check the splitting for that for this little C that I find here, because that inverting that is at finite and regular so it's running a regular so I, I only need to check the splitting for that C. Okay. So, and now, okay, so now the map. The R to Fv lower star R sending one to Fv lower star C splits. If and only if the harm set is subtracted if and only if the induced map home or Fv lower star R R. So, okay, let me just write this as multiplication, but after the C map to R, I harm basically I just I just harm this map back to R. I have this, this is surjective. Okay, but now everything is at finite so to track this map is surjective that you know you can pass to the completion. I guess this is an exercise but let me just just state the idea here. This is equivalent to this harm set our hat. Well, technically I should write, right so Fv lower star R tensor our hat, our hat to our hat. This is still the induced by sending one to Fv lower star C is surjective but this is isomorphic to harm. Again, this is because I find it. Right, because this is a finite general as an R module. So tensor with the completion is just the same as just I complete that model, but is a easy exercise to see that Fv lower star R completed is equal to Fv lower star R completion. So that's, that's okay. But then this is, but then this is the later, the latter is equivalent to like our hat to Fv lower star our hat, sending one to Fv lower star C splits. Okay, and so now. Let me just say we are done by the fear. Okay, so more precisely. The theorem tell us. So this theorem tell us since we have this C such that RC and our hat C are a finite and regular in particular, after local as a C, both rings are strongly and regular. The theorem tell us to check R is to check R is strong enough regular or respectively to check our hat is strong enough regular, you only need to check your existing E such that R to Fv lower star R sending one to Fv lower star C is split you need to check it for one C. Okay, and for that C, and now this analysis here tells you to check that C to check that single C right so. Like, you know R to Fv lower star R sending one to Fv lower star C splits, if and only if our hat to Fv lower star our hat, sending one to Fv lower star C splits. So to check that one is equivalent. And the, and you, and the way you do this is because you know, you look at the harm set the harm it's surjective, and the harm is surjective. If and only if the, it is surjective after base change to the completion. And so that's that's how it is. Maybe let me just comment. So the subtlety here is that okay, at the first glance, you can do this analysis for every seat, not just this, you know, particular so you can do this for every scenes at R. But then the subtlety is that, you know, you don't I mean, in the completion of our in priority at least at the completion of our. So, you have, you have many more elements that you have to check it is splitting. Right, because they are, they are much more element in the completion that in the ring itself. So, to check our hat is strong and regular, you have to check a lot of like Fv lower star C where C may not coming from R. So, this argument tells you like, you know, by the theorem, and this fact that I said here you only need to check it for one C and that and this is important that C is actually coming from our. And so then this way, you know, you, you easily prove that this, you know, strongly a regular of our and our hat are equal. Okay, so if no questions, so we're almost done. So, okay, so we, okay, so we want to, we want to show strongly a regular rings are comacotic so that is the, the last missing piece. And so far I told you like, you know, strongly a regular rings can pass to the localization and further pass to the completion. Okay, so basically, so now to the question is that to show strongly a regular rings are comacotic, you can assume it is local and complete. Right, so that's, that's why we need this, this reduction. So the next, I'm going to state a well known lemma. Okay, so this is independent of the characteristic independent of the character P or strongly those F singularities. So, this is a completely general fact, if you have a complete and equidimensional, for example, complete domain. The local ring of dimension D, and suppose is comacotic on the puncture spectrum, suppose our P is comacotic for OP away from the not equal to the maximum ideal on the puncture spectrum it is comacotic. So then I came the local homology module of R has finite length for all for the lower for all less than D. Okay, so I believe this is you should, you should be able to know you should know this before but just for come to compete it. Let me just give a short argument so this follows from local duality. So first of all, by Cohen structure theorem. Well, I'm assuming complete. I need to assume it's a homomorphic image of a currency ring I guess but anyway, I, let's just assume it's complete so it's constructions your miser. So, are you close to as my eye, where as is a complete regular local ring. Okay. And so now by local duality. The madness do of the local module R is the X module, where and I just endless the dimension of us. Okay. So this is that. So this is the funny generated as module. So, so this you can, we can use. We can use this hypothesis, our piece, that's just let me just start. Okay, so now this guy if I localize a P. X as P minus. Okay, I'm going to use P to denote both the prime idea of our and the prime in as that corresponds to this. So this is a little bit of notation. RP as P. This is X as P. Okay, let me just describe as P. Minus I minus the dimension of our market. So I guess here I'm using the. Okay, so what am I using. So here. Okay, so I guess everything is trivial I'm using that this dimension of SP. So this is just the dimension of our model P is this and because this is because as is right. Right. So P is a primary of us, our model P is just this as model P. I'm abusing notation I'm doing this P both as a primary of us, and a primary of our so this is just as model P. And then the height of P plus the dimension of as my P is the dimension of us, which is so this. So here I'm using that. I haven't used the actual dimensional hypothesis, but now I'm going to using it right now. So now I use local duality over this as P. Okay. So we have this axed as P, the dimension of as as P is still regular because of the localization of a complete regular local so this is the regular. So this one minus this I minus the dimension of our P. As P, if I do a lot of this, this is duality over as local as a P. So this is isomorphic to the I minus the dimension per mark P. So this local code knowledge. I'm using local duality or SP. Well, it's not complete but I'm, but it's fine because I'm, because I'm using the, the acts do to the local code knowledge I'm using that. Okay. Okay, so now, since we assume our is equidimensional. We have the dimension of arm of P plus the height of P or dimension of our local as P. This is deep for every for every P. So hence, if I less than D, which is this is the mention of our right. Then I minus the dimensional arm of P. This is going to be strictly less than the dimension of our local P. This is the point because this guy and this guy they sum up to this dimension of the ring. Okay, so, so basically. Okay, so basically the point is that one. So this number, if I is less than D. So this number is less strictly less than the dimension of arm of P. So this is zero, because our P is common colleague. This is our assumption. Okay. If P is not the maximum ideal, and I less than V, then HP RP, I minus the mention of harm of P, our local P. This is zero. The assumption said our P is common colleague. This. So now if you trace everything back right so we know this is zero. So that means this guy do it's zero. So that means this guy, or this is all isomorphism right so so this is zero so this is zero and so this is zero right so what you get. The discussion above tells us that this. So on the puncture spectrum this X module is zero. Okay, but that means this has finite limits. So this is only supported at a maximum ideal and it's finally generated so this has finite net. Then Matt let's do preserve the final answer that is just a h and r has finite. Okay, and this is this is what we wanted to prove. Okay, so for I less than D. All these lower local homology modules, they have finite that. Okay, so I, I believe this lemma should be well known to most of you but just in case I just as follows is a standard argument using local duality. Okay, so. And finally, let's just, let's just prove the main theorem. This is really the key. So, so let RMK at finite. And strongly at regular. So then the conclusion I want to make so then our is homocon. Okay, so let's just prove this. All right, so, as I said, we may assume our is a complete local domain. Okay, so first of all, it's a strongly at regular local rings so by the first lemma we prove on Monday. So our is a domain. And, well, sorry. Sorry. Okay, so first of all by this colliery. Okay. If our is strong enough regular so then the completion is run effects. So first of all, we may assume. R is complete because I mean, so our is running better than the completion is called the completion is as strong a regular and to check something is called my colleague. We can check the completion right so the call my colleague property is unaffected when you pass from our to the completion of one. Okay, so your first of all you may assume our is complete. And it's, it's a domain because it's strong enough regular and local. So it's a domain so you may assume our is a complete local domain. So in particular, you are we are in the right so we are in a good stage of applying this lemma here. So we can assume we are in this setup so we can assume we can use this. Sorry. Because he raised this I didn't put rescue. But this is lemma 16 in the lecture notes, if you can. Okay, so. All right, so the next thing. Okay, so I want to show our is called my colleague and now I want to use in that. So my induction on the dimension of our, we may assume. We may further assume that our P is called my colleague for all P on the puncture spectrum, right because strongly a frequent passage to localization so I know our P is called my colleague for every P and by induction of dimension, you know I know it is. I can assume our P is called my colleague for P on the puncture spectrum. Well, I guess the initial case is obviously because ours domain so whatever is normal so in dimension one, this is really a trivial statement but anyway. Okay. And so now I use above lemma so by the lemma above. I assume that this lower local homology modules HMR has finite length for all I less than D this is I'm going to use D to denote the dimension of. Okay. So we want to show our is called my colleague. So the induction we assume is called my colleague on the puncture spectrum. And now this is very general Emma tell us, we may assume the local the lower local homology modules have finite length. Okay, so, so now since this has finite length for all I less than D, I certainly I can take a non zero. It exists one zero C. Right, such that C times HMR is zero for all I less than D. So these are finite length modules. So they are the annihilators are sort of and I'm primary. Right. And I just look at the intersection of, you know, finally, many number of and primary ideals is going to be none zero. Okay, so, so this is, you know, you can always. Find the non zero C such that see a notice each one. Actually just pick and you can pick any element in a maximum ideal, some power of that, you know, now it's HMR for for each eye and then just replace that element by that power that power will do the job. Okay, so this is there. There are various ways we can argue this is easy. Okay, so, and now I'm going to. So I'm going to my goal is to show our is called my colleague. So I want to use this C and the strongly, we haven't used this assumption yet other than just the induction process right so I want to use this strongly of regular property to split off this C for some large E and then use that to argue that these lower local modules has to be zero. So that's the idea. Okay. And now since our is strongly of regular. There exists e bigger than zero. And then our linear map from a few others are too hard. Such that the. Okay, so I have our two at the lowest are our. This is multiplication by a few lower star C. So this composition is the identity. Okay, so this is the, this is the natural for being this is a natural map. And this is multiplication by this map. So this from here to here from the first one to the third one is the map from sending one to a few lower star C, but I'm just writing down everything in detail. And this is a splitting so which means this composition is that so we have a social thing. Okay, so now what I'm going to do I'm going to apply so now apply. I've local co homology function. Co homology. Okay, so let's see what we got we have HMIR maps to HMI. I feel or star or I feel or stars. This is more picking. Again, this is a natural map or two I feel or stars the convenience map so this is your natural natural map. And this is the multiplication by a few or stars C. And this is the map induced by the last I feel or stars. Okay. Oops, sorry. This is. Okay. This is the identity so this composition is the identity. Okay, but now the way we pick C is just basically saying like this middle map. Because here I remember I can identify, I feel or star are with our and right so let me just. Let me just say. So, if I identify this I feel or star are with our. So this guy is identified so identify I feel or star are with our. So this thing is getting identified with with edge. Oh, let me just be careful. So here I'm identifying with our so this, this, this maximum ideal is, well actually it's getting identified with am to the bracket Peter the e, technically speaking. Right. But, well, I don't care because local homology doesn't see the, the, you know, doesn't see the radical of the idea so this one, if I'm, if I might die, if I identify I feel or star with our. This thing is just HMI are this guy. So once I do this identification. So this guy is identified with HMI are again, and this is the multiplication by CMAT I'm just dropping all the actual star. Right. And if you drop this there's only one Saturday is that the ideal downstairs should actually be replaced by the convenience power but actually it doesn't matter because we take local homology local homology doesn't see the difference. If the, if the ideal had the same after ready. Okay, so this is a zero map. And this is here. I also have a zero map. Okay, so what do you get out of this diagram is that the identity map on HMI are factors to the zero map. Okay, but what does that mean that means this is zero. Okay, so what follows is that. So this is an identity or HMI are factors through the zero map. But that tells us. So this is the proof that strongly F regular rings are common color. And finally, let me just putting putting everything together. So if our to ask is a map of rings of character P. So IE R is a direct summit of S. If as is regular that are as common calling. So, let me just take so. So the, the very quick proof that I'm going to sketch here needs to assume that both R and S are at finite. This is a pretty mild assumption but let me just let me just quickly sketch the proof to have the proof assuming this assumption by me but if you're assuming this then this quarter is as I said it's just just basically follow from the older results we prove about strong F regular rings. Okay, so, I mean, if you have this assumption that as is regular implies as to strongly F regular but strongly F regular member pass to the direct summit if your rings are fine and so R is from a friend. But the theorem we just proved right right here, strong F regular rings are called McCarty. Okay, so, so this is the way you, you, you, you can get this quarter pretty, pretty fast from the theory of strong F regular rings. But of course, okay, so now there's a one annoying sort of assumption that, you know, we are assuming both rings are finite because we want to say like this, because we only define strongly F regular rings in the finite setting. Well, to do the general case, while you can just, you know, develop a theory of strong F regular rings beyond the finite setting, you're actually some work and type closure, I guess, and recently like this data and smith they also have some proposed some sort of definition of strongly F regular rings beyond the finite setting, you can do that. But you can also I believe that actually when I prepare this lecture I saw like this thing. This colliery at least also shows up in neo abstains lecture knows you can also prove this using the colon capturing property of of type closure so that's another way to do this, but I'm running out of time but actually in my lecture knows, you can, you can pretty use the techniques we prove like these. Okay, so this is a theorem, this photo from the cruise, and this is a theorem just to direct some an astronomy program, and this is a theorem that we just proved you can actually just examine the proof more carefully, you can actually prove the more general statement by essentially by reducing yourself to a setup that that both R and S are fine. And you can do that I carry out the details in the in the lecture knows, but it's it's kind of a little bit technical, but it's not too difficult. Let me just make a comment. So the general case, just reduce to the, I find it's you can easily. Well, not, I mean it requires a little bit work but you can reduce the final setting and then you basically run this argument and then you can prove the general case but I'm going to skip the details clearly the for for me I guess or like for this lecture. The, I find case is the main is the main case. Okay, so, but that's the sketch of the proof that the direct summon of regular rings are are coma colleague in cars EP, but finally, just a one final remark. So the corollary holds. Without assuming. R and S has character. So it also holds the direct summon of regular rings are always coma colleague this is true in character zero and also you mix characteristic. So this is the character zero is done by hoax of unity zero and the mixed character. Mixed character is the case. It's done in a paper of height man and myself. But it also follows. So all these follows quickly. I think this was from, from the existence of so called weekly fun toriel. They come across. I'm not, I think this. Maybe later on in this lecture series some, some people might talk about the weekly fun tora become a college at least in cars EP, but this. This is, this is holes in arbitrary characteristic now this is due to Andre. So this is a remark I want to want to make so I only prove this in cars EP but the, the corollary also holds in arbitrary characteristic. Okay, so I think I will just stop here and I mean, sorry I didn't know it contains a lot more than what I want to say but I mean what I can say here but anyway, so I will stop here.