 Okay, so I haven't shared the screen. I can, I can start at any time so. Yeah, we can start now. Okay. So, okay, so let me continue my, this, this lecture on direct summit of regular in so last time we, we define strongly have regular rings and we, sorry. We didn't say much we just proved two things. So first of all, strongly of regular local rings are domains and then we proved that strongly of regular is a local property. Okay, so, oh, and by the way, so I think I should say that they're, I just realized that there are some overlap of my lecture with the previous lectures by Neil and Kevin. I think one point is that my luck. I want to make this lecture like more like self contained so you can. You don't need to know a lot of type closure in order to understand the main theory. Okay, so let me. Okay, so I'm going to start today by recall this theorem of Coons. This is a theorem that says ours regular if and only the Frobenius flat ring R of terms of P is regular if and only if the Frobenius for why he or equivalent for largely so hard to. R. It's flat. Okay. And so, in particular. So, you know, so we will mainly apply this Coons theorem when our is local and I find it. If arm K, I find it local. And in this case, since I feel or star R is finally generated so flat is the same as free because the ring is no theory local so so then I was regular. And only finite free. So let me just give so I'm of course like so the really the hard direction in this theorem is actually the, if you're assuming the Frobenius is flat, then ours regular, but actually for us we're going to use it in the for only for the easy direction that is if. As far as regular so then the Frobenius is flat. So let me just sketch a very. So I'm sure I think Neil already did a proof of that but let me just give up another way of prove this one are is at finite. So, so, so suppose okay I'm good. I'm just going to try to sketch the proof of this in particular statement. So if you have a regular local ring, which is at finite. So then you know from this homological criterion of regular ring that every module has finite projective dimension so so you got. So, I mean if ours regular, then projective dimension of this. Plus the depth of this module is equal to the depth of the ring, but then. So, these two are obviously the same because the. So from from this you get the projective dimension. It's zero so in other words, I feel ours are is a projective is a finely generated and projective our module so it's fine at so. You know, the easy direction under the assumption that you your ring is at find so I'm not going to prove the other direction and we won't use it actually. Okay, so, so last time we define this strongly at regular ring. So, so the first, I mean, so the first thing we want to say is that regular rings are strongly a friend. So, and a finite regular ring of character B is strongly regular. Okay, so let me. So Kevin already proved this but let me just give a quick or remind you that the quick argument so. So first of all, last time we already showed that strongly at regular to both regular and strongly at regular local properties. So, you may assume. R is at find RMK is a finite. And local. Okay, just check this on local rings. And so then by this coons theorem by coons, but it's coons theorem above. We know that I feel ours are. It's a finite free our module. Okay, so this is regular tell us. Alright, so now we want to show that our is so this under this assumptions are is strongly afraid. Right, so remember for what is the strong effort I mean it means so let me quickly remind you the definition in words. Let me show that for every element for every C, not in any minimal prime of our. So there exists a large enough such that the map from our two I feel ours are sending one to I feel ours are C splits as a map of our modules. Okay, and so now. So we are, we are in the case that we have a we have a local. So it's a domain to saying that C is not a minimal crime is just saying C is not zero. So now for any C none zero. So, you wanted to show like, you can find the such that one to the map goes from one to a few hours are C splits. So now, but what we know now is that a few hours are R is a finite free our module. So came is that I came, so there exists e bigger than zero search that. So this, you look at this element F you lower star C, of course this belongs to F you lower star R. So this element is part of minimal basis of this free mode overall. Okay, so this is pretty easy to see is because otherwise. I mean if this is not part of a minimal basis, then what you know is that oops. Yeah. So if this is, if this is, if this fails. So then saying that this is not part of a minimal basis or minimal generator is just saying this element belongs to the maximum ideal times. If you lower star. Okay. But remember how we, how does this notation mean so if you, if you multiply some element inside our, if you want to pull that inside, you have to raise it to the Frobenius power. So this is equals to FB lower star and to the bracket P to the. Okay. So, if FB lower star C is not part of a minimal basis. So then you get FB lower star C belongs to FB lower star of this and to the bracket P to the E this idea. Okay, so for all right so we want to find we want to say there exists some a minimal basis and if this is not the case, then I feel or is our C belongs to this for all E. Okay, but quick but of course this is a contradiction because I feel or is our C belongs to I feel or is our end to the bracket P to the E means right so, but this means C belongs to enter the bracket P to the E for all E. So this is zero by the core intersection theorem so this is a contradiction. I'm just putting this in the premises I'm showing that there exists E such that FB lower star C is part of a minimal basis. Okay, but if this is part of a minimal basis then you can send it back to us. So, no, since I need to continue here. I feel or star C is part of minimal basis of FB lower star R as an R module over R. So the map R to FB lower star R. So think about this is just a free module now. So, one to FB lower star C scripts. Okay, so this is a theorem that follows from Coons theorem. I feel again I want to emphasize this really follows from the easy direction of the Coons theorem you don't really need to know the other the hard direction. Okay, so are there any questions. Sorry. Can you clarify again how you know that in this case your ring is a domain. Sorry, can you see the share screen now. Sorry. Okay. Can you clarify why this is a domain. Oh yeah, good question because it's because we reduce to the local case. Right, so okay I have a regular ring, I localize it I have a regular local and then there's the. I don't know how I just. Yeah, sorry. Can you see the share screen. Yeah, okay, thank you. Thank you. Yeah, so the, the reason we can assume are the domains because we were used to the local case and there's a, we know that regular local rings are integral domains. Okay, thanks. Any other questions. Yeah, feel free to just interact me or someone should. I can't see the, the, the chat so sorry, if you put it in the question in the chat. Please also just remind me that I can address the question. Okay, so. Okay, good. So we have, okay, at least we have one example of strongF regular is not, which is regular local ring, or regular ring in general, and so. So, I will, I will later today I will give you some more examples of strongF regular rings, but at this point, I first want to prove a property about strongly F regular rings. Um, I'll be a finite and strongly a regular ring of Kersic, Kersic P. So then the property I want to show you is that any module finite extension splits. So then R to S splits for any module finite extension S. So prove this. So, since assets module finite to track this one. So to track this splitting. It's enough to track this space at any primary of our so I mean the tracking the splitting is a local property. So we may assume. RMK is local. And strongly afraid. Okay, in particular it is a we may assume it is a domain. So in particular, we may assume R is a domain. Because last time, remember last time we showed strongly a regular local rings are integral domains. Okay, so now I have a module finite extension of a domain so. Excuse me, excuse me. So is there a typo in the statement of the theorem so you mean I find that and strongly F regular. Uh, let R be an F. Oh yes, thanks. And strongly. Okay. There's a type of heroes to all. Thank you. Okay, any, any other comments questions. Okay, so so far I did, I basically did nothing. I'm just saying tracking a finite extension splits. It is enough to check this space locally on spec R right so you can track. You can track you can localize the prime idea of our and check this split for every, for every prime P. And so then you can localize our at a prime and replace as by the corresponding localization, and to check that finite extension splits. So if you're going to do that, you may assume the base ring is local. And last time we showed strongly a regular local rings are integral domain so we may assume ours of a man. Okay, so this is the first reduction. And so the next reduction I want to say you may also assume as is a domain. So by just modeling or killing a minimal prime. The minimal prime of us lies over, you know, a minimal prime, just pick a minimal prime of us, you know it contract to zero hours of domain so contract to the zero. And so, and, and just kill that minimal prime by killing a minimal prime of us. We may assume as is also a domain. I hope this is clear right so the point is that. So you want to show like R2S split. And so now you pick a minimal prime that contract to zero so you model that minimal prime let's call it P. So now the, the point here is that I want to say it is enough to track this composition is split, because if the R2S split, then the first map split, right, because you can just, I mean saying that R2S might be split means you can map. There's a, you know, there's a, there's an arrow here that sends like one to one. But if, if you if you have a map here that said one to one, you can take the composition. Right so then as to R2S, you can also send like one to one. So this is just some sort of like a trivial observation you can compose, you can model a minimal prime of us so yeah hopefully I mean just maybe this is a small exercise or just very easy to track. Okay. All right, so we may assume both R and S are domain now. Okay, so now what's the advantage of doing this. So now, as is a torsion free R module. There exists an R linear map. So it's a finite and torsion free R module. So you can find a nonzero map from S to R. So there is an R linear map theta from S to R, such that theta of one is nonzero. So I hope this is clear. So you have a torsion free R module. So all I'm saying is that there is a nonzero map back to R. So the reason is that, well, so it's a torsion free and finite R module. So if you answer with a fraction field of R, you got a finite dimensional, you know back to space and then you, you know, you have a map back to the field to the field. So generically, if you're trying to read a fraction field you obviously have nonzero maps. Okay, but the every at a fraction field level everything is just based change from a map from S to R. So, you know there's a nonzero map from S to R. That's all I'm saying. Okay. Of course this is, this is not, I mean, this is not really what we want. We want to show that the map splits. That is, we want a map that's sending one to one. And what we know at the moment is that we only know there's a nonzero map. Okay, so now we have to use this crucial strongly of regular property to say that you can sort of modify this map. Or it's not, sorry, not really modify. You can create a new map that's sending one to one. Okay, so the idea is very simple. So now since R is strongly of regular, we can find E bigger than zero such that this R to, I'm just repeating the definition R to FER, sending one to FELOWERSTAR C splits. All right, so this little C here is the image of theta one that I define here. Okay, so now maybe it's just easier to just approve the theorem by drawing a diagram. For R to S, I have FELOWERSTAR R to FELOWERSTAR S. Okay, so I have a map from this map theta from S to R that's sending phi one to C. So in other words, well, I can just apply FELOWERSTAR here. So just use it. I have a map from here to here. All right, and this map sending one to FELOWERSTAR C. Okay, so this is by what we asserted here. I have a coordinate map theta sending one to C. So if you apply this FELOWERSTAR, so you have a map from FELOWERSTAR S to FELOWERSTAR R sending one to, or maybe I should let me just be very careful here. Maybe I should write FELOWERSTAR one here to FELOWERSTAR C. Okay, and so now we also know by this strongly of regular assumption you can map this to one because the map from R to FELOWERSTAR R sending one to FELOWERSTAR C splits. But what does that mean? That means you have a map the other way around sending FELOWERSTAR C to one. So now basically it's clear. So now what you know is that this composition splits. You have the natural map from R to S, and you have the natural Frobenius from S to FELOWERSTAR S, and this composition splits because I already tell you what are the two maps sending FELOWERSTAR one to one. Okay, and so now you use this trick I told you before, if you have a composition R to S to FELOWERSTAR S, and if the composition splits, so then this map splits. Okay, so let me just say the diagram shows that the composition from R to S to FELOWERSTAR S. So these are both the natural maps like sending one to one, and this is the natural Frobenius sending one to FELOWERSTAR one. So these composition splits. Right, but so this which implies R to S splits. I mean as a map of R-modus. I always say in splits as a map of R-modus. Okay, any question? Okay, so this is the first property that we talked about from African rings, and let me just mention that, well, just pause a little bit so there's a, okay, so we show that this very nice property is from African rings splits up from all the fine extensions, and you can ask about the converse, and that's actually an important open question. So there's an open question. So, let me say R is a, you can ask this question beyond the finite case, but let me just focus on that. So R is a finite domain, if R to S splits for all finite extensions, then S, the ring R, strongly afraid. So this is not known, so there are many partial answers that we know, for example, when R is Gorenstein, this was proved by Hochso-Hüdingte and SIN, extended to the Q-Gorenstein case, and there are some further evidence on this, but as far as I know, this question is open in general. And you might also, I'm pretty sure like the Neal and Kevin, they talk about the weekly, so the open question about strongly afregular and weakly afregular. And so I mean, it turns out this question, if, I mean, this open problem that I stated here, if this can be answered affirmatively, so then this also implies that strongly afregular and weakly afregular are true, because this property, so R to S splits for finite extension, it is enough to assume R is weakly afregular, other than strongly afregular. So. That property of splitting out of all the monophonic extensions, that's a splinter, is what it's called, right? Right, this is the splinter, yeah. Yeah, so I'm saying this question is, so this question implies the weak and strong. And this is a little bit of a conjecture because weakly afregular implies splinter, implies this property. And if you can show this property implies strongly afregular, then you also equate the weak and strong. So, and you know one thing that, you know, at least with some experts they, they, they prefer or they want to work with this. The stronger problem is that, so it's actually not very hard to see that this is a local property also. So what, again, last time I already said, right, so what the weakly afregular, we don't know whether that property localizes. Well, whether that's a local property, but for strongly afregular, I already showed you last time so strongly afregular is a local property. And this property, so like saying, split off from the finite tension, this is also easily, you can show that it passes to localizations. So to prove this, so to prove this problem you can, you can assume R is low, or you can even assume R is a, R is a, you know, complete local, you know, normal something like that. So I mean, this is this stuff. But anyway, the chat. First one, she asks, is there a general result where S is not a torsion free arm on. Is there a general result. Well, the, the, okay, maybe I should clarify this theorem here doesn't, doesn't, you know, this theorem here doesn't say as is torsion free overall, you could have torsion. But what I'm saying is that we can reduce to the case, as is torsion free. Is that, is that answer the question so the theorem itself doesn't say it's just a finite extension. She could have answered your question. Yeah. I don't see a response it but the next question is from Kamari and asks under this property. I assume this property is being splinter is our weekly afregular. No, that's not that's a good question that's also not. Right. So, maybe I should say the open problem is that if ours and finite domain has a key, if R2S space for all funny extensions, finding extension as so then is our strongly afregular or at least weekly afregular. So both, both are not no. Yeah, it's a good point. Yeah. Any other questions. Okay, so let me move on so. Okay, so there's a colliery of the results we've proved so far. Okay, we. So let me just go back a little bit. So we proved regular rings are strongly afregular and strongly afregular ring splits from all the finite extension. So putting this together. As regular. In terms of P. Then R2S splits for any module finite extension. So, Well, if you. Yeah. Okay, so this, if you put in the previous two theorems together, you already, I mean this. And pies this colliery when R is at finite. Right, because we showed, because I mean the definition I gave you for strong afregular rings, you know, have this implicit at finite assumption built in. And we show that at finite regular rings are strongly afregular. So, but, you know, you can easily reduce the finite case to prove the general case of this. So, again, so to show like this R2S splits for any finite extension, it's a local property on R. So we may assume R is local. R is regular local ring. And then we may further complete R and S to assume R is a complete regular local ring. We may further complete to assume R is a complete regular local ring. Okay, so see there's so this left is an exercise. So to track, basically the exercise says to track a finite extension or a finite map splits. You can compete at the maximum ideal to track that after completion, the map split. And if that's true, so then original map also split. Okay, so this is an exercise. That okay, so I want to reduce the module finite case but now. So R is a complete regular local ring of character P part of equal characters taking particular so by co-instruction theorem. R is isomorphic to a power series ring over a field of character P. And so then, well, this one is not necessarily a finite because this field may not be a finite, right. But you can, you can easily make the field I find we can I mean, so last time I already said, so for field, I find it just means, you know, K to the one over P is a finite dimension okay vector space. For example, any perfect field or any algebraic closed field is automatically a finite. So, so notice you have a split extension. You just enlarge the field to the same algebraic closure. And so it's an exercise I mean this is a pretty easy exercise. Track this, this map is a split extension. Okay. And so now, let me, let me, okay, this is, this is our, let me call this our children. Okay, so now you just track this commutative diagram r to s goes to our tutor goes to our tutor tensor r s. And of course this is your regular. Because it's just, it's a power series over over actually make a closed field. Okay, so now the point is that so now. Right. Okay, I'm going to use. I'm going to use that trip again and again. So, so now this ring is regular and I find it so it's strongly F regular. And this is a finite extension of this strongly F regular ring so this map splits. And this map also splits. This is, you know, because this is just a very down to earth map that just enlarge the coefficient field. So this map splits this map splits. So the composition from our to this our tutor tensor I would ask this composition splits, but then it factors through our to us so our to us also. Okay, so you can easily. So I omit the details, but this is how you, how you do this. Okay, so you have a splitting here. So let me just. And then you can just take the composition. I do this just just to, you know, I do this just to reduce to the case that we can use the theorem, we can cite the theorem we prove because we only show this regularly pass from a regular et cetera we only show this for in the finance so we just Okay. So that's the question. Let me, let me perhaps put a remark here. So, so in this colliery, we showed a regular rings split up from the module final extensions. But we have an assumption that a character key. So now let me just mention, here's a remark. So this colliery. The colliery holds without assuming the rings as character P. But this is a very difficult result. This only proved recently. Well, sort of recently by Andre in 2016 that actually this colliery holds for any regular ring, even, even does not for any regular ring even without assuming this positive characteristic or something. Okay, so, and then, okay, so, let me just say another easy colliery office. If our is a finite and strongly a regular. So then our is normal. In particular, one dimensional strongly a particular rings are regular. Regular rings are strongly a regular. And so the converse holds for one dimensional rings but it fails for two dimension or higher dimension already. So this is pretty easy. So, So suppose our is not normal. Suppose our is not normal. So then you can write a fraction so then there exists some A over B. Integral. Well, okay, let's say just reduce the local case let's just assume our is a domain because normal is a local property we can just maybe I should say that at the beginning but let's assume our is a domain if you like right so because we can reduce the local case as usual. So I'm okay so now suppose this is not normal so then there is a fraction like a over B. That is integral over. Okay. The element A over B itself is not in. Okay. So, so now we set our prime equals to our adjoin this A over B. Okay, so this is an integral element adjoining an integral element is a finite extension so then our to our prime is a module finite extension. So now we use the theorem we just proved for module finding extension it splits so there exists a splitting. So I either use a map from theta from our prime to our sending theta one equals to one. But now you need to see a contradiction basically this normalization if you have a partial normalization map or partial normalization map they they never split unless it's the identity so. Okay, so the point is that so therefore. So thus. Right. Okay, so I'm saying if you have a splitting from our prime to our that sort of forces this fraction A over B inside our. So the reason is that so now you're right okay be. So let's just think about what is B times theta of A over B. Well, but that thing should be since the theta is our linear so this is theta of a you can pull this be inside. But theta is a times theta one which is, which is, well, this is a. All right. And so, so basically this tells you theta over B, it has to be a over B. So in other words, so this is in our right because theta is a map from our prime to our so theta of anything should be inside our, but we also we also convince ourselves theta of a over B, it has to be over because B times that equals to a so it has to be over. So we showed like the over B is fraction, it itself has to be inside our which is a contradiction. Okay, so strong afregal rings are normal and we know normal one dimensional normal rings are regular so one dimensional strong afregal rings are regular. Any question. Okay, so the last theorem I'm going to talk about for in the second lecture, as a premise. So I want to give you some more examples of strong afregal rings so so far. Well, we define this rings we prove a lot of properties but so far the only example we have is this this regular rings right so we prove like regular rings are strong and regular. And I want to have some other examples. So let me just prove a theorem and using the theorem you can write down some a lot of more examples. Okay. So let R. And as the fight rings of character P. So, if our is a direct summit of us and as is strongly of regular. So then our is strong. So what does it mean by saying our is a direct summit well so saying like you have a map from R to ask that splits splits as a map of our models as our models. Okay, so this is what I mean by saying our is a direct summit of of us. So, um, Okay, so let's just try to prove this. So, okay, so actually the, the sort of the hard part in this proof is to reduce to the case that s is a domain. Okay, so let me just try to let me just try to skip that details. I'm just saying we may assume that both are in as our domains. Okay, so I mean, you can see the it's it's it's not really hard but it's just a little bit. You know, Nancy to do it here. It's like a one or two paragraphs proof but so okay so maybe let me put it's it's to show okay you can always assume our is, you know, our okay so so the way you do this is that you can you can first assume our is local because you want to show our is from a regular and we know this is a local property so you can localize that the maximum idea of our or prime idea of our, you can you can assume our is local. So then you replace our by say our localism P and as by, you know, the corresponding localization. So then you use the normal assumption on x right so the assumption is that as is strongly of regular. So you don't know whether as is local if you know as is local that you are reduced to the domain case because strongly of regular local rings are domain, but unfortunately you don't know as is local, even you, you reduce to the case are as local you don't know as is local, but the point is that since as is normal. So you can write as as a product of normal domains. And if our is a direct assignment of as, then you can show that actually our is a direct assignment of one of the factors. So that you reduce the case as is a, is a domain. But anyway, so this requires an argument, and it's, it's, it's not hard and I just, I think I'm running out of time so I will just omit the details you can refer to the nose. But after you're assuming this both aren't as our domains. Oops, sorry. Can you see the screen share screen. I think I'm back to the share. Yes, yes. Yeah. Okay, so let me just quickly finish off, finish this in like one minute so the after assuming like both rings are domains is going to be very easy. So, okay, so you want to show our is from a regular now. So the so just just think about a definition. Well, let's see again maybe it's just easier to to draw a diagram. So. So, you want to show for every scene out the animal prime of ours. You know, our to F you know, start our sending one to see splits, but ours domain says you just just focus on this non zero and C. Okay, and so now things as is from a regular. Right. So, e bigger than zero. Such that there exists, there exists a map fee from F e lower star s to ask, sending this F e lower star C to one. Okay, as you know, maybe again, as I said, I just draw a diagram you have our to us. And here we have F e lower star arm, and here we have F e lower star s. Okay, so this, this is so that these are the natural inclusion. So sending one to one and the vertical map, I'm going to send you one to F e lower star C. This goes to F e lower star C. Okay. So now what you know, let me just use it in blue, you know you have a splitting from here to here sending F e lower star C to one. Okay, so this is by our assumption that assets from if I remember assets of domain, it is important to see is not zero. So she's not in any minimal prime of us because as is a domain. So you can, you can map F e lower star C back to one, but our to us is split our direct so much so this, this map split sending one to one. Okay, so now you use that trick that I said earlier again and again. So, so now you have a composition from R to F e lower star R 71 to F e lower star C. And you have a further map to F e lower star as, and this composition splits. But then that implies the first map better be Smith also. So R to F e lower star are sending one to F e lower star C split. Okay, and that's the proof. The diagram implies R to F e lower star, sending one to F e lower star C, and using this. Okay, so now using this last theorem, you can actually write down a lot of examples of strongly F regular rings because while we know regular rings are stronger regular but then this theorem tells you direct summon of regular rings are strongly F regular. Okay, so example. Two easy examples. So for example, you can take R to be k x y z mod x y minus z square. If you don't like this. Let me just tell you this is as a morphic to K s square as T T square. Okay, so this is a direct summand of this is a second baronese. So it's, it's, it's a summand of this regular ring. So, and by this theorem we just prove R is strongly F regular. Well, okay, let's say K is a finite field, if you worry about that. But anyway, while there's another, let me give you another example. So, okay, of course, more generally you can say all the baronese sub rings of polynomial rings are strongly F regular because polynomial ring is regular so strong and regular and baronese rings are always direct summand of the polynomial ring. So let's run it right. So let's look at this K x y z w minus x y z w. So this is isomorphic to the or this is K AC AD BC BD. So this is a separate product of two polynomial rings. And again, so this is a direct summand of this as which is K, this is R. All right, so you can track this is a, this is a direct summand of a polynomial ring in four variables. And so by the theorem. So this, this R is strongly F regular. And more generally, you can also say the separate product of polynomial rings are strongly F regular. Well, again, let me just say K is an finite field, if you're worried about this, I find an issue. But anyway, this is some examples of, I mean, you can use this theorem that we proved here to write down a lot of examples of strongly F regular things. I think I want, I can, I just want to stop here for this second lecture.