 Great. Well, good morning. Good afternoon. Good evening. As I said before. So, in this hour, or in just 50 minutes. The plan is to cover the equivalent of section three, four and five of the of the notes. So we'll see. Yeah, so the first, yeah, section three is the shortest one, which is sort of a connect, which is a connection of regularity with the flatness of the convenience. The fourth section is, is called just called tight closure, I guess. And it's, of course, the longest section of the of the of the document. And section five is, is going to be about modified extensions of domains and how that works. Okay. So, as you see, I sort of pre loaded some some things. So, if I write in green, does this make sense? Okay, great. Okay. So, regularity and the flatness of the convenience, I wrote F flatness, but nobody says that. Anyway, um, so it might seem so regularity right the right what was a regular local range that if, you know, the, the maximum ideal is generated by D elements for D is the definition of the ring doesn't seem like it should have anything to do with the for Venus. But if you have a ring of characteristic P, it actually has everything to do with the for Venus. Because couldn't prove in I guess 1969. Yeah. So, that if ours are reduced the theory and ring of prime characteristic. And actually you don't even need reduced for what I've heard for that. But anyway, because it'll, it'll come it comes out in the, in the, whatever. But are we a reduced rate in the theory and ring of characteristic P. Then, R is regular. If and only if the previous endomorphism. R is flat. And by that means, you know, F lower star R is flat over R. I mean since we're reduced to say, R to the one over P is flat as a module. Okay, I'm not going to prove this whole theorem, because we're really only going to need one direction of it I think throughout the whole course. And that's the, the forward direction that regular rings have flat for being used that if you're a regular ring, then R one over P is a flat model. Now regularly is always reduced because it's, it's, you know, it's it's normal. And so it's a product of finally many normal domains. And so we can, we can talk about R one over P in this context we don't have to talk about reduced anyway, whatever. So, so we're just going to prove the, the forward direction and I'm going to sort of prove it in cases that kind of build on each other. So, so first of all, the regular rings that the people really, you know, care about our polynomial rings in some number of variables over a field. And it turns out that this is going to end up being a big case that we can build on. So suppose that R is such a ring. So that means it's K bracket X one through X D where K is a field of characteristic P. So now let's look at R and R one over P. Just take everything to the one over P about it. So hey one over P I mean it's actually going to feel the extension of K. Okay it's perfect it's just K but anyway. So, so this is the, this is isomorphic to the previous math basically. And well, and I claim that this map factors in the following useful way. First of all, you extend the coefficients of the polynomials. And then you throw in the all the peak powers of the peak roots of the variables. So let's call these maps alpha and beta. Alpha is free on a vector space basis of K one over P over K. Right. If you take a polynomial in, you know, so let's say you take a polynomial and s, you can you can sort of split out like and look at it as as a and you take a vector space basis of K one over P over K, you can express it as like as an R linear combination of elements of that basis. So, now beta is free. And by just saying that morphism is free I just mean that that the target is free as a module on the source on that on that on the basis. So beta is free on the basis. I won over P X to the I two over P dot dot dot X D ID over P, where each I J is less than is between zero and P minus one. So it's a basis of P to the D. So it's a free module. R one of these is free over S on on this P to the D basis elements. Right. If you take a polynomial in these in these variables over here. You think about it is you can express it as as a as an S linear combination of these guys. Because once you get to, you know, what was any of these coefficients get to like one, then then you can factor that out as an element of S. Okay, so therefore, therefore, since R one over P is free and flat as an R model, they basically combine in those bases. Okay, so this is very, very concrete hopefully. Okay, so, so that's the polynomial case very, very concrete. Now let's do the complete case. So, let's say R is a complete theory and local ring. Okay. Then by the common structure theorem. I assume you guys know that. But the co instructor among other things says that if you have a complete local ring that contains the field. It is actually and it's regular a regular complete local ring that contains the field, then it is actually isomorphic to a power series ring in that set of variables. Um, so ours actually isomorphic to a double bracket X one to XD for some. Okay. Okay, of course. Um, so, um, so now let's look at this. Okay, this looks sort of familiar hopefully K X one. Or it's just we're putting double brackets at a single brackets, it won't be free anymore actually, but, but it is, it is going to be flat, because this map is kind of a faithfully flat. Well, it's the completion of the following map. So, and that map is just a localization of the map that we know is flat. Right. So this is flat. And therefore this is basically flat. Okay, so now let's do the local case. So let's say that are is a regular local ring. What you have is that then are, then, then you can make a similar diagram. Okay, double bracket X one to XD. I don't have to write that that's true. So the one over P. It's easy to show that this is actually our one over P hat. This is flat. Basically flat. This is faithfully flat. This is faithfully flat. And therefore this is faithfully flat. And the general case. So, but so what's the general case that's where are is a regular ring. This is necessarily local. Then what you have is that RM is a regular local ring for all maximum ideals and local case. But this is the same work it out as our one over P. Yeah, since flatness local property. Basically, you localize. You go up to the complete case, you go down to the polynomial case. And then, and then you sort of go back to the background. So that's the easy direction I'm not going to prove the hard direction of this theorem, which is that if the previous is flat then it forces the ring to be regular. And this is, this is, I think all we're going to need. I think it's, yeah, it's really interesting that there's that there's this connection at all and it really comes from as you see the construction theorem. Okay, so corollary, let R be regular and then I enjoy the ideals of our then for any power q of p. So we have IQ colon jq equals I colon j here. Well, if you first call it and then you take bracket powers is the same thing as if you took bracket powers and then you call it very false in any other kind of range. So, um, this, the business is going to rest on the following proposition which I call burbaki flatness because, as far as I know this, this result was first stated as an exercise in burbaki is. It's a little commutative forgive my French pronunciation. And it's, you know, it really is, is, is an exercise that comes from the equation. So I won't prove this, but I think this is this should be seen as one of the basic ways to think about about flatness of modules over a rate. It's a version of it for non commutative rings but let's assume ours commutative so I don't have to worry about left modules and right modules and left ideals and so on. And we're assuming that ours if you're in so I don't have to say which ideals are finally generated there's that there but anyway, if you're a Ethereum commutative ring an arm and you have an arm module and it's flat if and only if for all ideals I and J. I'm colon J. Right, so the elements of them that James applies into I am the same as the, as if you take I colon J, and you extend it. So the right hand side is always contain the left hand side for any module. The left handed right hand side is actually characterization of planets of a module. And it's enough to take J being principal ideals. And in the non if you're in case you have to, you have to restrict J to being finally generated ideals. Okay, so proof of corollary so why does this prove that the colon power, they call that bracket powers commute with with Collins. So here's one. Okay, so. So first of all, this is always true. Well, you know, I'll just do it this way. So, so let's take you first. It's regular so everything's reduced so. So this is this is one of this is from exercise one. And then this is for bucky flatness. And theorem about and consists theorem says that well, good. I are one of repeating flat make means by duration are one of our queues as well because that's your rate of degrees. Okay. So this is the same as IQ colon jq. And then, and then you take your powers and then you have, you know, the result. You can work it out a little bit later if you want. I've given you the real ingredients. Okay, so tight closure. Craig did me the favor of defining it, or he defined it for, for ideal. So, okay, so, so first of all, the notation are not. You didn't define that notation. So another notation is that we're going to say min. Minimal primes of R and max R is the maximum ideals. Okay. So are not is are my is the complement of the union of the minimal primes of our that's more for you to set because the complement of the union of any set of primes and set. And if you have a inclusion of our modules and an element Z of M, then Z is in the tight closure of L and M. If there's a q not a power of P such that for all you bigger than or equal to q not. Sorry, if there's a C in our not and q not such that for all q bigger than or equal to q not C times C Q M is in L Q. So same thing that he defined for ideals except them taking bracket powers and bracket, bracket powers in a module. So, um, So, okay, so proposition type of the factorial closure operation on the category of and you want to say out of our modules I don't know that that's true in this generality for a nephirian rain, but it is true on the category of finitely generated or modules. It's factorial I just don't know that it's closure operation. It is true if there's a test element, as we'll see in the next hour. Moreover, the previous closure of L and M. Sorry, there's always the team. Okay, so what do we want to show we want to show that the fact that this definition gives you a sub module at all. And so why is that. But X and Y being L star M. And let R be an R. And when I show that X plus our Y is in L star M. And then there is exists a Q naught, but there's exists, let's say a C naught and C one and R naught. Q naught Q one powers of P. Such that for all Q bigger than or equal to Q naught. C X for C naught X to the Q is in L Q M. And for all Q bigger than or equal to Q one C one X C one Y to the Q M is in L Q M. So let's see what the product of C naught C one and let Q to be the maximum of Q naught Q one. And for all Q bigger than or equal to C times X plus our Y to the Q is what C X Q M plus R to the Q C Y Q M. And then both of these things are in L Q M. And so the. So for all Q is equal to Q two. Hence, X plus our Y is in L star. Okay. So sub module. So L is so. L contained in L star M. So, take C equals one and got to see equals one. So all then X to the Q M is in, you know, L Q at right anyway. And then for this, just use the fact that K to the Q is in L to Q. Okay. So, item posts. So this is the one where we need finite generation of the models. So we just have to show that the left hand side is at the right hand side. And what there's C and R naught and a Q naught, such that C Z to the Q, Q M. So, let Z one through Z N. Be the generate be a generating set for L star M. And there exists what Q one through Q N. And C one to C N and are not such that for all Q, figure they're equal to Q I reach I C Z I to the Q is in L. So, right, so let's see be the product, or let's see prime be the product of C times C one times that the dot times C N. First product. Q prime be the maximum of q naught q one dot dot. Then, you know what CD, you could show your CD C Q M is contained in what D times L star M Q M, which is then contained. But you see how we use the fact that it's, that's finally generated, that L star M is to finally generated, which it is because M is finally generated and L star is a sub module and ours in theory. So functoriality. So, let L contain an M and N be finally generated our modules. But G from M to N are linear. And let's see in L star M. And what C is an R not Z. Q naught is a power P and for all Q, Q naught C Z Q M is an L GM. So, C times G of Z Q. And by the exercises is C times epi star of G of Z Q M, which is in epi star epi star of G, I should say, equal to the star of G of C Q M C Z Q M. Which is in epi star epi star of G of L Q M, which again by the exercise is contained in I guess equals G of L. So what do we have C times this element Q N is in G of L Q M. So that's all for all Q for the some fixed C and hence G of C. G of L star. So that's functorial. And then finally for the previous closure claim. Let's, you know, use T equals one. Let me not. Okay, so it's Z is an L FM. That means there exists Q naught, such that Z is Q naught is an L Q naught M. And that means that for all Q bigger than or equal to Q naught, what we have Z Q M is equal to Z Q naught Q over Q naught. And that's a power of P epi naught star of M, which is then an L Q naught Q over Q naught F E naught star of M. And by the exercises, this is L Q M. So hence with C equals one that implies L star. Okay. So remark, the only place where we needed the module to be finally generated wasn't showing item codes. And we're going to just be able to dispense with that later with the test elements. Yeah, that's how tests on this. Okay, so Lana in our, like if you take the zero ideally are tight closure of zero and the type and the previous closure of zero are the same as the no radical direct. So, so let X be in the type of zero. And so we have C X Q zero for all. So there's C and are not such this is true for all the queue. But for all minimal prime. So P C isn't in P. So X to Q is in P. Thus, X is in the intersection of those keys, which is the no radical. So, and so the type closure is contained in the no radical and then I want to show the no radicals between exposure. So X is in the previous closure of zero, then there's a Q. And I want to show that X is in the X is in the radical, then it's in the previous closure. So there's a, I'm going to say that there's an n such that X to the Q is zero actually at a zero. So let Q be bigger than equal to n, then X to the Q is zero, which is the back of power of zero. So, type closure is contained in radical radical is contained in previous closure and previous closer to take closer when it comes to the zero ideal. Okay. So the next results are pretty technical so I'm going to call this thing a pack lemma pack for prime avoidance containment in results that I've read this isn't stated separately as a lemma but it's basically the tool that one uses to prove this thing. So if you let R be in a theory ring and takes any final collection of time deals that where there's no containment relations between them so generally you're taking the minimal primes but you don't have to for this lemma. Take nonzero elements of these residue class interval domains. Then you can lift each of those elements to the complement of the unit of those primes. And moreover if you take some elements in like all the primes but one, and take the linear combination of these ti's with these ci's you get inside of your outside unit of those primes. So it's just, it's probably it's I've got the proof in the notes. And that allows us to, as it turns out, a constructive proof, which again I'm going to not prove because of time. But you can believe that this is going to allow you to fix like this hopefully. So, let L and M be our modules and take an element C of M. And then Z is in L star M. For each minimal prime P of R, the image of C is in, so you can look at L plus PM mod PM. And you can think of that as a sub module of M mod PM, an R mod P sub module where tight closure is taken over for my Z bar is the image of C in. And it really does follow from this pack lemma. And here's another thing. It's one of these things that I think it's just by, you know, by hand anyway. So and then I have a proposition in the notes that says ways that you can take closure and like using all two and so on. Okay. So the next result is actually what we're showing is the tight closure to contain the interval closure I, I wasn't planning to introduce simple closure separately. But actually, in. So, but but it's anyway it's the important thing here is that it's in the is it in the radical. So, I'm going to say, I'm going to first reduce to the main case. Is it an I star X bars in IR mod P star for all P in R if the domain case, and you would know that X bar was in the radical of IR mod P in R mod P. So for any prime ideal containing P plus I X is in Q, but any prime ideal contains P and any I prime ideal that contains I contains some P plus I. So for all Q that contains I X is in Q. So therefore, X is in this. So let X show that. So there's an error in the notes. Okay. So. So now let's assume that ours in our domain. So let X be in. I claim that it's enough to show that prime ideals are are tightly closed, because the radical of an ideal is the intersection of the prime ideals of the primary unit. So let P be a prime ideal, and let X be in the tight closure P. And let's choose the DVR between R and its fraction field, such that the M such that M intersect R is P. And while you have there's a non zero C in R such that for all very big Q C X the Q is in P Q, which is contained in M Q. And so that means if you take this valuation value of C plus Q times the value of X is bigger than or equal to the value of M to the Q, which is just Q. And that means that the value of C over Q plus the value of X is just one for all the Q. But since this is these are all integers, that means the value of X is at least one, which means that X is in M. But that it's also been R. For any ideal I, I is contained, so I star is contained in the intersection of the primary of the P star such that P contains I. But then that's the intersection of keys that contain I. So, all right. The above actually shows. I is contained I stars contain in. So definition ours weekly of regular. If all ideals are tightly close. And ours regular if for all prime ideals. RP is weekly up regular weekly up regular. People have wondered for 42 years or something like that whether those are the same thing 37 whatever the number is long number. And but we don't know. That's rational. If for all parameter ideals are equal to their. And it turns out that any regular ring is that regular. And that's how the term came into being, you know that that that that all ideals being tightly closed in all localizations. So this is like a kind of a weakening of the regularity property. So proof, let R be regular. Then by Alcindor books from Sarah RP is regular for all time ideals. Hence, we may assume. R M is regular local. And you want to show all ideals are tightly closed. Okay, so let X be what I be an ideal in such a ring and take an element and it's tight closure. So you have then there's a non zero C and a Q naught such that C X and Q is in I back to you. Well, what does that mean that means that C is in the intersection of all these cues of IQ colon X Q, which by the flatness of convenience right. Is you can take the bracket power out of the colon. Okay. So, then, by the crow intersection theorem. It follows that I colon X is not contained in there. Because the corner section theorem says that if you that all the intersection of all the powers of m even all the bracket powers of m would have to be zero, but C is at zero and C is in there and therefore. It's a local ring. So I colon M, I colon X is equal to R. So, because that's the only ideal that isn't in M. So that proves that X is in I colon X is equal to R means one is an icon X so one times X is in. So, then there's this technical proposition about systems parameter. I had a question real quick, sorry. Sure. So, in your, your definition of rational just maybe wonder are we assuming the ring is like a finite or excellent or anything. I'm not in the definition of it. I guess, yeah, I mean, I'm assuming there is local. So, and actually now I think about it the way that I find out rational here is is that any ideal generated by a full system of parameters is slightly closed. So, but it turns out that that's the same as saying that any ideal generated by part of a system of parameters. It's tightly closed. So, the zero ideal is pretty close because that's that's a generated by an empty part of the system of parameters. So, or it's reduced. And then the other thing I wanted to mention here is that if for every maximum ideal of R. And primary ideals and primary ideal is slightly closed, then, then every ideal is slightly closed. And basically, the both the proof of both of these things use a version of the pro intersection theorem so maybe I'll just do that, and then stop for the hour. I'm trying to do but six and five six seven. Okay, proof. But x1 through xd is the system of parameters. And let I be generated by x1 through xt where zero is less than or equal to t less than d. So, for each positive integer you if you take x1 through xt, and then you take the rest of them to the youth power, you're going to get a system parameters. Hence, what do we have. I star is contained in the intersection overall these use of x1 through xt, xt plus one to the uxd to the star basically because take closure preserves containment, and I is contained in each of these ideals. But then these by assumption we're assuming so these are in primary we're assuming that in primary primary ideals are tightly closed. So this is intersection of the use of x1 through xt along these elements to the u, which is then contained in the intersection overall the use of i plus m to the u. Because each of these other elements, each of these generating elements here are in i m to the u. And then this is the pro intersection theorem. And to is similar. The primary and primary ideal is tightly closed. Take an arbitrary ideal, look at i plus m to the u for all you. Those are all tightly closed. I is the intersection of them. So I stars contained in all all those stars, but all those stars are themselves so it's 949. And I think, therefore, actually it's 950 so I think therefore I should stop there. Sorry, this ends up being faster than I was hoping that. Thank you. X Q M is at the upper star of I but by the algebra of bracket powers which you'll do as an exercise it is also kind of works nicely. And so lemma is for media closure is a factorial operation on the category of all modules. So I guess I have what four minutes left before questions. So I wanted to state one more theorem and that has to do with regularity. But maybe I should do that next time and instead I'll prove this. I'll prove this and I'll get to the regularity. So, so let L be a sub module of them. You want to show that these closure of them is in fact a sub module. So let X and Y be a sub module of these closure of L and M and let R be R. Then what do you have so X is in some L Q M, Y is in. Sorry, X and Q is an L Q M, Y to the Q prime is an L Q prime M. So, and then what do you have X plus R Y and Q Q prime by the algebra of these powers X, Q, Q prime M plus R to Q prime, Y to Q, Q prime M, which is X, Q and Q prime and something plus R to the Q, Q prime times Y to the Q prime and M to the Q and something. And then that's in L Q M, Q prime plus L Q prime M Q. That's L Q, Q prime M. So it's a module. And it contains L by setting Q equal one. Containment preserving is a claim clear. Basically because KQM is contained in LQM, very Q. So let's just do item potents and functoriality. So let X, let Z be an L, previous closure of M, previous closure of M, okay. Then that means that there's a Q such that Z Q M is in the previous, is in the Q expected power of the previous closure of M. So that means that there exists Y1 through YT in the previous closure of M and R1 through RT in R with Z Q M is equal to some RI, YI, QM. So each YI admits some Q prime, some QI with YI, QI and M is in LQM because these are in the previous closure of L and M. Should be a QI. Okay, so let Q prime be the maximum of QIs and also Q. So just take the maximum of all those things and call it Q prime. Then Z to the Q prime is equal to the sum from I equals one to the T, RI to the Q prime over Q, YI, Q prime. Q prime is a multiple Q, so that makes sense. And by each of these are in LQIM to the Q prime over QI, which is then in LQM, LQ prime. So it's in the previous closure of L. And let's just do functoriality. So let M be an R module, L and M, and G are linear, and let Z be in the previous closure of L and M. Then there's a Q with ZQ is in LQM. And again, by the algebra for various powers, G and ZQ is going to be in epistar of G of ZQ, which is in, that's equal to it, which is in epistar of G of LQM. And then by the exercise, this is contained in G of L plus G of Z is in G of L. So it's a nice functorial closure operation. I'm sorry I had to rush it. I'm sorry I did rush it yet. I won't say I had to. All right. Questions. We started a little late, so. Yeah, I have a question real quick. So, for starters, I like this phrase meditate on Frobenius. I think that should have been my meditations on Frobenius should have been the title of my dissertation. Anyway, I did have a question so as you were defining kind of these these abstract closures. So I can obviously see in semi-primality for ideal closures and functoriality for module closures. So I really, I can obviously see why functoriality is the right thing to look at for module closures. It makes a lot of sense that that's sort of very natural to run into. But for semi-primality it seems a little less obvious as to why that's the natural thing to look at to me. So what do you think about why semi-primality might be natural to look at here? Well, I mean, if you care about primary decomposition, then it has nice. I would say semi-primality is like the poor man's version of functoriality. And the equivalent saying that any R linear map from R to R preserves the closure in this sense. Yeah, it's actually equivalent because R linear maps from R to R, the same thing is home of things, right? The same thing is picking up R and, you know, the map is multiplied by that element. And this semi-prime thing says exactly that that's going to preserve that, you know, I mean, it basically says that element A times the closure of I is contained in the closure of AI. And, you know, you can make sort of crazy, crazy closure operations where that doesn't happen. But for instance, that happens in all these star and semi-star operations that the, you know, that's popular in the non-materials world and it's true in, you know, tight closure, reduced closure, interval closure, radical. I mean, I don't know, it's not really good reason to say that it's natural. But to me, it's like functorial on maps to R. And not every closure operation has like the obvious way to define it on modules like, for instance, radical and even interval closure takes a lot of work to define it on modules. And so, you know, the semi-primality is a standard for that. Functoriality on home of things. Neil, are you saying integral closure of modules is a semi-prime operation? I'm saying it's functorial. Okay. Yeah. I think there's a lot of definitions of interval closure of modules and like, but I think in all the, you know, the existing ones that they are actually functorial. There's a question in chat. What's the relation between Frobenius closure and integral closure? Oh yeah, I'm going to, so I was going to get into that later in the lectures. Frobenius closure is so pretty, interval closure is bigger, is the short answer. But in fact, you have a whole sequence of closures. For any ideal i, the Frobenius closure of i is contained in the tight closure of i, which in turn is contained in the interval closure of i, which in turn is contained in the radical of i. Tight closure has characteristic zero analog. Does Frobenius closure also have characteristic zero analog? I don't think so. I mean, I've never seen it. I mean, I guess you could try to do reduction to characteristic p. I mean, the natural thing would be to say that in all the characteristic p models, I'm sorry, that infinitely many characteristic p models, the sort of the the characteristic p version of the element would be either Frobenius closure or the characteristic p version of the ideal. But I mean, yeah, I've never seen anybody do that. I mean, there's reasons like, I mean, for a long time, people thought that, that, you know, tight closure would have this property that if, you know, during the tight closure in all in infinitely many characteristic p models that you're going to be at the tight closure in almost all of them. But then Brenner and Patsman have a counter example to that. So that's, you know, it's not true. But it was known from early on that that wasn't true with Frobenius closure. I have an example in my lecture. So, but yeah, I guess you could make that definition of Frobenius closure. But in general, Frobenius closure is considered to be more poorly behaved in many ways than tight closure. And that's a segue into another question that I have, and this I guess is more related to material from from Craig's talk so feel free to defer, but I'm very interested personally in my own research, studying Frobenius closure. And in particular, I've been trying to develop the right, you know, and maybe it's it's out there, but the right map theoretic notion to get a closure of ideal, or sorry, Frobenius closure of ideals. So is there a notion of, you know, maps to the right kinds of rings that when you can track back you're going to get the Frobenius closure. Oh yeah, I mean, let me just be careful. Yeah, so there's this thing called the, I don't know if it's maps to them. I mean, I will say there's this thing called the perfect closure of, of a ring which is the, you probably know about it, which is the just all the all the basically all the order the one over cues at once, like the union of the order one of cues in that ring K that I showed assuming the wings reduced. And so, and, and one way to find Frobenius closure is extend to that ring and to track that. And, and similarly, you know, you look at the program for a module, if you have a module M, you look at the national map from M to M tensor with it's called our curve. It's called our infinity that everyone says our curve now for perfect. So you look at the map and the M to M tensor our curve and you look at your sub module L and you look at the elements that that in that map land inside L tensor our curve. I mean that's the elements of the M that landed our curve. And you call and then that's that's the range closure of L again. But I don't know if that's what you, what you're asking for a little bit I mean so our to our curve as an example of a purely inseparable extension. So, I'm wondering if something like, you know, if you contract, I back from the recurrence of the extension of even a certain flavors that you'll get. Anyway, this is a research question that's not. That'll work. Maybe I'll follow up with you some other time. Thanks. Can I ask you something new. This is not returning what you covered but like, is this first part of your talk will broadly follow your paper on like closure operation in community value bro. Yeah, yeah, that's that's yeah, yeah, that's that's sort of where I, that's when I developed this this approach is. Yeah, that's that's that's in there. Some version of that, but I mean some version of that's in these lecture notes to go see. Yeah, you know, I wanted to do a survey and closure operations and I was the and I thought well how broadly defined are these things and like really brother they're even more broadly defined in this after more did his stuff. And like, you know, post that theorists said, okay, well you can sort of define this on any post that it doesn't have to be the, the post set of subsets of the given set can just be any partially or set and then you, you change your containment relations to, you know, the, the, the less than or equal to sign that exists in that post that you still. And then it's still, if it's a if it's a post that we're all arbitrary meets exist. You know, then, then you still get this one to one correspondence that more pointed out for 112 years ago. So yeah, I mean it's very, it's very general. And, and that's, you know, they use that it's somehow a lattice theory in a way that I don't that I've never looked at. Any other question. I'm sorry. But yeah, I'll happily defer your question to anybody that might know more about it than I did. I've looked at it seems like no one has written about. I mean, much less is written about previous photos, where I'm trying to get many. That's great.