 Right, so the last time Dustin started talking about like the relative solid theory and the relationship to eddy spacers and so I want to kind of continue with that. Okay, so I guess I want to talk about the relation between these basic objects that appear in Hoover theory that are called Hoover pairs and the kind of basic objects that appear in our theory that are called analytic rings. There was some more motivation, but now we've seen several examples, a pair of some light condensing A, a positive notion of complete condense A multiple. So the first one we discussed was the integers and solid or just stress that it's one over z, solid being grouped inside of all condensate integers, but then last time Dustin discussed the example where you take the polynomial algebra with the adjoint T and then within all condense adjoint T modules, well you can somehow just take the ones that are complete in the center of this one so you can amount to taking modules over the polynomial algebra inside of the solid in the integers. But then Dustin argued that it's actually quite natural to also isolate a stronger continuous condition giving the solid adjoint T modules, which when geometrically speaking, well this corresponds maybe to some kind of a fine line or this corresponds to those units. And in these cases the condense thing was actually just a screen, but this notion of completeness is still interesting. You could also, I don't know, take VG or QP or whatever and then take polyGP modules or take QP, polyGP modules. And in those cases these are actually just the ones that are with the underlying condensate being grouped as solid. There's no meaningful way to strengthen this. Okay, so the notion of analytic wrinkles and the acceptorized situation. On the other hand, yeah, and the other thing you learn in the serial panic state, then you run into this definition, I mean the basic optics there are these Hooper pairs. So let me recall what these are. There's a Hooper, although Hooper, of course, use different names. Hooper ring, the pathological ring, let me also just call it A for now. A, that admits an open subring, A0 and A, such that they exist some finally generated ideal. And because now it's not clear what ideal means because there are two rings in place when you create an ideal force, I0, such that A0 has the IID quality. Let me give examples in just a second, let me just finish the definitions. And then Hooper has a notion of the ring of integral elements. This is an opening in a Hooper ring, A. It's an open and integrally closed subring, A plus contained actually in the power bond of elements. Maybe I will also recall what they are in just a second. And third Hooper pair, Hooper pair, well, A plus Hooper rings, A, and the ring of integral. Okay, so what are the example situations in mind? And then it first gives some secret example, any discrete ring is a word. So in this case, you can take A0 to be A, until 0 to be 0. Suddenly, let's give it, when any other ring that's itself has an added ring for finally generating a DOI, it's a logical ring with the IID quality, or some kind of regenerated ideal of I. So in this case, you can say it's an A0 to be A, and I to be I. And maybe like actually interesting is when you have something that you buy out of algebra. So, I don't know, so for example, Qt, a Hooper, or any non-academic field of Hooper. And in this case, you take A0 to be Vt, and I to be the only regenerated, which really is only an ideal in Zp and not in Qt, of course. So basically, the idea is that Hooper rings are basically certain kinds of localizations or something like this of such a ring for some IID quality. And so also, whenever you have any kind of Bonnet algebra over Qt, or some other non-academic local field, there's also always Hooper ring. And as A0, you think somehow takes the unit ball in your Bonnet algebra. And as the idea, you can take the one that comes from a uniformizer. A remark is that it's a completion of any Hooper rings again, Hooper. And you will generally only consider the key examples. But completion now in the classical sense, like the law of converging perceived sequences, there's like, perceived sequences more than a no-sequence, is again Hooper ring. And in fact, if you start with such, it's called ring of definitions, the one which has the IIDX apology for some fanatical, very good ideal I, then it will be the case that the completion of A0 will be an open suffering of the completion of A. And this is actually just the IIDX teaching of it. And at least when one uses Hooper rings and Hooper pairs to do edict spaces, then the edict spaces associated with a Hooper ring and this is completion are definitely the same. So in this sense, it's non-complete Hooper rings have the most technical in the whole theory. So and we will only consider complete ones. From the following whenever I say people really put a pair and so on, I always assume that the underlying Hooper ring is complete. Okay, and so I'm here this last time Dustin discussed some notions of like the blotching uniform elements, power bond elements and so on. This can also be defined in here. So again, a Hooper ring, you can define the blotching uniform elements and power bond elements in A. So this is a set of all F and A such that F to the n goes to 0s, n goes continue, which once I assumed it's complete is really just a condition right now. And these are all the elements that the set of its powers is bounding. It's actually problem to say that it's contained in such a 0, some sort of a definition. Actually, a different way to think about this is to say, first of all, you have these canonical these are canonical objects, this is a certain double circle. But on the other hand, we all actually see the fear in the definition that there is some rate of circle ring of definition and some ideal definition of the same. And these are always because of power and elements and the ideals that all consist of topological uniform elements are in a 0 and A. So for example, in the example of Qp, Vp is actually the A3 and the ideal direct P is actually the blotching uniform. So in this case, these kind of conclusions are equalities. Of course, this can generally be true because it could also as an ideal definition here take it's the ideal direct by P squared, but then you don't have all the blotching uniform elements. But in some sense, this is as better as it gets. In fact, the collection of all such A0s and the collection of all such I say from filter collections and A3 is actually actually a filter column of all possible such A0s and also I sorry. And similarly, these topological uniform elements of the filter column of all choices of such ideal definition and rings of definition of A. All right, this was part one of the definition to give a little bit of discussion. Part two was that, oh no, I'm sorry, there is no I wanted to define, I wanted to say that if I also have a ring of integral elements and I can define it A plus, but as it's kind of weight is because sorry. Okay, so when we're learning some theory, I think it's very simple. At first, I think it's extremely hard to appreciate the significance of this ring of integral elements A plus. It is somewhat necessary for Huber to set up a theory, but it's kind of hard to feel why it's necessary. But it actually turns out that the theory that we developed, you think the solid model gives you a way good understanding of what it actually does. And in the precisely matches, okay, here's an example, VHNT, actually has several possible rings of integral elements. I mean, just to just quickly, in this case, the ring of integral elements is just some integrating for a subring. For example, you could take just V instead of VHNT, or you could take the full thing. And in our theory, we will somewhat use this and associate to this an analytic ring with a period of modules. It's somewhat only complete with respect to V. And to this, you will associate the role of the theory. And so the role that is a possible play is precisely the one of saving out which modules to be considered as complete. All right, so maybe I should, I guess it's definition of analytic rings. By the way, and I think I can make this a more notation. Huber uses a single letter A through the note that Huber pair are consisting of some topological ring and the ring of integral elements. And we will follow that conversation. So also when I discuss analytic rings, I want to use single letters to denote the analytic rings, but then they will have an underlying, well condensed ring in that setting. And we needed some symbol to denote the underlying condensed ring. And we didn't come out with anything good. So we chose to just follow Huber's lead. Okay, so let's say A triangle is now some life condensed ring. And then I want to say what is an analytic ring structure on the same? So one isn't allowed to have complete modules. It's equivalent to the one that doesn't get in the first lecture, but presented from a slightly more elementary perspective. So it's a full stock category that I would call the AMO. It's a full stock category of the condensing. It's implicitly nice. So this is a condensed ring group and together with a map from the tender product. The data was just set and now I will make a lot of conditions on this, but now those conditions that we had already seen before twice, like twice we stated that like solid to be induced and the solid to get onto modules, that they have a lot of nice properties and it was a long long list. And sometimes because we don't want to save this list also start me, we didn't make this definition. So first of all, it should be in the accessibility. So stable under kernels and poke idols, but it's also stable. In fact, under all limits and columns or extend them. So if you have an extended to complete guys, it should also be complete. And then there is a slightly pretty condition that you want to check. It's also stable under all x i from some n x i over a triangle where anything, any condensate triangle model or looking something and contains so we want modules. Dustin, did I have something? Thanks, sir. So can I speak? Yes. Okay. Does it imply maybe you can one can prove from this in some way? So for example, does it imply that it is a growth in the category where the condition that I alluded to is the existence of a set of generators. This is automatic under these conditions. What? He said yes. He said yes. Yes. Oh, this is a And does it imply that the x groups in the subcategories are the same as the x in the full or the I don't know what he said, but the answer is yes. Yeah, I know that answer is yes, but he didn't hear that. I mean, in this presentation, the drive completion might not stand to be zero. So if you really phrase it fully in the beyond the level, you have to be slightly careful when you say that. Right. Because it did not be the case that like in here in the second level defined to the right category of modules, and it's not to the right category of this opinion category. It's not. Does he hear what I say? Yes, I hear what he said. So you said that the drive category. Yes. So the second level defined a certain triangle category or stable and thin or whatever. As a full subcategory of condensed, the right category of those things. Yes. Which is a correct one, but in general, it will not be the same thing as the drive category of this opinion category, I don't say. So in this case, in this case, when you have this full subcategory, you'll say that you have a drive category, which is not the drive category of this subcategory. Yes. This is the heart of a T structure. Ah, this is how you do those with homology groups are in this. So it doesn't imply that the X are the same. Right, right, right. So it doesn't imply. Right. It does not. And sometimes it does. Sometimes it is true, but not the agenda level. Like in most practical cases, it will end up being true. Pull this generally. No, I guess. Okay, so I can press that. Yeah, you press that. Now gonna put me, make me confused. So I want to claim that there is an automatically elected on to this inclusion. Dustin, do I really need something for this? No, I think that's fine. What's the argument again for this? Well, I mean, it's again, this general principle. I think it's not quite trivial, but I think it's a general principle that if you have a presentable category and a full subcategory closed under limits and sufficiently filtered co limits, in this case, all filtered co limits, then then the inclusion has left the joint. So there's a very related statement that involves this sets the record nonsense as well. Thank you principle. But I think that you don't have some liners, right? Yeah, or no, something or maybe you don't have the sufficiently filtered co limits or something. I believe that the statement that I just mentioned, which I believe you also mentioned at some point earlier is just a fact in ZFC. Right. If not, then just assume that there is a left to join. Okay. What is the fact that there is? It's like if you have a presentable category and then you have a full subcategory closed under all limits and under all sufficiently filtered co limits, then the inclusion has a left to join. And the category is presentable? Yes. The big category, no, the big category, only the big category has to be presentable. The small one will automatically eat. Yeah, but it is not clear from the stability that they're telling the set of generators. No, that's right. That's it's a non-trivial claim. It's a non-trivial claim which is drawing and what was this? I think it maybe implies you can leave out sufficiently filtered co limits or something. Don't quote me. Don't quote me. Right. Okay. End of definition. Okay. So, okay. So, so, so first I want to explain some general spectrum of this. So, first on the opinion level, so the claim is that there exists a left to jump as inclusions that I will write as setting model m to the base change from a triangle to a. So, so much a right now is just purely intentionally, but I will think of it as a module over this analytic ranking that I call a. And so, it's such a basic standard function. The kernel of this function is the table ideal. And model a requires a unique symmetric analysis structure making the base change a symmetric analysis function. Let's sketch the proof. So, we already discussed the questions in the systems of the left to join there. You think it's formal nonsense. If it's not, then just make it part of the definition. Okay. Now the question is about that this is a tensor ideal. So, what does it mean to be a tensor ideal? So, so the left to join factors, it's definitely this one that we should all call in this. So, to show to tend to go up to show that if I have something in the kernel and that's anything, then it's still in the kernel. So assume n is a triangle module such that and then there's a triangle meaning it has no match to any module. Anything, any kind of same. Then we want to show that take the standard product. Then it's a zero. So this means step for all in model a. So the home from n-sender and the standard approach itself is zero. So this is what this function is by definition is left to join. But now using center home induction to see the result of the homes from n to the internal from n to l. But we assumed that this guy still comes in by functions. It was one of these functions that are stable under all internal likes in particular internal homes. But now if this is a mod a, then again, because I mean, this precisely the condition right, I can also write with it on the mod a, on the base stage, but this guy serves to make one structure. So, so the standard product on a has to be given by taking the standard product in, we'll see, just regards to misconducting modules and then competing again. The question is whether this, this makes it fun to make it more normal. And again, so let's check for all and then I can then first standard and then okay, or I can first keep all of them. There was a standard product where this was by definition. Okay, and now it looks really ugly. But okay, it's only only at the very beginning when you first set up the series, I think, to disambiguate all these things. But the point is just that this function here, this localization, the kernel of this is a tender ideal. And so, I think you don't use the kernel, but you just repeat the same time of the type of argument, homing this to something in the subcategory and using. I did this in the solid case and just do the same argument. And so same argument, let me not try to do this again. That's actually, I realized that on the ABL level, the better thing to say would have been that if I have a map, which is turned into an isomorphism under a, under a localization, that also if I tend to this map with anything else, it also becomes an isomorphism. And then the, which follows from the same argument as this one really. And then the point is that, for example, I have this map from M to its completion, which becomes an isomorphism of the localization because it's an unimportant operation. And so if I tend this with N, some of the same states through it. So there's some structure you automatically have an analytic brain. So there'll be a category of modules, some kind of localization of security of condensed modules that was the underlying ring. And it's quite required in terms of product. And now we pass to the derived categories, let me not just say that a, you know, the green structure on an underlying light and then seeing a triangle. Then I'm going to find the derived category of a modules. So the full subcategory of the derived category of condensed modules. The full subcategory is such that for all the subcategory of all, then we still just call them n. So complex of modules, the homology group. Let me think homologically, if I can hear it. All the homology groups. Line with subcategory. It's defined for thesis. And then, okay, so here it is already a warning. So there is a natural comparison factor of the derived category of what a computer is. But it's not always an equivalence. And essentially all, I mean, basically, all cases I'm aware of, it will come out to be an equivalence, but it's just not a general set. But yeah, so the good thing to really focus on is the thing that we simply call D of A. And so the previous proposition has an n was on the derived level. So D of A, sort of space cluster. A triangle, the triangle of the subcategory. So I mean, probably in one or two lectures, we will probably switch to the infinity categorical language where we would say stable infinity category instead. But now it's not really required, so let me just raise it in more classical terms. Stable under all. So again, in stable infinity cases, say stable under all limits and cold limits, but general limits and cold limits are not well behaved if you focus on the categories. But you can say something equivalent in the stable under all direct sums and all products, which are well behaved and to what they are supposed to do. I'm trying to say. The inclusion again has an episode that I want to call the derived center park. And again, this has the properties that if you have something that becomes nice in here, then if you turn it with anything else, it's still the same. But because there's no triangle category, you could actually phrase this according in terms of the problem. But so if you have something in the kernel of this, so you tend to have something saved in the kernel, so the kernel is the tender idea. And then again, this in the tender property and uses the tender property. So if one wants to do the previous type of argument for this fact, then one lands into the question of whether the, of course, there is internal home in the full derived category by unbounded and so on. But the question is whether if you have internal home from anything to something in D of A, then it lies in D of A. Right. And this is not because of unboundedness. I don't know. Of course, if you have a bounded complex, you have a spectral sequence. I mean, you still have to work with that. Here you have unbounded in both directions. I can see in one direction you have in the same way. So both of the light condensing is still replete. The derived category is left complete. And so I think you can control the question. Okay, let me do this in a second when I come to the proof. So the product, making your own space change. Yeah, so first of all, triangulators. So here you have M-tron to M-double-tron, M-tron-1, the triangle, and then same on yours. And that's the same two of them. And by shifting, it doesn't matter which two are in D of A. Then we want to show that M is also in D of A. That's just how M is also in D of A. But for this, we just look at the long and thick sequence. So we have HM of M-tron, and then we have H1 of M-double-tron here. And these are A. So if I have some portion here, and I have some kernel, kernel here, right? And some parts of the kernel are the same, of M-double-tron. This is a portion of M-double-tron. So these are both A-modules. And then this one is an extension, right? So here we use stability under kernels and functions, and then we use stability under experiments. And I think standing here actually realized... Okay, so stability under direct sums is used to the HIs. And I think standing here actually realized that the stability in the product countable ones, they definitely reduced to HIs. And then the uncountable ones, that was okay, went back where we were working the full condense setting. And when preparing the lecture, I overlooked it. There might be arguments in here, Dustin. Should I just assume that there is a left that's on the middle of the right categories, or...? I'm sorry, I was busy with the chat. What's going on? Why is it stable under all products? Why is what stable under all products? The... Stop getting your complete ones. Oh, all products? Oh, and instead of just countable products? Oh. Ooh. All products exist, but is it exactly your technique? Yeah, this is a problem. The problem is that, like, enlightening things, all products are not exciting. Yeah, this is a problem. Okay. We'll have to think about this. You want an actual issue in some sense, but I screwed up to the definition. So we should ask for the existence of a left electron. I mean, Dustin did in the first lecture, and I just threw it up when I prepared your lecture. For the existence of a left electron, do the things we do. Obviously, of course, I definitely won't say I knew the left electron, because the right electron is similar to all of this. So now I made this next thing actually part of the definition that the left electron exists. So the criminals tend to ideal. So let's say N is complete, and then in any concept, that's all right, what I want to show. N isn't a kernel, and then it's anything. Then you have to show that all the ones that are complete, all of them are pretty often then it's kind of one of those problems. And the center, and first of all, because left on the sense of what's known as a retreat, so of course, this means it's a countable limit of subjections for subjections. Noted that I was introducing my paper, the slugger's button, the provisional sign. And one thing we showed there is that this implies that any such, anything, sorry, for all, I don't know, I didn't use a letter K, right? So for any K, which is, for example, condensate triangle model, came as ethnomorphically to the derived limit of its truncations in degrees at most, and to some kind of possible limit. So in the usual derived category of being groups, I mean, it's somewhat true, right? I mean, if when you truncate it to some degree, and then just take a limit of these things, you're somewhat stable, I think, for an answer. In general, that's an issue because you're taking a countable limit here. In general, a countable limit is in the right category, and that's the same. But under this assumption, you can control it. So this means that I can certainly assume that L is bounded here, right? What I mean. So first of all, I'm going to show this. I can again use tender homogenesis, and I assume that M has trivial completion. So it's enough. It suffices to show that in general, all home condensate triangle models from N, so, I'm sorry, it's not clear. Because then you can rewrite this as a home from N into this guy, but I assume that N has trivial completion, so it doesn't have to anything in D of A. Again, so this I want to reduce to the BN level, where I kind of had this statement that I've had something in mod A and anything, then all the internal X on mod A. The issue though, as the government already pointed out, is that you really need to ask this condition for all possibly unbounded complexes. That's why I mentioned this fact, so this at least allows us to assume that L, here is, so, okay, let's assume L and E minus, as this goes to the right, because also all the truncations, they are still in D of A, but because the condition was just on, there are more of these. On the other hand, N can be written as a core limit of the truncations to the left. This is always true that there's a core limit of the truncations in the region, less than or equal to minus then, as this much here, because core limits are used to direct summons, these are always good. And somewhat similar, you can pull the core limit here out into a limit, and because we know that this is stable, at least undercountable products, I can also assume that N is in D minus. But I think now I'm in business, because once this goes to the left and this goes to the right, everything reduces kind of to D equal X. Okay, we'll sequence to some L, just being degree zero and also N, and then presiding the conditions that the internal X remains real. I think that's fine. And once you have that, the existence of the tender product is just the same formal diagram so that I didn't execute just previously, but didn't sort of paste. All right, so another thing I should have really mentioned as part of the general thing, but didn't, so let me put it now, is that the derived tender products are, sorry, yeah, so the D of 80 has a natural T structure making the inclusion of the T-exact. The left and right is not a general T-exact, as we've already seen in the solid base, where solidification could turn something to the left, but if, but still, since there are phase change, it preserved as a connected part. A G structure is that you have a notion of like things which, that you can have a notion of truncation of complexes kind of, so that you have a notion of like truncation complexes, which certain non-negative degrees complex, which certain non-follipant degrees, they satisfy all the usual properties. So lately, I sort of thought, we've definitely made this trying the subcategory which is stable under all the different patients, right? So, so this inclusion is T-exact, and then it's just a completely general fact that if you have a left adjoint to a T-exact function, then at least it preserves the connected part. Check whether this maps to anything which is considered on the right, but this is a left adjoint, so you can compute the morphisms in the larger category, but then this is still in this category. Right. And so, in particular, you can talk about the heart, and the heart, all this also, definitely, is just not A. And so if you take this and pop to the heart, this is this thing, and if you take the derived tender product and pop to the heart. So in this sense, the derived and the VN level are compatible, and then there's the other question whether if you start here and just animate all these constructions above the derived category, whether we recover those, but this is just not true in general, yeah? So in general, we don't give them recovery of A, but even if you do, there's separate questions whether you recover the product functions, and again, it's not in general. Although, I think if you do recover the product categories, the adjoint also is a correct one, just by functor reality, but the tender product is a subject. But again, in practice, it is true. The ski of A is just a derived category, and all these functions are in derived terms. All right, was that out of the way? I'm almost done with my lecture in front of me. So the next two comparisons list is Hubert rings. So when we had Hubert ring, we had these topological newpot elements, the top one at the moment, and so on. And Dustin already gave a variant of this principle, Hubert rings themselves, or Hubert rings in pairs. They're meant to condense rings, of course. They're all about the race. This is actually fully traceable. All my ones are confused. But actually, this is an ancient solid rings. All my rings are completely new. The ones where they underline condensate being used as solid. And maybe I should call those not exactly. All right, and so then Dustin already asked that he gave a definition that it's a triangle, a solid ring. We can define a subset, a double third and a third of the underlying ring, where then we're called a recognition. A double third was set up to those elements in the underlying ring, such that the corresponding map leads on T to a triangle, set E, T to F, that factor is actually important. There was a discussion about how much structure do you need to check here. And so the condition was that it factors as condense rings. It's actually enough to check that it factors as condense modules over this point. And then it's simply saying that there is a sequence, a null sequence, one F squared and so on until zero. But together, it was a small condition that some of F times the sequence of shift of the sequence, negative model. And so if you apply this to the case where just sort of being a rose from a Uber ring, then there's precisely a set of top-up of the elements of the Uber ring. And then a lot of second. And how about the elements? Those were defined from those elements. When you regard A3 as the VO2-IT module in this way, on the same way, that it actually comes up. And again, you can show that there's precisely the same thing as the worst condition of being powerful on it. And so then, Dustin Soplowski talked about that this is always an integrally closed subject. And this here is always an undue. Radical update. So what's the definition of Western X? Yeah, sorry. I mean, given F, I can again use on T20 served. Let me write it again. So why are we going to have to change T to F? I kind of regard it as a CO2-IT module. And then the condition is solved. I'm already speaking some of the modules over at the CO2-IT. They have the modules of the closed unit disk. So Dustin, what did I do last time? So this means that on this algebra, you have to have T should be at most one. So it should be a problem. You can also check it. OK, but now I can also make the following. So assume A is an analytic ring structure a solid ring, a triangle. So then I can also define an A plus. I realize I didn't find maps of analytic rings. So let me do this in just a second. Let's set the map from D to N, T to A. Same as always. T goes first. This map induces the map of analytic rings, D to N, T, solid. I'm going to be corresponding to solid D to N, T modules. So something that I should have said previously, but forgot. This is what I'm not going to be doing. From A, which is the pair of condensed rings. And it's getting your modules to some B, which similarly is the pair of condensed rings exiting your modules. Here's a map of unaligned condensed rings, which has the properties that if you use this to restrict the condensed B module to a condensed A module, then people will say completely. For a section of sphere, I think it's all here. So here you have the same module pointing onto this. And in this case, you can pass the left adjoints. By the way, I mean, once it's true on the A-B level, the same statement is true on the level of the right categories because they can check it on the level of bodies. And once you pass the left adjoints, the left adjoint to restrictions on the extent of scalars, the left adjoint here is what I termed the base change function. And so you get it also left adjoint here. And if you want, you can compute it by first base changing as condensed modules and then completing. And you also get a derived environment. Okay, so the claim is that, first of all, once you have such an unaligned structure, you can get a datum as in Hooper. And so I really need to check that it's actually automatically satisfies this list of conditions that Hooper puts on this particular element. Conversely, I'm not sure if I have time. I hope I can say it, is whenever you have a regular integral element in Hooper's theory, you can actually produce an analytic ring, which somehow the initial one in some sense. Okay, so, right. So first of all, I can also rewrite this as the following conditions. It's all those things such that one minus f times shift, which isn't in the morphisms, from the derived base change, p versus, so p I recall is always a speech based on the morphism. And we characterized being solid over zero and t by, well, being solid, but this we already asked. And that one minus t times shift over this is an isomorphism on this projective generator. So what's actually going to ask is that, if I look at this thing here, get an object in G of A, then this is nice. Why? So if I admit such a map, then this already happens here. Like already here, one minus t times shift becomes an isomorphism of this object, like definitionally. But on the other hand, because this precisely characterized the solid modules here, you can also show the progress. Like if I want to show that I have such a map, I need to show that all complete modules here restricted complete modules here, but being complete precisely means that if I hold one minus t times shift into there, it becomes an isomorphism. And so it just translates into this. So basically we are, like whenever you have any element of the underlying ring, you can ask this condition that one minus f times shift becomes it's an isomorphism of P and this will define for your analysis ring structure by this general, yeah, basically whenever you have an anomorphism of your compact project with object P, declaring that this should be nice and often always find an analysis ring structure. So you can take any subset of glory of this ring and ask this type of conditions. So I want to ask a question, I just asked Dustin, but it's not completely clear. So since you define the notion of map of analytic just in terms of models, so if you do this, you instead of does the derived tensile L, you send to, so this P tensile L means that you take the tensile product and you apply the left agent in the derived sense. But you can just take the tensile product and apply the left agent on modules. And so this is what does, this means P tensile Z A without the L I suppose. But then, so to check this condition, you get what I said. So if you try to check the in this map of an analysis ring, you can translate to this kind of statement, but with this factor on modules, but then it probably is equivalent to what you have written if you think about it, but in the derived way, but I just want to confirm that the two versions are equivalent is the one. Yes, right, because you can actually pretend analytic ring structures on the level of modules. It's enough to check it on the underlying, on the Indian level. Then it's also true, yeah. Okay, so it's equivalent. It's equivalent. It seems. I'm going to use a smaller mantra. So with the drive phone, I would be more confident, but I think the argument just gets this also true, so here. Right, so the point is that this satisfies basically all the, this subset also automatically satisfies the conditions of two versions. So it's always contains the top-locking new book in the islands, which are always an open subset of the series. In particular, it's open when it comes from Huberry. It's always contained in the power of all the elements. And it's always an integrally close suffering. So why is that? So, well, as F is an A double third, then we actually get a mapping Z power series T. But if I have a module that's actually Z power series T module, we'll call it Z. That's automatically. I mean, Peter, actually, I mean, this proof is just exactly the same. Yes, exactly the same as last time. Because all I used was arguments about modules being solid over one and plotting are solid over another and so on. Let me just say it again. Okay. So, okay, so these are actually all there, but this means that whenever I add a module here over A, so in particular, it becomes a module here, which is everything was solid. So it must be an underlying solid ring. And so it must be solid with the adjoint T. I should have said solid. If the adjoint plus, then I get a map from Z adjoint T solid to A. But in particular, this means that a random line condensing, which we always assumed is a complete module, by restriction becomes a solid to exchange on Z1. So F is an A. Right? So this is a proof-seeing solution up there. And it's engineering for a separate that's the same argument as in Dustin's list. So in fact, yeah, so the argument that Dustin gave there was already talking in something not just about a triangle, but about any module. And so if you just run this argument, you see that this is what it actually proved. Okay, so, right. Thus for a Huber ring, a triangle, I have the solid analytic instructive solid under the instructive one, which lives over a solid Z. And Huber consider this notion of the rings of integral. And I just gave you a recipe here that was taking some such, you know, the ring structure here and produced a ring of integral as a positive mean of view. And it's actually factorial. So you can actually show that one and the rings work to respond to it. Yeah. I mean, if you have a ring of the rings, then you get an inclusion of the ring of integral. And this actually has a left joint, I assume. So whenever I have a ring of the rings, I can produce a new ring structure. So it might be because of the solid. Sorry, yes. So it's solid, but it admits. And that's exactly what you need from the solid. Okay. Well, this solid is the Z. I mean, it's unique because the matter of rings is unique and then it's just the condition. So it's a condition that all the A modules are actually solid. And the different way to phrase this is to ask that one minus fifth has an n-emorphism of key tending Z, A. All right. So I wanted to say that it has a left joint. So if I had a rubber pair, then I can't send this to and then the secret A is associated with the rubber pair. Where mod A, by definition, all those, from then it's modeled over the underlying ring, such that all F and A plus what I take internal home from P into F. I have one minus F times shift acting on that. And that's this one. And because some of such a process are always integrally close, it's actually enough to ask for some kind of generating set. So whenever you have a subset of A plus that generates that there's an open and simply close up, the ring of integral elements, then it's enough to check it for the subset. So usually there's just one or two elements of something like this where you really have to check. So it's the same thing as those ones. So it's always for those two elements. So yeah, so to connect the spec at the beginning, so for example, z comma z is a rubber pair, and this is just called a z-module. If I take z dot t, that's all the pairs. So if I only put z here, then I'm only asking that some are solid over z. So I take all t dot t-modules and solid z-modules, but then when I take t dot t, here's your initial smallest, then this becomes the relevant source. And then if you have the time to let it just take one last proposition, so when you start with a rubber pair, and then go to, then if I go back, then this actually match back to A plus. So if I, right, so I have an analytic ring, and then I can take its, so I can take this thing here, and take its plus ring. So this is actually A plus. If we want to leave, so the left and right bunkers are definitely accessible from a rubber case into all other rings, but they also and yeah, I mean, really still quite struck by how closely the series that we saw that was the rings really matches Hoover's classical theory. So if you restrict to, and the rings where you only ask allow yourself to put conditions that one minus some element and shift operator on t becomes an isomorphism, then you're precisely getting those under the ring structures that are induced by rings of integrals and Hoover's sense. So which is kind of a very strong, a posterior motivation for this definition of ring of integrals. All right, I should stop.