 First speaker of the day, Mark Shardin. He will talk about residual intersection. Mark is a GCA in France. He's a big exponent actually of commutative algebra in France, unfortunately. We don't have as many. Mark is fantastic. He has done so much and organized so many conferences at Lumini and made France a point for commutative algebra as well. I met Mark almost 30 years ago, unfortunately. And I would like to remember that time, because it was a big time for tight closure. It was actually a conference, a CBS meeting in Fargo in July 1995. And Craig Unique gave 10 fantastic lectures on tight closure. It was my first actual introduction to tight closure. And Mel Ochster gave a very, very good talk. I still remember two of them on tight closure and characteristic zero. And the meaning of tight closure and characteristic zero. This was an amazing meeting. I met so many people. It was the end of my PhD. And Mark was one of the fantastic meeting I met there. And it started a very, very long friendship. So with big pleasure, I introduce Mark Shardin. We'll talk about one of my favorite subject, residual intersection. Thank you, Mark. Thank you very much, Claudia. OK. And as you will see, this is very much created to early work of Craig. But let me start. I think you have seen it yesterday with Bent already. So my subject today is residual intersection. And you already know what it is. But the first part of my talk will be very much very close to what Bent explained yesterday. But anyway, I decided to recall this somehow. So in all the talks, I will be a Macquarie local ring. But everything that I will explain extends to what is called sometimes a style local. It means a greater situation where you have a unique, greater, and maximum ideal. But this takes longer for the statesmen to hold out. But so I will stick to this local case, which is sufficient for most of the things. So the first part will be about years on linkage in Macquarie, like yesterday. So I remind the definition of years on linkage. And so two R ideals, I and J, are linked by a complete intersection ideal B. So it's an ideal B equal X. What would be my notations? Maybe it's better to put F1, FR, such that the co-dimension or height of B is equal to R. So that's a complete intersection ideal. And you require that J is equal to B, co-on I, and I equal B, co-on J. And so some remarks on that. So this implies, as Brent mentioned, that both I and J are mixed. So both have only associated primes of height R. And if, on the other hand, if I and J are mixed, are supposed to be un-mixed, these conditions are current. So these two conditions are equivalent, provided that R localize at P is Gornstein for all P that contains I plus J, and height of P is equal to R. So in particular, Gornstein could mention R, but also if the height of I plus J is bigger than R, this is called geometric. So if the height of co-dimension, I will use maybe two words. If co-dimension of I plus J is bigger than R, and this is called geometric, as you do in the section, because it means they do not share any associated prime, then you know that this one is equivalent to the other, provided they are un-mixed. And moreover, in this case, you have that the co-dimension of I plus J is equal to R plus 1, and the ring of I plus J is Gornstein. So that's a nice way also to produce Gornstein rings. Well, now if you iterate basically all, so iterating this, you get that you can start from an ideal I, then you rigged by the first ideal, convenient action B1 to an ideal say, maybe I have one, whatever, and you continue n times to ring to an ideal J. And then there is an old theorem of Dubray based from 1935 that tells you that if n is even, then the depth of arm of I is equal to the depth of arm of J. So in the Ingrid class, you have the same depth. And now if further R is Gornstein, this is Peskin and Spiro, by iterating, was published in 1974, tells you that if R is Gornstein, then arm of I is called Macquarie, if and only if arm of J is called Macquarie. OK, nevertheless, I mean, it's not true that the depths will be the same, but you can say things about the depths from one from the other. But I won't enter into that. But it's, I mean, for instance, if locally your coin Macquarie, then the local homology models that measure the defects of coin Macquarie will be permuted and dualized. So you can say, what happens? But it's a little bit more complicated. So what is the key point in these two terms, what gives the so-called JSON sequence, is that if you look at J mod B, then this is by definition, it's already divided by I mod B. And so you can write this as home of R of arm of I into arm of B. And here, the correspondence in element here correspond to the map that's n1 to x. So this map is defined if and only if x is in that. So this gives you this isomorphism. But now, since B is a complete intersection, this model is isomorphic to this x model into R. And you see that this doesn't depend on B, which is independent. And that's the key thing to relate. Because, I mean, B has no comorgy. So you get that the funds of the local comorgy models, with respect to the maximum, will be the same for I and J, except the last two. You have to be a little bit careful about the last two. But that's the key thing. And now, if further you are Gorenstein, this will be just the canonical model of arm of I if you're Gorenstein. So if further you're Gorenstein, it tells you that this, on top of being independent of B, it's also depending on the ring. Somehow, it only depends on arm of I. And that's very important as well. Now, another thing is that Dubray needs n to be even. And there is a nice example already in the paper of Piscine Spiro that shows that. I mean, if you take the ideal I, so this would be a projective example. So it won't be, you can localize, but this would be typical of the graded case. OK, so you take the following ideal. So this would be correspond to curves in P3. So you take a polynomial field with four variables. So you take x0, x3 minus x1, x2, and another equation, x0, x2 squared minus x1 squared, x3. And this ideal defines rational quadratic union two lines. And this is easy to see, because this vanish on the rational quadratic. It has a degree 6. This has a degree 4. And you have the two lines, so it's easy to show it's equal. And also these two lines, these are explicit. They do not intersect. OK. So first, I mean by Piscine Spiro, you know that R-modulo IC intersection I1 is con macore, because R-modulo I2 is con macore. But also as these are disconnected, you know that R-modulo IC is not con macore, because this union of lines doesn't give something like con macore. But on the other hand, R-modulo I1 is con macore. So this is just, I mean, one link in the con macore ring. One side is not con macore, where the other one is. So there is no chance for Dubro's term to be true. OK. Now, there is a nice term of Charles' water. Well, that says that this is not an isolated phenomenon. So this is in 93, which is 93. And they say that if you take C, it's more general, but I just give this, if you give C in P and a curve, any curve, which means you have no associated points, OK? Then there exists another curve, C prime in the same PN, such that both R-modulo IC prime and R-modulo IC intersected IC prime are con macore. So any curve could be linked to a con macore curve by a con macore curve. That is the union of the two. So there is no chance. If you look at, like, Yezong classes for by a con macore, everybody is the same Yezong class. It's easy to deduce from that, OK? On the other hand, if you fix the ring, OK, and you look at how you pass from one guy to another, that's a different story. But if you are allowed to ring by any con macore guys, that's OK. So if you ring by complete intersection in a con macore ring, many things are kept, OK, or transformed in a way that is similar, OK, for all even and all odd, OK? But this is different. You ring by con macore and then nothing is gone. OK. So now, to understand what happens, I'll explain. I mean, you uniquely introduced causal homology. So now let's assume your idea I, oh, this was bad. I mean, this should be fine. X i's were a good idea here. Sorry. No, these are just linked. And so this means that, I mean, this is geometric linkage. So you have that, of course, this is easy to see. When it's geometric, you always have that i c will be i of c union c prime, colon i c prime, and the other way around. So when you have a geometric operation, there is no problem in reversing or whatever, and the other way around, OK? So we are just, I mean, OK? So it's, I mean, the thing might not be reflective if you are locally on the intersection, I mean, you're really not a grandstand, even at the generic point. So, but if you do geometric linkage, you don't need to ask both. So this is something. So now, well, let's choose generators of i. So i equal f 1 f p, OK? And then consider the causal homology model. So this would be h i of the causal complex on the f's over the ring r, OK? So these are, I mean, well, these are m mod i models, OK? And then you put the definition as in fact yesterday, you say that i is stronger on Macquarie if these h i's are quite Macquarie for all i, OK? And, well, this implies, in fact, that these are all maximal Macquarie are modern models, OK? And they do satisfy some dualities, and this was proved by Herthard, OK? And now, what I wanted to say, yeah, so why introducing this one very simple way to see why this is important, maybe continue here, is if you look at the model h p minus r, so number of, I mean, you can choose different side of generators. Typically, you take minimum one, but it doesn't matter. This model is isomorphic to this model we have seen before, xr r, r mod i into r, OK? So knowing that this guy's got Macquarie, of course, we force things to be good, OK? But before stating this, I said this first important term of Munich in 81 that says that this property of being strongly con Macquarie is stable under even linkage. So that's, we have seen that here yesterday, I mean, so if i and j are evenly linked as strongly con Macquarie, if and only if j is strongly con Macquarie. So as Ben explained, this gives you all class of ideals that are the richie ideals, but because these are, of course, con Macquarie, because you're linked evenly, or all the way, that's the same two-compete intersection, OK? And this includes the perfect i2 and the grand shine con dimension 3. I mean, perfect of i3, whose resolution is self-due. So this includes important classes, OK? And there is a second very important term that will also finnacle in 85. This form that says that i strongly con Macquarie implies that j is con Macquarie. So this, if you have strongly con Macquarie on the other, you can pass, you have con Macquarie on the other side. And if, and that's important for residuals, if it's a geometric residual intersection, then the modules h i of i over the quotient ring s mod j, OK? So let's put i bar r con Macquarie modules as well. Why change ring here? OK, r mod j. OK, so this is what I will say about linkage, or liaison. And now I'll pass to residual intersection. So we have seen yesterday the definition of residual intersections, OK, but I will record that. So the second part will be about residual intersections. OK, so let me record the definition. So r is always my con Macquarie local ring. So I give an ideal r inside r, OK? And next, co-dimension r, way of co-dimension r, OK? And I will take an ideal a, generated this time by s elements, x1, xs. And I assume it's a proper sub-ideal of i, OK? And I will call j, j will be the column, a colon i, OK? So remind the definition, j is an s residual intersection. The co-dimension of j is at least s. And it is geometric. Further, you have that the co-dimension, same thing of i plus j is strictly bigger than s, OK? Means that no associated prime of j contains i, OK? So it's very important that the, I mean, s here is the s here and this is s here, OK? You assume that the dimension, I mean, the co-dimension of the colon is exactly what corresponds to the number of elements you have here, OK? So, yes, all correspond to s equal r, OK? That's good. OK, now let me say that, I mean, this notion was introduced by Craig Hineke, I mean, the definition like this. Why he was repairing, I mean, an argument of Artin and Nagata, who claimed, I mean, falsely that, I mean, if you take residual intersections of ideals whose link are coin-macquarie, first link, I mean, even odd classes coin-macquarie, then residuals should be coin-macquarie. The thing is that in the paper of Artin and Nagata, they look more as the composition of your ideal A as I intersect some, say, j prime, OK? So that the co-dimension of j prime is at least s, OK? And this is not uniquely defined, of course. And the mean point is on the non-unique defineness of that. And this gives you a very defined guarantee. So now the term of Hineke in 83 says that if I satisfy the foreign condition, well, is coin-macquarie first, is strongly coin-macquarie, and the foreign condition we have seen yesterday holds, and this is called GS condition. So it's a minimum number of generators of I localize that P is at most the dimension of our localize that P for all P containing I, such that the co-dimension of P is smaller than s, OK, then r mod j is coin-macquarie. And so Hineke for the inductive argument of Artin and Nagata and repair the gap that was there, OK? And to be able to do such an inductive argument, you need to have residual intersection up to write s and up to write s minus 1. You need them to be geometric, OK? And this is exactly the condition that is equivalent to the existence of such geometrical residual intersection. So if you don't have that, the procedure cannot work. But as soon as you have this, OK, then you can repair the thing. And using this thing, essentially, this star I'm here, well, this was published later, but the thing wasn't before. I mean, not only, I mean, I is stronger than McCoy, but you look at it over r mod j, you are still stronger than McCoy. And that's the key point in the interactive process, OK? Now, this was refined, Herzog was shown that a weaker condition is enough, OK? So that's the time. So that's the time of, so this is for Herzog, Pasconseo, and the RRL. This is from 85, OK? It says that if I satisfies condition Gs and you put some condition on the depth of, because you're homologous, but these are not as strong. You don't have them to become McCoy. So you have this condition siding depth, which means you are at the depth of h. We have seen that the hp minus r is a canonical model. So this should be of 8, at least the dimension of r mod i. But you are at this, if you look at the next one, it will be at least the dimension of r mod i minus i, OK? So the last one should be McCoy, the next not too far, et cetera. Then, OK, you can also assert that r mod j is McCoy, OK? Yeah, you can apply this. Of course, I mean, this do not depend on that. So this you have that. But what they prove, that's a good question, is that the converse holds. So if you assume, now, also they show the converse. So if you assume g infinity, of course, infinity is just the dimension of the ring, if you wish. Then, if any residue intersection is McCoy, it implies that siding depth is satisfied. So this is a minimal condition. If you want it to be true for any co-dimension, you need to satisfy this. Well, if you can make geometric residuals. So if all geometric residuals are co-in-McCoy, OK, and there exists residue in any co-dimension, then you should satisfy this. That's also what they proved, OK? So that under g infinity, that's a necessary and sufficient condition if you want that all residuals of all co-dimension are co-in-McCoy. But, well, this is not entirely satisfactory, because as a band, you've just noticed, I mean, you have a strong dichotomy. Either s is r, and you just impose something on r or die, or s is 60 bigger than r. And then you impose things on all the cosy homologies, OK? And what it is was to, oh, no, before I go to that, well, I can do that, but I will, because it's not anywhere in historical talk. So I mentioned this first, but yeah, why not? OK, so what Bernat introduced were condition on depths of power. So let me state this condition. So there is another condition, we could try to get some powers, and this depends on the number, OK, epsilon. And this you asked that the depths of r modulo i to the j to be at least equal to the dimension of r mod i minus j plus 1, for j equal to 1, it's just going macro, but you ask this in the range j from 0 to this number epsilon. So epsilon is sdp0 is just going macro, and s, and then you have more and more conditions, OK? And so what Bern proved with time of fruition, this is 94, I think, is that now assuming r is going to shine, OK? Then first, I mean, if is gs and satisfies your, sorry, gs, but I said it was correct, but not what I wrote, and satisfies this condition on power for s minus r, so s minus r first powers, OK? Then three things are satisfied. First, what I would call the as, I mean, you have that r mod j is going macro, OK? bs. So this j is always an s, as we do in the intersection, OK? The canonical model of r mod j is, I should erase something. Yeah, I think it's the first time to explain this. So it is what, well, it's known now to be the expected one. So this should be the s minus r plus 1 symmetric power of i, OK? And third thing, cs, is that, in fact, omega of r mod j has to be i to s minus r plus 1 first j mod j if it is geometric, OK? And he also proved the converse, namely, if you have a little more, so if you have gs plus 1, meaning that you have, I mean, geometric residue up to co-dimension s, OK? And condition as prime and gs prime are satisfied for all s prime at most s and s as prime as you do in the intersection, OK? So if you have things up to the difference between the two, OK, then you know that s, the second depth on power holds s minus r. So you should have this depth condition, OK? If there exists geometric residuals, OK? And the canonical model is the expected. And I have some time. So I wanted to mention another thing. And yeah, you have now, in all these statements, there is this, the proofs relies on the inductive technique of Artyn and Nagata. And so you need some gs condition or something. You need gs, at least, OK? Well, maybe you can do it with gs minus asking a little less. But this would be complicated. And anyway, but there is no hope, at least, to have no condition on the overall number of generators because it's an iterative process. But there was one case where, using the structure term on Rison, Benton and Craig showed the following thing. Is that if you're evenly linked to a nice Sonico and Macore gs ideal, then things are also fine. So this is Heunek and Ulrich in 88. Let's show that if R is Gornstein, i is evenly linked to an ideal that is Sonico and Macore and gs, then, well, R mod j will be equal in Macore for any S resolution intersection. And the canonical module of R mod j is isomorphic to the symmetric power S minus R plus 1 of R mod a. And furthermore, it will be isomorphic to this power. In the argument, it's important to trace, I mean, so they do some generic linkage and use their structure term to be able to deduce things by deformation. But as far as I remember, in this deformation argument, you need to keep track of the canonical module as well to be able to deform. So it's important to know what is the form of the canonical module. OK. So now let me come to. So in all this, due to the induction, there is this gs condition. And then my student, Hamid Asansali, tried to get rid of that and to see what could, I mean, things could be true without that. OK. So if you want the question, at that time was is this condition gs necessary for this to all? OK. So this will lead me to the third part. Well, I couldn't go on maybe homological methods. OK. And well, the idea is to build a tentative resolution for the ideal j for the residual. OK. So we have done something like that with band in good cases. So in the case, for instance, where i is a complete intersection, but even more generally. OK. And this comes from an old idea that traces back probably at least to camp for, or if you, or you can see that also in the book of Fulton on the residual intersection. Idea being that you take your things, but you work your idea, then you become principled, and you divide there, and you push it down. OK. In algebraic terms, that's the way you also construct basic or North Scott complexes. One way of constructing them. OK. So the idea was to use this kind of procedure. OK. And I will explain a little bit how it goes. OK. So now you have your ideal. I recall it's F1FP. OK. And you have your elements. And so your elements, XI, you can write them as some of some coefficients here, jfj. OK. It's not in a unique way, but you write them like this. OK. And now you look at this on the symmetric algebra. So you introduce a ring, which should be S, which is RT1TP, as many variables at the FIs. OK. And you have this symmetric algebra, which can be seen as a, if you give generators, it would be the quotient of that by the CGG relations. OK. And now the thing is that this ring is easier to handle than the Ries ring. In good cases, they coincide, as you have seen. But this is the symmetric ball of this is nice, because we have a tentative resolution for it. OK. Which is this azocimus vasconcellus approximation complexes. So you have the complex Z dot, the approximation complex of azocimus and vasconcellus. Well, which is, we have seen that a little bit yesterday. I mean, you have the causal complex of the vibrates of S. And inside this, OK, you can look at the, so I mean, the modules here are modules Ki of T of S. OK. But this Ki is, you can look at, yeah. Inside this Ki, this is just, I mean, wedge I of R to the P. OK. So inside this thing, you have the cycles. This will be the cycles of the causal complex on the Fs of R. OK. That's the sub module here. OK. And so this will be the sub complex that you have. I mean, so you take the cycles, you extend to the T. So if you want, you take the elements there, but you take coefficients that are cycles. OK. And this is indeed a sub complex, because when you look, when you apply delta T and delta F, this is minus what you obtain by applying delta F first and delta T. So this will show that this is indeed a sub complex. Now, by the very definition of this, you have, well, I should put a grading here. OK. It doesn't really matter. I mean, you have that by the very definition that h0 of the dot is just a symmetric algebra with this presentation as a quotient of that. OK. And also, the fact that is very important is that when you look at the homology of these things, these are SI modules that only depend on R inside R, up to isomorphism. Of course, you can choose many generators, but these modules, these homologies do not depend on the choice of generator, although the complex depends a lot. OK. Now, out of this, what you do is you study, you introduce the corresponding things will be corresponding things on the board. So let's call gamma i to be the sum of CIJTJ. And now, you will look at thing of work and project. And the right way to do that is that you look at k of the gammas, and you look at it on the symmetric algebra, but you replace the symmetric algebra by this tentative resolution. So you take this complex here. OK. And this will be complex d dot. OK. And now, if you look at the ingredients of that, the module d i is the direct sum of some modules e i, I would say what R is, that R modules that you extend to the variables here, and there is a shift of minus i. So that's important in the argument. It's the degree t shift. This is important that there is exactly this shift. That's the important thing when you want to project what happens, like in the Egon-Norskot, OK? And this e i has a simple expression. It's the direct sum from i, from j equal what? i minus s to i of zj to some power, which is s i minus j. So it's a bit complicated, but explicit, OK? And now, you push down this complex, like you do when you construct Egon-Norskot complexes, and you get a family of complexes like the Egon-Norskot ones, OK, z dot, they call z dot plus, OK, which is 0. Well, here, and this is the length of that is s, OK? And so this will be z dot k s, et cetera, OK? And this will finish. So these complexes are the expected length, s, OK? And you can describe what are the ingredients. So these are complexes of r modules. This is complexes of r modules. These are graded pieces of complex of s modules, but these graded pieces are complex of r modules, OK? And now, if you look at what are these ingredients, OK? This will be, well, just the module E i we have before, and s in degree k minus i. So just copies of these E i, OK? This will be for i at most the minimum of s and k. And for i bigger than k, there's something slightly more complicated. So you have E p minus i plus, oops, sorry, plus i minus 1 times the dual. This is, yeah, converging the t's of the polynomial in the t's. But this, I'll write as r of s in degree i minus k. What is it? Minus 1 into r, OK? So I mean, these are just some of copies of these E i, OK? And the maps are just the, I mean, the maps induced by the ones here, by Caussure. These are pretty simple, except one map. And the transition map here, OK? So you have a transition map that is more complicated. That is the one that goes from the piece in degree k plus 1 to the piece in degree kk, OK? Because this will be, what? This will be, I guess, E k. And this will be E k plus p. And this is the, what you obtain by this process is the transition map in the spectra-seq, so this is a bit more complicated to describe, so we'll return on that. But the homology at least is simple. The A0 of this complex is the k-symmetric power of I m of A, and that's easy to see, OK? Now, for I equal to 0, it's more complicated, and so I will return that. So, well, maybe with the term, I mean, it's essentially due to a sans-sardine part of it. So I will put the first term of proposition where I put the standard things about that, very important. I mean, the first thing is that about our thing, I mean, so J dot, which will be defined as A0, sorry, A0 of this complex, the degree 0, 1, OK? So this, by what we, by the form of this, this will sit inside R, OK? And this is always inside Rj, which is the column. This need not be a residual intersection, whatever. I mean, you take any, I mean, two finitely generated ideas in a commutative ring, and you have this inclusion, OK? And all the important thing is also easy to see that this only depends upon I and A. The second thing that is, I've already mentioned, I mean, is that the HGO, the K thing, is a symmetric power. And the third thing that is important, maybe, I mean, anyway, so it's, I mean, of course, I mean, some of these are not so easy, but, and the third thing that is important to know to start with is that if I is stronger than Macaulay, then you have some acyclicity, I mean. So there are many things about the acyclicity of these. I won't turn into that, but if I is stronger than Macaulay, then you have that these complexes are acyclic, OK, for all K in the range we want, OK? So that's also important. So this tells us if I is nice and that we can condition forcing this, then this will be acyclic in a good range, OK? So this will tell us things about, I mean, J and, because this is in J, and it will tell us also about the canonical model and all that, because we know some symmetric powers. That's what we expect to be the canonical model and things like that. So then using this, not all this at that time, but using this technology, Hassan Zali proved in 2012 the following thing is that if I satisfies conditions running in depth, then first you have that J star is going Macaulay, OK, and this ideal has the same radical. So I mean, it may differ by, I mean, somewhere, but at least this has the same radical. So I would say that now that we have all this done and that we explain later, this could be a better definition somehow for the residual, OK? But OK. And the second thing is criterion to know when these are equal. So J star localized at P will be equal to J localized at P whenever localized at P is cyclic for all P containing I plus J such that height of P is equal to S. So in particular, if you have a geometric residual intersection, this is always satisfied and this is equal to J. OK, so now let me turn to the last part. There are two questions that are interesting. It's first to describe which J star in general, OK? And second to know when it's equal to J, OK? Further than this example, these cases, OK? Besides the geometric case or some, so explain a bit this. I think I will have time at least for the first one, OK? So the first thing is for the first question, we, Hoa, Naregitan and myself, proven this was 17. If R is stronger than Macaulay, well, in fact, a weaker condition which is a little more than the starting depth has to be satisfied with this SD plus, then J equals J star always. So if you start from a stronger Macaulay ideal, then you always have this. So this means, if you want, that if you look at the composition of your J star, primary decomposition, so now you know it's a mix and it's called Macaulay, so you can write it like this, which these are primary. Your Q i is a P i, OK? And then let me put stars. And you can take the same type of decomposition for J. The radical of this is also P i. Under some condition, you can assert that these are mixed. But even if not, you take these, I mean, have the radical of this, OK? And then you can assert, due to this, OK? Then you can assert that Q i is equal to Q prime i, or Q star i, well, unless V of P i is inside the non-SD plus or non-strong Macaulay focus of your ideal i, OK? So unless the thing you link is really inside the focus with not strong Macaulay or not this, when you do have equality. So I mean, there's a little space for non-equality, but not that much, OK? And let me terminate in saying on what realize this theorem. So this relies on some duality properties that were also obtained in a very similar way by Eisenberg, Heineken, Berndt, OK? And it says the following. So that's maybe important. And I will just state it. So this is also Hoare and Iit and myself, OK? And I will state in the case where it's Gronchstein, because that's where it goes to the best, OK? Oh, I forgot a very important term. Maybe I should do that and not this one, OK? I forgot to say what is J star, OK? I think that's maybe more important. So I will explain. I forgot. I wanted to say what is J star, but I moved further. So let me, OK, this is also interesting, but I didn't describe J star, I think, OK? So to describe J star, you have to look. So I will do that in case. I mean, so you have your causal complex here over R, OK? You have your causal complex over R. This is a differential graded algebra, OK? So I would prefer to write this as R with this. It's a skew symmetric thing, where R of E1 to, what, Ep, where you put the derivative of EI is FI, OK? So you get this as a differential graded algebra. And now you look at the m and psi i equals sum of Vcij Ej, OK, that are inside this. And you look inside this, you have an algebra, subalgebra, which is the subalgebra generated by these guys, psi 1, psi s, OK? That corresponds, if you want, to your m and psi, OK? And of course, I mean, inside this, you have the borders, OK? You have the cycles, and all of these are in K, OK? And now Vinicius Boussa and Asansali describe the ideal G star in this setting. So this is the work, Boussa and Asansali in 2.19, and show that for any ideal A and I, and this works also for any ring, OK? You have first that the ideal A, you can see it as you take elements in what is generated by these elements, gamma, this psi i, OK? You multiply this by elements by borders, OK? So if you multiply this, and you look at the degree p part, this will sit inside Kp, which is r, OK? And this thing is equal to A, OK? So when you look at what you obtain by taking elements, so degree i and p minus i, and multiply them, you get A. And now the J star is obtained by taking this element here, but multiplying by guys in z, by cycles, and looking at the degree p. So this is the description, OK? And OK, and this is important, but it tells you, if you put this together, that your J star is equal to A plus what is generated, what you obtain by taking these elements here, and multiplying by, I mean, representative of the classes of the causal homology. So if you know the generators of the causal homology as a DG algebra, you will be able to give, I mean, universal description of the generator of the J star. I think I have no time to continue. OK, thank you very much. Let's thank Mark, please. Are there any questions? Questions for Mark? One of the consequences of Bernstein's theorem on the collinance of residual intersection is that result about the S2 property of residual intersection of U. Yes, that's another type of extension. You wanted to know, to have some results about river function of residual, when the residual is unmixed, we say S1, but S1 is technically more difficult than obtaining S2. So we pass to S2, and there is some connection. Yeah, OK, so you can ask weaker things. My question is also that theorem about residual S2 as the GS assumption. Can you use these techniques with the causal complex to remove the GS assumption for that one? It's not so clear. It's not so clear. Because to obtain this residual S2 thing, I mean, you really use the induction and some precise, I mean, some precise tracking in the sequences that Bern used in his 94 paper to track when you get exactly the, how you can get exactly the vanishing you need for the thing to be S2 from these sequences. I mean, the general thing is maybe more, I don't know. We don't know, that's a good question. We don't know how to do for, I mean, for instance, if you fix an S that is, I mean, typically much smaller than the mentioned, how to get just S residual, not everything from these kind of techniques. And that's not so clear. Thank you. Mark, I have another question. Here, you just need SD plus to get the quality. Is SD plus enough to give you comacoliness here in three you need I stronger comacoli. Can you weaker? No, no, no, I mean, much weaker condition enough. But I wanted to give this in order range. I mean, for the, for instance, for the aceticity of the zero one, SD is enough. That's why this holds. I mean, and then, I mean, you need some kind of SD or SD plus condition. It's a bit technical because, I mean, also here, R is comacoliness, so sometimes you need this on the canonical, sometimes you need on the ideal and the kind of depth, I mean, so I wanted just to say that under, I mean, very strong condition, you are all what you want. But you need much less, yes. Yeah, in fact, you need much less. Yes. Okay, so, Albert, and the other question, yes. How to say, I mean, certainly you get things for rich ideas, which is part of this, but strongly comacolic activities, I mean. Of course it's comacol, I mean, yeah, but it's, I think it's nice because it tells you things about all the, even linkage cars in any, I think that's nice, but I mean, not really an answer to your question. I would like to thank Mark again.