 Okay, okay, so let me start by talking about trying to combine our coefficients result with the rational rings. And so let me talk for a moment about Branson's go to theorems with coefficients in that rational rings. So, he and I were interested in that of course, once we had the rational result and so we proved the following serum in a follow up paper. It's a little technical so, or it has a lot of hypotheses. So let me write it out. Let's let R M be an excellent f rational. If you don't know what excellent is just think complete local ring. We have J contained in I a reduction. Okay, let's let L be the analytic spread of I let's assume that strictly larger than H, which is the height of I. Okay. So, the following properties. Let's see our mod I is equidimensional. So, that means our mod H minimal prime is the same dimension for all minimal primes. So that I is generically. So that means that of reduction number. At most one so that means that each minimal prime, the, the I localized at P has reduction number at most one. And so there's three that condition that condition, and that are mod I on mixed. So this satisfies Sarah's depth condition. S sub L minus H plus one so this is a condition that locally, we have a certain minimum depth. Then we proved I to the L bar is contained in J. Times I on next. Okay, and I'm not going to try to do any proofs for this little bit, but that that was what we got. We had a student Aline Cosry. And so we were able to generalize this theorem. So, let me take the statement and paste it over here and make the appropriate. The appropriate changes. Okay. So, so we were able to prove that you don't need excellent. You just need cone McCauley so afrational rings are cone McCauley under mild conditions. Okay, so you don't need to assume our mod I is equidimensional. You don't need to assume I is generically of reduced reduction number at most one. And you don't need to assume the depth condition. So you don't need to assume anything, but you can get a better result because for all W greater than or equal to zero. You get that the usual statement holds so over here we get to add a W plus one. And so one thing for for those of you who are are young and starting out. This is a really good lesson to learn is you can read other people's papers. And you can, you should give some real thought to what their hypotheses are, because there are definitely times when most of their hypotheses are not nest that most some of some of the hypotheses may not be necessary. But again, I'm not going to talk about any proofs here just just wanted to point out that we, we can actually combine these. So, I'm going to switch gears now a little bit. And I want to talk about what are called coefficient ideals. Yeah, I think that's pretty good advice. So let me talk now about what are called coefficient ideals. The terminology here perhaps isn't ideal. Because we've been talking about Branson scota theorems with coefficients. Now I'm going to talk about coefficient ideals and they're not quite the same thing. So, let me point out right that are in our earlier results. We've always exploited the difference between the analytics spread in the height of the difference. We looked at L of I minus the height of I. And so the, the equation that I wrote at the end of lecture one and the start of lecture two that that really has the most power so far that we've seen when we when when there's a difference in these two things and we can use that. So then, but, of course, I, that leads to the question what can we say when L of I is equal to the height of I, and these are usually called equimultiple ideals. And so the most basic example here is if we if I is contained in a local ring. And the radical of I is M, then we have that the analytics spread of I well it's bounded below by the height which is the dimension of the ring and above by the dimension of the ring which is the dimension of the ring. And so L of I is equal to the dimension of our is the height of I. Okay, so, so there's a situation where we might want to be able to say more, but we certainly can't exploit this difference. And so I think that leads us to wanting to define this coefficient ideal. We've got properties so I guess I meant to define the coefficient ideal. All right, so definition. Let's let R be no Therian and J and I, and are any two ideals. I'm going to give myself a little bit more room here. So let, so notice J is not I'm not assuming J is contained in I here. So, the coefficient ideal of I relative to J is we're going to use this kind of what's that a frack a I comma J. It is the largest ideal say frack be such that I times frack be is equal to J times frack be. Okay, and if, if we know we're talking about I and J, I'll drop the parentheses here, just call it a. Okay, so that is what the coefficient ideal is so let me go through some of the properties and these are all exercises. So, so say R is no Therian. Okay, and I this is, I don't need to be local or characteristic P. And now let's assume for the moment let's assume J is contained in I. Then, actually, I think I'm going to number these one. This coefficient ideal is always well defined. And to if a so again this is a if frack a the coefficient ideal contains a non zero divisor. Then the two ideals must be equal up to integral closure. Okay. And so, J is a reduction of I. Okay, and in this case, then there exists, and bigger than or equal to zero such that a power of I is contained in the coefficient ideal. All right. And now let's assume we're in the local case. So, if our is local. Let's assume J contained in I is a reduction. And let's set R equal to the reduction number of J with respect, sorry, I with respect to J. Let me define a sequence of ideals let's set a one to be the colon ideal J colon I itself. Okay. And for larger values, let's set a and plus one to be the J times the previous one colon I. Okay. So this is a sequence of ideals. Let's call this three. We have that I to the R is contained in the coefficient ideal. The coefficient ideal is contained in each of these a sub ends are all N. Okay, and this is a decreasing sequence. So, a N plus one is contained in a N. Right. And the fourth thing, if, if I is M primary, then the sequence stabilizes, you know, I E a and is a for all and large enough. Okay, so those are the basic properties here. In your statements to be who is are in the exponent you have in I are because I don't know. Sorry, sorry, sorry. Okay, so those, those are the basic properties of this, and I've left those as exercises. Okay, so here is the theorem, my lights automatically go out if there's no motion. And I guess my arm moving isn't enough. So, this here is a theorem on the notes it's 3.4 again this was proved by myself and Craig in 2001. Okay, so the following. Let R M be a regular local ring containing a field with I guess arm on M infinite. So, let I be an M primary ideal and J contained in I a minimal reduction. Okay, then for all W there's a little trick here. We're going to start with minus one instead of zero. If we look at I to the D plus W integral closure so that's the usual remember D, D is the analytic spread here. In this case, this is contained in J to the W plus one. Well, so that would be the usual pre instance go to theorem, but where can we put the coefficients we can put them in the coefficient ideal. So, as usual, the characteristic zero case follows by reduction to characteristic. So we are going to concentrate on the characteristic case. And so one, the really the biggest tool we want to use here is the following observation. Okay, so in in a regular ring. So two ideals say a and B contained in R. Well, so we can, we can form a colon B, and we can take the q bracket power. And that's obviously related to a to the q colon B to the q. In a regular ring they're equal. And again, that is really the flatness of Frobenius. Okay, in general, if you take the left hand side so if you take the colon and you take the q bracket power you're going to land inside the a to the q colon B to the B to the bracket q. And the right hand side is usually much larger I mean that that, you know, there, there's just all kinds of things that that should get in that colon, that don't start there, or don't come as q power. Okay, but in a regular ring. We, we get that those are the same. Okay. So, yep, I think the rest of my notes are just to do the proof. Okay. And then talk about a few other things. So, I'm going to do. So, most of the work is the w equals minus one case. So which we will. Okay, so we want to show. We want to show that I to the D minus one bar. Well this will be j to the zero. Hey, I J. So in other words, we just, we want the D minus first power to be in a. Right. So, um, So, what we show. So we will show that I to the D minus one bar is contained in a and all and and then right for and sufficiently large. equals. Right. Um, actually. So in fact, we'll show something slightly different. We'll show that I to the D minus one. Bar is contained in a and to the bracket Q. For all and and for all, I guess for all Q. Okay. So, um, And of course, notice that this. We look at I to the D minus one bar to the bracket Q that's contained in there. Okay, so, um, So then I to the D minus one bar is contained in a and star, which is equal to a and because we're in a regular local ring. So that's, that's really our goal. And one thing that this argument is going to illustrate in. In rings of characteristic P when you're using tight closure is that you notice we've we've already taken a Q power here, but we are going to prove this using tight closure, which means we're going to take the Q prime powers of the Q's powers and use tight closure. Kind of doubly then I guess. Alright, so let's see if we can go through that argument. Let's start a new page. So, for n equals one. We want to show that. I D minus one Q the integral closure of that is contained in J colon I that's to the bracket Q, right this, that's a one to the bracket Q. And as we said, that's so because this is a regular ring that's J to the Q colon I Q. Okay. So, in this case, let's let X be in I to the D minus one Q bar. And I guess I'm going to call it to follow the notes or be in I to the Q. Okay, and we just need to multiply. Well, first, we'll choose a C, not equal to zero as in exercise two. And this is for the ideal ID Q for that power of I. Okay. So, then Q prime so we'll introduce our new power of P. We're going to look at C, R X to the Q prime. Okay, so we're, we're trying to show that when we multiply by I to the Q, I, I to the bracket Q times this ideal, we're going to land in this ideal. And there's our R in here. There's our X in here. Okay. So where does that land. Well by the choices we've made. This is in. This is going to end up in J to the D Q, Q prime. Okay, that's where that ends up. This is in J. J is degenerated so that gets us in the Q Q prime bracket power. And the coefficient D minus one Q Q. Okay. Oh, and I don't, I don't really care about the coefficient. So let me erase that. I do this each time I've practiced as well. That's all I need. Right. And so therefore, we've got that R X is in. Well, let me write this as J J to the Q to the Q prime bracket power. So R X is in J to the Q type closure, which is J to the Q. Okay. That's what we wanted. Right. So we've done. We've done that a one case. All right. And so now, following the notes, the way I wrote it, let's assume the result for N minus one, bigger than zero, I guess the way I'm writing that. Okay. And again. So, what do we get we've we still got our X and our R those aren't changing. So, we've got C, R X to the Q prime. Okay now. This is in. Where is it. Okay, so we're still in J D Q Q prime. Okay, which is contained in J to the Q Q prime because J is degenerated, along with the, the remaining terms. So that would be J to the D minus one Q Q prime. Okay. All right, so that's in J to the Q Q prime. Certainly that's contained in I to the D minus one Q Q prime. And that is contained in J to the Q Q prime. So I guess we can take the integral closure, and that gets bigger. I did the D minus one Q Q prime bar. Okay, but now by induction. And where does this land. This is in J times a and minus one to the Q Q prime. Okay, so we are applying the induction hypothesis. We're applying the induction hypothesis on a bigger power Q Q prime. And that's what we needed. And so therefore, again, if we write that as in J to the bracket Q have everything right. I'm sorry. There's no J here. It's the the integral closure in the am minus one. So now we are in J to the Q a and minus one to the bracket Q to the Q prime. And so therefore, we have that R X is in J a and minus one star equals J a and minus one. And so X is in J a and minus one colon I, which is a and that shows the W equals minus one case. That's the end of the proof. Right. So the So if we if we want to try to understand M primary ideals a little further, we have to introduce these coefficient ideals and we we get nice coefficients. So I should point out here that that we we use that I is M primary pretty strongly. So if if I isn't M primary, then the we have we still have this sequence. Right. It it's certainly need it's not at all clear that it stabilizes. Okay, so I should say that again, this is another area that I had my former student Aline Hansry look at and we have a paper where we do generalize this result. In the complete case, one can use Chevrolet's lemma and and get some results, but for for non M primary ideals, but but there it's a little mysterious, otherwise. So we're sorry in our primary deals the only ideals for which these coefficients are known to stabilize. So that's what I wanted to say about that. And then I wanted to use the last 10 to 15 minutes to mention a couple of other places where Branson's go to serums are are relevant. So let me talk about cancellation results. So Okay, so so in in the most basic form, the Branson's go to theorem says, for instance, that I I to the L of I bar is contained in J. But in fact, often, a smaller power is already in J. So so especially if you're if you're thinking about ideals that are particularly nice they have small reduction numbers with respect to J. And so there's there's no reason you necessarily have to go up to the analytic spread value to get inside J. So, so that's a question. So that leads to the following question. What conditions lead to getting say an N less than L of I such that I to the N bar is contained in J. Right night up here, of course, that's a reduction of I. So there's a nice paper of Craig uniki in from 2000. So he has a paper 2000 on cancellation of ideals. Okay, so EG. If say a times B is equal to a times C. That implies B equals C. Right. So that's usually very false right for ideals. For my matrix for my linear algebra students, it's always true for matrices, no matter what I've learned, but but it's generally not true for ideals, but he came up with some nice criterias for for for when this really does work. Okay. So, so no if let me say it this way, if the reduction number of J. I is R, then we have that I to the R plus one is equal to J times I to the R. Okay. And of course this is times I to the R. So there's kind of, you can try to kind of think well is there any form of cancellation here, right can you cancel one of the eyes and does this imply anything like I to the R is contained J. And of course the answer is no, usually right you need, you need a lot more going on here. Okay. But let me just write down what the theorem is that we were able to show. I guess say what that theorem is. So let's let R M be an F rational Gorenstein. Okay, and so therefore in fact it's it's an F regular ring of dimension D. Okay. Say we have an ideal in R with H being the height of I and that we assume is strictly less than L of I. Okay. All that L and suppose that our mod I is cone McCauley, so you have a pretty strong condition upon the quotient. Then for any reduction, J of I, we get that I to the L minus one bar is contained in J. So we can do one better than the Branson's go to theorem in general will tell us. Okay. And my student Aline Hasry generalized this slightly. I didn't, I don't remember the actual result of off the top of my head. But I think we, I think it was about maybe our mod I doesn't quite have to be cone McCauley. So, so this is an interesting result. Something I spent a lot of time thinking about that I don't recall ever getting anything useful is what about coefficients here. So, you know, could we put something here under the right conditions to get the coefficients. And that's a great new result. Anybody could do that. So, you notice here so the analytics spread needs to be strictly larger than the height. But even if the analytics spread is much larger than the height. This results still only allows you to go down by one. So it would be really nice to know if maybe with stronger conditions. Right so so I'll write it as a question. Maybe maybe can one do better. So questions. So one, can we get coefficients on the right hand side. Okay. Okay. Okay. So if L is greater than H plus one, so not just greater than H. Can we get. I to the L minus two bar contained in J. So those are both both interesting questions that it would be nice to see answers to, and then then one can wonder about if, if one can answer the second question positively can you get coefficients right there's no end. There's no end to, to these questions. And let me use the last few minutes. I guess I have another blank page to talk about uniform grants and scota results. So, so uniform. Branson scota results. Well this is, this is coming from one of my favorite papers of all time. So in 1992. I think he published this paper on the uniform. Art and Reese's theorem. Okay. And what that says. So, if our is a nice, I'll just put it in quotes, a nice no theory and ring. And again, the conditions are in the write up. If you want to see them specifically. Then, given. And the sub module of M. These are finitely generated our module. Then we have, of course, if, if I give you an ideal there's an art and Reese, the art and Reese's theorem tells you something. But what he saw is that there is actually a uniform art and Reese. So it goes this way then there exists some number K, which depends only on and such that for all. And bigger than or equal to K. And for all ideals. In the ring. If we look at. I to the N M intersect and then we are contained in. I to the N minus K. Okay, now the art and Reese theorem gives an equality with more information on the right hand side. And that is fault. There is no uniformity there that can't be done. But, and I see a Rena's face showing up higher and she has done lots of work. And that is the art and Reese that I encourage everybody to read as well. So, this is a beautiful theorem. It stems from lots of lots of the ideas that came up in tight closure. Go into that. And it's certainly a paper I encourage anybody to read. If you haven't already. The results. That he proved is the following he proved a uniform. And he proved the same conditions. And he proved in the domain case you can generalize it a little bit but but really it's the domain case. It's where all the information is. Then there exists K. Then there exists K. Such that again for all ideals I contained in our. And all n bigger than or equal to K. If we look at I to the N bar. This sits inside. I to the N minus K. So this is a form of Brienne since go to theorem, right that you just there's there's a, but notice what this says this says this is true for all I, no matter what the analytics spread is. The integral closure sits in a pretty big power of the ideal itself, right that you only have to go down some uniform K, and this, this. Of course this K can be quite large. It's not connected to the dimension of the ring at all it can be much bigger than the dimension. But are there. Um, sorry again, are there any sort of effective ways to compute this number that you need for the appropriate things go to here. Um, I think a proof analysis would probably tell you what you needed to do you have to. You know, for instance in the car at least in the characteristic P case you have to know something about the test ideal. And so, so, I don't know if you can compute the test ideal then you're you've gotten part way. And, and I, but I'm. Yeah, I can't, I can't answer that terribly off the top of my head. That's that's actually what I wanted to finish with I hope for everybody that this that I've helped you to understand what the Branson's go to theorem is and what some of the ideas are behind getting these improvements on it. And I appreciate your attention. Are there any questions or comments. I have a quick question on. I wrote on page 10 for the theorem. Yes, I guess without like any strengthenings. There are cases where L minus one is sharp is that correct. That is probably correct. Yeah, if if H I mean I think if if L is h plus one. You probably can't do better. I mean you can probably come up, you know, you can create examples where you can't do better I should say. Thank you. I have a question here. Yes. The Branson's go to theorem for with coefficient idea. Do we ever get equality. Oh, I. So, so how often do we get an equality for instance in in the, in the, like in the regular case. Right. So, so for instance, you hear something like that. Yeah, because it will, if it was an equality, it will help us to calculate the integral closure. Yeah. I don't, I don't have a really good intuition about that. It strikes me as being unlikely. In general, that we're getting an equality here. You know, unless the reduction, well, yeah, even if the reduction number is small. It seems, it seems unlikely to me that we'd get equalities very often. I mean, basically I would say, you know, there's, there's, there are more coefficients that probably are very hard to quantify in terms of a theorem, but I'm, I wouldn't be surprised at all to know that this is in this, you know, times, the maximal ideal. Most of the time. It could probably happen if the reduction number is low. Yeah. Or is M primary. But then the right hand side, right hand side will give you powers of an ideal. Right well this, yeah, this would not apply this particular form wouldn't apply for an M primary ideal then we'd be wanting to look at at this version. Yeah, and maybe, maybe there's more one could say, I don't know much about it. So, I couldn't, I can't. That that aspect of it I can't tell you. And there is also a generalization of Bransco and Skoda by recent Sally. Where they take in a D dimensional local ring. Instead of taking I to the D bar. The product of these ideals and then the integral closure. The product of the ideal is contained in every joint reduction. Oh, the idea I to the deep bar is really high times I D times. Yeah. I looked at any collection of the M primary ideas and then showed that the, the closure of the product is contained in every joint production of these be ideal. So something like that. Right. Yeah. So, I wonder whether coefficient ideals are considered in for this generalization. That makes sense to me that there's probably something there. But I yeah I have never really done any work with with that situation so I again I don't know. No, that's a good, good assignment for a PhD student I'd say. Or at least something for them to think about. Okay, so are there any more comments or questions. Okay, so let's thank the professor, other back for his lectures. Thank you. Thank you. It's been a pleasure to be