 I'm ready. I just want to write this ahead of time. I'm ready. Should we go? Do you want to start recording? Sorry, I actually forgot to stop the recording when we took the break so it's still recording. Oh, I see. Okay, alright. Oh yeah, and thank you. I do vaguely remember that 9.41 is the, yeah, okay. Thanks for the reminder. Okay, so on the left side we still have the definitions of tea and comicalifiers of R. And so, and now I want to prove a weak uniform martin reese lemma. And at least there will be two versions, several versions that I'll prove. This one here assumes that we have infinite residue fields. We assume that we have infinite residue fields and the comicalifiers and the teas of our mod P are non-zero for all prime ideals P. And then, for any finely generated our modules we have the weak uniform martin reese lemma. I want to say something about we assume that the this team R of P and comicalifiers R of P are non-zero. And it's not clear why we need to bring in the internal closure. However, all the proofs that we now have do use that into inter closure to our money. And the other thing is that the, those two of our mapping to varna being zero and you know, weak uniform martin reese lemma holding seem to hold for the same kind of rings. So, they seem to be intertwined in some way and there are some results that hinted more than that, but that's where we are now. Okay, so for you prove this version use using the infinite residue fields part. So, here's the proof. Now, we did not prove all the ingredients. But this is sometimes you have to go make progress and you're big here while working in the background for all the hard background. All right, so here is this is the fun part. Okay. So, without a loss of reality, we've made these reductions. M is a prime ideal P, sorry, M is R and N is P, and I are primary to the maximum ideal to max maximum ideal. Now, we will go back and forth. Sometimes we will need an arbitrary N. And sometimes we will need an arbitrary ideal or more general ideal, but at least we can go back and forth. So, first, suppose, suppose that P is a maximal ideal, then for all ideals, I and all integers and I to the end intersected with key. Well, if I is not contained in this maximal ideal P, then these two ideals I and P are co maximal. So this is contained in I to the end times P. If I is contained in this maximal ideal P, then we, then I intersected, I to the entire intersected P is equal to I to the end, and that's containing I to the M minus one times P. So we always have this so we have the, at least if P is the maximum ideal we do have a uniform Martin Reese number equal to case equal to one. So here's easy case. This is not here in induction. So what does that mean we proved it for maximum ideals. Suppose we have proved it for all prime ideals up to something and now we want to prove it for the prime ideal, a given prime ideal and we know the result for all, all prime ideals. Let P be in spec R, and assume that the, the weak uniformed Martin Reese lemma holds for all prime ideals, Q strictly containing P. So now this is where the theory induction comes and we can prove it for this being we can then eventually prove it for all of them. All right. Now, I need more space. All right, so now we have this element see by assumption. Let's see. Assumption is that the coma qualifiers. This is a non zero ideal and T bar my piece not zero. So let's see in our not be an element of P that whose image whose image in our month P is both in the coma qualifiers and the T of our month piece. Okay, so our map is a domain so intersection of two non zero ideals is still not zero. So we, we know that it exists. And then a dilemma. Oh, so all prime ideals needed for the filtration. So we have this sub module P, P plus C are that's a sub module of our. So we take a filtration prime filtration of this, all of the prime ideals needed are strictly contained strictly contained. P still by no theory induction, no theory and induction. There exists a positive integer K such that for all integers and for all ideals I, I to the end intersected with P plus C. And is contained in I to the end minus K times P plus CR. So we use them here in induction. And now we want to fix an ideal I, and without loss of generality primary to a maximal ideal M. We are assuming infinite residue fields by, by the assumption on residue fields. This is a minimal reduction. J. I might be localized at M. Of, I might be localized at M. Okay, and say this. D is the dimension of our month P to them so the number of generators of J are not P is D. We may assume that the D generators of J. Well, J arm of arm mod P look less than M are elements of our, it's easy to take elements and unlocalize them just clear all the denominators and then now we're in our mod P will lift them to some arbitrary elements but they still have to be elements of the ideal I. And, and the other thing is, we can still guarantee that they formed the correct height. So elements of our that they are contained in I. And in one in one of his groups when he was lifting elements from mod keys, he, he showed a trick how to make sure that the height when you left is still okay. Well he did it one element at the time but he can still do it, a sequence of elements, they're containing I, and such that the height of this J, when we just go mark P is at least D. So. So now we are in the situation. Which I. We proved a lemma earlier for. That should scroll back. So we prove this lemma earlier we have an a theory and ring. We have an ideal generated by some D elements and the height of this ideal mod P is at least the number of generators. And, and we already know this part here. For all ideals, in particular, we know this result for our ideal J. So by the last lemma for all N. J to the end intersected with P is contained in J to the end minus K P. So, we would like this result here, the last thing that I wrote, we would like this for all ideals. I, but we now only proved it for some very special ideals J. However, even further back on Monday, we had another lemma. So we had ideals I J P J was a sub sub ideal of I P was prime. And we had some conditions here. And, well, so this condition here the first condition we, we just thereby we have this condition. So now we hope to prove this second condition here. Well, now what we need, what we really need for this is, we haven't yet used a two hours not zero. So that's what we need now. So, So, somebody answer that phone. So, we. So now, what do we have there, there now. Since C is contained in Well, there was some element T so T was an element of T of R. So, there exists a T such that C isn't actually T sub T bar. That means that the inter closure right to the end is contained in the n minus T power of the ideal I for all ideals I. And So, well, no, C is in our month P. So we only have these results not P, but that's not P. This is good enough. All we have to do to verify this part two is we have to go. Okay. Well, if T sub T of our piece not zero, then also T sub T of our month P localized that M. This is not zero and again C is the localization C is still an element in there because localization commutes with inter closure that's very easy. So what we thus have to do is at least locally. So, back in the old system for the uniform art in Islam, I really don't need inter closure. But, right. Apparently, this proof requires us to go through the inter closure. Well, we don't have a better proof. And if we stick C in front. Yes, then that's okay. But mod P and localize it and the inter closure right to the end is just the inter closure. Jay to the end. And now we use this is C times J to the end minus T. Our mod P to the arm of the localized that M. So that's, we could have done this also for the ideal I I could have ended up with I. The problem is that I want when I want to unlocalize and so on then we'll have problems. Okay. Okay, so what do we do now. So what that means that C times I to the end. Localize the M. I'm skipping a step there. Okay, so let's see if I can work it out. She had better not do this. See times I to the end are not be localized at P at M. This is why am I doing this. Sorry, I'm looking at my notes. Why do I get rid of the C. Do I have to get rid of the C. Oh right yes, I need to get rid of the C. All right, so this is the whole point of these elements of T of R, you get rid of the C. So there is no C here. So this should be deleted. Now, that's why I needed this extra step step. Okay, so this C is, of course it's containing J to the end minus T are not piece of them. But it's also contained in C are not be localized again. So, so I will. I hope you can remember this part here I want to scroll down. So this is where we are. So that means. Now let's go lift from arm out P localized to arm out P. So see I to the end arm at after localization. So C times, I to the end so no integral closure appears anymore. This, by what we just did on the previous page. This is in J to the end minus T. Plus P localized them intersected with C plus P localized them. And then of course plus P localized them there. There's a little bit of an overkill here. See, okay. So now this year, again we do the same trick. I can get rid of this, this P is a subset of there so we just have J sub n minus T localized them intersected with. No, no, no. Yeah, that's one intersected with P plus the ideal C localized them plus P localized them. But this we already knew globally without prior to localization. We knew that for every ideal not just for these primary ideals. This is contained in J to the end minus T minus K times T plus C. And again localized them plus. So this we could have done for all ideals. Except that we need some other results so this here is then J to the end minus T minus K intersect with C localized them. Plus PM. So up through this step, or even from through this step, we could have just use I we didn't need to use J. But the problem is that now we would not be able to handle this intersection with an arbitrary with an arbitrary ideal J with an arbitrary deal. Okay. But now the same thing is, let's impose another color. All of this is contained in C localized them. Okay, so I just ideal C localized them. So what we have now is this is J to the end minus T minus K. M. Oh, sorry. This was times. This is times this is not intersection this is times. That's the power. This is where we used our induction. So this is just C PM. So from this part here we conclude that I to the end are localized that M is a subset of J to the end, minus T minus K localized them plus P localized them, plus I to the end are localized them. But see, we can even do better. If we can replace this this is containing. It's a containing P colon C, which is just key. So what we just prove is that I to the end are localized them is in J to the end minus T minus K M was this is localization. So this is what we have. Now that lemma worked globally. Right. And so how do we get on globalize. So, so what we do now is by that lemma that we have. We conclude that I to the M are localized local intersexual P localized them. That's contained in I to the end minus to see the two K minus T minus one P localized them. And so this is only true locally. We had another lemma this is one of the reduction steps that we have, as long as you can prove that you have a global this constant here that we're subtracting, as long as stop. As long as we have a global constant. We, all we have to do is prove this inclusion locally for N primary ideals. So we can conclude that I to the end intersect with P is containing I to the end minus. This was supposed to be to T minus K minus one. So we have approved this uniform ordinary slamma now for all. We'll prove it for the next P but by material induction were done for all of them. Any questions about this one. So, if not, I want to prove other weak uniform. Uniform. Art and res lemmas. Art and res lemma, and I'll talk through them more than right because we are already skipping a lot of proofs. So, so this week uniform arteries lemma holds in the father for the following in the following cases. First case is when our is essentially a finite type over an Ethereum local ring. That would be excellent. Did I omit the excellent assumption. No, this is my theory locally. Oh, right. Yes, because we can make it excellent. That's the beauty of it. In fact, that's the whole reduction. So before let's state the other cases. Let me just indicate a proof here. So it's in the fear and local ring a. And what we do is, so it's in the fear and local ring with maximum ideal N. And first of all, let's add a joined a variable and localize at the extended maximum ideal extension extension of a. So this is a faithfully flat extension. And then we'll go even further will complete. So when you complete whatever ring we get this is also faithfully flat so the whole map is a faithfully flat extension, but the completion now has the following property. It's an excellent material local ring with infinite residue fields. So let's call this ring be them. This is an excellent Ethereum local ring with infinite residue fields. And what happens that we can pass to our tensor with be over a. This is a faithfully flat extension of our. And furthermore, this is essentially a finite type over this be which is excellent Ethereum and local. So we proved in our reductions is we proved. It suffices to prove the, you know, the weak uniform pardonaries lemma in this faithfully flat extension to conclude it for our. All right. So, so we might as well assume that ours essentially find a type over an Ethereum, excellent Ethereum local ring. And furthermore, we can assume that that has infinite residue fields, and all residue fields of our tensor be over a are infinite. Because they all contain that unity here. So we are now in the situation from the previous lemma, we have an Ethereum ring within the residue fields. Oh, then they're minor details. They're minor. These are giant details. We need to prove that we have to establish that these these two circled ideals are not zero. And when we go mod P for every prime ideal in here. Okay, so how can we establish that so this is where we are not feeling and filling in all the details. So here is. So, so let's, without less, less agenda, let's just call this car now remain. So, we have the theory and ring infinite residue fields, and that the other thing we can do by the core structure theorem. See, our is essentially a finite type over. This excellent over a complete excellent ring. And so what we can even do is this. When we go mod P. Let P be in spec are so then our mod P is well is essentially finite type or we can lift it unlocalize it is a homomorphic, or we don't have to unlocalize it's the homomorphic image of a regular ring so that's this complete local ring. Our is essentially finite type over it so we have a polynomial ring mod something that maps on to our. So it's a polynomial ring over be mod a prime ideal that's mapping on to our mod P. So we might as well take a kernel of this completion here. So, this beam might as well be a domain here. So, and every complete local domain is a homomorphic image of a regular Gorenstein domain or Gorenstein ring. So our mod P is a homomorphic image of a Gorenstein ring. So, well, then there are some properties. I did not prove I did not give you when it's called Macaulay. When is the ideal coma qualifiers non zero, and this is one of the cases. And so this, if you read the section that I made it, then this here is not zero. Well, it's not your right. It's, and, oh, and it's a homomorphic image of a Gorenstein ring of finite cold dimension well it is. This ring is a finite cold dimension because it's essentially finite type over here in local ring. So everything is satisfied. How about T of R T of R mod P. Okay, I will move this T of R mod P T of R mod P is also not zero. Well, in characteristic P or in characteristic zero, we have two different proofs in characteristic P we proved it, we proved it. And in characteristics zero it is Lipman satin, which I did not prove, but it is, and it's not even in my notes. So this is an outline of the proof of how the weak uniform Martin B slammer holds if ours essentially a finite type over in the theory and local ring. And then we just impose excellent local ring and infinite residue. Okay, that's, let me just state two more cases. I will not prove prove them. The other cases are is ring. R has R has characteristic P. And prime characteristic, and is a finite. So, in this situation, again, what this says by a result of course this has finite pro dimension, then we proved in this case that T of R is not zero T of R mod P is not zero. So, the proof reduces to the same things to need to know that all these common qualifiers and T of R mod P is not zero and then you use the previous lemma. Very similar proof. And then very similar proof also. Well, much harder proof are is essentially a finite type over the ring of integers. So a lot is hidden here. The reduction the characteristic P is the background work in order to establish the common qualifiers. But in, as far as the interval closure or uniform market in recent lemma is concerned, we don't see any type closure here, right. But again, the proof is, can we establish that all the comacolic fires and T of R mod P is not zero. Yes, we can similar proof. And the other thing is, oh, how about infinite residue fields. Well, infinite residue fields. If when you go mod P, if that contains Q, then you're set. But if you go mod a prime ideal P, then it's done these same things before how it can add extra elements. So, this is all I want to say about proofs of this uniform market in Islam. Do you have any questions. I have a question. Yeah. Yeah. Is this true for all good filtration uniform what in this week version. Instead of powers of the ideal. Yes. I mean, at least parts of the proof would require that we would have good results also. So, I think that's the code of theorem for good filtration and I don't think that we have that the interval closure part. So maybe the T, the T sub case for these iterations might not be zero. But I'm not sure about the common qualifiers because kind of qualifiers really are talking about talking about them the biology, but actually then cats had some results, but don't use type closure. I don't know. Good filtration are contained in the integral filtration. That's the largest good filtration. Right, but they're more general right. They are characterized by saying that the reasoning is finite module over the ordinary history. So it's continued in the integral closure. Yeah, so I think that the for these filtration so the T sub case here we're defined for the inter closures and the ideals. And so if we define the T sub case for the good filtration, I think you're right. Yes, but then that would be non one, probably prove that prove that that's not here and maybe somebody could do it I don't think it's been done. But I'm not sure what I'm saying is that I'm not sure that the coma qualifiers will work for your iterations. I'm not sure. Yeah, that's a good question. I have another section there it's. So, right now, all I didn't even prove the uniform art and we saw my right there's a lot of ingredients, but I think it was good to see the flavor of what the methods are. And then there's something else to fill in. But why would want to want to prove this is it just a cute result or can be used and yes, this is used it has been used so. Section eight is some more forms of the uniform. Versions of art and res lemma, and if this is supposed, possibly we will publish these notes and it would be good to have a complete list of currently known results either for the classical use classical uses of the art and res lemma or for the uniform art res lemma so please contribute that so you don't have to do the other exercises and maybe contribute that or let me send me references to where I can find more results. I'm understanding that you are not planning to prove the full strong version of the art and res lemma but can you say, you know, on a scale from one to 10 how much harder it is to prove the quality. It's not even clear that it's true right. So, it was proved for all maximal ideals. So, if you vary the eyes only overall maximum ideals that was Duncan O'Carroll and O'Carroll and there is also Giral and Lana, they have some results, the latter two for principle ideals. The difficulty for arbitrary ideals is that for one thing you cannot really go to filtration. It's, you lose too much information when you go to filtration so you really have to handle your arbitrary modules. And, in fact, there's, there are some. There's some if you look at the exercises, there are some counter examples. If your ring is not that good. Yeah. Yeah, so maybe exercise 9.6 is a good one to look at. That leads on extra assumptions. Thank you. This has been many weeks of intense lectures everybody. We're tight closure, even the first paper by he was very big we couldn't possibly go through it and this is the 10 lectures on tight closure by he went even more was developed so there has been a lot of information coming at us really fast. So, I didn't assign now that much homework. But, you know, the course is designed just like a graduate course three lectures. Every week. Yeah. But every set of three lectures as soon as a lot of background. Yes. I like to. I'm not always comfortable using the background without having gone through the proof so sometimes hard. Okay, thank you.