 So, okay, so shall we go ahead and start the introduction. Okay, so welcome everyone. And so, Florin is going to continue his lecture on a fresh and orange and rational similarities. Go ahead. Thank you. Alright, welcome back everybody. I'm going to continue where I left off last time, as I was looking at what I managed to present and what I prepared in the note that I posted, I realized that perhaps like that carried away with how much material I have. So, I'll try to cover most of it and for parts that I don't manage to get to you will find them in the notes. And the interesting theory and I just find it, I just had a hard time choosing what to pick at the expense of others also. All right, so, last time, we're in the middle of a theorem that was basically trying to explain why if you have an equidimensional homomorphic image of a chemical ring. Assumption that one system of parameters gives you a tightly closed ideal implies that every system of parameters gives you a type of. And we proved a number of parts so if you have the one. If the ideal is tightly closed then we will manage to prove that the ring is quite Macaulay and then every ideal consisting of a subset of this given program system parameters is also tightly closed. So, trying to show that if you have another system of parameters why one YTB where these the dimension of the ring are then the ideal degenerate is also try to try to close. All right, so let's, let's see how that goes. We're going to take the powers of the original system parameters index by T. And first we're going to show that this ideals are tight. We know that I itself is tight close but now, well, what about this ice up T. Well, we're going to use a start rather standard argument. If an ideal that is in primary is not that if the tight closure of it is not equal to itself, it means that this tight closure strictly contains the original ideal therefore when you look at the cycle more the ideal, it has to intersect the cycle in the non-trivial element. And therefore we can find the there exists an element V in the tight closure which is non-zero in the cycle of our mode it but now because it is conscious of the powers of the original axis. We can write the elements in the cycle in the form x to power T minus one times you where you is non-zero in the cycle of our mode I the original. Okay. So now let's write what it means that V is in the tight closure of it star means that there exists C in our zero such that C times V but instead of V we have this form of the element. Okay. So this this element belongs to it power bracket queue for q larger. Okay. Isolate the part about see you to power queue you get it to belong to the following column ideal. Okay. And using the calling my calling property because of the x is for more regular system a system of parameters therefore a regular sequence, you can by using inductively the regular sequence property, you get that this calling is containing next one to power q x d to power q. So now therefore, you itself is going to be in the tight closure of I, which is I. It gives a contradiction, because you're supposed to be known zero, the image of you were supposed to be known zero in our model. Okay. All right, so therefore, all these it's are tightly closed. All right, so now how do we made the passing to to J. Well, there is a T large enough such that this is inside J. And now both of these ideas are generated by system parameters that are regular sequences because we already proven that the ring guy is calling my calling. Therefore, we can apply the following exercise which is stated in my, in my notes and I'm not going to say to you, it's a, you know, it's a rather. It's a non trivial exercise. If you have an inclusion like this of two ideals generated by regular sequences, there exists an injection and the exercise in the notes explains how the injections obtained. R mod j into R mod I. Okay, like this. So you can find an injection like this which is given by multiplication by a specific element. All right, so if you take an element in J star, then the image of X under map above is going to go in it star. But that's the same as it, which in turn implies that X, because the map above is injection has to be in J. Okay. So, here's this little, little trick. I'm going to, once we start connecting the rational reason to local homology I'm going to give you an alternate way of, of justify this part that doesn't involve this exercise. But I would, I wanted to mention the exercise because it's stated in the notes and it's, it's a good thing to know. It has the passing to an arbitrary system parameters so therefore, now we know that if a ring is an equal equidimensional homomorphic image of primary calling rings and admits test elements. It's enough to check the, the fact that one system generated by one ideal journey by a full system of parameters is slightly close to came to conclude that the ring is a rational so that's very useful in practice. Speaking of practice, I think it's important to maybe get a sense of what would mean to check that you have a rational. Okay. It would be interesting. It's interesting to realize that in fact, the things are not necessarily trivial even for, for nice examples. Okay. So let's take this example was discussed before in the context of stronger regular rings by Lynch on this hyper surface is x square plus y q plus it fifth, and I want to show that the ring is a rational if the characteristic is greater than seven. All right, so notice that the partial derivative of the hyper surface is to x, and I made an observation at the beginning that the Jacobian ideal over K gives us test elements. The two is invertible, if the characteristic is greater than seven in the field K. So X is to X would be a test element or for X itself would be a test element for our case. So this is actually, I would notice. First of all, that this is not trivial. Okay, I'm using something. A theorem of Huxley, you know, get based on my study results that is not trivial but is very useful. Otherwise would be, I think it's kind of difficult to check elementary that this example is a version. Okay, so it would be useful to localize this I don't want to deal with this is a non local ring so one can actually notice that this is an isolated singularity so you just need to check that the localization of the homogeneous maximum ideal is a rational, and you can you complete that it's enough to check that this local ring. Okay, the powers for power series of in XYZ over K module of the same hyper surface is a rational. Okay, so we're using the same trick as before. So we're trying to show that this is a rational if the type why why is the form a system of parameters on this if you mode out by Y, Z, you get the dimension zero and it's easy to see that this originally has the mission to. The form a system of parameters, so if the type closure of YZ as an ideal is strictly bigger than the original ideal then an element in the cycle has to be in there in the type closure, but the cycle module YZ it's easy to see that is generated by X. Therefore, you just need the element X if the type closure is not trivial. Then X will have to be in the type closure. Okay, so this is a you could say that this element this argument is a little elementary because it's involves samples but you know. It's elementary but it's not necessarily obvious. If you see me for the first time. So, if this ring is not a rational then X will have to be in the type closure for easy. Okay, since we know that X is a test element, it will have to follow this relation. Okay, X and X to power P has to be an I to power P, comma Z to power P. Okay, so how do we. Okay, thank you. Alright, so this is X to power P plus one, which because peace odd can be written as extra power to K, which is X square to power K. And you know, if you take minus one out is minus one to power K, and then you have a sum from zero to K, some binomial coefficients. And why to power three is easy to power. Five times K minus. Okay. Now, because two K is P plus one means the case less than P. And we don't need to worry about the binomial coefficients, it's never zero. So we need to look at the distance. We have a sum of terms like this. Okay. Now you can actually short argument tells you that we have a sum of terms like this right but this you can regard this relation in the KYZ, in fact, and if a sum of terms like this has to it's in this ideal because this idea is monomial then all terms have to be there. And a little exercise shows that there exists an I such that both. Maybe I should just say J. Three days less than P, and five. Okay, so there's a little numerical exercise if you want to show so there will be at least one term not in I power bracket and I power P comma Z power P. Okay. So this is basically a direct proof of the fact that the hypersurface is a rational okay uses a few tricks and uses the fact that X is a test element and then a little argument with numbers. So you can try to reproduce this. So that tells you that X is not in the type closure of YZ so that tells you that YZ is tightly closed. Okay, so that makes it an irrational ring because we checked it for one system parameters and that's enough. Okay. You can read you the computation to see what happens if you have. Sorry, if you have this ring. Why can actually argue that this is a little easier. When X is greater than five so here you can actually show that this ring is not irrational. In some sense it's easier to show that something is not irrational if you have a good guess about what the element will be in the type closure, for example by looking at suckers of elements modular the system parameters, then you have a good guess about what should be in the type closure and then you, you can try for the C, which is an important part in type closure to the, the C the elements see the multiplier that it is in are not. You can choose whatever is convenient to prove that you have a type closure relation. So the other exercise, the original exercise that show you need to show that the ring is irrational, then it's a little harder because you don't know what C, you should use that's why it's very useful to know ahead of time what would be some test elements in the region. All right, so now to after I gave you the flavor of what computations direct computations mean in type closure, I will hopefully have time to get back to this to show you that this result, this ring can be obtained as a consequence of some more sophisticated results that would be short approves give shorter proves that the original the first ring is irrational in the second ring that I just wrote is not. All right, so like, I would like to move on now and talk about a different characterization of irrationality, which is called stronger rationality which is due to values. All right, so let's see. So what's stronger rationality. Let's take some notations. So, in this part we're going to take the ring to be reduced to begin with an aetherian and for an element D, not in any minimal crime. I'm going to have this map. I'm going to be divided by fiend xd comma q that sends one into D to power one over two. So basically the map the map sends little R into D one over two times R. And we say that this map is quasi split if it's injective and if he stays injective when you test so with our modi or is a parameter idea. And now what's this notion of strongly irrational. So irrational says that for every element D, not in any minimal prime, you can choose a q zero such that all these maps fiend xd qr quasi split as soon as q is greater equal than q zero. And we're going to show that this is a characterization, a different characterization of irrationality okay the first part of the theorem. The only way that I'm building the words is to show the strong irrationality implies irrationality okay so that's a little easier the converse it's a little harder. So, let's, let's see why a strongly rational ring that is a homomorphic image of my colleague is irrational. Okay, so you can make a quick reduction and assume that are is local and that's the case when I is generated by a system parameters. Okay, so now assume that you have something I want to show that this ideal is tightly close that would be enough right so. Let's take an element in I star. What do we have that there exists an element D in our not such that Cx. Power q belongs to I power bracket q. Okay. Well, if you take cute roots here means that D to one over q times x is an expanded to R one over Q. Okay. Well, but we know that this map is injective, because that's the definition is injected for large that's the definition of stronger rationality. So that means that access to be an eye. Okay, if the one over Q times x is in. I am sorry, I should be. Okay, so this is an immediate observation. Just going back here to the definition of stronger version rational if you compare that with a ration. Every rational says that if you have something like this. So every rational says that for every X for which that happens. There is a D. And that that thing implies that X is in I. Okay, the quantifier formulation for strongly rational definition says that it changes the point of view it says that for every D. You can find a q zero such that a relation of the type that I underlined here implies that X is in I. Okay, so it is some ways that I close your point of view looks at X. And for every X, you must have a D and a q zero. While the stronger rational definition says that for every D, you can find a q zero such that these relations implies X and I so it's a it's a matter of perspective and it's interesting that we will conclude that strongly rational rings are the same as a rational. So now, why, why do we want to look at the stronger rational property, first of all, it has an obvious connection to strongly of regular definition that link one and Neil talked about, but also useful it's very useful to proving this result, which is established for reduce a finer rings. If you remember an infinite ring is by result of gobble is an image of a finite regular local ring so in this theorem is that's actually needed. To have are an image of Gorenstein basically but regular would be enough. Anyway, so if you ever reduce a finite ring of rationality is the same as stronger wreck rationality actually that that's history in general that I miss stated this so this hypothesis is necessary to prove that the rational locus which is the collection of prime ideas that be where our localized piece of rational, it's open in Zarisky topology. That's a very useful property to have the rational locus to be Zarisky open, and that is true if the ring R is reduced and a finite and, as I said, this locus is open in the Zarisky topology uses the characterization of irrationality. Yes, strongly rational. Okay. All right, so as I said I'm going to prove the converse of that shortly. Now I like to mention the result about the formation of irrational rings. So, if you have a local ring that is a homomorphic image of Koyama calling, and you have a non zero divisor on our such that module at you have an average ring, then the original ring is a rational. So I like to sketch the proof of this. Okay, this is useful sometimes. You can find such a non zero divisor and then if you model by it you draw the dimension by one so it might be easier to prove that R mode Z is a rational. And then you get immediately the original ring R is a rational. Okay, so let's take a look at how the proof goes here. So the quotient arm or Z is a rational therefore is going to call in a domain. Okay, and also we get for from coin McCoy theory that are self is going to call it because it's going to call the module on a zero device. Okay. So, remember, because arm or Z R is domain that means that the principal ideal Z R is prime. Okay. So, now we can complete. Z is an element of R is going to call it so let's just complete Z to a full system of parameters in R. Okay. So I want to show that this, the ideal that is generated by these elements is slightly close. Okay. So let's start with that I close the relation. So, that means we can find C in our not such that this happens for Q large enough. All right, so now, because Z are is prime. We can write it in this form D times a power of Z, okay you can take powers of the out of C as much as you can until these not policy so we got the fact so this D is an element of R. I know that I'm denoting the index of the, how many elements and this is no parameters by little D but hopefully that wouldn't be confusing. So now if you take q greater than T. Then what do we have that I close your relation is this. Okay. The fact that we have a regular sequence, we can get here. Okay, you see, we can combine the powers of Z together and obtain this C times extra power q belongs to that ideal. And I'll go. Mod ZR. Modular Z are we have that the ring is a fractional. Okay. That doesn't mean it means that the image. So if you look at this relation that gives you a tight closure relation in the quotient if you go mode Z. So the image of X is in the image of X to comma XD in our mode ZR because that was a rational. So, what we get. I'm sorry, I think there is a typo. Is it the Z to the power T. Sorry. Yes, thank you. Thank you. Thank you very important. I appreciate it. This D doesn't disappear when we go more Z that's why it needs to be there. So what we're using essentially this part. When we go mode Z are we need to have a tight closure relation in the quotient. And this D is non zero. Thank you very much for the time. Yes. So, when you go in the mode Z are you have a tight closure relation. So the image image of X has to be in the image of X to XD in our mode ZR. So that means that X itself is in the ideal gender by Z X to XD. And that finishes the book. Thank you. So this is a very simple proof that shows that the irrationality deforms. And I've learned that this is in contrast with other type of deformation results for classes of rings in tight closure theory, strong of regularity does not deform by an example by an relaxing. And in general, if you really also doesn't deform so one has to be careful but luckily if rationality does. Okay. I'm going to skip in the lecture notes you can I stated the result that tells you what happens with a rational rings under faithfully flat maps. Basically, does the rationality property propagates to ring from R to S. And the answer is yes on the certain conditions which mean that the close fiber of the the morphism has to be nice in some way so if you have that the close fiber is if you have a local map between the things like this and the close rationality does behave nicely under faithfully flat local maps. So I'm not going to state the result I'm just going to indicate select your notes if you're interested I included there for completeness. All right, so now, let's move to a different point of view about irrational rings which is very important and that's in connection to local homology. So, let's fix some notations I'm just going to talk about local rings the Italian dimension is the and as before x one to x, the, you know the system of parameters. So, oftentimes we're going to talk about Frobenius actions on modules and the Frobenius action means a map that is additive. And when you apply it to R times M are to what comes out is out to power P. Okay, so that's a Frobenius action and most of the time. The module that we're going to discuss is the highest local homology of the ring I will support in the maximal ideal and that's an opinion ring, and I like to recall how there's no way of obtaining it so hopefully most of you have seen this, if not I suggest you look in the monograph by Bruce Herzog they describe it very well there. So there's a check complex for which it's called homologies are the, the, the local homology of the ring are. Okay, so. Okay. We're going to be mostly interested in the last one. Okay, which goes from here. And because is a, it's a quotient of our localized at x, where x is the product x one to x the module. And this last map in the check complex, we can denote an element in this fashion. Okay. And now, one observation would be that the original Frobenius map that you have on the ring R is obviously exists on all these localizations of our various elements, and therefore it's also compatible with direct sums. So you have a natural Frobenius action induced by the original Frobenius on each of this module in this check complex. Therefore, you have a Frobenius action on the homologies, therefore you have a natural Frobenius action on the local homologies. Okay, well, what does this Frobenius action if you track everything down. How does that work on this element data, basically raises z to power p and underline x to s. Okay, so that's how the Frobenius map works. Okay. So that's one description of the check. Sorry, of the local comology. Another one, another description that is useful for I'm going to do this for the, the highest local homology only is using the direct system of maps, obtained in the following way take the quotients are mod x12 power t, xd to power t, multiplied by x, remember x is the product, x underline is the product of all these elements x12 xd. And you end up in, so you can construct this map with codomain R mod x12 power t plus one xd to power t plus one. So this is organized in a direct system so you can obtain the direct limit of the system is known to be the highest local comology of the ring R would support in the maximum idea. Okay. So we can construct this local comology module as a direct limit. Okay, now, a few observations here. One of them would be that when the ring R is square McCoy all these maps are injected. This comes from the regular sequence of the property x of the, of the system of parameters x12 xd. So this map will be injective is R is square McCoy, and you can think of the highest local comology as a direct union, basically, of this type of you just have to realize that the transition maps are multiplications by this x underline. Okay. So now with this description, you can tell that if you have two sequences of parameters one j, let's say x12 xd and why I want to ID, then one can find an injection. I use this in the previous theorem from R mod j into R mod I want to party ID to party and the reason is as following R mod j here. j can be used to define the local comology right so this definition that they provided is independent of the choice of system of parameters. So if you use in the definition the elements x12 xd to get the local comology that's one way so R mod j injects into this local comology. But on the other hand if you use the other system of parameters. This local comology is a directed union if you want of this of modules like this. So for T large enough you can find this to embed into one of those, because this right hand side is what gives you the local comology for T large enough, as a directed union. So that's a different point of view and it's a different way of finishing the result that I stated earlier. Okay, so we have these two descriptions for the highest local comology one remark here if the ring R is quite my colleague, we only have one non zero local comology, the one level D. Okay, so this would be important for chemical ring and if you remember every rational rings are chemically so when you talk about rational rings you truly have just one or zero local comology module support in the maximum idea. Okay. All right. So, one important result that I didn't get to and you could say this should be you know is one of the main things about rationality is that in the present of the Gorenstein property. And an F rational ring is the same as a weekly of regularly and to recall what the weekly of regular means, means that tight closure of every ideal is itself. Okay. So, the example that I work with before that hypersurface hypersurface x square. So, for example, that type of result would. That type of ring would be where a theorem like this would apply. So we proved it's a rational therefore that ring is in fact weekly of regular as soon as we see a proof of this theorem. Okay, so let's let's take a look how does the proof go. So, we can localize here. And we can assume that our is local the reason that is, is because we can have regularity can be checked in the presence of Gorenstein, you can, you can localize the maximum ideals. So using Gorenstein there. So, using that reduction we can assume that the ring is local in Borenstein. So, let's show that every ideal is tightly closed. If R is a rational. Okay, if the ring is weekly of regularly it's automatically rational so you want the commerce. Okay. So how do we prove that if we if we know that the ring is a rational how do we prove that every ideal is tightly closed. A standard reduction in the closure to you it tells you that when you start with. If you want to prove that every ideal is tightly closed it's enough to look at M primary ideals. If you can prove that I is M. For an M primary deal. If the ring is tightly closed and it will follow the every other ideal is tightly closed and the reason is because in the presence of test elements, the type closure of an ideal I can be read as the intersection of the type closure of I, plus the maximum ideal power and and you take the interest, the type of those ideas and you take the intersection over all. Okay, so there's this little trick that allows you to reduce to M primary ideals. So we need to show that every M primary ideal is tightly closed. Okay, so now, because the ideal is in primary it means that this is our team. Arm or die. So, every opinion module can be embedded in a direct sub some of copies of the injective file. Actually, you can also complete the ring I forgot to say that so we can have a nice matters quality going on. The ring can be embedded a direct some of injective house of the residue field. And when the ring is Gorenstein. We know that the injective how is the same as the highs of a comology. Okay, and the lock, if you remember the one of the way we describe the the local comology was as a direct limit. So we have a lot of ideas like this and in this direct system the maps are injective because the ring is Gorenstein there for chemical. Okay. So, now, this is an Artenian ring. Okay, it's also not here right so in fact, in air in this injection, you can replace each copy. This is a homology by some quotient like this for T large enough. Okay. So, now, zero is tightly closed in here. We know that the ring is a rational right so every ideal generate by system parameters is is a rational therefore if you look in the module arm or the x one to party exit to party zero as a some module is tightly closed in this in this. So therefore zero is tightly closed in the direct sum. There we have here. Okay. Well, so therefore, zero will be tightly closed in arm or die. Okay, because if you have an injection and zero study closing the bigger ring then zero will be tightly closing arm or die. So that means that I was I start here. So, you can look at this proof and realize that this proof is it's like one of the reasons you can ask, why do we need tight closure for modules. Okay. And this proof is an example. Because if you look what we did. We actually use module theory right so in embedded arm or die we looked at the question about whether I was I start for an imprimary deal, and we converted from ideals to our model as a module and we embedded that in a direct sum of copies of the module using the fact that the injective hell is the hall is the same as the local call Moloji. We replace each copy of the injective hall by one copy of this form our mode x one to party exit to party. We rephrase everything as we notice that zero is tightly closing this direct sum of modules. And because it's tightly closing this direct sum of modules is tightly closing every some module they inject into it. In particular, it's tightly closing arm or die. And if you rephrase what that means it means that I was I start. So, for example, this would be the reason why I don't want to consider one of the many reasons why this useful to consider tight closure for modules as well. All right. So now we know that for Gorenstein rings of rationality is the same as week of regularity. All right, so I like to finish in the last five minutes of this lecture to with a proof of the fact that an irrational ring is actually strongly rational. Okay. So, let's set up the notations for that so I'm starting with an irrational ring I want to show you strongly rational, and I'm going to denote x one to x d to be the system of parameters, and I'm going to take see an element not containing any low prime. And I indexed T as being the idea generated by x one to 40 x d to 40. Okay. So I'm assuming that the ring is a rational so under these conditions I'm claiming that for this seed that I have, there is a q zero such that this map, our mod I power bracket T into art one over q module it times are one over q. The sense one into C to power one over q is inject. Okay. So basically, if you think about it we know that this is where it's injected for I, because the ring is a rational and we want to show it's injected for every idea of the form it so the crucial part is this. This is always injective for a given C and a given I, in this theorem, this Q zero is independent on T, that's the claim. Okay. So, if you fix it, there is a Q zero but we want to use your the works for every T. Okay, the only thing that is fixed here is the system of parameters at the beginning. Okay. So, again, I'm, I don't have time. So I'm going to skip the proof of this is not trivial, but it's not very complicated either so it just uses the fact that we can understand the circle of an L of a ring like or for module like this are mod it is. Okay, so if you look at the activity of this map you can look to trace it back to suckle what happens to suckles and then using some neat arguments about. Artinian modules basically you can you can show we show that there is a Q zero that works for all T. Okay, so I'm going to move on from here so we know that there is a Q zero that works for all this ideas it. Okay. Now, the second part is to to to get from here that. In fact, this map of local homology modules is injected. If I have a rational ring, I want to show that there is a Q zero such that for every Q greater equal than Q zero, the map at the level of local homology modules that is given by multiplication by C to power one of a Q is injected. Okay, so now the proof here. The proof here boils down to this diagram. The vertical I have the direct system, the map at the bottom is the map that I care about and okay. So, so basically the previous result allows me to find the Q zero that makes all these maps injective at every level in the direct system. Okay. Now that's why it was important for the Q zero that I chose previously produces all these horizontal maps for every T to be injected. Well, the bottom map is a consequence of these horizontal maps. So you will get a get the injective injectivity propagate at the very last line. Okay. And then, I have, if I can just use a minute to finish this. Yeah, I think so, I think so. Yeah, go ahead. Thank you. So finally, how do we use all this to show that a rationality implies stronger rationality. So, are is a domain so it's enough to take scene on zero. Okay, and we need to show that if I is parameter ideal, then there exists q zero such that for all q greater equal to q zero. This map is injected. That's what we want. Okay. We need to realize that we prove this. This follows immediately from the rationality for full system parameters. How do you do it for partial system parameters. Okay. So, let's say that I generate by x 13x k where k is less than the dimension. Okay, we're going to complete the sequence to a full sequence of parameters. As before, we're going to take the end powers for n greater equal to zero. Okay. For the C that we have the local co homology result shows that there exists a q zero, such that for all q greater equal to q zero, the map. The level of local homologies is injected. Okay. All right. So, for that, for any q greater equal to q zero you have the maps that we care about. All right, so these are injective. This is injective, and this is injective. It means that this is injective. Okay. So that means that if you have any time you have something like this, you're going to get that x is in line. Okay. So now if you take, not D. So, if you take C times extra power q and eyeball racket q, you get the C extra power q is in I am by natural inclusion. Okay. And you get that X is in iron by this by this observation. So in fact, X has to be no data station of all I am. But there's just a. Okay. All right. So, this proves that strongly, sorry that an irrational ring is strongly rational. Okay, so I'm going to stop here for my second lecture. And guess we're going to have a quick break. Let's see, are there any quick questions for foreign.