 Okay, welcome back for lecture two on test ideals. Again, same disclaimer as in the first lecture. If you can't read something I write or understand something I say, please speak up. I can see the chat. So feel free to ask there if you like. All right, so I want to pick up right where we left off last time. Right, so we ended the first lecture, which is really a lecture aimed at showing the existence of test elements and talking about test elements for ideals and things like that. Right, so we ended with the following sort of corollary of our proofs and definition. Right, so we defined an ideal to be uniformly f compatible. Only goes on the right side. Interesting. All right, so we defined an ideal to be uniformly f compatible. If it is compatible with all of the potential for being a splitting for all iterates and for being us. All right, so take a potential for being a splitting and you have to be sent back into the ideal you're talking about. Right, so we saw in particular that the finiteistic test ideal, or at least it was an asserted exercise, right, is uniformly f compatible. So to show that test elements existed, it's suffice to find an element in all non zero uniformly f compatible ideals and that was what we did last time. Right, so and as a corollary, you get something else for free. Namely, there always exists a unique smallest non zero uniformly f compatible ideal. Right, so and if you take that is your definition of the sort of big or non finiteistic test ideal here. Okay, so we're defining in this thing this non finiteistic test ideal right here. Okay, and so I had a lot of trouble in some sense trying to decide exactly what to do for these lectures. So I was asked to give lectures on test ideals and the problem is that to me, there are so many different flavors of test ideals and things to remember that it's hard to know exactly what to say about what. Right, so I have attempted to in the notes put a lot of details down that I'm not going to have time to say but I want to say a little bit about. So some things that follow sort of immediately from the definition and some other interpretations of the definition that you can see sketched in the in the exercises that are not necessarily all that straightforward. All right, so let's do an easy one first. All right, so we also define last time what it means for a ring to be strong there for regular ie what it means. So the definition was that if you take any non zero element of the ring after taking sufficiently large for Venus push forwards you can find a potential for Venus splitting that sends that element back to one. Okay. Not immediately straightforward but really is very similar to the arguments that we did in the lecture last time. All right. It's it's one of the exercises I think it's meant to be turned in are strongly if regular, if and only if this new ideal, this non funny to stick test ideal, right, is equal to our is trivial. So is it not not completely straightforward, or pretty easy but not completely straightforward it needs to you need to go through some of the arguments we did last time. Alright, so, but all the things I'm going to put here in blue now our exercises, right for you to look at in the notes. Another one very important here. So the, the, as I sort of mentioned in the last lecture one of the reasons for looking at these other conditions is that we saw very, very quickly from the definition of the finiteistic test ideal that it wasn't so clear that it had any good properties at all. So basically it didn't commute necessarily with localization. It didn't necessarily pass to completion after you localize to the prime. Right. So, however, pretty straightforward to check. Right that this non funny to stick test ideal is compatible with localization and completion. Right. Okay, so when I say it's pretty straightforward to check there is still some things to check. Alright, so I was pretty I tried to be pretty careful. In the last lecture about it may seem very technical about going through the arguments about what you need to produce a test element, but being able to produce test elements that work in multiple rings is very important for many of the proofs. So in particular, right. What you will want to do to see this in many ways is you want to be able to produce a see a test element that will work, not only for our but also remains a test element after I localize, or after I complete a so called locally completely stable test element. Okay, and that comes out of the proof that you can produce such things. Right. So it's important that this proof is very has a lot of sort of surprising consequences, which is that the source of why the theory has really been helpful. Okay, so these two exercises are very important, but not all that hard. They're quite tangible. The next ones. I've explicitly listed this challenging. Right. So these are not meant to be turned in, but are historically sort of very important and also help to explain some of the notation here that I'm using. So let me mark this as challenging. Right, so the first one is okay so I've called the small test ideal or the finiteistic test ideal. We saw this was the intersection of all all elements that multiply tight closures back into the original ideals for all ideals in the ring. Right. And the exercise is to show, well, I know in the first lecture series. A definition of tight closure for sub modules of amount module was also introduced. Right. And you can check that in fact, if you look at all inclusions of our modules and inside of M, where the quotients and mod N is finally generated. Right, you could also look at the colon ideals here for the tight closures of N and M back into N. And the exercise is to show, okay, so if M is R and then N ranges over all of the ideals of R. So this is clearly something which is a priori a bit smaller because I've intersected over more things than a finiteistic test ideal. But it's possible to show that this gives you all of these elements that knock all type closures of modules back into the original sub module. When you're looking at situations where the quotient is finally presented. Okay, and, okay, so you can sort of see some finiteistic information here. The comparison theorem for the non finiteistic one. Right, so the non finiteistic test ideal then is take the same thing, but let's just copy it. Take the same thing, but forget about the quotient being finally generated. All right, right, so here, if you remove all the fun in this conditions you get the sort of non finiteistic test ideal, right, and this heuristically should help to explain at least some of the notation and why I've tried to be so careful about writing these two things down. Of course, part of the hope here, and I'll say this in a second is that you don't have to be so careful, and that they're really the same object but I have no idea. Okay. All right. So, and sort of lastly, so this is maybe I'll make one more remark about this last equality here, right so as from my definition of the non finiteistic test ideal, it didn't have anything to do with tight closure, right so another way to see this last quality is it gives a tight closure characterization of the non finiteistic test that deal was which was defined in terms of uniformly uniform compatibility with all for being a splitting so define in a completely different way. Okay, but again, if I would have defined it this way, I would run into the same problems that I ran into with the finiteistic test that deal I have an infinite intersection, and I have tight closures that I don't know that localize right so from this presentation. It's a little it's hard to see nice properties that you do have for the non finiteistic test that deal. Alright, and so let me let master mentioned one other characterization of the non finiteistic test that deal here. So let's say that our M is local and even complete. Right. So there's another sort of infamous characterization of the test ideal which is very important. Namely, this non finiteistic test ideal. You can think about the following way okay so it by the previous equality knocks all tight closures for all sub modules back into themselves, but the claim here is that you only need to look at exactly one. All right, so that in fact this is the annihilator of the tight closure of zero inside of the injective hole of the residue fields. All right, so again, completely not obvious so okay so it's clear from my presentation here that that annihilator is certainly contained or contains the intersection above it appears in the the intersection straight above it. Okay, but it's not clear that that's the only thing to check. Okay. And so I've sketched that as an exercise as well. Right, but the way I think about even showing this is to go through and show that it agrees with some of the definitions involving the existence of test elements right so it's not completely straightforward here. Okay. Any questions about any of these exercises that I've listed. The first two which I think I assigned as official exercise to be turned in are a bit more straightforward. Everything with challenging and below really means challenging so right so requires quite a bit more work to go into this. Okay. All right, so. So as I keep going here, right so let me also just mention. The a couple of open questions here so the non-finitistic test ideal, which we've defined right so again I'm thinking about this sort of as another flavor, we saw that this was contained inside the fun it is stick one. And the question here or one thing you could ask is are they really the same thing. Right. So here, this would solve some of the problems I asked in the first lecture, we don't know if the finalistic test ideal is compatible with localization and completion but the non-finitistic one is so you would get some things like that. Right so and this is really an avatar of another question that came up so is strong F regularity the same as week F regularity. Right so, and again, if you had this equivalence then you would also get localization and completion for the weekly F regular condition. Okay. All right. Okay, so with that, let me sort of proceed with a little bit more theory here. So, we've defined the non-finitistic test ideal to be the smallest non-zero uniformly F compatible ideal. Right. So you might ask, what are other ways to produce some uniformly F compatible ideals. Right. Well, one, one reason that you might want to do that. Is that from the definition, anytime you can produce such a uniformly F compatible ideal, then you get the non-finitistic test ideal is smaller than that. Right, so it allows you to bound where that thing is. Okay. So, let me sketch sort of an important set of ideas, right, which gives some other ways in which uniformly F compatible ideals show up. So, and this also is related to sort of the idea of solid closure which came up in the previous series as well. Right, so let's just say that B is any R algebra. Okay. What I can do is I can create some ideal which sort of looks like the finiteistic test ideal for sort of be closure of ideals. Right, so let me just write it down. So look at the intersection of all ideals I and R right and what I'm going to do is look at the closure operation which is given by expansion and contraction for this R algebra be so expand up to be and then contract back down to get an ideal inside of our right. I think about that as some sort of be closure of ideals. Okay. And then you could look at all of the elements that send all be closures back into the original ideal. Right. And the claim again is that this ideal is always uniformly F compatible. Okay, now running to exactly the same problems we saw before for the small test ideal, or the finiteistic test ideal. I've written down some horrible gross infinite intersection again. All right, completely not clear that this has anything that's non zero in it. Okay. All right. So let's say you're trying to show that something non zero is in it. Well there's an obvious ideal which is contained inside of it. Right so we call it the sort of trace ideal. Right so this is the by definition is the image of the evaluation at one map from the home set. Right and these trace maps will show up here more throughout this first lecture. Or today. All right so look at the map from home be our back to our which sends homomorphism to five to five one. Okay, so that image is an ideal inside of our. Right. The trace ideal here is trivial is equal to our if and only if there exists an arm module retraction from R to be ie R to be is split. Right so that just comes out straight from the definitions. Right. And it's pretty easy to check that this trace ideal is always inside of this be closure, sort of test ideal up on top. Right. And again, sort of the magic of Frobenius gives that this ideal here is also uniformly f compatible. However, whereas the first ideal this beginner section on top it was a little hard to see whether this thing would be non had any non zero elements in it at all. It's very easy to say when the trace ideal. All right has something non zero in it at least at least to define it as such. Right so the trace ideal is non zero exactly when be is a solid R algebra. So, again, what's one reason you might be interested in doing this is that if both of these are uniformly f compatible. Right so, and let's say be is solid. Right, then we know that they're both non zeros so they both contain the non-finitistic test ideal. Okay, because the non-finitistic test ideal is the smallest one. Okay, so for any be algebra, at least solid be algebra, you get some pretty direct ways here to bounds the non-finitistic test ideal from above just using the definition. Okay, any questions. Yeah, so are there any restrictions here on what kind of our alphabet is. If it's, if it's not solid, right, then it's possible the top two ideals are zero. Right, and then I'm, I'm out of luck, but assume it's solid, then there's no other restriction. Right, so then you at least get a bounce. Right, so the sort of the idea of solid closure is then to intersect over all bees so that it's solid to create and we'll see whatever the theorem there is that you get the non-finitistic test ideal again. Or you get the finite statistic test ideal as we'll see in a second. Okay, so sort of before I sort of transition to a new segment I want to sort of mention a couple of big theorems, if you will, related to this. And so the first one is really a result builds on a important results of Hoxter. Right, so, but the formal statement appeared in a recent paper of myself on rank a data. Right, so, and, in fact, let's say they're in a complete local setting. Right, then the claim is that there exists a, okay, big balance Comicali and hence solid are out ruby so that, in fact, the constructions right above all give you the finiteistic test ideal. Okay. Right, so from sort of for many years, the existence of the big balance Comicali Algebra, if you will, have been related to tight closure and test ideals, right. And so one of the later lecture series is targeted exactly at studying these things and sort of ties in with looking at the uses of tight closure and test ideals to approach some of the so-called homological conjectures and things like that. Right, so, but this really builds off of a pretty difficult theorem of Hoxter showing that there is some big Comicali B that captures all tight closure relations for all ideals in your ring all at the same time. Okay. All right. So in the second one here. The second theorem gives another instance in which the constructions of these bees give you, again, the test ideal, but it does even better in this case, it gives you both the finiteistic and the non-finiteistic test ideal all at once. Right, so the claim is that if R is, okay, so to be careful here, while you can certainly extend it, but let me be precise and say, if you know what a Q-Gorenstein ring is, I'm assuming that the ring is normal and if you don't assume the ring is normal, is Gorenstein, right, and then you'll get something out of the theorem still. Okay. If you take a Gorenstein or Q-Gorenstein ring, right, then in fact that there exists a module finite domain extension, right, so that the finiteistic test ideal is equal to the non-finiteistic test ideal is equal to this trace ideal of S over R. All right, so the evaluation at one map from the home. Okay. In addition, right, if you're in the complete local setting and you look at what's called the absolute integral closure of R, or R plus, right, then you could forget taking something module finite over R and instead passed the co-limit of all such extensions, right, so, and then in fact you get the trace ideal of R plus. All right, so sort of plus closure test ideal equals the test ideal in these settings, right, in some way. Okay. So, again, this really builds off of sort of the deepest results that's used to prove it are some very intricate arguments of Hewneke and Zubeznick, right, but the formal statement here, at least in the setting I put down comes out of work of myself with Manuel Blickel and or Carl Schwede. Okay. All right, so again, these theorems are really just features featured advertisements, right, for lots of things to look for. There's much that I can't say in sort of the introductory lecture but I'm trying to get you excited enough to go look at some of the papers. All right. So I want to change gears a little bit at this point. So, and sort of the right way to say this is over the years, partly because some of these questions about test ideals remained open for such a very long time, right. There are just a huge number of additional flavors of test ideals, right, for me to tell you about. One of the most important ones. And you can sort of already see that coming up in some of the theorems here, right, are things that are best motivated motivated by looking at duality. Right, so I want to tell you a little bit now about so called parameter test modules and ideals, right, which are sort of a dual version of the test ideals that were already introduced. All right, so. So, so throughout this setting and I'll tell you exactly when I remove this assumption, right, so let's just assume for this segment that are as come a colleague, right, you can do a much of this. You can try and extend things to the case where it's not come a colleague but you then don't have duality so you really should just least is your first attempt through this material go through it. So R is come a colleague. Okay, I don't actually need normal. And what I'm saying here so I've fixed the notes to get rid of that. Okay, so R is just an f finite comma call a domain. Right. I'm going to assume something that's slightly non trivial but not too bad either. Right, so. Okay, so as we've said, my rings are all a finite. Result of Gabber says that R is the quotient of an excellent regular ring. So in particular, it's a quotient of a Gorenstein ring, right, so R has a canonical module here. Okay. I'm going to assume something slightly stronger than that, which holds so long as R is so called sufficiently local. Right, so this statement always holds locally, and also holds if our feels like it's local enough so for instance when R is a polynomial ring which I think about as corresponding to contractable space. Right. So I'm going to assume that so we have that R has a canonical module omega R. And sort of the magic starts happening when you make a non trivial identification coming from duality. Namely, if you look at the harm set. If you look at the harm R, F to the E lower star R into omega R. Right, duality. Right since R to F floor star R is a finite morphism locally is going to tell you that that that gives you the canonical module of F floor star omega R again. Right, or F floor star R. Right, so I'm going to assume that that duality holds globally. And that's really just a simplifying assumption to make sure I don't have to worry about twists by line bundles everywhere anyway so much of what I'm saying you can get rid of. Right, so but for simplicity, assume that sort of holds everywhere. Okay. And we'll we'll see an example here for the polynomial ring of exactly what that looks like. Right, so this is sort of my black box assumption. Right for this whole segment. Okay. All right. So as I sort of get started. What does this give you. Well the whole point is that because I'm working with Coma Collie ring I have Coma Collie duality. Right so what this means is okay so I can look at the sort of growth and dig dual. Right or the Coma Collie dual of a module which is take the module and then hit it with harm blank omega R. Right. If I do that twice, I always get a map into the double dual harm harm and omega R omega R. Right as written here. Right so I have the evaluation it an element map. Right so and if M is Coma Collie. If M is maximum Coma Collie, then then this will give something that's the same thing is what I started off with. Right. So I have this dualizing functor. Right or Coma Collie duality that I can play around with. Right so and much of the theory here is comes out of just sort of naively applying this functor to some of the basic constructions. Right so the sort of key map that shows up everywhere is something called the trace of Frobenius. Okay, so again, my standing assumption was that I could identify the Frobenius push forward of omega R with the HOM set HOM F to the elorsa R comma omega R. And if I can do that, I get a very special distinguished map called the trace of Frobenius. Right. And again, we saw this before and looking at sort of the solid closure segments. Right, this trace of Frobenius map is just the evaluation at one map but it's a sort of God given distinguished map inside of this HOM set. All right so I can take phi and send it to five after the elor star of one. Okay. Right so and if you like take that as your definition of the trace of Frobenius. Right. Now the magic here really is that once I've made this identification. F to the elor star omega R comma omega R right is isomorphic to F to the elor star R again right and is as an F to the elor star our module IE is generated by the trace map. Right, so let me write it this way. Right so the HOM set. So I take any map from F to the elor star omega R back to omega R and the claim is that it's always a multiple of phi E this trace map so it's sort of the one map to rule them all. Okay. So and maybe to say it even more precisely right so what does this mean very explicitly this says that you take any such map right call it little phi. Now look at that map there exists some elements are little phi inside of the ring so that any map here can be written as a pre multiple of the trace map by some element are phi. Okay. Are there any questions about the definition of phi E or what it means for this to generate the HOM set which are really the two properties I'm trying to stay here. Okay. And again, I think it's somewhat instructive. I've listed. So I'm going to go back to the notes and heard so chapter three is sort of a prerequisite set of docket things for these talks. Right, but if you're not quite familiar with everything in that chapter which certainly I use as a reference book myself. Right. So here you just have to assume this isomorphism of flora star omega r with the HOM flora star comma omega r. Right, and just naively assume that taking duality on maximum homo calling modules produces the same thing over and over again. Right. So for instance, let's just see that real quick and this is kind of a free form segment if you will. Right. So, let's look at some things well one. I know that's from duality. Right if I look at HOM omega r omega r. Right. Well this really is the, the double growth and the dual of our itself. So this just gives you our back itself again. Right. So, and, you know, you can convince yourself, or one of the sort of main tenants of duality for finite duality for the Frobenius morphinomorphism says something a little bit sort of stronger or different here. Right, so you could ask what happens when I look at Frobenius push forward. Right. And then take its dual. Right. Let's say M here is a maximum homo calling module. Okay. Now, there's many ways to say this. Right, but the push forward functor is an exact functor. Right. And if you're familiar with the formalism of how these things work. Whenever I have a left exact functor, it is a right adjoint and whatever is I have a right exact functor it's a left adjoint so in fact this functor has adjoints on both sides. Okay. So, we saw one of them in the previous lecture series, namely F, the upper star, right so, and that is the left adjoint of FD lower star, right, but here I need the right adjoint or upper streak. Right so without sort of saying too much about that. It's easy to convince yourself that sort of using some form of adjointness. I can take this home and move sort of some of the linearity onto the right hand side. Right. And when you do that you get exactly this module home FD lower star R comma omega R. And that's really why that isomorphism I'm using is is so important is this shows up over and over again. Right. So, our assumptions on our precisely give that this, this right entry here, right, is FD lower star omega R again. Okay, right so I just use this one isomorphism that we assumed as part of our assumptions here. Right. Okay, and now if you buy my mantra from before whenever you see F lower stars and everything it's just the decoration to keep track of the linearity and nothing else. So, this is the same thing as FD lower star of home M into omega R again. Right. So in other words, right, what's the punchline, the punchline here is that if you look at the dual of a push forward. It's naturally identified with the push forward of the dual. Right, so there's this beautiful compatibility with duality that you get out of sort of finite duality. Right. And so in particular, right. I could apply that to the set of maps from FD lower star omega R back to omega R. Right. And right, the push forward commutes with the dual, which we saw already was FD lower star R again. All right. Like sense. All right, great. Okay. Here we see the abstract statement that this home set is isomorphic to FD lower star R. How do I get that it's really generated by the trace map. Right. Well to get that it's generated by the trace map you just sort of use duality over and over again. Right. And it's sort of I leave that to you. This is an exercise to try and do these sort of next segments. Okay. And instead of doing more, right. Let me tell you something here about the most important case to keep in mind. Namely, the case where our itself is a polynomial ring. Right, so let's say that case are perfect fields of positive characteristic. Right. Look at the polynomial ring and some number of variables. In this case, you can show explicitly all right well we know our here is regular. So we know that after the lower star R is in theory projective. Right. But in fact, you can write an explicit free basis to check that it's actually globally a free R module. Right. So it's a free R module on the set of push forwards of monomials where you can't pull out any of the variables any further. Right. So we're all the exponents showing up have exponents that are strictly less than P. Okay. Okay, so I'm going to write something here I wrote something here which I'm already kind of quivering here. Right so I'm working with global canonical modules. So it's natural to think of omega R as being isomorphic to our here. Right. But this really is the community of algebraist definition of a canonical module and it makes me quiver inside with fear. Right so let me just for the experts in here, say that here I'm forgetting the grading right which you never want to do. Right so but just forgive my sins. Okay, so here identify omega R with our will right so and then what I'm talking about in this duality second is you're looking at the homeset then from of all potential for being a splitting is again. Right. And what this is saying is that every such map can be written as a pre multiple of some generating map. And in fact the generating map here is very explicit right or a generating map it's only ever defined up to pre multiplication by a unit anyways. Right. As all these isomorphisms are right and that's the problem with all duality in some sense. Right so what you can do is you can take this fight to be the the projection on to the factor corresponding to the monomial with the exponents right so the dual basis projection on to f to the lower star product of all the variables to the p to the E minus first power if you will. Okay. And so I've listed this is an important explicit exercise as well for you guys to do. I should check a sort of immediately that, in fact, the, if I look at Phi E capital Phi E, that in fact we've talked about how iterate for being a splitting that in fact Phi E here is Phi one iterated e times. So, in this case it happens on the nose, right so, and in general, it's always true up to multiplication by units, right so alright so what does that mean. Well one is I could tweak my fight you to take into account the units to always make it true and when I'm taking images under of maps and things like that you'll never get any different image so nothing, nothing really goes right. More generally, it just means you have to be a little careful with twistings and things and worrying about how things glue and a lot of the theory, but morally, it's true all the time. Okay. That helps. Right so but that's sort of the most important example to keep in mind. Alright, so let me give you so sort of the next sort of important definition here. Right. What is the parameter test sub module. Well this is going to be something which very much mimics. This is something we had before. So, tau of omega r. So I'm defining this guy here, right, is the unique smallest non zero phi one compatible sub module of omega r. Right. What does that mean. Again, that just means that if I look at just this generating map by one, it should send a total typo. Phi one should sends tau back into itself. Alright, that's all I'm requiring. Right and so here I've done it sort of by the definition from this one generating map, but note that since the generating map of phi one to the e is identified with phi e right. So it's equivalent to say that that holds for all e as well. Right so for any of the generating maps, or more generally, it's equivalent to say that you're compatible with again all of the maps inside of all of the these sort of hum sets for all e, because they're all generated by the fight ease up to pre multiples. But again, sort of the magic here that's coming out in the sort of parameter test sub module world, right, is that there's one map that governs everything. Right this trace of for Venus. Again, just like if I were to have naively defined the big test ideal this way. Completely not obvious that there is such a unique smallest non zero phi one compatible sub module, ie, it requires justification to say that this just this definition is well posed ie that there is any such non zero guy. And again, adding to the list of perhaps surprising consequences. It's not too hard to show that the previous arguments we have for the existence of test elements, again produce this sort of by magic. All right, so, at least in this case magic here has a very precise meeting magic is just duality. Okay. All right, so. I've sketched this in the notes, right. But let me just say that you can check that if you take any file one compatible sub module of omega, then in fact the colon ideal of omega r into n right so the things that multiply omega r back into n is a uniformly f compatible ideal. And the previous sense so again, what does this produce this produces yet another way of randomly producing a uniformly f compatible ideal inside your ring. All right. In particular, this says that this non financial stick test ideal since will always multiply omega r back into any phi one compatible sub module, right. So, and I don't really have time to go through the proof here, but let me just say, what do you have to show right so you have to check again that you take any f to the Elora star r to our map and you have to show it sends this colon ideal back into itself. Right. And the proof is to just use duality again. Right so and I've sketched this in the notes. Right. But again, it's kind of instructed to go through on your own all you really end up needing to use this one identification again of home f to the Elora star r omega r with f to the Elora star omega r. Right. And everything else just comes out of playing with duality. Okay. Okay. All right so what's the consequence here. So the consequence really is that this parameter tests a module is well defined is well defined, ie. You know that the big test ideal times omega r is always inside of any uniform or any ideal which is phi one compatible. Right. Now, if I iterate over all of the different multiples of phi and just do that over and over again, I produce an ideal which is manifestly phi one compatible. Right. And hence, I get that the parameter test some module can be explicitly described. Right. By the some on the right hand side. Okay. All right. So, I want to sort of link up real quick back with like closure and maybe we'll do this to sort of end the second lecture and I'll really move on here in a second after picking a taking a short break. Right. So, give another definition here, the parameter test sub parameter test ideal, the parameter test ideal of our right is for me just by definition, one of these colon ideals that showed up in our proof. Right. So, we saw that or the argument was that if you look at the colon ideal of omega into any phi one compatible ideal, that that was uniformly f compatible. We know that tau was the smallest sub module that was phi one compatible so look at this colon ideal where you take in in the previous description as tau of omega. Right. So that's the right hand side. Right. And my definition is that that I'm going to call the parameter test ideal. Right. Why would you want to do that. Well, or what happens when you do that. Again, it links up back to the theory of tight closure. Right. If RM is local. Right. Then in fact, you can show that the parameter tests ideal of the ring. Right. So defined is in fact the set of all elements that multiply all tight closures of parameter ideals back into themselves. Okay. So and maybe this is perhaps the much more classical way that the parameter test ideal would have been defined. Okay. All right. If I were to do that using the classical definition once more, you run into all the same problems, namely, not so clear from at least the description I've written down here that what you get is compatible with localization and completion everything else. In this case, you can do it by brute force arguments. And this really goes back to some intricate arguments of Hoxon-Hunike from the origins of tight closure. Right. And again, you might also have noticed that I haven't put finiteistic or non finiteistic anywhere on the board. Right. One of the reasons is that in for the sort of parameter test versions, the finiteistic and non finiteistic versions all agree. Right. So they're all the same. Right. So again, in the notes, I've gone through and tried to sketch how all this is. So if you'd like take these, these statements as a advertisement, right, for things to come later on or things that you can return back to. In fact, one of the later lecture series is about effrational rings. Right. So our ring is said to be effrational if and only if the parameter, if and only if all ideals generated by a system of parameters here are tightly closed and all localizations if you will. Right. So, and we see from this that our ring is effrational if and only if the the parameter test ideal is the whole ring, or equivalently the parameter test sub module is equal to omega. Right. At least in the Komakali setting here. Right. Where I've defined things. All right. So, but this is this is a big topic again. This is a whole nother topic from the later lecture series. Right. So some of these propositions really are not not so trivial. Okay. All right. So it's 50 some minutes after right so I'm going to take a short break now here redo my notes to say that we're starts of lecture three and come back here in a second.