 All right, so welcome back next section module finite extensions and plus closure. Um, so in this section are will be integral main, so we don't have to worry about equidimensionality like what was talked about in the questions. Um, so, um, So let I be an ideal and let X be an element of our we say X is in the plus closure by if there exists a module finite domain extension are into s and what I mean by that is that s is an integral domain, and you've got this injective ring homomorphism from our into s such that s is finite as a normal finite as a module with respect to that the structure homomorphism of that as a module, I mean this homomorphism. Um, such that, so it's called a speed, such that the of X is in the of is. And so we write, usually write X is in is intersect R. Right, because if you say X is in the pre image of that about. That extension. Okay, so proposition plus closure is a plus is is a closure operation. Okay, so what do we do so let X, Y be a plus and that are, you know, R, and I want to show that X plus our Y is there so. And so what we want to do is we want to find a common module finite domain extension of s and T. And so what we want to do is we want to find a common module finite domain extension of s and T. And you can always do that. And I noted, but not how to do it. It's like you take the, the, it's a, it's a, like a push out construction. So hence, X and Y are both in you intersect R. And so X plus R Y is to um, so I plus is an ideal. And I claim that extension and order proper order preservation are clear. And so, if X is in I plus plus. It's quite a potence. Um, so you have X is in I plus s for some module plane extension s. But why one through Y K generate I plus but T one through T K, the module finite extensions with Y, I, Y J in I T J find a common module finite sensor of s find a common model to find extension you find each Y is in I U. So X is in I as well. Okay. So, now let's say you have module finite extension of domains. Take any non zero element D of s and then there's some R linear map module map call it G from S R such that G of D is non zero. This is what Mel Hoxter later called solidity. So instead of saying that s is solid. So what do you do? So let W be the non zero elements. And so then Q is equal to the fraction field of R is equal to R W. And for all W and W, we have W D not equal to zero. So D over one isn't zero. SW I guess I should say right so you have R W into SW. And so. So what do you have so then SW is a Q vector space. It's a finite dimensional finite dimensional new vector space with the over one zero. So, then it's part. So then the over one is part of a minimal generating set that is part of the basis. So we'll let the SW to Q the projection onto like that. So, and that's a Q linear transformation, right. So the set of Q linear transformations from SW into Q is what home R W SW R W. And then since S is finally presented as an arm module. That's the same as home R S R localized W. So there's a R linear map from G from S to R. And C and W such that the is equal to G over C. And then that's the same thing as saying that G of D over C is V of the other one is one, which is C over C. And since we're in domains that's the same thing the same. So it was not so. So, why do we care. Well, if you have a modifying domain extension and I as an ideal of R, then I claim that I star is equal to is star. And actually equal is true but all I need is containment. Sorry, I need this. That's my plus. So, proof, let X be an IS intersect R. Let D be a non zero relevant of S for S to R. Zero. Actually, I'm going to do it this way. It's not zero. And q not a power of P, such that DX the queue is in IS to Q, which is I QS for all Q, not, then what you have G, S to R. That's the same thing here. With G of D is not zero. Then by our linearity, what you have C X to the queue is equal to G of D X Q. So that's our layer that's just what we have, but then that's the same as G of DX to the queue, because X to the queue is an R. And then G of IQS, which, again, by our linearity, this is contained in IQG of S rational local ring. Then I claim that R is a normal domain. Okay, so first one shows that R is a domain. So assuming that you knew there was a paper domains. So I'm going to show you the actual bill of R, let alpha be integral over R. Let S be our bracket alpha. It's going to be a lot of finite. Right, that's, that's one of the equivalent definitions of an integral elements, and we'll close your rings. So let's say alpha is equal to a over B. So a and B are in R and it's not zero. So a is in BS intersect R, which is contained in star, which is equal to. That's because, you know, be the principal ideal is a parameter is a parameter ideal of, you know, any non zero element ring certainly after dimensional, so it's okay. So where you have a equals BC for some CNR. And so you have the alpha is equal to BC, and therefore by cancellation is S. There's probably a slicker proof of that somewhere, but I'm colon capturing test elements and persistence of the power of tight closure theory comes from this colon capturing stuff, which is allows you to treat a system parameters as if it were a regular sequence up to tight closure. So theorem. R be equidimensional quotient, Y bar, the system of parameters for R, then two things, actually a whole lot of things, but at least these two things for all J less than the Y one through Y J colon Y J plus one is contained in Y one to Y J star. And to T for all natural numbers T to take Y one to Y D to the T. You call in with Y one to ID. T minus one is single element. You get inside of the tight closure of Y one to Y. And so this certainly would be true for like variables over over a, you know, how long we're over a field. It's true for regular sequences without the tight closure and so but then the nice thing is that it's true for services and parameters in this very general context. So proof. Well there's proof one. So, first, reduce to the main case. Let P one to P K be the minimal primes of R by equidimensionality. I'm assuming that are as I could mention all the dimension of arm odd J is deep for all J. That's what I could mention is. So the image of one bar is a system parameters in each or larger. So, Y one to Y J colon one J plus one over R. And for each. Yeah, I just die. Y one to Y J bar colon over arm odd. And then with Y J plus one bar is by the domain case. Below containing the Y one bar to Y J bar tight closure and P. So then by reduce to domain case proposition above phase in the tight closure of these matters in. Okay, so there are we may assume now that ours of the main. Okay, so let S in the common quality local with R equal S mod J. And J is prime because ours of the main. But TV dimension of S, then like home the colonists, the height of J. So let's call that H is equal to T minus T. So let X one through X H the elements of J, such that in S localized at J. They become a system of balance for S localized J. So, then X one through X H S J is J S J primary. So, now, now just one question just just for clarification. Just for clarification, you say that your your sequence leaves in J so and you are working in S mod J. And when you are modding always mod J this is become a system of parameters but this becomes zero priority. I'm not working in S mod J. I'm working in S localized. Okay, okay. Okay, okay. Okay. Thank you. So there is. Yes, yes, sir. So there exists an M such that J S J. The end is contained in X one to X stage. That's J. The maximum primary primary and maximum ideal means that some power of the maximum ideal is contained in your ideal. So there exists a C in S minus J with CJ to the end is contained in X bar let's call it X bar. So we're kind of working in S here. So let's choose, and I'm calling it C suggestively, because this is going to end up because that the residue class of this is going to end up being sort of testing, or parameter ideal stuff. So choose q naught bigger than or equal to n, that's power p. Then, you know, CJ to the q naught is also. Okay. Um, so what do we have before we have this system parameters in R. So, let Z one through Z D. Yes, the listings of why one true. Then I claim that for any power q of P X one through X H, along with Z one to the q comma Z D Q is a commutable regular sequence in S. Right, because it's a system of parameters and a comma call a local ring and so it's a permutable regular sequence. So, so now to prove one. So now, let a be an S with a bar in Y one through Y J colon Y J plus one. Then what you have is that in in S of R, then S Z J plus one times a is in Z one through Z J plus capital J. So that means that for any q bigger than or equal to q naught. So, Z, Z J plus one to the q A to the q is in Z one to the q comma Z J to the q plus CJ to the q. But CJ to the q is contained in the X is right so this is contained in the one Q, CJ to Q X is. That's what you have C A to the q is in that sequence. comma the exit this colon Z J plus one to Q but this is a regular sequence of colon doesn't make any difference. And that's contained in the one to CJ to the Q colon or plus capital J. So you have C bar a bar to the Q is in Z one to the Q. Sorry. And that's for all and that and C bar is a non zero element of the integral the main R. And this is for all q bigger than or equal to this q naught, and therefore a bar is in Y one, one J star. So that's the one that's the sort of classic colon capturing and the proof of two is very is very similar modulo and exercise. And a remark, any excellent rings the homework the image of a comma collie ring, how a sock to prove that in 2002. And so this is true for any equidimensional excellent local. To similar. And as a corollary that the menomial conjecture in character, but but I didn't prove to so I won't state that. So the corollary to what we did above is an effrational rain in this circumstance has to become a colleague. So that are be effrational and the homework image effrational equidimensional. So proof of a comma collie local ring. Actually I don't even have a colleague. So proof by an earlier result is a normal domain. So the corollary to the national, but why one through YG parameters, and what you have Y12 y j colon y j plus one is contained in Y one through y j star, this is colon capturing, but then this is equal to Y12 y j rationality. Okay, and that's what the common collar. And the system parameters is back to record. So section seven, test elements of tight closure. Okay, so like I said before, this is something that really that makes tight closure very powerful because choosing the C is generally a persnickety difficult thing in a general ring. It's what makes tight closure in some ways more difficult to deal with than my previous closure where you just take the power and it has to be the correct power of an ideal. So right, you haven't really made tests and you don't really know how to test it because you don't even know what elements try to multiply by. Well, so definition. Let's see the element of R naught. And we say C is a test element, respectively a big test element. If brawl ideals i and brawl x and i star, C x to the q is an i bracket q brawl q, respectively brawl submodule inclusions L and M and brawl C and L star M, C z q M is an L q M brawl. So right, so big test element that's, there's, it's like tests for big modules as opposed to just ideals. And then there's this weak test element thing as well. So, and that might seem like a lot to ask for. And in some sense it is, but there's this big theorem by, you know, Hartzmann-Huecky and then there's part of it is in this thing of shark. So let R be a reduced ring, that is either essentially a finite type over an excellent local rate, or F cure or, so that middle one is, is so sharp as theorem 2010, such that R1 over P is finally generated as an R module. So that's the condition called F-finance. Then, oh, not just that there exists test elements, let's see the any element of R naught such that Rc is regular. So such an element C always exists in an excellent ring of spring, basically by definition of excellence, which has to do with the regular locality being over. Then some power of C is the test element. Is a big test element, C Tucker's lectures, proposition in, and then a proposition. If R mod P has a big or just a test element for every minimal prime P and R is reduced. So you really just have to do the main case, really. So I guess I should mention something about this theorem. You might not know what essentially a finite type over another ring is, so what that means is that there exists an excellent local ring A, and variables x1 through xn, w, a multiplicative set in the polynomial ring such that R is as far to A bracket x1 through xn, localize the w. So that's what's essentially finite. Okay, so I'll just one question. In your previous definition, maybe you can allow that this, in the previous definition you have given about R, maybe you can allow that you can also mod out by an ideal or is like this definition. Oh, sorry, yes, sorry. Sorry, yes, you're right. Or something, you can mod out some ideal there. Yes, you're right for now, I'm sorry, yes. Yes, there's an ideal i such, yeah. So you extend polynomial variables, localize, and mod out by an ideal. And so that's what essentially finds. Okay, good, good. So let's just prove the non-big base of this, that p1 through pn be the middle primes of R for each, let di in R mod pi, get test element, choose ci, ti and c as in the pack element, and c is a test element. And so let i be an ideal and x and i start, and for all i, x bar is in iR mod pi star. So for all q, di times x bar to the q is in iq plus pi, is in just iq plus pi. So ti, ti is in all these other prime ideals, all the other middle primes, is just one example of how this pack element works, and that's contained in i to the q since R is reduced. Right, because it's in i bracket q plus the intersection of all the middle primes of R, but all the middle primes of R hence cx to the q is in iq, since c is for the proposition 7.4, tight closure is a closure operation in such a way. So let R be a ring satisfying, I'll just say a ring with a big test element, then tight closure is a closure operation, not just the finite ones. So proof, we need only established item codes. Let z be in the tight closure, of the tight closure L and M, in M, let c be a test element, and for all, we begin with the q naught, so I'll draw q really, cz to the q is in L star qM, but also c times L star qM is contained in L qM, because it's true for all the elements of L star M. So c squared, cqM is contained at c times L star qM, L star, which is then contained in L qM, but c squared is also in R naught, thus z is in L star. So I said test elements and the persistence of tight closure. So I claim that with this theorem on test elements, you can then show the tight closure persists. So that's like better than functoriality, that means if you're actually changing rings, you can maintain that tight, that if you're in tight closure on ideal, then the image of that element is in tight closure, but the image of that ideal. So let G from R to S, your ring homomorphism. We say tight closure persists across G for ideals, respectively for modules, if for all ideals, if X, let's set it this way, G of I star is contained in, respectively for all sub modules inclusions L and M, G of L star M, let's say G prime, L star M contained in L S star tensor S. So what do I mean by that? So L S is gonna be the image of L tensor S, M tensor S. G prime is the sort of the natural map from M to M tensor S. So you take the closure of L and M and you take the natural thing induced and you take the closure of L and M and you take the natural thing induced from G and you apply it, you get inside of what happened if you first went to sort of tensor everything with S and took the type code out there. So that's persistence. You can see how that would be a useful, and the big theorem here is that is that existence of test elements in the way that we've said implies persistence theorem. Let's see from R to S, your ring map, such that R satisfies the conditions of the theorem on test elements, then type closure persists for ideals of M. So proof is for ideals. So basically you wanna reduce to the case where R and S are domains, G is subjective. It's subjective. So the reason you can go to that case is basically, okay, you reduce to the domain case to show that R and S are domains. If it's an inclusion of domains, then you're okay by an exercise. And so every map between domains factors through a surjection and then an inclusion. So you've got a surjection from one domain onto another and then there's gonna be an induction on the heights of times. So assuming that we're okay with all those reductions and it's all detailed in the notes. So let P be the kernel of G. Prime be the integral closure of excellence. All right, we're assuming satisfies these nice conditions. R prime is module finite. And so by integrality, you can choose a P prime and spec R prime with P prime intersect R is equal to P. And again, by excellence or by integrality, P prime has height one. Excellence of R prime since we know it's a theory and stuff. The regular locus of R prime is open. So there's a C in R prime minus P prime. That's true, but I skipped something. So R prime localized P prime is a DVR. Hence, right, because R prime is normal. It's a very normal domain. So by excellence, the regular locus of R prime is open. So there's some C in R prime minus P prime such that R prime localized at C is regular. Hence by the above theorem on test elements, C has power and C to the n that is a test element for R prime. So now let I be an ideal then X is in the tight closure of IR prime because persistence is okay for inclusions of domains. So for all Q, CX of Q is in IR prime, right, Q. So going mod P prime, you have C bar X bar of the Q is in IR prime mod P prime tight closure. And C bar isn't zero in R prime mod P prime. So that means that X bar is in IR prime mod P prime. But R prime mod P prime is a mod of finite extension of the domain. So when you have X bar is in IR prime mod P prime intersect R mod P star, which is then, but R mod P is S and that's what we wanted to show. So sorry for skipping all the reduction steps but they actually do end up being quite straightforward. So this has some important corollaries and I'll end with those. So let S be a complete weekly of regular local ring and R a complete local subring such that IS intersect R is equal to I for all ideals I of R, for instance, R is a direct sum and of S will always apply that. This is called secret period. Then R is weekly of regular, it's approved. So I is contained in I star, which is contained in IS star intersect R. This is by a persistence is a direct sum by some end of a complete ring to complete. Oh, actually I didn't even say that I said R is complete. So it's an excellency stuff. And it's done anyway. So this is equal to IS intersect R which is equal to I since S is actually a regular. So now proposition S for regular and R and Ethereum subring such that R to S split and R is common color. And if R is locally excellent, it's also normal. Yeah, and I'm out of time. But it comes from persistence because you can, well, it comes from this corollary above. You've got these basically, once you reduce to a nice case, you reduce to the complete case. So S is a complete regular local ring. Therefore it is weekly F regular, in fact it's F regular and therefore R is going to be. And you also get F rational or general. And so F regular is twice weekly F regular. And that ends up giving you a new proof of the Hoxter-Roberts theorem that if S is a polynomial ring, K of field, then G is a linearly productive group. Then the ring of invariance S G is called the collie. That's because it's direct summit because there's this thing called the Reynolds operator which splits the inclusion from S G into S. Okay, again, sorry for rushing. I guess I put too much in these lectures but it was hard to know what to believe out. I did believe out a lot of things. Anyway, thank you. If you have any questions, I assume there's probably a lot. I have one question in the previous corollary. Of course, if you have that R inside S, if you have that R inside S is faithfully flat, this also works, right? Because the condition I S intersectional F is automatic. This is more general decision. Of course, if this is a direct summit, this is another situation but I think faithfully flat is a bit more general sometimes. Well, I mean, they're independent properties. Yes, yes, yes, you're right. It doesn't imply it's a great summit and direct summit doesn't imply a faithfully flat but they both imply what's called secret purity which is this property. That's right, that's a good comment. Yeah, so that's a good thing. Yeah, that we kept regularity among complete local ranks, descends along faithfully, sorry, along faithfully flat maps. Yeah, so you got all this power from being able to pass tight closure. Katie, you have a question? Actually, there's a comment. Cindy Hochscher, I would say, you're assuming characteristic of the field doesn't divide the other two. Yeah, yeah, I'm assuming that the characteristic, I think that that's part of being linearly deductive, isn't it? Yeah, maybe I need to assume. Yeah, and then the other question in the chat was answered in the chat. Well, it's not quite true. So normal, I guess the definition of normal is different from what was said in the chat. So the question is, what does it mean for ring to be normal? What it means is that whenever you localize at a maximum ideal, you get a ring that's integrally closed, or that you get an integral domain that's integrally closed in its fraction field. But it turns out that if you had an inferior normal ring, it's going to be a finite direct product of an inferior normal ring. So that's in the basic theory of integral closure of rings. So that answers those chat questions. So for instance, if you're local, then you're really close in your fraction field. Then normal is the same thing, it's a neural domain that's integrally closed in its fraction field. And that the invariant stuff and the direct summand stuff, I think Lin-Chuan Ma is gonna talk more about that in his lectures, most likely. So in your proof that an efferational ring is normal, it seemed like really, I mean, is this factoring through, yeah, sorry, that was quite a while ago, is this really factoring through somehow in fact that for principal ideals, tight closure is equal to integral closure? Yeah, yeah, I mean, I guess that's equal. Yeah, that would be another way to say it. I tried to make these lectures without having to mention a closure of ideals. So yeah, I mean, it's essentially because of that. Yeah, there it is. So I guess it's something like a further name and a domain where every domain where every principal ideal is tightly closed than your normal line. So yeah, so here's the sequence of the implications. And some of this is gonna be done in the Leonson-Skoda lectures. I forget who's doing those, maybe Ian Averbach. So for principal ideals, tight closure is the same as integral closure for those ideals. And so, and in general, whether you're characteristic key or not, a ring is integrally closed in its total ring of fractions, if and only if every principal ideal generated by a non-zero divisor is integrally closed. So therefore it follows that an effrational, not an effrational, it's an effrational local ring. But yeah, that's right, that's right. I mean, the thing is that the integral closure, but that also factors for the fact that the integral closure of a principal ideal is in some sense a much simpler sort of object than the integral closure of a general ideal because an integral closure of a principal ideal is just the same thing as go up to the normalization and the trackback, about to the integral closure of the ring in its total ring of fractions and the trackback. And that's not true for non-principal ideals in general, like two-generated ideals. It's like really false a lot. And but so this like principal ideals are simpler than other ideals, no surprise there. Yeah, thanks. Okay, so we want to thank Neil for his nice lectures. Well, thanks for bearing with my pacing issues, but I hope this is useful stuff to understand the basics of tight closure and how to work with it and the other lectures going forward. Thank you, Neil. I'm saying it's factorial. Okay. Yeah, yeah. I think there's a lot of definitions of integral closure of modules and it's like, but I think in all the existing ones, that they are actually factorial. Okay, there's a question in chat. What's the relation between Frobenius closure and integral closure? Oh, yeah, I'm gonna, so I was gonna get into that later in the lectures. Frobenius closure is so pretty, integral closure is bigger, is the short answer. But in fact, you have a whole sequence of closures. So for any ideal I, the Frobenius closure of I is contained in the tight closure of I, which in turn is contained in the integral closure of I, which in turn is contained in the radical of I. And just like tight closure has characteristics zero analog, does Frobenius closure also have characteristics zero analog? I don't think so. I mean, I've never seen it. I mean, I guess you could try to do reduction of characteristic P. I mean, the natural thing would be to say that in all the characteristic P models, I'm sorry, that in infinitely many characteristic P models, that sort of the characteristic P version of the element would be in the Frobenius closure of the characteristic P version of the ideal. But I mean, yeah, I've never seen anybody do that. I mean, there's reasons like, I mean, for a long time people thought that tight closure would have this property that if you know, through the tight closure in infinitely many characteristic P models, that you're going to be able to tight closure in almost all of them. But then Brenner and Katzman have a counter example to that. So that's not true. And it, but it was known from early on that that wasn't true with Frobenius closure. I have an example in my lecture that was at the end. So, but yeah. I guess you could make that definition of Frobenius closure. But in general, Frobenius closure is considered to be more poorly behaved in many ways than tight closure. That's a segue into another question that I have. And this, I guess, is more related to material from Craig's talks. So I feel free to defer. But I'm very interested personally in my own research is studying Frobenius closure. And in particular, I've been trying to develop the right, you know, and maybe it's out there, but the right map theoretic notion to get integral closure of ideal, or sorry, Frobenius closure of ideals. So is there a notion of, you know, maps to the right kinds of rings that when you can track back, you're gonna get the Frobenius closure? Oh, yeah. I mean, let me just be careful. Yeah. So there's this thing called the, so I don't know if it's maps to them, but I mean, I will say there's this thing called the perfect closure of a ring, which is the, and you probably know about it, which is basically all the R to the one over Qs at once, like the union of the R to the one over Qs in that ring K that I showed, assuming the rings reduced. And one way to define Frobenius closure is extend to that ring and you track that. And similarly, you know, you look at the, for our module, if you have a module M, you look at the natural map from M to M tensors with what's called the R-perf. It used to be called the R-infinity, but everyone says R-perf, you know, for perfect. So you look at the map from M to M to M tensor R-perf and you look at your submodule L and you look at the elements that in that map land inside L tensor R-perf. I mean, that's the elements of M that land in L tensor R-perf and you call, and then that's the Frobenius closure of L again. But I don't know if that's what you're asking for. A little bit. I mean, so R to R-perf is an example of a purely inseparable extension, right? So I'm wondering if something like, you know, you contract I back from the recurrence of the extension of even at certain flavors that you'll get in Frobenius closure. But anyway, this is a research question that's not made enough for me to decide. I bet that'll work. Oh, okay, yeah, interesting. Yeah, maybe I'll follow up with you some other time. Thanks. Can I ask you something, Neil? Of course, I can. Yeah, this is not regarding what you covered, but like, is this first part of your talk will broadly follow your paper on like closure operation in community algebra? Yeah, yeah, that's sort of where I, that's when I developed this approach is, yeah, that's in there, yeah. Some version of that's in there, but I mean, some version of that's in these lecture notes too that they go see soon enough. Yeah, you know, I wanted to do a survey and closure operations and I was, and I thought, well, how broadly defined are these things and like really broadly, they're even more broadly defined than this. After Moore did his stuff, then like, you know, a post set theorists said, okay, well, you can sort of define this on any post set. It doesn't have to be the post set of subsets of the given set. It can just be any partially ordered set. And then you, you change your containment relations to, you know, the less than or equal to sign that exists in that post set, you still, and then it's still, if it's a post set where all arbitrary meets exist, then you still get this one-to-one correspondence that Moore pointed out in his paper 112 years ago. So yeah, I mean, it's very, it's very general. And that's, you know, I think they used that somehow in the lattice theory in a way that I don't, that I've never looked at. Any other questions? Yeah. I'm sorry. But yeah, Kyle, I'll happily defer your question to anybody that might know more about it than I do. Any more questions? I've looked and it seems like no one has written about, I mean, much less has written about Phrenius Bogey's closure, which is where I'm trying to make money. That's what he's doing. No, it's great.