Congratulations! You've done it. After all the years of hard work and studying, you're ready to go out into the real world and start that career you've been working so hard for. Now that you're about to take your place in the working world, it will soon be time to begin paying back the student loans that help make your education possible for you. You've got some decisions to make and some important things to understand about your student loans. In the next few minutes, we'll go over the basics of what you need to know. Ready? So here's the answer to a big question I bet you have right now. You don't need to begin paying your loans right away. In fact, you have a six month grace period before your first payment is due. To make it easy to remember and to make certain you won't miss your first due date, the Direct Loan Servicing Center will send you a statement telling you the amount you're expected to pay and the exact date your first payment is due. Then, just 30 days before your first payment date rolls around, you'll receive an invoice. It's really a bill and you pay it the way you would any other bill by sending your payment in the envelope we'll provide. To make it even easier and smarter, since you'll also save a quarter of a percentage point on your interest if you sign up for the electronic debit account program, you can even choose to allow the Direct Loan Servicing Center to automatically debit your payment right from your bank account. Now, how much will you have to pay?  The Direct Loan Servicing Center knows that everyone's situation is different, so they give you the flexibility to choose the payment plan that's easiest for you and your budget. The first is the standard repayment plan. If you choose this plan, you'll make the same monthly payment for up to ten years until your loans are completely paid off. You can also choose the graduated repayment plan that allows you to make lower payments now and higher payments later. If you're a new borrower with at least $30,000 in loans, you can also choose the extended repayment plan and you'll have up to 25 years to pay back your loans and you'll pay the same amount every month. Or you can choose to make lower payments now and higher payments later with the extended graduated repayment plan, yet another choice offered by the Direct Loan Servicing Center. Your payments will be lower in the beginning and it will increase, usually every two years for 25 years until your loans are paid off. And if you choose the income contingent payment plan, you would make payments based on your annual income, your spouse's income, the size of your family, and total loan balance. If you meet the requirements, you make monthly payments that may change as your income changes over the course of 25 years. You can find out more about all of these plans by going to the Direct Loan Servicing Center website or by talking to your financial aid counselor. You're probably wondering just what your payments might be under each of these plans. If your loan amount totals $40,000, your payment under the standard repayment plan would be $460. Your payment under the graduated repayment plan would be $316 and would increase every two years. Under the extended fixed repayment plan, your payment would be $277 every month. And with the extended graduated repayment plan, your monthly payment would be $230 for the first two years and would increase every two years as your income goes up. If you choose the income contingent plan designed to help those with low annual incomes, your payment would vary based on your family situation and your income, but it's likely to be significantly lower than with other repayment plans. I know it's very tempting to choose the plan with the lowest payments right now and that may well be the right thing for you to do, but the important thing to understand is that you'll pay more interest the longer you take to repay your loans. But the Direct Loan Servicing Center knows that starting out in the real world isn't cheap. You may be moving to a new apartment or buying a car and if you're like most of us just getting out of school, money is likely to be tight for a few years. One of the most important things you can do over the next few months is to learn how to make smart decisions about managing your money, including the importance of setting up a budget for yourself and sticking to it. Okay, managing your money, a very big subject. To get you started on the right track, the Direct Loan Servicing Center has prepared this booklet. It's called Learn the Basics and Manage Your Debt. It covers everything from how to choose the right repayment plan for you, to how to giving you a simple framework to help you set up a monthly budget. The biggest thing to remember, and this is really important, is that your loan payment is a regular monthly cost, just like your rent or car payment, and as an adult it's your responsibility to pay it. But what if you just can't get the money together to pay your loan? Sometimes not so great things happen in life and the Direct Loan Servicing Center knows that. So if you're having trouble making your payments, the most important thing to do is to reach out for help by contacting the Direct Loan Servicing Center. How can they help? Well, you might qualify for a deferment that would temporarily allow you to stop making payments on your loan. Or if you don't qualify for a deferment, a forbearance can also allow you to temporarily stop making payments, make smaller payments, or even extend the time you have to make your payments. But here's the important thing to remember. If you don't contact the Direct Loan Servicing Center, they can't help you. And the reality is that if you don't make payments or get a deferment or forbearance, your loan will go into default. Now what does that mean? Well, it means that the full amount of your loan would immediately become due. You may also be sued for the full amount of your loan, plus any attorney costs or collection fees. You might have part of your wages or federal tax refunds taken. And the Direct Loan Servicing Center will report your loan default to all of the major credit bureaus. It isn't pretty, is it? Especially considering that in today's world, keeping good credit is more important than ever before, and the serious dent on your credit rating could mean that you can't rent an apartment or buy a car, or even get that job you've always dreamed about. So please, and this may well be the most important thing you hear today. If you're having trouble, reach out to the Direct Loan Servicing Center for help, and don't allow your loan to go into default. Because if you do, and there's just no way around it, you'll find yourself living with the consequences for years. There's another way the Direct Loan Servicing Center may be able to make it easier to afford your monthly payments. If you have more than one loan, as many students do, it could make sense to consolidate your multiple loans with just one loan. So, is consolidating your loan something you should consider? When you combine your multiple loans into one, you'll make just one payment. And because you're also extending the terms of your loan into a longer period, your payment will usually be lower than the multiple payments you might have to make if you didn't consolidate your loans. What's the upside? A lower monthly payment. Something the Direct Loan Servicing Center knows could be very important when you're just starting out. On the downside, you'll ultimately pay more interest on your loans because they will be extended for a longer period of time. Let's take a look at a specific example. If you had two student loans totaling $40,000, your combined monthly payments on a standard repayment plan might be $466. And, over the 10-year life of your loans, you would be paying a total of $15,929 in interest. But, if you consolidate your loans into one and extend the terms to 25 years, your monthly payment would be just $280. That's $186 less each month, but your total interest would be $44,000, or approximately $28,000 more. You'll need to weigh the advantages and the disadvantages carefully and make your own decisions. Lower payments now versus greater interest in the long term. But, it's easy to get the help you need from other places. Just call the Direct Loan Consolidation Center or visit the website to find out more. Another great source for information is, of course, your financial aid counselor. Along with helping you make the choices that are right for you, your financial aid counselor can also tell you more about some of the ways the Direct Loan Servicing Center can save you money and maybe even forgive a substantial portion of your student loans. You're probably ready for some good news about now, right? Here's something easy. You've probably already qualified for and received an upfront rebate on your interest. To keep it, all you need to do is make your first 12 payments on time. If you're late or miss a payment, the rebate you've already received will be added into your loan balance. So be sure to pay right on time for the very first year. You can also reduce your interest by a quarter of 1% with the program we mentioned earlier. You'll be able to use the Direct Loan Servicing Center to electronically debit your monthly payments from your bank account. You won't have to write checks or worry about mailing your payment and just as important, you'll be sure your payment will reach the Direct Loan Servicing Center right on time. Just check your monthly statement for information on how you can sign up for this plan. Here's some more good news.  And if you teach in an elementary or secondary school, serving a low-income community, the Teacher Loan Forgiveness Program could reduce your total loan balance by up to $17,500. To keep the good news coming, there may also be additional forgiveness programs available if you choose another type of public service work. Be sure to ask your financial aid counselor or visit the Direct Loan Servicing Center website for all the latest and greatest information on any programs that may be available. Going out into the real world means accepting real-world responsibilities and paying back your student loans is one of the most important of them all. We've given you a lot of information in this video, but don't worry about remembering everything. Be sure to ask your financial aid counselor for a copy of this brochure, the Exit Counseling Guide for Direct Loan Borrowers. You'll find all the facts from A to Z inside, as well as information about how to contact the Direct Loan Servicing Center by web, phone, or mail. And while you're talking to your financial aid counselor, be sure to ask for a copy of your borrowers' rights and responsibilities if you don't already have one. It spells out everything you need to know about your student loans. And remember, keep it and any other paperwork related to your loans in a safe place. One thing more. If you move, please be certain to let the Direct Loan Servicing Center know on their website. All the information we've shared with you today is important, but if you take just one thing away from this video, we hope it's this. If you're having trouble at any time making your monthly payments, please get in touch with the Direct Loan Servicing Center. They can't help you if you don't ask and defaulting on your loan isn't good for you or for anybody.  Go get them.


That last time we approved this theorem that f signature Determines whether the ring is stronger of regular or not. So f signature is positive if and only if are a strong refragment so last time we showed the direction from positivity to a regularity is easy by contradiction if it's not a regular we easily showed just by taking negative of the definition that f signature is going to be zero So the more difficult definition is to show positivity starting from stronger friglarity So let's start working on this and the first thing that I want to say is that we can assume we may assume that r is a complete domain. So with some work for which we don't have time you can show that stronger friglarity passes to completion and then we showed yesterday that anything stronger friglar has to be well and local has to be domain. So this is how we get here Now once we know that it's a complete domain we can use a theorem of coin and gather so coin and gather So coins theorem tells us that any complete ring containing the field will contain its Coefficient field right so a field isomorphic to the residue field. So there exists a Which is a power series subring Inside r such that this is a finite Extension and what the addition of Gabber says is that we can choose this a so that this containment is generically separable Generically separable means that if we pass if you localize as a fraction field of a We are going to get here a field here. We are going to get a field and those fields are separable extension The consequence of this from separability of fields is that well purely in separable can extension and separable extensions that is joined so if we Lift it up. We are going to get that there exists a non-zero element C inside a so that C is going to map the ring of per roots inside the ring Generated by p roots of eight right so this is algebra right subring of this one generated by p roots only in it and We can also show that this is isomorphic to the tensor product of our and So this comes From this generic assumption right same statement holds for on the field level and then we just using finally generated Collecting in interest and we get see Yeah, thank you. Yeah, this is for all e Okay, so what can we do with this so we have a magic Constancy element C inside a provided by Cohen-Guyberg's theorem now we assume that are is stronger if regular So let's try to use the definition for this element C. So we will find some integer is zero for which This condition holds so there is this the map also. There is an integer. There is a map phi inside home Such that Phi of f well, let me write a period one E zero is going to be equal to one and from this data We will construct Splitings to show that the signature is positive right so we are going to construct a big sufficiently big number of Three some months so that we will get actually positive So our starting point is that a itself is a regular ring All right power series ring and we know then then the ring of periods is going to be free We know that a one over P e is three Over a and they ever know what is the basis so the basis let's say be It will be consistent of elements lambda and then x1 a over P e XD over P e So this lambda is the basis of the field extension P over K and ai's So this is a D. This is a one Adjust somewhere between zero and P e so we know an explicit basis right and Using this explicit basis in a we will start constructing free some months in our using This formula so first of all we can do the tensor product to start getting something free over our right We know that by tensoring this our tensor a one over P e is going to be Free over our and the basis is just we get it by tensoring right the elements of this basis, right? in particular Here because this is the basis will have projection maps right we have maps Let's call them psi beta so that if I apply So beta and B and let's call B and B prime I inside the basis So if I apply this projection on the basis element B to any other basis element B prime What am I going to get I'm going to get either one or Zero one if B is equal to B prime and zero otherwise right because of the projection map How do I get free some months? I just project onto basis, right? So I'll get the same maps induced here, right? It is our free and I can get you explicit presentation as a free model just by tensoring those Homomorphism that they have here the projection once So this was first stage now. We need to use that our element C is Going to be Magical so what are we going to get we are going to get that if we have multiplication by C right on our So it's not going to go in our one over P. It's going to go there So our tensor a one over P e and this is going to be this appropriate number of Appropriate rank free mode, right? so what is What is left is to twist this map, right? So what did I say? What did I say? okay, so What do I get here? So I get that this one was our free, right? And I want to multiply our into this is a problem is that I'm not Surjective here, right? So this is what I'm doing right? okay, so So what do we do so we want to combine this with multiplication? So if I make now maps I need to let our pie Pi B right combined with multiplication of C with this So multiplication of C with this maps one tensor Psi B Right, I'm going to get maps so that as I send my basis element by be applied to be prime right because This element B prime is still element inside Inside it is element inside the periods of a but periods of a contained in our right So I still can apply it here and you see that by this definition the result of my I mean the result is the same as what I get here multiplied by C, right? So this is going to be either C or 0 depending whether B is equal to be prime So the benefit is now this map is defined on our one over P Okay So What is left is that I don't like C over here, right? Because this is not going to split right? This is a map which sends some element inside our one over P to our right? But I want to get Surjective map so that I can split and I guess the subjective map using the map fine So what I can do is that they can take our one over P plus E naught, right? And then I will map by what they have here direct sum of pi B one over P zero and then I'm going to Then I'm going to apply phi and the end result is now a subjective map on the appropriate number of On the free model of appropriate rank So this last step applying phi Maps as this element C right here one over P E zero right because I took perutes here Right, it will send it to one. So all those maps are subjective and they were disjoint by the construction So long story short. What did we get? We got that this number of so f signature of r is going to be Greater or equal as in the limit as e goes to infinity and Here we get this number of splitting so number of elements in B e and This is rank of f lower star e plus e zero over r and then well, we know actually that this As this basis had how many elements it was a generic rank of F lower star e Right of just as a perutes of a one over P. So the same as rank of r one over P here So if you do the consolation we see that this limit is positive So it's going to be well equals to one over rank of r one over P zero So this is positive Does it make sense? All right. So essentially from the coin gobbler theorem We can very explicitly construct the number of Splitting's which will give me positive f signature. All right so now We close this chapter and we start new chapter. So we are going to Start actually working to prove this theorem that I promised at the beginning that both f signature and Hilbert Kuhn's multiplicity at the value one they detects they detect Whether the ring is regular or singular as a starting point. I want to Prove that they satisfy localization property So what do I mean by localization property is that if I localize my invariance behave in Appropriate way because we know that localization of a regular ring is going to be regular. So I if I If I need to if I want to get the theorem that shows that my invariant detects singularity It should behave in the same way, right? So Let's start with a finite case. So suppose that r is in a finite local domain So then I can show and P is a prime So what I can show is that f signature Localized of the localization is going to be greater equal that a signature of the original ring and Opposite inequality works for Hilbert Kuhn's Multiplist if I localize it's going to get less or equal again If we claim that this value being one should say if it be a regular then all other values at the other point should be also one Okay, so this is a very easy theorem because we have this geometric definitions So what do I have? I know that to compute Hilbert Kuhn's multiplicity I will need to look at the minimal number of generators of a flower star e r, right? So this is a sum sum finitely generative module and what do I know if I take minimal generators in r and I localize at the prime they still will generate the localization, right? So the minimal number of generators here is going to be greater or equal than the minimal number generators of localization Just because The original generators still work But then I look at the number of minimal number of generators So it may decrease and the opposite inequality works for the splitting number if so what is the splitting number? Right f lower star e r is equal to some free some month right plus Some model m what happens if I localize it may happen that localization of m is going to get another splitting, right? so mp May split and I will increase or it may not split and then the number of free some months is going to stay the same So it's easy just from the geometric definition the only downside is that While this level of generality a finite local domain Works perfectly for a signature if I know that the signature is equal to one It's going to be positive and in particular r is going to be a domain This level of generality on the other hand is not very satisfactory If you want if you want to prove a statement about Hilbert-Kunz multiplicity because we don't need to be a finite We need to we don't need to be a local domain to talk about a signature to talk about Hilbert-Kunz multiplicity So what I want to do now is to remove these assumptions for Hilbert-Kunz multiplicity And I want to also do it in a way so that I can showcase this uniform convergence methods that I was Selling to you the other day. So what we are going to do is to prove theorem 2. So I take local ring Yeah, it doesn't need to be a finite. It doesn't need to be a domain and I'm going to take a prime ideal such that a Dimension of the localization plus dimension of quotient are going to adapt to the dimension of r So I claim that the same inequality To prove this theorem I will need To provide you a lemma that will allow me to go from r to r localize that p so first Yeah, so So actually maybe let's take it. So let me Under the same assumptions so same assumptions I want to prove to you that if I take Hilbert-Kunz multiplicity. So let me Now let me say that not the same assumptions. Let me take prime p Let me take co-dimension. So Let me start the proof. So let me I Can prove the statement by induction on the co-dimension, right? So I can do it by further localization. So by localizing further I can take p and some prime q containing m, right? And I can localize if I know that the inequality works for m and q I can localize that q improves the inequality for p and q so by induction I can assume that dimension of Armot p is going to be equal to one I just take maximal chain from p to m and they do it once at a time using localization so now in this situation so dimension of armot p is equal to one I Want to prove to you that the limit as n approaches infinity of Hilbert-Kunz multiplicities of p plus Xn where in this condition I take x such that x plus p is P prime is m primary so I take a parameter modulo p I lift it to r so now I get Element x such as ideals that generate together with p is going to be m prime So in particular the Hilbert-Kunz multiplicity of ideal p plus xn is defined I can compute it and then if I Divide by n I can compute the limit of this Hilbert-Kunz multiplicities as Well in this case it's just going to be equal to the Hilbert-Kunz multiplicity of X r mod p times Hilbert-Kunz multiplicity of RP so let's try to prove this theorem 2 and then see how it applies to theorem 1 alright, so How I want to prove Theorem 2 I want to so this is proof of theorem 2. I want to consider the sequence Let's call it a yeah, let's call it a and e which is given as the length of R modulo p bracket p e plus x n N P e r So this is a by sequence dependent on two parameters parameter n and parameter e I know that the limit that I wrote over there is a double limit as I first take limit of e and I take Limit of n right so if I compute limit as e approaches infinity of a and e I'm going to get Hilbert-Kunz multiplicity of p plus x n r divided by n right I Also know that if I take limit as n approaches infinity of this sequence Then what I'm going to get I'm going to get here nothing else as Hilbert-Sammel multiplicity, so it's a limit as so I get multiplicity of So X P e on The ring R mod P by properties of Hilbert-Sammel multiplicity. I can actually erase this P e I Mean it just goes outside right it's P times multiplicity of element X So I cancel one copy and the numerator and one copy and denominator so I will get this right and Then I have associativity formula and I get that this limit is equal also to what to the sum as P minimal primes of maximum dimension of r mod i because Not I and here I will get multiplicity of X r Mod P and here I'm going to get I Lend of our localized at P Modulo Frobenius bracket power Dimension of r minus 1 does it make sense? So the usual associativity formula tells me that the multiplicity of a model is going to be the sum over Minimal primes of r mod P and here is this co length of a model, right? So this is a co length of localization of this ring at the prime P, right? The only minimal prime so if you take the limit of this as e approaches infinity Look as a denominator is appropriate dimension for our localized at P, right? So the limit of this as P goes to as e goes to infinity is going to be exactly what I told you Over there multiplicity of e X r mod P Hilbert-Kuhn's multiplicity of RQ right So you see that the theorem 2 is equivalent to the statement that two different iterated limits I That I can interchange the order of limits because C over here What did I do? I computed the limit as an approaches infinity then I can compute the limit as e approaches infinity And I found that this limit is exactly what I claimed right e X r mod P Hilbert-Kuhn's multiplicity of our localized at P, right? And the statement Sorry, ah other way around This is this thing So this is the ones that we have there is a limit of Hilbert-Kuhn's multiplicities and there is this is the limit that I claim It's equal to so what do we have we? Reduce the statement to proven that we can exchange the order of two limits When can we exchange the order of two limits when the convergence is uniform? This is the standard theorem of calculus is that we can change the order of two limits if Single limits exist and converges that one of them is uniform So it remains to me to show that the convergence that we indeed have uniform convergence And I will want to get it from there right remains To prove that we have uniform convergence, but I told you Where is the convergence estimates come from they just come from a lemma that bounds Lens so I want to Prove that if I take a model and This is such type of prime and they take bracket powers here and I add xn P em Then this is less than some constant C and PD Yeah dimension of M so if I have such Convergence estimate which refines the original bound right on the con on the Estimate if I didn't speak if I didn't specialise this parameter n Then you can go through the convergence machinery through the proof of existence of Hilbert-Kunz multiplicity and This lemma will then imply that I will get Uniform convergence, I will get that P plus X minus this length right xn are going to be bounded uniformly by C and PD minus 1 Because the constant I told you that this constant so convergence over there is just a multiple of a constant Or that we get over there for torsion models, right? So I will get this dependence on parameter n here But if I divide both sides by n now I get uniform convergence, which does not depend on that So it's all comes from this limb if I have this lemma then just the Convergence estimates that we set up when we showed the existence of Hilbert-Kunz multiplicity will show me Uniform convergence that convergence of this limit as E approaches infinity does not depend on n But this lemma is very easy to show because we have Just an exact sequence which allows me to prove such a statement by induction. I can have So what they have I have M. So let's not call it M. Let's call it. Let me just say that say n Is this quotient model? M model of PE P to the power P M M So I have exact sequence and Modulo xn plus 1 and so n Modulo xn and The usual filtration exact sequence. I have this map just because this is a smaller ideal and What do I have in the kernel? You can show that in the kernel? I will have n and here xn Plus annihilator of xn inside them, right? but this is a Homomorphic image of n mod xn, right? So from this exact sequence if you compute the length you can show by induction right that the length of n Modulo xn n is going to be less or equal than n times Length of the original quotient by x. So just induction here If I know the statement for n, I'll get the statement for n plus 1 because here I add one cooperate Okay, so this finishes the proof of theorem 2. We start with this lemma From this lemma we feed the output of this lemma into the convergence estimates We did before we get from here uniform convergence as E approaches infinity independent on n and then the as a general calculus fact all of us to exchange the order of limits and From here we get that We compute double limit one way we compute double limit the other way And this is exactly the statement of theorem 2 and that was theorem any questions so overall this is the strength of This uniform convergence approach right that we need to do very little to get various Uniform convergence estimates like the ones that we did here. So let me raise this and now we can Actually, did use theorem 1 from theorem 2. So I will need One more little ingredient So I need to prove the lemma So if you take a local ring and I take I am primary Then I can show that Hilbert-Cons multiplicity of I is going to be less or equal than Hilbert-Cons multiplicity of R times length of the quotient of our montaig. So this is Often called filtration boundary So how is this proved? So I look at what is the What am I going to do? I'm going to take a filtration Giving me the length of our modi. So it's I modulo I1 and so on Sorry, this is strict inequality and then I'll get I L and This is equal to R right because what is the equivalent? What is the length of the quotient? I can think about as the length of the largest Saturated chain of ideals, right? So L is the length of our modi Because this is a saturated chain what I can tell about the quotient of I let's say K plus one Modulo IK Well, it needs to be isomorphic to our modem, right? Otherwise, I could have saturated it more and then what I can do is that I can just put bracket powers to this chain so bracket power Q bracket power Q and so on bracket power Q and Well, I could write IK plus one because this is isomorphic to IK R mod M. There is only one generator Of IK plus one modulo IK so I can write it like this So then what happens if I take bracket powers? I get that Here bracket Q here you to the power Q modulo IK Q I Cannot say that the annihilator of your Q is equal to M bracket Q But I can say that it contains right so I will get You can easily see that M to the bracket power Q still annihilates you to the power Q So I'll get a surjection onto this from our Modulo M bracket Q Just by sending one to the element UQ and if you just Take Q powers of the equation Like this right you are going to get this surjection just because I annihilator it may be bigger We don't know it, but I just need an inequality so from this we will get that as a Length of R mod I bracket Q is going to be lesser equal than L Right times a length of our modulo M bracket Q And then we take we divide everything by Q to the power D We take the limit and that's precisely the assertions that I have here So we just apply Frobenius to the filtration realizing the coolant of I Okay, very good. So now I can finish fair and warm So I reduce the statement to the case where dimension of R mod P is equal to 1 and I still need to show this inequality so What do I know? I know that the by theorem one limit as an Approaches infinity of Hilbert-Kunz multiplicities of P plus Xn R Over n is going to be equal to Hilbert-Kunz multiplicity of EXRP times Hilbert-Kunz multiplicity of actually R localized at P So this is theorem one the theorem two On the other hand, I also have the lemma which allows me to We have the lemma which allows me to estimate this Hilbert-Kunz multiplicities. So by the lemma Hilbert-Kunz multiplicity of P plus Xn is going to be less or equal than as a length of R mod P plus Xn times Hilbert-Kunz multiplicity of R Right and then I remember that I had the limit so I had the limit as an approaches Infinity divided by n. What do I see here? I have a length of Some ring Quotient out by Xn divided by n. This is the definition of Hilbert's sum of multiplicity So this is equal to multiplicity of R mod P Hilbert-Kunz multiplicity of R. I Look here. I Look there and what do I see I see the inequality of Multiplicity times Hilbert-Kunz multiplicity of R and multiplicity times Hilbert-Kunz multiplicity of R localized at P So it follows that if I cancel those two Multiplicities it follows that I get the inequality between localization and the regional ring that I wanted Okay, are there any questions? so we use The uniform convergence machinery to get this formula Right This is a key part of my proof because this formula allows me to get from m primary ideals to something in the localization, right? So this is a power of theorem 2 Okay Good So now I know this localization Properties I know that Hilbert-Kunz multiplicity if I localize gives the right inequality and I know that the signature if I localize gives the right inequality From this I will start Building as a theorem that I want to show I will start building that those and I will start proving that my invariance detects singularity by Essentially using inductive approach So the general scheme is as follows So if I assume that Hilbert-Kunz well if my let's call it invariant F, right? F of R is equal to 1, right? Then by the localization it follows that F at any point other prime ideal is also equal to 1, right? Because the values 1 are extremal values both for Hilbert-Kunz multiplicity and for a signature and the inequality is going the right way And then I use induction over here on dimension to say that R localize that P is Regular and from here. I want to use that R is regular How can I do this? Well, I will do this by understanding the aqua multiplicity condition so deduce that R is Regular from two statements that RP is regular and that My invariant doesn't change So this is equimultiplist So this is the general scheme of the two proofs that I'm going to give to it So we will start with doing this proof for a signature and then we will do it for Hilbert-Kunz multiplicity But the overall scheme is going to be absolutely the same. It just is Equimultiplicity condition right is different depending on what invariant you take but the first part is absolutely the same So I'll start with a signature and I will start By proving a lemma about splitting so suppose that a signature of R is going to be greater than 0 and suppose then I can split let's say E plus whatever it is Let's say n0 So let me fix just one of the possible direct sum of the compositions with some module m that I'm interested in and Some module n which I'm not interested in and then I claim that there exist decompositions e For all e so here is there exist so for some e0 and the statement is I can get a decomposition for all e so that I actually Split a number of copies of m which is going to grow Very fast as e approaches infinity such that the limit as e approaches infinity of Maybe e over the rank is going to be positive. So let me denote a e to be this maximum of well the maximum rank right maximum Rank Of a free sum and then I can decompose a flower star e plus e not by Taking this decomposition and applying a flower star e not to it. So it's going to be direct sum right of a e copies of f lower star e not r plus well, let's call it me so plus A flower star e not just because I take this decomposition and apply Frobenius push forward to it And I get precisely what they have here and then I can decompose it further Given what it was my assumption So I can say that this is direct sum of a equal piece of m and then I have some other stuff whatever it is right and then I claim that the assumption as the Assertion holds because the limit as e approaches infinity of a e over the rank of f lower star e plus e not of r is going to be It's going to be equal to f signature over the rank of the original F lower star e not r. So it's positive does make sense. So we just iterate think right big if I can Split a copy of m from from r right f lower star e not r And I get a lot of copies of our then I just use this copies to split more and more So it's a very easy limit But from this very easy lemma. I can actually get the equal multiplicity statements that I want so the theorem that I like to call rigidity theorem tells me that Under my assumption. So our m local F signatures positive so strongly of regular domain to have equity multiplicity condition for a signature is equivalent for the entire Sequences to be constant. So a e is this right the maximal number of direct summons that you can get So if signature is the limit of the sequence, right? but what I claim is that It's only possible for the limits to be equal if and only if the entire sequences are constant So this is why I call it's rigidity. There cannot be any variation initially, right? It is very very strong property, right? Something very not intuitive So recall that the signature of r is going to be a e r. So limit a e r over the rank of f lower star e r But f signature of our localized at P is going to be again limit as e e localized at of RP but the rank is the same right because what is the rank is just the rank at The is this is the dimension of the vector space, right? That's a generic point But localization I mean R is a domain the generic point is the same So these two sequences only limit only differ in the numerator, right? So this statement exactly tells me that the limits are equal if and only if the entire sequences are constant And the proof is two lines. So what does it mean that? So one direction is clear, right? If my sequences are constant Well, do not differ than the limits are going to be equal. So this is trio the other direction By contradiction so I can write So suppose that a e r is going to be strictly less than a e of RP Right because this is the only way that can happen, right? I know that the number of free summands can only increase. So what does it mean that it increases? I can write the flow over star e r to be Well, let's write that are Plus the lift over pit So what does it mean that the number of free summands increase it? It means that if I localize m e at p is going to give me one more split right? so m e localized at p has a Free summand So these two guys I mean all those guys are localized they're free right and how do I get that the number of free summands increases? It should happen from the localizing of me, right? But then I can just use the lemma. So let me say that this is zero So that I'm in the assumptions of the lemma Right at some is zero. I get a strict inequality and then the lemma tells me That I can build further direct some decomposition right of f lower star e r into something which with m So so m is zero So let's say right and here we'll get Be right at and here we are going to get some number of free summands and so on, right? So I get this direct sum decomposition which is given by the lemma I know that if I increase e I can split more and more copies of this original model m is here, right? This is what my lemma tells me And from this it follows that a e of r localized at p is going to be what is going to be greater equal Than this be copies of m because they're going to give me Each each will you're going to give me at least one free summand plus the original a e of r, right? So this free summands that they had over here. They're not going to disappear anywhere, right? I just get extra ones from getting copies of this model And what do I know of from the assertion that the limit of this guy is not going to be zero So I must have strict inequality between f signatures, too So this is very very easy just coming from the lemma all right, so one second to finish The statement about the signature if s of r is equal to one then r is regular So what am I going to do? I'm going to say well if r is regular then of course the signature is equal to one We did it. It's just going to be every model, right? So the sequence is constant one but on the other hand To prove the interesting direction if f signature of r is equal to one Then I note that f signature at the generic point is also equal to one, right? Because I know that if f signature is equal to one then r is stronger flagral So it's a domain. So there is generic point and f signature is a generic point Well, since it's a domain it's going to be also equal to one And then I use my theorem to deduce from here that a e of r is just equal to the rank at the generic point, right? But this is a generic rank So I get the entire sequence is equal to one and therefore I can use the coons theorem that tells me that Regularity is detected by a single ae, right? It tells me that my model has to be free already, right? Okay, thank you. I think I should stop