 Well, welcome everyone to day five of the USS. This is Paul Melbourne. I'm one of the steering committee members and you're in a Swanson as you know is the other one. I just want to make a brief announcement. You then and I have prepared a survey to discuss your experience your engagement your concerns and so forth from the first week of this summer school. So that is not yet online but it will go out to you it's a survey monkey. At the end of the day, we would very much like your responses by Saturday evening if possible if not then certainly by Sunday noon, and we will try to act on those. So that's all I have to say, and welcome and kill, take it away. Thank you. So, yeah, let me set up the screen chair. So, yeah, so I guess, right so so last time we introduced introduces construction of the, the pediatric numbers. So last time. So I introduced the field QP for prime number P. And so let's just recall that this was. So this was a complete. So, this is a complete non Archimedean field so sorry so I mean that QP is equipped with some sort of it's a it's equipped with the pediatric absolute value so an extension of the pediatric absolute value on the rational numbers. So this is a complete non Archimedean field. So with an absolute value. And right so it was, it was defined as the completion of the rational numbers with respect to this absolute value sort of analogous sort of defined as somehow analogous to the real numbers. The way the real numbers are defined as a completion with respect to the usual Archimedean absolute value. And right so we also saw last time. So you can sort of represent an element of QP you can you can represent it by a by a pediatric expansion. So expand elements of QP via pediatric expansion. You can write an element of the form, the sum of a a sub I times P to the I, as I ranges over all integers but it sort of starts at some minus and for some, and large enough, and goes to infinity. And the AIs are between zero and P minus one. Yeah so QP is is constructed as a means constructed as a completion of of the rational numbers. And it's supposed to be kind of like kind of analogous to the construction of the real numbers as completion of the rational numbers. And in general whenever you do sort of when whenever you have a completion process it usually becomes much easier to solve solve equations in a completion, because how there's more more room to maneuver. So for example it's much easier to solve polynomial equation or to show that there is a solution of a polynomial equation over the real numbers, then over the rational numbers. So over the real numbers, if you're given a polynomial. So given a polynomial f of x, in R of x, then you can detect roots well let's say let's say your polynomial just has simple roots. And you can detect at least simple roots by looking for changes of science. So you can detect existence of roots by looking at sign changes. And when we're over the periodic numbers there's no longer going to be an ordering so I say in fact there's there's no way of ordering QP because minus one is going to be a sum of squares. Similarly, but well somehow analogously. There are ways that you can you can look for. You can look for solutions of polynomial equations over the over the periodic numbers, and these come from from from looking at Congress's. So that's sort of a general tool for, for, for producing solutions of polynomial equations over the periodic numbers. And so this tool is called Hensel's lemma. I'm sorry. So there's some drilling that just started in the background and so if it becomes noticeable please let me know and I will try to switch devices in such a way that may or may not help with that. Okay. Right, so Hensel's lemma is the following statement. So let f of x, the polynomial in ZP brackets x, and Hensel's lemma is going to produce is going to give a criterion for, for the existence of a solution of f of x. So let alpha be an element of ZP. With the following properties. So, so first of all, it's sort of an approximate root of the function f so f of alpha as pediatric absolute value less than one. But the derivative is not too small. So the value of the function is absolute value less than one, but the value of the derivative. So f prime of alpha has absolute value equal to one. So the definition is there exists a unique element beta in ZP such that alpha minus beta is less than one. And f of beta is equal to zero. It's saying that if you have you have sort of an approximate root of of the function, function f, and its derivative is not too small, then you can refine that uniquely if you're within a disk of a one open disk radius one to an actual solution of your equation. Okay. So, another formulation of this. So if you take f of x is a polynomial over the patic numbers, but you can reduce f of x to get a polynomial f bar of x, which lives over, which lives over f p so so so you can reduce reduce my p and you get a polynomial with coefficients and f p. So this polynomial has a simple route, which I'll call alpha bar. So alpha bar is the reduction of alpha on p so so saying that f of alpha is less than one is saying that f bar of x has a has a root of alpha bar, and the second condition is saying that the root is simple. Then this route lifts uniquely to a root. So let's call it beta in zp of the actual polynomial that we started with f. So if you have a solution if you have if you have a solution of your equation mod p. And it's a simple, simple route so the derivative doesn't vanish at that at that point, then you can, you can upgrade that to a solution in in the patic integers, and it's unique within its congruence class. So this is Hensselstam. In fact, there are many different formulations of Hensselstam and this is one, you know, one formulation or one, one case of it. Okay, so I want to give sort of a proof sketch of Hensselstam and the proof sketch is basically use Newton's method. So use Newton's method to refine alpha to the root beta to alpha sort of an approximate solution. And then you're going to sort of iterate the this process. And then it's going to converge in the limit to to the solution beta of the equation. So explicitly what what this means is that you define a sequence. So define a sequence of patic and address alpha zero alpha one to dot dot dot. Alpha zero equals alpha. And right so inductively, you're going to define alpha and plus one so so you're going to use Newton's method so that means that I mean you have your sort of approximate solution alpha and, and then you want to sort of wiggle it a little bit to get a better solution. And the amount of the amount you're supposed to change the solution or the approximate solution is is given by linearly approximating your function at that point. So the first formula is alpha n plus one equals alpha n minus f of alpha and over f prime of alpha. And so notice here that we're doing a division. And so we want it. So in order for this to work well we want to be dividing by something which is not just small, which is exactly this condition that you know we're starting with an element alpha such that f prime of alpha has a value one it's a patic unit so we can sort of divide by it without causing too many problems. And so then, what's going to happen is that. So this is exactly right so this is your new method. And then what you find is that the patic absolute values of FN are going to converge to zero as n goes to infinity. And in fact alpha n is a is a Cauchy sequence, which is going to converge to the limit beta. And that's going to be the root of your polynomial. So, so essentially the, you know, so the way the way that you prove this is that right so I mean, the way that you prove this is essentially you prove that f of alpha and n plus one is is going to be less than the absolute value of f of alpha and I think even you can put a square here, maybe less than or equal to. This way. So it's going to converge very rapidly and the reason is that you're going to use the Taylor approximation of F near. So, right so specifically you're going to use that f of alpha n plus one is equal to f of alpha n plus f prime of alpha n times alpha n plus one minus alpha n which is minus f of alpha and divided by f prime of alpha and plus big O of the difference squared. So big O of f of alpha and divided by F prime of alpha and squared. So, I think that means we get to we get to say this in our inequality. And this is this is this is I mean this is the Taylor expansion so so there's no issue with the expansion because you're just working polynomials. And I mean you have to check a little bit that the Taylor expansion is okay that that you have this and the denominators and the Taylor expansion sort of cause problems. But essentially what you're going to use is that if you have a polynomial minus minus the linear approximation at a point you're going to, you know it's going to be divisible by like x minus alpha, alpha n square point. Right, so then you observe that these two terms, the first two terms are just going to cancel. So this is big O of F of alpha and divided by F prime of alpha and squared. And so then that's going to that's going to show that since you started with something where the value of F was less than one, it's actually a converge very rapidly the F of alpha and are going to converge very rapidly to zero. Okay, so this is really I mean this is really again this is really the construction of Newton's method. Probably seen over the real numbers. But in fact that can, you know all these questions involving the convergence and so forth are, are a lot easier in the setting because one is that you're working with polynomials but I mean, more silently is that you're working in a non or comedian setting. So instead of the triangle inequality for for bounding absolute values you have an honor comedian property. Okay, so, yeah, so maybe that's a leave it at that I think. So on the problem set you'll explore this I mean there are many there are actually many ways you can formulate the proof of Newton's method and you explore this a little bit more on the, on the homework. But I mean just just sort of to give an idea of how this is going. Essentially, what what you want to do is that this is a this is approved by successive approximation. And that's sort of made very clear in this in this procedure via this method, but you could also sort of formulated in a, maybe slightly more. Well, in a slightly more direct but maybe less efficient fashion, which is that what you're trying to do is you want, you want to beta, such that f of beta is equal to zero. And so beta is given by some some patic expansion so so beta has a patic expansion. And, well, that you're trying to find beta such that the first patic digit or the zeros patic, I guess it's the same as the spirit of patic digital alpha. And after that, well after that you have to find the rest of the patic digits. And, right, so the observation is that the first and patic digits of f of beta depend on the first and patic digits of beta. And then you sort of inductively solve for each of the patic digits. And then you sort of inductively solve for the patic digits of beta. So again this is not unlike how you might, you know try to solve such an equation over over over the real numbers you sort of, you know you first work out that first end digits of the decimal expansion and then you work out the next one and so forth and you sort of do the same thing in this in this patic context, except that the expansion is going in the opposite direction. So it really is very analogous to how you, you would try to solve some sort of equation over the real numbers. Okay, but so let me give some examples of this. So first let me give an example which, which came up on the exercises yesterday. So let me assume that actually let me assume that p is greater than two. And let's consider the polynomial. It's one minus one. Right so so this is this is a polynomial and in zp brackets x. And right so when you reduce mod p, it's, it takes zero at all the classes of fp. So if you have any element of fp, it's, it's p minus first power, or sorry, if you have any non zero element of fp it's p minus first power is equal to one. It's possible that you to the p minus one is equal to one. If you use any element of fp cross, exactly crosses a cyclic group of order p minus one. So, any element x with which belongs to zp minus p times the piece of any element which is a patic unit has satisfies that the absolute value of x to the p minus one minus one is less than one. However, the derivative is not zero. So but the derivative, but the derivative is p minus one times x to the p minus two. So this is a patic unit. And so then what Hansel's online saying that you have this sort of approximate root of the polynomial extra p minus one minus one is in fact any element any patic unit is a sort of an approximate root of the polynomial extra p minus one minus one. Oh, sorry, so this means a set theoretic difference. So this means all x and zp such that the patic absolute value of x is equal to equal to one. Okay. Yeah. So a conclusion is that in each congruence class of of zp, except the zero congruence class in each congruence class mod p, except zero. So there is a p minus first root of unity. So, I mean, I guess a way of saying this in terms of in terms of groups is that you can take the patic integers and take the group of units. And this maps to the units and FP cross, which is the mod p minus one. And in fact, this produces a well canonical section of that map. So that's saying that you can, you know, that that's saying that for, for this element, you know, for each, each congruence class might be except zero, you're producing a p minus first root of unity. And in the pre-medge. And in fact that's unique by the uniqueness part of Hansel's online. Right, so I in fact I didn't, I didn't actually prove the uniqueness part of Hansel's online I think the existence proof is more less going to imply that but in fact you're going to do that on to see that on the exercises so. Okay, so conclusion is that QP contains the peanut P minus one P minus one, a primitive P minus first root of unity. And you see that by looking at this polynomial X to the P minus one minus one that has that has a solution that has solutions want to be in fact every every element not zero of FP is a solution and it's a simple route. So you can lift that to roots in ZP. So I think on the problem set. Yesterday there's there's sort of another way of producing it explicitly. So, on the problem set, there's an explicit way of producing this which is that you start with any element x which again is in ZP, ZP cross. And then consider the sequence x, x to the P, x to the P squared dot dot dot. And this converges periodically to a P minus first root of unity, which is congruent mod P. So in this case you can really mean you can really produce this. You can produce this via an explicit limiting process by sort of raising to lots of P powers but Hansel's Lambo works sort of very generally. So what's another example of this. So let P be a prime number such that he is congruent to one module of four. So, then the square root of minus one. That belongs to QP. So I guess yeah so technically this is a this is a special case of the previous. Previous example. So, so this is a special case of, but right so we can also do this very explicitly so we can consider the equation x squared plus one in FP. And this equation has a root. Because P is congruent to one month four, it's it's necessarily a simple route, because the derivative is 2x, and then you apply Hansel's Lambo. So, right. So in fact, sort of thinking more about this and thinking more about Hansel's Lambo, it lets you completely determine which elements of the periodic numbers are squares. So, Hansel's Lambo gives a complete determination of which elements of QP cross. So which non-zero periodic numbers are squares. And in particular determines the structure of the group QP cross modulo QP cross squared. So this is this is a group we want to sort of have access to if we're going to think about quadratic forms over QP. Okay, so let's, let's start to do this. So let me do this when P is greater than two. So, so this this determination is going to be a little bit more complicated when P is equal to two. Because then when you take the derivative of something like x squared minus a you pick up a factor of two which is not a not a two attic unit. But when P is greater than two, we can do this pretty explicitly as as follows. So if we're given an element you in ZP cross, then Hansel's Lama implies that you admit the square root. If and only if the reduction mod mod P, which lives in FP cross admits a square root. Similarly you apply and so summer to the polynomial x squared minus you. Okay. Right. So, right. In general, if we're given some element of QP cross. So if we're given some element in some non zero periodic number, we can always. We can always write it as some power of P. So let's say P to the I times a pediatric unit you cross so where I is an integer, and you isn't CP cross. And right so if this is going to be a square then necessarily I is even because necessarily if you have a square, then the pediatric valuation has to be even. So what we've seen is that sort of is a necessary and sufficient condition in order for this element to be a square which is that I should be even and you should be a unit, you should be a square mod P. So this is a square in QP cross. If and only if I is even. And let's say you bar in FP cross is a square. And so the conclusion is that we've determined the structure of the group QP cross mod QP cross squared. Namely, it's Z mod two cross Z mod two. And where do these factors come from well on the one hand you have P. So PS, P is not a square, because P at evaluation is one. And over here you can take any not quadratic non residue. Let's say some you in ZP cross, such that you bar is not a square. So you bar in FP cross is not a square. And so you've got the structure of QP cross one QP cross squared as the mod two cross Z mod two in this case. Right, so I think there might be a question. Oh, yes, please. Um, so for this reduction of a periodic and number to the field FD, how does that work. Right so sorry so you mean this construction you bar. Yes, exactly. Right so for this. Okay so if you if you consider the ring ZP, we can take the quotient so so ZP is is is going to be the the sub ring of QP, consisting of elements of Norma most one. And inside ZP we have the multiples of P so these are equivalently the elements that have patic up to the value less than one so P times ZP. And this quotient is the same as FP. So if you think about patic expansion so right so explicitly if you think about patic expansions and any patic number. It has a patic expansion and we're just remembering what happens in level zero. Sweet. And so this, like we avoid the possibility of dividing by P, since we're not working with QP as a whole but just with the subring as EP. So exactly so much. So any patic integers of the form, some of a I times p to the I where I goes from zero to infinity. And then we're mapping this to FP by sending this to the class of AI, sorry, a zero. Right. And so this is, I mean this is sort of, I guess, very fun to work out explicitly because, like if you want to find if you want to actually find a square root. You can, you know, you can try to, like, however you might try to calculate a square root, the decimal expansion of a square root you can you can do that in this case as well and sort of inductively work out the Okay, so Hensel's Lama completely determines the structure of this group QP cross mod QP cross squared. It's a little trickier for p equals two. So what you can show so it's not necessarily true that it's not so easy to see, or it takes a little bit more work to see when a two added unit is a square. So Hensel's Lama is going to run into problems now because you have squares and so when you take the derivative it's it's zero mod two. But what you can show is that I think you'll see this on the problem set that any element you and Z two cross such that you minus one is divisible by eight is necessarily a square. So you can see this either using sort of a slightly more general version of Hensel's Lama than the one that I that I stated just earlier where where you don't necessarily have a simple root module of P but instead like the value of your function is a lot smaller than the value of the derivative. So this, well, it's sort of equivalent by sort of changing variables in a clever way. So, the, right, so the conclusion, in this case, is that Q to cross mod Q to cross squared is isomorphic to Z mod to cross the mod to cross the mod to, and as generators well, you have to so because the two added numbers necessarily even. And here you can take minus one and five to be to be generators to be sort of basis elements for this F two vector space Q to cross my Q to cross square. So, right, so it's a little bit more complicated at at the prime to when you're talking about squares in this way. But you can still, you can still work it out pretty explicitly and in particular something that's kind of nice about the periodic numbers is that if you look at units modular squares you got a finite group. So that's one way in which the periodic numbers are I mean that's also true the real numbers. I mean that's one way in which the, you know, the two added numbers are a lot simpler than something like the rational numbers, because whether or not something is a square sort of a finite condition. Okay. Sorry maybe I should pause for more questions here. Sorry. So yeah, so now let's apply this to to quadratic forms, quadratic forms over the field QP. So let's try to understand what are quadratic forms over QP when can classify them, you know when are the isotropic and so forth. And so the first step in somehow doing that was to think about units modular modular squares in QP. So, right, so let's fix a quadratic form. Let's call this brackets a one through a and over QP. The AIs are living in QP cross. And let's imagine we want to try to sort of classify these. So the first thing that we can do is we can, well, if we wanted to classify something like this we can we can multiply all the AIs by square classes. And right so. So in particular we can multiply everything by even powers of P. So, by rescaling. We can write it in the form. Say, you one through you are direct some PV one through PVS, where the UIs and the VIs live in ZP cross. So by multiplying by some even powers of P everywhere we can assume that the AIs are all either patic units or like P times patic units so we can sort of break it up into into two pieces like this. Right, and so it turns out that if P is greater than two this sort of breaking that breaking it up in this way and using the classification of quadratic forms over over FP is going to give a complete classification of quadratic forms over QP. So P is greater than two, which again is is the assumption that makes everything much simpler. When you're talking about quadratic forms in this context. We can use this to completely classify over QP. So, namely, the idea is that if you have a quadratic form over QP. Well, you can break it up in this way, and then you can reduce each piece mod P. And so you can reduce the UIs mod P and you can reduce the VIs mod P to give you quadratic forms over FP. And if P is greater than two, then this essentially gives you a sort of more or less canonical decomposition of of quadratic forms over QP. And since we've been saying things in terms of, I mean, right, so since we introduced this this bit ring, I want to, I want to try to formulate things in terms of the bit ring. Or in terms of this. So, what's going to happen is that the bit ring of QP, at least as an ambient group is isomorphic to two copies of the bit group of FP. Okay. So, right, so I want to make the following construction. To see this fact, what I what I'm essentially going to do is to reverse the above construction. So let you one bar through you are bar be a quadratic form over FB. So let's say I have a quadratic form living over over the field FB. And then what I'm going to do is I'm going to produce a quadratic form over QP. So you can produce quadratic form over QP by lifting UI bar to a periodic unit UI. And then what you do is you send you one bar through you are bar to you one dot dot dot you are. So note that, right, so there's there's some non uniqueness. There's some non uniqueness in the choice of flips. So, there is non uniqueness in the choice of the UI, but all the unique all the non uniqueness is up to scaling by square factors. Right, because if you have, if you have, if you have two different choices of you one, then their, their, their quotient is congruent to one month P and therefore it's a square since he's greater than two. So, so yeah so this is. So the choice of the UI is not relevant. And in fact, this gives a well defined, well defined map from, isomorphism classes of quadratic forms over FP to ISO classes of forms over QP. So, so this is not this is not sort of immediately obvious because in order to define this map I've sort of chosen a diagonalization I said let's, let's take a quadratic form over FP. Let's, let's put it in this diagonal form you one bar through you are bar. And let's lift it to this diagonal form you one through you are. But in fact, one has to argue that this is actually independent of this is actually well defined up to isomorphism it only depends on the isomorphism class of quadratic forms over over FP. In fact, this is something that now we have the tools to do using the chain equivalence theorem. So I guess in this case you could, you probably just check it directly because isomorphism classes of quadratic forms over. I think in this case you could, you could sort of check it directly. But in fact, you can see this using vets chain equivalence theorem. I mean if if you were to change the, the sequence you one bar through you are bar in such a way that you get the same quadratic format P, then the only way that can happen is that if you do this particular sequence of moves on the you one bar through you are bar. You know this particular sequence of moves that we saw like you can multiply something by a square you can interchange to consecutive elements, or you can do this one. You can do an extra, extra construction that comes from a suitable change of basis. And the observation is that if you if you do any of those moves on the you one bar through you are bar, you can also do those, you can do the exact same moves on the other side, you one through you are. And so if you think about this it's it's saying that the moves that you're allowed to do on the, you know, on the, on the diagonal form over FP, you can, you can do them on the on the right hand side as well. All right, so there's a question, right so how do you lift. How do you lift UI bar to UI on so any, any, any lift is okay. So, right so I mean, the lift is coming from ZP mod P times ZP is equal to FP. And so if we have an element in FP cross, we're saying let's let's let's pick some element of the pre image so some element of ZP cross, such that not P it's it's that element of FP. So that's not unique. So that's that's not going to be unique but if you have any two such choices they're going to differ by a spread their quotient it's going to be a square. So for the purposes of writing down the associated quadratic form, it's not going to affect the isomorphism class. Right so this produces a map of right so this is going to produce a map in fact it's a it's a ring so it's going to produce a map of commutative rings, which goes from the bit ring of FP to the bit ring of QP. And again P it's greater than two for this. Because in order for this to be all fine we need we need this, we need to take P, P odd. And so this this map is sending. Sorry, I guess I should say GW here because we said isomorphism classes so this is sending brackets you one bar through you are bar to brackets you one through you are. Right, so this this produces a map of commutative rings which is some sort of lifting, lifting construction. So let's call this map fee. We can also produce another map. P times fee, or let's say yeah P times. I guess brackets P times the I should say, which sends brackets you one through you are to brackets P you one. So it's exactly, it's exactly the previous construction but you've you've rescaled everything by P. So it's very well defined because it's so so it's it's literally brackets P times times the math fee. And so one obtains a map from GW of FP direct some GW of FP into GW of QP. Sorry so there's a question. Just to confirm so the map fee is sort of a section of that restriction down to the finite field, but the map brackets P fee is not a section right. Right so it's not. I guess I'm not sure I would say it that way because there's no map so this is kind of a funny map because there's no. There's no map in either direction between QP and, and FP. I mean there's a map from ZP to FP. And I guess we did not talk about we did not try to define growing the rings for something which is not a field. So you could, you could indeed phrase it as saying it's, I guess I think you could phrase it this way that it's, it's something going from the growing the equipment ring of FP to the growing the equipment ring of the ZP appropriately defined and then you're composing that with the map to the growing that ring of QP. But I think explicitly I'm just sort of thinking about an ordinance like you have a you have a quadratic form over FP, and then you lift that to like a diagonal form you lift that to a quadratic form over ZP, and then consider that as over QP. And it's not completely obvious that that's well defined and you know independent of the various choices that you made along the way but in fact that is the case and that's using that piece greater than two. Okay, so, right so you obtain a map like this and this is the map fee and P times fee. And so this map is so so this map as we've seen is basically so this. Because if you have any quadratic form over QP, while you, you know, you can break it up as as it's over here you can sort of break it up into a diagonal form involving a bunch of copies of petic units and P times a bunch of petic units. So that's going to tell you that this map is surjective. It's not quite injective. And it's not injective because brackets one minus one is the same thing as P times that so it's the same thing as brackets P minus P these are both the hyperbolic plane. But somehow that's the only redundancy. It turns out that to the theorem of Springer, and right so in fact GW of QP is the quotient of GW of FP direct some GW of FP by the, well, I guess by the hyperbolic plane. So minus hyperbolic plane, or the VIT group of QP is isomorphic to the VIT group of FP direct some the VIT group of FP. So, I mean essentially what this is saying is that yeah so this is saying that a quadratic form over QP so if you like an anisotropic quadratic form over QP, and an isotropic quadratic form. So what this gives you to give an anisotropic quadratic form over QP is the same as giving a pair of anisotropic quadratic forms over FP. And namely what you do is you, you lift, you lift each of the anisotropic forms over FP to ZP over QP, and then you multiply one of them, you rescale one of them by P. And that's essentially the sort of classification of quadratic forms over QP when P is greater than two. It's in terms of essentially a pair of quadratic forms over FP modular this this redundancy involving the hyperbolic plane. And in particular there's following. I guess it's sort of a corollary or you can prove it directly. Which is that any quadratic form over QP and at least five variables is isotropic. There is in fact an anisotropic quadratic form of, of dimension four. So the you invariant of QP is exactly four. Right, so, you know, so why, why is something like, so, so let me, maybe I should answer maybe let me end by at least explaining the proof of this. So you can prove this by sort of examining the above proof if you have a quadratic form in at least five variables, well then you break it up into like sort of a quadratic form involving petic units, and p times a quadratic form involving petic units. And if you have at least five variables and one of those two pieces must have at least three variables. So if you have a quadratic form in three variables over FP, we know that's isotropic. And so in particular contains a hyperbolic plane. And that, well, that's going to mean that for QP it also contains hyperbolic play. So, this really follows from the previous arguments. But I also want to explain I mean this is sort of equivalent but I also just want to explain this kind of directly. So explicitly. The quadratic form say in five variables. So let's say your quadratic form. I mean again you, it's a diagonal form a one three five, and by sort of rescaling by even powers of P you can assume that it's at the form let's say brackets you want you to brackets PV one PV two PV three, where the you guys and the v is our, our petic units. And what we want to do is we want to see that this is want to see that this is isotropic. And to see that this is isotropic. Well in fact, just this last part this PV one PV two PV three is going to be isomorphic isotropic. So in fact, and that in fact well that's equivalent to saying that v one v two v three is isotropic. Because you're solving a homogeneous equation. And to see this in turn, what you need is a solution of v one x one squared plus v two x two square plus v three x three square equals zero. So this equation has a solution over FB by Hensel's I'm up. So this has a solution let's say x one bar x two bar x three bar over FB by sorry not by Hensel's I'm up because because quadratic forms in three variables over FB or isotropic. And this is a right so this is not a not this is a non trivial solution so so one of the outside bars. And then you can lift this to a solution over over ZP or FB using Hensel's I'm up. Right, because so so one. So, well Hensel's I'm up was formulated as a one variable statement, but essentially so let's suppose x one is not zero. So let's let's suppose x one bar is in this solution p is not zero. And then we're going to consider this equation, we'll just pick any x to an x three that lift x two bar next three bar and consider it as a function of x one bar. And so it's a one variable, you know, then it's really just a one variable polynomial, and it has a root mod p it's necessarily a simple root because, you know, because it's a quadratic polynomial. And then using Hensel's I'm like you can, you can lift this to a solution over ZP using Hensel's. So just fix two of the variables. Okay, so this is this is really. So this is really a fundamental feature of QP, which is the quadratic forms and at least five variables are isotropic. So I guess I didn't quite get to prove that there is an isotropic quadratic form of dimension force and maybe I'll pick that up on Monday. I also write so so so far I've been working for p greater than two which makes things a lot easier. But in fact this result is also true when p is equal to two. So that's this quarrel areas also true when p equals two. But you have to work a little bit harder to prove it. You have to work a little bit harder to prove it, because for example being a square is a more complicated condition, and because applying Hensel's I'm is a little bit trickier, because you're going to have repeated roots month two. So you're going to have to apply for example some sort of refined, you know some sort of more general version of Hensel's summer, which you will explore on the exercises. But in fact this, this corollary is is still true over over Q2 with a little bit more more work. So I guess I will, I will stop here for today, and we'll continue I guess I think we'll go into sort of more details. I'll try to put some more details of these arguments and what you need for this on on the problem set, because a little bit rushed at the end. And otherwise I'm going to, yeah, going to continue on continue on Monday so. Yeah, so I stick around a little bit for questions. So there's a question. Right so, so in this, the proof of this corollary I considered a quadratic form, the form you want you to PV one PV two PV three, but the same argument would also work if you have you want you to you three and then PV one PV two, I guess, because you would say that the first three is isotropic. So this works. Similarly for you want you to you three and then PV one PV two. Namely, you would show that this particular, you would show that the first, first piece of the quadratic form involving the three patic units. This part itself is already isotropic. It's isotropic mon P. And, well, it's isotropic mon P and then you can lift that using counsel so much to show that's actually isotropic in QP. Okay. Next question was what why does where does the argument fail for just three variables over QP so. Okay, that's a good question I'm going to, I'm definitely going to put more details on this on the problem set. But if you just had three variables, right so let me give the example or even a four variables of something which is not isotropic. So, so maybe I should, yeah, should be more. So, so let's say one comma you one comma you bar is an isotropic over over FP. Right so, ie minus you bar is not a square. So let's let's pick some anisotropic form of dimension two over FP. And then let's lift you bar to you in ZP cross so again doesn't matter which left because it's all going to, you know it's all going to vary up to a square. And then the claim is that the form. One comma you P comma P you is anisotropic over QP. And again, assuming P is greater than two. Right and so, so I'm going to put this on the, I'm going to put this on the problem set but I mean so let's let's suppose. You know so let's suppose that you had x one squared plus you times x two squared plus P times x three squared plus P you times x four squared equals zero over QP. So let's suppose you had a solution of this equation over QP, and we want to get a contradiction. So, so without loss of generality can assume that the Xi's actually live in ZP right so you can so by rescaling your solution you can assume that it's a vector that lives in in ZP. So you can assume that not all the Xi's are divisible by P, right so you have x one x two x three and x four, they're they're all in ZP but not all of them are divisible by P. And now you sort of have to split into two cases. So suppose, for example, that x one is a pediatric unit. And you just, you just reduce the thing on P. So if you reduce mod P. You get that x one bar squared plus you times x two bar squared equals zero. And this is a contradiction, because, because this doesn't have any non trivial roots. So if either x one or x two is in ZP cross, then already you get a contradiction because you can look at what happens my P, and you find that this quadratic form reduce spot P has a has this has a non trivial solution which we assume does not happen. Conversely, suppose x one and x two are divisible by P. One of x three and x four lives in ZP cross. And then you do the same thing, but you reduce. Well, then you do the same thing but you first divide everything by P. So, then if you call this star, then the left hand side of star is divisible by P. If you divide by P, it's not zero. So it's, well, but it's not zero mod P square. So if x three or x four is not zero mod P, then you observe that when you consider the sum of four terms the first two terms are divisible by P squared but the second, the sum of the second two terms is cannot be zero mod P square because again, it's not isotropic mod P. Okay, sorry. So I think I'm going to put this, I will put this on the problem set so can work through it in a little bit more detail but yeah so this is how you construct an anisotropic format for variables that are QP. Yeah, okay, so there's a little bit more discussion in the chat, which I think is good. More questions. And I can ask you a question please. Yes, please. Sorry, I don't always see the hands so it's just. Oh, sorry. So basically I wanted to. Whoops. I still hear. Yes. Okay, hope everything's fine. So, I just wanted to double check basically whether I, I understood the, your, your proof about why the U invariant of QP is less than or equal to four. So the idea is that if I have a form of dimension five a quadratic form of dimension five over QP, then, well, we have this that we have this orthogonal direct sum where you know I have a part which reduces to you to F P star and a part that reduces to, well, zero modular FP. So I have this you, you, the UIs and the PV eyes, and one of those and one of those two parts must have dimension at least three. So if I reduce it to FP, then because you you invariant of a piece to, there's a hyperbolic plane that split. And then if I lift that I still have a hyper hyperbolic plane that split, therefore I have isotropy in my original form, right. That's exactly right. Yeah. Yeah, so you can also say it without using the words hyperbolic plane which is, I mean so you can also say it using hence the summer which is that you have a, I mean you have a solution when you've reduced that, like, you have a solution, let's say the first piece has dimension at least three, and then that has a solution on P and then you lift that to a solution using some you can also phrase it in terms of hyperbolic. Yeah, I see, I see that's great. Thank you. So, so for example, this is going to tell you that if you have a this this arguments maybe I'll put this on a home site and so if you if you have a, if you have a field, if you have a field of characteristic not to then the you invariant of the power series of the Laurent series field over that field is going to be twice the you invariant of the field that you started with. We can see how this could relate to the case of why you have FP is to and you have QP is is for because we can think of that as some kind of Laurent series in the case of QP for with a pediatric expansion right. Yes. I see. That's great. Thank you. Yeah, so I guess the more generous you would be like if you, I mean so so FP Laurent series T and QP, and what's sort of analogous about them is that they both have complete discreet valuations. So discreet, I mean evaluation and we saw like the pediatric valuation, you can also do a tea attic valuation for FP Laurent series T and being discreet means that the value the valuations are all integers. So it can really formulate it as a statement about complete discreetly valued fields. So this would be the devaluation from the exercise set from yesterday. Is that the, that's great. Yeah. Thank you. More questions. Thank you. Thanks. And thanks everybody for wearing PC my t shirts for our special day. Has everybody gotten one. Yeah, some of you are wearing it. Yes. I did not get mine because you were sent to a wrong address. All right, well, I'll see you in my office hours. See you to all either in office hours or Monday. Have a nice weekend.