 back. I also just want to encourage everyone to go to the TA sessions. I think it's really, you know, it's really good to take these sort of problem sets. Some of the problems are really just sort of reinforced or some of the discussion topics are to sort of reinforce the material lecture and somewhere to go a little bit further. And even if you haven't started the problem sets or you have lots of questions or you want to work on a previous problem set, you're very much encouraged to go to the TA sessions and the TA's are very happy to talk not only about the most recent problem sets but previous ones or answer any questions. Okay, so let me set up the screen chair. Okay, so yeah, so I want to sort of pick up from last week. So right, so last week or towards the end of last week, we introduced in the field of piatic numbers. So this was a suitable completion of the rational numbers with respect to the piatic absolute value. And right, so I guess we discussed proof of are we discussed Hensel's lemma that gives a criterion for when you can solve equations in the piatic numbers. And at least for p greater than two, we discussed the classification of quadratic forms over QP. And so let me let me just recap how that how that went. So so the main result was, well, so you can you can say that in terms of bit rings, for example, so as a classification of anisotropic quadratic forms over QP when P is greater than two. And the statement is that the the bit group, so let's say the bit group of QP is isomorphic to two copies of the bit group of FP. So in other words, an anisotropic quadratic form over QP is a sort of equivalent data as a pair of anisotropic forms over over FP. And so specifically, right, so where does how does this correspondence come about? Well, if you have an anisotropic quadratic form over QP, well, you can write it in diagonal form. And then you can assume that it's, so you can assume it's it looks like, say, you one through you are comma PV one through P vs where the UIs and the VJs are in ZP cross. So by sort of multiplying by even powers of P, you can always remove the diagonal entries of their piatic valuation as either zero for a periodic units or one, so it's P times a Patic unit. And so the class of this is going to go to the class of U one bar us, sorry, you you are bar, so you sort of reduce mod P, and then V one bar through VS bar. Right, so okay, so, so, since you're a quadratic form over QP, what you can always do is you can, you know, you can diagonalize it and then you can put the entries so that they have Patic valuation zero or one, and then you sort of reduce it mod P, reduce the ones that are Patic units mod P, and then you reduce the ones, you give the ones that are P times Patic units, you divide by P and then reduce those mod P. And so this gives a correspondence between anisotropic quadratic forms over QP and pairs of anisotropic quadratic forms over FP for P greater than two. So let me just write that again. And yes, so this is what this is a special case with your move Springer. And so this was this was really proved using Hansel's line, you needed Hansel's line because somehow you're, you're taking square roots. And, well, it's easier to take square roots when, when in QP for, for, yeah, so thanks. So, right, so where did we use P, P is greater than two, well, we use P is greater than two. So the fact that this is even well defined, so it's not obvious that something like this is even well defined, first of all, because, you know, you can, there are many different ways you can write a quadratic form in, in diagonal form. And so it's not obvious that this kind of construction is even well defined, begin with. I guess, in fact, the way we really constructed it was more as a map from right to left, that if you have a quadratic form over FP, then you sort of lift it to a quadratic form over ZP, and then QP. And, and so that you could see was sort of well defined by really by using Hansel's sum up, because so, so for example, you're really using the structures of squares, right, so maybe I should write this so this really used the structure of squares in QP. So in particular, so we use the fact that any element, let's call it X, which is congruent to one mod P, is a square in QP, right, say in ZP. And so that's going to imply that the map from right to left, which is I guess maybe the most more natural way of actually defining the map is well defined. Right, so this is again, so this is for P greater than two. Right, so this, right, so this gave a pretty sort of concrete, but yeah, so the conclusion was that you have a pretty concrete sort of parameter parameterization of anisotropic quadratic forms over QP. And in particular, a corollary is that any anisotropic form over QP, again P greater than two, has dimension less than or equal to four. Right, because we know that any anisotropic quadratic form over FP has dimension at most two, any three dimensional form over over Q over FP is isotropic. So that was, oh yeah, so that was a little bit of a recap from Friday. And the first thing, yeah, the first thing I wanted to do today was to try to sort of revisit this, this type of statement. And in particular, I mean, this is really for QP for P greater than two. But in fact, there is, there is sort of, so this particular statement isn't quite going to work in in characteristic two or sorry for P equals two. For example, we haven't even defined the bit, the bit ring of F2, we've only considered bit rings in an odd characteristic or zero characteristic. But in fact, there is sort of a uniform, a somewhat uniform description of the bit ring of QP that works even when P is equal to two. Well, at least sort of modular filtration. And in fact, it doesn't just work for QP. It also works for any local, any local non or comedian fields. So I mean, the first goal today is so to describe the bit ring of QP in a more uniform way. Sorry, so yes, yes, thanks. So in the question, right, so yeah, so let me write it this way. So in other words, the un variant of QP is at most four. Well, actually, it's equal to four. It's equal to four because you do have an anisotropic quadratic form over over FB. And then you can lift that. And then you can, you know, add that to P times itself. And so by this, you get, yeah. Okay. Right. Right. So and I guess another thing that you can say is that there are exactly 16 isomorphism classes of anisotropic quadratic forms over QP for P greater than two, because the bit group of FB has carnality four. I guess that was something right that we saw earlier. Okay. So first is also I want to say something about, you know, like the PI like numbers in general and right. So, yeah, so in fact, it's not it's not just going to work for for the fields of PI numbers, but it's going to work for any any local fields. So I want to make the following definition. So definition, a local field is a field E with an absolute value. So with a non-trivial absolute value, well, up to equivalence, right. So we said that two absolute values are the same or equivalent if they give you the same topology or one is a power of the other. Right. And such that E is locally compact for the induced topology. So as a metric space, it's it's locally compact. And so in particular, it's complete. Right. So for example, we've seen that that QP is an example of a local field. So I think this was on the this was on one of the problem sets. So if you take the the unit ball or the unit desk, so that's the piatic integers in QP, that's a compact space. And well, also the real numbers is a local field. Well, again, the real numbers are locally compact. And similarly for the complex numbers. Right. But so in fact, there's a classification of local fields. So essentially, there are three different types of local fields. So, right. So let E be a local field. So the first statement is that if E is Archimedean, so if you have a local field that comes with an absolute value, and you can ask if that's Archimedean or non Archimedean. So Archimedean, the non Archimedean meaning that you have the stronger absolute value of X plus Y is at most the maximum, the absolute value of X and the absolute value of Y and Archimedean, the condition that that is, well, that's not true. So if E is Archimedean, then E is isomorphic to the real numbers or the complex numbers. So that's sort of a first type of local field. Yes, isomorphic as a topological field or as a field with an absolute value with just to find up to it. Yeah, thanks. Okay, so if you have an Archimedean local field, then it's it's the real numbers or it's a complex number. So that's great. And then second is the case if E has positive characteristic. So if then E is isomorphic to the field of Laurent series over a finite field. So another way that you can cook up a local field is that you take a finite field and then consider all Laurent series over that field. And so the field of Laurent series over a finite field, well, it has a natural absolute value, which is the tiatic absolute value. Where instead of, well, I mean, right, so where you look at so the valuation of the absolute value of something is going to be in terms of how many powers of ti divided or what sort of the order of the zero or pole of this Laurent series at a t equals zero. And right, so that's going to give you that's going to give you an absolute value. And in fact, it's a local field. So for example, if you're in a situation like this in the unit disc, the elements of norm at boost one is given by fq double brackets t, the ring of formal power series over fq double brackets t. And that is given the tiatic topology. So that's given. But if you think about it just as a topological space, it's a it's a product of copies of fq. So as a topological space, the unit disc looks like a product of copies of fq, accountable product of copies of fq. And that's that's compact. Well, I guess it's a special case of taken off serum. It's a product of finite sets. But in fact, it's accountable products, it's really sort of maybe a more explicit case of taken off serum. Okay, so then this is this is a case of positive characteristic. And right, and then finally, there's the case of characteristic zero, but but non Archimedean. So if the characteristic of E is equal to zero, but E is non Archimedean, then E is a finite extension of QP of some QP. So so the periodic numbers are an example of the local field. But in fact, if you have any finite extension of the periodic numbers, any finite field extension. So for example, you could take something like QP, and at a square root of P, or at a P through unity or something, and you're going to get some sort of finite extension of QP. And then it's a basic theorem that you can always extend the absolute value to this finite extension. And it's going to become an example of a local field. So for example, I mean, there's essentially, there's going to be a unique way of extending the absolute values of the absolute value of the square root of P, for example, is going to be square root of the absolute value of P. And, and then it's going to write so it's going to be a local field. I'm sorry, so it's a question. Isn't there with respect to what you just said, just to, to see whether like this stems from a general theorem, isn't there a general result that says about this extension of the the absolute value that if I have an algebraic extension of a complete field, which has some absolute value, then I can uniquely extend it. I think it's like cruel extension theorem, right? Or not exactly sure who would. Well, that's true. That's correct. So if you, if you have a, if you have a complete field and even algebraic or finite extension, you can always is a unique way of extending the absolute value. Thanks. It's not unique. If it's not complete, but you can always do it. But if it's complete, then yeah, thanks. Yeah, thanks. Okay. So, right. So I guess I wanted to kind of, you know, sort of say this explicitly for a couple of reasons. One is that some of these results are going to really work uniformly for local fields. So not just for QP. And so it's good sort of stadium in this generality. But also, like if you, if you haven't seen QP before, then maybe this sort of also helps to sort of orient orient yourself because the idea is that if you look at, if you look at cases two and three, the idea is that QP, you should think of QP as sort of analogous to the field of formal, formal Laurent series over a finite field. And so in general questions involving, you know, various types of questions, you expect them to be sort of easier for formal Laurent series than for QP. So QP, I think of it as sort of maybe slightly more difficult version of formal Laurent series. And for example, it's more difficult because, well, if you think about piatic numbers, right, I mean, so you sort of informally a piatic number is something like a formal Laurent series in the variable P. But when you add piatic numbers, you have to carry. There's some sort of carrying involved. Whereas if you add formal Laurent series in the variable T, there's no carrying involved. So, so informally QP is like Fp Laurent series T, the field of formal Laurent series in a variable T, except the variable that quote unquote variable is P. And somehow when you add and multiply the rules for adding and multiplying is going to be a little bit more, going to be a little bit more complicated. I mean, in practice, it's probably best not to think of QP as literally sort of Laurent series in P is this really more of an analogy. I mean, really you build QP using this completion procedure. But there's just sort of an analogy. So what I wanted to do is to try to describe the bit ring of, so in general, there's sort of a description of the bit ring of any local field or at least a description up to sort of associated graded terms. And yeah, so I wanted to sort of explain that. So in general, there is a pretty explicit. But to explain this, we need to introduce one more ingredient, which is called the Hilbert symbol. So this is going to use an ingredient called the Hilbert symbol. So I guess my next goal is to sort of explain that. But maybe let me sort of state one of the main, so let me start by sort of stating like one of the main sort of outputs of this classification. So theorem, so let E be a local field. And so first of all, I should say that if E is the real numbers or the complex numbers, we already know we saw sort of really in the first lecture what quadratic forms look like. So let's say E is a non-archimedean local field. And let me assume E has characteristic greater than not two. So Q2 is okay. Q2 is characteristic zero. But I don't want to consider F2 Laurent series T because we haven't really talked about quadratic forms in characteristic two. So then what is always going to be true is that any quadratic form in at least five variables is isotropic. And there exists a unique anisotropic form of dimension four. So in particular, the univariant is going to be equal to four. Okay. So with this in mind, I want to try to say something about the bit ring of E. So goal is to say something of the field E. And so in general, if F is any field, we can consider the bit ring of F. So this is the ring of all isomorphism classes of quadratic forms, modulo the hyperbolic forms. And the bit ring of F has an ideal, has a natural ideal sitting inside it, which is the ideal of forms of even dimension. So I is additively generated by the forms, well, I'm going to call them the forms brackets one minus a, as a ranges over the elements of F cross. So just sort of for reasons of normalization, I'm going to put in a minus sign here. And this ideal I is additively generated by these two dimensional forms. Because if you have any two dimensional form, you can write it as sort of the difference of these things. Well, if you have any even dimensional form, you can write it as a sum with signs of these things. Okay. So, right, so you always have this ideal, this is sometimes called the fundamental ideal. And right, so moreover, what you can do is you can describe, you can describe, so here's a basic fact about this ideal, which is that there is an isomorphism, which is that i mod i squared is isomorphic to F cross modulo F cross square. So that, so if you take i modulo i squared, then it's isomorphic to the nonzero elements in F, modulo the squares in F. And this isomorphism is coming from the signed determinant or the or what I guess what I or the the discriminant. Right, so sorry, so there's a question. How do we know that over E we can always represent one? So we can't, I mean, I guess the statement is that if you have, right, so I mean, if you have some form brackets, you know, B comma C, then you're supposed to write that as a as a, right, so sorry, so this is this is saying that as a as an abelian group, it's generated by brackets one minus brackets a. So this is also in the bit ring, this is the same thing as, as one, one minus the class of brackets a. And so if you have, if you have some even form, which is like a comma B, or brackets a minus brackets B, you can write that as a difference of these times these types of elements. Yeah, so, so explicitly this this isomorphism sends the class of an even dimensional form, let's say V comma Q. Well, it's going to go to it's going to go to the determinant of V, but it's really going to be up to a sign. So it's going to be minus one to the dimension over two times the determinant of V. And the claim is that this induces an isomorphism between I mod I squared and the units mod, mod the square of the units. So, yeah, so the reason for the signs basically is that the is that you're working with a bit ring instead of the grown G at that ring. So, so, so the bit ring is you're taking all quadratic forms, but you're allowed to add on a hyperbolic form and a hyperbolic form is not going to change anything. So the hyperbolic form has determinant one, but it has dimension two. And sort of because of that normalization, because if you add on a hyperbolic form, you'll change things, you'll change the determinant by a sign, but you'll change the dimension by two. You define the sort of signed determinant or also called a discriminant. And it's well defined on the ideal of even dimensional forms in in in the bit ring of F. So, so this. So this is a map called the discriminant which goes from I to F cross mod F cross squared. And now I squared the square of this ideal I is generated by products of elements in I so it's generated by things of the form one brackets one minus A minus B, AB. And these classes have discriminant one. So it actually factors over a map which goes from I mod I squared to F cross mod F cross squared and then one one checks that that's an isomorphism. And so the map in the other direction is right to the map in the other direction which goes from F cross mod F cross squared into here is sending A to brackets one minus A. So this is something that's true for any field for any field. For any field this augmentation ideal modulo I squared is automatically isomorphic to F cross mod F cross squared. So, so let me just sort of recap that by saying the general facts. So for any field F, the idea is to consider the augmentation ideal I sitting inside the bit ring W of F and the iatic filtration. So and the sequence of powers of this ideal I so you have the bit ring of F that contains the ideal I and that contains the square of the ideal and so on and so forth so you have this you have the sort of descending filtration of the bit ring. And you always have, for any field F, you always have the formula that I mod I squared is isomorphic to F cross mod F cross squared. Well, first I should say that W of F mod I, well, I is the ideal of even dimensional forms so W of F mod I is Z mod 2. And I mod I squared is F cross mod F cross squared. So now I want to explain what this sort of gives you if you have a local field. So for a local field, in fact, you have the formula that this filtration is actually finite. So for a general field, this filtration might be infinite, but for a local field, what you always have is that I cubed is equal to 0. So for a local field, let's call it E, you always have that the cube of the of the augmentational ideal is 0. So this is, I guess, closely, I mean, this is sort of coming from the fact that five dimensional quadratic forms are automatically isotropic. So this is coming from the U invariant. The U invariant of E is at most, well, it's equal to 4. And in fact, and in fact, I squared mod I cubed is a Z mod 2. So for E a local field, when we have this finite iatic filtration on the VIT ring and the associated graded terms. So in general, if you have an ideal, I mean, if you have an ideal and some commutative ring, you can I mean, you can look at the the terms of I to the n mod I to the n plus one. Right. So you have W of W of E containing the ideal I containing the ideal I squared and then containing zero. So I squared here is equal to Z mod 2. The the the VIT ring of E module I is also equal to Z mod 2. And I mod I squared is equal to E cross mod E cross squared. So there's this description, if you have any local field. So there's not sorry, right. So sorry. So there's a question which is, do we already know that local fields have you invariant for so no, I have not proved that yet. So what we've proved is we've proved that for QP. So yeah, so thanks. So so just just to be clear about the logic here, I guess I'm not probably going to be able to prove everything in this lecture. But I just want to describe. So what I'm going to explain is going to be stuff that we've sort of entirely proved for for QP for P greater than two. And I want to sort of try to explain how this so we haven't proved it for Q2, but it's not actually too hard to prove it for Q2. So for Q2, yeah, right. So so for Q2, this is kind of right. Sorry, I think I should have experienced earlier. So for Q2, I think I will put this on the problem set. Yeah, so just just to back up a little bit. So yeah, sorry for that. So for Q2, I guess I want to sort of put this on the on the problem set. And so it's fun to actually work work all this out pretty explicitly. So can show directly that every five dimensional form is isotropic. And that any anisotropic four dimensional form is isomorphic to the following, it's brackets one minus five to minus 10. So so for QP for P greater than two, you can work this out sort of abstractly using Henselzama. But for Q2, you sort of have to get your hands a little bit dirty. And you have to you have to sort of play around with, you know, what can you have to play around with the different possibilities for like five dimensional forms and four dimensional forms. And the key thing that you need to do is you need to essentially look at congruences instead of looking at congruences mod P, you need to look at congruences mod eight. So if you want to prove something like every five dimensional form is isotropic, you need to look at congruences mod eight. And that's because any element of z two, which is congruent to one mod eight is a square. So in fact, the squares and z two cross are exactly the elements which are congruent to one mod eight. So z two cross squared is exactly those x and z two with x congruent to one mod eight. Yeah, so sorry, so thanks for asking. So I'm not going to try to give. So in this lecture and in tomorrow's lecture, I think I would not try to give these proofs in general for a local field. So it's actually right. So it's not too difficult to prove it if you have a local field, which is a finite extension of QP for P greater than two, more or less the same proofs over QP for P greater than two are going to work. But it is going to be a little bit tricky to prove these types of things completely directly. If you have if you have an arbitrary finite extension of Q two. And a lot of it is really subsumed under under local class field local class field theory and all this really sort of follows as a special case of somehow of local class field theory. But for Q two, I think it's a good idea to actually check directly that every five dimensional form is isotropic. And you just sort of have to get your hands a little bit dirty by thinking about congruence class is not eight and so forth. Okay, sorry. So there was a question. Yes, that's right. And so, right. No, so, right, sorry. So the key feature of any local field is that there's a unique an isotropic four dimensional form any any non Archimedean local field. Right. So the you invariant is foreign. There's a unique example of an an isotropic four dimensional form. And in the case of Q two, you can write it down explicitly. It's it's this particular form. Yeah, so I'm going to put this on the problem set for today. Okay. So sorry. So there's a question. Is it difficult? Right. I mean, I guess so hence was almost so true. It's just that when you when you when you have squares, then two is not a two attic unit. So it's not true that anything congruent to one mod two is a square you need to be congruent to one mod eight. Okay. But so just to go back. So there is a description. There is a description of the bit group of any of any non Archimedean local fields. So up to this filtration. So although you don't have this nice description as it's a direct sum of the bit group of FP plus a direct bit group of FP, when P is greater than two, you always have this finite filtration. And you always have that I cubed is equal to zero. And you always have that the associated graded terms are Z mod two, E cross mod E cross squared and Z mod two. So sorry. So just to recap, the bit group of E is a local field has a filtration where the associated graded terms are w of E mod I is given by Z mod two. I mod I squared is given by E cross mod E cross squared. So for example, you work, I mean, I think in the exercises, maybe in Friday's exercises, there was exercise of working what this is, especially when he is Q two. And then I squared mod I cubed is equal to Z mod two. So you don't have a just you don't have a direct description as, as an abelian group of the bit group of E as like a direct sum of two things. But you always have this, you always have this sort of three step filtration. Yeah, sorry, question. So I was wondering, looking at the filtration, if we think of it, if we think of all those things, not as well, rings, but just as abelian groups, and we think of this as a, as a actually as a normal, as a normal series. Well, in general, this thing is not going to be a composition series, right? Because this E star modulo E star squares and going to be simple. Is there any, like in the case of QP, for example, for P greater than two, we have Z mod four Z. Is there any interest in well, extending this to to a composition series? Like, is there any interest in treating this as a as a normal series and extending it to a composition series? Sorry, what do you mean by the difference between a composition series and a normal series? Well, I mean, in this case, because all the groups are abelian, I guess that well, subnormal and normal, they coincide in this case, right? So I'm just I'm just looking at this filtration and thinking of it as basically a sequence of abelian groups and not just rings. And I'm wondering, well, this, this factor, this E star modulo E star squared, that doesn't have to be simple. But of course, we can refine. If we look at it as a series of abelian groups, we can refine it to a composition series. Would there be any, any information hiding there? Like, is that something that's investigated in, in the bibliography? So at least if I understand your question, I think it's not going to be very canonical, I think it's not going to be very canonical because these are F2 vector spaces. So there are many different ways. So I mean, E cross mod E cross squared is going to get quite large when E is like some big extension of Q2. But so I think what what really is what one really wants to do is to understand the multiplication. So the thing is that like the exact, right, so I think is what I think what you said might not be the composition series is not necessarily going to be very canonical. But what you do have so that there is some extra structure here, which you do want to understand. So, so this is right. So this is this is as an abelian, what I've described is, is at the level of abelian groups. And so, so I cubed is zero. So, so this is a sort of a complete sort of filtration as abelian groups. But there is a little bit more information here that you want to understand, which is a multiplication. So the multiplication, right, so there's a map from i mod i squared times i mod i squared into i squared mod i cubed, which is also just i squared, which is the mod two. And so, right, so if you want to understand the the bit group, you want to understand some of the multiplication. It's, it's given by, well, you can try to understand it by looking at the multiplication on associated graded terms. And so this is a pairing, which goes from E cross mod E cross squared times E cross mod E cross squared into Z mod two into plus or minus one this this multiplication law. And this turns out to be super actually sort of a fundamental importance and it's this thing called a Hilbert symbol. Sorry, so question. So that's a Z mod two. Yeah, it's not, not, yeah. Okay, above, sorry, above where? Yes, like your multiplication and E square and then what's the other term? So it's i mod i squared. Sorry, you mean here? Yes. Yes. Sorry, it's the same thing. It's i mod i squared. Thanks. All right. Okay, so, so, so there's a natural pairing that this gives you and this pairing is actually sort of really fundamental for studying quadratic forms over local fields. And it's, it's this construction which is called the Hilbert symbol and it's sort of a general feature of local fields. So definition, let E be a local field. So given elements a comma b and E cross, we're going to define the Hilbert symbol brackets a comma b to be the following. So it's plus one if z squared equals a x squared plus b y squared has a non-trivial solution. So ie if brackets one minus a minus b is isotropic and it's going to be minus one otherwise. So this is called the Hilbert symbol. And it's just a sign. So the Hilbert symbol is just a sign that you can associate to any elements of, of E cross. All right. So, so for example, so, so maybe let me list some, so just have a few minutes off. So maybe let me just list some, some, some properties of the Hilbert symbol. So explicitly where this is coming from is that the Hilbert symbol of a comma b, so sorry since I've been talking about bit rings, is the following. Well, you form, you consider brackets one minus a, tensor brackets one minus b. And so this, this lives in the bit ring of E. So this, this lives in I, and this lives in I. And when you multiply them, this lives in I squared, and I squared is exactly z mod two. And so the Hilbert symbol is exactly, is exactly the sign that you get out of this. So brackets a comma b is equal to, well, you can, you can equivalently say it this way. So it's, it's one if the form brackets one minus a minus b, a b over e is hyperbolic, which, well, you can check this is, so I'll put this on the, on the homework that's the same as brackets one minus a minus b being isotropic. It's the same as this being hyperbolic. And it's minus one if the form brackets one minus a minus b comma a b is an isotropic. And so recall that there's exactly one choice of an isotropic form of dimension four over a local field. Okay. So, so, so, so if you have elements a and b in your in, in the group of units of your field E, you cook up either this three dimensional quadratic form or this four dimensional quadratic form. And then you ask whether it's either in the first case, whether it's isotropic or not, or in the second case, whether it's hyperbolic or not, sort of equivalent. And then that, that is encoding this construction called a Hilbert symbol. So the Hilbert symbol is right. Okay, so yes, so there's a question, why, why does E need to be a local field? Why, why can't we do this just very generally? Right, so we can do this very generally. But it's not necessary. So you can make this at least the first definition or we can make either these definitions very generally. But it's not going to be a very good definition for the following reason. So, right, so let's, let's, let's write, let's write down some, some, some properties of the Hilbert symbol. So the Hilbert symbol is going to satisfy various properties. So for example, the Hilbert symbol of a comma one is always one. Because, well, right, the Hilbert symbol of a comma one minus a is equal to one because you can always solve z squared equals a times one plus, well, because you can write one is equal to a times one plus one minus a times one. A comma minus a is equal to one. So, so you can write down a bunch of basic properties of the Hilbert symbol that do not use the fact that E is local and you can, you can make this definition. But what's actually kind of sort of special about the case that E is local is that this is bilinear. And so it gives a bilinear pairing, it gives a symmetric bilinear pairing from E cross mod E cross squared times E cross mod E cross squared into z mod two, into plus or minus one. And in fact, it's non-degenerates. Okay, so maybe this is where I should sort of, I should sort of wrap up. So, so, so what's key about a local field? So I'm going to have to come back to this next time. What's really key about a local field and about this construction of the Hil, of the Hilbert symbol is that by looking at whether or not you have a solution to this quadratic form, you get a bilinear pairing from E cross mod E cross squared with itself into z mod two. And this bilinear pairing is actually non-degenerate. So I think in, you know, I'm going to try to put this on the problem set, but this, so you can, you can try to work this out very explicitly for q p. And when p is greater than two, this is going to give you essentially the quadratic symbol or the Legendre symbol. And for q two, it's going to be a little bit more complicated. For q two, of course, it's going to, you're going to have to try to work out some conferences mod eight. But right, so, so I guess the main thing that I, you know, sort of wanted to explain is that you can think about this in terms of the vet ring. So this Hilbert symbol is something that is encoded in the vet ring, because it's exactly the map from i mod i squared times, it's exactly the multiplication squared, the multiplication map from i mod i squared times i mod i squared into i squared mod i cubed, which is z mod two. And i mod i squared is E cross mod E cross squared, i mod i squared is E cross mod E cross squared. And this, this sort of multiplication law in the vet ring is exactly the Hilbert symbol. Okay, so sorry to end in, I say the awkward place, but yeah, so thanks. I'm going to, I'm going to pick this up tomorrow and say more about that. So stick around for the questions. Is there, okay, can I ask a question? Yes, please. So is there a reason why, like, even dimensional, the ideal of even dimensional forms is important and like not like dimension multiple of three? Oh, sorry. Oh, that's because, okay. Yeah, exactly. Right. Yeah, sorry. That's, that was obvious. Yeah, sorry. So the question, there's a question. Do we need E to be a local non-archimedean field of characteristic not two to define the Hilbert symbol? So the Hilbert symbol, sorry, so I guess we need the characteristic not to be not two. But in fact, yeah, right, so you can define the Hilbert symbol over the real numbers as well. So, so this, this, this definition that I wrote up here of the Hilbert symbol, it works over any, any local field of characteristic not two. And it's going to be, it's going to have this bilinearity property. So, right. So for example, you can define the Hilbert symbol over the real numbers. And then it's going to be minus one, if and only if A and B are both negative. If you're over the complex numbers, it's just going to be, I mean, C cross mod C cross squared is zero. So there's nothing there. But yeah, right. So yeah, so that's, yeah, that's a great question. I mean, that's exactly the, sorry. So the question is, you know, what does this have to do with Galois cohomology? I mean, this is exactly the subject of the Milner-Conjecture, the theorem. So, so in fact, the, the, so I will try to tie this together next time. But since I'm not going to say much about Galois cohomology, in fact, what happens is that for any field, so we can write this for any field. So, right. So for any field F, one considers this iatic filtration, one considers the bit ring of F, and one considers this iatic filtration. And so in fact, the, I guess the Milner-Conjecture, so a theorem of, in this case, Orlov, Rijek and Riewadzki, is that if you take this associated graded ring, so after that ring, so this is the direct sum of i mod i to the n mod i to the n plus one, then this is isomorphic on the one hand to Milner k theory, mod two, which is what I would talk about tomorrow. And also to the Galois cohomology of the field F with coefficients in, yeah, C mod two. So in fact, that's exactly, exactly that. Well, I guess that's, so there are actually two Milner-Conjectures, and I think that the relation with Galois cohomology was the, was, yeah, was proved by Riewadzki. And maybe I think it was proved first. Right, sorry. So the question is, right, so how do we extract information about a ring using knowledge about its filtration? Right, I mean, I guess it's an example you would give in an algebra class. Um, I don't know, I mean, right, so if you have a, I mean, if you have a filtration on the ring, it doesn't quite determine, it doesn't, if you know the associated graded ring, it doesn't quite determine the ring. But, I mean, for example, it will tell you, I mean, it will tell you how to compute a product modulo terms of higher filtration. So I don't know if I have a great answer, I don't know if I have a great answer to that question. Um, I mean, sometimes, right, I mean, so for example, if you know, if you want to show that a product is nonzero, then if the product is nonzero and the associated graded, then it's nonzero in the ring to start with. But for example, it's going to be hard, if you just know the associated grade, it's going to be hard to tell whether something actually is zero to prove that something is zero. Because the associated graded is going to let you prove that something is zero modulo higher filtration. Thank you. Thank you. Okay, so I'll see you, see you tomorrow or at the office hours.