 Now, I want to echo what Irina said about attending lectures and so on, I think in general should have the goal of understanding something or getting some idea out of it, not necessarily the goal of understanding everything, which is an ideal one can rarely meet. I'm speaking for myself, based on personal experience at least. So, right. So in some sense, after a slight digression into K2 and others a question already in the chat. The clever trick. Yes, you can find it in this book of Milner's, which I already referred to. Which is also listed in the, I put list of references in the other a given outline of my lectures. And it includes some books that are relevant to the material been discussing and Milner's book is listed in that in there. Yeah, thanks Freddie yeah exactly. Yeah thanks for it okay perfect everyone's chiming into the relevant links and everything great. Okay, so we're in some sense going to come back to our roots here and talk about quadratic forms. So, so well, our main character today is going to be a quadratic form over the rational numbers. So just you can think of it just as a homogeneous polynomial of a degree to homogeneous polynomial with rational coefficients. And, well for every place V. And then get a quadratic form over the completion QB. Now I recall that a place V is either a prime P, and then QP is the P addicts, or it's just a formal thing called infinity, where Q infinity is the real numbers so these places are just a way of comprising the possible completions of the rational numbers that we've been discussing. But so if you have a quadratic form over Q you get a quadratic form over QB just, you knew sorry, just take the, just take the same coefficients, so to speak in your homogeneous polynomial. So these ones will have rational coefficients but now we're interested in them as quadratic forms over this bigger field. So if you have zeroes or a number being represented. Well, in other words, we're allowing the variables, for example, to live in Q and you and not just in Q, even though the coefficients are still in Q. Okay. And the main theme of possum and Kowski is a theme called the local global principle, which is not a, you know, a general principle that's always valid, but it is valid in some examples. So if you're including in the case of quadratic forms and that will be essentially the house and Kowski theorem knowledge. I'll say it, you know, FB gives you knowledge of B. Sorry, F mu knowledge of F mu for all new gives you knowledge of F. So if you know how your quadratic form behaves locally that is at every completion at every prime and ending prime infinity. You should know a lot about your quadratic format. So that's the general framework in which the possum and Kowski theorem operates this local global framework. And now let me state it precisely. And there are actually three different forms. Three different forms as in quadratic forms of three different. Well, theorem, three parts, let's say, three different manifestations of this global local global principle which the theorems as are satisfied. And the first one is that if F mu is isotropic for all new, then F is isotropic, recall that isotropic just means that it has a non trivial zero. So the sum of genius, a degree two polynomial has a non trivial zero. So if you can find the zero in all of these q news, then you can find a rational zero and this is actually kind of shocking because what we saw you have very various techniques for producing zeros in the, in the qps for example Hensel's lemma. So if you have an analytic process like this Newton's method, for example, for funding zeros and R, you're allowed to use analytic tricks and you're allowed to write infinite convergent series to produce your solutions. And you can do it completely independently in all the places and then somehow this theorem is saying that magically guarantees you that there is some rational solutions like you can do analytic work at every place. And then magically there has to be a so called global solution, a rational solution as well. So that's quite shocking and powerful. The second part is, well now, now instead of just having, oh no, yeah, now instead of just having my quadratic form F, I'll also take a rational number a in q. Let's say if F new represents a for all new, then F represents a. So that means that you can solve this means you can if you can solve the equation F new of your variables equals a, or the variables line q mu for all new you can solve that. Then you can solve it in q as well. So any kind of thing except instead of asking where F is equal to zero you asked it when when F is equal to some a right so it's a generalization of one. Instead of asking just about the zero sets of your polynomial you asked about the level sets. Yes, so. But about what you just said however we cannot say that we can get one from two, because when we say represent we also allow the zero vector to represent something. So I guess one wouldn't follow from two right. You wouldn't be able to apply two for a is equal to zero and actually get one, because represents. Indeed, yeah. Yeah, so you can go from two to one technically yes yes yes indeed thank you. Yes. So a lot of Ferris is saying is that, well, if a in the case where a equals you wanted to use this one a equals zero but when you ask it to represent zero you that doesn't include the condition that it's a non trivial zero of F. Yeah, so thank you for that precision and in fact what we're going to see is that one implies to and that's the correct implication, but not for F you have to change the form but yeah. Yes, and then the third form so instead of just having F we also have a G so F and G or two quadratic forms. So if, so there are those correspond to quadratic spaces right and we have this notion of isomorphism of quadratic spaces and so if F mu is isomorphic to G mu for all new, then F is isomorphic to G. And that that just this notion of isomorphism of quadratic spaces just corresponds to saying that you can get from F to G by some linear change of variables. And saying if there's a linear change of variables which turns F mu into G mu for all new and again the linear, the linear variables are allowed to be in tune you and you can select them completely independently of each other as new berries. Then there has to be some rational linear change of variables that connects F to G. Okay, so those are the the Hassam and Kowski theorem. The first thing I'd like to explain is that and it comes from something that Achille already taught us that the form number one implies both form number two and form number three. So Achille taught us this general principle that the question of whether a quadratic form is isotropic or not is the sort of most basic question you can ask about a quadratic form. So let me just go through the argument Achille gave but in this specific instance. So, so the remark is that one implies two and three. So let's start by showing that one implies two. Well, well recall in general that F represents a if and only if the quadratic form you get by just adding a new variable let's call it Z. So I'm looking at F minus AZ squared is isotropic. So, if F mu represents a for all new then F mu minus AZ squared is isotropic. It's just a general field yeah not not just over the rational numbers, and then F mu minus AZ squared is isotropic from you therefore by form one of the theorem F minus AZ squared is isotropic over Q, and therefore F represents a. So, it's just this general trick of reducing the representability question to an isotropic question but with one extra variable. So let's go back to the chat. Don't we need Z to be non zero. Yeah, I mean, I'm not saying it's entirely obvious that this equivalence holds, you need to do you do need to check some cases and so on. But but no indeed this equivalence is is literally true as stated. Yeah, so even is giving the argument thank you very much. Yeah, so you do have to split into cases in the proof but nonetheless, when you look at the cases you see everything's okay. Yep. Now for one implies three this is a little more complicated and it used these with these theorems of bits, though these cancellation theorems and hyperbolic and nonsense about splitting off a hyperbolic and so on. So let me see if I can get it right. So, so right. Well, so F F and G or two quadratic forms over Q. So, well note that F represents a for some a, and we can even take a to be non zero you just choose some random value of a of F right. Random non zero value of that. And then this tells us then that. Yeah. F minus AZ squared is isotropic. And that means that we can write by vids cancellation theorem, no no by vids hyperbolic theorem. So it's an isotropic one so we can split off a hyperbolic form so we have F minus AZ squared is isomorphic to some H direct some f crimes. Now, this form is isotropic. And think I think I'm going to get a little confused about this so better open up my notes. He just tells me I'm going to get confused in the course of this proof. All right. Right F minus AZ squared is isotropic, but F is isomorphic to G over over Q new for every new. So this also implies that, you know, F mu minus AZ squared is isotropic for all new, just because genius isomorphic to F mu, and therefore by part one, this implies that G minus AZ squared is isotropic over over Q. So now here we're using part one. We can use the bit theorem again and we get a similar statement for G so G minus AZ squared is isomorphic to hyperbolic plane direct some some G prime. Okay. Now we're going to use induction on the number of variables. So we're going to assume the statement is true for note note that the hyperbolic plane has dimension two and this has dimension one. So this has to prime has dimension one less than F or the number of variables is one less than F. On the other hand, so F is isomorphic to G. Therefore F minus AZ squared is isomorphic to G minus AZ squared. Therefore H direct some F prime is isomorphic to H direct some G prime. Therefore by bit cancellation F prime is isomorphic to G prime. Oh, I mean a over over mu for every new. But then by induction we get that F prime is isomorphic to G prime over Q. And then that means that then H direct some F prime was isomorphic to H direct some G prime. And then by bit cancellation again we can cross these terms out and we get that F is isomorphic to G. So, right, so we have F direct some something is isomorphic to H. And that's the same and these guys are the same so by bit cancellation the question of whether F is isomorphic to G is equivalent to the question of whether F prime is isomorphic to G prime. And that holds over both Q and all the confessions you knew. So in this way, you look at it and you see that you can conclude by induction. So yeah, if you're a little confused by the logic of it, you're not alone. I'm also a little confused I have to write it down every time. But you should just remember the general principles yeah the bit cancellation, and the trick of representability of being equivalent to isotropic. Yes, there's a question of various. No, oh, oh, there isn't. Oh, your hand is not. No, no, no, I didn't have a question I was just thinking. Oh, I thought there's a marker that says your hand is up. Oh, sorry, I forgot to lower it. Sorry about that. No problem. Okay. So, I want to make another remark. This is special to to degree two equations. Well, or it doesn't hold this theorem I mean it what I mean to say is it doesn't hold in general for higher degree equations. And in fact there's an example of Selmer that if you switch from degree two to degree three there's failure. So if you look at three x cube plus four y cube plus five z cube equals zero. This has non trivial solutions in q new for all new. But no non trivial solutions in q. So this is the failure of the local global principle. So, it's in general easier to detect whether or not you have solutions in q new than in q. So, this part showing that you have non trivial solutions in q new that you can do. So, the case that you have for example, what's it called the Hensel's lemma. Hensel's lemma gives a way to approach this part and in fact I put it as an exercise on the problem set to show that this equation does have solutions in q new for all new. On the other hand, showing that it doesn't have any solutions over q is is quite difficult so I didn't put that on the problem set. The correct context for this example is actually the theory of elliptic curves. This Selma equation gives something which is called a, it's a, a torsor for an elliptic curve over q which becomes a trivial torsor over q new for all new but is not trivial over q. It represents an element in what's called the Tate Shaffer average group of a certain elliptic curve and that's, and then the general theory of elliptic curves gives a way to, to check things like this. I mean it doesn't give a, it gives a potential way and you have to check in any example whether the algorithm actually terminates or not but in this case the algorithm does terminate. And it's conjectured to always terminate. So yeah, this is the general context for Selma's example is the theory of elliptic curves which is certain cubic equations is one degree of complexity higher than the quadratic equations we're talking about here. That's just to situate this in the massive realm of further mathematics. Okay. So, the next thing I want to remark. And again it's not a yeah it's just sort of a meta remark. The proof of this theorem is difficult. And besides a number of clever tricks you could say it also relies on two big theorems, two big theorems in number theory so one is, well the Hilbert product formula. Hilbert reciprocity, which we proved that well which we almost proved last time. And I see that the sun and one me again. Okay, there we go. And the second thing is Dirichlet's theorem on primes and arithmetic progressions. So I don't want to write it down but I'll say it in words what Dirichlet's theorem says, is that if you have a positive integer m, and you have a congruence class mod m, which is relatively primed. So if you have a congruence class of integers relatively prime down, then there does exist at least one prime number, living in that representing that congruence class. So for example, there's always going to be a prime congruent to one mod m there's always going to be a prime congruent to minus one mod m, and you know m is a prime and there's always a. And in fact it says something more it says there's infinitely many such problems. It says something even more than that but let me stop there. And proof uses analytic number theory, it's a beautiful proof and we'll have some of it at least some of the tools involved in giving the proof by the end of the next week. So that's my ambitious plan, we're going to be talking about Dirichlet L series at least and those are one of the ingredient but what the main ingredient in the proof of Dirichlet's theorem. Yeah, and it should be not quite obvious how you're ever going to use these two theorems to prove a local global principle so it all comes in the, you have to really look into the intricate nature of the argument to see why these two ingredients could possibly be used to prove such a statement to produce a rational solution to an equation, at least not obvious a priori to me that these would be helpful. Okay. Right. So now, before diving into the proof, I want to collect a few lemmas that are going to be useful in the course of the proof. So the proof will actually proof will be by induction on the number of variables. So the question step, you know, and to make the induction run you're going to sort of split up your equation into, you know, you can assume it's diagonalized right so you just only have terms Xi squared in it, and you're going to try to split it up and with a grouping of two, you'll separate out two of the variables and then the remainder of them and you'll try to use some induction of that form. So the crucial case is going to be somehow the low variable cases like two variables and three variables essentially to make the induction run. There's just going to be a couple of lemmas concerning. Yeah, such low variable, low variable equations. So, right, so here's one lemma. And this is a local lemma. So, let the be a place of you. A be in q new such that, yeah, I guess a x squared plus b y squared is non degenerate. So, and let also see in q new cross like I guess I should just take a that's the same saying the same thing so just make everything non zero and you new cross, then A x squared plus b y squared represents C, if and only if you have a certain equality of Hilbert symbols. So a comma B sub new is equal to C comma minus a times B sub new. So for the proof. Well, a x squared plus b y squared represents only if, well, a over C x squared plus B over C y squared represents one. And that's the same thing as saying that a over C x squared plus B over C y squared minus D squared is isotropic. And that by definition of the Hilbert symbol the same thing is saying that the Hilbert symbol a over C comma B over C and you is equal to one. Now we're just going to expand out this equation using the bilinearity of the Hilbert symbol. So, so this is the same thing as saying that a comma B I have to get it get it right right you have to be you know it's a little bit subtle this bilinearity. So the A B comma B C comma C equals A comma C B comma C. I should it should be C comma B but the Hilbert symbol is symmetric already B comma C. So these are kind of the cross terms there's like a paired with B and then C paired with C and it doesn't matter if it's divided because we're taking values in a group of order to so I keep it the way it is. So it turns into this when you use bilinearity. But now we can use that this is equal to a B comma C new, or maybe for purposes of proving the precise statement I had. Well, I should write it as C column AB. Anyway, and this here is equal to we can write it as C comma minus one times C comma minus C. But this term always drops out right. That was a general fact about Hilbert symbols that you know x comma minus x always gives you something trivial, or it's a general fact about Steinberg symbols in fact that he'll explain. And now we can move this over to the other side and use bilinearity again and we get that it's equivalent to AB new is equal to minus a A times B C new so a comma B new is equal to minus AB comma C just again by bilinearity. Okay. That's, that's the proof of that. And the other lemma is actually something that a keel explained but maybe not in full generality. So it's again about just degree two equations. So, well, now it's going to be over an arbitrary field because we'll be applying it for both the rational numbers and for the completions of the rational numbers, maybe I should say characteristic different from two. And again, I'll take a and be in k cross. Then, um, a x squared plus b y squared. My Z squared is isotropic. If and only if a is a norm from the quadratic extension K adjoin square root of B. So it's in the image of the norm map from K square root of B cross to K cross. So, well it's not necessarily a quadratic extension deep could be a square, in which case, K adjoin square root of B is K and this condition is vacuous, but you can also see that if he is a square than this condition is back to us so the, the key cases when he is not a square. And this really is a quadratic extension. And I don't want to give the full proof because it doesn't involve breaking up into cases and such. But let me just give the, I mean, let's just let's just say a say B is not a square so this is quadratic. And assume this is isotropic so then you have again you can solve a x squared plus b y squared equals one. And I'll tell you then that a is, you can just solve for a right it's one over x squared minus B times y over x squared. And then you write this, and you recognize this as a difference of squares. One over x minus b y over x minus square root of b y over x times one over x plus where it'll be y over x. And that's exactly the same thing as the norm of the element of one of our x plus word. So the norm is the of an element in the field of a quadratic extension is the product of the element by its conjugate. So, so this, this give the baby and it's not too difficult to reverse the implications law so this is just I don't want to give the full details in this, but just to give a basic idea of what's going on. And the reason this is a useful statement is that this, the norm map is a homomorphism so this subgroup of K cross is closed under multiplication. And that's what we're going to use. So we're going to be fixing B and letting a very and we now we know that the set of collection of a for which this form is isotropic is actually a subgroup of K cross. So that's helpful. That's the non trivial conclusion we get from this one. Okay. So this is another set of lemmas we're going to need but instead of introducing them at the beginning I'd rather only talk about them when they're when their need arises. So, let me get started on the proof. So, so proof of us and Kowski. So, well we have a quadratic form over Q. So we can write F as, you know, we can diagonalize that was the one of the very first theorems we saw about quadratic spaces you can always choose a basis where it's a diagonal. In the language of quadratic forms that means after a linear change of variables you can get rid of all the cross terms. So in all these ai's and Q cross. Now I want to just at the very beginning use in general reductions we can simplify, we can, we can massage this a bit to restrict what the ai's can be to make our task easier. So we're called it well, we can always modify each ai by a square. Right that's just by another linear change of variables or you just scale the corresponding variable. So using that you can in particular clear denominators right because you can clear denominators by multiplying by the square of the denominator that will certainly clear the denominators. So you can assume all ai's are actually integers, non zero integers. But then again, we can for free to modify the ai's by squares. We can even square free angels. So their products of distinct primes, we can assume all of the ai's are products of distinct primes. Oh sorry I meant to just a sec, I meant to do it. This was, I meant to do a reduction first we can always assume that ai is equal to one if we want. We're only trying, we're only trying, we're proving the first part of Hasan Minkowski's we're only interested in whether F represents zero non trivially so we can always divide by a constant and it doesn't affect the question. So by dividing by a one at the beginning we can assume a one is equal to one. And then we use this argument for the rest of the variables and we can assume we see that we can assume that the rest of the variables are square free integers. Okay. All the rest of this argument was just variable by variable. Yes. I was wondering whether you could please quickly explain this, this reduction where the first, well, where this a one can be set equal to one. I'm not sure that I, that I fully got it. Yes. So, the forums F and and a one inverse times F. The question of whether they have a non trivial zero is equivalent. I get it. I understand it. Thank you. Yeah, yeah, you just use. So now we're again we're going to proceed by induction on the number of variables but the induction is such that such that we're actually going to have to do the first three cases by hand, or even the first fork it up. We're going to have to do the first three cases by hand sort of an induction induction is that complicated. Because of the form of the tricks using the inductive step. So we need to handle and n equals one and two are not too difficult so for n equals one. Well I just said we can assume that f is x one squared right. And, well, there's no non trivial zero over any field. So it's vacuous right there's no, there's no one, there's no form that satisfies the high form form of a form in one variable which actually satisfies the hypotheses. So there's nothing to prove. So n equals two is the first real test. So then we have f is a x one squared plus a two x two squared, and a two is a square free integer. So now what does it mean that f represents zero. Well, if that is isotropic over q new, then that tells you just by solving for zero equals x one squared plus solving for a two in the expression that you get by setting this equal to zero. So you get that a two is a square, non zero square and q new. And that tells you so there was maybe some activity in the chat, is it keel here or not, because that affects whether I just, oh, it seems like everybody I haven't read it in detail but it seems like people have resolved their confusions in chat. If there's a confusion in chat that doesn't get resolved by the other participants and just do please bring it to my attention because I actually have trouble following the chat when I'm lecturing. So, right. And that tells you then, well, first of all that a two is positive. That's what you learn when new is equal to infinity. So the squares and the real numbers are just the positive numbers, but you also learn that the pediatric valuation of a two is even for all crimes P. So, you have a positive square free integer, which has even pediatric valuation for all P. So it would be one. So it's a product of distinct primes. If there are any primes occurring and the pediatric valuation for that prime would be would be one right, which is not even. So, yeah. And yeah, so it's a plus or minus one times a product of distinct primes but since it's positive. Yes, there's a question. I didn't see who asked but I saw it appear. Sorry, should be negatively to or negative. Oh, yes, thank you. Yes, of course. Yeah. Yeah, thank you so much. Thank you very much. Yeah. So, a two is then equal to minus one, the only possibility or minus a two is equal to one. Yeah, and that tells you that f is the hyperbolic plane, which represents zero over any field. In particular over the rational numbers. I mean it's isotropic. Right. So, there was some activity going on here right basically the claim is that a rational number is a square if and only if it's a square and q new for all of you. And there's something going on there right it's not completely trivial but on the other hand it's fairly elementary. So that was the n equals to face there was some activity but not much. So the real the real work is going to start when we look at the case of three variables. So, so again we'll have x one squared plus a two x two squared plus a three x three squared. Let me rename since we only have three variables let me rename things a bit. Let's say that this is a, I don't know, a x squared plus. I could have made this minus one if I want. So I'll put it in the form we had earlier. I could have made the first coefficient minus one if I wanted. And again, a and b are square free integers. Now we're going to. We will prove the desired statement by induction on absolute value of a plus absolute value of B. So, so what can we make this. What can we make this change of relations a x equals to minus a. It's because when the argument I gave for the general reduction I said you can make the first coefficient equal to one. But you can actually just as well make the first coefficient equal to any non zero scalar, you could also make it equal to minus one. It's just the same argument you divide through by minus one and it doesn't change the question of whether the form is isotropic. It doesn't directly follow from what I said but it follows from the same reasoning we can make one of the coefficients, any arbitrary thing we like in particular we can make it minus one. And then we do the same argument for the other two coefficients to make them square free integers. Yeah. Okay. So, to do this right. All right so we're doing an induction so I should probably handle the base case. I guess the base case is when absolute value of a plus absolute value B is less than equal to well, it has to be equal to two. Because these are non zero integers. Then there are only four cases right so a and B are both equal to plus or minus one. And, right, and the only case. So there's not a solution in Q is where a equals minus one and be equals minus one. But then there's no solution in our. So, so the local global principle holds and in fact you only need to look at the one place the real numbers. So we're assuming that a plus B is greater than three. And the key claim where the dilemma I want to prove is that. Yeah, a is a square mod B. So right. So, B is a product of distinct prime since the square free. So by the Chinese remainder theorem. So the choice is to show that a is a square mod P for all crimes P dividing B. Now, to do this we're going to use the hypothesis that a x squared plus be y squared. Equal zero has a solution. Sometimes I'm going to say has a solution when I mean has a non trivial solution. I hope it's not too confusing in QP. Well we can always this is a homogeneous equation. So we can always clear denominators and get that we have a solution in ZP. And then we can further again because of homogeneity we can further arrange that. The solution is primitive so. So where, say VP of X, you know, either, or VP of Z is equal to zero. So we can always, just by multiplying by powers of P, you can always arrange this kind of stuff. So primitive solution. But now, now recall that P divides B. So, we see if I can arrange this properly. So, we have, we have that P divides will be times why and therefore P divides a x squared minus z squared. What I want to say is that then. I want to say that you can divide. So in other words, a x squared minus z squared is congruent to zero mod P. And I want to say that this or a x squared is congruent to z squared mod P and I want to say that this implies that a is a square mod P. So, and you're done. So I need to justify that you can divide through by x, but so I need to justify that x is a unit in ZP or in other words that x is not divisible by P, but if x were divisible by P then Z would be divisible by P. And then, if Z is divisible by P, then. So you would, you have to look at it a little bit to figure it out, but you get a contradiction to primitivity then you'd be able to cancel up further P from all of the three variables here. So, again, I don't want to go into all the little details of the group, but this is the main idea here. So you get some, since P divides B you get P divides this which gives you exactly the kind of problems you're looking for. And you need to make sure everything really works but it does. So that was the proof that A is a square mod B. And now let's write out what this means explicitly. So this gives you a B prime in Z, such that. So A is congruent to, sorry, such that B prime in Z and an X in Z. I shouldn't use X. I'm sorry guys. What should I use that T. So A minus T squared is equal to B times B prime. Okay. And now we're going to use this lemma we had about this, this equation having a solution over any field this equation is going to have a solution. If and only if A is a norm from Q adjoin square root of B or actually I'm switching the rules of A and B for the purposes of this argument. So, don't note that we can recognize this as a norm. So this is a, again, as a difference of squares we can write this as a square root of a minus T times minus squared a, what is it? Why am I having trouble writing differences of squares. Sorry, let me flip it around and write T squared minus a T squared minus a so I don't get myself confused. So that's T minus square root of a times T plus four of a. So this thing is a norm from Q and U adjoin square root of a. So then that tells you that the fact that this form here is isotropic implies and this fact is in equivalent to saying that the form AX squared plus B prime y squared minus Z squared equals zero is is isotropic and Q and U for all new, but you can also run the same argument over Q and see that the question of isotropy over Q for these two forms is equivalent. Yeah, so if there is there's a question. Yeah, it's actually just at the end of your proof of that that intermediate lemma about how a has to be a square mod B at the very end. So you we have this equation AX squared is equal to Z squared mod P and then you divide by X squared on both sides. But isn't it possible that well VP of X is not equal to zero. I mean we said before that VP of X is equal to zero or VP of Y or VP of Z. Yeah, that was just my that was my very primitive way of saying what a primitive solution mean but if you look at the specifics of the equation so first of all. So if you look at the equation so. So if P, if P divided X, then he would also have to divide Z, right. Yeah, because it divides P as well. So yeah exactly. Yeah. And so then. Then this would have odd valuation and this would have odd valuation. So if this were if this were the one that was a unit. So the only possibility to rule out is that if this is the one that's a unit, but then if this one is a unit this would have even p-atic valuation and this would have even p-atic valuation. But then this would have odd this would have odd p-atic valuation because p only divides b once and then you'd have a contradiction to this being equal to zero which has even p-atic valuation. So you need to look a little bit about the specifics of the equation and use the fact that p is only dividing b once but I believe me it I didn't explain why but it checks. So basically in what you said if if we assume that p doesn't did that p divides x then we would get on this equation a x square plus b y squared minus s squared is equal to zero we would get p-atic valuation equal to one on the left hand side but well not equal to one on the right hand side right? Yeah exactly so yeah I mean you'd be able to you'd see that there's a p in all of x, y and z basically and then you'd be able to cancel that p and make it more primitive yeah yeah I see thanks sure right so what I was just explaining this trick that I call the bb prime switcheroo. So if you have an equation bb prime equals t squared minus a then your questions of isotropicity of these quadratic forms for b and b prime are equivalent by this lemma about the norms we stated earlier. So we can actually transfer our question from b to b prime and now the claim is going to be that we've made our thing we're inducting on swap smaller by switching from b to b prime and to show this you just you know solve for b prime and yeah and well we have to choose t a little bit more carefully right so we're we're making an we're trying to solve an equation mod b we can always assume that t is less than or equal to absolute value of b over 2 right by choosing a minimal or residue class representing the square but if you if you choose t like this and you solve for b prime you will learn that absolute value of b prime is smaller than the absolute value of b and then you conclude the argument by induction. So since we know the Hasan Minkowski theorem for this equation because it's smaller because the coefficients are smaller we get that we have a a non-trivial zero for this one but that's the same thing as having a non-trivial zero for this one by the b b prime switcheroo. So that is the essence of the argument and I it looks like I'm not even getting to n equals 4 here which involves kind of a different set of tricks. So I'll tell you what why don't I continue continued next time and it's not as bad an idea as it sounds because I was kicking some of the crucial lemmas to the exercises so this way you'll be you guys couldn't be doing some of the legwork for me. There's an interesting set of theorems which goes by the name weak approximation maybe I'll give you a little introduction to what's on the exercise since it's not coming up in the lecture so eventually we're going to have to go from some local solution of an equation to some global solution of an equation right. And the forms of the equations are such they were always free to modify by squares and so that well there's a there's a theorem called weak approximation so which is all about going from a local local thing to something global so let me just state the theorem so if I'll state one form of the theorem if s is a finite set of places of q and x v in q v or nu sorry epsilon v in r greater than zero and you're given such things and x nu and q nu and epsilon nu in r greater than zero for all nu and s are given then there exists a global thing in x and q such that it's epsilon nu away from x nu for all nu in s or in other words the map from the rational numbers to the product over nu and s of q nu is dense so the inclusion of q and the product of finitely many of its completions is always dense so we know this for one if s has one element because by definition these q news were completions of q so q is dense in them but the claim is actually even if you just if you take finitely many it's still dense so you can approximate yeah like this and yeah I did also didn't I told you that so that's there are variants on this theorem and some of them appear in your problem set and we'll be using them in the next lecture to deal with the higher variable cases and in particular I haven't yet used you'll notice in n equals one two and three we haven't used silver product forming we haven't used Dirichlet's theorems on primary arithmetic progression so we haven't somehow hit the the key points yet and they'll be thanks to theorems like this that will be able to use this the product formula but this is this is the way one way you can go from q nu to q and the fact that you only get it up to an epsilon is not going to be a big deal because another exercise on your problem set will tell you that if you're a if you're epsilon if you have a fixed number x nu and you have anything else epsilon close to it if then that one will be a square if only if well if you know sorry if you have yeah then then they'll differ by squares so multiplicative please and then with our quadratic equations we can always replace things by squares without changing things so well I feel like all of this will make more sense in the next lecture okay so that's all for today are there any questions soon the last the last thing I wrote let me share my screen again I didn't I didn't quite hear your question so if you could repeat it I'd be grateful uh the norm is the new the new norm yes thank you so much it's the new attic norm yes exactly yep uh yes I'll affairs just you know thinking about this this weak approximation theorem and how maybe I'm just wondering if there's like some kind of similar more general result in the sense that like if I have uh in the general like for general metric spaces if I have a you know a dense isometric embedding of a metric space into its completion isn't it would would it or would it not be true in general that if I well basically if I took an I don't know it would be an embedding if I basically took a product of those embeddings where I send that space to like one two three of its completions and in the first coordinate it's the first isometric embedding and in the second coordinate it's the second one and in the third coordinate it's the third one would that map also be dense just make it a bit like as a generalization of that no no this is not a general but let me give you two examples which are both very important in number theory um so one example would be if you take just uh integers with the prime p inverted uh as your your metric space uh well no I mean I guess you have two different metrics on it right so what I want to say is that if you complete this with the periodic metric you still get all of qp you don't need all of q to fill out qp it's enough to have integers and then be able to divide by p right so that that is also so this is still dense in qp on the other hand it's also still dense in r um so both qp and r are completions of z bracket one over p however z bracket one over p sitting inside qp cross r this is a very good exercise uh this is not dense uh in fact the the quotient is is house dwarf and in fact also compact that's so it sits very nicely in there so and this is analogous to z sitting inside r so z sits beautifully inside r and the reason is they're not the integers are not too scrunched together so uh the quotient is house dwarf and um but they're still relatively close to each other so the quotient is compact so that's why z sits beautifully in r and in a similar manner z bracket one over p sits beautifully in qp cross r a very good exercise you can also generalize it to finite sets of problems um now let me give you another example again important in number theory um so let's take uh the ring of integers of the quadratic extension to join but like you got by adjoining the square root of two just for example um um this uh is again dense inside the reels so well because you can make rational numbers arbitrarily close to the square root of two basically um let's take that as an exercise as well z bracket square root of two is dense inside the real numbers but it's also dense inside the real numbers in a different way but via a different injection a different ring homomorphism where you send square root of two to minus square root of two so you compose with the the conjugation map you have on this ring before you include into the real numbers so so in other words this quadratic extension q square root of two we say it has two real places which differ by uh by Galois conjugation so it sits inside the reels by just by definition but on the other hand you're algebraically speaking you can just develop a square root of two by minus square root of two and then that gets a different embedding in the real numbers and again we have the similar situation so z bracket square root of two sitting inside r cross r is uh yeah uh is discreet and discreet so the quotient is house store and co-compact so the quotient is compact so if you I encourage you to plot this it's very beautiful you can make the z bracket square root of two you can it looks all scrunched up when you just have one copy of the real numbers when you have the other copy of the real numbers you get a beautiful lattice and I encourage you to plot it it's it's quite fun uh but you'll see all of a sudden it pops out in forms of lattice and now you can really see the elements of z bracket square root of two and this is extremely important for for studying the arithmetic of these uh real quadratic fields um yes so there is and just to be clear by the quotient you mean the quotient of the target of the target space modular the image of the of the map yes and the map is an injection so um yeah so that is uh I see oh thanks you're welcome by the way you could ask what the analog is so for the rational numbers so apparently it's not enough to just take finitely many places you know you don't get a beautiful picture of the rational numbers to get a beautiful picture of the rational numbers you need to do something more involved called the Adele's I won't tell you what it is but you can look it up it's a very clever way of combining all the chaotic numbers for all p with the real numbers and here again it's discreet and co-compact and that is why the Adele's are essentially why the Adele's are so useful that's how you want it if you if you want to visualize the Adele's so to say I've visualized the rational numbers so to speak in the same way as you visualize the integers you have to use the Adele's replacing the real numbers but of course so it's difficult for everyone to visualize the Adele's so this is kind of um yeah yes Dustin yes uh maybe an even simpler example uh would be the rational's embedded in the reels and then take two copies of that so the rational a rational number will now be embedded in reels cross reels by sending it to the diagonal delirium the rational itself or you could compose it with reflection in the reels and then it's on the anti-diagonal yeah clearly not dense so um yes indeed yeah so I was giving examples where what the embedding is not dense but yet it's still fundamental and very useful so that was doing more than was required by the question yeah there are also these other kinds of examples from that you shouldn't expect then for the in general thank you yes thank you I had another question about the Hassan Minkowski theorem yes is it also possible if you are given the solutions over all pediatric fields uh to like construct a solution over Q or is it only telling something about the existence it's only telling you something about the existence so um yeah I don't know I'm not one of the kinds of people who study these questions about making things explicit you know what I mean so all I can say is that if you look at the argument it doesn't tell you how to find the zero it just proves that one exists I don't know if you're sufficiently clever and have a sufficiently detailed understanding of all the ingredients in the argument whether you can with a lot of work turn it into an algorithm that is very much not clear to me whether that's possible or not and in particular I I don't know the whether someone has done it I maybe a kilo do you know this kind of thing uh sorry algorithm for what I oh like like finding a rational solution when you which you know one exists uh I yeah I don't know sorry okay um I guess maybe yeah I mean I guess I mean you can always just search I guess you can always search for a solution with brute force right like but you know order by height meaning you know absolute value of the numerator plus absolute value of the denominator just brutally and if you know there's a solution you're gonna hit one eventually but I don't know if there's an algorithm that's guaranteed to terminate in finite I mean in amount of time that you can a priori bound um yeah okay thanks no that's a question sorry was there another question uh yes so this like uh this weak approximation is about like embedding q and two this product of this finite product right so like is it possible to like combine all of the like because right as as it sounds now it's like a statement for each finite subset so is there a nice way to like combine everything into a single statement or um I mean in general finite subsets often arise and so it's like the you know you use flexibility in choosing the finite subset but there is a way indeed again using the Adele's so there's a if you can you can look at the definition of Adele's and you can also take the Adele's but remove one of the factors like say q5 or something like this and then the another form of weak approximation will tell you that you know so q is just sits in discreetly inside the Adele's but if you remove just one of the factors then it sits densely inside um any one of the factors so this is kind of uh um but then again you have this arbitrary one factor you're emitting and so it's just in some sense just as bad as your arbitrary finite set I don't know um yeah I know I think some other yeah I don't think there's any um I don't know I basically don't know okay so I guess the lesson is that Adele's are like a nice way to think about are like things like this I think so but they take a lot of getting used to and there is something to be said for keeping things I don't know I think I think you can spend your whole life getting really comfortable with the Adele's and then some people have done it and now they write all their papers with Adele's you know I don't know okay thank you sure yeah also Matthew is asking if there's a proof of possum in Kowski which is less ad hoc and Kiel says if you yeah Kiel says if you do it for all feel a number of fields at once you can use class field theory to help um but I don't really know any argument that I would consider not ad hoc um well I guess I just meant that then you for example you're not using uh Dirichlet's theorem um and I guess maybe you're also not using this sort of inductive argument at n equals three though um but yeah but I guess you're still sort of inducting on the number of variables and then there's a but then you can say okay are there any proofs of the theorems of class field theory that are not ad hoc and that's that's a whole another question but uh and yeah um yeah so there is I also wanted to ask why is the weak approximation theorem called the weak approximation theorem it's because there's something called the strong approximation theorem which I believe is the statement I just said with the Adele's I'm not so up on the terminology uh over that noise that one's still weak uh let's google it here let me google that for you know sorry I'll do it too thanks no you can let me you google strong approximation theorem and see what comes up uh I think it's the same one I just explained I think strong approximation lets you control um at the places the the finite places for for a finite subset still um but also allowing you to have um something with norm less than one at all the other places or less than equal to one okay so this is a statement of this kind of forums on your problem set so um yeah uh what is weak approximation what is strong approximation anyway I'll let you guys google it it's been a long time since I've looked into this but yeah I guess that's exactly given you um an adelaic statement right because then at all but finally many places you have a pediatric integer and I'll just the number yeah there's a question then of like um how well all but finally many can you get because in the in the form I have in the problem set you can actually make it all except one and that uses Dirichlet's theorem on primes and arithmetic progressions but yeah if you just have it for all but some unspecified finite number then it is the same I guess the same the same for the Adele's um by what you just said yeah oh yes uh singh daa and uh I just want to ask uh so uh the last place of what you're doing so uh so uh you you said q's it's inside the finite product uh uh of progress q come q new so uh I want to ask how does it say what is the explicit map the the map from q to the product over new of q new yeah yeah yeah yes good question so on the new factor it's just the inclusion of q into q new also uh so the lattice of uh z root two uh you do in r cross r uh I suppose uh you're taking so z root two is a two dimensional module over z so I suppose what you're trying to draw there is uh you're taking one of the basis as root two and one of the basis is one and then you're going to break it forward some like it's yes indeed so there's another way of thinking about this instead of yeah I mean yeah you know that every lattice indeed fits inside a real vector space of the same dimension and you can just tensor up and you get it right so if m is free of rank two and you take m tensor r then you get a beautiful real vector space in which it sits and then this can be rephrased as identifying what z bracket square root of two tensor r is namely is just two copies of r um to I mean as a ring even right so as it's r cross r as a ring the of these two different real embeddings of z brackets square root of two so yeah there's two different ways of of looking at this I actually do prefer the way you you said where you you have the tensor product with r a priori and then you try to ask what it is explicitly um yes I think that's the best best perspective I think a cast yeah to answer Akhil in the comments I think castles for like is the canonical reference and since it's been the canonical reference for so long probably no one's tried to do anything better than it at least they haven't to my understanding or to my knowledge I guess authors are there like aside from this some cubic analog of Hasan Minkowski not actually holding because of zelmers example are there like other cases where local global principle just fails it's more you shouldn't rather be asking if there are more cases where it holds that as well because it it fails as a as a general rule so it's really only in specific cases that it holds but in there are also other cases where it where it holds um so uh yeah but I'm not I'm actually not the right person to ask this I'm not yeah um why don't you look at I think maybe there's a good discussion so Poonan has no yarn Poonan I'll type it in the chat Poonan has notes um oh uh rational points on varieties I think I think that's what it's called um and anything Poonan writes is just exceptionally clear um and this is exactly about the topic of the general study of rational points on varieties which is kind of what this question is about and I think there's a section in there that discusses um Hasan Minkowski uh someone who's founded in Linkwood thank you very much Elperius yes um and I recommend those notes for all sorts of purposes um but in particular for this question I think I think there's a discussion of what in general holds I think thank you yeah and Akhil's putting pointing out another good example yeah yeah yeah right there is this yeah there is some statement describing W of Q in terms of the W of the Q News some analog of Tate's calculation basically yeah I don't know I don't know it off the top of my head but yeah but again I don't think that yeah maybe I think Akhil touched on this earlier I don't quite know how to go from that statement to Hasan Minkowski oh yes uh Garrett okay you you sound like a robot yeah I can't hear it's a little better or no it's not okay I don't know what's going on but it sounds like you're speaking robot language I'm sorry it's really weird I wish you could hear yourself maybe you can type your question in the chat the great right great what's so special about quadratic forms you can go from I don't know uh well I don't know how to answer that question the proof works is kind of the silly answer I don't know yeah uh I mean the quadratic forms are simple enough that you have a hope I guess I don't know I'm with any really have to check and see I don't Akhil do you have a better answer I'm sorry I was responding to an earlier question um well I guess quadratic forms uh I mean I guess they are like torsers over the orthogonal group and I think there are these general statements for for torsers which are supposed to be generalizations of that but yeah I also don't know what the sort of state of the art on Hassan Minkowski um for more general um varieties is is Garrett is your question why quadratic forms in particular and not say cubic or higher forms don't apologize Garrett it's not your fault I'm sure and it was kind of fun um any more sorry would that question be like um Fermat's last theorem in when we have exponents too then we have many solutions but when exponents are bigger than we don't is that what's happening here well I don't know if there's a I don't know if there's a relation between those phenomena um maybe I guess just one time go ahead I'm sorry just to go back to like this homological I mean maybe one idea is that quadratic form since they they form a nice algebraic structure this bit ring which is closely related to like the homology of the field whereas if you're studying some you know other general class of varieties then there's sort of no analog of the bit ring and the bit ring I mean the I mean the Hassan Minkowski can I guess so if you formulate it for all number of fields at once it is equivalent to a statement about the bit ring and then sorry it's going to follow from a stupid book sort of Galois homology which is part of the the class field theory setup but I don't know that that was just thought right that's a really good point yeah um but there's one general setting in which you do have a local global principle and it's um Galois homology so you can kind of well at least you don't always have it literally in that statement you can at least understand the discrepancy between the global and the local this you can say things about and there are beautiful general theorems describing what's going on there so as opposed to the case of algebraic varieties points on varieties which is there aren't uh general descriptions there's just a vast zoo phenomena so I guess what Akila is saying is that one should hope for a local global principle in the setting of varieties if you can rephrase the existence of a rational point um in terms of some kind of Galois homology or some similar thing like some similar invariant like a bit wringled whatever that's yeah that's I like that answer a lot thanks Akila thank you for all the answers for everything okay so see you in some other activities I do have office hours right now but I don't have evening office hours on Friday and I'll change them for next week you'll find out see you guys