 So, hello everyone. The topic of this lecture is going to be one of my favorite theorems and mathematics, both for the statement and for the proof. I think I'm not going to go into full details on the proof but I do want to say enough to give you an idea of what all the ingredients are and why the various. There's some various funny things that arise and I want to make them kind of make sense at the very least. So, but first let me just tie up some loose ends from last time. So, so we had D was a fundamental discriminant negative. And we had that the set of strict equivalence classes of positive definite binary quadratic forms of discriminant D. And there's one different interpretation of this in terms of quadratic the quadratic field F, which is a Q adjoining square D. So this was invite this that was naturally in bijection with a set of ideal classes. So, non zero ideals. In the ring of integers of F. I equivalent to J, if and only if there exists some non zero scalar alpha and F cross such that alpha I is equal to J. And this is so called the ideal classroom. Or well. And well actually yeah. So we got a little better than this we got a statement without just passing to the equivalence relation we actually described all strict equivalences between binary quadratic forms in these terms, and in particular I also that gives you a statement on automorphism so you know the special orthogonal group of any such final quadratic form will be isomorphic to the will okay module isomorphisms from the corresponding ideal to itself. And then you can see you can show that this is the same as the okay module isomorphisms from okay to itself. And this is the units in okay. And I don't want to go into all the derivations of this but this is also just the same thing as a group called okay. Oh, F became K. Well, you know what, that could happen more than once. So let me say that F equals K equals Q adjoining square root of B. So, the only field we're going to be talking about is is this one so I hope it won't cause any confusion. Right, so it's the group of roots of unity in K. It's independently of the ideal or another in other words independently of the quadratic form you talk about. And this is actually a very small group so it's plus or minus one if D is different from minus three and minus four. You know, Mu three so that I mean six the group of six roots of unity is minus three, and it's new for the group of fourth roots of unity if D is minus four. So it's some very explicitly understandable thing. So describing the automorphisms, but okay this is, this is kind of a parenthetical. Let's go back to the main point here. So we have these strict isomorphism, a strict equivalence class of binary quadratic forms is the same as this thing over here, the so-called class group. And the important thing about the class group is that it's an abelian group. So the set of strict equivalence classes of positive definite binary quadratic forms of discriminant D has a natural abelian group structure just induced from on this side, the product of ideals. It's actually a little bit complicated to translate it back over to the length of binary quadratic forms, although it can be done. Now I wanna say that this abelian group structure sort of explains a lot of, or gives the answer to a lot of natural questions that arise in this theory of binary quadratic forms. So there were some questions we had earlier that we can kind of answer in terms of the group structure. So question one. So given F and G binary quadratic forms, positive definite binary quadratic forms of discriminant D, when is F not strictly equivalent, but just non-strictly equivalent. So there's a priori more natural notion of just being isomorphic under an arbitrary linear change of, or invertible linear change of variables, not necessarily one line in SL2Z. And the answer is if and only if either, so, well, so either they are strictly equivalent, right? That's one possibility. If they're strictly equivalent, then they're conjugate under GL2Z. Or in other words, if their classes in this finite agreement groups are equal, then this happens. And the other way, the other only way it can happen is if the classes are inverses in terms of the abelian group structure. And I didn't put this on the problem set, but it is a very nice exercise. So if you're interested in this and you wanna think about it, you can do that. So in other words, the set of, so the set of isomorphism classes of binary quadratic forms without the strictness or without the orientation condition is just a quotient of the ideal class group, modern equivalence relation, which identifies every element of the group with its inverse. Now, this is a funny equivalence relation. So it's not modding up by normal subgroups. You don't get a group structure anymore once you do this. And so if you were to not do SL2Z equivalence, but GL2Z equivalence, then you'd lose this really awesome thing in this abelian group structure. So you, so this gives one justification for why you wanna look at strict equivalence and not utilize the morphism. Because this fundamental structure, it is just kind of not visible if you impose too many, to identify too many objects. Okay, and the other question, so given FG, again positive definite binary quadratic forms of discriminant D, when are F and G in the same genus? So that means that they're isomorphic over the piatic integers for all primes P. So we're kind of asking about the, yeah. So being in the same genus, but not being isomorphic is kind of like giving a counter example to the integral form of the Hasselman-Kauski theorem. So it's a very natural question about this. And this is also gonna be answered very nicely in terms of the group structure. So if and only if, so if you take F, that F is equal to G times some X squared for some X in a class group of OF. So if and only if F and G differ by a square. So actually, so that means that sort of the genus class group is just equal to the class group of okay modulo of squares. So that is a finite opinion group. So classifying according to genus, strict equivalence according to genus is kind of a nice and reasonable thing to do. And this is again, that I wanna emphasize that the thing, this kind of classification is the one that generalizes better to quadratic forms of higher number of variables. So this was just to give, I don't wanna explain why either of these is true. This one's actually quite difficult. So I don't take this as an exercise unless you're a Gauss, I don't know. But yeah, I just wanted to say that this group structure gives a lot of information. But now let's move to the actual public of the lecture which is your place class number formula, which gives another, the proof of which also goes through this correspondence but doesn't quite make use of the group structure per se but makes use of the alternative interpretation in terms of ideals. So your choice class number formulas. So from now on, there exists a version of it for arbitrary D but I'm just gonna take the simplest special case. So for now on take P a prime congruent to three mod four and we'll let D be minus P. So we're taking the case when D is one mod four and it's square free and the simplest kind of square free number is a, well, simplest kind of square free negative number is just minus a prime. So that's what we're looking at here. Then the theorem is as follows. It gives a closed formula for H minus P. So recall, this is the number of elements in the ideal class group of, so that's by definition. So this number is equal to and it gives a closed form expression for it, which I think is very remarkable. So I'm sorry. It gives a closed form expression. Yeah, so let me, okay. Let me write mu minus P for the number of roots of unity Q of going square root of minus P, which is just equal to two if P is different from three and six if P is equal to three. So, but just to have a uniform freedom. So then the correct thing you want to find a formula for is not H minus P in isolation, but H minus P divided by mu minus P. And I'll make a comment about why that's the natural thing in just a second, but the formula for this, Dirichlet's formula for this says that it's equal to minus one over two P times the sum from K equals one to P minus one of the Legendre symbol K on P times K. So this is just something you can just, you know, bang out by hand, right? I mean, given your prime number P, you just calculate all the Legendre symbols, multiply them by weight them by the integer, the least positive residue and then divide that minus one or two P times that. It's a priori and integer, right? It's not obvious that it's greater than or equal to zero. I mean, greater than equal to one actually, but it's positive. And I'm not aware of any proof, any sort of independent proof that this quantity is positive other than showing that it equals, or no, not an integer, a rational number of course. I mean, up to the, I mean, when you cancel the mu P, so how should I say? Yeah. So it's a very, very strange, very strange kind of expression. And so I want to make a remark about why the right thing, the right, the thing that should have a formula is the thing on the left-hand side. So just a small remark. So H minus P over mu minus P is what's called the groupoid cardinality of the ideal class groupoid. So you don't just have an ideal class group. You have, you know, that's where you take ideals up to this equivalence relation, but you actually have a groupoid where you have the objects of your groupoid are ideals and the morphisms are these equivalences given by multiplication by alpha, alpha and F cross. And for a general groupoid, the groupoid cardinality is, so for a general groupoid C, a groupoid cardinality, a finite groupoid, is the sum overall, so X in C representing isomorphism classes. So you take a sum over a set of representatives for the isomorphism classes and then you wait by the automorphism group of that representative. And it's this number that sort of behaves well, like has good general properties in the world of finite group oids. So I don't wanna explain too much about why that's the case, but this is kind of just some general theory tells you that this is the correct thing to be looking at. And in the special case of the ideal class group, all of the automorphism groups have the same cardinality, namely this mu minus P. And so this just reduces to the count of the number of isomorphism classes divided by that common cardinality of the automorphism group. But again, when you move to higher numbers of variables, you're not gonna have a group structure anymore, but you still do have a group void of binary quadratic forms up to strict equivalences in the same genus. And the thing you have a formula for is exactly this group void cardinality and that's called the Siegel mass formula. So yeah, so this is kind of the correct general perspective here, but that's just a remark, we're not really gonna do anything with that. Right, so now, where does this formula come from? It comes from combining three separate ingredients based on the following analytic definitions. So the first definition is of the Dedek and Zeta function. So Zeta function F of S, it's the function of a real variable, let's say a function of a real variable S defined by an infinite sum. So you take the sum over all non-zero ideals in the ring of integers and one divided by the norm of the ideal to the S. And you should compare with the Riemann Zeta function, which is the sum overall, natural number is greater than or equal to one of one over M to the S. Which you could write, I guess, as a Zeta Q of S. So if you take the analogous definition for replacing F by the rational numbers, you'll just get this here. And the other definition, the other analytic function that's gonna be important for us is the so-called Dirichlet L series with respect to the quadratic character, chi. So this will be the Legendre symbol. So chi is the Legendre symbol viewed as a function from Z mod PZ cross to plus or minus one, and you extend by zero to a function from Z mod PZ to minus one, minus one comma one comma zero. So you define it to be zero on the missing residue class here, I mean zero. So you say it's declared that it goes from, that it sends zero to zero. Then the definition of the Dirichlet L series, it's like a twisted form of the Riemann Zeta function. So it's the sum overall N greater than or equal to one of chi of N over N to the S. And all of these converge absolutely, or both of these I should say converge absolutely for S bigger than one. Okay, now Dirichlet theorem is a consequence of three separate ingredients which relate to these analytic functions here. So ingredient one is that the Zeta function of F, oops, Zeta function of F is the product of the Riemann Zeta function with this quadratic L series. Ingredient two is that, well you may know if you've studied the Riemann Zeta function, that it has a simple pole at S equals, so it converges for S greater than one, a real part of S greater than one if you wanna use complex variables. And it has a simple pole at S equals one with residue equal to one. So let me rewrite this kind of in real variable language. So the limit as S tends to one of S minus one times the Riemann Zeta function is equal to one. And now let me write down the analogous expression for the, this quadratic field F. So this is, so let me just recall F is Q adjoining the square root of minus P. So this is equal to two pi divided by the square root of P times exactly this quantity, class number divided by mu minus P. So this is kind of really where this, where this expression appears in the whole story. And then the ingredient three is that, if you take the limit as S goes to one plus of just the Dirichlet L series, L S chi, you get an expression that is, now I have to look at my notes to make sure I get it right. Yes. Minus pi divided by P to the three halves, summing from K equals one to P minus one pi of K times K. Okay. Now this ingredient three, this is your problem set today. So I split this, this is a big long calculation which I split into five pieces and put those as the five problems on the problem set. Yes, so there it goes. In the ingredient three, where does S go in the limit? One, always S goes to one. One plus, right? From the rest. Yeah, one from above because it's only defined when S is bigger than one. Yeah. I see. Thank you. Now, in fact, the series defining, I mean the series is actually convergent at S equals one. So I could write this as L of, I mean this is actually equal to L of one chi but you always have to be aware that it's only conditionally convergent, not absolutely convergent. So you have to be kind of careful with this notion here. And just for uniformity, I'm choosing to write it as the limit of values at which it's absolutely convergent. Okay. So why do these three ingredients imply an interest-based formula? Well... The ingredient one is for what values of S? Is it for S as that real part of S is greater than one? Yeah, yeah, yeah. For S greater than one, yeah. Yeah, so it's a quality of like series, a formal series really. And then it's valid wherever everything converges. Yeah. So well, it's because, so we know this limit here is equal to this number here but given this expression here, we can calculate that limit in a different way. It's the same as the limit as S goes to one of S minus one times Zeta S times L of S chi. And then that limit will be one times this value. So you learn that this has to be equal to this. And then you cancel the pies, magically the pies cancel, the square root of P knocks this down to just a P. And you get the, and the two goes in the bottom, right? When you solve for this and you get exactly the Dirichlet class number formula. So as I've kicked ingredient three to you guys, so I'll just talk about ingredients some one and two. So for ingredient one, we defined it as this sum over non-zero ideals of the norm of the ideal for the minus S. But there, well, let me remind you actually something about the Riemann's data function first. So this is the sum of N greater than or equal to one of N to the minus S, but there's this famous Euler product that you can rewrite it as a product overall primes of one over one minus P to the, and now I should make sure I don't get it wrong. Yeah, minus S. So this is a famous identity for the Riemann's data function that follows if you expand this out as a geometric series. So this is one plus P to the minus S plus P to the minus two S plus dot, dot, dot. And then imagine just in your head, expanding this infinite product out. And what you'll find is that as a consequence of unique prime factorization, you'll be getting precisely these terms here by just applying the distributive rule or foil or whatever people call it. Yeah, for multiplying sums together, you have to match all possible terms with each other and you're exactly building a prime factorization of some number and then raise to the minus S power. Like this will be P to the minus two to the S and so on and so forth. So that's the Euler product factorization for the Riemann's data function. And the reason the Dedekind's data function is defined as it is, is so that you have an analogous Euler product factorization because we know that every non-zero ideal is uniquely a product of prime ideals or maximal ideals, I should say non-zero prime ideals. So you get the analogous Euler prime factorization overall maximal ideals of the ring of integers. And then also you use that the norm is multiplicative, but so we're trying to prove a product formula. I mean, a product formula for Zeta FS as a product of two things. So it's good that we now have Zeta FS self-written as a product. So now we group maximal ideals P according to, to which prime P they contain. So each maximal ideal, this was more or less on your problem set from last time, each maximal ideal contains a unique prime P. And so there was, yes, and there was this, yeah, so and if P splits, then there are two primes, P one, P two, as we say lying above P, which means that the ideal P generated by P is P one times P two. And if P is inert, then P itself is a prime ideal. So P is its own prime ideal factorization, even in this larger ring and P ramifies, then P is equal to the square, sorry, in the ideal generated by P is equal to the square of some ideal there. And now let's look at what contributions to this infinite product you get from each of these three cases. So from when P splits, so then the if P splits, then the norm of P one and P two are both equal to P because the norm of their product is the norm of P, which is P squared. So you get the Euler factors. So one over one minus norm of P one to the minus S times one over one minus norm of P two to the minus S. This just becomes, you get the Euler factor for the Riemann Zeta function and then itself again. And if P is inert, then well, it's P is a prime ideal and it has norm P squared. So we get one over one minus norm of P to the minus S, which is one over one minus P to the minus two S, which is one over one minus P to the minus S times one over one plus P to the minus S. Okay, sorry for making that a little. So then here again, we find we recognize the Euler factor for the Riemann Zeta function and then this multiplied by some extra term. And if P ramifies, then there's only one prime in consideration, but it has to have norm equal to P by the same reasoning. And so we actually get just one over one minus P to the minus S, which I'll write as one over one minus P to the minus S times one. So here again, we recognize the Euler factor for the Riemann Zeta function and then this term here. But you also saw in your problem set that the question of whether P splits is inert or ramified is detected by a character mod D. And in the case where D is minus P, this character is exactly the Legendre character mod P. So P splits if and only if the value of the Legendre character. Oh no, I shouldn't have been using P. Oh no, sorry. P was supposed to be our fixed prime at the beginning and now I'm using it for a variable prime. Well, sorry guys. So the character of P, if and only if the character of P is equal to minus one, oh sorry, the character of P is equal to one. P is inert if and only if the character of P is equal to minus one and P ramifies if and only if the character of P is equal to zero. And when you plunk all this together, you get that the Zeta function of S is equal to the Riemann Zeta function times the product over all primes P of one over one minus chi of P times P to the minus S. So it just matches up. And this by the exact same unique prime factorization argument and the multiplicativity of this character chi, this is just the L series. So there's also an Euler factorization for the L series and that is the explanation for ingredient one. It's kind of just a repackaging of this study of how primes split in this quadratic extension. Right. Okay, any questions so far? So now let's move on to the second ingredient. So ingredient two, this is kind of the, I guess maybe this is the, I don't know, they're all the main point in some sense but this is the one where this expression H minus P over mu minus P actually appears. So we want to know that the limit, oops, limit as S goes to one plus of S minus one times Zeta F of S is equal to two pi over square root of P times H minus P over mu P, mu minus P. And I'm not going to give again full details in this analytic evaluation, but I want at least to make clear where all of these factors come from. So whereas the product formula, the first ingredient was proved by making a multiplicative analysis of the Zeta function based on the Euler product, this will be based on an additive kind of analysis of the Zeta function. So we'll actually just directly use the definition. So Zeta F S is the sum overall non-zero ideals of one over norm of the ideal to the S. And now naturally enough, we're gonna partition the ideals into their various ideal classes. So we'll let I one up to I H minus P be representatives for the ideal class group. So that every ideal is equivalent to a unique one of these I sub Ks. So then this is equal to the sum over sum from I equals, little I equals one to H minus P of the sum overall non-zero ideals in the same class as I sub little I of one over norm of that ideal for the S. Okay, so it will suffice to show that, you know, if you multiply this by, so the limit of limit is S goes to one plus of each of these terms of S minus one times this is equal to the same thing without the H minus P term, right? So two pi over square root of P times new minus P because of each of these things has that same behavior on the limit is S goes to one plus if each of these has the same residue, you know, then when you sum them up, it'll just multiply that number by H minus P and that will be where this H minus P factor comes from. Okay, now I'm going to make my life a little bit easier and only explain what's going on in the principle ideal class. It's not substantially different for a non-principle ideal class, but it's just a little more technical to explain. So I wanna cut to the chase. So I'll just analyze the principle ideal class. So we'll see that limit as S goes to one plus of S minus one times the sum overall non-zero principle ideals over pi to the S is equal to two pi over a square root of P times new minus P. Okay, okay, now we're gonna make this factor go away because look, what is the condition here? We're in the principle ideal class that means we're generated by some element alpha, but alpha is not uniquely determined. A generator of an ideal is not uniquely determined. It's only determined up to multiplication by units. So let's just note then that what if we remove that, right? So what if we just summed overall alpha in okay minus zero of one over norm of alpha to the S? Well, then we'd be counting each principle ideal. This is the norm of same as the norm of the principle ideal generated by alpha, but we're now counting each principle ideal more than once. Namely, we're counting it by waiting it by the number of generators for that principle ideal, which is the exact same thing as the number of roots of unity. So this would be equal to mu minus P times the guy we're interested in. So sum I not equal to zero, I is equal to principle ideal generated by alpha over sum of alpha, one over norm of the ideal to the S. So actually it's suffices to see that sum of all alpha and okay minus zero, one over norm of alpha to the S. And then I have to multiply by S minus one and take the limit as S goes to one, that this is just equal to two pi over square root of P. Okay, right. And now I wanna say it's gonna pay to abstract just slightly. It's not gonna be important anymore that this okay is the ring of integers in a imaginary quadratic field. That's not gonna be relevant at all. The only thing that's gonna be relevant is that it's a lattice inside the complex numbers. So let's make a more abstract claim for any full lattice lambda inside the complex numbers, i.e. L is some Z omega one plus Z omega two, where omega one and omega two are linearly independent. So it's just, I mean, we have some picture like this, right? I mean, it's just a lattice as you would imagine it in the complex numbers. So it could be a square lattice or not, but in our case, it will be the ring of integers of our imaginary quadratic field. Then if you take the limit as S goes to one plus of S minus one times the sum over all nonzero elements in the lattice of one over, I mean, I can still write the norm of that element, right? Again, norm of alpha just means alpha times alpha bar or absolute value of alpha squared, right? So it makes perfect sense for an arbitrary complex number as well. And what's this gonna be? It's gonna be two pi divided by the area of the fundamental parallelogram. So if you take a fundamental domain for action by translation by elements in the lattice, and you get some rectangle like this and you calculate its area and you take two pi divided by that. So what is this abstract thing and apply the theorem? So to see this finishes the job, we need to know that area of the fundamental parallelogram is equal to square root of pi for lattice being O, a square root of P, sorry, O being a Q would go in square root of minus P. But in general, there's a formula for this area. You can calculate such an area as a determinant. So if you imagine, so there's this first basis vector, which I guess was omega one and the second basis vector called omega two, then this is also, then you get this fundamental parallelogram by applying to the standard unit square, the linear map, which is omega one in the first column and omega two in the second column. So this is also just the determinant of omega one, omega two. And we have an explicit basis for our ring of integers. So we can just calculate the determinant and it's just a small calculation that I'll avoid doing, but that's how you can check that you indeed get square root of P. In general, for D, you get square root of, I mean, minus D, I guess, square root of absolute value of P. Okay, so now we're down to something kind of, it doesn't look as scary, I suppose. So this abstract claim. And in fact, this abstract claim can be proved very similarly to how you show that the limit is, S goes to one plus S minus one times the Riemann Zeta function of S is equal to one. So the idea is, well, so the key claim is that you can, approximate the sum by an integral. So, right, so more specifically, I mean, I don't wanna get into too many details here again, but the claim is that if you do this here, this is close enough to, or actually if you do this and then multiply by this area of the fundamental parallelogram, that's close enough to the integral, sort of a plane integral over all alpha of one over norm of alpha to the minus S and then kind of, I guess, DX, DY. So now this is an integral over all of the complex plane. And the reason this should be believable is you can imagine, you can sort of approximate this function on all of the complex numbers, I mean, maybe minus the origin, but the, I mean, the integral will be, it's an improper integral, but it'll converge. As I said, there are some analytic details that need to be looked at here, but you can approximate this function, one over norm of alpha to the minus S, defined on the entire complex plane minus the origin, by kind of a step function, which just says, okay, if you're in this parallelogram, then just be constant on value given by the value of your function at this lattice point. And if you're in this parallelogram, so you always select the lower left, always make a constant function with the value of the lower left integer lattice point. And then, so you make some kind of two-dimensional step function. And on the one, that will be fairly close to this, but on the one hand, when you take the integral of that step function, you're exactly gonna pick up a factor of the area of the parallelogram times the sum over the values on those lattice points. Okay? And then you can actually calculate that the limit, well, you can actually calculate this thing. I mean, oh, shoot, I put a minus, I did the mistake that people always make there. They do, you know, there's always this choice where they write one over N to the S or you write N to the minus S. And I swear, I've seen like a million people out, right, one over N to the minus S just kind of had their hand making a mistake. And I've done that too, right? Yeah, so you can actually calculate this by, let me just explain the technique. So you integrate over circles. So you're doing a two-dimensional integral over a plane, but you do it instead by integrating over the circles. And on the circles, the value of the normal function is constant. And then you'll pick up a pi factor from the fact that these are circles. And then it'll actually reduce to the exact same calculation you have for the Riemann Zeta function. Sort of when you then integrate over the remaining variable, which is the radius of the circle. So you'll have like one over R to the S and then you're gonna be picking up some pi factor from integrating along the circle. And then you'll essentially just reduce to the claim for the Riemann Zeta function. Yes, so there it is. A bit of a, maybe a very simple question, but when you say we integrate, you integrate in circles, you mean that we just do that integral in like polar coordinates, right? Yes, exactly. Double integral, okay, thank you. Yep, yeah, so I should just say, I mean, I don't know why I didn't say that. I mean, that's something that people study in calculus, right? So you change coordinates and they're great using polar coordinates. That's what I'm trying to say. Yeah, thank you. Yeah, and then you'll pick up a pi factor and then you'll get the same calculation as you have for the Riemann Zeta function, which again, is just you can evaluate the integral and see that the claim holds. So this was only for the principal ideal class, but in fact, the non-principle ideal class is also reduced to the very same abstract claim. So it's just for a slightly different lattice. Yeah, the lattice is given by the inverse fractional ideal of the ideal representing your ideal class, but it's still just a lattice and you can again understand what the volume of the parallelogram is and everything works out. So my special gift to you on the last day is that we're ending early. So I wanna say also thanks to all of you. This was a really great experience for me having such a tentative and sharp students and contributing to the environment and making my lecture so much better. So I really do appreciate you guys. And yeah, I'd like to say thanks, but also to the people who haven't said anything. Thank you also for paying attention or not paying attention showing up. I mean, you know, I do appreciate everybody. Yeah, so Therios. I mean, I guess we should clap. Yeah, there's a- I saw you guys clapping. It's okay. You don't have to unmute and clap. Let's have questions instead. Don't mean to interrupt that, but I do have one tiny question. So we have this formula, right? This I think it's called the Dirichlet formula for the involving the class number in terms of this. Well, this M minus B, that's something that we've started in the tutorial sheets. And I'm just wondering how, like, how do we use this formula, right? Do you have any, like, are there any applications of that to compute stuff? Like, how can I use this to, do we use this to compute a class number or? Yeah, you could. I mean, it's a girl. You did it by hand, right? For the first two fundamental discriminants. Yeah. You can compute class numbers. So I wonder which one's faster. You can also use this formula to compute by hand. I mean, the nice thing with the by hand computation using the tools from lecture 12 that you gave to us is that we can also have explicit reduced forms. So I don't just get a class number, but I also get some reduced forms. And then we have that theorem by which primes are represented. So I get this extra nice stuff. Yeah, you get more from that version yet. But I think that probably this one is just faster. I mean, if you just want to know the class number, but it's true that you get more from the other approach. I don't know. First of all, I'm not an expert in this. Like, so I'm kind of, it's a little embarrassing that I've chose to lecture on this topic because it's more just a love I had as an undergraduate. I thought I'd give it onward to you guys undergraduates. I mean, it was one of my favorite things as an undergrad. So I thought it makes sense, but I'm not an expert on it. So I don't, I mean, I'm not good at giving answers to questions like this. What is this used for? Yeah, I don't know. No, that's fine. But I do, I do know that the, yeah, I don't know anything that it's actually used for. It's just sort of a gem, a little gem sitting there. And of course it also, I mean, these kinds of formulas like for the residue of the, you know, formulas for residues of Zeta functions or L functions or special values of L functions. This is a whole huge world in arithmetic geometry. And there are some amazing fascinating conjectures about them, which are, I mean, for the most part, completely out of reach. But they're kind of a driving, like generalizing this kind of formula a driving force for a lot of arithmetic geometry and with a number theory, it's a huge area. So I think it more, it's more like a finished product in a sense than a thing that you kind of use to do other stuff, to my understanding. It's like a, wow, what a cool formula. Let's try to do that in other situations as well. I see. Thank you. Yeah. But again, I'm very much not an expert. So don't trust my answer. That's just based on my limited understanding. So essentially in the lecture, we have shown a closed formula for this, finding the class number when the discriminant is minus a prime. The prime is equal to three more. So I was just wondering, because this class number one formula, class number one problem for negative discriminant seem to be a difficult problem. You said at the story, right? I wouldn't even have find solving this. So I was wondering, probably there doesn't exist a trivial formula or a simple extension of this formula for other discriminants, right? Which are a lot of alternatives. There does. Oh, there does. Okay. There does. And actually you can see using the result from the previous problems that describing non-trivial two torsion in the class group, that if you're not minus a prime P three, one four, then you can't have class number one. So in fact, I mean, this formula, it would be all you need to know to attack that problem. But I don't know any way of using this formula to attack that problem. It's, yeah. Yes, Eleftheres? Could you please repeat that last thing that you said in your answer about the class number one? So if P is not prime, then H minus P cannot be one. Right. Because P is not minus a prime or minus four, yeah? Then, oh no, it could be four, okay. Or minus four. Minus eight also has one. Yeah, or minus four times a prime, I should say. But you can actually rule that out. There's a way to rule that out as well. But that's not gonna be obvious from what I say. But if D is not, okay, let me say if D is one mod four and minus D is not a prime, then it has more than one prime factor, right? And then you saw in your problem set that if you could just take that, then that prime will ramify. And so it's P becomes, you know, a math frack P squared. And then that math frack P is gonna be non-trivial in the ideal class group. So it's gonna, yeah, generate, in fact, those prime advisors will generate two torsion. And the only relation between them is that when you take a product of all of them, you get one in the ideal class group. So as long as you have more than one prime factor, you get non-trivial two torsion. I see, thank you. Yeah. Yeah, what about four times a prime? No, no, wait, no, no, it was D, D square free. The condition was D square free. Wait, no, no, I'm confused. I know, yeah, oh, right, right, right. No, D mod four square free. But then you look back, your D is a product. I know you look at the primes dividing D. Okay, so indeed, no, no, the argument does handle everything. So I was just getting confused because the argument shows that if D has more than one prime divisor, then you get non-trivial two torsion. So this only leaves minus P, P congruent to three mod four minus four or minus eight. Those are then the only possibilities, yeah. Yes, so there it is. So we define this genus class group, right? This class, well, the ideal class group of F, modulo the squares. And I mean, if I recall correctly from the last lecture, we had that the ideal class group was, well, technically, I think I should say isomorphic to the Picard group of the ring of integers. Is there any, like, do people do that as well when studying the Picard group? Like, is there an interest in Picard group modulo the squares? I don't think there's any interest in it above, like the same interest you would have for Picard group modulo the cubes or Picard powers. I mean, just on abstract grounds, one way to understand the feeling group is to understand, you know, how many twos there are in it, so to speak, or how many threes there are. So not that I'm aware of is the answer. So the, I mean, this is, again, something very special to this binary quadratic case that the genus can be expressed, genus classes can be expressed in terms of, oh, you don't have the ideal class group really for higher dimensional quadratic forms. You don't have a group. So I mean, again, some sort of special phenomenon. I see, thank you. So I think in the last problem set, we have this black box that the class group modulo the class group square is an abelian group of order two power r minus one there. Yeah, yeah. And today you were talking about this. So basically those represent the isomorphic classes of quadratic forms that are isomorphic over ZP. So how does our come up? It comes up because you can detect, you can detect genus classes by means of a certain homomorphism to, well, so if you have a binary quadratic form of discriminant D, you can ask which residue classes mod D it hits, copram to D, let's say. And it can only hit the ones for which the value on which the value of this character chi is trivial. So that's something we more or less saw. So, and then it turns out you actually do get a homomorphism from the ideal class group to the kernel of that homomorphism modulo, the subgroup which is the image of the principle norms and which is the kind of the residue classes that are hit by the principle of formal discriminancy. And so, and then you can just, you can just calculate the size of that quotient group and you find it's the same thing. It's again, this two group of size, two to the number of prime factors of D minus one. And then you give a separate argument to say that the homomorphism has kernel exactly the squares of the ideals. And then that's how you kind of, so this is, yeah. So it comes up by, you figure out whether you're, so yeah, you figure out whether you're in the same genus or not by looking at, basically by looking, not at ZP for every prime P, but just directly looking modulo D essentially. And then you can cook up some invariance there. Is more, is the simple version of the answer. So being isomorphic over ZP for every prime P is the same as being isomorphic mod N for every N for every natural number N, that's some kind of Chinese remainder theorem from pathness argument. And then it turns out the QN to look at is just D. And that's where the kind of D pops up on that side as well. I see. Thank you. You're welcome. And I think in this book, primes of the form expert was M. Weisberg, which I now have looked at. It gives another way of understanding this genus theory in terms of more class field theory. So like a lot of what we said here is kind of amounts to class field theory for quadratic extensions of Q, but then you can understand this genus theory in terms of quadratic extensions of your quadratic field and applying class field theory to that. So yeah, so if you want a more advanced perspective on this kind of subject, you can look in that book, which is in the list of references. Cox theorem, yeah. Right, theorem group. Thanks, thanks, Freddie. That's the more low brow, like Gauss genus theory version. But yeah, the class field theory version, I think some of us were looking at it yesterday actually, including from Montagion, I think that's like chapter five, if I remember correctly. So he does the, he does both the lower and the high brow versions, yeah. It's amazing that Gauss could do all this stuff kind of by hand, just binary quadratic forms and has like a 19 year old or whatever. I don't know. He wrote this book when he was 21 or something. It was just because he only said with Medicaid, I don't know. Don't feel too bad guys, we haven't reached those heights yet. Yeah.