 So, the title of the lecture is the ideal class group. So let me remind you what we talked about last time. So, um, we studied a so called strict equivalence classes of binary quadratic forms integral positive definite of fixed discriminant D, which should be converges zero or one mod four, and negative, so that we do indeed have a positive definite or a definite one at least. And we saw that this set of strict equivalence classes is finite. So we're using reduction theory. So we found a sort of a canonical representative in each equivalent class the so called reduced form, and this is following Gauss's reduction algorithm. So that's all not all well and good but today we're going to be doing something different. We have a totally different perspective on this finite set, which will elucidate a lot of a lot of the properties that we care about it will be very, very helpful for understanding this set of strict equivalence classes of binary quadratic forms. And, well, the idea is what the phenomenon is that you can basically linearize the problem. So, instead of looking at something quadratic. So quadratic forms over Z, you know binary quadratic forms over Z will pass over to something else, which will be some linear problem. So we have some data over not Z, but some sort of quadratic extension of Z, called the ring of integers in the number field. Q a joint square root of D. So, the basic onsets is that instead of looking at a degree to problem over the integers will replace it by a degree one problem essentially over a slightly more complicated ring which is a degree to extension of the integers. This is of course not a precise statement and I also want to issue a warning right at the beginning that, well this this imprecise description might make it seem like it's an instance of some general procedure, but actually this is very specific to the case and if you replace Z by some other ring or you try to go to more variables than the correspondence we're describing just breaks, breaks down. So, I'm describing something very special in particular here, which when it exists which is only here basically then it's really helpful, but you shouldn't get the idea that it's um, that it's an instance of some general principle. Okay. Yeah. And one other word. This correspondence, it doesn't really exist. Well, if I really use the ring of integers well we'll talk about what the ring of integers is in a second but to make this correspondence. There's no further assumption on D. So, from now on, we'll assume. So, beside D besides being congruent to 01 one for and being negative will assume it's a fundamental discriminant. This is the bit of terminology. So that means that. Well, so, if D is congruent to zero mod four, then you want to ask that D divided by four is square free. And if D is congruent to one mod four, then you ask directly that D is square free. And if D has some square factors in it up to the four that it has to have in this case is just slightly more require slightly more notation to take care of it's not fundamentally well, I don't know. You can make something like this work outside of this case but it's just much easier to restrict to this most basic case here. Yeah. So in this theorem of Hegner that you gave us the end of yesterday's lecture, and you said that the class number of D is equal to one, only for nine values you meant fundamental discriminant right because we found some more in the tutorial which were not fundamental. Thank you so much. Yes, yes, yes. I was, yeah, and also there was, yeah, there was one other claim that I made, fortunately I scratched it out but yeah I was implicitly the only thing about fundamental discriminants at some point although I hadn't formally introduced that convention thank you so much I'm happy you guys found the the mistake there thank you very much and I'm very happy to hear that you guys worked through that and discovered that. Yes. Yeah, my mistake. Okay, so I should have been maybe this convention should have been even introduced in the previous lecture but it was sort of irrelevant to everything I was literally discussing the lecture so I forgot about it. Okay, right. So now. So, what is the basic observation behind this correspondence here. So, and I'll give it in the, or, well, our favorite simple case D equals minus four and our basic form x squared plus y squared. Well, the observation is that you can rewrite this expression x squared plus y squared. And recognize that as a difference of squares, as long as you allow yourself to work, as long as you give yourself a square root of minus one right so you can write this as x plus i y times x minus. So here, I'll say x and z here, it's x and y here are integers. So in other words, I could, I could say this is a alpha times the conjugate of alpha, where alpha is an element in the sort of integral analog of the complex numbers so Z bracket square root of Z bracket square root of minus one so Z bracket I. So if you want to picture. If you think of the complex plane as usual as R2, then this is just the standard lattice inside the complex plane. And there's another name for this expression, which is the norm of alpha is alpha times alpha bar. So, and so, in other words the numbers of the form x squared plus y squared are the same as the norms of elements of Z bracket I. Oops, I went off the screen sorry about that. I can say explicitly this is just a set of all x plus i y with x and y in the integers. Yes. And actually already this observation tells you something. So there's a corollary here which not quite obvious from the beginning so that this set of numbers is closed under multiplication. Well, that's because this expression alpha times the conjugate of alpha, that expression is obviously multiplicative and alpha if you have alpha beta times alpha bar beta bar. Well, complex conjugation is an automorphism that commutes with multiplication and multiplication is commutative so you can move things around and you get the norm is a multiplicative function. And that's not obvious. I mean it's not obvious I think that if you take the product of two numbers of the form x squared plus y squared it's also a form x squared plus y squared. Of course you can unwind this and figure out the algebraic identity that makes that true so I guess like x squared plus y squared times Z squared plus W squared is equal to. And so you know. Yeah, something or another you could just do them. Do the product in the just unwind what it says you'll find some algebraic identity which proves that a sum of the products of some of two squares is also a sum of two squares. But you know you say maybe have to be a little clever to discover this but if you, I don't know, well maybe have to be clever to discover this to the point is that this new perspective can make things which looked ad hoc and complicated can make them look rather natural instead. So this idea of passing to a quadratic extension. So that's the kind of basic observation is that quadratic forms arise from norms, norm functions. So now we're going to. We're going to now look at this from a more systematic perspective. So, yes. So again so D again will be a fundamental discriminant. And it's negative as always. And we're going to let K to note the field you get from the rational numbers by joining a square root of D. So it's just I guess the set of numbers of the form a plus B square root of D, where a and B or rational numbers. So this is a degree to field extension of the rational numbers. So you know there's addition and multiplication and they work just like you think you just multiply and or add these expressions together and simplify. And there's a, it's actually a Galois extension degree to extension which is Galois and the, the automorphism is, you know, the unique non trivial automorphism is denoted alpha goes to alpha bar, and it just sends a plus B it switches the square roots of D that you have right so this will go to a minus B square D. You can also if you like think of this inside the complex numbers, right. So D is a negative number but it's square root is a complex number and we can do this set instead of thing as a that is an abstract thing we can do it as a subset of the complex numbers. And then this really is just induced by your, you know, usual complex conjugation on the complex numbers. Okay. Now, what do we want. We want a submarine. Called okay subset K, which should be as analogous as possible to, you know, the integers sitting inside the rational numbers. And now it'll be called the ring of integers of K. And so the other, I guess a theorem which gives characterizations so you let alpha in K, then the following are equivalent. So the first, I'll write them all and then I'll make some comments. We're not going to completely prove this theorem but I'll make some comments that will maybe elucidate some aspects of it. If you look at the the subring generated generated by alpha inside K. So just take all polynomials in alpha inside K with the integer coefficients. This should be finally generated as an appealing group as a Z module. So the first condition would be that alpha is the root of a monarch polynomial. Monic means leading coefficient equal to one with the integer coefficients. The second condition says, well, okay, that's very abstract and everything but we, you can you can just write down a polynomial with rational rational coefficients, which alpha is the root of just sort of directly. So if you take the polynomial T minus alpha, and then T minus alpha bar note that this actually lies in. It's a polynomial, a degree to polynomial with rational coefficients, because it's equal to. T squared minus alpha plus alpha bar T plus alpha bar. And both of these expressions are invariant under conjugation. So, so then the things which are invariant under conjugation line the ground field rational numbers. Of course, you can also just, I mean, nice thing about quadratic fields is any abstract claim that I make you can actually just sit down and check it concrete right you can. You can just calculate alpha plus alpha bar for any A plus B square B and just see yes indeed it lies in the rational numbers and same for off the time. You don't have to believe the abstract dialogue theory explanation. You can just check it yourself using the definitions. So that lies in qt that's not the third condition that's always satisfied for any alpha, but the condition is that this polynomial actually lies in zebra T. In other words, I guess it's saying the same thing that both the trace of alpha and the norm of alpha are integers where the trace is the word for alpha plus alpha bar and the norm is the word for alpha times alpha bar. Okay. So, why is this the definition of what it means to be, sorry, then I should say, the definition of alpha such of alpha satisfying these conditions is a is our definition of the ring of integers. So, and alpha satisfying these conditions is called an integral element of K and the set of integral elements is called the ring of integers. So why are these, why is this a good definition of an integral element. Actually, my in my opinion the most intuitive one is maybe the first one I don't know why I think this but I don't. So, we're trying to say that you don't have any denominators in some intuitive sense for your element alpha. And what happens when you have a denominator think of like a rational number that has maybe a five and the denominator, five and the denominator. When you take powers of that the denominator gets bigger and bigger and bigger you get a worse and worse denominator right. And then there's no way they could all live in the same finitely generated Z module because if you choose any finite number of them that they some bound on the amount of fives that can occur and you know the power five that can occur in the denominator. It wouldn't keep growing without it grows without bound and you know the suffering is not going to be finally generated. So, so the analog of this does characterize the integers inside the rational numbers and it, in my mind, capture some intuitive sense in which the denominator. There can't be any denominators because the denominators will get get worse as your place alpha by its powers and we have to be infinitely generated so I like I like this condition. And then these there's an obvious appearance of a condition that something which is a priori a rational number is actually an integer. So, that kind of makes sense to from my mind a little bit less convincing than number one but still. So then why are all these conditions equivalent. Well I'm not going to give a proof like I said but I do want to just say a thing or two. So first of all, Well, for one implies to for example I mean, you can look at all the powers of alpha, and then there has to be some relation between them. Because it's finally generated, but with a little extra argument you can see that in fact the relation has to be of the form that alpha to the end is equal to is in the span of some previous set of alpha to the n minus one and minus two and so on and so forth. And that will exactly give you a root of a monocolonial with integer coefficients, and that three implies to is just obvious here is a monocolonial with integral coefficients and alpha is evidently a root. What about two implies three. Oh yeah and going to implies one is right so it's the then that we are once you hit the power of alpha equal to the degree of the polynomial that your alpha satisfies then you're in the span of the previous ones you're generated by all the previous ones. So one is equivalent to two and why does three imply, why does two imply three if you're a root of some polynomial why do you have to be a root of this one. Well, I guess, you can argue that if alpha is the root of some monocolonial with integer coefficients then it's minimal polynomial has to have some integer coefficients, and then the minimal polynomial divides. Well, yeah you have to do some sort of argument here. And then I guess these are. This is just three promise just a rewriting three. Okay. So, it's not obvious that okay forms a ring, ie is closed under addition and multiplication. To be a ring you also need to be closed under negation but actually it is obvious that it's closed under negation. But it's not obvious that it's closed under either addition or multiplication. It is true. And kind of on somewhat general grounds as well. But again in the case of quadratic fields you can actually just check this by hand. In fact, on your problem set. You'll explicitly describe. Okay. And in fact what you'll see is that. Okay is a free Z module of rank two with generators. One and square root of D over two, if D is congruent to zero mod four. And one, and then I guess, well there's, of course, there's a lot some freedom in choosing these generators right like, but here's a particularly nice set of generators. One, and then one minus square root of D over two, if D is congruent to one mod four. And then once you, and you can, you can prove this by just, you know, writing alpha as X plus square root of dy and just looking at what these conditions mean about those two rational numbers alpha X and Y that appear there, and just analyzing the possibilities. And then once you've explicitly described okay you can just check by hand that it's closed under. Well it's as clear from this that it's closed under addition but then you check by hand it's closed on the multiplication. You just have to multiply this by itself and see that it's in this in the set. It's a linear combination of these two. Right. Yeah, yes, there is. Adam, when I looked at this square root of the over two I kind of thought about those primitive polynomials that you that you mentioned yesterday. And you said there's something special about them will they be norms like x square plus y squared was a norm of something. Well, they also going to be norms of stuff. Congratulations you predicted the very next thing I'm going to say so. Sorry about that. I'm very happy. Right. So, yeah, what, so what does it mean that Oh X is the free Z module range from these generators it means that every alpha and okay is uniquely of the form. You know, X plus square root of d over to why with X and Y integers in the first case or alpha is X plus one minus square root of d over to why in the second case. Yeah, let's, yeah, let's follow up on a theory of this comment. So, yeah, so what is then the norm of alpha in the first case is in case one, the norm of alpha where you just calculate it you multiply alpha by its conjugate so you take X plus square root of d over to why multiply by X minus squared d over to why, and you get x squared minus d y squared. And then the second case, you do the same thing, and you get x squared plus x y plus one minus d over four. So these were exactly our fundamental forms of discriminant D. So that's from this some this theory of quadratic fields and ring of integers we recovered. Well, at least the most prominent of the binary quadratic forms of discriminant D. What about all the other ones. We're going to get there. But before we get there. I want to do two preliminaries, just to set us up. Yes, there is. The first of the equal to zero mode for should it not be minus the over. I don't know why I don't know why I got it wrong my notes first of all and then why I didn't catch it when I was writing it here. Thank you. Thank you so much for your comments and corrections. Yes, it should be the over four. All right, so we're just going to do a couple of preliminaries and one of the parameters will have to do with the theory of. Well, this norm function and the other preliminary will have to do with a theory of binary quadratic forms and then we'll meet in a minute right. So, two preliminaries. So one about the norm. So it's going to be our most important function for this lecture so I think it pays to give, you know, several different perspectives on it. So well first of all I should maybe say that it doesn't. I mean, it actually lands in the non negative integers. I said it's, it's given given by these positive definite binary quadratic forms so that follows but it's also alpha alpha bar for some alpha in the complex numbers so it's the absolute value of alpha squared. So the two different ways of, I guess, seeing it. Right. So, lemma. So that's alpha and okay. Norm of alpha. Well it's equal to alpha alpha bar I already said that. But it's also equal to the determinant of multiplication by alpha from okay to itself. And, well, this is a, well this is remember that this is a free Z module of rank two. So, and I have a, this multiplication map is a linear map so I'm getting a linear map from a free module of rank two to free module of rank two it's given by some two by two matrix. And I take the determinant of that. And it's independent of the choice of basis of the choice of isomorphism with Z direct some to because yeah because determinant is ice, invariant under change basis. So determinant makes sense for an abstract endomorphism of a free module of finite rank. So I'm applying it to this particular endomorphism free module of rank two. Now, that's not too difficult to show but the next one is, I think very interesting because it's a completely different expression for the norm. It says it's actually a good measure of the size of alpha. So, it's the same thing as the number of residue classes of okay modulo alpha. This is a. So, so this okay mod alpha okay, I'm taking okay modulo the ideal of okay generated by alpha. So this is just the set of residue classes. Okay. So you identify two elements of okay if they differ by a multiple of alpha, meaning if they're the form alpha times beta where beta lies in okay. You ask how many, how many you get. Okay. So let me give a little example first of all. So suppose alpha is actually equal to an integer. The usual integers sit inside. Okay, right. So we can calculate all of these expressions so well the norm of the norm of K is equal to K times K bar but K is fixed by this that's K squared. I mean yeah K is fixed by conjugation so that's K squared. And the multiplication by K map from okay to okay is diagonal. So that's the two entries. Okay K, so the determinant is also K squared. So that's good. And then. Okay mod. Oh no, and I realized of course that K was a very terrible choice for an integer when my quadratic field is also denoted K. I'm going to switch. Okay so up to this point I hope it hasn't caused too much confusion but now I'll switch to M. So this is Z direct sum itself modulo m times the direct sum itself. So that's that's the mod mz direct sum Z mod mz. So that also has size and squared. So it all kind of kind of checks out. So the value of n squared really that's the same as n squared. So it checks out there. Right. So what about the. So why is it true in general well, so why is alpha alpha bar well again you can you can just calculate yeah we gave a basis for okay. So right right alpha in terms of coordinates for that basis calculate the determinant this matrix note that it equals this quantity here no problem, but there's also a good abstract reason why this first equality holds. It's to calculate a determinant you can base change to whatever field you like as long as that field contains your, your ring. And if you base change to the complex numbers then you'll find that you can write this as a diagonal matrix with entries alpha and alpha bar. You can choose a different basis. So that that explained this and the second the last equality is much more interesting. And I want to give a little picture which I think which is a proof right it's kind of a geometric proof of this. So, recall that well the absolute value of the determinant of a map is the number in which areas are scaled, and when you apply your transformation. That's the kind of geometric interpretation of the determinant. So now we have kind of sitting inside the complex numbers here complex numbers we have okay, which is some lattice. And then we have alpha times okay which is going to be some, some bigger lattice right. And how much bigger is it well, we can take a fundamental rectangle for okay, just spanned by the basis vectors that we had. And when we multiply by alpha we know that the area of this fundamental rectangle is going to grow by the factor equal to the determinant of alpha. So what is okay mod alpha times okay, so alpha okay is the sub lattice, then you're going to see that the integer points inside this blown up parallelogram you get by multiplying by alpha are exactly giving you the residue classes. So then you use the fact that the number of integer points in a parallelogram is the same as the area of the parallelogram because every integer point you have a copy of the parallelogram so you just choose the corners, so to speak. So, this is, yeah this is not supposed to be a convincing argument. It's maybe something to ponder at home. So the magic eyes, like a fundamental parallelogram for okay hit it hit it by multiplication by alpha you've got a fundamental parallelogram for alpha times okay. So I claim that the integer points inside that fundamental parallelogram exactly give you the residue classes mod alpha. So this is similar to you can do a similar picture for the integers right you take the integers and you scale them up by three. And then you look at the integer points inside that fundamental interval, and those are the residue classes module three. Okay, so. So let's, let's take this lemma note an interesting consequence. So corollary. So if I'm sorry I should have said non zero, a non zero element of okay. I mean, for the first two equalities don't require alpha to be non zero but the last one does because when alpha is equal to zero. So this is infinite right. So, right. So now I've. So if you have any non zero ideal and okay, so that means that I is a Z sub module on okay sub module. So it's closed under. I mean, and closed under multiplication by alpha for all alpha and okay. So that's I'm just reminding you what an ideal is. Then. Okay mod I is fine. And what the proof is I guess pretty simple. So if I is non zeros, then we can choose a non zero alpha inside I. And then, well, then the ideal generated by alpha. So alpha times okay, will be inside I inside okay. And then so, okay mod alpha okay, we'll subject on to okay mod I it's a finer okay mod I is a finer quotient and okay mod alpha okay. So since this is finite this will also be finite. So that's not really what I wanted to talk about that's a preliminary to what I want to talk about is that we can extend the norm function to non zero ideals or two ideals. So, definition. If I not equal zero is an ideal, you know, okay, then you define the norm of the ideal to be the number of residue classes modular that ideal. And so, so our lemma says the same thing as saying that if you take the norm of the so called principal ideal the ideal generated by the element alpha. So that's the same thing as the norm of alpha. So yeah, so in this sense the notion of norm of elements can be extended to the notion of norms of ideals. And here's a non trivial fact, which is important. If you take the product of two ideals. Then the norm of the product is the same thing as the product of norms. And what does. What does this mean. This is the. This means the ideal generated by the alpha times beta for alpha and I and beta in J. So you take all paralyzed products and then you might have to close it under sums. In other words, yeah, it's also just a feeling group generated by alpha times beta for alpha and I and J. And let me just remark that this is consistent with the previous multiplicativity property of the norm because if you take alpha okay times beta okay. That is indeed alpha beta. So the product of norms is is compatible with the product of elements. Okay. So that was, that was a little preliminary about norms. Yes, so there is. So, looking at this multiplicativity of norms that would technically enable us in theory at least to compute the norm of any ideal right by unique factorization. Yes. Yeah, this is a good way to go. We have we're not going to be we're not talking about prime ideals and unique factorization yet though that will come in them. Maybe tomorrow. But, but yes this is going to be in a very important fact, moving forward as well because of this thing called unique prime factorization of ideals. Right. So, this is the binary. Is the basis free approach to strict equivalence of binary quadratic forms. So, if you have a binary quadratic form. It has a function from the standard free Z module of rank two to Z. So this is just coordinated with X and Y so X is will be the first coefficient of the first quote, the first factor and why will be the second factor, and this is a quadratic form in the abstract sense so this f of you is going to be lambda X equals lambda squared f of X and f of X plus Y minus f of X minus f of Y is bilingual. Now, yeah, now you should be careful because now I mean X is now supposed to be an element of this thing so maybe for clarity I should denote it by something else like M. So little M. M plus M M M. So, M is in the Z to X and Z. But the notion, so there's a standard notion of now that we're in this abstract language so we can abstract this to a map from a free Z module of rank two. So this is the definition of what it means to be a quadratic form and now we've got a basis free notion of quadratic form which we already discussed, but the notion of isomorphism between basis free quadratic forms the natural notion is that you have to give yourself an isomorphism of a billion groups, which makes the diagram commute right so you have your f in your f prime and you should give an isomorphism alpha so that if you do alpha and then f prime it's the same as doing that natural isomorphism. So that corresponds to not strict equivalence but usual equivalence of, you know, f x y's. So, ie up to GLNZ, GL2Z, not SL2Z. Because yeah it's an arbitrary just an arbitrary invertible if you put coordinates on these, which is how you go backwards, and it's just an arbitrary invertible thing so to get something that corresponds to will consider oriented a free Z modules ranked to equipped with a quadratic form. So, and an orientation on a free Z module of rank two you can read it on the corresponding real vector space of dimension to so it's just, it's just giving an orientation on that real vector space so clockwise sort of a notion of counterclockwise or whatever. And that's because you know to go from GL to the SL to its equivalent to giving given an element of GL to to say it's an SL to is the same thing is saying if you do it as a real matrix then it has positive determine. So, you can read it off the data on real numbers usual notion of orientation real numbers. But then when you do this. And into every binary quadratic form, their corresponds a oriented free Z module of rank two with a quadratic form on it. So this is just some naive expression right f x y is a x squared plus b x y plus C y squared. So isomorphisms are strict equivalences between binary quadratic forms, exactly match up with isomorphisms of oriented isomorphisms of these guys. So it's just isomorphisms as before, which have the extra property that they respect the orientation. So, the respect all the battle. They also go backwards so for every oriented free Z module of rank two if you choose an oriented basis to a basis where it goes counterclockwise or goes according to the orientation. Then when you write it in coordinates you'll get a binary quadratic form and so it will be such that oriented free Z model ring to you associate to that is isomorphic to the one you started with. So, so we have a corresponding sort of on objects and on equivalences and so if you want to use fancy language. So you actually have an equivalence of group points. So called equivalence of group points. So, there's a groupoid of binary quadratic forms where the, the objects are binary expressions of this form, and the isomorphisms are changes of variables with determinant one. That's a very concrete group or you just have concrete objects and concrete maps. Now we have this abstract group where an object is an oriented free Z module of rank to a quick with a quadratic form in this abstract sense. The isomorphisms are the just the maps for you know I said the bijections preserving all the structure. Those two group points are equivalent. So, in particular, isomorphism classes are the same. So, and also when you fix the discriminant. The equivalence classes of binary quadratic forms of discriminant D can also be understood in the abstract language. You also get to the automorphisms are the same so these special orthogonal groups of binary quadratic forms are just the automorphisms of the objects in this group point here so you can see them abstract as well. So it's going to be more convenient for your, your dear lecture. I'm not going to claim it's more convenient for you. I'm just for me to work in this abstract language here because I get confused when I write down formulas, and it's harder for me to get confused when I'm dealing with abstract data. That's just me. So we're going to. So the correspondence between binary quadratic forms up to strict equivalence with linear algebraic data which I'll explain, we'll give an equivalent with this kind of thing here. And to make it concrete you have to choose a basis for your, for your free Z module rank to but I don't want to use a basis because it's kind of, I think it obscures the picture. Okay, so, so for example. So let's take okay, and this norm function is fundamental thing we had. This is an oriented. Well, it is a, an abstract binary quadratic form. So it is a it is a quadratic map from this free Z module of rank to Z. And we can orient it by. Well, let's use the standard orientation. Let's look at the numbers if D is congruent to zero mod for, and the opposite, if he is congruent to one month for. There's no special significance to this choice seems a little fun. But it just happens to correspond to on the basis I wrote down earlier for this free Z module rank to saying that that basis has the correct orientation. Two basis vectors and asked about orientation there and they're in the standard orientation of D is zero month for an opposite of D is one month. So that's just how it worked out. But there's no special significance to this convention of how we orient okay. Okay, so that that is what corresponds to our principal forms, or the strict equivalence class of our principal forms now has this canonical abstract representative. So I'm going to say how to generalize so how to get all forms. So here's the definition. Let I inside okay, be a non zero ideal, be fine. F from I to Z by f of alpha, we take the norm of alpha. And we recognize that there's a common. There's a sort of a common factor to all of these norm of alpha expressions, namely the norm of the ideal itself. So we divide by that. So, yeah. So this, yeah. This uses that since alpha is inside I that implies that alpha okay. Is a sub ideal of I, and then there's a non trivial fact that if you have a sub ideal, then, then you have to be a product of I with some other ideal J. This is also non trivial, right. So, and we will orient I, as before. So, so the counter clockwise in C, if D is congruent to zero mod four, and opposite otherwise, then let me state the theorem. Are you serious. I'm probably misunderstanding something but wouldn't this definition of F that we're giving end up being constantly the norm of capital I, because the norm of alpha by the thing you wrote in the parentheses. Well it's equal to the norm of the ideal generated by alpha, which is equal to I times I. So it would constantly be getting norm of I times I over norm of I right. Okay, wait. Oh, it's j. Okay. Let's, let's take the case where I is equal to okay. So the unit ideal, then the norm of I is one, and we're just getting our principle quadratic form that we had before. I see. Thank you. What, what is the thing in the parentheses. Is that a justification that it's an integer. Yes. Wait, doesn't that just follow from. Okay, over alpha. One of them being a quotient of the other. Of course, thank you so much. That's a much better argument. Yes, yes, yes. Let me explain what a sorry, I didn't get your name was it Kenzo. Sorry. Yes, I just popped up on the screen briefly. So let me give tens of arguments that which is much better. So, right, so we know the norm of al why this is an integer so norm of alpha is the number of residue classes mod alpha and norm of I is the number of residue classes mod I. And yeah, since alpha is in I okay mod I is a further quotient of okay mod alpha. And so, yeah, so I don't know what to say. This ratio will actually just be the same thing as the number of elements in the kernel of this so that we're in other words the number of elements in, in I module of the ideal generated by alpha. So, yeah, I probably you were about to explain it better than weekends. No, I think that was good. Sorry, yes. How did you say it. I just said it's a quotient so it's just a factor. Yeah, so it's a, so a number of elements it's a factor yeah so if you have a quotient of a group. Right, that's a good way of saying if you have a group and then you have some quotient of the order of that quotient group is always a divisor of the order of the group. So, in fact, the thing is to take the order of the quotient and multiply by the order of the kernel, then that's equal. I guess it's essentially Lagrange's theorem, but yeah. Right in reverse. Yeah, exactly. Lagrange's theorem right in reverse. I think, yeah, there is you also have a question or comment. Oh no I just forgot my hand up sorry. Okay, so now let me state the main theorem which I guess will continue discussing next time. So, so. Well, F is quadratic of discriminant D. So it's a quadratic form of discriminant D. The second thing is every quadratic form of discriminant D is strictly a equivalent to some. One of these guys, one of these norm functions that I'm calling F, maybe I should call it F sub I when I want to stress the comments on I. And the third thing is for an abelian group isomorphism so for a Z module isomorphism. So fee from I to J. So if you have just an isomorphism of abelian groups from an ideal I to an ideal J. So the following are equivalent. So, condition a is that fee is an isomorphism from this oriented quadratic form to this oriented quadratic form. And the second condition is that fee is an okay module isomorphism. So it commutes with the scalar multiplication by okay. And the third condition is the concrete one. So, so there exists an alpha in K, not okay, such that fee is given by multiplication by alpha. So every quadratic form is strictly equivalent to one of these eyes, one of these ideals with this norm form. But more than that, if you want to understand strict equivalence is so you so if you're only interested in quadratic forms up to strict equivalence which we are, you might as well assume from the start that any quadratic form you have is exactly equal to one of these f is right. But then you still want to know when any two sets are strictly equivalent or more generally what are all the strict equivalence is between two such. And now this says it's a very simple linear condition. Just commuting with the scalar multiplication by okay. So this is this linearization that I'm talking about. Well, I guess, somehow, yeah, I don't know. All right, so I guess in the language of group words, you know, I don't know. Well, maybe, maybe we're running out of time so I'll. We'll take this up again next time. And I want to lose the date this statement a little bit more and get some consequences of it as well as proving it, I suppose. Okay. Sorry. Did you define orientation. Yeah, I always have trouble defining orientation. I mean, I guess, like, I searched up and it's just isomorphism with the high six year product. But like, I mean, I guess that makes sense. So for a Z module, there's that trick. Yeah. So it's a it's a nice and morphism. The interior power with the integers. So that's fairly, fairly nice. Yeah, the problem is we're, we're kind of using this also using this observation that the notion of orientation only depends on the corresponding real vector space. Like, when I just, I want to say that like orienting I and orienting okay or the same thing. And this is less obvious from the perspective of like if you give that purely Z module definition that's a less obvious claim. Maybe it's not too hard to prove there's some square factor or something somewhere, but, but still, yeah. So yeah, I mean an orientation, but then you could say an orientate if you like this abstract language you could say an orientation is a choice of connected component of the top exterior power of your real vector space. But minus zero. Yeah, you take the top exterior power real vector space minus zero and choose a connected component. That's an orientation. So you tensor of R. What? Yeah, yeah, yeah. So if you have a finite dimensional real vector space, you can take the exterior power as one dimensional. So when you remove the origin has two components you could say an orientation is a choice of one of those two. Okay, yeah, that makes a basis up to positive scalars. Yeah. Okay, that's, so that only works if it's integers, because, right. Yeah, but which one of its integers. Does that, doesn't that definition only work for Z modules, since that you tensor of R. And so, or does that work in general. I mean the notion of orientation is is kind of specific to the real numbers, although I guess you could always define it as like a, I mean maybe in the, if you're thinking abstractly, maybe you want to say that an orientation is like, a vector space or where a field maybe you want to say an orientation is a choice of base basis in the top exterior power up to squares or something like that. And then it makes sense for an arbitrary field that's probably like them. I think for people doing like motive a common topic theory which is not what this whole thing is about. I think they use that kind of definition. Like, classically speaking the notion of orientation is for real vector spaces. So, like, we take it for granted without really giving a formal definition of what it means. Oh, it's like counterclockwise or clock. Oh, okay, what the heck does that mean. Yeah, I mean, it's a very good question and I don't really think I have a good answer. Oh, I thought the R vector space explanation was nice. If you like that, then, yeah. Yes. On this orientation discussion just to kind of verify whether I get things right or not. So, for the case where that we're working on right now where we have like rank to see modules and we can, I guess, regard them as we can kind of take the real case, like the orientation of a real vector space and take that down to to Z. So, could we say that, like, I guess the kind of more elementary definition that I'm probably familiar with would would also work right that if I have a basis, then, and I say that this basis has positive orientation arbitrarily than any other basis has the same orientation I believe the the matrix that takes this basis to the other one has positive determinant. So that would be like the definition I keep in mind that is a good definition. Yes. So, yeah, indeed. So you can think of it as a basis of the vector space up to some notion of equivalence. But then there's some of this more refined thing, the basis of the top exterior power of the vector space up to a different notion of like that somehow. I think in some sense that's preferable, because, because like the definition is fun toriel for longer. I mean, what am I, when you pass data and giving the basis of the top exterior power. So the amount of equivalence you have to keep track of in your head is less when you talk about the top exterior power first. Then if you talk about a basis then it's up to a huge notion of equivalence positive any transformation positive determinant whereas when you pass the top exterior power is just positive scalars. And then this top exterior power functor court thing is a functor so it like carries all the struck anything you want along with it. And so that I think that's, that's why that might be kind of preferable on abstract grounds but yes it's hard for me to decide which definition I prefer in general it probably depends on the context. Thank you. Just make a very small comment I guess here we're only using orientation the notion of orientation for a two dimensional object so really is sort of a notion of counterclockwise. Yeah, but what does that mean. Well, I mean, you could say that if you have two vectors I mean it, it's, you know it's saying. It's a it's a map which assigns every pair of linearly independent vectors of plus or one. And that's basically the same thing that's what we, you know, or as what a serious explain this exterior power, or we're giving a basis of the dual to the top exterior power. Yeah, but that's, that's actually I like that a lot so yeah. I mean, I guess it's essentially, it was equivalent to what it was serious that it's, it's like you're giving your partitioning the set of all bases of your vector space into two classes. Yeah, yeah. Right. I was also wondering and this maybe goes back to the intersection of today's lecture with the with the previous one. Because I mean in that exercise we had the tutorial sheet it for computing some some class numbers and finding some the reduced forms in that X sub D. So that was an exercise that I mean for each integer that we had to work on it was like fairly okay to to handle and it was, it was a bit, it was quite fun to be honest. So I was wondering like, has anybody undertaken this task like make a table of those things or write the program or something where you can just, I'll give you a discriminant and surely many many programs exist. I think party GP certainly, which is the only program I've ever really used, although I haven't used it in some number of years, knows how to do this. I don't know, maybe some do the TAs or a keel. Yeah, are there favorite programs that people use nowadays I mean my knowledge of this kind of stuff is like 15 years old. I have a specific program at the table. There's the LMF DB, which is just like a really large database of various sorts of number theoretic objects that people interested in it will tell you, you know all about their various and variance. So maybe most relevantly they have like a section on number fields. You can look at number fields and then, right, as we've like just seen every like number field has like some ring of integers and then attach this ring of integers you can like make sense bits discriminate. And so you can like go through here and then see lots of information about number theoretic objects. I don't think this one says anything like directly in the language of a binary quadratic forms so maybe there are other resources. One could point to this. This is I think a good website to be aware of. The great thing about LMF DB is like you can download stuff in multiple forms so like you can download it for to be used in like magma sage at least maybe party as well. For for quadratic orders, I know, because because I've needed it at some point I think it's Mark Watkins has some huge list of like, we're like you know you have some algorithm to just generate like all the data to like I think he's gone up to like 100 but I think maybe somebody else has done it. Maybe John Voight, I might be wrong there for like, assuming GRH for like a even larger bound and you can go in and use that stuff. Sorry, I don't know if this is a easy question to answer but what do you gain if you assume GRH in this context. Yeah, exactly. I guess it's something about like you want to say that you your list of. Let's just say like Mac maximal orders of class number, like quadratic orders imaginary quadrant quarters of class number like up to some bound is complete. GRH comes in somewhere I actually actually completely forget where it is. Does anyone does anyone else know the answer to this. I could look into it in GRH doesn't that give us like corollaries about like prime number distribution. I mean that sort of thing has like implications basically everywhere. Right. I'm guessing it. I'm guessing at least like, but I'm guessing assuming GRH at a very least let's like optimize the algorithm for determining it right. I doubt this that person would be able to write down that much out by hand. Right. I'll look for the exact paper yeah I'm just, I'm not sure of like the exact way that it comes into play. Okay, thank you. And I think I have another another question and again just going through my notes. So we gave this convention about how we how we orient this this basis that we have this kind of canonical basis that we have for the for the ring for the ring of integers for okay, we said that so near the end of the lecture we talked about how we should orient okay. So that orientation was chosen to make sure that our canonical basis of okay is positively oriented right. Exactly. Okay, I see. Thank you. I mean canonical basis, and the reason I chose that basis is, is it the one that makes the norm for me reduced representative for that strict for that strict equivalence class. And the reason I gave that basis specifically because it wasn't the one that matches up with the reduced representative. Now this notion of reduced binary quadratic form while beautiful and and lovely and has great applications computationally and so on it's not really canonical I wouldn't say I mean it's going to use all, you know, well maybe it's I don't know. You mess around with it a little bit like instead of having to be greater than or equal to zero that's B to be less than or equal to zero I mean I don't know, because it's not clear, except for convenience it's not exactly clear in what sense the that for that precise set of equivalence classes is canonical. I don't know, maybe it is. But our making choices like for example that saying that in the boundary cases be is is is non negative. But here's that's maybe that would be a question in my head. Why would it like have you have you encountered a case where it would be expedient to fiddle around with its definition, like, I mean, change those arbitrary decisions. I mean, but because I think the only point is to be able to write them small representatives. And so you could fiddle with it and choose a slightly different set of small representatives but the purpose would be the same and in particular you wouldn't get anything more out of it. Everyone might as well just use the same one right just have a convention. If you go to higher dimensions like from SL to Z to SLN Z and so on then their people have different choices they use for fundamental domains and it's not it's really much less obvious how to select representatives for the set of equivalence classes. Okay, thank you. So in particular there it's quite evident that there's nothing that's canonical in any sense. I think it's all just some convention. So if there are no more questions, I will get to uploading everything and so on. See you guys later.