 We're going to be continuing the discussion for last from last time. So let me just give the setup again. So we have capital V is a fundamental discriminant. Either D is converted to zero mod four, and D divided by four is square free. Or D is converted to one mod four, and D itself is square free. And we have the quadratic extension Q adjoining square root of D. And we have its ring of integers, which is generated as an abelian group freely by one and square root of D over two of D is zero mod four, or one and one minus square root of D over two is one mod four. Okay, right. And we are interested in strict equivalence classes of positive definite binary quadratic forms of discriminant D in terms of this ring of Q adjoining square root of D. The basis for this is the following definition. Oops. So, if I inside the ring of integers in Q adjoining square root of D is a non zero ideal, define the quadratic form. We're going to be calling f sub I some map from I to the integers by f sub I of any element x is equal to the x. So x times its conjugate divided by the norm of the ideal. So, just write out what both of these mean x x bar, this whole field lives inside the complex numbers so this makes sense. And then the norm of the ideal I recall is the same as the index of the ideal inside the ring of integers. So it's always finite last time when I was a one zero ideal. Then we had the theorem that we were going to that we stated last time that we're going to prove that says that well this is, let's say what it says so first of all. So fi is a positive definite binary quadratic form of discriminant D. So if you're interested in binary quadratic forms, you might as well assume that your binary quadratic form is one of these guys associated to some ideal so we only have to care about the form of discriminant D is strictly equivalent to some fi. So if you're interested in binary quadratic forms up to strict equivalence you might as well assume that your binary quadratic form is one of these guys associated to some ideal so we only have to care about these guys now up to strict equivalence and the last thing you want to understand is what are the strict equivalences between these binary quadratic forms, and the result can be stated as follows. So for a Z module isomorphism I j between non zero ideals, the following are equivalent. And I'll move to the next page. So part a says that. fi is a strict equivalence, or really since we're using abstract language here I should say it's a orientation preserving and respects the quadratic form or intertwines I and FJ. So in other words if you do fee and then FJ, it's the same as doing just fi. The abstract notion, which translates in district equivalence once you choose an oriented basis for I and J, actually write out an explicit quadratic form and two variables. And the second condition is that fee is okay linear. So it's actually an O F module isomorphism not just an abelian group isomorphism, or another yeah in other words it can need to a scalar multiplication by O Q a going to score and D. And the third condition is that fee is equal to to multiplication by alpha for some alpha in some unit in our field. And the fact that we already said that fee is an isomorphism does put restrictions on what alpha can be. So, um, Well, maybe I'll make those kinds of remarks after after we prove the theorem. Right. So this, this tells you that you can understand everything you, everything you want to know about binary quadratic forms of discriminant D, just by studying some linear data over this ring of integers all you need to know is ideals. And then you can even forget about these f eyes it's just ideals and then I owe linear O F linear ideals in this ring and the question of when they're isomorphic as O F modules. Yes, so there is. So just a quick question so when we when we say in the statement of the theorem that this f sub i is a positive definite binary quadratic form with discriminant D. Does this mean that this non zero ideal i that we take is isomorphic, I guess is a Z modules and a billion group to Z two. Indeed that is true. I mean in fact this follows on sort of abstract ground so if you have any finite free Z module and of rank n. In this case n equals two and any finite index subgroup. Then there's a theorem that says that that is also free a billion of the same rank. Yes. Okay. Right. So yeah so this translates the problem of studying binary quadratic forms into the problem just studying ideals and the question of when they're isomorphic or equivalent of the multiplication by some non zero scalar. Okay, so let's just before exploring the consequences of this of which there are many. Let's give the proof. So proof. So part one. So part one was the claim that fi is positive definite of discriminant D. Well, we actually already saw this for I equals the whole ring of OF. In fact, so then this fi is just the norm. And we saw we get the principle of binary quadratic form of discriminant D. Now we're going to reduce to this case by means of a purely abstract claim. So, suffices to show it as a lemma. If. So if you have M isomorphic to Z direct some Z. The map from M to Z is a binary quadratic form of discriminant D. Then for any finite index subgroup M crime subset M. The map from M crime to Z, given by F crime of M is equal to F of M divided by the index of M prime inside M. So that's a binary quadratic form. Also discriminant D. So the same discriminant. Well, there was also the claim that it's positive definite, but the claim about being positive definite is only about what happens when you send scalars to the real numbers and these guys differ by a fire up by just finite index and so on used to extend scalars to the real numbers you get the same thing. So note that what I'm trying to say is that M over R is isomorphic to M prime or well up to a positive scalar constant, at least. So, so the positive definite thing will be automatic. So the real crucial part is this lemma saying that it's got the correct discriminant. And in turn this lemma falls from two sort of calculations. Well, you can break this up into two self claims. So claim one is that yeah I guess by the way the fact that it's a binary quadratic form is also pretty immediate because if you just were just modifying by a scalar right and if you just look at the definition of what a binary quadratic form is if you don't buy a scalar it doesn't. I mean you still get a binary quadratic form so this is this is easy. So again this this part is the crucial part. So claim one is that, you know, if f is a binary quadratic form. Then discriminant of lambda times f is equal to lambda squared times discriminant of f for all scalars lambda. And the second claim is that if f is a binary quadratic form and M crime is a finite index subgroup, then the discriminant of f restricted to M crime is equal to x squared times the discriminant of f. Oops. So if you compare these two things and you'll find that when you restrict D to M crime you're picking up this extra scalar but then you're exactly dividing by it when you multiply the values of the quadratic form by the inverse of that scalar. So the two effects kind of cancel each other. Now both of these claims are fairly easy to prove so well at least claim one is fairly easy to prove so when you multiply the quadratic form by lambda. This multiplies the bilinear form by lambda. So it multiplies the matrix which calculates the discriminant by lambda, but that matrix is a two by two matrix. So when you multiply every entry by lambda that scales the determinant by lambda squared. So, as for this one, well you have to be a little bit more clever so you have to use a result. That's an abstract result from the theory of a billion groups that oops and this should be. And there exists a basis of M, I guess I can call it e comma F of M, and, and, and, you know, scalar numbers so N comma M, the natural numbers, such that N times E M times F is a basis of M crime. So this is a consequence of what's called the Smith normal form for matrices over the integer so I don't want to get into this abstract stuff but it's just a result in the theory of doing it. The analogous thing is true for finite free Z modules of arbitrary rank when you write it right up for just rank to get this. Then what is the index M and M crime is equal to M times M. I mean small m times small m. So the amount of the first vector scale by multiply by the amount of second vector scale by because when you take the quotient you get Z mod M time Z mod M. It's a toughening component wise. But on the other hand when you calculate the bilinear form. So if you look at the matrix of the bilinear form you're just going to get, you know, and if your matrix was a by ABCD before you're just going to get, I think it's n times a n times B and times B and times D. And the whole thing, the determinant is multiplied by an M as well. So, in other words, the determinant is multiplied by the index. Um, so that's just, I mean you just have to do the little computation of what the quadratic form restricted to this basis looks like, or the matrix associated the quadratic form looks like and it's either this or maybe it's and I forget but in any case what happens the determinant implies by n times M. And that's equal to the index so that proves the claim claim wanting claim to which in turn proved that prove the first claim of the theorem. So that indeed these are quadratic forms do have the correct discriminant. Okay. Now the second claim. So, every F is strictly equivalent to some fi. So, sorry, just came to have a question. There's a pop up box here. Yes. Yes. It wasn't the second claim about like the discriminant being multiplied by the index squared. Oh, Yeah, wait, I must have messed up the calculation somewhere. Thank you. Yeah this is what I get for not doing it in my notes. And I'm not going to fix it. And the reason I'm not going to fix it. Thank you for catching the mistake is because I'm worried I'll just keep messing it up if I keep trying and I don't want to waste time but I hope the strategy of the proof is at least clear such that one could carry it out on one zone if one wanted to. So you just I would do it after class just to make sure everybody gets it right. So we look at the expression this expression here and and just plug in F on it. Sorry. So you just basically okay let me just maybe I should do it now yeah so if we before we had f of x, y. Now we're going to have f of nx, and why. And so that will change. So let me just yeah so if f of x, y was a x squared plus b x, y plus C y squared, then this one will be equal to and squared x squared plus b and m x, y plus m squared y squared. And then yeah when you calculate the determinant of this you pick up a sorry and m squared indeed so so I messed up my explanation in terms of matrices but if you just do it in terms of a good in terms of the coordinates then you can see it. Okay, thanks. Right, so every F is strictly equivalent to some fi so let's just take a random f and I'm again going to be just very explicit about explicit about this. So you have a x squared plus b x, y plus C y squared. So ABC and Z, and discriminant is B squared minus four AC. So note. So, well, B, B squared is congruent to D mod four. So if D is congruent to zero mod four, then B is even. And if D is congruent to one mod four, then B is odd. So, in any case, B plus the square root of D over two will lie in our ring of integers. Why, because the question of whether B plus the square root of D over two is in the integers, one way to explain it is this only depends on the parity of B, because if you add to to be, then you're just adding one to this expression. So the x lies in the ring of integers if only the x plus one, because it's a ring. So, but then when if so when B is zero we're saying square root of D over two is in OF which is what we had as our basis vector in this case. And when B is odd we're saying one plus square of D over two or one minus square root of D over two that's the conjugate, which is what we had as a basis vector in this case. So, this is kind of nice. So, without making a case distinction we have this sort of elements in OF. And now I'm going to consider the Z sub module of OF span by by one and all right B plus or minus square root of D over two, where the so plus if D is your mod four, and minus if D is one month for this will be just. Don't worry about this plus or minus too much it's just going to be to get the orientation right according to the conventions I had in the previous lecture. Right. So I just look at all integer linear combinations of these two elements of OF, then the first claim is that, well this is a one and B plus or minus square root of D over two is a is a Z basis. And this is free of rank two as an abelian group, and this is just because the only, I already said it's spanned by them so the only extra thing I need to check is that these elements are Z linearly independent. But actually it's, oh no sorry one out one a, I'm sorry guys a, I have to use a as well. But, well actually they're even we're linearly independent over the real numbers right because one has an imaginary part in this one doesn't. So you think I'm in the complex claim one I'm sticking up and the other one sticking just right. One of them sticking either up or down at least a little bit up or down the other one sticking to the right so they're even linearly independent over the reels. Okay, and the next claim is that the sub module is an ideal. So to check that you need to check it's closed under multiplication by an arbitrary element of OF, but it's actually just enough to check. It's closed under multiplication by square root of D over two, if he is zero mod four and multiplication by one minus part of D over two. D is common to one mod four, because these two elements of this ring generate OF as a ring, because the other basis vector already lies in the integers and this is already a Z sub module so. And then just do it by you so this you can just do by hand. You multiply this by either of these two expressions and just write it as a linear combination of these two, it can be done I'm not going to, not going to do it in front of the camera. Okay, so we've successfully produced an ideal and even produced a basis for the ideal. And then the next claim is that so let I denote this ideal, a B plus or minus squared D over two. The claim is that the f sub I function on you take up work according to this basis so x a plus y B plus or minus squared of D over two is, is it is just equal to a x squared plus B x y plus C y squared. So it is does give our original quadratic form. So, oh, I forgot to make a remark. Sorry. Sorry before before doing this thing let's make another claim of pre claim. So the norm of this ideal is equal to just a, and the purpose. Well, so, well, OF is spanned by. I said it was spanned by one and, you know, either squared of D over two or one minus squared of D over two but it's actually spanned by one and be a plus or minus squared of D over two, or have basis so has basis. And you can just, you can reduce B by two every time, and you still get a basis, because you can always write it in terms of the one we know. And then you get down to the case that we're familiar with one square of D over two or one one minus square of D over two. And then, only depends on the modulo two. So this has basis this and our ideal has basis a B plus or minus squared of D over two. So we're in this same situation as we had previously we're just one of the vectors is scaled by a and the other vector isn't scaled at all. And so the index is the product of the scalars and that's just a. In fact, we have something sort of more refined so OF mod i is just isomorphic to Z mod AZ. Okay, so now let's go back to this claim. So we can just calculate this using the definition of fi so this is equal to, by definitions we have to take the norm of x a plus y B plus or minus for D over two. So this is divided by a sorry this is getting a little complicated here. And you could just write this out right it's just the product of this guy, with its conjugate. And here you see that the plus or minus doesn't matter because the conjugate will just have the other one so. And you'll see that on the x, you get x squared times a squared. And then you can also so that you get a squared over here and that's good when you divide by a you'll get a. So that you get this term here you can also see without too much effort that you get this B term here, and then the C term is a little more annoying to calculate but actually you don't have to calculate the C term. Because we already know a priori that these binary quadratic forms have the same discriminant. They both have discriminant D. So it's since D squared equals, you know, some formula in terms of BA and C you can always solve for C in terms of AB and D so once you see that the A's and the B's match up which you can see by expanding pretty quickly. Then the C has to be correct as well so that's why we didn't have to use C in order to sort of specify our ideal once we fix the discriminant and C comes along for free. So it's really just a bunch of calculation to verify this, this claim. All right. Now the most interesting part is like, actually already this is kind of interesting but let's, let's move to the maybe, I don't know if it's the trickiest part, but let's do part three. Let's give ourselves two nonzero ideals, I and J, and a Z linear isomorphism. So Z linear, yeah, just an isomorphism of dealing groups. So we wanted to claim that the following conditions are equivalent so Yes. So, since we now observe that we need to ideals inside O, Q, D is free of rank two, aren't they automatically Z-module isomorphism instead? Aren't they naturally isomorphic as Z-modules? No, they are. So this, this is not a hypothesis on I and J, this is the specification of the data that's under discussion, we're discussing an isomorphism. Oh, I see, okay. Yeah. So just to see, because maybe I misunderstood something in what we proved in the second, for like the second part of the theorem, we ended up showing that this F that we started with was equal to this F sub I. Well, I mean, F was an F of X, Y thing, right? F sub I was an abstract thing, so we produced an I and a basis of I, yes, such that we get an equality, yeah. But that's, but that's no more than the claim we started with, but that's the same thing as the claim we started with that F is strictly equivalent to some F I. Exactly the same thing as saying there exists an I and an oriented basis of that I such that when you expand it out on that basis you get the quadratic formula you started with. You shouldn't be surprised that we were able to do that. I get it. Thank you. Mm-hmm. So yeah, so we're giving ourselves this evening, we want to understand when it's a strict equivalent. Right, so it won't, it won't necessarily, I mean, there will be cases when it's not a strict equivalent, right, but yeah. Right. So the third condition is that he is OF linear. And the third condition is that he is a multiplication by alpha for some alpha and cross. So let's, let's first show that B is equivalent to C. So for C implies B. This is clear. Because multiplication also, if you have lambda in OF, then lambda alpha x is equal to alpha lambda x, because, you know, F is commutative under multiplication. So if you're given my multiplication by a scalar, then that will commute with multiplication by all other scalars, including ones in OF, and that means that this math has to be OF linear. This is a scalar multiplication by OF. What about B implies C. Why is it that just knowing that you're off linear you automatically get information about precisely what the map can be. Well, I guess there's probably several different ways of explaining this but let me give a maybe a somewhat abstract one so we have I inside OF of finite index. That means when you rationalize you get an equality. But this is just F. So this means that any OF linear isomorphism from I to J rationalizes to an F linear isomorphism from F to F. And now we're working, you know, so now this is a free F module of rank one and F linear isomorphisms all just given by multiplication by alpha for some alpha in F. So that alpha should be non zero scalar here but it has to be non zero if this map is going to be an isomorphism. So, so I can build a little extra condition that alpha is non zero. Okay. Yeah, that's a little bit abstract. Right. So now let's do a is equivalent to be. So for B implies a. We can actually just actually no I should, I should maybe do a is equivalent to see. No, a is equal, I could say a is equivalent to B and C we already proved B and C is equivalent. And instead of doing B implies a let's do C implies a. So let's write alpha equals. I don't know x divided by y with x in x and y and OF. We have y and j. And we have multiplication by. We have a map is given by multiplication by alpha by assumption. We can then also multiply by why here to some and we get some J prime, namely, J times y. And this map will also be an isomorphism. And the positive then will also be an isomorphism and it's given by multiplication by x. Are you following what I'm saying so far. So, if we want to show that this map gives an isomorphism, we want to show that this map gives an isomorphism of binary quadratic forms. But then it'll be enough to show that both of these two maps given isomorphism of binary quadratic forms because this is one of them composed of the inverse of the other. So we reduce to the case where alpha is in OF. So I'm saying that the claim for alpha follows from the claim for why and claim for X. So now we have, I'm mapping to alpha times I by multiplication by alpha, which is an isomorphism, and we want to show that this preserves orientation and that it preserves the quadratic function. So we have multiplication because multiplication by any complex number preserves the orientation. Right, we know that geometric picture for multiplication by complex numbers. It's a rotation followed by a positive scale or followed by a scaling. Those operations preserve orientation. So the thing you really need to check is that the F maps are intertwined so fi going to Z and then F alpha. So we need to know that F alpha I of. Sorry, F alpha I have alpha times X that's going around this way should be the same as fi of X. But this is equal to norm of alpha times X divided by norm of alpha I. And this is equal to norm of X divided by norm of norm of I. So this is the multiplicativity of the norm which recall holds for both elements of the ring and ideals in the ring. So this is just a norm of alpha times norm of X. And down here we get norm of alpha times norm of I. And then these cancel. And so we get the desired equality. Okay. So that was B and C implies a. And the last thing we need to check. And the most complicated one is that a implies B. In other words, if you're an oriented isomorphism or quadratic forms you necessarily communicate with scalar multiplication by over. So the key is going to be the following. Instead of thinking about an I and a J and isomorphism between them, let's actually just think about an eye. You can single out the sub ring. So if you look at all z linear endomorphisms of I that's a matrix ring, it's a non commutative ring under composition and addition of matrices. So it has a commutative sub ring, given, which is given by OF acting by scalar multiplication on the idea. So, but I want to say that you can describe this subring intrinsically in terms of the quadratic form. And I'll say exactly how. But before I say how I want to explain why this, this key claim actually implies the, the, the result. So, let's assume we have some way of doing this. This is somewhat vaguely phrased. But, but well I hope it's going to be, I hope it'll make what follows. So, why is this good enough. So now suppose you have an isomorphism of quadratic forms. This is the z linear isomorphism which, yeah sends fi to fj. We're not going to use the oriented hypothesis just yet that will come in at the very end. Then you get an isomorphism by conjugation you get an isomorphism from. Well just from the fact that it's a z linear isomorphism, you get an isomorphism of rings by conjugating by alpha. You get an isomorphism of rings but moreover, it has to preserve the subring. And you get an isomorphism by the key fact. So since I say that you can single this out intrinsically in terms of the quadratic form, you have an isomorphism of quadratic forms then it has to preserve this, I mean it has to send the OF to the OF. So you get an induced isomorphism from OF to OF. And you get such isomorphisms. Identity and complex conjugation. And if you unwind what this is saying this will tell you that alpha is either OF linear. That's in the first case, or sort of OF semi linear, meaning it. I wasn't calling it alpha, I wasn't calling it alpha, I was calling it phi. Well, anyway, I'll call it alpha now. So alpha of lambda x is equal to lambda bar times alpha x for x and I, and lambda in OF. But this one will be orientation preserving. And this one will be orientation reversing. So our hypothesis that our map is orientation preserving will rule out this possibility. And then we'll get the good conclusion. Yeah, so, that's a bit of an abstract kind of argument, but I don't know. So really the key is to figure out which isomorphisms are a scalar multiplication just in terms of the quadratic form. And then we state the criterion. So, more precisely, so given A from I to I, a z linear map, A is multiplication by lambda for some lambda in OF, if and only if the following two conditions are satisfied. So, this is going to be a weird that is as a constant independent of x and y. Oh, sorry, what am I, what am I trying to say. No, so that was that was going to be the second condition so this should be equal to x y over. Oh no, I messed it up guys I'm sorry let me try again. So, sorry. If you do the inner product of a x and a y and divided by the square root of fi x times fi y. I'm going to find fi ax fi. Ah, geez, sorry guys. Okay. Let me try again. Inner product of a x, a y divided by fi ax fi times fi a y square should always be equal to the same thing without the A's. This is going to this looks mysterious but I'm going to it's going to make total sense when I get done with the argument. This quantity here doesn't change when you apply a. The second condition is that fi ax divided by fi of x is a constant independent of x. So it's an integer constant. Sorry, is there a square root missing on the right hand side of. Thank you. Thanks. It's very mysterious at the moment but it, well, you'll see where why I put these exactly these two conditions in just a second. So, well actually what is this expression here. Sorry, you couldn't see what I was pointing out what is this expression here. This is exactly if this is exactly the angle between x and y. So I think in terms of, so we have a positive definite binary quadratic form we have an associated inner product. When you have a positive definite binary quadratic form you can think of it as measuring the squared length of a vector, and then the associated inner product behaves just like you're used to for inner products in our square. And this is the usual expression for the angle between two vectors. So what I'm saying is that our linear transformation should preserve angles. I'm also saying that it should scale up vectors by. So, and this is so this is again the norms, the norm of a vector squared this fi of x. So I'm also saying that it should scale norms of vectors by a constant factor. So, so the claim is suffice it suffices to show. So we have to look at multiplication by lambda for some lambda in the complex numbers. We don't have to worry about arranging that it's integral. And the reason is that. Well, again, indeed, so you can rationalize a. You don't have to be given by the same constant multiplication by lambda but now this is equal to F, and this is equal to F. And since you have one in here. That would tell you that lambda times one is an F, which will tell you that lambda is necessarily an F. And then lambda, but lambda is also an then multiplication by lambda will be an endomorphism of I. It's a lambda is the root of the characteristic polynomial so I'm going to go very fast here. The characteristic polynomial will have integer coefficients because this is a finite free Z module of. Yeah, it's a finite free Z module of rank two. So, on the other hand, lambda is the root of the characteristic polynomial so that by definition of OF will tell you that lambda is an OF. So once you know it's just a complex scalar. It's an OF by some little argument. So, and now to check that it's a complex scalar we can actually extend scalars to the real numbers, and we reduced to a geometric fact about the complex plane. So, an R linear isomorphism, or an R linear map from C to C is multiplication by lambda for some lambda and C, if only if it preserves angles, and multiplies norms by a scalar. Well, and how do you prove this well you just look at where one goes, and then you see that where I goes is determined by that, according to these two facts. So, right. This I realized I went through this very very fast. And I think the nice thing about this argument is that it doesn't really matter what these conditions look like all that matters that it's intrinsic in terms of the quadratic form and then you can run the other argument which is kind of cute, I think. Okay, so in conclusion we have proved that theorem. So, now I want to explore some consequences. These are crazy consequence so well, I have to remind you some abstract ring theory fact so OF is a dedicated domain. And I won't remind you what that means but what's more important is the consequences so. So one consequences that every non zero ideal is uniquely a product of maximal ideals. A unique prime factorization of ideals so to speak or prime ideal prime is replaced by maximal ideal. It also implies that. Yeah. An OF module is isomorphic to some non zero ideal, if and only if it's invertible under tensor product tensor product over OF. In other words, if only. So if the module is M, then there should exist another OF module and such that you get. You can transfer them and get the unit for the tensor product. So that's a characterization of the ideals and abstractly as OF modules. And this tells you then that combining with the theorem we deduce that so corollary. The set of strict equivalence classes of binary quadratic forms of positive definite of discriminant D is in by Jackson with the set of the invertible. OF modules up to isomorphism, or I could write it in a more concrete way it's non zero ideals up to. But this is like the nice abstract one here. So this is this is called the this is something which makes sense for any commutative ring, and it's called the Picard group. And this is what's called the ideal class group. So, and these are actually the same. And I say group because these are indeed groups. This is a group under a billion group under tensor product. You have two invertible modules you tend to them together you got another invertible module, or if you have two non zero ideals, you can just take their product. And this passes through the equivalence relation he gives a group structure on the class group. So that's a little bit crazy, because the group structure from this perspective is quite opaque, although gas did sort of successfully define it by hand in fact. And this group structure is what's remarkable and it elucidates all sorts of questions, binary quadratic forms. So I think, maybe I'll have to postpone discussion of the implications of this to the next lecture. But I don't know this I this I find completely striking, I mean that this, these classes of quadratic forms have a group structure on it kind of a billion group structure in fact. Again, so in the in the start of the next lecture I'll explain how you can view these, we had these further equivalence relations so there was strict equivalence. That's this SL to z thing but then we also just isomorphism or equivalence, that was the GL to z thing. And, and then we also have this notion of being in the same genus, and we'll see that all of these courses are equivalence relations can be explained in terms of can be explained group theoretically in terms of the class group. And it's quite beautiful. Yes, there is. So by what we proved a couple of lectures ago the Picard group and the class group of ring of integers of some of the fields that we're considering now they're both finite right. Yeah, because that set of strict equivalence classes is finite need. And they have the same number of elements. So the number of elements in the Picard group is the same as the class number of D. And in particular, yeah, and there's this also this abstract fact about dedicated domains that. Well, if in terms of the class group perspective, yeah, what does it mean that the class number is equal to one it actually means that every ideal is principal so generated by a single element. And that's the case when this unique prime factorization by deals actually implies unique prime factorization of elements in your ring. So this list of nine examples where the class number is equal to one is also the list of all of the rings of integers and quadratic fields where you have unique prime factorization for the elements themselves. Another thing that I wanted to ask so we have a group structure I mean quite naturally in that that Picard group, and we know that it is in bijection with the strict equivalence classes. Is there a way to, I mean, if we suppose that we use the reduced form representatives for the strict equivalence classes. Is there a way to carry that product that we have in the Picard group, and somehow see it as a, as a way to multiply those classes, like carry it back to a product of the classes that looks. Yes, but I think the only way to do it is just to write out what the, you can take the corresponding ideals like I gave the formula for going from explicitly from a binary quadratic form to an ideal right with this a and b minus the word of the over two. So you can take the corresponding ideals multiply them find a z basis you get a new quadratic form it won't be reduced. So then you have to apply reduction theory again to make a reduced representative. So in terms of that, yeah very specific set of representative the group structure is actually highly opaque. I don't know any kind of direct description. Sorry Dustin but is this I mean is this the, this is related to this like composition of quadratic forms. Yes. So this was how it was many classically considered before the ideal. If you, if you do this you'll find you can write down a formula. I mean, for the, for the product of two things but what I'm saying is in any case this product won't preserve this fundamental domain will preserve the, the reduced condition so if you think just in terms of those reduced representatives it'll be very opaque. I see. Thank you. And see there have been some on some questions yeah, but it feels answered them. Yes, so various. So, if I understood the argument that we discussed couple of minutes ago correctly, the, the class number is equal to one. Is that an if and only if if and only if, or, I mean definitely we have one direction right that if the class is equal to one, then we have a trivial class group. So we have a PID. So we have a UFD so we have unique factorization in the ring of integers. It's the converse true the converse is true but I know this as a fact and I don't remember ever having seen a proof. So I can't give any insight into into it. Thank you. It's obvious to me at least right now, but I do know it as a fact like I read it somewhere. That's fair. Yeah. Thank you. Yeah, I think that's true for a dedicated domain that is proof, because we could try to write, we could, we could make ideal divide a principle idea by getting out of product so I think that is true for a dedicated domain. Yeah, if you remember it also being true in that abstract context yeah. So I can domain and you have a unique factorization if it only if it's a PID I think. So right. Right, I guess that's the claim. Yeah. I think it is something like, if you have a UFD, and you consider some maximal ideal, then you like pick some element, you're like maximum ideal, and you can like factor into a bunch of primes. And then, you know, it'll have to have one of these, like irreducible factors in there because it's because it's prime. And then maximum ideal will actually contain, like some prime element. And then this prime element you get will be will I generate the whole ideal. Like this like prime element will generate some prime ideal sitting inside of your maximal one, but because you're a dedicated domain then like every non zero prime ideal is already maximal. So the thing it generates must be the whole thing. And so then even amazing just proof right on right on the spot. Wow. Okay.