 Okay, well, I'd like to begin by thanking the organizers that This is not something I do these days very much giving zoom talks It's not something that I thought I would really want to do ever again I at least a few years ago. I would have had that perspective But this is a rare privilege and I'm very happy to to be here talking to all of you I see a number of familiar names in the participants list and Well, it's good to see you virtually So I'm going to begin by saying this is a talk about non unique factorization But I'm going to do a quick recap of what I mean by unique factorization to set things up So let me start with an integral domain D Then a non zero non unit element of the domain is called an irreducible if it can't be factored as a product of two non units and D is called a unique factorization domain if every non zero non unit can be written as some product of irreducibles And this expression is unique up to order and up to unit factors and because it'll be important a little bit later in the talk I'll spell out what I mean by up to order and up to unit factors If I have one product of irreducible say the PI I equal to another product of irreducible say row J then Uniqueness means that there should be the same number of terms on both sides so K should be L and the notation I have here and After rearranging the terms you can make PI I a unit multiple of row I for every I it was one through K So this is what is standard to mean by a UFD Now in a first portion algebra you meet many examples at UFD's and Here are the ones which may be your most likely to meet in a first portion algebra These are the ones that we talk about in our first undergraduate portion algebra at the University of Georgia There's the ring of integers the ring of rational integers I would expect given the audience that this is maybe everyone's favorite example of a UFD There's the ring of polynomials over a field F F bracket X and there's the ring of Gaussian integers a plus VI If you trace the history of this it's it's sort of interesting so Euclid got quite a ways towards stating unique factorization But he never quite does it so it's kind of interesting if you look at the various statements that he makes How close he gets and yet how he never takes that last jump So if you want to find the first formal statement of unique factorization Together with the first complete proof it actually doesn't seem to come until the 19th century So with Gauss and his disposition us and the proofs for F bracket X and Z bracket I also seem to be due to guess Okay now if you only see examples of unique factorization domains it's easy to be lulled into a false sense of security and There is an antidote for this so there is a very common example that is again seen in introductory algebra courses So it's become an almost canonical example of a non unique factorization domain This is the ring Z join with negative five So in this ring we have two times three is one plus root negative five times one minus root negative five and well, okay These Two sides they certainly look different But you have to check and you can check that this is a genuine example of non unique factorization All of two three one plus root negative five and one minus root negative five They are irreducible in Z root negative five And there's no chance that the irreducibles on the left are somehow secretly the same as the irreducibles on the right Up to units because you can also check that the only units in this ring are plus or minus And so you're not going to get one plus root minus five out of two times a unit Okay So zero negative five is certainly not a unique factorization domain One of the points I want to make in this talk is that there's a sense in which it's close to a unique factorization domain And you should be asking well What in the world does he mean by that? Okay So i'm going to talk about maybe What I don't mean by that before I talk about what do I mean by that but before I get to that Let me first zero in on the class of rings. I'll be thank you So the rings that will come up in this talk are all rings that arise out of the fields So a number of field just means a finite extension of q to each number field k There's a ring called o sub k Which we zero in is the central object of study. This is the inter enclosure of z and k just It's the set of all elements in k that are roots of some monocolonominal of integer coefficients We've already seen a couple of examples of rings of integers in this talk So the ring of integers of q a join i is the ring z i that already appeared The ring of integers of q root negative five is the ring z root negative five Here are a couple of other examples the ring of integers of q a join q root of two is z a join q root of two the ring of integers of z of q a join root five Is z a join one plus root five over two Okay now If you're thinking about these rings of integers, we've already seen that z a join root minus five is an example. That's not a u f d and Well, if that was the end of this talk, it would be a very sad way to end the talk and it would also be a absurdly short talk. So that's not the end there's more to the story and Well one chapter which I think is well known certainly well known to the people in this audience is that Okay, you don't always have unique factorization into elements But you always have you unique factorization into ideals So this is a famous theorem of Dedekind that if you look at any of these rings o sub k any ring of integers in a number field Every non-zero ideal of okay always factors uniquely as a product of non-zero primates So if you look in old-fashioned books, then you'll see this statement called the fundamental theorem of ideal theory I think if you look in new books, they don't give this a name. They just say it's true But I like to call it the fundamental theorem of ideal because it just sounds fancy okay so yeah as I said, this is a talk about non unique factorization and Well, I said that if you think about ideals, then you don't have these problems with non uniqueness That's too easy. So I don't want to think about what goes right. I want to talk about what goes wrong That's kind of the focus of this talk and I want to talk about how much goes wrong when things go wrong So in in terms of element wise factorization Things can go wrong, but how badly can they go wrong? And there is already a standard way of quantifying this which is discussed in first courses on algebraic number theory And I'm going to give my take on kind of the the standard narrative about this So the first thing is if we're really interested in thinking about factorization of elements Well, it was already apparent in the statement of unique factorization that we didn't really care about units So let's go ahead and remove units from the picture entirely And I can do that by just replacing elements with principal ideals So I'm going to let int print of k denote the collection of all non zero principal ideals of the ring of integers okay and then int print of k Makes perfectly good sense as an object under multiplication. It's a monoid under ideal multiplication And it's essentially just the non zero elements of okay under multiplication, but I've removed the influence of the units Okay, so I have a monoid under ideal multiplication int print of k And saying that okay is not always a u f d says that this monoid does not always enjoy unique factorization But there's another natural monoid I could consider which is just the collection of all non zero ideals of okay And that's bigger than int print. Okay And that monoid does always enjoy unique factorization. That's exactly that it can see So one way to quantify non unique factorization is to think about okay. How far is the first monoid away from the second monoid? And because the first monoid lives inside the second you might think okay A good measure of that would just be to take a quotient of some kind And that's basically what you do So I'm lying a little bit because really what you want to do is you want to look not at the integral ideals But at the groups of fractional ideals generated by the monoids on the previous slide So I'll let it k be the group of non zero fractional ideals. Okay. I'll let print k be the group of non zero principal fractional ideals And I'll define the class group as that quotient it k mod print k And yeah, this is typically how one talks about measuring the failure of uniqueness of factorization And yeah, this way of measuring things has some nice properties. So for example The class group is trivial. So this quotient is trivial precisely when okay is a ufd Okay, again, that should make some intuitive sense. Hopefully from the way I motivated these things That should say these two monoids are more or less the same And idk came from something that we already knew had unique factorization. So it's not surprising maybe when I put it like that Another beautiful and important fact from algebraic number theory is that if you take this quotient You're always getting something finite Okay, and so well, that's comforting because If you have something finite you can point to it or you can point to the size of it and say that's my measure of non unique factorization And so it's very common to hear In first courses on algebraic number theory that the class group measures the failure of unique factorization And it's true, but that's not what I want to talk about Okay, so for me, it's not that I don't believe that it's that I do believe it I believe the class group knows everything about the failure of unique factorization but I wanted to actually tell us What it knows so the hard part for me is getting it to speak and to give you an example of what I mean I want to go back to this zeroed minus five thing So we saw already that unique factorization fails in zero minus five because of this famous example here Something that is less well known though. I think well known to many people in this particular audience Is the unique factorization actually fails, but only halfway in zero negative five And what do I mean by that? Well, I'm going to go back to the conditions for uniqueness that I had up on the earlier slides I said that whenever I have some product of irreducibles equal to another product of irreducibles I want there to be the same number of terms in both factorizations And I want the two to match up up to unit factors after rearrangements Now in this example, I can't make the two sides match up up to unit factors after rearrangement But yeah, the two sides do have the same number of factors. They both have two factors And it's tempting the first time you see this example to say this is silly This is just because you were looking at factorizations of the number six If I had maybe taken a different example in zeroed minus five I would be able to cook up a case where I had a different number of factors But you can't in fact It's always the case that no matter what non-zero non-unit element of zeroed minus five You start with factor it into irreducibles, whichever way you want You'll always end up with the same number of terms So domains with this property have a name and they're studied in certain segments of the commutative algebra community The main goal I have with this talk is really to advertise What some of these commutative algebra is you're thinking about to the wider number theory community because I think There could be greater penetration of these ideas than there has been so far And then towards the end of the talk as you'll see I will advertise some new results But really I'm giving this talk to get out this perspective maybe more widely than it than it's been appreciated Okay, so a domain is a half factorial domain if every non-zero non-unit can be written as a product of irreducibles And if any two factorizations of the same non-zero non-unit have the same number of irreducible factors There's a beautiful theorem of Carlitz from 1960 that if you start with a number of field k Then the ring of integers is half factorial precisely when the class number is one or two And this explains what I said about z root negative five because the class number of q root negative five is two As a further illustration you could look at like q root negative 23 That has class number three. And so according to Carlitz it should not be a half factorial domain and it's not So here's an example if I look at 27 I can factor it is three times three times three and I can factor it is two plus root negative 23 and two minus root negative 23 You can check that all the factors involved are irreducible and well, they don't have the same number of terms the left hand side has three of the right hand side Okay, so Carlitz's theorem uh started well a number of Different investigations again mostly in commutative algebra Maybe about 20 years later. There was Uh a realization that you could pick up the thread of Carlitz's theorem and take it a little bit further So instead of just talking about When the two sides always involve the same number of irreducibles you could measure Basically the stretchiness. So if you have one factorization in the irreducibles and another factorization in the irreducibles How much can they differ? okay So I want to explain this and this also explains the strange title of my talk So let me let d be a domain where every non-zero non-unit factors into irreducibles This is a pretty weak requirement on a domain. It holds for instance for all the rings we'll talk about it holds for any no theory in domain For any non-unit non-zero non-unit alpha in the domain I'm going to define the link spectrum of alpha as just the set of all positive integers k for which alpha has a factorization as a product of k irreducibles And then I'm going to define the elasticity of alpha as Well the supremum of the link spectrum divided by the infimum of the link spectrum Now the link spectrum is is a non-empty set of positive integers So instead of infimum, I could have written minimum there that would have been fine But there are pathological domains where this numerator can be infinite. So that's why I write soup And the domains we'll care about the numerator will be finite, but there are pathological domains where the numerator can I'm going to define the elasticity of the domain as the supremum of the elasticity of alphas Taken over all non-zero non-unit alphas And then if you think through these definitions What you see is that actually Asking for d to be a half factorial domain is entirely equivalent to asking for the elasticity of the domain to be Okay, so this elasticity is somehow measuring something more general than what carlotte's was looking at You look at any two factorizations of the same element and you ask What is the the biggest possible ratio somehow between the number of terms that are enforced? and Well the main theorem I want to advertise is an answer to this question. So This theorem has kind of a funny history. The results were assembled over a number of years It may be that the first paper on this Subject to be written was a paper of valenza, but it actually was not the first paper to appear So I think it appeared in 1990 But if you actually look at the tiny footnote in the article in the journal number theory You can see that it appeared in that it was submitted in 1980 So it's something funny. There has to be a story here that I don't know but this article somehow took 10 years to appear But it was worth the wait So what is the theorem? So first let me let k be a number field. Let me assume the ring of integers is not a ufd if it is a ufd. Well, that's great then It's a ufd in particular. It's an hfd. The elasticity is one. I don't need a formula for the elasticity If it's not a ufd, then I have a formula for the elasticity the elasticity is half the davon port constant of the class And now whether or not you consider this to be reasonable theorem depends on whether you know what the davon port constant is So let me quickly review what that means So for a finite abelian group g by the davon port constant of g What I mean is the smallest positive integer d so that if you take any d elements of the group Then some sub-product has to have Uh product being the identity So every linked d sequence of elements of g has a non-empty subsequence whose product is the identity That's the davon port constant Now if you haven't seen the davon port constant before of course the the properties are not immediately apparent But I'll just quickly mention a couple that Are useful for orientation. So if you have a finite abelian group of size n Then the davon port constant is always in most the size of the d of g is in most n Any quality holds if g is sick In the opposite direction the davon port constant of a group of size n is always at least the log base 2 of the size of the group Any quality holds there if n is an elementary abelian 2 So you have d of g sandwiched between these two functions of n and you have equality on both sides in certain cases So one thing that is somewhat remarkable is that Uh, you might think that It would not be so hard to write down a formula for g of d of g in terms of structural invariance of g So there's this famous theorem that every finite abelian group is a direct sum of cyclic groups And you know you can make it unique if you impose the right condition So there should be a formula for d of g maybe in terms of the invariant factors But no one knows such a formula There was an old conjecture about what such a formula might be And the greatest progress we have is now we know that there are infinitely many counter examples to that formula But we don't actually have a conjecture for a correct formula Or uh, I mean even a guess we don't know what's going on. It's an interesting problem Okay, so as I said my main goal in this talk is really to advertise this theorem Which is not my own and I think in order to advertise that I should give you some sense of how it's proved So I'm certainly not going to go into all the details But I want to sketch a proof of half of the theorem So remember the theorem said that that the elasticity of the ring of integers is equal to half the Davenport constant Let me try to convince you that it's at least half the Davenport constant And for this I'm going to use a theorem of Landau which says that every element of the class group is represented by infinitely many non-zero prime ideals of the of the ring of integers So nowadays this is a theorem that would be thought of as A result in class field theory or a quick consequence of results in class field theory and it is but of course Landau is working before Class field theory was really a theory So let me let d be the Davenport constant of the class group So d is the smallest positive integer so that if I take any d elements of the class group Some sub product gives me the identity That means that it's possible to pick d minus one elements of the class group and have no sub product to be the identity And I do that and then I pick primes p1 through pd minus one from those classes using this theorem of land So now I have d minus one prime ideals where no sub product of those prime ideals is the identity in the class group In other words, no sub product of those prime ideals is a principal idea Well, I want to get a principal ideal out So now I'm going to choose a d th prime to be in the inverse class of p1 through pd minus Okay, so that way p1 through pd the product of all the pis will be principal And I can write p1 through pd as say pi times. Okay, so pi is a generator of that principal idea Then I claim that the pi I've just constructed has to be irreducible Why well if I could write pi as a product of non units alpha and beta Then I could look at the prime ideal factorizations of alpha or the prime ideal factorizations of beta They would each have to factor into sum of the ps one of them would involve just p1 through pd minus one And the one that involves just p1 through pd minus one Well, that gives me a sub product of p1 through pd minus one which multiplies to a principal ideal either alpha or beta And that's not supposed to happen So that means that I've constructed an irreducible pi Whose ideal factorization involves d prime ideals Okay And now how do I continue with this? Well, I'm going to go ahead and choose another set of d prime ideals In the classes inverse to the classes of p1 through pd So q1 is in the class inverse to p1 q2 is in the class inverse to p2 etc And now you can easily check that by the same argument if I were to multiply all the qis I will again get a principal ideal and it will again be generated by an irreducible. So this irreducible. I'm going to call row Okay, and now to get back to elasticity. I'm going to look at the element row times pot On the one hand and this is not the surprising part of the argument If you multiply the two irreducibles row and pi you get a product of two irreducible row pi Okay, that I think people believe The more interesting part is if you multiply row pi and look at the ideal it generates Then you just get p1 q1 p2 q2 all the way through pd qt So you get the product that p is the product of the qs And each pi qi we arranged that q was in the inverse class of p So each pi qi is principle. So pi qi is say the ideal generated by gamma i And that means that up to unit factors. I have some sort of equation row pi is gamma 1 through gamma d And I set up to unit factors But in fact all the the row the pi the gamma i's they were only determined up to units anyway They were generators of ideal. So I can make this a true equation just by multiplying row by units So this is a true equation now in okay And now on the left I have a product of two irreducibles on the right I have a product of d non units the gamma i's are certainly non units because they have prime ideal factors And I could break the gamma i's maybe further up into irreducibles But that would only increase the number of terms on the right And so I'm going to get a product of two irreducibles on the left Equal to a product of at least d irreducibles on the right after I decompose the gamma i's And that means if you remember how I defined elasticity It means the elasticity of this particular element Sorry is there a question I think someone's microphone got turned on by accident. Okay. That's right Okay, so Uh, yeah, if you remember the definition of the elasticity of row pi that means that the elasticity of that element is at least d over two And so the elasticity of the entire ring of integers, which is the supremum the elasticity of elements is also at least Okay, and that is half of this beautiful theorem that I wanted at And if you're interested in the other half, I would recommend that this paper of narkovich Which is something like two or three pages and you know does the whole thing Okay so That was the first half of the talk and the second half of the talk. I want to move into Well, the new results, but again, it'll take some time to to set up what I What I really want to talk about So the theme is something that a lot of people have looked at Which is how badly does unique factorization fail as you look across families of rings that come out of numbers And Unfortunately, there's not a lot known here rigorously Despite substantial effort. So this is a very old problem. You could even trace it to work of gauss Gauss was not really looking at quadratic fields, but he was looking at some equivalent questions in terms of binary quadratic class numbers of binary Everything I'm going to say in the rest of this talk is about what you could think of as the simplest case This is the case of quadratic fields Uh It's the case that is easiest certainly for me to study and I'll contend that it's it's hard enough. So I don't feel bad about just thinking about this case There are so many hard questions there that uh, you know, I'm I'm happy to just be thinking about this particular special case Okay, so let me review some some of what's known for imaginary quadratic fields We know that unique factorization holds only finitely off So the largest d in absolute value of larger square free d in absolute value for which the ring of integers of q root d is a unique factorization domain is d equals negative 163. This is a famous theorem of Hegner And we also know from work of hyaluron that as d gets more and more negative as d goes to negative infinity The size of the class group the class number tends to infinity and so if you think about Factorization or failure of unique factorization is being measured somehow by the size of the class group factorization gets worse and worse If you think about the failure of unique factorization is being measured by the elasticity You also get factorization getting worse and worse because the remember we had a lower bound on the elasticity of The class group in terms of the logarithm of the class group. So if one goes to infinity, so does the other So anyway, the point is factorization gets worse and worse in imaginary quadratic fields as the discriminant gets more and more negative Now for real quadratic fields, we expect the situation to be rather different And here we actually think that the class group should be trivial infinitely often in fact, there are these Beautiful heuristics of Cohen and Lester that predict that the class number of the fields q or join root p So just to join the square root of a prime number that this should be one for a little more than three quarters of all primes p So there should be infinitely many real quadratic fields with class number one. And in fact, they're not even that thin on the ground. They're pretty common Unfortunately, this is conjectural and even though there has been a lot of work understanding the Cohen Lester heuristics in recent years There's this is a beautiful project of arithmetic statistics, which has made a lot of progress The methods we have don't prove theorems like there are infinitely many real quadratic fields of class number less than a million Or even less than 10 to the 10 to the 10th I think for all we know the class number of q or d could tend to infinity with d We don't think it does but that's something that I think no one knows how to disprove So it would seem that if you really wanted to look at to get infinitely many u f d's out of rings coming from quadratic fields You'd be out of luck because there's no hope for imaginary quadratic fields And even though you should have infinitely many u f d's coming from real quadratic fields. No one actually knows how to prove that you do So what can you do? So I got into the the topic of this talk I guess the reason I'm giving this talk is at some point I stumbled across a paper by a mathematician who is Not too far away geographically from where I am. This is a commutative algebraic Jim Coykendall at Clemson University And he asked the following question. Let me not think about u f d's. Let me think about h f d's Can I find infinitely many half factorial domains? By wandering in the land of quadratic fields And the first time you hear this question you think well, no Because if you want your ring of integers to be half factorial Then carlis's theorem says that the class number has to be in most two And I just said that's only going to hold for finitely many imaginary quadratic fields And for all we can prove it also only holds for finitely many real quadratic fields So it seems hopeless to produce infinitely many a quadratic half factorial domains But Coykendall said ah, I didn't say ring of integers What if you look at sub rings of the ring of integers? Okay, so this brings me to quadratic orders So let me let k be a quadratic field An order in k is a sub ring of the ring of integers that properly contains z So the ring okay itself of course is the biggest possible such thing and that's called the maximal order And it's well known that the orders in a quadratic field are in one to one correspondence with positive integers f For each positive integer f you can look at the elements in the ring of integers Which are congruent modulo f to some rational integer So if you let o f be the set of alpha and okay alpha is a mod f for some integer a Then that gives you an order and every order is of the form o f for a unique f and that f is called the conductor of the order now non maximal orders cannot be u f d's this kind of Game that we're playing looking at non maximal orders is not going to give us any more examples of u f d's because Well, they're they're smaller than the ring of integers And so they can't be integrally closed and u f d's have to be integrally closed But they could still be h f d's and In fact, this is the kind of thing quick and all was interested So in this 2001 paper quick and all writes down what I'll call two conjectures There's a weak form in a strong form So the weak form which i've listed as part a is there are infinitely many half factorial domains Where you allow yourself to vary over all quadratic fields and all orders contained in those fields And then conjecture b is a lot more specific Don't vary over all quadratic fields. Just look at this one field q rejoin root two And then there are infinitely many half factorial domains just as you vary over the orders in that specific field q root two Now The point of these conjectures or at least certainly conjecture a is not to write down a new statement that you know Is surprising conjecture a is certainly true because as I said there should be infinitely many u f d's that appear as maximal orders of real quadratic fields But that seems very difficult to prove. We don't have any shot at that right now But maybe we have a shot at proving there infinitely many h f d's Where we allow ourselves to vary all of these parameters And the new result I'll announce is that yeah conjecture a is true and conjecture b is also true if you believe grh okay So the rest of this talk is really Explaining where these things come from Now before I do that I should say something about elasticity of non maximal orders and I don't want to say very much because Uh, I think the results known here are not as satisfactory as they could be So I said that for the maximal order, there's this this beautiful theorem that As long as the maximal order is not a u f d the elasticity is just half the Davenport constant of the classroom Now for non maximal orders, there's still a class group And you might hope that maybe the elasticity is just half the Davenport constant of that class group But that's not the case so let me give you a troubling example here Let's look at z adjoint 5i So z adjoint 5i this is an order in the gaussian field q adjoint 5i It's the order of conductor 5 And I'm not going to prove this but you can check that the class number of the order is 2 What's the elasticity? elasticity is actually in That's pretty weird. This surprised me the first time I saw it that the elasticity of z adjoint 5i is actually in and Well, I'm not going to say what what this is true if you're interested in solving this exercise You can look at the slides in detail. They'll be posted on my website or I guess this talk is being recorded But basically you can look at 5 to the k times 2 plus i Check that that's always irreducible no matter what k you pick Same for 5 to the k times 2 minus i that's also always irreducible But if you multiply those two irreducibles you get a product of 5 this huge number of times 2k plus 1 times So you get a product of two irreducibles being Well, a product of 2k plus 1 irreducibles and that means that the elasticity is at least 2k plus 1 over 2 But I could have done this for any k and so the elasticity isn't Okay, what's going on here? Well, there's a theorem of halter cock that if I'm looking at the order of conductor f Then you get finite elasticity precisely when f is not divisible by any prime that that splits completely in k So what goes wrong here? Well 5 is divisible by 5 and 5 splits in z bracket i is 2 plus i times 2 minus Okay, so this is just to convince you that elasticities of non-maximal orders are a little bit weird as an objective study But you can still say something and in fact there was an arithmetic characterization of half factorial orders in quadratic fields elasticity one orders in quadratic fields that was obtained independently by halter cock in coordinates So I'm going to say a little bit about the classification the full classification is just a little too complicated for me to want to get into now But I'll say something about the classification here The imaginary case is actually very easy and has a very simple beautiful answer If you're interested in non-maximal orders in imaginary quadratic fields There's only one that's a half factorial dummy It's z adjoin root negative three, which is the order of conductor two in the field q adjoin root negative So there's precisely one non-maximal quadratic half factorial domain z adjoin root negative three and that's a great result The characterization in the real quadratic case is not as simple So I'll just say some results here. So suppose I have a real quadratic field k Then in order for the order of conductor f to be half factorial The following conditions are necessary. So first the ambient maximal order. Okay has to be half factorial Which by charlotte's hysteria means the class number has to be one or two The conductor f itself has to be either a prime or twice a prime And in the case when it's twice a prime that prime has to be odd And the prime I've just been mentioning has to be a prime that's a nerdy k Furthermore, if we zero in on the case of orders of conductor p I just said that to be half factorial. Okay has to be half factorial and p has to be a nerdy k If those conditions are satisfied then op is a half factorial domain If and only if the two class groups coincide the class group of the order and the class group of the maximum And what do I mean by saying these two class groups coincide? Well, there's a natural Surjection from the left hand side to the right just mapping the class of i to the class of i and I want that rejection to be an isomorphism Okay, so What can I do with this? Well, I'm more of an analytic number theorist than an algebraic number theorist So this sort of condition that I have here doesn't speak to me directly But you can reformulate it a bit if you actually analyze what doesn't mean for this natural Surjection to be an isomorphism Then it turns out that there's a rephrasing in terms of the units of the ring of integers of k So suppose k is a real quadratic field and suppose I have a prime p that's a nerdy k then This equation of class groups that I'm asking for is actually exactly the same as asking that the smallest power of the fundamental unit That belongs to the order of conductor p is the p plus 1th power So epsilon to the u where u equals p plus 1 should be the smallest power of epsilon Which belongs to the order of conductor p And this equivalence between the equality and the condition on the right hand side Follows for instance from the class number formula for orders Or you know analyzing the map that was hidden on the previous slide that's rejection Okay, I won't go into why this is true as I said, you know You can look at the proof of the class number formula and you'll see it But I do want to point out that it is at least easy to see something that maybe is not apparent If this is your first time coming to the subject It is at least easy to see that epsilon to the p plus 1 does land an op I'm asking for p plus 1 to be the smallest exponent for which it lands an op But it's at least easy to see that epsilon to the p plus 1 does land an op And the reason for that is that op consists of the elements of the ring of integers that are congruent to a rational integer mod p And if you take epsilon to the p plus 1 and you reduce it mod p You get epsilon times epsilon to the p but working mod p epsilon to the p is just the Frobenius acting on epsilon and because p is inert the Frobenius is conjugation So you just get epsilon times its conjugate, which is the norm and that is a rational integer It's plus or minus So epsilon to the p plus 1 is plus or minus 1 mod p. So it's an op Okay I'm going to reformulate it a little bit more. So when is p plus 1 the smallest positive integer where epsilon to the u is No p. That's the question we're after Well, let me view z mod p as a subfield of the ring of integers mod p This is just the usual thing we do in algebraic number theory with residue fields Then asking for a u for u equals p plus 1 to be the smallest exponent I want is actually just the same as asking that epsilon be a generator of This quotient of unit groups. So I look at okay mod p that unit group there I look at the subfield zp that unit group there when I quotient it out I'm going to get a group of size p plus 1 and asking my condition on epsilon says that epsilon mod p should generate this group gp Okay, so What's the upshot here? Let me let k be a fixed real quadratic field of class number one or two Then op is a half factorial domain for infinitely many primes p If and only if there are infinitely many primes p that are nerd and k for which the image of epsilon mod p Generates this group gp depending on p Do we expect that there are infinitely many of these primes p Well, it was pointed out by alan that you need one extra condition The norm of epsilon should be negative one the norm of the fundamental unit Otherwise, you can show this as hopeless you can show that the order of epsilon Will will not be p plus 1 of p is odd will divide p plus 1 over 2 So let me go ahead and assume that extra hypothesis the norm of epsilon is negative one Then are there infinitely many primes p and nerd and k for which the image of epsilon generates this this group that varies with p Well, now this is a question that an analytic number theorist or at least one with the same inclinations as me Can really wrap it their heads around Because it looks an awful lot like an old conjecture of a meal art So ardent said, okay, let me take an integer not negative one and not a square I'm going to ask whether there are infinitely many primes p where generate g generates the multiplicative group z mod p This is a very similar question in my problem. I have a fixed something. I didn't call a g I call it epsilon. I have a group that varies with p. I'm asking Do I have a generator mod p as p varies infinitely often? Now the bad news is that ardent's conjecture is still open But it settled under grh. This was done by holy in 1967 And so this maybe gives one some hope that the problem I was interested in could be settled under grh And indeed it can be and that's essentially what I announced earlier So I'm not working in the integers. I'm working in a quadratic field So you need a quadratic field variant of ardent's conjecture not ardent's conjecture itself But these have already been investigated So I've mentioned the names here are a few people who have thought about this And there's a paper of yenmei chen which gives almost exactly what I need So there's one tiny tweak I have to make to her argument But it's almost all in her paper and well you get a nice theorem out of this Okay, so let me just announce more fully what follows from These methods on ardent's conjecture So I'm going to call the real quadratic k viable if k has class number one or two and fundamental unit of norm negative one Then if you assume grh Every viable real quadratic field has infinitely many orders a prime conductor inside it, which are half factorial domains And the scope of this theorem is best possible in the sense that if k is a real quadratic field Which is not viable then in most finitely many orders in k are half factor And that follows from results of halter concord can all and all and that was already known And okay as people who've worked on these ardent conjecture problems will know When you can prove a theorem like this in principle, you can also compute a density When k equals q root two you can actually show that The proportion of primes p for which the order of conductor p inside zero two is a half factorial domain Is half of a where a is this funny constant i've given here is an oriola product called the ardent constant Okay I also mentioned some kind of unconditional result on quadratic hfds I said that you could prove without any hypothesis that there were infinitely many Hfds as you varied over all quadratic fields and all orders contained inside them Where does that come from? Well, again, the inspiration is from work on ardent's primitive root conjecture Now unconditionally ardent's primitive root conjecture is is still open as I said before And in fact, we can't even point to a single specific integer g Which we know generates the multiplicative group mod p for infinitely many primes p For example, we don't know seven is a generator of z mod p cross for infinitely many primes p So that's terrible But the good news is that that somehow even though we can't point to a specific g which we know works We know it works for many g that we just can't point to So what does that mean? Well, there there's a remarkable Cain of events that was it was set off by A paper of murdi and gupta And i'm not going to quote the result of their paper But after their paper there was a paper of murdi schrinivasan and a paper of heath brown Where they proved a theorem of the following form independently So there's some constant m. I'll say what m is in a second Such that if you list any m different primes at least one of them will generate Z mod pz cross for infinitely many primes p And heath brown has the the better constant m using stronger server methods. He gets the m is three So if you take any three primes at least one of them will be a generator of z mod p cross for infinitely many primes p Okay, so this looks like the kind of thing that could be useful in The problem we were looking at again, we would need a quadratic field analog And well again problems like this have already been looked at there are some articles by joseph cohen in the early 2000s looking at porting over these sorts of techniques to quadratic fields What he does his theorems are not quite applicable in the context that that I want so you have to rework the method but it goes through and You can prove the follow so Let me take any 46 viable linearly disjoint real quadratic fields viable was the condition on the previous slide Linearly disjoint means that I want my fields to be have as As much nothing in common as possible So I want the composite to actually have maximal possible degree 2 to the 46 If you take any 46 viable linearly disjoint real quadratic fields There has to be one of them in that list which contains infinitely many hfd orders inside it In particular this implies after a short computation That if you look at the real quadratic fields q root d Then there's some d less than a thousand for which the ring of integers has infinitely many hfd orders and so Okay So Just in the last couple of minutes. I want to say something about elasticity is larger than one Everything so far has been about half factorial orders. What about elasticity is larger than one? so If you're going to ask that about this and Maybe make some sense to think about what are the possible elasticities That you could even see And you can show that if you take a quadratic field k Then Well, I'm not going to go into this because this description is not so important. I'll just say twice the elasticity Is the certain is a supremum of a certain set of positive integers Okay, so twice the elasticity is the supremum of a set of positive integers Which means that the elasticity itself is either infinite Or it's half an integer Okay, so that's what I really want to emphasize here the elasticity of any order inside any quadratic field Is either infinity or it's one of the numbers one three halves two five halves three seven halves, etc So what have I been looking at lately? Well, let me call the quadratic field k universally elastic If The full ring of integers is a ufd That's the only order in k that could be a ufd the full ring of integers I'm going to ask that it is And then I'm going to ask that everything in my list of possible elasticities One three halves two five halves all of those and infinity those should all occur as elasticities of orders in k But not just occur once occur infinitely often And then What I can prove is that yeah, if you assume grh Then well, there are universally elastic quadratic fields and in fact q root two is universally elastic So in some sense you see every possible factorization behavior as measured by elasticity Just among the sub sub rings of z join root two And okay This theorem here is maybe a little bit weaker than what you would expect That maybe the correct condition here is that every viable field k Which viable was just defined really to ensure elasticity one but infinitely often but I suspect that every viable k is actually universally elastic That I can't quite prove so that would follow from grh and some plausible hypotheses on Some on these primes that are somehow like v for it primes and issuance But I think that's what I wanted to I'm going to skip this I'm going to say that's what I wanted to say So thank you for your attention