 First of all, of course, I want to thank the organizers to for giving me this opportunity. So this is. Well, this is my first foray into number theory seminar. All right, so let me start this. Oops, it's not. Okay, so I will start with some rather old questions, which motivate the investigation in the stock, and there are two questions so we start with a field of algebraic numbers, possibly an infinite extension of q, and we have a ring of integers of that field. And what we want to know is, is there first order definition of the other string of integers in the language of ranks, and is the first order theory of the ring of integers undecidable. If you may be unfamiliar with all this terminology and I will explain it shortly. First of all, what's the first order language of rings. The short version is that it is a language of polynomial equations. You could think of a sentence in this language as being a conjunction of polynomial equations and it's prefaced by some collections of quantifiers. So, if I'm thinking about a sentence as opposed to formula, so all variables will be in the range of some quantifier. So the coefficients of the polynomial, maybe in Z, or we can augment our language by adding countably many constant symbols to name elements of the ring. There are some silly sentences you can write. Okay, so the first sentence is clearly true for while I'm thinking about community friends with identity, nothing strange. And the first sentence is obviously true in every ring. The second sentence is true over some rings. And the third sentence is both in all rings. So, I mean, it's natural for us to want to know which sentences of the language are true over a given ring. So the collection of all sentences that are true over a particular ring is called the first order theory of the ring. And we say that the first order theory is decidable if we have an algorithm which can decide given a sentence whether it's true or not. Otherwise, we will call the theory of the ring undecidable. So let's use language of ranks to define subsets. And the basic idea is that you leave one variable outside the range of your quantifiers. And then the set you are defining is the set of values in the ring for which the sentence is going to be true. So there are some silly definitions and there are not so silly definitions. So, for example, this is the definition of the set of even integers over Z. And it's obviously very easy to tell what's going on. On the other hand, if you write something random like some random polynomial of random degree. If you want to know what kind of a set you have defined, it may be quite difficult. For example, I have no idea what's defined over Z by the formula by the sentence I've written. So easy to see that the difficulty of deciding what you have defined depends on the ring you're looking at. So if instead of Z I look at the ring of all algebraic integers, then I can tell what's going on in particular that have defined just the ring of all algebraic integers. So it could be that one of the sets you're interesting in defining is actually a subring of a given ring. Now, why would you want to do that? Well, there is some connection between definability and decidability. The connection is if you start say with the ring R1 contained in the ring R2 and the first order theory of R1 is undecidable and R1 happens to have a first order definition over the bigger ring R2. I conclude that the first order theory of R2 is also undecidable, because if it were decidable, then we could use the algorithm for deciding the truth of sentences over R2 and the first order definition of R1 over R2 to construct an algorithm for deciding the truth of sentences over R1. So the first ring undecidability result, though is due to Hilbert Bernays, that somewhere like they had several volumes on logic it's buried somewhere there, I understand, and it's also was in a paper by Rother. As you see, these results are quite old and of course the ringer question was Z. So it was shown that the first order theory of Z was undecidable. And that, of course, is the beginning of all the other results about undecidability. So if we apply the reasoning from the previous slide with the smaller ring B and Z, then we will conclude that if the answer to the first question is yes, then the answer to the second question is also yes. So the answer to the first question will give us actually more information than the answer to the second question. Sometimes we can give a definition of a set which is uniform of a collection of objects of rings. So what we mean by that, because the notion of uniformity can vary depending on the context. Okay, so let K be a field of algebraic numbers, L a collection of extensions of K in its algebraic closure, and a subset of the ring of integers of the algebraic closure. So we say that a has a uniform first order definition of a rings of integers of all fields and now if there's a single polynomial equation. It has two sorts of variables T, and then there are axis. And only T is of course is not in the range of any quantifier and the formula QT gives a definition of the intersection of our set a with a particular ring of integers save the field L for every field L in our collection. So, so one polynomial works for every field in the collection. There is one very easy set to define uniformly and that's the set of units. Right, so let you be the set of all units in the Q bar. In any field K containing Q bar, we have the units of the ring of integers of that field. I defined by very simple formula. So, now, in describing the language of the rings we stated that given a formula in that language we can add both universal and existential quantifiers. Now, we can restrict ourselves to using just existential quantifiers. So the result of this will be a so called existential language of rings. And the corresponding theory will be the existential theory of the ring. And then the question of decidability of that existential theory is essentially Hilbert step problem for the ring. And there is the same connection about existential definability and existential decidability as the connection between first order definability and first order decidability. Okay, so a bit more history. So, the story for us starts with results of Julia Robinson, 1949, when she produced a definition of the four rings of integers of every number killed, and thus showing that the first order theory of the strings is undecidable. Her definition was of a simple form it just had one universal quantifier followed by some existential quantifiers but it explicitly used the degree of the extension. Later on she produced a uniform definition of the across the rings of integers of all number fields, but the uniform formula was more complicated. Of course, by now we have existential definitions of Z of rings of integers of many number fields. So do you know for the first person to show that Z has an existential definition of our every about to break into just a number field, and he was looking at the dimensions of degree two. This is his result was followed by many more some results of his and ellipse and other people. And so here's a list of people, not in historical order but just an alphabetical order, who produced some definitions existential definitions of Z over number fields. In 2010, Maisa and Rubin showed that Shafarevich state conjecture implies that there exists an existential definition of the ovarian rings of integers of any number field. And later on in a paper of myrti and past and another paper of a stand it was shown that other conjectures will reach the same conclusion so what this means that assuming any one of those conjectures is that Hilbert step problem is undecidable over a ring of integers of any number field. Okay, so, um, due to results of Julia Robinson, we have a pretty good understanding of the first order theory of rings of integers of number fields. But it's quite different when we look at infinite outbreak extensions of here. For one thing there are infinite extensions where the first order theory of a ring of integers is decidable, and there are other ones where it's not decidable. For example, the results of wrongly and one address show that first order theory of a ring of wall algebraic integers is decidable. And it's also more annoying that sometimes the status of the first order theory of the ring of integers, and its fraction field can be different. For example, the first order theory of the field of all totally real algebraic numbers is decidable so that's a result of read Quran and folklore. And the first order theory of the ring of all totally real algebraic integers is undecidable and that's the result of Julia Robinson from 62. And Julia Robinson was the first person to produce example of infinite algebraic extension of q with the first order theory of the ring of integers is undecidable. So she developed a method for constructing first order definitions of Z for rings of totally real integers and a method for constructing a first order model of the and also thus proving undecidability for some rings of integers that are not totally real. Julia Robinson's methods were used and generalized for totally real rings of algebraic integers and rings of algebraic integers of degree to extensions of totally real fields by a number of mathematicians so. So here's the below is the partial list of people who have produced such results. So moving on to our contribution to the literature. And so, in our case, the decidability or so our undecidability results do not so much depend on whether the field is totally real. The extension of degree to a totally real field, but whether the field is not big. So what do we mean by a big field. Well, one way you could think about it. So the field is big. If it contains a sub extension with a degree over q divisible by N for any integer and so. There is one of our main theorems and so if a is a collection of all non big fields of algebraic numbers. There exists the first order formula of the form so to universal quantifiers followed by some existential quantifiers uniformly defining the other rings of integers of wall fields in a. And of course in particular that the first order theory of the ring of integers of all non big fields is undecidable. And we also have a consequence which pertains to fields. And the consequence of if K is the non big Galois extension of Q, then the first order theory of the field K is undecidable. And that follows from the fact that in the Galois Galois extension, which is not big and such an extension. You can give a first order definition of integers of the field. And then if you remember from earlier slides that tells you that the first order theory of the field is undecidable. Okay, so now I'll try to give you a flavor of Okay, so the proof starts with a very simple idea right. If you're in any ring. Well, let's say ring of characteristic zero though. It's still the sentence that's still true and positive characteristic but we are mostly we're discussing rings of characteristics zero only so might as well see in that. So, this is a simple formula from algebra x to the n minus one divided by x minus one is equivalent to n modular x minus one. All right, so you take this formula and take it to the next stream if you wish. So you start with some algebraic extension of Q and let you sub K be the group of units of the ring of integers of K. And then you look at the set define in the following fashion so X is on the set top set of the ring of integers. Even on the if for every unit different from one there is another unit such that X is equivalent to this ratio modular the original unit minus one. So, you know we are motivated by that simple formula just showed you before, because we know that the set are sub K will contain all positive integers, because we can always take delta to be epsilon to the power n. Now, the fact that I denoted this that by our sub K of course makes you suspect that the sudden fact is a ring. And it is I mean so there's some not very difficult exercises that show that the site the status happened to be closed under addition and basically multiplication of deltas in the formula ends up being correspond to sums of access so and of course it's enough to prove it for two X one for two elements of our sub K. And it's just silly algebra exercise that you see that it works for two elements of our sub K. And it's also the set of K is also happens to be closed under multiplication. It's also not particularly difficult proof. So if you have X and Y in our sub K right and for every epsilon use of K you need to satisfy certain congrats. So you start with observing that since access in our sub K I have direct with the delta in use of K to satisfy the congrats. But then since why is also in our sub K then I can use the original delta as the epsilon and find another delta delta prime to satisfy the congrats with epsilon replaced by delta. So the end result shows you that the product is also in our sub K and the final exercise is to note that minus one is also an element of our sub K and that will complete the proof that that's in fact during again it's just an exercise with some congruences. But the end result is that minus one is equivalent to epsilon to the minus one divided by epsilon minus one modular epsilon minus one. Okay, so now now so we have a ring of integers and we can define a sub ring of that ring of integers using the units. So what do we get. Well, of course, it depends. Now, if we start with imaginary quadratic fields. Then the situation is not very difficult simply because the imaginary quadratic extensions of Q do not have that many units and you can directly see what all the possible outcomes. The first proposition is not difficult to see. Second proposition is a lot harder. So here you want to know other any circumstances where the ring you have defined is actually the original ring. And it is the case if every polynomial of the form x to the d a times x to the d minus one plus B has a root in your field. Now it's clear that if F is algebraic closure of Q that it will satisfy this condition for having roots to these polynomials. Of course, a much more interesting question is are there other fields with the same property and we do not know the answer to that. Okay, so in general what we have so if K is an algebraic extension of Q that is not big. Then either R sub K is B or R sub K is an order in some imaginary quadratic field F contained in K. Now it follows from this theorem that if K say contains two different quadratic imaginary extensions of Q. Only one of this extensions can potentially contain R sub K. Now, this asymmetry seems entirely unnatural. I mean how does R K select which quadratic imaginary extension to lend in. So that leads us to suspect that actually what happens is that if we start with something which is not an imaginary quadratic field then R K is going to be Z always whether K contains an imaginary quadratic field or not. It's just too strange that the process of constructing R sub K would discriminate between quadratic imaginary extensions of Q contained in K. Okay, now our initial goal was after all to define Z and so there could be an impediment right we cannot show that R sub K is always going to be Z even for non big fields the ring of integers of non big fields. So we have to do some work to compensate for the fact that we can land in a quadratic imaginary subfield of our original field instead of Q. Unfortunately it's not that big of a deal because you can modify the original results of the nerve to show that there is a uniform existential formula defining the ring of integers of any imaginary quadratic field. And basically what you need to do you append this uniform differential definition to the definition of R sub K and then you will get Z. So so far it looks like to define Z of the ring of integers of a non big field, we need just one universal quantifier. So an element X and OK is actually in Z if and only if the function of the following statements is true so the first you recognize that's a construction of R sub K. And then we append some existential statements in case we do land in a quadratic imaginary extension of Q instead of Q. But in this case appearances are deceiving if you look more closely. So the problem is we're not allowed to quantify over the set of units. So the best we can do is a statement of the following sorts. So you start with two elements of OK. And either you're looking at units and more specifically units not equal to one. And then essentially you do what we did in the previous slide. Or you don't have a unit. So that's the other term in the disjunction. So in order to get rid of the second universal quantifier for the case that's always not a unit, we would need a uniform existential definition of non units. But we don't know how to construct such a definition or even if it exists. At the moment we are stuck with two universal quantifiers. All right, so there is another interesting field where we can say something using this methodology and the field is called QUP is the largest Abelian extension of Q. So the question of decidability or undecidability of the ring of integers of QUP is also a rather old one. Still unanswered. So what do we know? We know that we know that the real number of integers of Q. There is a sub extension of degree two of QUP, the field we call QUP plus. It's the field of all totally real number of numbers contained in an Abelian extension. And that ring of integers has an undecidable first order theory by a result of Julia Robinson. So if you look at the first order definition of the ring of integers of Q. Then the ring of integers of that totally real extension has a first order definition over the ring of integers of QUP. Of course, the existence of such a definition would now imply that the theory of QUP is undecidable. So before we get to QUP, we need to revisit some properties of the rings we construct. Now, the idea here is that instead of using the full group of units to define O sub K, we can actually use a subgroup of units. It will also generate a ring, but essentially by the same argument as for the full group of units. But of course it can very well be a different ring depending on the field and the subgroup. So under some circumstances, the real extension of the subgroup of units can lend you in a subfield. So if it happens, okay, so if F is a subfield of K and VK contains elements of infinite order, and the subgroup itself happens to live in the subfield, then the ring we will produce is a subgroup of units. If it is in the subfield, then the ring we will produce will be a subring of the rings of integers of F. So we will lend in a subfield. And the other modification one can make, well, maybe I should say one can choose a particular subgroup VK. And the subgroup will first of all consist of nth powers of units, and all those units will be required to be equivalent to one module of sum ideal. So what happens in this case is you get some other ring, but the original ring times n will be sitting inside the new ring. Now it takes some work to show that the first part of this lemma is true, but the second part of this lemma is not hard to show. And we will use some properties of our rings. So what we want to show is if we start with an element of the original ring R sub K, then n times this element will land in the new ring. So let's choose an epsilon in the subgroup. Now since element X was an R sub K, there must be a delta in use of K such that the required equivalence is satisfied with that delta. Of course, we're not done yet because delta is only in the group of units. It's not necessarily in the subgroup. If you remember, delta has to be an nth power of some other unit. Now the requirement that delta is equivalent to one modular selected ideal, that will be satisfied automatically if we select epsilon in the subgroup because then epsilon itself will be equivalent to one modular that ideal and it will force the same thing on delta. All right, so what we can do is use our addition lemma, which basically tells you that multiplication of delta leads to our addition of Xs. Here's a reminder of that lemma. And we will apply this lemma so with all Xs being equal to X and all deltas being equal to delta. So the end result will be so we add all Xs and we'll have n of them and we multiply all deltas so we'll have delta to the power n and we get this new equivalence. So delta to the n is now in the required subgroup. So what we see is that n times X is going to be in this new ring. Okay, so how are we going to use in our situation? So if k is an extension of degree two, a totally real field and k itself is not an imaginary quadratic field, and f the maximal real subfield of k, then what will happen is that our ring R sub k is going to be sitting inside this totally real subfield. Well, why so? First of all, let's choose a subgroup and the subgroup will contain only squares of units of the original field and every unit will have to be equivalent to one modular some integer bigger than one. Then, okay, that's an exercise to show that units like that are actually totally real. I mean, basically the argument depends on the fact that if you go from totally real number field to a totally complex extension of degree two, the rank of unit groups do not change. So that's so. So now we will use our lemma on subgroups on some fields. I just want to remind you what it was. This lemma, this part, the second part of this lemma. In our case, n is going to be two. So what we conclude is that two times original ring R sub k is a ring generated by the subgroup V sub k, but the subgroup sits inside the totally real field. So we conclude that two times R sub k is in the real subfield. Well, then, of course, R sub k itself has to be in the real subfield. So if we are looking at an extension of degree two, totally complex extension of degree two of a totally real field, then R sub k is going to be contained in this totally real field. So if we apply this argument to OQAB, what we will see that OQAB contains a first order definable totally real subring. Now this totally real subring is not necessarily a ring of integers of any field. It may be smaller, but it takes very little effort to actually convert it to a ring of integers of some totally real field contained in OQAB. Now Julia Robinson conjectures in 1962 that the first order theory of any totally real ring of integers are nondecidable. Of course, this conjecture would immediately imply given our result that first order theory of OQAB is undecidable. But in fact, we don't need to conjecture that much. It's enough to conjecture that the first order theory of the ring of integers of every totally real abelian extension of Q is undecidable to deduce the undecidability of first order theory of OQAB. And or alternatively, we can try to figure out exactly what we have defined. What is this ring R sub VQAB that is defined using units and try to see whether the first order theory of that is undecidable. And as usual, I finish ahead of time. Thank you.