 Since I'm talking to so most people here don't I'm not very familiar with my subject meta Perhaps it's up directly with a problem at hand and then backtrack to tell you where it came from All right, so the start with the objects which I'm play so the number fields then global function fields and more generally function fields Of transcendence degree one of our arbitrary constant fields Of course, this is a bit misleading because the constant fields are arbitrary So the total transcendence degree could be arbitrary And I will be Discussing our immediate discreet variations. So for our purposes is just a map from a product for the field into Z plus infinity satisfying the usual conditions and If K happens to be a function field Then we will assume that the variation is trivial on the constant field So the values of the validation is zero Okay, so then Devaluation rings, that's just the rates of elements of the field where the value of the valuation is positive That ring has a unique maximal ideal Sorry, so the ring is the set of all elements where the variation is non-negative and the ideal is that will consist of all the elements of the ring which where variation has a positive value and We will often identify the variation with the ideal So if we want out by the ideal we get the residue fields in the case of a function field It's always a finite extension of the field of constants And if this extension is actually degree one, we say that the variation is of degree one We will also look at integral closures of variation rings in the infinite algebraic extensions of our product formula fields some Okay, so what's the problem? Let K be product formula field though. It's infinite algebraic extension infinity So first for the product formula field itself, we would be looking for a polynomial Fv which has M variables and a parameter T such that The sentence where EI is Some quantifies either for every or there exists So that's the sentence is true if and only if T belongs to some intersection of valuation rings for some set of variations of the field Obviously often we will just look at one variation ring but in principle we can look at arbitrary number and and We will look also into defining the integral closure of our valuation Rates in an infinite algebraic extension So these are the two problems which will come up Often in our talk So This question is a part of a more general series of questions Which has to do with Language of race So that's the language we all use and basically It's the language which underlies Statements which are essentially polynomial equation. So you start from zero and one of course you can get all the integers by iterating addition And you also have multiplication You have logical connectives But in general you can rewrite any sentence a formula of the story pulling on quantifies in front So I will only be concerned about sentences or formulas where All the quantifiers are pulled out just to simplify life so so this is a Typical formula of the language so some variables about my quantifiers someone out And of course if every oops And of course if all the variables are within the range of some quantifier you get a sentence. So just as a true or false So the general problem Define ability problem for countable rain just you can also consider uncountable ranks. You just have to be more careful Is To see if there is an algorithm to determine what a sentence of this ring language is true or false Of course, it's all going to depend on the ring and also what is the power of the language in other words What can one define? using this language so Quantifiers may range of the whole ring or a subset of the ring sometimes You want just existential quantifies with just universal quantifies or the women the number of certain quantifies So there's an infinite variety of questions there Okay all right, so Where does it all come from? I mean why do? People study these things. Well, there are some yes ethical reasons obviously and also historical ones So one point of departure was the Hilbert's test problem Which was a question about algorithmic solvability polynomial equation equations. So that was posed circa 1900 and Of course at the time there was no Motion of the algorithm. So this is a modern restatement of the problem And then there was another result Which is somewhere between the listed references? I was never able to pin down exactly which one contains what but it's somewhere in there. So Which showed so that result showed that the first quarter theory of Integers is undecidable in other words even so there is no algorithm to determine Where there is sentence of the sort where again you survive at just universal or existential quantifiers is True when variables are ranging over Z. So that was the first kind of definability type result Talking about earlier Then I guess preceding chronologically Julia Robinson Shared that Z is definable by first-order formula over Q So that's using universal and existential quantifiers And so the first-order theory of Q in the language of rings is also undecidable and Later on she extended her results to all number those Now finally the question of Hilbert was answered in 1970 the last piece was put in by Yuri Monticevich Building on work of one day, so repart now and Julia Robinson, but they actually proved a much stronger to find ability results They showed that all recursively innumerable subsets of Z and the same as the Daphontan sets so recursive sets or For our purposes will be subsets of I didn't say subsets of integers Where we have an algorithm to determine membership of the set or a program if you wish to Determine membership in the set Then there's another class of sets which are recursively innumerable So a subset is called recursively innumerable if there is an algorithm to list the elements of the set now the listing can go on forever The set is infinite and it's a classical theorem of recursion theory that there are recursively innumerable sets that are not recursive So Daphontan sets Which one can define using polynomial equations So this is sort of for more traditional number theoretic versions. So or as I think You can also see that Those steps as projections of algebraic steps or as I said as existentially definable in language of brains So what followed from Marti Savage, Dennis Robinson, and partner result is that there are undecidable Daphontan subsets of Z That immediately implied that Hilbert's steps problem was undecidable So or positive existential theory of these undecidable positive theory first of the fact that some may consider equalities We're not looking at Statements which say that something is not equal to something else We will deal with that shortly It's easy to see how this corollary will arise so Consider Daphontan definition of undecidable Set if Hilbert's test problem is decidable that for each T We can determine if the polynomial of Tx bar equals 0 has solutions of Z. However, this process will also determine whether T is an element of our Undecidable set which of course produces a contradiction not Before proceeding further some brief notes of some elementary properties of Daphontan sets So intersections and unions of Daphontan sets and Daphontan unions always you can just multiply out Daphontan definitions Intersections are Daphontan over not algebraically closed fields because In order to write Daphontan definitions and intersections we need to combine polynomials and that requires The field not to be able to break the closed I mean if we wanted to I need that condition would have to allow finite systems of equations instead of one polynomial equation and So as long the field is not algebraic with closed having one equation to find out the many is the same thing and Here's an important property of Daphontan definition in particular with Z and in general overall into really close submarines of number fields well global fields and Their algebraic extension is that the sets of not zero elements of Daphontan so that property allows us to Actually write down that something is not equal to something else Okay, so you can ask an arbitrary so you can take an arbitrary recursive ring R and buy recursive ring here Just me in the ring where you can tell the element sign how to do multiplication and addition and Then you can ask well, is there a polynomial to determine whether this equation has solutions in R? and as it stands now the two most prominent questions are probably the issue of decidability of this version of Hilbert's test problem over Q and the ranks of Integers of an arbitrary number field Okay, this is light is to eliminate Confusion which often arises when people think about Q versus Z for a long time so It goes easily one way, but not the other so if In other words if we have and now if we had an algorithm over Z We would have an algorithm over Q by just rewriting all of rational numbers as ratios of integers and Requiring that denominators are not zero. This is where we would use the fact that This element zero elements is that Fanta. So perhaps Hilbert thought that that algorithm existed and He stated quote unquote the harder problem But it doesn't go the other way the fact that there is no algorithm over Z at least not does not at least directly say anything about Q right, so one old method of showing that Hilbert's test problem is undecidable of our Some range of characteristic zero is Constructing a Davonta and definition of Z over the ring So this is a quick proof that that will do the job So suppose you had such a Davonta and definition of the some ring are So and you wanted to know whether your polynomial H Have Solutions in Z So you would set up a system Over your ring are and observe that the system has solutions in your ring even on the of the original equation solutions in Z so Since we don't have an algorithm to determine whether polynomial H has solutions in Z follows that We don't have an algorithm to solve our system in R. So in general, we don't have an algorithm polynomial equations in R So before proceeding further, I'll let me just quickly say what variations we have over Q so we identify them generally with prime numbers and Define the valuation via the order function in an obvious fashion And then the valuation ring R sub P corresponds To the set of all elements of Q whose reduced denominator is not divisible by P Now we can rephrase the problem of defining defining Z inside Q Via Davonta and definition as a problem of defining as extensually intersection of all valuation rings All right, so unfortunately This is probably not going to work and that's a conjecture of Boris which has an unfortunate consequence and Just rephrasing our problem in terms of Intersection of valuation ring doesn't add much to our knowledge about the matter Now We can also look at so partial intersection of valuation rings. So just like a subset of prime numbers and look at intersection of Primes in that set So for those rings you also have a problem of whether Z is definable of this ring and whether the rings themselves are definable of a Q and As it stands now we can define Z only if we were to find out with many primes and we Can't define the ring from Q only if we weren't all but finally many primes So these results are actually were actually contained in the original work of Julia Robinson though She was not at the time interested in existential definability. She was looking at first order definition but in the process of doing that she did Produce these two results and it took us actually some time to realize that these results were there So another way to show our decidability First order existential theory is to construct a model of the theory of a very good question So you can map An integer into some subset of the ring. So the Union of all such set has to be definable and so should be the Graphs of addition and multiplication. All right. So what else can one do if you have a Definition of a valuation ring Okay, so if we're looking at number fields, then we can also identify valuations of a number field with prime ideals. So just Look at the ring of integers of the field and use order again to define the valuation So here's a result of More or less result of Pune I mean I say more or less because it's kind of a restatement of what he did. So the difference being that Put the neutral element of an electric curve at infinity personally, I prefer to put it at zero Makes at least for me easier to understand what's going on so The electric curves were one of the devices used to generate Sets of integers for the purposes of constructing and I find a definition of Z of a ring in question I mean in general you look at any kind of Group you can define but elliptic or perhaps one of the first Objects that come to mind this connection and So the Trick is to make the coordinates be divisible by sufficiently high order of Power of the prime in question and then the last Yes, so the this lost and equality provides a way to generate integers so so So the sort of an example of an application of this guy consider an infinite algebraic extension of a number field And assume you happen to have an elliptic curve Which is the fountain stable and of positive rank? In other words this out of points over the infinite extension is the same as over a number field Just for convenience sake unless you know it's normal them on you know, I generally don't need this requirement But just makes explanation simpler so the client would be that if you have a sentence And for every Y and K infinity, so there are points being you on the elliptic curve such that that ratio belongs to the integral closure Of some Brian from the from belongs to me to closure of some variation ring from K, then X would have to be in K and the second claim is that If X happens to be a positive integer then the sentence will be true for X So there are only as you see Well, there are two variables, and they're also quantifier touch to P and Q So strictly speaking, of course, you would rewrite it in terms of other variables, but it's easier to think about it this way so I said um ultimate reason is that given an element of Well, any element of outbreak closure of Q it will live in some number field and it will be You know in its order any Brian is going to be fine So that's the underlying reason for Why this works on a more sort of specific basis? Are you just a look okay, so take an element X from your infinite extension and look at some number field containing a regional field and that element X and Take its Gala closure inside the K infinity and pick a country good any country good of your element X now since There's expression the sentence have to be true for any Y Infinity we can assume that Y is okay for the moment and then you observe then of course if they That expression is an RP infinity for X the same happens for X hat So then if you take any prime above your chosen prime P in M then from The expressions two and three you'll see that order of the difference between X and its conjugate All the K has to be eagering will that order of Y and Q But Y was an arbitrary element of K So the only way this can happen for any Y is for X hat to be equal to X and Since X hat was an arbitrary conjugate of X okay Leave that X and K. So it's Not a very complicated argument. Okay, so and then the other part of it is that you can satisfy the condition if you're starting with an integer and Basically just choose P to be a multiple not sure if I wrote them backwards. I know take a multiple of your points and Then the Things should work out. Let me go back and just remind you they will work out because of five So then the order at the difference will be bigger equal than the order at P Which in general can be made arbitrary, but in particular we just need it To be positive We need it I guess depends on why so We needed to be bigger than the order of Y at P. So but we can arrange that so if we choose P with large enough order and then the corresponding multiple of P this will work Now why is it enough to justify Order of positive integers Well, I mean we observe of course that Any element of K Can be written down as a linear combination of some basis elements in K over Q Coefficient and Q and all elements of Q are ratios of integers So in fact what we have is the first order definition of K over K infinity So we can now use the result of Julia Robinson stating that the first order theory of any number field is undecidable and Then reach the conclusion that the first order theory of K infinity is undecidable because within The statements of the K infinity we now have all the statements about K. All right, so the next question of course Do we have such a limited curves? We do We do and whether they actually happen Over the fields where we can define variations. So yes, we have lots and lots of examples, but Not yet a definitive description What class of fields will be covered by these results? All right moving on to function fields So we Again have a very similar situation We use valuations so most of them, you know, you choose some polynomial ring inside This case our rational field then all the valuation will correspond to prime ideals except the valuation Attach the degree which will correspond to an ideal in K a joint one over T And then we have a similar story in algebraic extensions as before the difference with of course with number fields is that There is no designated ring of integral function. So But it doesn't matter the definition will work either way So we take the integral closure of our original polynomial ring and then all the valuations will come from prime ideals of that Except of course the valuation which used to be the degree So that will come from the integral closure from some prime ideals of the integral closure of K a joint one over T okay, so There are a couple of results which on the surface are not related to anything I've been talking about but that play a role together this definition of valuations in in Investigation of the first order existential theory of function fields One of these things is the result of Julia Robinson on addition and divisibility. So so here we do not have a language of rings we have kind of Minimized version of it. So we're missing multiplication and instead of my multiplication We have divisibility so we can say that something divides something else So if we deploy all possible quantifiers, then we can define multiplication in this language Now then there is another result which is due to feed us So this time We have so the same language as Julia Robinson had and we also have a peculiar Relation so where x p divides y is equivalent to x being y times the power of some prime fixed prime so if you have this kind of a language you can define multiplication Existentially so just using existential quantifiers Over Z. All right, so What has that to do with function fields? so if We can define Though this is there is a type of there So that be should be set of p to the s powers in other words y should be x to the power of p to the s Not x times p to the s So in other words if we can describe the set yeah of p to the s powers of all elements of the field so if you just have Description a first-order description of p Then you can construct a first-order Model of z using Julia Robinson's result Now if you also have Diffont and definition of some valuation ring then you will have an existential model of z Over your field and then you will show that Hilbert's dance problem is not decidable in the field I Guess in principle this could be used over function fields of characteristics error, but Seems to be only practical Function, but it seems only practical to do of a function fields of positive characteristics So using this two ideas Kirsten and I Showed that the first-order theory in the language of the rings of any function field of positive characteristic is undecidable and The existential theory in the language of rings of any function field of positive Characteristic is undecidable as long as the field does not contain the algebraic closure of a finite field now We can't define the p-th powers everywhere and Actually Hector has a definition which does not depend on the characteristic But the problem is the order Okay, so as far as the open questions about variations are concerned So one of them is to define fully Valuation rings in the case where the constant field is not algebraic over a finite field What we did to get Our existential and decided the answer is out was to define a subset of that valuation ring And that was enough, but it would be nice to be able to Define the complete valuation ring and of course the thousand dollar question is to define Evaluation ring when the constant field is Contains the algebraic closure of a finite field. So those things are mysterious as it stands now Okay, so what happens when we'll look at function fields of characteristic zero We use a little curves again so here we can get away with one just was one prime because Z will be part of the constant field So if we can define so if we have an elliptic curve of rank one and if we can define The valuation ring Then we can construct an existential model of Z Well, depending if you have a first-order definition of the valuation ring You'll get the first-order undecidability and if you have an existential definition, you'll get Existential Okay, now we have plenty of elliptic curves of rank one so they exist everywhere So that's not a problem What? So what is a problem is defining valuation rings. So there is the first result of this kind Was by Kim and Raj since 1995 So they define variation Rings for the prime of degree one when the constant field was either formally real or can be embedded into a finite extension of QP now This paper has been used inside it a million of times, but I'm willing to bet no one Wrecks for this completely because the paper is completely unreadable However, okay, so We do have so we do not have a better version of the proofs there are high hopes now So Kirsten and Morabai They independently constructed an existential definition of a subset of a valuation ring in an algebraic extension so that allowed them to show that existential theory of finite extensions of Rational field is undecidable assuming the field of constants were is described in the theorem of Keeman-Rouch and I after writing Keeman-Rouch finally constructed an existential definition of a variation ring for an arbitrary prime and find an extension of KT but We should be able to do more than in the Keeman-Rouch theorem, but it hasn't been done more but Okay, so So the open questions Function fields of characteristic zero so is to figure out when the evaluation rings are definable essentially a first order and again The $1,000 question here is what happens when the field of constants is algebraic enclosed So this is so bad that we don't know anything about even first-order theory of the rational fields if you take at an algebraic closure of Q or C T we don't know anything about those fields. So and yeah, the problem being that the field is algebraically closed and we do not know how to define valuations in those cases Okay, and of course if it happens to be the case that valuation rings are not definable and actually there is no firm Believe one way or the other about this about this question. So then we of course would have to find another method for showing that the Corresponding function fields have undecidable existential first-order theory. I guess we believe that the theory will be undecidable just something else would have to be done now I must say that if the Transcendence degree of the constant field is greater than one Like if you have two or more variable then We know that we have results which say that it's due first to Kim and Rauch and then to Kirsten that The theory is undecidable existential theory is undecidable and that method does bypass the definition of order So we might have to do something like that for For Algebraic So for Q tilde where Q tilde is algebraic closure of Q eternity or CT So, how is it done? Well in general, it's a pretty nasty exercise. I will just give you some ideas Sort of where the story begins and also there are many variations on the theme I think The original method goes back to Julia Robinson use quadratic forms and I will show you in a second the idea behind it But first some sort of elementary observation, so if you Have so let's okay, so We can reduce the question of whether something in the valuation ring to Divisibility of water by a given prime So let's say P is a prime whose valuation ring we're seeking to define and Q is some other rational prime Now if you look At this expression so order P of a is one so that's crucial And if we if you look at the expression and this is also one of many variations on the theme Expression is X to the Q over a plus one over a to the use power It's easy to see that it's equivalent to zero what you even on the F Axis integral at P. So indeed if X is the order of X at P is bigger equal to zero And then the order of X Q over a is bigger equal to minus one and the second term one one over a Q will dominate so the Order of the sum will be minus Q, which is of course zero amount Q however, if order of X at P is less than zero and then the first term will dominate the situation and The order will not be zero Okay now so this is the original quadratic form used by Julia Robinson and Reused on many occasions ever since Okay, so this form Over Q, let's say that is a tropic even on the F If they exist a prime P such that order of B At P is odd and C is not a square mod P or the other way around The form is clearly symmetric and B and C So fix a brand P and see this not equivalent to a square mod P and again let order A at P be one and substitute for B the expression we were talking about So then this equation has no trivial the form equation has no trivial solution Then order of B and P must be even and therefore order of X and P must be bigger equal to zero So that is the easy part the messy part is to Make sure that we do have not trivial solutions to our form equation if our X is Integral at P. Okay, so the main To here is Hassan Minkowski local global principle so basically to show that we have Non-trivial solutions. We have to make sure that we have non-trivial solution in every completion But fortunately, it's usually not that big of a deal Basically because I've had so slow and so but you have to walk around sometimes you end up with a couple equations You do something to coefficients And again, there are plenty of variations on the theme, but that's generally what people do Question the form equation the quadratic form equation to see that you're actually looking and a norm equation norm of degree two and Once you observe this can occur to you that actually it would be useful to get away from degree two because there are situations Where degree two is not desirable for example when you're working in characteristic two or When we're in an infinite extension and the local degrees of factors of p are divisible by arbitrarily high powers of two so wrongly was the first person to Just make this activation. He was working of global function fields and He set up a norm equation to Define order So that was instead of Hassan Minkowski use Hassan norm principle To make sure the equations are satisfied When C does not Have a negative order at P and again, it's the same story you just look at things locally and You might end up with a couple of norm equations to account for all the possibilities, but locally things are easier Because of Hansel's lemon Okay, I'll stop here