Since I'm talking to, so most people here don't, are not very familiar with my subject meta,

I'll hop stop directly with the problem at hand, and then backtrack to tell you where it came from. All right, so the, I'll start with the objects,

which I'm playing, so the number fields, the global function fields, and more generally, function fields of transcendence degree one over 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 comedian discrete variations, so for our purposes, just a map from a product forming field

into z plus infinity, satisfying the usual conditions, and if k happens to be function field, then we will assume that the valuation is trivial on the constant field, so the values of the valuation is zero.

Okay, so then, the valuation rings, that's just the rings 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 valuation is non-negative, and the ideal is the, will consist of all the elements of the ring, which, where valuation has a positive value, and we will often identify the valuation 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'll say that the valuation is of degree one. We will also look at integral closures of valuation rings in the infinite algebraic extensions of our product forming fields, some of them.

Okay, so what's the problem? So, let K be product formula field, or its infinite algebraic extension, K infinity. So, first, for the product formula field itself, we would be looking for a polynomial, so FV, which has M variables and a parameter T,

such that the sentence where EI is sub-quantified, so either for every or there exists, so that 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 we will look also into defining the integral closure of our valuation rings 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 rings. So, that's the language we all use, and basically, it's the language which underlies statements which are essentially polynomial equations.

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 sort, pulling all quantifiers in front.

So, I will only be concerned about sentences or formulas where all the quantifiers are pulled out, just to simplify life. So, this is a typical formula of the language,

so some variables are found by quantifiers, some are not. And of course, if every, oops, and of course, if all the variables are within the range of some quantifier, you get a sentence, which isn't true or false. So, the general problem,

definability problem for countable ring, just, you can also consider uncountable rings, you just have to be more careful, is to see if there is an algorithm to determine whether 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 over the whole ring or a subset of the ring, sometimes you want just existential quantifiers,

or just universal quantifiers, or limit the number of certain quantifiers, so there's an infinite variety of questions there. All right, so where does it all come from? I mean, why do people study these things? Well, there are some statistical 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 equations. So, that was posed circa 1900. And, of course, at the time, there was no notion of the algorithm,

so this is a modern statement 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-order theory of integers is undecidable, in other words, even. So, there is no algorithm to determine whether a sentence of this sort,

where again, these supplies are just universal or existential quantifiers, is true when variables are ranging over Z. So, that was the first kind of definability-type result I was talking about earlier.

Then, I guess, proceeding chronologically, Julia Robinson showed that Z is definable by a 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 range is also undecidable.

And later on, she extended her results to all numbered values. Now, finally, the question of Hilbert was answered in 1970. The last piece was put in by Yuri Matissevich, building on work of Michael Deiss, Hilary Potner, and Julia Robinson.

But they actually proved a much stronger definability result. They showed that all recursively enumerable subsets of Z are the same as the Dafontin sets. So, recursive sets, 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 of the set. Then, there's another class of sets which are recursively enumerable. So, a subset is called recursively enumerable if there is an algorithm to list

the elements of the set. Now, the listing can go on forever if the set is infinite. And it's a classical theorem of recursion theory that there are recursively enumerable sets that are not recursive. So, Dafontin sets, the sets which one can define

using polynomial equations. So, this is sort of more traditional number theoretic versions. So, or, as I think, you can also see that those sets as projections of algebraic sets, or as I said, as existentially definable in language offerings.

So, what followed from Marty Savage, Davis, Robinson, and Putnam result is that there are undecidable Dafontin subsets of Z. That immediately implied that Hilde's step problem

was undecidable. So, or positive existential theory of Z is undecidable. Positive theory refers to the fact that when we consider inequalities, 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 Dafontin definition of undecidable set. If Hilbert's step problem is decidable, then for each T, we can determine if the polynomial F T X bar equals zero 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. Now, before proceeding further, some brief notes on some elementary properties of Dafontin sets. So, intersections and unions of Dafontin sets are Dafontin unions always.

You can just multiply out Dafontin definitions. Intersection are Dafontin over not algebraically closed fields because in order to write Dafontin definitions of intersections, we need to combine polynomials and that requires the field

not to be algebraically closed. I mean, if we wanted to omit that condition, we would have to allow finite systems of equations instead of one polynomial equation. And so, as long as the field is not algebraically closed, having one equation to find out the mean is the same thing.

And here's an important property of Dafontin definition, in particular, of Z. And in general, overall, integrally closed sub rings of number fields and, well, global fields, I think, and their algebraic extension is that the sets of non-zero elements are Dafontin.

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 by recursive ring here, just mean a ring where you can tell who the elements are

and 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 rings of integers of an arbitrary number field.

Okay, this slide 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, in other words, if we have, if we had an algorithm over Z, we would have an algorithm over Q by just rewriting all the rational numbers as ratios of integers and requiring that denominators are not zero.

This is where we would use the fact that the set of non-zero elements is that fountain. 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, so. Right, so, one old method of showing that Hilbert's test problem is undecidable

over some ring of characteristic zero is constructing a d'Avante definition of Z over the ring. So, this is a quick proof that that will do the job. So, suppose you had such a d'Avante definition of some ring R and you wanted to know

whether your polynomial H had solutions in Z. So, you would set up a system over your ring R and observe that the system has solutions in your ring even on the F, the original equation has solutions in Z.

So, since we don't have an algorithm to determine whether polynomial H has solutions in Z, it follows that we don't have an algorithm to solve our system in R. So, in general, we don't have an algorithm for solving polynomial equations in R.

So, before proceeding further, let me just quickly say what variations we have over Q. So, we identify them generally with prime numbers and define the variation via the order function

in an obvious fashion. And then the variation 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 Z inside Q via de Fontaine definition

as a problem of defining an extensually intersection of all variation 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 variation ring doesn't add much to an average about the matter. Now, we can also look at partial intersection

of variation rings. So, just select 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 over Q.

And as it stands now, we can define Z only if we invert finitely many primes and we can define the ring from Q only if we invert all but finitely many primes. So, these results 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 a decidability of first order existential theory is to construct a model of the theory of a ring in question. So, you can map an integer into some subset of the ring. So, the unit 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 Poonen.

I mean, I say more or less because it's kind of a restatement of what he did. So, the difference being that he put the neutral element of an elliptic curve at infinity. Personally, I prefer to put it at zero. It makes, at least for me,

easier to understand what's going on. So, elliptic curves were one of the devices used to generate sets of integers for the purposes of constructing a Difonta and definition of Z of a ring in question. I mean, in general, could we look at any kind of group you can define?

So, elliptic curves, perhaps one of the first objects that come to mind in this connection. And so, the trick is to make the coordinates

be divisible by sufficiently high order power of the prime and question. And then the last, yes, this last inequality provides a way to generate integers.

So, this is sort of an example of an application of this guide. Consider an infinite algebraic extension of a number field. And assume you happen to have an elliptic curve

which is Difontan stable and of positive rank. In other words, the set of points over the infinite extension is the same as over a number field. Just for convenience sake, I'll assume it's normal, but in general, we don't need this requirement,

but it just makes explanation simpler. So, the claim would be that if you have a sentence, and for every Y and K infinity, so there are points P and Q on the elliptic curve, such that that ratio belongs to the integral closure

of some prime from the, belongs to the integral 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 attached 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, why Sam? Ultimate reason is that given an element of, well, any element of algebraic closure of Q,

it will live in some number field, and it will be, you know, in its order, any prime is going to be finite. So, that's the underlying reason for why this works. On a more sort of specific basis, you just 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 Galois closure inside the K infinity, and pick a conjugate, any conjugate of your element X.

Now, since this expression, the sentence had to be true for any Y in K infinity, we can assume that Y is in K for the moment, and then you observe then, of course, if 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 over K has to be bigger than 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 over K, you can leave that X in 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, oh no, that's okay.

Take a multiple of your point, and then the things should work out. So, let me go back and just remind you, they will work out because of five. So then, the order of 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 Y. So, we need it to be bigger than the order of Y at P. 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 just define 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, coefficients in Q, and all elements of Q are ratios of integers.

So, in fact, what we have is a 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 elliptic 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 of what class of fields would be covered by these results.

All right, moving on to function fields. So, we, again, have a very similar situation

with variations, so most of them, you know, you choose some polynomial ring inside, in this case, a rational field, then all the valuation will correspond to prime ideals, except the valuation attached to 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 functions, 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 they'll play a role together with definition of valuations 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, here we do not have a language of rings. We have kind of a minimized version of it. So, we're missing multiplication,

and instead of multiplication, we have divisibility. So, we can say that something divides something else. So, if we deploy all possible quantifiers, then we can't define multiplication in this language. Now, then there is another result, which is due to FITAS.

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, oh, there's a typo there. So, ZP 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 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 a 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 problem is not decidable in the field. I guess, in principle, this could be used over function fields of characteristic zero, but it seems to be only practical of a function,

but it seems only practical to do over function fields of positive characteristic. So, using these 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 define the P 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 valuations are concerned, so one of them is to define fully valuation rings

in a case where the constant field is not algebraic or a finite field. What we did to get our existential undecidability result 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 a valuation ring when the constant field contains the algebraic closure of a finite field.

So those things are mysterious as it stands now. Okay, so what happens when we look at function fields of characteristic zero? We use elliptic curves again. So here we can get away with one,

just with 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 the existential undecidability.

Okay, now we have plenty of elliptic curves of rank one, so they exist everywhere, so that's not a problem. So what is a problem is defining valuation rings. So there is, the first result of this kind

was by Kim and Rausch in 1995. So they defined valuation 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 and cited millions of times, but I'm willing to bet no one grabs for this completely, because the paper is completely unreadable. So, however, okay, so we do have,

so we do not have a better version of the proofs, and there are high hopes. Now, so Kirsten and Morebai, 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

was described in the theorem of Kim and Rausch. And I, after rewriting Kim and Rausch, finally, constructed an existential definition of a valuation ring for an arbitrary prime and finite extension of KT.

But we should be able to do more than Kim and Rausch theorem, but it hasn't been done more, but okay. So the open questions, function fields of characteristic

zero, so is to figure out when the valuation rings are definable, and substantially in first order. And again, the $1,000 question here is what happens

when the field of constants is algebraically closed. So this is so bad that we don't know anything about even first order theory of the rational fields. If you take 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 variations 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 belief

one way or the other 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 you have two or more variables, then we know that we have results which say that it's due first to Kim and Rausch, 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,

like you're in T or C, T. So, how is it done? Well, in general, it's a pretty nasty exercise. I will just give you some ideas of 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, who used 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 evaluation ring to divisibility of order by a given prime. So let's say P is a prime,

whose variation 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 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 Q's power. It's easy to see that it's equivalent to zero mod Q, even on the F, X is integral at P. So indeed, if the order of X at P is bigger or equal to zero,

and then the order of X Q over A is bigger or equal to minus one, and the second term, one over A Q will dominate. So the order of the sum will be minus Q, which is of course zero mod Q. However, if order of X at P is less than zero,

then the first term will dominate the situation, and the order will not be zero mod Q. 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, is anisotropic, even on the F, if there exists 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, in a sense. The form is clearly symmetric in B and C. So, fix a prime P and C that's not equivalent to a square mod P, and again, let order of 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 at P must be even, and therefore order of X at P must be bigger or equal to zero. So that is the easy part.

The messy part is to make sure that we do have no trivial solutions to our form equation if our X is integral at P. Okay, so the main tool here is Hasse-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 of Hansel's lemma. 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?

Okay, the form equation, the quadratic form equation to see that you're actually looking at a norm equation, norm of degree two, and once you observe this, can I go 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 you're in an infinite extension and the local degrees of factors of P are divisible by arbitrarily high powers of two.

So, Ramelli was the first person to just make this observation. He was working over global function fields and he set up a norm equation to define order. So, that was instead of Hasse-Minkowski, you use Hasse 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 lemma. Okay, I'll stop here.