E Artin's 'Über die Zerlegung definiter Funktionen in Quadrate' (part 1, contd.)

0:00 So this is the one that will construct the star, which will be in fact a graph of some analytic function of a peak. So this would be, in fact, a connected component bigger and bigger is not too big, it's E. So this line is in EN plus R, but R is the dimension of B. Well, then he uses here algebraic geometry saying that if you have a non-singular variety of co-dimension R sitting inside 2R plus 1, you can project without introducing additional thermodynamics. So this gap can be, you can reduce it to 2R plus 1. I guess. I guess that's the right thing you have to do. Followed by the other paper, the one just like Arden, on the representation of positive semi-technic functions and we'll begin that one with danielle gondar she will do the first part of that and give an overview and then i'll do the second part of it the group of the door accepting problems and then there are other parts of that paper which we won't do it all today and then friday we'll discuss it more so uh we begin with anna thank you so uh i'm going to go of Artin et Schreyer, algebraic construction of a real field. This is a classic, maybe

2:30 the classic. You know, this time certainly not true, the content of this paper. So the main interest will be to see exactly how, what kind of statements we find in, and maybe those we find, maybe don't expect to find in the paper. So we are just going through this paper and to look, I will stress the organization of the paper which I think is very significant. So, we have it only in German and French. No, you must, you are supposed to have a translation, I will explain now, of Van der Parten. No, you? No. Algebraic, are you real quick? I come to it, it's almost, it's a translation in fact. So, as the point is raised, in fact, and it is important, the confirmation, it is an illustration, I will give you the proposition on the wall. So, in fact, it is interesting to see that, in fact, the presentation of Van der Waarden in Moderna Gebra of 1930 is exactly a reproduction of the paper of Artin Schreiber, word to word. With few exceptions. So, I realized that late, but that means that if you want a translation of this paper, just look at the translation of the Grandner Harden book, which is easy to find. Okay, it's really true, with few exceptions, I will point.

5:00 Not added. Their statement transforms as exercise. So, I'm starting with the introduction. The paper has three parts. three distinguished parts I'm not counting the infodiction it's not even without title so let's say four parts an infodiction the first part which is definition and main characteristic of bureau field the second part which is existence and unique theorem and examples and application okay so don't in the paper of Van der Garten because you won't find exactly this composition. I will come to that later. It's the same from the introduction. The introduction of Art in Schreyer is not in Van der Garten. So I will present it now quite precisely. So if you want to follow you have to look at the German translation or to the French translation translation and she will make some English translation soon so it is important this introduction is quite important so it's significant that it's not in Van der Waalden. Artin and Schreier are saying that they are following Steinitz's paper of 1910 about their theory of corpse. Fear, sorry. And they are concerned. So the claim of this paper is to include real algebra, in the usual sense, and parts of algebraic number theory, in the usual German sense, into modern algebra.

7:30 That's the progress. It's a kind of translation. How to do that? The way to do that is to give an axiomatic presentation of real algebra, of common real algebra. But for that, Artin and Schreiner are stressing three conditions. conditions. Those conditions are, and they are very important. First, and I'm quoting, an absolutely algebraic field, for example, can be said to be real only if it is isomorphic with an algebraic real field in the usual meaning. So the axiomatization is correct if it satisfies this first condition. Can we talk about this? No, it's in German. Yes, yes. John, I have an absolute impression. Ok, I think I'm in that enumerated as well. No, no, it's my enumeration. That's what I want to be able to follow. Yeah, you're right. The second condition is to give purely algebraic proof of the existence of real field. And, but we can have more real fields in the sense that they are true real fields. Okay? Third condition is to prove some real, to prove the real algebra theory. That means to give some algebraic proof of those common real theorems.

10:00 What were the common real theorems? Well, I don't know that. You have a list of them. He seems to know what they are, and we are going to have a list of some of them. Is this coming? Is it on the paper? Yeah. Yeah. We will get this point. So, and he says, at the end of the first paragraph, there is a mention of the schlomster, which is one of them, but we will have to close it. End of the first paragraph. Yeah. So there are the three conditions of each of different kind. The definition has to use only operation plus and ratification and not, it is explicit, we prefer that to use the notion of order. It is mentioned and excluded. And the characterization is this one, that if you have the sum of squares which vanishes, that means that each term is null, which is equivocated to minus one is not the sum of squares. And, last aspect of this introduction, it is said that, thanks to this theory, we are exposed to the 70s Gilbert problem. OK, that's the content of the introduction. So, maybe let's go now to the first part. Yes, I wanted to say something. They are also saying that this notion that to be a real field,

12:30 that minus 1 is not taken up as a sort of square, is coming with, well, Française, c'est le condition de l'imposé, so, in this condition, a beard appeared kind of, you're right, inevitably, you're right up to. Now, the point is that, I should have mentioned it, you're right, that Artin, in 24, did a work, It seems to be like a student exercise to characterize the subfield of the algebraic number which has co-finite dimensions. That means if you have the subfield of algebraic number which by a finite extension are algebraically closed. And doing this, he arrives with the characterization as minus one is not a sum of square. So this is the characterization of a little close... Yeah, this condition appears like that. That's the frame where these conditions are stressed. What? Can you blow up a little bit smaller? I can't keep cold. Make me a little bit bigger. It's small. Yeah, yeah, I know. it should be bold we can wait that the Sun is lower it's better I'm sorry sorry I'll show you my computer. It's black. So, well, we are coming to the definition. So this port is now inside Van der Waarden. Okay, the statements are the same as Van der Waarden and...

15:00 Ah, but where do you have any? Yeah, the point, yeah. Okay, so what is the differential of a real field is when minus one is not expressible as the sum of squared. One more time, the point is not to, you won't find new definitions. What is interesting is to see what is given. Don't expect surprises. So now a field, the definition of an order field. And it is given by axioms, which are those axioms you can see, and you know. I would stress that you have no cone defined there. The order is defined axiomatically in this sense, not with a cone. Then, we have, we come to Theorem. Sorry, first, he mentioned that a field pay is real clause if P is real, but no proper algebraic extension of pay is real. That's the definition of real clause, without surprise again. Then comes the theorem 1. Every real truth field can be ordered in one and only one way. And the argument uses the fact that every element in p, not null, is either itself a square or minus a is a square. okay that's the center of the curve so we are still we are now at the theorem 2 which is both an example of classical real algebraic theorem here proved algebraically which

17:30 is that every real closed field sorry no sort of to that I skipped to that a real close field in a real close field every point of the degree has at least one root okay So it's clear that Van der Varden is real translation, except that in Arpinsteier paper, it says real into quotes to mention this abstract notion, and Van der Varden translates it as formally real. Correct. That's one of the distinction, which is significant. And another one is that Van der Varden considers only comfortable field. deals with countable fields. Is that necessary to... Yeah. Later, later, later. No, but later. Later. Maybe. We come to it. We will see it later. Then, theorem 3a, which say that in an order field k, every positive rayman possesses a square root, and every polynomial of all degree has at least one root. then the field obtained by adjoining E is algebraically closed, and he gives two proofs of that, one which you have in the translation, on the paper of, in the book of Van der Varden, and another one dealing with Galois theory, which is very simple, I don't reproduce it, it's not in Van der Varden, I can show you after if you want, brief and elegant Galois theory proof of this statement with using just silo theorem and you have it. Then now we came to the link mentioned by Marc-Francoise with the theorem 4 which say

20:00 that an algebraic closed field with zero characteristic p, a subfield of omega from which omega is generated by an elementary extension, then p is real cloud. Okay. Then, theorem 5, we have another statement. Then, the theorem 5, which is Aswander, Arlenkollib, Nuschelenzatz, or Bayer-Strauss. So then theorem six, what we're expecting, a list of real algebra theorems which are true in this context. So theorem 6, you have the theorem of real algebra are true in a real closed field, and it is important to mention that you cannot give a precise statement that, and those theorems are polynomials are uniformly... In the first proofs... This is not in Van der Vaarden. This statement is not in Van der Vaarden, sorry. It is one of the statements... I'll come back. Yeah. In the first proof of a field which... where squares, where positive... where squares... Another field where square positive and where polynomials of odd degree add as root, have a root, adjoining the square root of minus one, gives an elevated closed field. The first proof, he says that it is Gauss, but it is Laplace.

22:30 It is Gauss, the problem is that it is Gauss, I think that's probably the last thing. Was Laplace satisfied by ghosts? There is a discussion because in a way Laplace is saying that there is a field where all the routes fly. So this view from Laplace was really worked by ghosts. It was the same idea, but there was a discussion about the red, because at that time you had no speaking theater like that, and so it was pretty good. Then part one ends with this theorem stating some real algebraic theorem which can be proved purely algebraically. One of them, which is quite important, is that you have a bound for the roots of the phenomenon. That finished to satisfy the first condition we mentioned in the introduction. Okay? To be able to reproduce the common real algebraic statement in this frame. I don't understand. The fact that the binomial are reformed contiguous. Still, I don't see what it has here. Uniform contiguity of polynomial, because I, well, following, I understand, all the range from the steps that will be used afterwards, why uniform contiguity of, why, as they mentioned, The polynomials are uniformly continuous, so I cannot figure out.

25:00 Why is it there? Why is it there? Yeah, it's nice to learn. Sorry. Uniform continuity is better than continuity. it's not you you have it sorry i remove it it's there it's the in the beginning the theorem of the first one is polynomial polynomials are uniformly continuous in every interval. So this one is not in non-elarmament. This particular, as far as I know, no. And now we come to the second part which concerns the proof of the existence and unicity of real closed field. First, the third rank 7, so I think it's the same numbering here in von der Marden translation. Let K be a real field and omega algebraic closed field over K, then there exists at least one real proof of A between K and omega, so that omega is the expansion of p. For the curve, we use a well-ordering of the element of omega. So here, certainly in Van der Parden, you must have countable in Van der Parden. some consequences of this theorem were known so maybe I will skip them now the theorem 8

27:30 which states that the difference between in a white-white, on the bottom of the table is some scrupulous about the choice. He replaces choice by dependent choice. He is replacing choice by dependent choice. The construction of the real closure works with dependent choice. And the factoring polynomials were very careful. So there is a note, after I've been found out a little bit, the slight modification, the above theorem, and also values, and the basis of the rule of order. In a note, in Wonderland, it's a note, so he's putting countable in the statement, and then he says immediately that it was uncountable, and said that I'm going to use some well-ordered. He does it in two steps, where he says, if it's true for a well-ordered set, and in fact, well if you're willing to assume Dermilo's back from the fence. Yeah, one thing that doesn't appear in the theory of 7a, that doesn't appear very clear, is the fact that, you know, there is one real closure exactly for each order of the field, because he's talking about the form of the real field, so there are possibly many others, so there is one real closure. it's not that there are eight yeah yeah the uniqueness of uh i know it's okay yes it wasn't for it seems to me that for a long time even in brilliant minds it was not very clear

30:00 That's why it's interesting to look at the paper, because here it's clear. It's there. The theorem 8, and it's really the main theorem of this part. So you have this uniqueness extension, and it is where the Sturm theorem is used to have this unicity. So, it is state, first, he proved this lemma, that you obtain the real closure just by hiding all the roots of positive elements and he established the isomorphism between various real closure by using storm error saying that you can count the roots in k and not in the extension So they will be the same for all. And the same for the order. You can prove that order are observed because it is possible to express that roots are, that one root is bigger than another, considering a polynomial with which roots are such that their square are those three. So, you can express that order is defined using polynomial and Sturm-theoray. Like that, you have the preservation of the order. And now we come to the part you won't find in Farnia Pardone, which is then maybe less known. It's, so you can find it just on the screen or on the German edition, it deals with Archimelian

32:30 fields. So he defined a subfield considering G an order field and K a subfield of G. He he said that an element alpha of g is infinitely large relatively to k if alpha is greater than c for every element of k why infinitely large just larger than every element in k okay and the same for infinitely small and then another field containing k is called to k if it contains no infinitely large element relatively to k okay well there are many commentaries but it's not the moment to say and you have the definition of maximal archimedian relative to k expected which is what we expect. And then you have the theorem 9 saying that considering a real close field P and a subfield K of P then the maximal Archimedian subfield of P are equivalent extension and they are real close. The existence is not a problem. The proof of that is the fact that if you have a root of a polynomial in A, it will be already in, the root will be in A because of the condition that roots, you have a bound for the roots of a polynomial, and then as the field is Archimedian, the root will be in A. The central ID in this Archimedian field is because roots

35:00 are bound, you are sure that all the root will be in the field when the field is a maximal Archimedean. And, well, I don't go into the detail, I just, what I wanted to say about issues of maximal Archimedean, and then you have to prove that they are all isomorphic. Yes, which is important also, so I say a few words, because to have this isomorphism of the maximal art media he used a representation of those fields considering the gamma which showed which is all the non-infinity large element of P and making the quotient by the set of infinitely small elements. And that's the representation, in fact, of all the maximal Archimelian field relatively to take. Then you have this unique representation, then you have the isomorphism between all of them. Okay? But the spirit of the proof is different than the rest of the paper, and it's interesting to mention that all this is steeped in Van der Harden's presentation and in many presentations of this paper. It's an interesting choice because it is not useful. If K is ordered, so the Archimedean closure is well defined. The meaning of the line is not so evident. It says that Archimedean of k inside the pi are k isomorphic. Extensions de k equivalents, c'est-à-dire qu'elles

37:30 sont k isomorphes. Yeah? They're k isomorphic. There is a k isomorphic from one Archimedean for zero and another one What is that? Can it be done to this later? There is a variation of the valuation associated to an argument, and what it is before it was there, and at the same time. A little later. We are at the same time. None of them are there. There are 26, 27, and 20 seconds. But it's in here. And what is maybe worth mentioning is that one of the differences between the two editions of Van der Baarden, the first and the second, is precisely the vibration. It's stressed in the second edition because of algebraic geometry. the development about vibration, which were not present, or very few, in the first edition. So, it's interesting to mention, maybe. The main point of the third part is theorem 10, where we establish what corresponds to the condition 1 I mentioned in the introduction. That is that an absolutely algebraic real field k, absolutely algebraic means algebraic over q, over rational numbers.

40:00 So, an absolutely algebraic real field k is always isomorphic to a field of real numbers. Real numbers here is a subfield of the real, the real reals. and reciprocally every order k generates a unique isomorphism between k and the field of real real numbers so that the order of k is conformed into the natural order of the real field in the German So that articulates the usual acceptation of real with what Van der Varden calls the forward real. But we have a statement, which is also in Van der Varden, you find this statement. And I just mentioned that at the end, he's dealing with a theorem element I skipped, but it's interesting. And he gives, at the end, two counterexamples, which I don't know why he's concerned with the fact that order fields are not necessarily separate. and stress this point then we cannot use separability in previous proofs and it's not clear for me because why he has to make this point maybe well at the end It's the, it's, yeah, it's the last, uh, the penultium paragraph. Okay. Which is the Schleswig? What's this? Oh, here. Again, it's concerned. I don't know why it is so important for you to establish that.

42:30 To establish what? Sorry, to establish that... To reestablish these... No, in order, if you have not the hypothesis that the real field is closed, you don't have separation. separation you were close fielded earlier but if you don't have the close condition you don't have separation then we give an example of that i think he mentioned in the second paper when he proves over 17th problem that he needs certain hypotheses on his ground field either it's archimedean or or something like that and otherwise you cannot separate two roots of a you cannot find a rational number between two consecutive roots and here's an example of that. Yeah, that's... But because maybe it could seem like the natural proof of it. Yes, the natural, I would like to make it clear. Which is the natural proof he has in mind. Or he seems to refer. But he gives a reason here. He says, since these two rules, the physics or disequation, cannot be separated from all things. This example shows... Yeah, that's a counter example showing that you don't have separability. And that the proof of theorem A cannot be done using separation of rules. Yeah. Yeah, but why is it so concerned, it is a problem for me, to have the right to use separability in this proof? Because he tried very hard using separability, and he concluded that that didn't give the deal. And then he had to record this without. If I remember well, in the first edition, there was some problem, that's important, so that's an important one.

45:00 So I am in charge to introduce Artin's paper, and I am not too big, I did just one slide, I'm going home. So the Artin's paper, you have a German version, a French translation. So for the Artin's paper you have the German version, I think, and the French translation, and Charles Delzell is doing the English translation. So the title would be in English, On the Decomposition of Definite Functions in Squares. It was published in 1927, as said, the last on the bottom. And so here is more or less the story which led to the end, to the Herb Tid's paper. will read the introduction and give some indication of the first part. So, in his Paris lecture on mathematical problems in 19, Hilbert posed the following problem, which is known as 17th Hilbert's problem. Sorry, Hilbert's 17th problem will be better. Monsieur Hilbert. So let f of x1, x2, and xn be a rational function. It's called definite. It does not take any negative value for any system of real values at x i. So the question is, He says, if you take such a definite function with rational coefficients, can it be decomposed as a sum of squares of rational functions with rational coefficients? So he asked the question of Hilbert's 17th problem on Q . So that was 19th, but he was thinking of this kind of things from long, because you can see there is a paper in 1888 where he already discovered that you cannot do that with using just polynomials. You really need rational functions. Afterwards he has got some results

47:30 for two variables and in 19 he asked the question because he has in mind the book he wrote later on the foundation of geometry, where it needed something on the construction with what is called, maybe it's written here, the straight edge on what we can call length transporter. So starting with a point in Rn, in a Qn, sorry, with n component, rational component, what What can you construct as point with what coordinates do you get using just straight edge and length transporter? So that's why he was interested in writing positive definite function as sum of squares. Because with these tools you can construct the square root of sum of squares. That was the reason why he was interested in it. So afterwards you have several papers by Landau, one showing that for one variable you can just use polynomials, and even that you can bound the number of squares needed by 8. And Landau also has done some work which is of interest for the paper, which is the decomposition of totally positive algebraic numbers in sum of squares. So the word totally positive appears here in Landau's work and will be used after. So after, there is the book by Hilbert, as I already mentioned, and then the paper you just have a report on. And now we go to Artin's paper. Well, the foundations of geometry, at least the first edition, was around the turn of the century. Yeah, probably, yeah, because he was already interested in these things. I do not have the right date for the first edition, so... How much?

50:00 So, the skeleton of the paper is the following. In the first part, the character is those elements of an abstract field which can be written as sum of squares. And he uses the notion of totally positive, already used by Landau, and he can find again the result from Landau. The second part is devoted to the proof of what is called Artin's theorem on Elbert's 17th problem, which Chip will report on, and the third part will return to the original question of trying to decompose things into squares, not only of rational functions, but of polynomials, and it's shown that you can do a little, just be polynomial in one variable, you can choose as you want. And also there are some things, some little facts on algebraic functions. So in the first part, general criteria, the first remark is that the question is of no interest if you are in a non-real field, because in a non-real field, minus 1 is being a sum of squares, you can write everything as a sum of squares, so there is nothing to do. And even if the characteristic is 2, you can also manage the things. So now it's only inside real fields. The problem is interesting. There is a little mention that it's clear that some product of squares are against sum of squares, and also that if you make the quotient of two sum of squares, you can get a sum of squares. That's more or less, well, two pre-orderings, I will say, because it's the closure on the sum product, and it does not contain minus one, and so on. So it's not written pre-ordering, but more or less it was in mind that the sum of squares were pre-ordering. And so afterwards, there is a first theorem, which says that a totally positive element of K are exactly the sum of squares in k, k being a real field, and there is a proof for that, so maybe we had

52:30 a quick look because I have to let time to charge for the proof of the main theorem, to the proof of that. So, sorry, now let k be a real field in which some alpha is not a sum of squares, and you You take omega, an algebraic, algebraically closed extension of K, and then using the same kind of arguments as in the theorem 7 of Artin and Schreier, you prove the existence of a field P between K and omega with the following property. Alpha is not a sum of squares in P, but it is a sum of squares in every proper algebraic extension of P. And afterwards, you know that P must be real, and you deduce with some little computation you can see on the paper, that minus alpha is a square in P. Otherwise it does not work, you get that alpha will be a sum of squares which is false. Okay, so So minus alpha is the square in P. And therefore, if alpha is the sum of squares in an algebraic proper extension of P, and minus alpha is the square in P, then alpha divided by minus alpha will be minus 1 is also a sum of squares. And then any proper algebraic extension of P will not be real, so P is real closed. And then, using Artinschreyer again, you deduce that there exists some ordering in which minus alpha is positive, hence alpha will be negative in the original field K. So this is, if you have an alpha which is not a sum of squares, it has proved that this alpha is negative And afterwards, he deduced the notion of totally positive, as in Landau. It's something which does not take any negative value for any other... Sorry? So, to make the point that it's not the same thing as in Landau, like that later...

55:00 Not... And there is a... I think that's an important change with respect to... For the totally positive? Okay. Maybe next question. No, that's the same, well, that amounts to the same, but that's, I think that's quite different point of view. And that's important in the difference between point of view between Landau and Martin Martin and Schreyer is an important point of view. I think we deserve a complete analysis of the presentation of the work of Landau to be able to make a comparison with Artin Schreyer and Artin. I have not done this work, but I think it has to be done. Because it's important. It becomes positive when it's for algebraic numbers. Yeah. And says that total equality means that in every field of mathematics, if for every real and many, it is mapped to a positive real. Just in two words, the difference is that in the Lando definition, it says one order field, which is a field of mathematics, one order you don't, something which is given, one order, and you want that in the field of your list, it's order, okay? Here, the definition of . The important point is that you can put any order, and you have to consider the totality of orders that you can put. Actually, for Ajit right now, it's amongst yourselves, but that's a different part of you, on one hand, on the one hand, you have one fixed order, which is the order of reference, which is the field of wheels, and I think it's an important point that you can vary the order, you want to take it as a structure, which is an order field, and you can choose the order that you wish that, Instead of saying you fix the order in your career, you could say you fix the order in your observation. And you define totally positive with respect to this order, which means order grading is a real closure.

57:30 But Richard said only one order in your career. Yeah. Maybe it's significant, look, that in this paper, algebraic numbers are absolute, so maybe we are not playing with the order of two like that. When we call them absolute. The idea to play with the order of two, maybe is... Yes, a base field, it depends on ordering, and then you call an element, an extension field, you call it totally possible with respect to this ordering, if it remains possible for all the patients, it's really close with respect to this ordering. Yes, but before, between Langdau, the first work of Langdau, and Artin, you have the work of Steinitz, precisely, developing the modern theory of the earth. I was about to ask this question. Is it in the 99th paper of Landao? Yes, but that's why we have to look closely. And I say that the first paper of Landao... I do not admit. In Martin Schreier's earlier paper, he did show the equivalence of an embedding into the real numbers for an algebraic number of fields and an ordering on the algebraic number of fields. I guess for Arden, they're already the same concept, twice by 1926. But he stressed at page 103 in the German original, he mentioned that after the definition of totally positive elements, mentioned that due to the term 10 in Schreier, this is the same thing as a totally positive number in London. Yeah, just the beginning of the 1 and 2 and 3. The fact that two notions are the same in the case of algebraic numbers, it's not immediate. It's a consequence of their term.

1:00:00 And I think that the shift between the two definitions, I think it's important to shift. And he has to establish the continuity between the two. That's one aspect of the paper, to establish his continuity. Still a question, algebraic numbers for a long time in 1919, is it something which is not in C? Which is not in C? I mean, what is NIACHI right now, it's completely just defined by the polymium with no . I can give you the papers. I have not the papers. How are you satisfied? No, no. As you know, it deserves a true study. so afterwards there is in part one some generalization of the theorem one on totally positive elements which is the theorem two which gives the characterization of totally positive with respect to air so i will not give the details here because i need to to let time for the core of paper to Charles on the main theorem on how it's proved and I will use the computer again so I have to well I might not have time to do the entire proof of Chardin's main theorem it's notoriously difficult or it's the part that's let's say it was superseded by simpler modern proofs and so now everybody has forgotten this whole proof history conference. Maybe we should go through it carefully and try to understand it. I hope you can read it all. I think it's too far back now. There, that's it. Okay, so I'll read most of it here. So we've already proved the theorem that every totally positive element in any, I guess, any field whatsoever, characteristic not two, is the sum of squares. So if you're that a certain rational function is a sum of squares, you must prove that it is totally positive. So that brings us to theorem 3 in Arkin's paper, which I read. Let R be a given, in the usual sense, real number field, that is, a subfield of the real numbers,

1:02:30 and K be the field of rational functions in n variables, x1 through xn, with coefficients in R. For K, furnish an arbitrary but fixed ordering, extending the natural ordering of R. Now, continuing this theorem three, given any system of finitely many functions, b1 of x, b2 of bm of x, where x is all those variables, in k, where m is arbitrary, and in what follows, the symbol x denotes all of the variables, there are n rational numbers, a1, a2, an, such that the functions phi sub nu of x are defined at the point where xi equals ai, and the function value of phi sub nu of a has the same sign as that of phi sub nu of x. This is the heart of what is now known as the Arden-Lange theorem, and it says that if you've got a bunch of functions that are ordered in some way, some are positive, some are negative in some big ordering, then you can specialize them to some point A, a rational point where the function values have exactly the same signs. And later he'll show how from this it is trivial to prove the Hilbert 17th problem. So he makes some commentary. These sign conditions mean more precisely that if in the given ordering of k the phi nu of x is positive, then phi nu of A is positive. And if phi nu of x is zero, then phi nu of it's just the zero operational function, and if phi nu of x is negative, then phi nu a is negative. Our theorem is trivial for n equals zero, since then the functions phi nu x reduce to constants, that is, there are no variables. Therefore, we assume it proved for n variables, and we shall deduce its correctness for n plus 1 variables. The carrying out of the proof will show that one may let the induction begin with n equals zero. First, we have to deduce consequences from its truth for n variables. Let p be the algebraic real-closed field over k whose ordering extends the ordering of k. The existence of p is guaranteed by the Archen-Schreier paper theorem number 8. You mentioned that the theorem 3 is what is now in the Archen-Schreier paper. Yes, I did mention that. Yes, but it's not because here, it's insisting that there are fractional numbers. Usually now, when this is stated, it's a real number of fractional numbers.

1:05:00 That's true, and we now call it a homomorphism from the ring of polynomials into some real closed field. So it's, maybe this is, in some ways it's stronger than what is called the Arcanine theorem, but at some point it's weaker. So we've got this real closure, P of K, where K is the field of rational functions. Now we consider functions F of T comma X that are polynomial in a new variable T with coefficients from K, and we agree upon the following terminology. A property E of a system of functions, f1 of tx, f2 of tx, fk of tx, is called specializable if there is a system, phi1 of x up to phi m of x, of rational functions in x with the following property. If A1 through AN are rational numbers such that all the functions B nu X are defined at the point where XI equals AI, and have the same sign, here I add my commentary, the same sign with respect to the ordering on K as their values B nu of A have with respect to the ordering on R, then all the functions F nu T X are defined at the point where XI equals AI, F1, T, A, F2, T, A, and FK, T, A also has property E. So he does not define the word property, and it's not necessary for us to understand what he might have meant by the word property, because he'll give only two properties that are in any importance here, and you'll see from those two properties what it meant. Maybe next Friday we talk about these properties in more detail. But basically, it's given some functions in n plus 1 variables, you must construct some functions in n variables that have certain signed properties. Now, here we can skip some of this. He said clearly a property assembled from specializable properties is again specializable. I don't think he ever uses this, so it's not worth trying to figure it out. Moreover, there always exist rational numbers AI that realize the specialization by the

1:07:30 inductive hypothesis. You see, we're doing induction on this Martin-Lang theorem to find specializations, and we're trying to prove this theorem now for n plus 1 variables, and this specializable property produces some functions of n variables, and those things have certain signs with respect to the ordering on k, and according to the inductive hypothesis, any set of functions and n variables can be specialized through some values that have the same sign. So he now has two lemmas, the famous specialization lemmas. They're famous to people who want to criticize the proof. They say those difficult specialization lemmas. So here's the first one, lemma number one, the property of a function, a single function now, I've said several functions, f, t, x, to have exactly r real roots, has a function of t in the field p, this real protein, is specializing and here is the proof remember our goal is to construct some functions p of x so he writes the given polynomial ftx is polynomial in t but rational in x he writes it expanded in powers of t like this with coefficients psi of zero and so on and he lets capital f of t be the general polynomial of degree s for some new coefficients, a0 up to a s, capital a s. Those are indeterminate. Now, according to Art and Schreyer theorem number 6, all theorems of real algebra hold in this real post field p. So there is, therefore, as is generally well known, a chain of finitely many polynomial functions, capital phi nu of capital A of the capital A-I with rational coefficients such that for special A-I from the real close to B-O-T, the sine distribution in the change phi-nu of A gives information about the number of real roots of the function f of T specialized for these A. Now, it doesn't, well, it gives some explanation. The existence of such a chain of functions, capital phi-nu, follows easily from Sturm's Theorem, or also from the theorems on the Dezouchians, which are also valid here. He's using not only Sturm's Theorem, but this upper bound on the size of the roots of a polynomial in terms of the coefficients and some rational function of these capital AIs. So, you use Sturm's Theorem

1:10:00 to count the number of roots from the lower bound to the upper bound of all possible roots and then give you a sequence of polynomials, a Sturm's sequence, conditions that determine how many roots this thing has. And I should say, in the definition of a specializable property, I go back to it, it's understood, I think, that the original system of functions must have the property E in that big field of all rational functions. It says a property E of a system of functions, and that's with respect to the ordering on the field of rational functions. We're trying to deduce that their system has that property once you specialize it to certain suitable rational numbers. So how are we going to construct our set of functions b of x based on this lemma 1 to get the guarantee that the number of real roots remains r. Well, we throw into the set of functions b two types of functions. First, we throw in all of the coefficients, psi of x, of the original function, mtx, so that one obtains something meaningful upon insertion of rational numbers. Because this condition of specializable is that all the rational functions should be defined at the specialized points, A. And secondly, we throw in all of the finite-to-many expressions, capital phi nu of x, phi nu of psi of x. So we evaluate the Sturm chain at all of these rational functions, giving the coefficients of the original polynomial. And now he says that is the right set of functions, phi of x, to prove specializability, because if AI is now chosen so that in the above-determined set of functions, each little phi nu of A has the same sign as little phi nu of x, then F of T A is defined and has just as many real roots as F of T X, since nothing at all has changed in the sines of the accompanying chain, big V nu of sine A. So that is the proof of level one. I know I'm going quickly and I know it's hard to digest. What? Pause? Burp? How do you swallow all this?

1:12:30 No, that group smells of elements of cell decomposition. It smells very nice. What do you say? Do you see it as cell decomposition because there's one distinguished variable after n variables? Yeah, and then the composition of sensors he's doing between these explicit functions of the n variables, x, that he has, and the signs of the l of b, x, etc. I'm not saying that it's important, I'm saying that it smells. Yes. But it's not quite that simple, because although he says the property, his definition of specializable is a little bit misleading, because he gives f1 of tx, so that one allegedly distinguished variable, but really there are several distinguished variables that are hidden inside the property e. Like the expression to formulate as a formula in a language order brings the statement that a function, a polynomial, has exactly r, your root, requires not just one quantifier, but maybe r plus one quantifier. So you're implicitly eliminating multiple quantifiers and not just one. So it's not a simple cylinder of one dimension. It's a higher dimension of cylinder, I guess. But, okay, so, fine. So he has proved level one that this property of having exactly R real roots is specialized. Okay, next, level two. That's one property. You weren't sure what a property is. Now you have some idea. Here's another such property. Suppose a finite sequence of functions f1 of tx, f2 tx, fk tx, each f nu tx has a certain root alpha nu in p, the real closure of the rational function field, and that these selected roots are in the proper ordering, alpha 1 less than alpha 2 less than alpha k. This property of the sequence, he says, is specializable. Okay, so we must again produce some new set of functions, p just of x, not involving t. And here's the proof. First, we adjoined to k, the field of rational functions, all the alpha nu, which

1:15:00 are these roots. And I remind you, back to the definition of a specializable property, in order for this, I guess it's assumed that the original system does have the property in the real closure of k of x. So there are some roots up there. The question is whether there are similarly chosen roots in the smaller real force field so anyway one adjoins to k all the alpha nu as well as for each integer pair where i is greater than j the square root of alpha i minus alpha j which is a positive quantity which should have it should have a square root the resulting field is real and arises from k by the adjunction of a real quantity xi, despite their primitive element theory, and invariably many adjunctions can be made with one single adjunction, xi. This quantity, xi, satisfies an equation of big F of t, x equals 0 that is irreducible in k. Actually, I think he means irreducible in some polynomial ring k bracket t, but anyway. Okay, so then each of these adjunctions, the alpha i's and the square roots of the differences, must be expressible in terms of xi, and indeed must be polynomial in xi. For example, we may say that alpha sub i is equal to g of i of xi and x, where g is subpolynomial, and these square roots are h of ij, h sub ij of xi, x, where hij is another polynomial. involving a polynomial in t with coefficients in k. Now, the composition fi of gi of xi of x, x equals 0 holds, because the gi, remember, was some root alpha i of fi, so if you plug gi into fi, you get 0. And therefore, the composition function fi gi with t of x, or t comma x, x, must be divisible by big F of t comma x, the minimal polynomial for C. So he factors his composition, little fi or little ti of t x x, he writes it as big F t x times some other polynomial, g i t x.

1:17:30 Similarly, the identity g i of c x, which is alpha i minus g j of c x, which is alpha j minus the square of hij of xi x, which is that square root. That difference is all equal to zero for i greater than j. And therefore, if you change all the xi's into the indeterminate t, you get some polynomial in t that is divisible by this big f of tx. So gi minus gj minus hij squared equals big f of tx times some other polynomial, Ij of Tx. Okay, so we're algebraizing these orderings. Finally, H sub Ij of Cx is not zero because it equals one of these differences of the alphas, which are not zero. Its inverse can also be expressed as a polynomial in C. Namely, 1 over little hij of Cx equals big Hij of Cx, standard field theory. And so we have an equation. If you change all the Cs into Ts, we have Hij Tx times big Hij Tx minus 1. It must be divisible by the minimal polynomial big F Tx. So the quotient he writes is big C or big psi ij tx. So there we have lots and lots of polynomials in t. So in the set of functions phi sub nu of x that we must construct in order to prove the specializable property, one first admits a system of functions that will give us merely the certainty that in the case of insertion, in the identities just constructed, everything will remain defined. Namely, one admits all coefficients of the polynomial functions of T, of every polynomial function of T that we have mentioned here, little f-i-t-x, big-f-t-x, little-t-i-t-x, h-i-j-t-x, big-g-i, big-p-i-j, big-h-i-j, big-c-i-j, All functions of t, they all have coefficients involving x, and we throw in all those coefficients, so if they are defined, then everything is going to be defined when you specialize in some value of A.

1:20:00 Thereupon, admit that set of functions, b of x, given by level one, that will ensure that big F tx has a real root even after specialization. It should have at least one real root. That can be done by level one. This suffices, he says. Here's the proof that it suffices to prove that this property is specializable. Why? Well, if Xi equals Ai is one of our systems of rational numbers satisfying the sign conditions, then the equation big FTA equals zero has a real root C bar. Now, this real root D doesn't say where it lives. It's not in the original subfield of the reals, but it's in the real closure. It has a real root C. I think there should be a bar. Let me see, I have a more legible copy. This is page 107. Yes, this is my copy. Okay. If you have a smudgy copy of page 107, you don't see the bar. This is near the top of page 107. There's a C bar. And the top of page 107. The xi itself was in this real closure of the rational function field, whereas this xi bar is in the real closure of the subfield r of the real number. It's an actual real number in the traditional sense. So anyway, using that real number xi bar, we plug it into all these little functions here. For example, you plug it into g-i, and we get alpha-i-bar, which is now another real number, algebraic over r, and if one inserts c-bar for t and x equals a in equation 4, which you can't see on my screen, but on the paper, you get that f-sub-i of alpha-i-bar, comma-a is zero. Because what was fi now? Oh, yeah, yeah, alpha-i was chosen as a root of fi, and so alpha-i bar is another root of fi in this smaller field. In equation six, gives the little hij of xi bar a times big hij of xi bar a equals one for every i greater than j, whence, in particular, hij of xi bar a is not zero.

1:22:30 And therefore, equation five implies that this difference, alpha i bar minus alpha j bar, is equal to the square of h i j of c bar comma a, which is positive. So that's nice. Now, we've descended. We have these big roots in this big real closure of the rational function field that were ordered in a certain sequence, and now we have found some real numbers that are also roots of this thing when you specialize, when you make the X is equal to some AIs, satisfying the right sign condition. Now we've got the new roots, alpha I bar, which are also in the same ordering as the original alpha I. The ones with the bars have the same ordering as the ones without the bar. And that's the end of the proof of Lemma 2. I guess I led you to, my tone of voice should have been going down to indicate the inclusion of the proof, and I kept going, and I'm still moving. So, do you need a recap on Lemma 2 before we complete the proof of theorem 3? No, I think you have to say something to your lemma before, because you mentioned the discussion I think the main point is that it doesn't need any kind of quantifier because you can express it by the storm. And that's what is used here. Well yes, you start with several quantifiers in the original expression of the property. Okay, yeah. And then you find an improvement expression without quantifiers. And the question was, have you eliminated only one quantifier, and so does it really resemble a cylindrical algebraic decomposition, or have you actually eliminated multiple quantifiers, and so it's a more jazzed-up cylindrical... The cylinder, you have said, as many roads as there are, actually there are many roads because to partition the parameter space in pieces was a number of pieces concerned, and so that's amounts, I guess, more or less to what's done in the US. On Friday, I plan to try to give a precise definition of a property, because Arden says

1:25:00 he refers to properties that may or may not be specializable, and I will give a definition of property such that all such properties are specializable, in this sense. And I'll prove it using the Tarski Elimination theorem, and the way I'll do it, a property will basically be a formula in the first order language of order rings that has n free variables which will correspond to the x1 through xn and then possibly any number of valid variables and that property will somehow include the one distinguished variable t is one of the bound variables but there may be other bound variables like when you try to say this polynomial has exactly six roots. Well, you need seven quantifiers. There's no other way to state that without seven quantifiers, except for after you use the theorem. Anyway, so now we have two specialization levels. You've seen two properties, and using these now, we go back to proving theorem three, and maybe I'll scroll back to theorem three just to remind you what We are proving 0, 3. It's the Arkin-Lang property, so-called. Here it is on the screen. Given a subfield R of the real numbers, and let K be the field of rational functions in N variables with some given fixed ordering, then given any system of functions phi 1 of X through phi M of X in K, there are N rational numbers A1 through A N such that the functions phi sub nu of x are defined at A and the function values at A have the same sign as the functions themselves as elements of the ordered field of rational functions. So we must find now a rational point A. And now I scroll to the end of the proof of the limit 2. Okay, there. After these preparations, the proof of theorem 3 can easily be brought to conclusion. easily is not so easy for the average person, but we denote the n plus first variable now by tau in order to make it stand out, and we again denote the field of rational functions in the remaining n variables by k. Thus, let there be a fixed ordering of k of tau, so we're doing induction on the number of variables, that preserves the ordering of r, and let

1:27:30 Let p prime be the real algebraic closure of k of tau that extends this order, and let p be the real closure of k contained in p prime. So you've got two real closures, one of the n variable function field and one of the n plus one of the variable function field. Now, if f1 of tau, x, and f2 of tau, x, and fk of tau, x are rational functions of tau and x, then one can assume, without loss of generality, that none of them vanishes identically. It is to be shown that one can insert rational numbers such that the function values have the same signs as the functions. For this purpose, we decompose f sub i of tau x into irreducible functions of tau. Let, for example, p1 of tau x up to pm of tau x, let those things be all the distinct irreducible polynomial functions coefficients 1 occurring in the numerators and denominators of these fi tau x's, and let psi 1 of x up to psi s of x be the other factors, those that are independent of tau, that one needs to construct the fi of tau x from the pi of tau x. So they would be maybe the leading coefficients of the numerators and the leading coefficients of the denominator. Evidently, it suffices to prove our theorem for this system of functions, the P1 of prod x's and the psi 1 of x's, because if one can specialize so that in this system all of its signs are preserved, then this holds for the original system, because the original functions were just some multiples or products of all these things, so their signs would be well preserved also. Now, in order to do this, one considers the functions P1 of Tx, P2, Tx, and PM of Tx over K and over P, where K is the field of rational functions, and P is the real closure of the limit. Let the totality of all the real roots of all of these functions of T be ordered according to their size. So we have many roots here, alpha 1, alpha 2, up to alpha L. And let, for example, alpha I be a root of, it's a root of one of these PIs. is denoted by the p-sub-lambus of i of t-x. So some of these p-i's, p-lambus of 1 might be equal to p-lambus of 2,

1:30:00 some repetitions now on our list of p-sub-lambus of i. And we now specialize as follows. In the function set, p-x to be constructed, we admit the functions z1 through psi-s, psi 1 through psi s, which are these extra coefficients that you need from the numerators and the denominators. And we throw in the set of functions b of x given by lemma 1 that guarantee that each pi of tx has exactly as many real roots as pi of ta. And finally, we throw in the set of functions b of x given by lemma 2 with a sequence p sub lambda 1, p sub lambda 2, that ensure that each piece of lambda i of ta has a real root alpha i bar so that those alpha i bars are also lined up in that sequence now the inductive hypothesis of theorem three implies that there are such rational numbers ai that is i think what he means is there are rational numbers that make all these functions phi of x have the appropriate signs because they don't they and they involve only n-verg. Since we have admitted psi of sub nu of x, the preservation of the signs of these functions is already provided for. It therefore suffices to concern oneself with the p sub nu sub t-a, for those signs to be the same as the p sub nu of t-x. Since p nu t-a has exactly as many real roots as p sub nu t-x, the total number of real roots of all the polynomials, piece of new Ta is again L, so that alpha 1 bar up to alpha L bar are all the real roots of our polynomials. The ordering of these roots is, however, the same as that of the alpha i because of the third condition. That was this third condition up here on my screen. We chose it from lemma 2. These functions bx will preserve the ordering. We make sure that the ordering of the roots, the smaller roots, is equal to the ordering of the bigger roots. field p prime which is the real closure of the n plus one variable functions the functions p nu tx decompose into linear and quadratic factors which i think he means to say irreducible quadratic factors monic where the latter come from the complex roots the same holds for piece of nu of t comma a that is if you specialize it those also factor as a product of linear and quadratic

1:32:30 Now, if one inserts for t any value from p prime, make real closer, the quadratic factors always give positive contributions to the products, because the quadratic factors are just sums of two squares, so that the sine of p nu tx depends only on the linear factors, i.e. only on the root interval in which the value of t lies.