A Tarski's 'Decision method for elementary algebra & geometry' (parts 1 & 2)

0:00 And by a source, I understand a paper on which the later development depended. And in this sense, I'm talking now on a paper which is not a source, because nothing depended on this. It was overlooked in a sense. but nevertheless it could have been important and so we will go through this paper and for this reason we will not only read this paper but also compare it with the other development and see what are the links. So the aim of the paper from Crivin was actually to prove theorems from functional analysis. So it was not in our area, it was from functional analysis. And he was certainly inspired by Stone's approach from 1940, where he suggested in functional analysis to forget the integration and try to develop the theory there more algebraically. So if you have an integral with a positive measure, then it gives you a linear functional, which, of course, takes a function which is non-negative to a positive or non-negative value. And so he only wanted to study this, and then later the so-called moment problem appeared there, which asks for which linear functionals taking positive functions to positive values really come from a measure. And I think the paper of Crevin is essentially along this line. And in order to develop this, at the end, And I think the main theorem he gets in this paper is about integration. It's at the very end. And you can also see, if you look at the paper, all theorems, with the exception of two, have a number.

2:30 Two theorems do not have a number. They are just called theorems. The last theorem, at the very end, this is one about integration. And for this, probably he developed all what is before. theorem, which doesn't have a number, it's the real Nullsternsatz. So he considered it as an important application of the theory he develops. And so what Stone did in 1940 was he gave an axiomatization of the ring of continuous function, of real valued continuous function on a compact house of space. And this axiomatic characterization was, in this axiomatic characterization, he considered this ring as a ring together with a certain pre-ordering. So he already introduced, without calling it a pre-ordering, he introduced the notion of a pre-ordering in 1940. Actually, there he considered only the so-called Archimedean pre-orderings. And Graveen's paper is based on this. He also works with, his main object is to consider so-called Archimedean pre-orderings. But since he was influenced by logicians like Kreisel, and he wrote a lecture note together with Kreisel on the model theory of real closed fields, for instance. He did not start immediately with Archimedean pre-orderings but with arbitrary pre-orderings because this can be treated in model theory while Archimedean pre-orderings cannot be treated. And though it's not surprising that at the beginning he develops the theory as general as possible and in doing so he developed already things which later have been introduced and proved by other people also who didn't know about Graveen's paper and I think his paper was simply overlooked because it was in the area of functional analysis and one didn't expect such things there.

5:00 Okay, yeah, let me just, so I'm going to use the blackboard, and I think there's no translation of the paper into English, so we have to use the French one. So, let me first give the content of the paper. So, first he introduces pre-orderings, and I'm of course using his notation and not the one which may be common today. So, he introduces a pre-order ring on a commutative ring with one, no other condition on the ring. And then he considers the maximal such objects, the maximal pre-order rings, and puts them together to the so-called maximal spectrum. Of course, he did not call it the maximal spectrum because he didn't know the other spectrum. So let me write it this. It is maximal spectrum. And also in functional analysis, you are only interested in the maximal one. And it also, I will explain later why he was, in a sense, automatically led to the maximal one and open. And his approach did not naturally come to the spectrum, to the real spectrum which we know. So, then in the third part, we will talk about the envelope and the radical of a pre-ordering. So far, it's still general, completely general, and then comes the restriction to Archimedean pre-orderings, Archimedean pre-orderings, and the so-called representation theorem, The representation theorem, which characterizes, in a sense, a commutative ring with an Archimedean pre-ordering.

7:30 It gives a homomorphism from this object into the ring of continuous functions on a compact house of space, describing the kernel and describing the image. and also the pre-image of the canonical pre-ordering on the ring of instantaneous functions. And then he passes to the object, which is interesting for him. These are measures on the spectrum. Measures on the spectrum. And finally, and this is the most important there, he gives applications in Banach algebra. So the ring then is Banach algebra and he gives two important implications. applications in Banach Algebras, Banach Algebras and our idea is to to talk today about the yeah I think the first four topics or maybe three and a half topics and then next time we talk on the representation theorem and briefly mention the measure the just one theorem from there and then the application I do not really understand all of them so we we are not going into details of this because it's also not our area okay yeah this this is the paper and now let me just start by not really reading the paper but giving you the main you can read it I while I write on the blackboard but I will select certain topics I can we are not really going linearly through through the paper yeah maybe I should before I do this I should write give a little list of other papers which are related to this so I already mentioned the stone paper which has

10:00 mainly influenced this from Krivine. This was 1940. And then there was Cadizone, a student of Stone. He extended the theory. This was in 1951. And then, chronologically, there was Grévin, who based his work on, essentially, Stone's work. This was in 1964. And after this, but independent of, Grévin came Dubois, with the Representation Serum, in 1967. And finally, in the paper of Becker and Schwartz, the representation theory was proved as general as possible, of its T-sit. This was in 1983. So today, I will not talk about this, because this is mainly related to the representation theorem, and this will be the topic on Friday. There's something which surprised me, that the bibliography is very short, and in particular, he mentioned mazes, stones, no, no, no... This is right, this is right. But I asked him once whether he knew this paper from Stone, and he said yes. Yeah, yeah, I... Uh, yeah, it's, see, the paper, if you look at the paper of Stone, it's not, uh, he, there he, this is more programmatic. He says one should do, in functional, in functional analysis, one should try to, uh, go, uh, algebraic, to make it algebraic, this approach, and he suggests this. In Stone, in Stone's paper, there's not a single proof. in a sense you could understand as making the things precise in particular in stone the spectrum is it's not this compact house

12:30 of space which is maximal spectrum is not clear he doesn't describe it I don't know exactly where the Du Bois also didn't do it. It was made clear then in Becker and Schwarz and in Krivine. So, but, but, more or less, yeah, but it was more restrictive. It was not just a commutative ring. It was a real algebra. Yeah, yeah, yeah. Okay, so, but this is not my topic for today. Today, I would like to talk on this part of Corinne's paper, which is general and does not refer to the Archimedean ones. But nevertheless, I start with the definition of this so that you can see the difference. So, I assume that not everybody is familiar with this part of real algebra. It's actually not real algebraic geometry. These are more algebraic problems, and so not everybody may be familiar with this. So, I do it in detail as it is done in Craven's paper. So, first, we talk about the pre-orderings. Pre-orderings, now what is it? It is a subset, it's essential, you think of the positive cone of an ordering, then this, it is closed under addition, it is closed under multiplication, it contains every square of the ring, and it does not contain the minus 1. Now, this is the definition of pre-ordering, and when you have a pre-ordering, you could, of course, also introduce a partial ordering by saying a is less or equal with respect to omega to b if and only if b minus a lies in omega. This gives then a partial ordering, but it's very partial because you cannot even say that

15:00 if a is less or equal b and b is less or equal a, then a is equal to b. Not even this is true. this is true if and only if the intersection of omega with its negative consists only of zero, yeah, and also it's not a linear ordering, but you have transitivity and compatibility with your arithmetic operations, okay, because sometimes he uses also this notation here. Now, let me start by giving two examples, which at the same time yields as an excuse why we overlooked this paper. It is the following. In real algebraic geometry, one of the main bases is the Artin Schreier theory, which we saw this morning. And there, with this Artin Schreier theory, 17-silver's problem has been solved, and the canonical preordering, which is, the canonical ring, which is considered there, is a field. It's the field, say, I do it over the reals, and it's a function field or a rational function field in n variables over the reals, and the canonical pre-ordering which you consider is just the sum of squares. So the omega, which you consider here, is the sum of squares, let me write this, the sum of squares, the set of sums of squares. Now, this is concerning real algebraic geometry, one of the basic prior drinks you are interested. In functional analysis, you will be interested more in this ring, the ring, and this is stone, the ring of the continuous functions from a compact house drop, the real value continuous function on a compact house drop space. And the canonical pre-ordering you are considering there is just, so the omega are just those continuous functions which never take a negative value. It is to say it's a function from x to r squared, where r squared of course is a set of squares and not r cross r.

17:30 Okay, so these are the functions which never are negative. Now, this is an Archimedean one. I'll give the definition in a moment. And this is typically non-Archimedean. And this is the excuse why we did not look at this paper, because he was mainly concerned with Archimedean ones. And our pre-orderings are mainly non-Archimedean. Only later, when Schmittgen's theorem came up, we realized that also in the semi-algebraic geometry, Archimedean pre-orderings turn off. Okay, now, and what is... I don't understand your argument because your argument is correct. Yeah. The person should still be ignored. What? If these arguments, those arguments are correct, the definition will still ignore? No, because, as I just mentioned later, in real algebraic geometry, also Archimedean pre-orderings occur. They occur, they do not occur if you consider fields, and in particular over the reels. but if you consider certain just polynomials and then in the in the ring of polynomials you will find also Archimedean period rings but they have been discussed this discovered only very late for the first time in 1991 okay Okay, now, just a short question, is, so omega is the C of, so the botanous function from X to, to the positive, to the positive, positive reeds, yeah, non-negative reeds, yeah, just, Okay, so now, what is Archimedean? Archimedean just means the following, so the definition is, so this is Archimedean, if in this partial ordering which I just erased, every element can be exceeded by natural number.

20:00 to be documented in omega see it's not a let me write it first if for every a there exists an n such that n minus a is in omega now this and this in the other writing this would just mean a less equal M. Now, as I mentioned, this is not a linear ordering, yeah? So, you cannot compare A with the natural numbers in general. Now, this is a very strong condition, yeah? But here, for instance, you see this, now you see that this is Archimedean, because if you have any a, a is a function, if you are given any function, it takes its maximum on this compact space x, so take a natural number which is greater or equal to its maximum, then n minus this function is never negative, so it belongs to this pre-authoring. It is never Archimedean, because otherwise, except you have not, yeah, no, it's never, because you cannot have an ordering, because if this would be Archimedean, one remark, if Omega is Archimedean and you have a bigger preordering, it remains Archimedean, because it contains this. So every ordering sitting above this, this means in particular every ordering at all here, would be Archimedean if this would be Archimedean. But this field does not have a single Archimedean ordering. So, this notion of pre-order and Archimedean pre-order, is it clear from Creven's paper? Yeah, it is in Creven's paper. I don't know, but you can define here, or you can... So... Was it beautiful before? Implicitly in Stone's paper. Stone has an axiomatization of the ring of continuous function,

22:30 and there he talks of a subset with certain conditions. And these conditions are editively closed, multiplicatively closed, the squares contained there are in there, I think minus 1 is not there, and every element can be exceeded. But it's one more condition, which he needed really to give a complete characterization of this ring. If he only wanted an embedding into this ring, you could skip the other condition. You can look at this. It doesn't give the name, but it is essentially there. Okay, now, let me see. Yeah, yeah, you have to go a little further, you have to go, yeah, okay, when we talk about this i can already mention this this is on page 320 on the real banach algebras and see then he he says there uh theorem 16 uh in in a banach algebra in a real banach algebra which is defined two lines upwards just minus ones not the sum of squares uh every pre-ordering is archimedian And this is quite easy, because if you take an element from the Arnach algebra, you can prove x, that its norm is greater or equal to x. And this means greater or equal in the sense of this pre-organ. This means that the norm of x minus x is contained in the pre-organ. And this is because of the completeness of this algebra. proof there three or four lines so every pre-ordering a banner algebra is Archimedean and this is the what he's mainly interested this is the main

25:00 the main example and of course this is about algebra here this example okay Okay, but this is, I'm mentioning this only to tell you, to explain to you that this is his main interest, while the other things, the more general things, he's only doing, yeah, to be as general as long as it is possible. I would now like to talk on this more general situation and next time come to Archimedean, to the representation theorem for Archimedean 1 but I wanted to point out this difference ok, now we can start reading the paper so I already gave you now where's my Yeah. So, I will not discuss every series because this will be too boring. So we know what is a pre-ordering, and after you have the pre-ordering, you hopefully remember, here you have it in the paper. He introduces the ideal which corresponds to this, it's written like this, it is just omega intersected with minus omega. And this is an ideal, this is easily checked, this is an ideal in A. Okay, and then, once you have this ideal, you can factor your ring through this ideal, this canonical epimorphism, to A mod, this object, and then you get A, we denote it by A bar, and in A bar, at the same time, you can take, consider the image of omega. image of omega is now a pre-ordering which sits in here but since you have factored through this ideal it has an additional property it has the property that this intersect with minus d is just zero though the partial ordering i mentioned earlier is not not so partial as it was earlier

27:30 now you know if x is smaller than y and y is smaller smaller equal to x then they are equal But still, it might still not be linear. It will become linear once you extend this pre-ordering to a maximal one. And this isn't the next object he is interested. But now let me first talk on the extendability, which is very easy. though CRM2 is, there exists, there exists an extension of omega, which contains little x, another element, this is if and only if minus 1 is different from omega 0 plus omega 1 x, for all omega 0, omega 1 in omega. Now, this is well known to everybody who ever was looking on pre-orderings. So the proof is just one line. And most proofs are to be not just a few lines. You just define this to be omega plus x omega. And if you check it, this is pre-ordering. And this satisfies all the condition of a pre-ordering, except perhaps that minus 1 is not in there. So this is guaranteed by this, by the condition. So this is the first theorem, which is quite easy. Now, the other theorems which come then are also, for the moment, I will skip them. I will skip number 3, 2, number 3, number 4. For instance, number 4 says every pre-ordering can be extended to a maximal one. It's clear. It's on slammer. And, ah, yeah, okay. Now, now I come to the maximal one. And this is the next important thing.

30:00 And also, this is quite interesting there. It is theorem six. Before he does six, he defines something, which I'm going to define also in the theorem. is quite interesting because he characterizes, he gives a nice characterization of the maximal ones. And since this characterization is so nice, he may have overlooked the fact that one can do the theory a little bit more general and what we know now as the real spectrum. I will explain it in a moment. So let me take this away. So, the theorem 6 then is that omega is maximal, this is maximal if and only if the, now I'm, the conditions are in the, in, here, in, in this quotient tree. Now, if there, omega, together with minus omega, is the whole ring. Of course, if you pull it back, then you have the same condition on A. And you remember this is one of the conditions in the real spectrum. And in the real spectrum, we also know, we consider, for maximal ones, you usually prove that the support of this is a primar deal. But it need not be, of course, not a maximal primary deal, and also the preordering, as a preordering, need also not be maximal. But what he does, he gives a condition, is that the preordering really is maximal. And this maximality in particular implies, then, it should come out, of course, that that this is an integral domain, so this means that this support is a primary deal, but at the same time, the pre-ordering is maximal there, and the condition is the following, and I must confess that I only realized it yesterday that he has this somehow strange condition, I never looked at this in detail.

32:30 So for every element, non-zero element in the residue ring, there exists Y, another element, such that the product of this, minus 1, belongs to this. or to so to speak this means the product is greater equal to one yeah that's the condition there and this if if a ring this is a ring if a ring satisfies this condition he calls it a pseudo field pseudo field and the first observation we know we have we know that the part is an integral domain the first observation is that from this condition you easily get that it is an integral domain let me give the proof it's a nice proof so note it's also of course all in the paper note that A bar omega bar is a pseudo field Pseudo field, then A bar is an integral domain, integral domain, okay, let me prove it, prove, we have given two elements which are different from zero, in the ring, and we want to show that the product is also different from zero, So, now we know, to x, we get an x prime bar, such that this is greater or equal to 1. To y, we get a y bar, y prime bar, such that this is greater or equal to 1. And from... Okay, and from this, it follows, I give it a moment, that xy times x prime y prime is greater or equal to 1. Now, of course, you will say this is clear. But don't forget, this is only a partial ordering. So far, we know almost nothing about this. So we have to prove it. But it is, of course, easy. If you have an element which is greater or equal to 1, and you have an element which is greater than equal to b, b greater than equal to 1, these are the product.

35:00 Then, of course, what you get is that 0 is less or equal, and now I use the properties which are really allowed. This is a minus 1, b minus 1, and this is a b minus a minus b plus 1, and this, of course, implies that we have AB plus 1 greater or equal to A plus B and each one, A and B, is greater or equal to 1, so this is greater or equal to 2 and now if we cancel, we get AB greater or equal to 1. And that's all, the operations have all been allowed. So, we have seen that this is an integral domain Of course, this cannot be zero from this condition. Okay, so it's an integral domain, and then, let me see. And now let's go to the proof, let's look at the proof of theorem 6. Is there a typo because it seems that it should be omega, the absolute value of omega? No, it's notation. Yeah, this is notation that he uses. Because it's getting 0.5 and 0.5. Can I? Okay, we remember this, that it is an integral domain. The quotient ring is an integral domain. And so what do we have to prove now? Now the proofs are all very elementary, so we first prove this implication. This means if it is maximal, it satisfies this condition. I'm not going to prove the first one, because this is as usual. When you have a maximal, as we know it from many books, when you take a maximal pre-ordering, then, of course, this condition,

37:30 and you have the, I'm not using this, that the support is a primary deal. This I'm not using, but I'm using this. So let me only show this condition. Now, this condition, he proved in the following way. So, we have an A bar, which is different from zero, and then the A, of course, is not in the support, so it's not in omega intersect minus, it's not, yeah. And then you have to make a case distinction, say A is not in omega, and if it is not in omega, it does not, there is no extension of omega which also contains A. So, there is no omega prime containing the omega together with A. But we have just seen in our theorem number 2 that then minus 1 can be written as omega 0 plus omega 1 A. And what does this mean? we have omega 1 a equal minus 1 minus omega 2 and this means it is we have omega 1 a is less or equal minus 1 ok now yeah we go down downstairs we have omega 1 bar a bar less or equal minus 1 And that's exactly what we want. We have found to A bar a Y, which is minus omega 1 bar, because this, of course, can be written as, is there? Yeah, it has to be greater or equal to 1. So this means A bar minus omega 1 bar greater or equal to 1. Yeah. So this is our Y. So we have proved it. It's really very, very elementary. So we, from a, from maximal one, you get this condition and you get this condition. Now what is interesting, when you have these two conditions, why is it maximal?

40:00 So in particular now, we also know that this ring is an integral domain, because I showed you from this condition, you get that it is an integral domain. So this ideal is a prime ideal. You should not think that if this is maximal, this ideal automatically is a maximal ideal and therefore a prime ideal. This is not true. So it's a non-trivial observation that it is a primal field. Okay, now the converse is also almost as easy as this. Yeah, let me do the converse. So, the converse, what do I have to do there? I take an x from A, and x is not in omega. have a bigger omega prime i could take the x from omega prime but not in omega so this is an arbitrary x which is not in omega and then we have the following this is not in omega than I r. So this implies, if you go down now, that the x-bar is negative. It cannot be, since it's not in omega, it's not in the intersection. And downstairs, we assume also that this is linearity. So if it's not in omega, it has to be negative. Now, to this element, there exists, there exists, but if it's negative, it's also different from 0. And if it's different from 0, you get a y, so that we have minus x bar y bar greater or equal to 1.

42:30 So here I'm using the minus x instead of x. Doesn't matter, it's also different from 0. To this I get a y such that this holds, and in particular this of course implies that the y is greater than 0. Okay, now the y is in omega. Why is it in omega? ah, yeah, because it's greater than zero yeah, it would be sufficient that it is greater equal to zero, this means just that it is in there so this is in omega, and we have minus xy minus one is greater, so, is greater equal to zero, which means this is in omega, okay, now this to omega plus xy. Now this is in omega, the y is in omega, and the minus 1 is a combination of this, though the omega cannot be extended, so it's maximal. Yeah, so that's a very nice characterization of the maximal ones and when I saw this condition I was wondering what it really means and now let me explain it in my how I see this this condition what it really means so you remember and it's the same time when I explain this I'm also we talk with you we look also it's a case where the is Archimedean. So maybe I erase now everything, everything here, and we have the following remark, we can also erase this. we have the following remark this is now not in in Krivine's paper but just to explain you this

45:00 but it will help us to understand the theorem which is proved later in Krivine assume is a prime ideal and omega union minus omega is the whole ring. So this just means in our language that omega is in the real spectrum, in spare A, as it is used nowadays. These are the conditions for being in the real spectrum. Okay, now in this case, we have that A bar, now in Polos, that A bar is an integral domain, in this case. And now, assume for a moment that this condition which I just wrote down is wrong, which characterizes the maximality here. Assume it's wrong, then this means there exists an x different from 0 such that for every y, we have 1 is bigger than x bar, y bar. So this means the condition is wrong. Now what does this mean? If this is an integral domain, then we can take the quotient field, quotient field of A bar. And since omega bar intersected minus omega bar is zero, the ordering which we have on A bar can be extended to this field. Just by saying that, say, A bar divided by B bar is greater or equal to zero, we just define it by saying A bar times B bar is greater or equal to zero in the ring, which just means it lies here. This is a true ordering on the quotient field. Now, let's draw the quotient field. Here it is, k, and here you have 0, and here you have your arbitrary y, for every y, and then there exists an x such that, Now, read it in the Gaussian field. This means 1 over x bar is greater than this. So, 1 over x bar lies somewhere here. 1 over x bar lies somewhere here, beyond every element of the ring.

47:30 So, if I take the convex hull of the ring, if this is the convex hull of A bar in K, yeah, you see that there is something outside, infinitely large. When this is infinitely large, then it tells you that the x is infinitely small, so the x-bar must lie somewhere here, very close to zero. It's infinitely small. Now, if you take, define the set of all infinitely small elements, then this is a convex ideal here. It's actually a valuation ring, this, and this is maximal ideal. So, let me call this the maximal ideal. now if you cut this together to one point and you can do it with every residue here cut down to one point and you get the residue field here O mod M and you have a canonical ordering on this and this is a field if you not now take the pre-image of this positive thing it is an ordering which is bigger so it was not maximum yeah so the moment such an element exists it gives you the possibility to construct a bigger watering so a bigger pre-ordering so let me just write it down so that you have it together so we I have to come on I have the canonical map from a to a bar this is contained in this O because I and now here they have the canonical map from O to M, and you have an ordering here, which is, and then call this homomorphism just tau, and if you now take omega prime to be the pre-image

50:00 of the ordering down here, which is just rho, oh, this is rho, It's just rho from omega bar, if you go down and then lift it up, then you get something which is bigger than the original preordering. So now the original preordering was not maximum. it's just a contraposition which I have proved earlier yeah because you get something this element for instance lies in there now this is just to indicate but what you can see from this now so from this remark is And now assume for a moment you're preordering here, or actually this was already a max, no this was an arbitrary preorder. Now if you're preordering is Archimedean, then this means every element here can be exceeded by natural number. then this O is just the convex hull of the integers, yeah? And if this is convex hull of the integers, this is an Archimedean-ordered field and therefore embeds into the reals. So, if you have a maximal pre-ordering over an Archimedean pre-ordering, automatically maps into the reals. The maximal spectrum, then, corresponds exactly to the homomorphisms to the reals. That's what we can see from this observation. And this is proved later. This is proved in... Let me give you the reference. It's... So, on page 314, he introduces pre-orderings, Archimedean pre-orderings, and then he proves in theorem 13 that if you have an Archimedean pseudo-field,

52:30 and the A bar now is Archimedean as well and it is a pseudo field because we had the maximal ones and then this is nothing else than an Archimedean ordered field so I will therefore I will not go through the proof which is given there I hope that this gives you the idea that the proof is the theorem is correct and And the proof there is also correct. So let me go back now to the general theory. So after we have, he has characterized the maximal ones, he gives, yeah, he introduces the spectrum. And this, of course, is the maximal spectrum now. And I think he did never consider, there was no reason for him to consider the case where the support is just a prime ideal, because he was only interested in the maximal ones. There's another reason, too. There's, I mean, there's a very rich theory of partial order agreements, which is quite old, and they, when they thought about representations, they, it was just a tradition to look at maximal ideals. as rings of continuous functions into the units. Okay, so now what did I want to say? Wait a moment. Now you confused me. The spectrum. Ah, the spectrum, yeah. So he introduces the spectrum of A together with preordering. These are just the maximal objects. this is maximal. So this is the maximal spectrum. At this point, he does not consider any topology on this. It's not, for the applications he's then giving, it's not necessary. He looks only at the topology when he restricts himself to the Archimedean case. And then you get canonical homomorphism to the rings

55:00 And then the maximum spectrum, it's always a compact house of space. Okay, now, and then he gives, what he now does is using this maximum spectrum and giving two examples of what, of a more general theorem, which is called the abstract positive Sternsatz. the more very general version which of course you know, he gives two cases which is the true positive and the real and I will continue to talk on those two topics and then Mark should continue So, yeah, he puts in theorem 7, he proves not positive Sternsatz, but a negative Sternsatz. But this is no problem. The negative Sternsatz says x is negative on the maximal spectrum. So this just means the x bar, when you go down, is always negative. And this is for all the elements from the maximal spectrum. Now, you cannot, this bar actually should have a prime, which refers to this prime then. But it's clear what I mean by this. If you have this, this is if and only if minus omega x is greater or equal to 1 for some omega greater or equal to 0. Now, let me, before I give, the proof is so elementary that perhaps we can skip it. So here, we are equal to x with respect to omega. Yeah, below 2 of you. Yeah, that means below 2 of you. This means belongs to omega, yeah, belongs to omega. So the corollary which we can draw is x is positive on this spectrum of A omega.

57:30 this is if and only if omega x I'm just writing this replacing x by minus x and then taking it to the other side no wait a moment let me write it first this is 1 plus omega 1 for some omega 1 in omega usual. This is the abstract positive Stehrensatz, which we all know, yeah? If you have an element which is positive on the maximal spectrum, then you get omega times x is equal to 1 plus omega 1. This is the usual way. And this is the abstract version of the positive Stehrensatz, which, in the concrete situation, was later proved by Stengler. And if you use this abstract one and add Tarski's transfer, you get also Stengler's theorem. I should point out that Stengler did it completely different. He did not consider pre-orderings on rings. I think, as far as I remember, he deduced it from the real nullsternsatz, of this but I'm not quite sure is much easier we use all cones yeah but but this is trivial when if it's strictly positive on the maximal it's strictly positive on all. This is too weird. And actually this would be a weaker assumption, so a strong theorem, but it's only seemingly stronger because you see easily that from the maximal one it always goes down to the other. So that's the reason why he did not have to introduce the that's the full spectrum just the maximum one okay now to prove it should I yeah what it may be possible to find it but

1:00:00 But it's not always true that d omega is the intersection of all maximum pre-order, of all maximum pre-order expectation of omega. This is not necessary. I know this, but nobody claimed it. Ah, okay. Okay, now, should I give the proof? It's just a few lines, I think. We... Just one, it's actually just one line. give the proof this is clear because why is it clear you know because we are in the residue of the quotient ring is a pseudo field and so you can work it's an integral domain so you can work in the quotient field and then you immediately see this that from here from this condition you immediately see this can only be greater equal to 1 if the x is negative yeah this is clear so we have only have to prove this but here here we use the fact that it is that we already have an integral domain that I had not more it's only needed to have an an integral domain. That A bar is an integral domain. Okay, now this is again there is no omega prime over this of omega together with x. Why is there no such? Because if we had one bigger one yeah then it contained omega and since it is maximal we could go to a maximal one then the maximal one would contain minus omega and minus omega so it contains omega minus x it comes x and minus x but it is strictly uh bigger than than uh it's different from zero so it's not in the intersection. So we have this cannot be true. And then, as usual, we have, hence we have minus 1 can be written as omega 0 plus omega 1 x. And this, of course, just

1:02:30 means that omega 1 x is less or equal to minus 1, yeah, if I throw away this, but then this of course means omega minus omega 1 x is greater or equal to 1. So we have found, what, what, that's what I want now this is really very easy and then he proves the real Nullstern this is this does not have a number it is on it's another application oh yeah Yeah, application on page 311, and it is an application of CRM-8, and CRM-8 is another instance of the abstract positive stellensatz, the more general, the general abstract positive stellensatz. It's an instance of this. And then he proves the real nullstellensatz as it stands there. I'm not going to write it. I should only mention that this is the weak version of the real nullstellensatz. It says, it tells you, when finitely many polynomials do not have a common real zero, and if you want to get the real Nullstein-Satz, which you are more familiar with, when you have a polynomial which is zero, where finitely many others are zero, you have to use Rabinovich trick. So this is no problem at all. So this is, as in the algebraically closed case, it suffices to prove this. This is just an application of the theorem 8, which again is an instance of the general abstract positive stellen or stellen sats, and this he uses, and then what he does is when he, I just give the idea, when you go down to A bar from a maximal ordering,

1:05:00 The A bar is again an integral domain, and in this integral domain, you can have an ordering, you can go to the quotient field, and from the quotient field, you can go to the real closure, and then since you are dealing with polynomials, you can use task as a transfer principle. that's the usual way one proves such things I think I should not go more into detail as it is usually done it's the abstract case plus usually it's done now by example when they stare at the CFP1 when they stare oh yeah usually I mean now done now in your book and in my book and... Yeah, yeah, yeah. That's right, yeah. And was it proved... Did he prove it before Krivine or after Krivine? Much after. Much after. 1970 or 71. But it seems also that it was us to use our carton, which is the reason why we brought And I was also, I spoke recently with Jean-Pierre Cahane, and Jean-Pierre Cahane mentioned, asked me if I know this wallpaper of Clivine. Yes, and he said that he was in connection with Clevin at that time, I don't know why. Because he was the thesis director of Clevin. And it seemed that there was no... It was kind of so. Okay, now you can talk about envelope, and yeah, that will be finished.

1:07:30 Well, the next that I'd say to me defines what he calls the envelope and the radical of a pre-order. He operates in vast generality. So I'm going to indicate his definitions and his characterizations briefly. And then try to link it to what Giguard did later on, and to what today is the actual, you know, the actual used version of the real radical of an idea. So, he defines two objects, two things, the envelope in a frame, we continue with the representation, and the ratio. And the envelope is simply this, I'm going to write it here, is the intersection of all maximal pre-orders. I use pre-order for this pre-order, And the radical, and, well, let's call this, just to make brief on the envelope of sigma, and the radical is simply what, well, this is a pre-operable, of course, no, containing sigma, omega. And the radical is simply the envelope of its, you know, what's the support, what is today, nowadays called the support of the, you know, what's the intersection, the C4 intersection, I'm going to omit the,

1:10:00 Okay, and then this is done in a completely abstract generality, and then she characterizes these two things algebraically, which is a rather interesting thing. And I'm going to write down the characterization, okay, and then connect it with the radical of an idea, the real radical of an idea, sorry, as we use it nowadays, which is sort of one step further down in the level of generality. The radical, well, he uses, okay, so he uses pi, let's say, in order to go with the same notation, omega, okay, then the radical is the set of all, well, we can write it this way, an element x, or x in a, x is in the value, for all lambda in a, will exist a little pi in omega, such that pi times 1 minus lambda. That means the difference of these two things is in omega.

1:12:30 It is some kind of real Jacobson radical here. And so this is what he goes, let me see, just to go on with it, pointing out the things that they are here in the paper, theorem 9. And that there is an algebraic characterization, even in theorem 10, of the envelope. So x is in the envelope of omega, if and only if. But there are two conditions. They look somewhat similar. lambda, there are pi 1, pi 2, pi 3 in omega, that is supposed to be non-negative, such such as pi 2 plus pi 3 lambda times 1 plus lambda x minus pi 1 lambda. And then there is another condition, okay, slightly different, which I'm not going to right, and not to be overpowering. But he does this, in very sort of general terms. And of course, as it happens in these cases, the more abstract you become, that's the general rule, the shorter the proofs become, but the more sophisticated. In some sense, at the end, it is a big simplicity, but it is a sort of second or third order of simplicity. Okay? Well, now, let me try to link this with the more theory of the real radical of an ideal. Okay? So, I have an ideal on the ring A commutative with U A, and, you know, the real radical is defined, one way of defining it, let me put it this way,

1:15:00 is an intersection of all prime ideals, real, prime ideals, and I am going to define real ideals, of A, of course, containing I. And an ideal I real means that A over I is real, here bringing the sense of okay if I know what's if I don't leave a minus one is not a sum of squares and a okay so the real radical is defined this way For Morley Page, you can see this in the book of Alex and Schiff itself, is called like this. Now, what happens is that without much work, actually, more or less straightforwardly, one has to work a bit. One can prove that in reality, if I is real, if I is real, then a real idea, then the real radical of I is simply the rad of A over I, and here the sum of squared. You know, as a parallel way. It's that, you know, you can, you can, you can prove this as long as this is a real ideal. The point is that if it, if I is not a real ideal, then this will come out simply to be the whole ring, to be the improper ring. So there is no damage in this

1:17:30 hypothesis. Well, there is some work to be done here. Now, how does this characterization relate to the usual characterization of the real radical? Because it is a purely algebraic No, no, no, no, okay, sorry, no, I'm explaining, this is including, this thing, and I'm trying to explain how, how does it connect with the, no, real ring is included, but real ideal, know okay but since we use the most used notion in it yeah sure oh sorry yeah you're right I should have been a slightly more clear in that sense but they I'm good I'm adding these comments because they are the actual notion that we use is real radical of an ideal okay and I don't have to be to come out from that. Now, there is another point which is not so obvious that it comes out. Let me keep this characterization. And it is not so obvious because notice that here we are working, okay, with the quotient ring, but we are working with sums of squares. He is working in complete full generality, Okay? So this characterization may not link so easily with the usual characterization, which says that an element A, let's say, of the ring is in the real radical of I, If and only if there are an integer e, and there is a, how shall I call it, sigma sum of squares in a, such that a to e plus sigma is nigh. That's the usual, the simplest characterization of the radical, the real radical of an idea.

1:20:00 And the proof that these and these are the same appears in... It's not just there, I guess. It's everywhere. It's everywhere. And as a matter of fact, the only slightly difficult point here is to prove that this thing, how is it, that the set of these things, okay, the set of all A that satisfies this condition is close under some. That's where you have to work a bit, okay? The rest is more or less easy. It's relatively straightforward. So, but I'm not quite clear yet how in actual computational terms, how to pass, even in that case, from this characterization, for omega equals this, okay, and A over I, okay, to this. You know, it is true that I haven't tried hard enough, okay, probably if I think several hours I will get it, but it is not so obvious, and probably this characterization, It's, you know, it takes its form because of the abstract character, because here we have any pre-order, which is much worse than the sums of squares. But even trying to bring it to that case is not so easy, I would say. It involves some working somewhat. So that's a point I wanted to make. And, well, the other characterizations of the envelope that I wrote there look even in that degree of generality. Although the proofs are very simple or look very simple, the characterization is even more daunting. But he does it. He proves it. You know, the proof is complete there in the paper.

1:22:30 So, that was for that point, okay? I have a question. Yeah, sure. You take a cone in your ring, and the intersection of the maximal cone is the same thing as the intersection of five cones, no? containing the cone. You take a cone, okay? How is the question? You take a cone, the intersection of prime cones, maximal cones and prime ones. No, I don't have... There are two notions. Yeah. That's why your initial observation of the... Two notions coincide, no? You don't know. That's why your observation about this looking like the Jacobson, rather than the NIL radical. I feel it's bigger than the radical in the meaning of intersection of Franco is containing No, this is maximum. The radical is the intersection of the ideas corresponding to all maximal pre-orders. So this radical is bigger than the usual radical. It seems to be bigger. We don't know. Usually it is bigger. Just take a code or a plane, such that the maximum ideal is convex, and there's just one point in the maximum spectrum, but there may be many, many points in the real space, so if you take the intersection of the points in the real space, you have something much smaller. Yeah, I am. Well, in general, if you take your thing to be a convex subring of real closed fields, let's say, the intersection will be zero. The intersection of the prime ideas, whereas there is only one maximal one. Sure. But yeah, that was a good remark. Perhaps one should look more at the Jacobson, which looks a little bit like this, Jacobson, the algebraic

1:25:00 characterization of the Jacobson radicals of a ring. I want to make one remark also, because as I pointed out already, there's a very much period of partial order brains, which is much older than real algebra as we do it and real algebraic geometry. I think probably would be a good idea if you really want to have a sound historical perspective to include somebody who is very familiar with that. Because this discussion of pseudo-princement, this is something which has been familiar for a long, long time among these people. And many other ideas that are in this paper here, I think, are very strongly influenced by the traditions of this group. So, something we are... Most real algebraic geometries are not at all familiar with the theory of partial or equivalence, and they are not familiar with our things. There's a large overlap between the two areas, but the people seem to come to the point. I think if you really want to put a historical perspective, you should include somebody from that area. Oh, yeah, there are some names. I mean, of course, yeah, I can give you some names. Matt Henriksen, for example, is quite old now. I don't know if he's still up to it. But he's very knowledgeable about this. He developed, actually, so he knows a lot. Okay, and somebody else who's quite familiar with the history and is younger is George Martinez from Florida. University of Florida. He knows a lot, too. You see, the point about So, Mark Henriksen is, he studied a lot of rings of continuous functions, and that is what it takes, and George Montevideo is also more familiar with art, things that are not necessarily very similar, so these are two names, and there are many more, I think.

1:27:30 If I think for a minute, sure, I can come up with some more from here, or ask John Gisbell. Gisbell. So the notion of pseudo-field was known before. I don't know. I don't think the name was being used, but the fact that you can characterize the maximal Well, combs, pre-ordered in this way, they had combs for a long time. So you are, I mean, none sort of the same sort of... Yeah, we continue on Friday. Friday morning. Today is Tuesday. What? Today is Tuesday. Yeah. Ah, we continue ourselves. Yeah, yeah, yeah. I would like to say, most French people will get a treat with us because they need to be paid off. If you want to be remboursed of your trip, you'll need to fill the papers. How many papers do you want to fill the papers? Scandale. I'm obligated to fill the papers. I'm obligated to fill the papers. So... So... Yes, of course. Yes, of course. Very good. Really not. Yes, that's right. No, but it would be embêté to not understand the paper.

1:30:00 The paper, right? Ah, yes. Thank you. I can't sleep. I can't sleep. I can't sleep. Do you have to sleep? I can't sleep. Directeur, dormez quand même. Moi, le café ne m'a jamais empêché de dormir en pratique. Maintenant, quand j'ai mis un peu plus tard, je ne peux pas aller les deux à la nuit, faire ce soir, parce que je ne peux pas avoir tout les délireurs. En un ou deux, je reste presque au bout.

1:32:30 You know what I mean? Thank you. Thank you. Thank you. Yeah, yeah, yeah, yeah, yeah, yeah.