 Okay, so I'm very pleased to introduce for our second lecture course by Tim Doktschitzer and he's going to tell us about the inverse Galois problem. And Tim has set up a very nice homepage for his course which is in the chat and maybe I'll just repost it for people who came in late. I'll repost in a moment and okay, thank you, Tim. Thank you very much. So, yes, I put it also here on a slide on the first slide the link to the homepage and this is where I'll try to keep everything. So, there will be exercises so things which I do know how to solve but they should be instructive. There's quite a few research problems which I think I don't know how to solve but I think that interesting questions. And there will be lecture notes, links to the videos and things like this. And most importantly, a magma package for working with families of Galois groups, because I really hope that during this week you will have opportunity, not just to learn the theory but what Henry referred to in his last lectures pressing buttons. So, and have an opportunity to actually construct families of Galois groups and see how you can work with them. All right, and there will be also problem sessions and teaching assistant for this course is Shiva. So please, I'm sure you'll see him in the problem sessions, which I'll also try to drop by it. All right, so let me start by just reviewing a bit of background from Galois theory. So, let's start with a field, a little K, which I will not probably use very much starting from the second lecture because it will always be a Q or Q of T. Now, let's think of the rational numbers. This is where the problems that we want to talk about is most interesting. So, a little K bar will be itself right closure. And suppose you've got a poly, you've got yourself a polynomial in Kx of some positive degree D. Let's look at its roots alpha one alpha D in K bar. And in what I'm going to say all polynomials are always seemed separable. So in other words, all these roots are going to be distinct. As I'm sure you've seen before, this is just equivalent if you want from one explicit side to F having a co-prime being co-prime to its derivative. But it doesn't have to be reduced. It doesn't have to be any polynomial and we're interested in Galois group. So recall from Galois theory that you can construct what I'll call capital K to be the field which you get by joining to little K the roots alpha one to alpha D. This is a larger finite extension of K. And this is a finite Galois extension. It's Galois because you can see it's finite. It's normal because I'm joining all the roots and it's separable by essentially by my assumption here. So it has a Galois group. So, which I will always denote by G. So G will be the Galois group capital K or small K, which is defined as a set of automorphisms capital K over small K. And this is a finite group whose degree whose order is that the degree of this extension here. And if you want to think in terms of polynomials and their roots. Well, what you get from this construction as a Galois group for polynomial. And that's a group because it comes with this Galois group here comes to the natural permutation action on this route. So recall that if alpha is a root of your polynomial f and then you hit it by an automorphism, then you get again, another root. The Galois group acts as permutations of alpha one alpha D, it acts faithfully, so only identity fix all this roots, pretty much by definition of capital K. So in other words, what you find is that this group G acts on the set alpha one alpha D by permutations. In other words, it can be viewed naturally as a subgroup of the symmetric group SD. So I will try to reserve the letter D for the degree of the polynomial. So my symmetric groups they will be SD, and the degree of little f that's, that's this little D here. So, so G is naturally a subgroup of the symmetric group SD. And if you, I mean, of course I picked a completely arbitrary order of this roots alpha one alpha D I just said, well, let's look at all the roots as D of them, because the polynomial is separable in k bar. Let's order them in some way. If you order them in a different way. So this order is a is an element of SD if you want. And what you find what you find is that the image of gene SD. If you reorder the roots it just changes by conjugation to this connotation the roots just conjugates this G to another subgroup of SD. If you want to forget about the ordering and define something canonical, slightly more canonical object, then what we are saying is that to polynomial of degree D, we've associated a controversy class of subgroups, or if you want to a subgroup in SD up to conjugation. This group is what's called transitive. So in other words, it has only one orbit on alpha one healthy to remember it some permutation group on D points. So it can have a bunch of orbits, but you can immediately see that if you put an orbit, then grouping these roots together will give you a polynomial, which by Galois theory again has an original field. So orbits, they correspond to irreducible factors of F over the original field came. So, it's an easy exercise to see, really, this is, this is the case and in particular, what we want to observe is that G is transitive. One orbit on alpha one up to alpha D, if and only if F is reduced. And very often, people, even though it's not always convenient, try to restrict themselves to just consider an irreducible polynomials and their Galois groups, as you will see in a second, this is not really a loss of generality. So explicit point of view. This is sort of nice because transitive subgroups of SD, they have been classified in other words, enumerated if you want, and listed and stored in a database for all reasonable size D. And at the moment, the database, you know, I think gap magma, it goes up to D 48. So for each degree up to 48. There is a complete list of transitive subgroups of SD up to conjugation. In other words, if you and then number. So for instance, if you look at transitive groups acting on two points there's only one cycle for the two. If you look at transitive groups acting on three points, there's two of them simply group of orders three and the full symmetric group S three. And if you look at for example of transit for four, again up to conjugation in S four, there's five of them, C four C two squared D four A four and S four. And these numbers do grow especially for certain degrees. So for instance for that's as far as a classification goes, the degree 48 is about 195 million of them. But nevertheless, if you ask for instance magma, what's a transitive group of 48 comma, I don't know whatever number between water 295 million, it will give you a specific transitive rotation group. Firstly, if you give it a transitive rotation group, it will return for you the 48 and the index. So in other words, if you want to think of. If you want to say that this specific or useful polynomial has this goal or group, then to quantify this is quite nice to go by transitive subgroups SD, because it encodes not just the abstract group, but if you want how it acts on the roots by just see if there's any questions in the chat, let me just open it so do feel free to drop by questions in the chat and now I'll try to pay attention. So, okay, now this does cover every finite group. So every finite group G is a transitive group of some SD. So how do you see that well one way to see this and essentially the only one of the only way to see it is to say that well every group acts on itself by left multiplication. So if you take the set to be the group itself, and let G act on itself by left multiplication that gives you an action, which is called regular action, and it is transit, because if you multiply by an element of G, you do hit every element. There's only one orbit, and therefore every finite group G is a transitive subgroup SD for for some D, namely D equals to its order. Now see there's question in the chat list of transitive subgroups. Yes, so they're classified up to conjugation in SD, that's a little bit different than isomorphism and I'll give you an example in a second. And does it mean there's this number of groups of 48. No. So, if you look at these groups for example. There's a difference between the order of the group, and the number of points on which it acts. So if you look at for example, this groups of degree four to my D is equal to four. In other words, if you're interested in gala groups for the moment also degree four, the largest group that you can have is a group of four factorial is a group of size for factorial, and the smallest group you can have is a size for you can't have a transitive group of smaller than that. What it says is that if you look inside S 48, which is a huge group of order 48 factorial, there's this number of groups in it, and the order varies between 48 and 48 factorial. This is presumably C 48, although I didn't check. And this is S 48, the full symmetric group of 48 letters, there's loads and loads of group in between that act transitively. And these ones, this sort of baby example of degree four, but sort of on a much larger scale. Okay, so now this difference, which you kind of alluded to in the question between groups, sort of between going by order and going by this transitive degree, these two classifications are a little bit different. So in other words, they have different transitive actions. Every, so most final groups, they will occur several times in this list here. In other words, they can act on different. They occur as different transitive groups. So in other words, most groups G, they have several different transitive actions. So let me look why this is the case. The result that says that in group theory that says that if you're interested in transitive actions, any finite group G, then they're in one to one correspondence to conjugacy classes of subgroups, H and G. The correspondence is very simple. If you take a subgroup H and G, you can look at the cosets G over H, or just left cosets G acts in them. So that gives you a set of G, a set with the G action, not necessarily faithful but nevertheless set to the G action. And conversely, if you start with any transitive G set, in other words, any set of points on which G acts transitively, then you can pick a point, look at a stabilizer of that point G, and that stabilizer is going to be a subgroup of G, let's call it H. And if you pick a different point, you get a conjugate subgroup, because they're all commuted the actions transitive, you'll see that you will spend the whole quantities class of subgroups like that. That's one to one correspondence. So if you want, and this is what we're looking at, we're looking at not just so a transitive action of G is a homomorphism from G to SD, such as the image of the transitive group. What we're looking for, we're looking at inclusions of G into SD. We're looking at faithful actions. And for that, what we need is we need this action to be faithful. We don't want any element to act trivially on this left cosets. And if you think a little bit about it, this corresponds to the fact that this subgroup here is what's called core free. So it has a trivial core where a core of a subgroup is defined as an intersection of all conjugates of that sum. So you take a subgroup H and G, and if you look at all of its conjugates and you intersect them, if you get the trivial group, then this is good, and then this is called subgroup was trivial core, and these are the ones which correspond to transitive faithful actions. Let me just see if I didn't miss anything. Just see where am I. I think she's keyboard because my mouse just jumps. Okay, so, alright, so what I was trying to say. So let's give you an example of this. So for instance, let's take the symmetry group and three letters G equals S three. There is a trivial subgroup. There's a whole group. There is a normal C three. And there are three C tools which are generated by different transpositions. And these are all conjugate. So as far as subgroups up to conjugation, there's only C one C two C three and S three. So these two subgroups to have trivial core. So there's a transpositive there is a cycle before a to generate by transposition, if you intersect with its conjugates you clearly get just a density, and similarly there's a trivial group itself. These two subgroups that these are two different subgroups at the conjugation, they define two different faithful transit actions on different number of points so the number of points the index of the subgroup in G. So the group S three occurs twice in our classification of transitive groups, it occurs as a group three T two, you know that second transitive group on three letters. This is the usual action on three of S three on three points, where the stabilizer is a transposition. Also, as I said, every group has also regular action as to leave no exception, so it can act on six points, namely on the group itself if you want by left multiplication. And this is called 62 in the classification of transit groups. Every group may occur several different times, and the number of times it occurs is the number of these core free subgroups of G up to conjugation. So this is, maybe sort of looks like a bit kind of abstract mamba jumpers of group theory, but from point of Gala theory. This is a very clear what why this happens. I'm sure you've observed before that sometimes in Gala theory, you have, you can have inside a given Gala extensions, you have different fields which have exactly the same Gala. So let's take an example here. Let's take a random S three extension. So what I took here is a splitting field of X to the six plus three. Alternatively, it's the same as a splitting field was x cube minus three. Let's let's think of it this of this in this way. So let's take a polynomial x cube minus three over Q. There are three roots. There's a cube root of three. And there's a cube root of three multiplied by various security. So the normal, the splitting field is this extension here. Q join Zeta three cube root of three. And this is an S three extension of Q. Now, it is a Gala. You can see this field in two different ways. You can see it as a normal closure of this field here. You can take this degree three polynomial x cube minus three, but it can also take a primitive element, which is quite easy to construct here, because this is a joining square root of minus three. This is joining cube root of minus three. So if you join six root of minus three, you in fact get the whole few. So if you look at that, this polynomial here and you just to join its roots, you get the whole thing. If you join even one of its roots. You get the polynomial of degrees three of which this is splitting field and this is a polynomial of degree six of which is splitting field. And, and that's precisely our situation so the Gala group here as three acts on the roots of heat of this polynomial as what we call three T two, and it acts on the roots here as what we call six T two is a different because this condition to have a field whose Gala closure is the whole is the whole field is the whole Gala field that we are considering to remember by Gala theory, every field is cut out by subgroup. In addition, that the Gala closure of this field is the whole thing, and not some smaller normal sub extension of this field. It exactly corresponds to the fact that this subgroup here has trivial core. In other words, it doesn't have any doesn't contain any knowledge. So this is so sometimes when you think of Gala groups. If you think of them as abstract groups, there is just S3. But if you want to realize them as groups of polynomials, then you have to think a little bit that you can have different actions, and sometimes you can have even different actions on the same number of points. So the same group can occur even know twice, for example, in the same degree. So there's a slight distinction. The observation is that some groups will somewhat unfortunately, they only have a regular action and nothing else, because, for example, every subgroup of them. G is normal. And this happens if, for instance, she's a billion. So she's a billion, then it's very easy to see that there are no core free subgroups, except the trivial group. But there are some interesting other examples such as generalized quaternion groups. I'm sure you've seen a quaternion group q eight, and it lies in a family q eight q 16 q 32 and so on which all have a very similar definition. And these groups also have a property that every subgroup of G is normal. They also don't have any core free subgroups. And that means that this group only acts transitively on number of the order of G points, and nothing, and nothing less than that. And that means that if you want to realize this group as a galore group of an irreducible polynomial, then this irreducible polynomial has quite a large degree. It must have degree, which is equal to the order of G, which is kind of quite unfortunate. It means, for example, that if you want to realize a sickly group of order, I don't know, 71, then you have to look at polynomials of degree 71. There's nothing else you can do, even though for some other groups such as symmetric groups. You can realize a group of order and factorial active on endpoints. So, these are kind of two different orderings which people who work in explicit galore theory have to go between sometimes, sometimes it's convenient to order groups by their order. It's convenient to group to order them by, let's say, the smallest transitive set on which acts, and therefore by that transit. So both in gup and magma. There's what's called a small group database, which orders groups by their order. So you can ask what's the fifth group of order 16, six group of order 16 and so on. And it also has this transit group database which I already mentioned. It's different. So for example, if you sort groups by their order, I just checked this morning that if you want to get to a six. There is 10 million groups that you have to consider before you get there, because there's so many groups of more than 512 and 256 and so on. But if you order groups by their order, then it takes a long, long time to get here. But on the other hand, you know you come up small sickly groups quite quickly. And if you order them by their transit, and that tends to group identification, then you get to a six pretty quickly, because I'm not that many group, not many, not so many groups in degree four, five and six. But you'll never get for example to cycle before the 49 because as I said, only groups of what up to up to transitive degree 48 have been have been listed. So this is not a sort of in any database. This is a bit. This can sometimes is a little bit of an issue but I just want to mention that there are two classifications like this. Are there any questions so far. Okay, I'll keep an eye on chat. All right, so now let me get sort of to the core issue now. One of the biggest unsolved problems in theory is what's called inverse a lot of problem and there's a very respectable collection of people who consider this problem interesting, starting from Hilbert and need to know that browser, people like that. It asks a very natural question which I mean it's kind of obvious once you started to really learn gala theory and you learn about gala groups as being symmetry groups of polynomials whether every symmetry group occurs in this way. So this is a conjecture which is called the inverse gala problem. There are two versions. There's one over Q and one over Q of t, I'll mention both of them, because this will be sort of particularly important for some of the sound the constructions that you have, and also for the package package that you're invited to play with. So one says that if you have a fine group G. The second structure says that there is always an extension gala extension cube, which has that group G is a gala group. And the second says that the same is true over Q of t, except you have to be a little bit careful, because you don't want boring sort of constant examples otherwise to be the same conjecture. If I just said there is an extension of Q of t with a given gala group. So then, then of course, if you can construct one over Q, you can take the same constant extension over Q of t which doesn't vary with T at all. So, you don't want that you really want this extension theory, and more generally, and as you will see later this is quite a metro, what's called a geometric assumption. You don't want any subfield of this extension to be constant over Q. So that's why there's what the regular here. So the conjecture says that there is a regular sometimes called geometric find gala extension K over Q of t, which has a given gala. So what this regular means is that one very concise way of stating the definition is that K has no algebraic elements. So K and Q bar is equal to Q. There is nothing in K. If you take any element it's always transcendent over over Q, and it's very easy to see that it's the alternative is to say that there is no subfield inside this large field K, which is constant except of course kids. Now, there's no reason we stick to one variable T here, you can take n variables T one up to TN, and sometimes will be interested in such families. So families over Q T one up to TN, which again I define regularly the same way as having no algebraic elements. And with the exception that, especially with a magma package in the example cell keep, I won't call variables T one up to TN, but I'll call them a B and so just by the usual letters usually not going very far. Just that if you see a and B and so on. This is, you know, these are just variables such as T1 and T2. Let's solve the inverse Galois problem for G equals S3. So, what does it mean, we want to prove these two conjectures when Jesus in that three letters, like what do you need to do for this. So for the first conjection, you just need to give one example, which has Galois group S3 and this is not very difficult to do I already gave it on a previous slide, you can take the polynomial x cube minus three, and by the usual Galois theory, you know, it's Eisenstein reducible. Therefore, the goal of the order of the Galois group is divisible by three and it also contains studies of unity so it must be the whole thing. You see that the Galois group is the whole S3 and therefore the inverse Galois problem is true for S3 or Q. I'll use this curly I to denote inverse Galois problem for given group on Q. So if you want to do it over Q of T. And by the way, in practice, this is very nice, but in practice not very useful. If you play with any sort of number theory. So for instance, you want to study class groups of cubic extensions of Q or anything like this. So having one example of a cubic extension is of very limited use what you'd really would like to have would like to have a family, and moreover you would like to have a family that doesn't have very specific properties like this which makes it non regular. So if you look at the family x cube minus a, so I call it a family, because you can think of it as a bunch of cubic extensions, kind of one for every able to. So you can think of, it's either a field extension of Q of T, or Q of a, if you call the variable a, or you can think of it as a family of extensions for you. And that's why I'm going to call such a thing usually a family over Q a. So for example for non regular, or if you want to know regular extension, because the splitting field of this polynomial contains Q zeta three and Zeta three is algebraic or Q, or in other words, if you want in this tower that I had before when you have an S three extension, and it has a quadratic extension over it. So this extension does not vary with, does not vary with T, or does not vary with a is completely constant. While on the other hand, if you take more or less any other example. So you take x cube plus some polynomial in a times x plus some polynomial in a. In other words, we take a more or less generic looking cubic polynomial whose coefficients depend on a, and it's most likely to define a regular family. So here, for instance, if you take x cube plus x plus a then this is a regular family. And how do you see this. Well, in this case the quadratic extension, which it contains, as you probably know if you have a cubic field like this and you want to see which quadratic, it contains it it's got well cool but the computer is discriminant, the discriminant of this by the famous for a cube was 27 b squared formula. In this case is this minus 27 a squared minus four. So this field varies with a it's not a constant extension q, and it's a nice exercise to reduce from here that in fact, the whole family is right. There are many other constant subfields. And therefore, if you wish, just this particular family proves the fact that the inverse Galois problem is true for S3 over QA. In this. So whenever I talk about inverse Galois problem over fields such as QA, I always mean that I want to regulate. I don't want to question whether such a field is unique. No, certainly not. We don't expect any sort of uniqueness here in a sense that for instance if you say I want to have quadratic extensions of q. There's lots of them q two q three q five and so on. And similarly, if you look at quadratic extensions of q of t, then you take any rational function t, you're doing it square root, and you get a different extension of q of t, unless you do functions of q by square. So, there is lots of them. There is and I may mention some sort of search sometimes for what's called generic families, which is kind of one family, possibly in many variables, such as any extension you can think of it's always its specialization. But this is very difficult problem except for like very small groups to find such things, and I will not talk probably very much about that. Okay, so now, before kind of going on sticking to only these two fields q and q of t completely. Let me just say a few words what happens if you do change them, because you can ask yourself. You ask this question of the cure qft. What if you ask it over a different field, different base food. And it is, there are some famous examples where this question is, is it known to be truly false or known to be true, or like in the case of cure for number field is really the most interesting case which we don't know the answer. Well, first of all, if you start with minutes clear that this inverse Galois problem, it cannot have a positive answer but any field, because if you take for example an algebraically closed field, well it doesn't have any fine extensions at all. So if you ask does it have any I know a five extensions of course the answer is no. If you take k for instance to be our C. Well in this case every extension of k has degree, but most two, if you took our or just trivial if you took C. So apart from C one and C two. This little cyclic groups you can't realize any, anything as a global group so the inverse Galois problem is trivially false. There are other examples where it is also false. For example, if you take a finite field, then every extension of a finite field is always cyclic so about from cyclic groups, which you can realize. No other group is realizable as a Galois group of q. And similarly, over adiatic fields. So again, if you ask yourself well does q three have a five extensions then again well the answer is no because it's not a soluble group. It's an interesting question, which soluble groups you can realize over given pediatric field, but at least it is clear that. So the problem, which Galois groups you could possibly have it certainly still has substance, but it's certainly true that you don't expect an arbitrary Galois group to be there. So for example fields where we do know the inverse Galois problem to be true. So, over CFT. It's essentially consequence for Freeman's existence theorem that you can construct. Well, the remand surface which has any given Galois group on P one. And in terms of fields, it means that the field CFT has extensions any Galois group is descending into q, which is, which is a problem. So there is a difficult theorem by Hobbiter, which says that the same is true if you replace C by QP. So, over QP of t, again, you can realize any group as a Galois. So the most interesting case is the case when case q or more generally a number field but Q is difficult enough. And it is generally expected that the inverse Galois problems with these conjectures are true. So in other words, any group you can realize the Galois group q or any number field. And similarly, as a regular extension of qft I don't know, in the case of qft or k of t where case a number field, if anyone has stated this as a conjecture. There are many questions, many books like Sarah's topics in Galois theory, for example, is a bit more carefully phrased a question, because we're not completely certain that we have enough evidence for that. But I think most people. Most people believe that this is true. So what is the result for k of t where case of arbitrary algebraically closed field. Well, I think in characteristic zero by sort of left its principal sort of nonsense. It's probably the same as over the over the complex number so it's probably the same as this question. In positive characteristic. I actually do not know. There are also over f p of t, which by rub correctly they follow from Harbiter's theorem, but whether they just say you can always do it, or there are some issues because separability and things like this. I don't quite remember. So, it's a very good question, but I don't, I don't know. I'm pretty sure it's not, it's not. And, well, this sort of being an explicit methods and number theory course. What I would also like to do I'd like to do two things or first of all, just to review some of the many methods that we have that really, and it's an interesting problem, because there are many methods, and many of them they look completely disjointed. Some use very difficult group theory coming from great groups and things like this. Some use very advanced number theory, some rely more on algebraic geometry. And some just one ad hoc constructions, and many of them, they, they're able to do sort of to construct many groups, but there is no universal picture in some sense, and none of the methods seem to in some sense work generically. So we're just going to review some of them in these lectures, and also because this is really workshop and explicit version if you want number theory, we will be very interested in the question, which is the explicit version of this. So we have a lot of problems, how to construct this fields K, in the case where you expect the answer to be. Yes, which is already as many people have discovered it's even much more difficult problem than proving theoretical results such extensions exist and many results that we have non constructive. So this is basically what's going to be the focus of this course. So let me give you a very, very brief history. So, I think it's kind of generally accepted from what I understand that this whole study started with Hilbert, who has a very influential paper where he proves what is now known as he was ability where he proves in particular that if you can solve the second conjecture for a given group G, then it implies the first one. So in other words, if you can realize G as a girl group of Q of t, even without the word regular, then it implies that you can also realize you. And in the same paper, he proves that groups such as SN, you can realize regularly if you wish, over Q of t, and also a very interesting example which are alternating groups and his constructions are explicit. And I will read you briefly of what's going on there because it's one of the general techniques that we have. This is basically a for soluble groups. This is a very famous story of share which is theorem so there is a result. Going back to 1937 by Schultz Reichard, who proved that odd meal potent groups are realizable as galore groups of Q. It's basically an inductive construction but as I'll explain later, it's difficult to set it up in a way that it works. It works for groups of order, and this was some gaps which were lately fixed extended to all soluble groups by ship ravaging 1989. So we now have a clean proof of that every soluble group is a galore group of a Q, but it does not extend to Q of t. So over Q of t, this remains a conjecture that every soluble group can be realized as a galore group of a regular extension of Q of t, even for our groups so for groups of prime power order, and some of the smallest examples are there are three groups of order 64, before which this, this conjecture is not known of Q of t. And secondly this proof is actually non constructive. So it doesn't give you specific extension. Now on a completely opposite side of the spectrum, there are some soluble groups, there is a famous method which is called rigidity, which has been applied to simple groups and it mostly applies on to simple groups. There are many authors associated to it for Sheikh for groups as PSL, and then Mali, Matzad, Belly and Thompson, looking at a lot of sporadic groups. Using this method, a lot of simple groups are known to be galore groups of Q and even Q of t. And sometimes these methods can be made explicit really. And for example, all sporadic groups, except M23 are known to be galore groups for Q of t even a monster group which is so large, probably still don't know its character fully, but we do know that this group is realizable as galore group Q of t remarkably enough. It's a very interesting method, but it's, it has only very limited applications. And if you order the groups by their transit degree, which if you want is a natural way to go about things if you're interested in, you know, galore groups of polynomials reduce polynomials to be 234 and so on. There is a theorem that all transitive groups of degree less than 15 are galore groups of Q of t, and I have a constructive version of the theorem in the sense that I have some families in the same degrees. And basically this course I'd like to present some of the methods which, you know, which makes these sort of things possible. The inverse galore problem for all galore groups for all groups which, you know, act transitive on sets with the most safety elements. And of course, a lot of people have collected a lot of data over Q and over Q of t and let me just mention a few results here there's Jones and Robert database of number fields that's made by Smith, who looks at families over Q and over Q of t, and there is a big databases, Robert Clunus and Maler, which is now I was told is in Paris, and in LMFDB, which collects lots and lots and lots of fields, especially with Q, with specific galore groups. Okay, so now, let me start by saying a little bit about the general methods that we have, starting with Hilbert and his new disability theory. There are various versions of this result. Let me start with the simplest one which is easiest to understand. It says suppose you have a polynomial. If you should to variables t and that's just like a different role. So we view it as a polynomial over the field Qt, but with the variable X sorts of polynomial D let's say in the variable X, but it's coefficients depend on T. So what we would like to do, we would like to specialize this coefficient so it's called Hilbert specialization theorem, by taking a random rational number R and substituting it in place for T. Then you get a polynomial and generically of course the degree will not drop. So it will be a polynomial still of the same degree. And the question is, is it true that if you start with an irreducible polynomial over Q of t, it remains irreducible. At least sometimes, when you specialize them, when you specialize it, and Hilbert's reducibility theorem says that this is true. And in fact for infinitely many Russian numbers aren't you this fellow specialization is going to have a degree and it's going to be irreducible. In fact, there are, there's a lot of stuff study, the set of exceptions, and the study of so called thin sets. When, you know how often this polynomial is irreducible, and it makes sense to say that for most are we can quantify this, this polynomial is going to be irreducible, you know, be careful here. For general version, you can replace one polynomial by a collection of n polynomials, you can replace one variable t by m variables t one and two TM, and access similarly by n variables x one up to x and so there is a capital and m and n, which are all equal to one in this particular version, which says, you know, exactly the same, go a bunch of polynomials and irreducible, then there are infinitely many specializations t one equals R one and so on TM equals RM, such as the resulting polynomials and n variables are all irreducible. But this more general version just is formally implied by this version here. So, it is, it is not, in some sense, more interesting, of course, in practice you do want to use it. And so Hilbert proved this theorem over Q and nowadays fields over which this theorem is true. Let's say the first version because the second one is implied by the first one, and they're called regression fields and examples are rational numbers, number fields. And there are some other interesting examples, for example, maximum a billion extension of Q and more generally maximum a billion extension of every number field unknown to be Hilbertian. So they still have the same properties here. But it's not true for every field. And here is an example where you can immediately see that it's not true, let's say for a finite field of Q. It's a family of, let's say C two cross C two extensions, which I take here from from the package that I advertised on my first slide. So if you take a polynomial x of four plus T x squared plus one, it gives a family of C two cross C two extensions over over Q. And in fact, yeah, so I'm, yeah, exactly. So this is a good question. I'm exactly answering this. So, no, there are no finite regression fields. Well, okay, I think such as 90%, I don't know about 90%. But certainly, what is true is that finite fields are cannot possibly be regression, always polynomial such as this one, for example, such that because it has gala group C two or cross C two over F Q of T, if you want to specialize it, it can't give a gala group C two over C two, across C two over F Q, because F Q doesn't have C two or C two extensions. Remember, it is only quadratic extensions. And from here it's immediate with this polynomial automatically be reduced. So every specialization of this with T is an F Q will always be reduced. And it is not difficult to deduce from this, your usability theorem, just by taking a primitive element in your field, and asking, let's look at its minimal polynomial and asking whether it stays irreducible. Well, if it doesn't the gala group has to say the same. So it's not very difficult to prove this corollary is that if you can realize the group G as a gala group Q T one T and then by specializing T is so by choosing a specialization for T two and for T and carefully, you can make sure that such specializations give you a gala group over Q. And this most where I'm talking about here is for example, implies that as a risky dense subset of these T is will give you specializations, which is a full gala group Q. There's more than you can say. So, in other words, we have sort of one implication, and a little bit more than that. If you know inverse gala problem for G over Q of T in its regular incarnation so remember this stands for being able to construct a regular family for G over Q of T, then it will specialization thing you get G as a gala group over Q, but moreover, because remember regular family has a property it doesn't have any algebraic elements in it. So you can base change it to any other number field and it stays regular was the same gala group. So if this is a regular family, then it gives you a regular family over cake for any number field cake. That's why it's amazing to be able to have regular families that's why it's such a very good notion. So, therefore, this conjecture implies the corresponding conjecture for G over K for any number field cake immediately. So in other words, if you can construct one regular family over Q of T, such as we did for S three on the previous slide, then suddenly you know that every single number field is infinitely many as three extensions. It's a very, very powerful result. Some construct such three products and so on. Similarly, you really need this notion of regular. Okay, let me see if I can do something in my remaining minute. Yes, so let me just do it like this. So as a corollary of this result. I showed that SN is a Gala group for Q, for every n greater than one, in other words, this inverse Gala problem for SN over Q. And the proof is you may have seen this in the course of Gala theory. It uses this rational function fields. So you first realize SN over a rational function field with n variables. And in determinants alpha one up to alpha N, which you think of the roots of some abstract polynomial, but they're just variables. SN acts on them by permutation. So in other words, SN acts on this field to join alpha one up to alpha is the field in N variables, and SN just X by permuting the values. So when you hear it was Gala theory, you say well I have a final group of water morphs field which fixes Q. So let's look at the field of invariance and the field of invariance as a famous result in this case is. So K fixed by SN in this case is a field of elementary static functions in the rules. So it's again a field in N variables, where these n variables are no different. They're not alpha one up to alpha and the a one up to a and where the AI elementary symmetric functions. For example, some of the roots dot product of the roots. These are the functions which are invariant on the G, that's quite clear. But more importantly, every other rational function, which is invariant on the G, it can be expressed in terms of them. So in this case, this field of invariance little K is again, like the field you started with a purely transcendental extension. And that means that we have just realized SN as a Gala group over Q a one up to a and what is such a purely transcendental extension, and by Hilbert's usability theorem, you can now specialize this AI is in such a way as to get Geo. Okay, I think there is an answer to a question that if case algebraic close to every fine group is a Gala KFT. Yeah, thank you. So for example, FQ bar of T, this statement is true. And it is used hard beta. Does use hard beta somewhere along the way. Okay, and finally to end with. There is Gilbert in the same paper he also proves that alternating groups and is a Gala groups over Q for every and greater equal to one. And his time I think this is very difficult to prove, but in our times. This is a bit easier. So I did leave it as an exercise, even though it's broken into steps on the homepage. Okay, I think it's a good place to stop here. So I guess I will see you tomorrow. Okay, let's thank him for that beautiful lecture.