 Thank you everyone for the introduction and thanks very much to everyone for being here and to the organizers for Inviting me here. So it's a pleasure to finally be able to speak in front of an audience I'm not used to this anymore. So please help me by asking many questions as you did for Sorry To bend it. Ah, is it better like this? Yeah, it's good Okay. Thank you. So as I was saying, I'm not used to giving talks in front of audiences anymore So please help me by asking as many questions as you like as you did with other lectures Also, the so the notes at least for the first two talks are written and Available on the PCMI webpage also the exercises. So I encourage you to go to the exercise session led by Diego with there And So, okay, I will be talking about the inverse Galois problem. So let me start by introducing this and Also, let me know if I write too small or anything. So my story starts with Galois theory. So 200 years ago So Galois explained how when you start with an irreducible polynomial FFT with coefficients in Q If it's irreducible Galois explain how you can associate with it a finite group That was in 1830 and Perhaps we forget how much of a revolution it was at the time people knew how to solve equations of degree 2 3 4 They were looking for solutions in degree 5 by doing computations and trying concrete things and then Galois came up with this very general theory Which you completely changed the landscape. So in fact, it's not even obvious how to associate with a polynomial It's group of symmetries. So today we understand this very well We look at Q bar we look at the subfield generated by all of the roots of F and We took we take the automorphism group of this field but at the time it was really not something obvious and it took many years even for other mathematicians to understand To understand this association in Galois theory And even just this is full of of computational challenges. So if you Feed the polynomial into a computer it can in theory give you the the Galois group back But in practice it will do this only if the if the polynomial is small enough It's it's a hard thing to do okay, so 200 years later We understand things better, but not so much vexing problem. We still don't know the answer to the question What finite groups do you get in this way? So or precisely Do all finite groups appear in this way that's the question They appear sorry in this way. So that's the inverse Galois problem. So can you reverse this arrow? So let me point out that it's a fundamentally number theoretic problem You can ask the same question starting from other fields than Q But then the answer so well depends pretty much on the field if you start with an algebraically closed field Of course, it's no with the reels. It's no with qp It's not so obvious, but it's no because you only get salt solvable groups And so on and so forth. So if the answer to this question over Q is yes It's really because of the precise nature of Q. It's it's a it's number of theory So The hope is that the answer is yes, but it's very much open So there's a very long list of positive results. Let me mention just a few for some positive answers so Well the first groups I guess where the symmetry groups and the alternating group So this was due to Hilbert's and an important theorem is Shafarevich's theorem that you can do it for solvable groups So the case of a billion groups is an exercise. I mean, it's it's in the list of exercises you will do solvable groups Shafarevich So then if you can do solvable groups, it's it's natural to look at simple groups So there's a classification So there are some infinite families and out the 26 sporadic simple non-a million groups So among them all of them have been realized except one so all sporadic groups Except M23 So here there's a long list of names. I'm not going to write them all. I mean they're they're in the notes And so but for example the monster group was realized by Thompson And okay some some various Sorry various infinite families of simple groups. So there are so many results. I'm just going to state one of simple groups So for example PSL 2 of Fp So for all prime speed, this is known to be a Galois group over Q So in this journey T, this is due to Zouina. There were many partial results before okay, but Even for such a simple group as PSL 2 of Q where Q is just a part of a prime and not a prime the Problem is open It's very open in general Open so even for PSL 2 of F 27. So that's just one example so in this course what we're going to do is to Discuss some of the methods that have been used to attack this problem. So again, there are many results also many methods I'm not going to be able to talk about all of them or even a Significant proportion of them, but I've tried to Select the most most important topics and also I I should say that as as is often the case in mathematics The proofs are sometimes more interesting than the than the results I mean, this is a fundamental problem, but I mean, there's no use. There's no direct use application for this but but many of the methods involve lots of Questions or objects that are interesting for their own sake and that have been studied and lead I mean independently of the inverse Galois problem So that's the actual goal of the Of the course to to this to discuss these topics connected to this but which lead to other other directions as well okay, so This particular problem Has no direct application, but there are lots of variants some of which do have applications So let me mention one example of a variant So it's the so-called Grunwald's problem So from now on I will not look just at q but at any number field It makes no difference for everything I will discuss but you can stick to q if you like So that G be a finite group and K a number field So some some notation I will denote by omega the set of places so if you're looking at q this would be the prime numbers and the infinite place and As in Bianca's talk, I will denote by Kv the completion Of K at a place V so What the group what Grunwald what the Grunwald sorry what the Grunwald problem is about is In this situation, can you not only realize G as a Galois group of a K? But can you also? Realize it by an extension whose completion at the finite set of places is prescribed So let's introduce a finite set of places s and For any place in s. Let's fix some Galois extension of the completion so capital Kv over Kv so Galois a Galois extension So if you start with a galois extension of K with group G if you complete it You don't get in general a galois extension with group G but only with group a subgroup of G So here I'm going to assume that the galois group is a subgroup of G. So Okay, and let me say this in a very weak way that the galois group can be Embedded into G. Okay, and then the question is is there a galois extension of K Capital K over K with Galois group G and Whose completion at a place above V is capital Kv for all sorry. So this is for V in s. I fix this for For V in s fix capital Kv and then okay For any V in s The completion at V is Kv. Okay. Is the question clear? Yes I'm sorry. You ask if it's a place in capital K or small K Yeah, so I mean here really is a place of small K But here what I mean is you take a place of capital K above V you complete there And then you should recover this thing which I denoted capital K sub V And I mean this does not depend on the place you choose because it's galois Okay, any other question No, okay, I mean you can give a more precise statement and this will come later, but for now I'm just it's just in a loosened isomorphic to it But yeah, and you can be more precise of course Okay, any other question Okay, so I mean this okay, this is the Grunwald problem It has an interesting history. It started with a paper of Grunwald who in fact proved the That the answer is yes if the group is a billion and this was used by many other people And it was even given another proof by someone else and 15 years later a counter example was found So This happens fortunately the the applications were saved So the counter example was given by Wang and so the answer is no So even for K equals Q For S you take just the prime two and and the Extension of Q2 that you can that you consider Sorry, the G G Z mod z mod 8 Z very simple And the extension you consider at 2 of Q2 is the unique and ramified extension of degree 8 and ramified of degree 8 So when proof that if you start with a cyclic extension of Q of degree 8 You can never by completing it at to get this one Okay, I'm not explaining the proof it's part of the exercises You so you will see it So the answer is no but Wang was able to fix the mistake and so this gave the so-called Grunwald Wang theorem Which gives a complete answer for a billion groups a complete answer for a billion G so I'm not stating the theorem because it's a bit technical but In summary the answer is almost always yes You only have this this problem with respect to the prime two And in general so if you take any finite group G The expectation is that the answer should be yes for any finite group G as long as The set of places S avoids a certain set of exceptional places so hope for any G Yes If S does not contain any place which which lies above one of the primes dividing the order of G as Contains no place So let me say dividing the order of G at least no country example to this is known Okay, so this is one useful variant there are other variants So for example embedding problems and yet other variants that you will see in the exercises But let me not discuss them now so Yeah, I'm sorry. I couldn't hear you. I'm really sorry. I didn't understand the question Is it okay? Okay for every finite a billion group we can already realize it is a Gallo extension over Q. Yes That's easy, and it's just used for a cover term. Yes. So what I didn't understand when you say it's complete answer Abelian G. I think you just meant for this completion method, right? Yeah, yeah complete answer to Grundwald's problem Okay, yes because it's kind of confusing. It's sorry. Sorry. Sorry. Thank you for the For the question so first I discussed the inverse Gallo problem So indeed for a billion groups the answer is always yes And in fact even for solvable groups and here here's a variance Grundwald's problem It's more restrictive and so there's a complete answer to this To Grundwald's problem. Thank you For a billion G. Yeah, so it's really this you you fix places Extensions of the completions and you you look for for a This capital K here. Yeah, so the answer you cannot you cannot always do it Even over Q But at least this theorem tells you exactly when you can do it and when you cannot for any number of fields and any Abelian group and any I mean any given set of places and and completions and and extensions of the completions Okay. Thank you. So Today I'm going to talk about the first approach that was suggested for attacking the inverse Gallo problem Ah well Okay, that's was just the hope Sorry, so let me repeat it. The hope is that Grundwald's problem has a positive answer for any finite group G as long as you avoid these exceptional places so To discuss this first approach with which was due to Hilbert and let's let me let me talk about Torsos so Torsos and their fibers so I'm going to employ a geometric language because it's it's Really useful to understand the situation So let me define what the torso is so So fix a finite group G and K a field So a G torso. It's going to be a morphism of varieties algebraic varieties defined over K So, let me add some words finite oops Finite et al morphism, so let me call it pie From Y to X. So here why an X are varieties over K? So I mean variety in a very loose sense. I mean it's something defined by algebraic equations I'm not assuming that my varieties are connected for example, so let me write it Here so XY Are varieties of a K? possibly disconnected so Finite is going to imply that the fibers are finite So why an X have the same dimension and et al well it means there's no ramifications So if if you're working in And if K is a subfield of C the complex numbers you look at the complex points. It means it's a local isomorphism Anyway, if you if you don't know what this is you just ignore it in characteristics. Oh, it's going to be a consequence of what I write next so An et al morphism with an action of G on Y of G on Y such that Well first the morphism is equivalent. So It's invariant under G pie is G equivalent and if you look at the fibers of the of this morphism When you look at the say the K bar points K bar is an algebraic closure of K The action of G on the fiber is simply transitive the action of G on the fibers of the map between K bar points Induced by pie is simply transitive so Another way to formulate this is to say that first of all G should act freely on on Y Well, of course because it acts on the fibers and it acts freely on the fibers and It induces an isomorphism between the quotient of Y by G and X and pie induces an isomorphism From this quotient Y quotient it by G to X Okay, so it looks like a rather abstract definition But we will see examples and I mean you will understand the motivation for introducing them So here's a trivial example trivial action Thank you. So here. I really just mean that If you compose if you pre-compose by an element of G you get the same morphism Thank you Okay, so here's a trivial example, so that's the trivial torso It's just a distant union of copies of X as many copies as there are elements of G so X cross G With the projection This is not very interesting, but it's a G torso Okay, so Why are we interested in this? So here's a remark. So we're interested in Galois extensions of K with Galois group G So and the thing is Galois so field extensions a capital K Over small K of with Galois group G. Well, in fact, that's the same thing as Connected G torso over the point Connected G torso Over the point so the point is the variety consisting of just one point if you like it speck K over the point speck K and Because of this we're really Translating our question into a question about torsos. So here connected means really the variety Y the right is connected it's not the distant union of Non-trivial distant union of varieties for example, this would be the case here or is a distant union of copies of X So why is this? Well, let's think about what's the right hand side. What is a connected G torso about the points? well So you have some Y to X X is the point This is a finite morphism. So Y is going to be an affine variety so it's speck B And so B is a K algebra It's reduced because the morphism is et al It's finite dimensional as a vector space over K because it's a finite morphism and I'm assuming it's connected So it's not a product of algebras in a non-trivial way Well, these three things imply that B is a field. It's an integral domain. It's a field. I mean that's elementary fact about in commutative algebra So B is a field. Okay, so that's one thing but Why is it a Galois extension with group G? Well, that's because of This characterization here So if Y is affine with ring of functions B and the quotient is affine with ring of functions the invariance of B of G on B. So Y over G that's speck B G So we are assuming that B G is a small K and that of course tells us that B because it's a field It's a Galois extension of K with group G Okay So, okay, this is the explanation of of that Okay, and then you can look at G torsos over the point which may not be connected and what are they well, then you have a you have several connected components and each of them is Going to be a field extension of K which is Galois with group a subgroup of G By the same type of considerations, so more generally a G so you have a Correspondence between G torsos over So I will I will just say over K instead of over speck K So G torsos over K. That's the same thing as Galois extensions with group a subgroup of G Galois extensions of K with Galois group a Subgroup of G Okay, and because I had a question about Subgroups fixing subgroups and so on so to be perfectly precise here to have absolutely in equivalence here I should say G torsos with a choice of a connected component Okay, but this is not going to be important for us. I mean otherwise. It's just a country casey class of subgroups right, so What's the point of this? Well, the point is the following observation which is due to Hilbert's So now we're looking for this for a connected G torsos over the point Hilbert observed that you can start With a G torsos over some variety given a G torsos Y to X If you want to solve the inverse Galois problem now All you need to do is to find the rational point of X Whose fiber is connected? Because if you take the fiber of a G torsos, it's still a G torsos Given a G torsos over K To solve the inverse Galois problem It suffices to find a rational point of X Such that the fiber is connected So let me call it pi here so pi inverse of X is connected and why isn't it is it interesting? It's because it's it's in fact easy to exhibit G torsos like this where X and Y are very very nice varieties As connected as you want and then it becomes a problem of a slightly different nature. You're looking for rational points on the base and What's Hilbert's? Proved is that you can find such fibers in in many situations So this is Hilbert's irreducibility theorem so assume that X is a an open in P1 over K K is a number field and you fix a G torsos From Y to X a G torsos G is any finite group a G torsos You assume that the total space Y is connected. Otherwise, you're not going to find such fibers So Hilbert says then X contains rational points whose fibers are connected X of K Such that the fiber is connected So usually this is stated in terms of two variable polynomial Which is irreducible use you can specialize one of their variables such that the polynomial is still irreducible This is the translation of this statement So There's a useful refinement of this which was proved I think by Iqadar It's that in addition you can choose X to be as close as you want to local points of Of P1 at any finite set of places in addition can choose X close to X V Any any X V you like at any finite set of places For V in S any finite S. Sorry. I keep losing the eraser a Corollary of this is that in fact the the same statements are still true if you replace P1 by Pn That's actually in the exercises Excuse me Yeah, I'm over here I just wanted to ask I'm a bit confused on that second ball from the right What happens if the point that you take is also rational for a subfield of K. Do you still get? The Galois extension of the right group or just with a subgroup Sorry, if it's also a rational point of a sub extension of K you say yeah So I'm not exactly sure what you mean. I mean it could happen that if you go to a An overfield of K the fiber is not connected anymore But then if you apply the theorem over the overfield it's not going to give you this point but another one So Yes, it could happen to be defined over a subfield, but that doesn't matter Okay, okay, maybe that's just my algebraic geometry, which is a bit too confused. I Mean what the theorem is telling you is that You can find some point which maybe is defined over a small subfield, but that least viewed as a k point It gives you a fine a Galois extension of K with group G Okay, so corollary the same same same statements are true if you can replace P1 by Pn the same holds with Pn For any M Okay, so This is interesting because now we are looking at a slightly different problem. So a corollary of all of this If you want to solve the inverse Galois problem, it's enough to find a faithful action of G on some variety such that the quotient Is an open subset of Pm to solve the inverse Galois problem for G over K enough to find a Variety Y and an action of G on Y faithful Such that the quotient is so I will not say an open subset of Pn, but I will say a rational variety Such that the quotient is Rational so let me recall what rational means here So a variety is said to be rational if it contains an open subset, which is also an open subset of projective space so that means Ie contains So it contains an open subset That is also an opening Pm Okay, and so why is that? Yes, sorry to be to be sorry to be close Yeah, so I mean you you you can fix a finite set of places and at each of these places you can fix a local point Then you can choose you can fix also a neighborhood in the via dik topology Of this point in X of KV then you can choose X such that it belongs to these neighborhoods Okay So it's a weak approximation as in Bianca's talk you can do that You can do weak approximation on Pn That's well known and you can do hebertary disability, but you can do the two at the same time Okay, so why is this well the thing is if you take a variety like this and a faithful action of G So maybe the action is not free, but it's going to be free on an open subset because the group is finite So you just have to remove some finally many sub varieties to get a free action so proof G acts freely on Why not an Open and Then take the quotient It's a torso. It's a G torso, but now the the base is an open of PM So you can apply this theorem to it and the top space Okay, I want okay connected variety a connected variety So I assume it to be connected and then then we can just apply Okay, so this is really a different problem. So example This is not a big deal, but it illustrates the The corollary if you take G equals Z mod 3 Z and you let it act on P1 by the automorphism given in homogeneous coordinates by XY maps to Y X minus Y You can check. This is an automo. Sorry. Why Y minus X? Sorry, why Y minus X you can check this is of an automorphism of all those three and it's faithful So you apply the corollary and what's the quotient? Well, the quotient is a curve which is dominated by P1. It's a uni rational curve of genius zero There's no other option it and it has rational points because there are lots of rational points coming from P1 So the only such curve is P1 but just abstract reasoning, you know This is P1 and so you can apply the corollary and so in this way you get Cyclic extensions of degree 3 over any number field Of course, it's not a big deal. You can do this directly, but it illustrates. You can do it without thinking Okay, so this leads us to Nutter's problem So that's section 3 So Nutter asked so she saw this and she asked the following It's a question which makes sense over any field. Okay a field G finite group Let's embed G into a symmetric group In some way then let's make the symmetric group acts on a fine space by just permuting the coordinates by permuting The coordinates This is the easiest example of a faithful action of G on a variety which is connected So you can ask does it satisfy the hypothesis of the corollary in other words is the quotient rational? In very concrete terms. This means is the field of function of this variety purely transcendental in other words If you take the field of rational functions k of t1 tn And you take those that are invariant under the action of G by permutation of the variables Is this truly transcendental over k? So of course if the answer is yes by the previous corollary you get G as a galore group of a K So the first example of such a such a situation was given by Hilbert himself When you take for G the symmetric group so example Hilbert Take G the symmetric group Then we're looking at rational functions, which are invariant under all permutations But this is well known Such functions Our polynomials are sorry our rational our rational functions in the elementary symmetric polynomials So yes so you get the Function to your rational function field In the elementary symmetric polynomials and so that's it. We've already Realized SN as a galore group over any number of fields without thinking So okay, this does not give you concrete equations. This is another topic, but at least we know that there exists An extension of k with group SN another example due to Fischer is The answer to this question that was problem is yes if the group is a billion and K is the field of complex numbers This is not so hard. It's part of the exercises another example Due to Maeda the answer is yes over any field If the group is a 5 the alternating group a 5 so this is any any k Of course here also it was any k and I encourage you to look at Maeda's papers is really quite astonishing. I mean the proof is completely explicit He gives a trend of a transcendence basis for this field by I mean explicit polynomials And he proves that it works Surprisingly enough the problem is completely open over any field for the other alternating groups open Fisher statement is how is g embedded in any embedding of Sorry, so here I mean g equals the sun in fissures. I mean fissure. Yes any embedding any embedding. Yes Thank you for the question. Sorry. The question was does it depend on the embedding of g in SN here? And no, you can take any embedding Okay, so here I was saying the not us problem is open even over the complex numbers for any other alternate Every all other alternating groups Even a six so that's a challenge. Maybe you can just find an explicit transcendence basis. Who knows? Yes a 3 and a 4 are easy. Yes. Yeah. Thank you. Yeah. No, I mean the yeah the bigger Alternating groups okay, so For a long time people thought that the answer to not as problem would be yes in general and that would be the way to attack The inverse got a problem, but in fact it was not so 50 years later negative answers were found So I will so next I will So my next goal will be to explain In the simplest way one negative answer So this leads me to the notion of versatility so a If you have a g-torso Pi from Y to X So for any finite group G and if over any field k we say that it is versatile If you can get all g-torso's over all field extensions of k just by looking at the fibers of pi if for any field extension k prime over k any g-torso Over k prime Comes from a fiber of pi over a k prime point of X So the torso knows about all the also over fields field extensions of okay So the word vessel comes from a universal but to remove uni because there's no unicity of this X and Okay, an example is the torso that is considered in the test problem So you embed G into SN And you let it act on I find space Like permuting the coordinates you look at the open space the open sets where it acts freely So let's call it Y So that it's the open Where G acts freely so the points whose coordinates are per was distinct and Then this also is versatile So why goes to why not G? So this is not Completely obvious. It's a consequence of Hilbert's theorem 90. Well, it's part of of the exercises But you can take it as a black box Okay, and why is versatility interesting for us? It's because of the following criterion so proposition K a number field So suppose you have a versatile torso a Verso G torso Y to X Defined over K Such that so with why connected and X rational For example, this is the case if you have a positive answer to not us problem So what I claim is that then Grunwald's problem has a positive answer for G over K Sorry for G over K and for any finite set of places for G K any S So why is it so? Let me just explain in in 10 seconds So sketch of proof So recall what what are we trying to do? We have a finite set of places We have fixed Gallow extensions of the completions KV whose galore groups are subgroups of G and we try to find a Galore extension of K whose completions are these ones Okay, so recall Locally we have these galore extensions. They give us tall source over KV Torsals for a subgroup of G But as I've explained these are the same as G Torsals which may not be connected so the local Given extensions give possibly disconnected G Torsals Over the completions KV, but then by versatility These G Torsals come from rational points of X over KV. They come they are by inverse of X V for some X V in X of KV, but now we are on a rational variety So we know we can do with approximation We can approximate all of these X Vs by a rational point X and Hilbert's irreducibility theorem tells us We can find one rational point whose fiber is connected and as I told you Iqadal tells us we can do the two things at the same time So by Iqadal You can find X in X of K close to the X Vs Such that the fiber is connected But then we're done the fiber is connected So it gives us a Galore extension of K with group G because it's a G Torsal. It's a connected G Torsal And you just have to convince yourself that if you choose X X close enough to X V then when you complete at V you will get the same Torsal and Okay, that's a variant of Krasner's lemma if you move a little bit the Equations of the coefficients of the polynomial it will not change the the fields generated by the roots Okay, but that's just a sketch so I I Don't enter into details here. And so from this We deduce a positive answer to Grunwald's problem for any finite set S for the symmetric group For a five So we can do more than just inverse problem here. We can interpolate local extensions but more importantly From this we deduce a negative answer to not us problem for Z mod 8 Z I'm sorry, we will need to stop pretty soon. Yeah, I know I just write this and that's it So Corollary of this so you just apply it because we know Grunwald's problem has a negative answer for Z mod 8 Z over Q That's once come to example the contrapositive tells us that We have a negative answer to not us problem. So Q So using one So Q of t1 t8 If you take the field of invariance under Z mod 8 Z acting cyclically under by permuting cyclically the variables So this is not Purely transcendental Okay, and that's it for today. Thank you very much Okay, we have time for a quick question or two anybody have a question