 Welcome back to our lecture series Math 42-30, abstract algebra 2 for students at Southern Utah University. As usual, I'll be a professor today, Dr. Angela Missildine. In lecture 34, we reached the climax, the fundamental theorem of Galois theory. Okay, now there's only a couple of, you know, a handful of theorems in the Math-Max curriculum that actually get the name fundamental. It's a big deal, right? And so really, this fundamental theorem of Galois represents perhaps the climax, the pinnacle of all of the topics we've been studying, not just in Math 42-30, but also the prequel series Math 42-20, abstract algebra 1. And so we're going to present that one in the next video. And so before we present the fundamental theorem of Galois, I want to introduce the idea of a fixed field, because the fundamental theorem of Galois primarily, I mean, it says a lot of things, but the primary thing is it establishes a correspondence, a one-to-one correspondence between the subgroups of a Galois group with the subfields of the field that's connected to the Galois group there. And we have to go back and forth because it's a one-to-one correspondence. That is, we have this bijection between the subgroups and the subfields. And it's an order reversing correspondence. Like I said, we'll talk more about that in the next video. Right now, we've already established how you go from the fields to the groups. You take these Galois groups, so this automorphism group, the Galois group of E over F, right? For which we can do this for any field extension E over F. There are extra properties, of course, when E over F is a Galois extension, but we can discuss these automorphisms and these automorphism groups in general. What we want to do now is reverse this process, that if we have a subgroup of the Galois group, that is, we have a set of field automorphisms, we can actually construct a field from that, the so-called fixed field. All right? So without further ado, let's define our concept here. We have some field extension E over F, and I'm making no assumptions about it. This could be a finite extension, it could be an infinite extension, it could be Galois, it could be not Galois, not normal or normal, separable, maybe inseparable. No assumptions, just a field extension right here. Okay? And so as we've in this lecture series defined, the set Galois of E over F, this is the set of all field automorphisms on E, which leave the subfield F fixed. So that set makes sense without any further assumptions about the field extension there. All right? And suppose you have a subset of this Galois group. So it's some collection of automorphisms that fix F. I'm not even claiming that there's any algebraic structure here necessarily, that is, it might not be a subgroup, it might not be a coset, it's just some set of automorphisms. Okay? So given that set S, we can define another set E sub S is the notation we're going to use. When we defined Galois groups previously, I mentioned how there's not a universal notation for defining these automorphism groups for fields. The same is also true for the fixed fields. I feel like the notation is even more diverse when it comes to the fixed fields of these things. We're going to take the very simple notation that E sub S is such. Another competing notation, I think some people use a superscript in that situation. And like I said, there's a lot of other notations here. But for this lecture series, we will adopt the notation E sub S as the set of all things from E. So E is this field. So E sub S is all of the elements in E, such as say alpha, so that when you apply an automorphism sigma from S onto alpha, you get back alpha. And this needs to be true for all of the automorphisms of S. So when you look at the set E sub S here, this is the collection of all things in E, such that every automorphism in S fixes those elements. So these are the elements which are not moved by the automorphisms inside of this collection S. Okay. Now, the first thing we're going to show here is that, like I said, S doesn't have to be a subgroup or anything, it just has to be a subset. But it turns out that the set E sub S will coincide with the set E sub, the subgroup generated by S. So because of this equality here without the loss of generality, we can just assume that S is a subgroup because if S generates a subgroup, which it clearly does, that will produce the exact same fixed set. And we're likewise going to show that this collection of objects is a subfield of E. So it's a field. And as such, we call it the fixed field associated to the automorphisms in S, which again, we can assume by the end of this proof that S is a subgroup itself. All right. So let's prove the first one, the first statement here that E sub S is equal to E sub the subgroup generated by S, that the two things are one of the same thing. These are sets after all. And so we're going to argue that they are just subsets of each other, containment in one direction, containment in the other direction. All right. So let's consider the following here. So note, if we have these subsets S and T, so S sits inside of T and they both sit inside of the Galois group E over F, you have that containment. So T contains potentially more automorphisms than S. This actually does give a correspondence to this fixed set. So be aware that if we take tau as an arbitrary element of T right here, and we apply tau to the element alpha, where alpha is an element of E sub T, that'll of course produce back alpha by definition. E sub T is the set of all elements which are fixed by tau. Okay. Now be aware that since S is a subset of T, if you take an element sigma inside of S, that means sigma is inside of T. So likewise, since alpha belongs to E T, we're going to get that sigma of alpha is equal to alpha as well. So sigma will fix the element alpha, and it's alpha is an arbitrary element of E T, we get that every element of sigma is going to fix every element of E T. And so that then gives us a very important direction. ES contains the set E T. And so notice what we had here that if you take a smaller automorphism set, that gives you a potentially larger fixed field. We haven't established it's a field yet though, but we will do exactly that. So the fewer automorphisms you have, the bigger fields you get. And in particular, these inequalities are facing the wrong direction. That is, they switch directions. So when your set S is less than T, this will coincide with ES is greater than E T, like so. And this is actually the heart of the fundamental theorem of Galois, which we'll see in the next video here. There's this order reversing relationship that the inequality isn't preserved. We could call this a decreasing map because it switches the order, we could call this order reversing. Some people refer to this as a Galois map. Like a Galois map would be a map between ordered sets, partially ordered sets that reverse the order because it's an analog of what the fundamental theorem of Galois tells us. So this is the embryo of that fundamental theorem of Galois we're going to see in a little bit. So getting back to the proof of this statement here, this is true for arbitrary sets S and T. In particular, if we take T to be the subgroup of the Galois group generated by S, that will necessarily be larger. Well, I mean, if S is already a group, then it will be the same thing, but it can't get smaller, right? So S is contained inside the subgroup generated by S. And so that then gives us the first direction. ES will contain E sub, the subgroup generated by S because S, as a generating set, belongs to the subgroup generated by it. So that gives us the first direction. We have this direction, we need to go then backwards now. So on the other hand, let's take an arbitrary element of the subgroup generated by S. So let's call that element omega. So it's an automorphism on E, it fixes F that we do know. But in particular, because omega belongs to the subgroup generated by S, then there exists some finite list of elements from S or their inverses are from S. Without the loss of generality, we can assume that S contains inverses to each of the elements in there. But that's not necessary here. So the thing is, we have a bunch of these elements, sigma one, sigma two up to sigma N, so that omega here is a product of these things that are from S directly. And be aware that it is the possibility that sigma inverse is one of these elements here. That is, when you look at this right here, it's a product of things from your generator. Yes, sometimes in group theory, this is referred to as a word, which is actually why I wrote omega as the symbol. I know omega is a Greek letter for O, but it looks like a W to compare to the Roman alphabet here. So omega here is a word of things from S. Now it's important to note that if you have an automorphism sigma that fixes the element alpha, this is going to happen if and only if alpha is equal to sigma inverse of alpha. That is to say, the automorphism sigma will have the same fixed set as its inverse. And so the fixed set doesn't depend on whether it's the automorphism or its inverse. So the inclusion of inverses into S doesn't change anything whatsoever. We can do that without the loss of generality. And so, like we were saying a moment ago, omega is a word of all these elements from S, which if you didn't include the inverse of an element of S, you can throw that in there. It's not going to change the fixed set. It won't change ES whatsoever. And so we have this word omega with all these S's here. So in particular, if you take an individual sigma i, it fixes the element alpha since alpha to alpha. And so by induction, if you go through this statement right here, if you look at omega of alpha, this is going to look like omega of, excuse me, omega of alpha is going to look like sigma one of sigma two of sigma three all the way through sigma in of alpha. But then by induction, sigma in of alpha will send alpha then sigma in minus one will send alpha to alpha. Going through all the way, you end up with sigma two will send alpha to alpha then sigma one will send alpha to alpha. So if omega is a product of automorphisms that fix alpha, then their product will also fix alpha. So we see that omega of alpha will equal alpha as well. And so since alpha was contained inside of ES, right, we can then infer from that that alpha also belongs to E sub the subgroups area by S, thus giving us the reverse containment right here. So these things are in fact equal to each other. And so without the loss of generality, we can suppose that the set of automorphism is always a subgroup of the Galois group. So why is it then why is the fixed set of a subgroup of the Galois group necessarily a field? That's what we're going to prove right now. So we're going to prove that this fixed set E sub H going on now H is a subgroup of the Galois group. We're going to show that E sub H is closed under the four field operations. It is a sub set of a field. So we need to show it's closed under addition, subtraction, multiplication, division. So suppose we have two elements alpha and beta that belong to E H. And suppose sigma belongs to H as well. So note here that alpha and beta are going to be fixed by every automorphism in H. And so we're going to take arbitrary automorphism sigma inside of H. See what happens. So let's look at what sigma does to a sum. So sigma of alpha plus beta. Since it's an automorphism, alpha sigma of alpha plus beta becomes sigma of alpha plus sigma of beta. And since sigma alpha that is alpha is fixed by sigma, sigma alpha is equal to alpha. Similarly, sigma beta is equal to beta. So we get that sigma of alpha plus beta is the same thing as alpha plus beta. Therefore, sigma fixes the sum. The exact same argument applies to the difference of these things as well. No distinction there. The same argument is going to apply for the product. If you take sigma of alpha times beta, well, since sigma is an automorphism, this becomes sigma of alpha times sigma of beta. And as sigma of alpha is alpha and sigma of beta is beta, we get that sigma of alpha times beta is alpha times beta. So sigma fixes the product with quotients assuming beta is nonzero right now. We're going to get sigma of alpha divided by beta. Since it's a field automorphism, it'll preserve the division operation. So you get this is sigma of alpha over sigma beta. And since alpha and beta are fixed by sigma, this reduces to the fraction alpha over beta. So the fact that sigma is an automorphism to the field, and since alpha and beta are fixed elements, those properties combined show you that the sum difference product and quotient of alpha and beta, again, in the latter case, assuming beta is nonzero, each of those binary operations will be preserved by the field automorphism. And thus, the sum difference product quotient will be fixed. And so each and every one of these elements belongs to E of H here. So since alpha beta belongs to E of H, so does their sum and difference. Because alpha and beta belong to E H, so does their product. Because alpha and beta belong to E H, so does their quotient. We get that. What else do we have here? There are a few required elements that have to belong to the set. What about zero? Well, every field automorphism sends zero to zero because they preserve the additive identity of the of the ring there. So zero is a fixed element. And field automorphisms also fix the number one, because this is the multiplicative identity, the unity of the ring. Every field automorphism is going to do that as well. So this shows you that zero and one are fixed elements. These are the two required elements to belong to a field. So they belong to E H. And so this is then going to show us that E H is in fact a subfield of E. It's closed into the four field operations, and it contains zero and one. So we've now established that a Galois subgroup fixes a field. And we have this connection here. For every subgroup of automorphisms, we can construct a fixed field. So I also want to prove a very quick corollary of this result. Because after all, we can turn field extensions into groups, and now we can turn automorphism groups back into fields. What happens when you do this statement twice, right? What if you do these operations twice? So let K be Galois over F. So unlike the previous theorem, which we just required, it was a field extension. We're requiring this be a finite, excuse me, not finite, we required to be a Galois extension. So K is a separable normal extension over F. In this situation, if you take the fixed field of K, whereas the subgroup, you take the full Galois group of K over F, this fixed field is turned out to be F. So be aware that if you compose these operations, you have this operator of the Galois group and this operator of the fixed field, when you double compose them, it's just the identity on the original field F. And this is going to establish that these maps, in fact, are bijections because there's an inverse to each other. Now proving the other direction is an inverse is a little bit more complicated. We're not going to do that in this video. All right, so let's start off with, let's simplify things. We'll call the Galois group K over F. We're going to call it G for the sake of simplicity. And so this field right here then becomes K sub G. All right, so we want to show that K sub G is equal to F. Now, in particular, we know that K sub G will sit in between these fields. How can we do that? Well, clearly, K sub G is a subfield of K because by definition, K sub G is the subset of K, which is fixed by the group G. All right, so that's pretty easy. Why does K sub G contain F? Well, G, by definition, is the Galois group of K sub F, for which every automorphism in G fixes F by definition. It's possible, though, that there are larger fields, larger sets that are also fixed by G. So we will prove that's not actually going to happen. But by definition, there is that possibility, right? So F is contained inside of KG and KG is contained inside of K. So we have a field extension and then we have another field extension. So using the multiplicity of field degrees, we get that the degree of K over F is equal to the degree of K over KG times the degree of KG over F. Okay? And so then, this is where the Galois component comes into play here. Since K over F is a Galois extension, the sub extension K over KG is likewise a Galois extension, right? So we have these fields where K is Galois over F. Well, if K is Galois over F, then it will also be Galois over any intermediate field. That's a fact we've established already. So we get that this is a Galois extension as well. Now, what we know, of course, about Galois extensions is that the order of a Galois group is equal to the degree of the extension. Okay? So for example, when you look at the Galois group K over F, like so, the order of this Galois group is equal to the extension there, K and F. Okay? So those are the same quantity there. And so likewise, if we were to look at the Galois group of K over KG, like so, by this exact same result, we're going to get that this is equal to K dot KG, like so. All right? So those things are the same thing. It's also important to note that these two Galois groups are actually one in the same thing. How can we establish that? Well, because of the containment we have before, F sits inside of KG. If you take an automorphism that fixes KG, then it has to fix F as well. And so very naturally, we get this direction happening right here, that those automorphisms that fix KG will necessarily fix F as well. So you get that you get that containment. But then why do we get the containment in the other direction? This comes back to what is G in the first place, right? G, the group, is the Galois group K over F. So this right here is, in fact, the Galois group. And so if we, so by definition, KG is the set of things fixed by G, right? And so in particular, if we take, because we're trying to go this direction right now, if you take some automorphism over here, all right, Sigma, it's going to fix KG by definition. And therefore, Sigma would have to then belong to this set, the set of automorphisms that fix KG. So we do get that these two groups are one in the same thing. So if these two groups are the same, then their cardinalities, their orders are going to be the same. And so then we establish this inequality, excuse me, this equality right here, that K colon KG is equal to K colon F. And so then when you compare this equation right here, since these two are equal to each other, that's going to force this last one here to be one. And therefore, we get that KG is equal to F. Now be aware in this situation, I am making the assumption that these were a finite Galois extension. You can prove the same result for infinite Galois extensions, but that involves a little bit different approach to this thing, for which that's not how we're going to approach it for this lecture series.