 If E is a field extension of F, let sigma be an automorphism that belongs to the Galois group of E over F. I'm not assuming that E over F is a Galois extension. This is actually true for any field extension, where the Galois group of E over F is just the set of automorphisms of the field E that fix the base field F. It can be shown that sigma induces a permutation on the set of zeros, the set of roots of the minimal polynomial with F coefficients for any element alpha. This sigma has to induce an automorphism there. For the following reason, let's take our element alpha that belongs to E, and let's take its minimal polynomial. Let's call that polynomial with F of X right here. It has coefficients that belong to F, so we'll call them A0 plus A1X plus A2X squared, all the way down to A to the NX to the N, and we're viewing this as an F polynomial. Alpha is a root of it, so we have that F of alpha is equal to zero. What I want you to do is then hit this with the map sigma, because as sigma is an automorphism of these polynomials of E and F, I should say, excuse me, it's an automorphism of the field, this actually induces an automorphism on the ring of polynomials by just leaving X alone and just changing the coefficients. So we could look at the polynomial, I'm gonna denote this as F to the sigma. What this means is you're gonna take sigma of A0, you're gonna take sigma of A1X, you get sigma of A2X squared, and all the way down to sigma of ANX to the N, and this actually works. I mean, sigma we can view as a map from EX to EX, these polynomial rings in this manner here, but with F, the thing about F here is that all of these coefficients of F actually belong into the field, capital F right there. So these automorphisms, since we belong to Galois of E over F, don't do anything to the coefficients. So you end up with just an A0, an A1X, an A2X squared, all the way down to ANX to the N. So the polynomial is just gonna be F of X here. The coefficients don't change when you do this. The polynomial F is fixed by sigma as well. In particular though, the thing about alpha is that if you take A0 plus A1 alpha plus A2 alpha squared, all the way down to AN alpha to the N, this is of course what we mean by F of alpha. If you then hit this with sigma, dee, you hit this with sigma like so, what's gonna happen is that all the coefficients are gonna be left alone. And we end up with A0 plus A1 alpha to the sigma. I'm just using the superscript here to be the image of it. That's a common thing we do here. So then you're gonna get A2, A2 times alpha squared sigma all the way down. You're gonna get AN and then you're gonna get alpha to the N to the sigma for which you can actually move all these things around. So like in this situation where we had the square, because it's an automorphism, the image of alpha squared is actually alpha sigma squared. And same thing over here, this happens for all of them since it's an automorphism, we get that this is actually alpha to the sigma N. So this is actually why we write the images of function sometimes as exponents that actually behaves and exponentially when it's like an automorphism or homomorphism. And so this is just equal to F of alpha sigma right here for which we know this is equal to zero. So this shows that the alpha sigma is also a root. So this sigma induces an automorphism on the, excuse me, the sigma induces a permutation on the roots of a polynomial, a minimal polynomial, or a reduced polynomial there. And so this is what we call the conjugates of a polynomial. So we say that two elements, let me clean up the screen a little bit. We say that two elements of a field are conjugates if they have the same minimum polynomial. So then what this obsesses earlier here is that if you take the conjugates of an F polynomial, sigma induces a permutation on those. You can scramble up the conjugates of the same minimal polynomial, but that's the worst thing that you possibly can do, right? To be conjugates of the same minimal polynomial does give you an equivalence relation, right? So suppose that alpha and beta suppose we have these elements in E here, the conjugates, over some polynomial in F, then there is an isomorphism from the field F join alpha to the field F join beta. And this is the isomorphism induced by the map alpha maps to beta, okay? One can prove that this is going to be an isomorphism. And this comes from the fact that the two elements alpha and beta satisfy the same minimal polynomial. And so we get this field isomorphism like so. Now, every field isomorphism can be extended up to, it can be extended in a unique way to the algebraic closure. So this map, we'll call it sigma here, can be lifted, it lifts to a map sigma on the algebraic closure. This is something we proved previously. And that then gives you a field isomorphism on the algebraic closure here. Because F join alpha is a subfield of F bar and F join beta is a subfield of F bar. So that field isomorphism lifts to a field isomorphism of the closure. And like we said before then, and this field sigma is going to fix F, right? So sigma restricted to F is just the identity in that situation. So sigma then is an automorphism of the algebraic closure that fixes the base field F. It's an automorphism, therefore it'll permute the conjugates and therefore a field automorphism induces a permutation on the conjugates of the same irreducible polynomial. And that's where we are getting this equivalence relationship from. These permutations can actually use to create automorphisms. So like if we're looking at, for example, let's take the field, for example, the Q join the square root of two and the square root of three and we view this over the rational numbers. Well, who are the conjugates of square root of two? That's gonna be plus or minus two. Who are the conjugates of square root of three? It's gonna be plus or minus square root of three. So if you think about like, what are the possible permutations you could do here? You could leave this one fixed. You could swap those two. You could swap these ones, not that one. You could swap both of them, let us times them by their negative there or you could do nothing. That gives you like four possibilities and we'll actually see that those are in fact the four possible automorphisms. We're gonna do an example like this one just a moment. But I do have to caution you that because automorphisms produce permutations on conjugates and for these type of extensions, we often form that by adjoining roots of polynomials to a field. So we know kind of what the conjugates are gonna be. We can move them around, we can conjugate them. But you have to be warned that not every possible permutation is gonna work. As we have these four elements here, you can't necessarily get every possible permutation. There are some restrictions, so one has to be a little bit more careful. But it's a good starting place. What you often can do when it comes to computing a Galois group is you list all the possible permutations and then argue that the order of the Galois group is such and such and I listed all the possibilities. It's the same number, boom, I'm done. And it's a very, very nice technique. Let me try to demonstrate that to you in this next example. So consider the example, I mean, I almost did this one in the previous slide. It wasn't quite right, but it'll work here. Take E to be the field where we've joined the square root of three and the square root of five. It doesn't really, this example doesn't matter which square roots you add as long as they're algebraically independent of each other. So I take the square root of two different prime numbers. The same example will work there. Take F1 to be the field where we only joined the square root of three and we take F2 to only be the square root of five. Sorry about the typo there. It's a crude fix, but nonetheless, I feel like we need to have proper grammar on these videos. Anyways, so note that these fields, the two algebraic elements we joined, their respective minimal polynomials over Q are gonna be X squared minus three and X squared minus five. All right, when you look at X squared minus three, if we factor this thing in the splitting field, you're gonna end up with X minus the square root of three and X plus the square root of three, which of course the splitting field of this polynomial is exactly F1. And so we see that the conjugates of the square root of three are plus or minus the square root of three. Likewise, X squared minus five, if we factor it in its splitting field, we're gonna get X minus the square root of five and X plus the square root of five. Of course, the splitting field for this polynomial is of course F2. And so we see the conjugates of square root of five are plus or minus the square root of five. You're always a conjugate to yourself, but we get another one in the situation. Clearly, if you're a root of a degree in minimal polynomial, you're gonna have in conjugates, one being yourself. These are quadratic extensions, so we get two conjugates plus or minus. That's basically what always happens with a quadratic extension. It's not always plus or minus, but think of like how the quadratic formula works, how you get negative E plus or minus the square root, yada, yada, yada. You get two options in that situation. And honestly, the quadratic formula applies to every field. As long as we're not working mod two, because after all on the bottom, you have like this two A, which means you divided by zero, destroy everything. Don't do that. But for the most part of the quadratic formula works in all field extensions, as long as we're not characteristic two. So consider the following. Let's define the map sigma, which is gonna belong to the Galois group of F1 over Q. In the following manner, what sigma is gonna do is you're gonna take A plus B square root of three and you're gonna map it to A minus B square root of three. So basically you're mapping sigma to its inverse. That's really what I want you to think of right here, because if you took A to be zero, you took B to be one, that means you would map sigma, excuse me, sigma would map square root of three to negative square root of three. Sigma right here is just the linear extension. Sigma, this is the map where we fix all of the rational numbers and we just move square root of three to negative three. This is the linear transformation that's induced by that. Fix the rational numbers while swapping square root of three with its conjugate. It's the natural construction in that situation. We're gonna do the same thing with tau here, where tau is gonna belong to the Galois group of F2 over Q, where we're just gonna swap the square root of five with negative square root of five. So if you linearly extend that so that this becomes a linear transformation on the vector space structure, we have to leave the rational numbers fixed because we think of those as scalars. And so A plus B square root of five becomes A minus B square root of five. So so to speak, we take the conjugates of each of these maps, each of these elements. And that's what these automorphisms are doing. You can verify that in fact, these are field automorphisms. Now, each of these maps can be extended above, right? Because sigma is a map on F1 and tau is a map on F2. Can we map it up to E, the whole field? Q would join the square root of three and the square root of five. Well, we can extend sigma to be a field extension, excuse me, automorphism, belonging to the Galois group E over Q. And we can do that so that Q is the unique linear extension of the original Q so that it fixes F2, okay? So a typical element of E can look like the following. You take A plus B square root of five plus C plus D square root of five times square root of three. So what you're doing here is you're thinking of the following situation. I can think of E as an extension of F2. So my coefficients here are coming from F2. A plus B root five, that's an arbitrary element of F2. C plus D square root of five, that's a generic element of F2. And so if you extend that by adding the square root of three, you can think of it, oh, E over F2 is a degree two extension. So E F2 here, the degree is two. And so we get exactly this situation here. So sigma is gonna leave F2 fixed, much like how the original sigma was leaving Q fixed. We're gonna leave F2 fixed, but just replace sigma is just gonna replace the square root of three with negative square root of three like so. And so we can extend sigma to be an automorphism of E. Now tau, we can do a very similar thing, right? We're gonna extend tau, which was a map from F2 to F2. We can extend it up to E by leaving F1 fixed. So instead of tau fixing Q, it's gonna fix all of F1. Again, by analog to what we're doing right here. And so now we've constructed two automorphisms, sigma and tau that belong to the Galois group of E over Q. Now since sigma fixes F2, it's gonna fix Q because Q belongs to F2. And tau fixes F1, Q lives inside of F1, so it's gonna be fixed as well. So sigma and tau belong to this Galois group. Who else belongs to that Galois group? Clearly the identity map, I'm just gonna call that one because it's the multiplicative identity. We also get that sigma times tau will belong there as well because when we compose these together, sigma times tau, if we take a rational number, just like A, tau is gonna replace any square roots of five with a negative square root of five. This doesn't have any, so you just get back A. Sigma is gonna replace any square root of three with a negative square root of three. We don't have any, so we just get back an A, which was our rational number here. So the product sigma and tau fixes every rational number. I want you to notice here that the intersection between F1 and F2 is Q. And since sigma fixes F2 and tau fixes F1, their composite will fix the intersection of their fixed fields, which in this case is the rational number. So sigma tau does belong there. And if we look at just these four elements together, they form a Klein four group. I want you to try to convince yourself of that. This would be the Cayley table of this Galois group because in fact, these are the only four elements of that Galois group. That's an argument we'll show a little bit later. Notice that E over Q is a degree four extension because it's a biquadratic extension, you get degree four. The Galois group has also order four. This isn't a coincidence. This is actually how I know to stop right here. I've come up with four distinct field automorphisms. And since there could be only four, that then tells us that we're done. But we'll prove that a little bit later.