 Consider the following two very important definitions here. I guess maybe three, although two of them seem very connected, which is why I didn't originally separate them. The first one's actually called a separable extension, believe it or not. Imagine we have a field extension. So E is a field extension of F. We say that E is a separable extension of F if all elements of E have a minimal polynomial of F with no repeated roots. Okay, no repeated roots. And so this seems sort of like a strange, such a strange definition to take on, mostly because we haven't seen a lot of examples of an inseparable extension. Now the good news is, is that if F of X is some irreducible polynomial, it's an irreducible polynomial, because after all, when we talk about minimal polynomials, those are by definition irreducible. So if F of X is an irreducible polynomial over a field of characteristic zero, so if the field F has characteristic zero, or if it's a finite field, so those will have prime characteristic, but if it's a finite field or characteristic zero, then in that situation, the extensions will always be separable, always, always, always be separable in that situation. And so that's really, as you look at our lecture series, this is the setting that we mostly have lived in. We've looked at fields of characteristic zero, so things that extend the ration numbers, such as like Q join i, Q join the square root of two, Q join the square root of seven are the complex numbers. These are all fields of characteristic zero. And then our very last unit, as we talked about BCH codes and such, were applications of finite fields. So the places that we've been playing the most with field theory is exactly the situation. So this is a non-issue in that situation. Now, when it comes to fields of characteristic zero, why are these extensions always separable? Well, it comes down actually to this derivative operation. When it comes to a polynomial ring like F join X, we can still talk about the derivative. We don't need calculus to introduce the derivative. We just define it in the usual manner, right? If you take the derivative of some power function X to the n, you define this to be n times X to the n minus one, and then you extend the derivative linearly. That is, the derivative will distribute over sums and differences. You can factor out scalar coefficients. And so with the so-called power rule plus the linearity of the derivative, we can take the derivative of any polynomial, just like you would in a class like math 1210, calculus one, here at Southern Utah University. The derivative makes sense in that setting. You can prove that the Leibniz rule applies, the so-called product rule, the derivative of the product of two polynomials, F times g. So F prime will be the same thing as F prime g plus F g prime. You get that. This Leibniz rule applies in that setting, just like it does in calculus one. Now you can prove it, of course, just by following the formulas of what the derivative does. It can get a little bit tedious and messy, but it can happen. I mean, for the sake of my students watching this video, they had to do this as a take-home question on an exam. So they know exactly what I'm talking about right now. At least, hopefully, hopefully they know what I'm talking about right now. The derivative can be used for any polynomial ring whatsoever. Now, speaking of that take-home test, I also had the students prove the following statement that for a field of characteristic zero, right, you're not assuming prime characteristic because the kernel of the derivative changes depending on its characteristic. Finite fields will have positive characteristic. So for fields of characteristic zero, the derivative, since it's a linear map that has a kernel, that kernel will be all that all those polynomials whose derivative is zero. When you are characteristic zero, it turns out the only derivatives which map to zero, I should say the only polynomials that map to zero under the derivative will be constant polynomials. And as such, you can show using the product rule that a polynomial has a repeated root if and only if that repeated root divides both the polynomial and its derivative, right? So let me clean up my screen a little bit. So if you have a polynomial f of x with that repeated root, so you have something like x minus r, raise it to the nth power, then there's some other polynomial g of x. By the usual product rule, you can also argue the chain rule in this situation. I won't go through all the details of that. But yeah, those usual rules, the usual rules for taking derivatives of polynomials apply here. The derivative is going to look like n times x minus r to the n minus one times g of x plus x minus r to the n times g prime of x, whatever that turns out to be. And you'll notice that the derivative, since n is strictly greater than one, we see that in particular it's at least two, right? We get that this is at least one, this is at least two. So there's a factor of x minus r that divides f prime, x minus r also divides f, so the function and its derivative have a common divisor, which is that root. And you can argue that this happens if and only if the derivative and a polynomial are co-prime, then in fact they have no repeated roots. And so with irreducible polynomials, because the polynomial f and the polynomial f prime both reside inside of f of joint x here, this is in fact a principal ideal domain. Therefore, GCDs exist and GCDs can be written as linear combinations of the two elements here. So whatever the GCD of f and f prime is, you can form it as a linear combination of these two. And so we can pull out that polynomial x minus r basically. So if you have a repeated root, you can pull it out. And therefore that shows that your polynomial is not irreducible because it has a GCD with some other polynomial, where these things don't divide each other, but they have a common divisor that's a smaller degree, okay? So you couldn't be irreducible if you have a repeated root when your character is 6-0, and this comes from using the derivative. Why are finite fields necessarily going to be separable extensions as well? So imagine we have a finite field, so it's orders p to the k, and so we can view this as a finite extension of the field zp. And you can also argue that if you have fields f contained inside of e contained inside of k, if k is separable over f, then it turns out that e is also separable over f as well. So all the intermediate fields have to be separable as well. In fact, we can also show that k is separable over e. We get all of these things here as well. So I'll leave that as an exercise to prove that one, but we have a finite field. It's going to be an extension of zp necessarily, okay, finite extension in that situation. Notice that f is the splitting field for the polynomial x to the p times k minus x, which we view this as a zp polynomial, and we'll call this polynomial f of x. By previous work, we have shown that every element of f is a root of this polynomial. So if you view it as a polynomial with f coefficients, this thing factors as the product of x times alpha here, where alpha ranges over the elements of f, and so it splits in that situation. This is the splitting polynomial. Now this accounts, this factorization here accounts for exactly the p to the k mini roots of this polynomial. There's no repeated roots in this situation. And so if you take a typical element of alpha, excuse me, a typical element alpha inside of f, its minimal polynomial over zp is going to be a divisor of this polynomial. Since this polynomial doesn't have any repeated roots, the minimum polynomial of alpha doesn't have any repeated roots either. So alpha is a separable element. So you can make this into an element wise thing, right? An element is separable if its minimal polynomial has no repeated roots. Now every element of f satisfies this. So therefore f over zp is a separable extension. And hence, f is separable over zp. Any intermediate field would also inherit this because of how chains of separable extensions work. And so this does, in fact, argue for us that every finite field can be viewed as a separable extension over any subfield. So in those cases, you have characters of zero or finite fields, this idea of separability is not a big deal. Well, when might you be inseparable then? Okay, let's take as our, let's take some element, some transcendental element, right? Transcendental element. We'll call it t over the field. We'll just make it simple here. We'll take z2, okay? We take some transcendental element in that situation. I want you to take as your field E consider the field z2 adjoint t. Now we've done a lot of extensions that are out to break, but this is a purely transcendental extension of z2 here. This is actually going to look like the set of rational functions, the set of rational functions involving the symbol t, you know, with symbol t over z2 here, okay? So if something like 1 plus t over 1 plus t plus t squared, this would be a typical element of this field, something like that. Those are the things in there. So we've talked about polynomial rings, but you also talked about a rational function ring. This is going to be a field in this situation. Take E to be that field, and then take F to be the subfield z2 adjoint t squared in that situation. And then consider the polynomial f of x equals x squared minus t squared and view that as a polynomial in f adjoint x, okay? There is no root of this polynomial with f coefficients, because if I only adjoint t squared, I don't have a t. There's no way to algebraically produce a t from t squared, because t squared is transcendental. It doesn't have any algebraic relations except those mandated by the field axioms. So in particular, the elements t does not belong to the set, which would be, of course, the root of this polynomial. Any other root, you'd also get that. So when you see this as a f polynomial, you don't have any root. But if we view this as an e polynomial, this thing factors, because t is now a root, but in fact this factors as x minus t quantity squared. Freshman exponentiation applies in that same situation here. So this polynomial factors as x minus t squared. Now this polynomial can't factor because it doesn't have a root. The only root is actually t. So it makes an irreducible polynomial. It would be the minimum polynomial of t squared in that situation. But it does have a repeated root in the splitting field, which is going to be e in that situation. And then if there's nothing special to right here, you can replicate this example for any prime characteristic essentially. It's just it's easier to prove it's irreducible because it's quadratic. So if it doesn't have a root, then it's irreducible. So there are some situations in field extensions of prime characteristic for which you do get inseparable extensions. This long discussion is just to tell us that over the fields we really care about, finite fields and characteristic 0, we don't have to worry about it. It's sort of an automatic thing. So with that long description about separable extensions, we can then define what a Galois extension is. So same thing as before, we have a field extension, e extends f. We say that this extension is a Galois extension if it is separable and it's normal. Remember what did normal mean? We defined that previously. We say that a field extension is normal if it's the splitting field for some collection of polynomials and that collection could be a single polynomial. The fact that we're required to be separable, then this tells us that e is the splitting field for some family polynomials for which those polynomials don't have repeated roots. It could be a single polynomial, but it could also be infinite, but none of those polynomials have repeated roots. That's what makes it a Galois extension. In the setting that we care about, the separability doesn't make too much of a difference. But yes, in this case, in the modular case where you have these infinite fields of characteristic p, one does have to be very, very careful about separability. It's not going to be a concern for us. So for us, basically, Galois normal extensions are one of the same thing, but that's not exactly true. The separability does come into play. And I'm pointing out when we need to use separability, why does it matter? It seems like it's a non-issue for us, but it is a big deal. A Galois extension, you have to be normal and separable. Sometimes, because of the setting, we think so much is just a splitting field for a polynomial that we forget that we need separability to. It needs to be the splitting field of a polynomial with no repeated roots. So imagine we then have a field extension, e over f, drop the assumptions from before. We then define the Galois group of this field extension. In our lecture series, we'll denote this as g a l of e over f. Some people use other notations. Some people just call it g of e over f. Some people might use commas. So g comma e comma f. Some people use more calligraphy or cursive g's or something like that. There's a lot of different notations you could use. In this lecture series, we'll use this one, g a l short for Galois e over f here. And so the Galois group is the set of field automorphisms of e, which fix f. So this is the automorphism group of e, but they have to fix the subfield f. Now, I should make a comment here that some people don't refer to this set as a Galois group. They only refer to it as an automorphism group, and they reserve the term Galois group to describe the automorphism group of a Galois extension. But this definition does make sense even when you don't have a Galois extension. So I'm of the more liberal side here that we call automorphisms of field extensions, Galois groups. That's the common convention there. And the Galois extension doesn't have to be there to be inside the definition, although we are going to care about Galois groups of Galois extensions. And many of the things we prove about them are only applicable when it's a Galois extension. But for the rest of this video, we're mostly just we, well, for the rest of the lecture series, when we talk about a Galois group, it doesn't have to be a Galois extension. That's an important caveat there. If you have a chain of fields of some kind, so f is a subfield of E, which is a subfield of K, then it turns out that the Galois group that the Galois group of K that fixes E is a subgroup of the Galois group of K that fixes f. So notice the reverse chain there, right? So f is smaller than E, but the Galois group that fixes E is smaller than the Galois group that fixes f. Because after all, if you fix E, that fixes f too because f is inside of E. But potentially there could be an automorphism that fixes f, but there are things in E that don't belong to f that might get moved around a little bit. So there potentially could be more in that situation. And this is at the heart of Galois theory that we have these order relations that get reversed. f is smaller than E, but the Galois of E is less than the Galois group of f here, K over f and E over f. There's this inverse relationship, sometimes called a Galois correspondence. We'll define that some other time. I want to give some examples right now of some Galois groups. So consider the complex numbers and let sigma be complex conjugation. That is the map that sends a plus bi to a minus bi. I will leave it as an exercise to the viewer here to prove that complex conjugation is a field automorphism on the complex numbers. The properties of the complex conjugate will preserve all the field operations, addition, subtraction, multiplication, division. It's a field automorphism. Note that if you look at sigma here, and if you strict the domain to the real numbers, then you actually get the identity map. The complex conjugation doesn't do anything to a real number because if you take off the imaginary part, you just get A mapping to A. So restricted to the real numbers, you just get just the identity. And so sigma is a map that fixes the real numbers. Clearly, the identity also does that. But we'll get to that in a second. Sigma belongs to the Galois group of C over R. Complex conjugation fixes the real numbers. And like I mentioned, the identity map also fixes the real numbers because it fixes everything. And one can show, I'm not going to go through all the details right now, but one can show that the Galois group of C over R is in fact just these two automorphisms. There's the identity and there's complex conjugation. This does, in fact, form a group. It's a group of order two. So it necessarily has to be the cyclic group of order two. And you will note that the extension C over R is degree two. And the Galois group also has order two. This is not a coincidence. This is actually how one argues that these two sets are equal because you would show that this contains two elements. Oh, I've accounted for those two elements. Done. I have the Galois group. And this is something we're going to prove later on in this lecture. This is a very, very important theorem.