 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. Hinch and Missile 9. In lecture 29, we want to talk about the titular topic of splitting fields. So, as we've discussed field extensions previously, we know that by Kroniker's theorem that given any field, any irreducible polynomial over that field, we can always build a field extension that adds a root to that polynomial into the larger field. The algebraic closure is then essentially the union of all of these processes. That is, you take all of the roots of every possible polynomial put together. But in some sense, this is sort of like, well, as we've proven, this was a maximum algebraic extension that makes all of the polynomial split. Is there a minimal algebraic extension that makes certain polynomials split? That's what we mean, of course, by splitting field. But I keep on saying polynomial split. What is that exactly that mean? Let's make these definitions explicit here. Suppose that we have a field F. And suppose that S is a set of polynomials from the polynomial ring F for join X. And say that E is a field extension of F. We say that the set of polynomials S splits over the field extension E. If every non-constant polynomial that belongs to S, which I should mention that throwing constant polynomials into S is completely pointless. So for the most part, we can assume there's none in there. But if there is, our definition accounts for it. But the set of polynomials splits over E. If every non-constant polynomial factors into linear factors in the polynomial extension ring, E join X. So we say a polynomial splits if factors into linear factors. That's a set of polynomial splits if each polynomial factors into linear in that situation. We say that E, this extension field, is a splitting field for the set of polynomials S. If S splits in the field extension E, and this is the important part, it splits in no proper subfield of E. In other words, E is a minimal extension of F where S splits over E. This minimal extension is very, very important when it comes to the definition of a splitting field. Many students sometimes ignore this. So like, oh, E is a splitting field because this set of polynomial splits over E. I mean, it splits, yes. But to be a splitting field, you have to be a minimum split. This minimum algebraic extension in that regard. Now related to the idea of a splitting field, you have the idea of a normal extension. We say that E is a normal extension of F if E is a splitting field for some set of polynomials over F join X. So splitting fields and normal extensions are essentially the same thing. But when it comes to a splitting field, if we say that E is a splitting field, we're referring to some set of polynomials S. When we say that E is a normal extension, we're thinking of it as an extension of the field. But the set of polynomials is implicit in that situation that are splitting. So it's the same notion, be aware, you're a splitting field if and only if you're a normal extension. But being a normal extension, of course, depends on the base field. If you choose a different base field, might not be normal anymore, normal extension. We'll talk about these things in the future. And so the two notions, of course, are very related. Why do we use the adjective normal? Unfortunately, that's something we're going to have to talk about in the future. There is a reason for that terminology here. Believe it or not, it has something to do with normal subgroups. But I get a little ahead of myself here. Now of great importance is going to be the situation where your set of polynomials is actually just a single polynomial. Now if S is just a single polynomial, we often don't refer to it as a set. We just refer to it as an individual polynomial. And so we would say that the polynomial f of x splits in e or that e is a splitting field for f of x in such a situation. Of course, you know, e is in fact the splitting field for the set S in that situation. Now, I should mention that if your set of polynomials is a finite set, then really it's no different than just being an individual polynomial. Because if you have a finite set of polynomials, what you can do is you can build a polynomial S of x such that it's the product of all of the polynomials inside of your large set S, which is finite, but it contains several polynomials here. After all, each of these elements f of x belongs to the polynomial ring f of join x. And as such, the product of all of those polynomials is a polynomial in f of join x. And so a set of polynomial splits over e if and only if every root of every polynomial inside of S belongs to the field e. But all of the roots, if you look at the collective roots of the polynomials inside the set S, that's exactly the roots of the polynomial S, right? Because if, you know, let's say alpha is a root of f in the field extension e, then as f of x is a factor of S of x, then alpha is also root of S in that manner. And this goes in both directions. So the roots of the set of polynomials S is actually identical to the set of roots for the individual polynomial S right there. So the set of polynomials S will split over e if and only if the individual polynomial S of x splits over e as well. So when it comes to a set of polynomials, if that set of polynomials is finite, then without the loss of generality, you can just assume it's a single polynomial. So you'll often hear us talking about the splitting field of a polynomial. Now, of course, if your set of polynomials is infinite, then there isn't necessarily a single polynomial that can do all that lifting there. For example, you could take S to be the set of polynomials x squared minus p, where p is a prime number, p is a prime number inside of the integers. So we can think of these as a set of rational polynomials. Like so. This is an infinite set, because we know, of course, there's an infinitude of prime numbers. And there's no single, there's no single polynomial that will contain all of the roots of this polynomial, of these polynomials here. Because the splitting field for this set is in fact going to be q adjoin the square root of a prime, which this is an infinite extension. This is an infinite extension of q, for which if you have a fine, if you have just one polynomial, the field extension will always be finite, because you only have to adjoin finitely many roots for which each root worst case scenario extends the field by a finite amount. And so yeah, you basically have two options when it comes to sets of polynomials you want to split, either a single polynomial or infinitely many polynomials. Oftentimes, we'll be very interested in this lecture series, of course, just in the single polynomial situation. Now, of course, we should mention that when you look at the algebraic closure of a field, every polynomial in f adjoin x will split over the algebraic closure, f bar there. But usually speaking, f bar is not going to be the splitting field for a specific set of polynomials. And this is where the minimality comes into play here. Yes, every polynomial splits in f bar, but is f bar the smallest algebraic extension for which that set of polynomials splits. Now, there does exist, of course, such an example where f bar is a splitting field. You know, if you take s to be the set of all polynomials, like so, then f bar is in fact the splitting field. It's the splitting field of, of course, this set right here of f adjoin x. So if you take literally every polynomial in f adjoin x, then f bar, yeah, is the splitting field of that because no smaller field can do it. I mean, that's the whole point of the algebraic closure. Every polynomial splits in that situation. So yes, f bar is a splitting field for a set of polynomial. In particular, that means that the field extension f bar over f is a normal extension. That's a very important observation because it's a splitting field for some collection of polynomials. But be aware that just because every collection of polynomials splits in f bar does not mean f bar is the splitting field for every collection of polynomials. No, no, no, no. f bar is not the minimal extension there with that regard to make the polynomial splits. So it's not the splitting field for that set of polynomials. Although it is a splitting field, but implicit in the definition of a splitting field is a specific set of polynomials. f bar is not the splitting field for every set of polynomial. But it is the splitting field for the set of every polynomial from the whole plan words there. Alright, so we mentioned some examples of splitting fields, but let's look at some finite examples here that is for some finite sets of polynomials. Take for example, the polynomial p of x, which will take it as this quartic polynomial x to the fourth plus 2x squared minus eight. And we're going to view this as a rational polynomial. Now this quad, this quartic polynomial, excuse me, it's not irreducible, it does in fact factor into two irreducible quadratic polynomials irreducible over the rationales, of course, p of x will factor into x squared minus two times x squared plus four. And feel free to pause the video right now to verify that multiplication if you would like to. So we took a single polynomial p of x, but if you wanted to, you could take as your set of polynomials s these two individual irreducible polynomials x squared minus two times x squared plus four. So if one wants to we can always consider the pulse the set of polynomials s to be a set of irreducible polynomials. Because if you had a polynomial in s, that was reducible. Well, it'll reduce into irreducible factors. And that's because f of join x, it's a Euclidean domain, in particular, it's a unique factorization domain. So every polynomial has a unique factorization into irreducibles. So if you have a reducible polynomial factor into irreducible factors, and then you can separate the factors as if they were different polynomials in the set. So when it comes to the set s, you can make some important assumptions. If you want to, you could pretend, since s here is finite, that it's just one polynomial to rule them all. And in the darkness bind them. Or, of course, you can also segregate all of the irreducible factors. And you can consider you can consider s a set of polynomials that are only irreducible, you know, if it's beneficial for you, it won't make much of a difference really in practice. But just want to be aware you have those op, there's our options on the set s here, right? Given a splitting field, it could actually be a splitting field for multiple sets of polynomials. But if it's a, if it's a splitting field for any set of polynomials, that makes it a normal extension. And that's why we introduce this idea of normal extensions. Because the splitting being a splitting field, like I said, it could be a splitting field for multiple sets of polynomials. If it works, if it splits for one polynomial, then it's going to be a normal extension. Alright, so getting back to our polynomial p of x right here, it factors as x squared minus two and x squared plus four, for which we can factor this easy enough that if we have the square root of two, then x squared minus two will factor as x minus the square root of two and x plus the square root of two. And this is a factorization in the, in the ring q adjoin the square root of two, square root of two, adjoin x like so. So that polynomial would factor in that setting. Then when you look at x squared plus four, we can also factor that one x squared plus four. This is going to factor as x minus the square root of negative two times x plus the square root of negative two. And so of course, this would then factor inside of the ring q adjoin the square root of negative two, adjoin x. So p of x has four roots, those four roots, of course, are going to be plus or minus the square root of two and plus or minus the square root of negative two. And so the splitting field is going to necessarily have to contain these four roots. And so basically the splitting field means we're just going to take all the four roots of the polynomial and adjoin them to the rational numbers. So we're going to adjoin the square root of two because if you get the square root of two, then you'll get negative square root of two. We adjoin the square root of negative two. And so this gives us our splitting field because we threw in all the roots of the polynomial that will make sure the polynomial splits. And why is it a splitting field? Well, if we were missing any of the roots, then it wouldn't be and wouldn't split. And so in some essence, that's how one builds the splitting field. And we'll talk about that in just a second. That's how you build a splitting field you throw in the roots of your polynomial. Now we can think of it also as the splitting field for this polynomial is going to be adjoin the square root of two times i. Maybe this is a little bit more preferable here because notice the square root of negative two can be factored as the square root of negative one times the square root of two that is its i times the square root of two. So clearly this field contains the square root of negative two it clearly contains the square root of two. So this we go in that direction. And you can also show that both of these fields are in fact degree four extensions of the rational numbers. So these are the same field and thus they're both the splitting field of p of x. The splitting field is unique. This is the same field with two different representations of that splitting field. Alright, let's look at a degree three polynomial this time. Take q of x to equal the rational polynomial x cubed minus three. Okay, now clearly one of the roots of this polynomial is the cube root of three. Now, and that's pretty easy to see here, of course, if we take if we take this polynomial and evaluate it at the cube root of three here, you get the cube root of three cubed. Well, by construction, the cube root of three is when you cube it is three. So this gives you zero. So the cube root of three is a root. And we're of course, taking the real cube root of three as the root we mean right here. But if we adjoin it to the rational numbers, you do get the factorization that q of x will factors x minus the cube root of three there, but you'll also get x squared plus the cube root of three x plus the cube root of nine, which be aware that the cube root of nine is just the cube root of three squared, which does belong to the field q adjoin the cube root of three. So this gives us a factorization and you can make the argument that this is in fact, irreducible, it doesn't have any other roots. So our polynomial q of x doesn't split over the field q adjoin the cube root of three. Now this was a big difference between the quadratic polynomials we saw a moment ago, like when we adjoin the square root of two, the other root of the polynomial was negative square root of two, which is its additive inverse. So because the square root of two belongs to its to a field, its additive inverse had to also belong to that field as well. So if we were only looking at the polynomial x squared minus two, then if we take q adjoin the square root of two, we get one root of the polynomial, but we also get the other root of the polynomial negative square root of two. And so by adjoining one root, we got the other root. That was also true for the for the polynomial x squared plus four. If we adjoin the square root of negative two to q, we automatically get the other root negative the square root of negative two. So you get one root and you get the other automatically. That's not always the case. This is actually sort of a special observation here, because in these two quadratic situations, the two roots you have are in fact, additive inverses of each other. Does that always happen? Well, it turns out for a quadratic extension, the answer is going to be yes. Because by the quadratic formula, if you would get a root of a quadratic, that's not rational. The other one will just be its conjugate and up to, and since your field will contain the rational numbers up to field operations, it really comes down to adjoining the square root of a number, which is not a perfect square, which case its conjugate will be added in verse, you always get the second one for free. But for arbitrary extensions, that's not always the case. If we adjoin the cube root of three, we don't automatically get the other cube roots of three, for which what are the other cube roots of three? Take zeta three here to be the principal complex third root of unity. So we get e to the two pi i over three, like so. So this of course is cosine of two pi thirds plus i sine of two pi thirds. This is a complex number, in which case all of the complex cube roots of unity will be powers of zeta. So you have of course zeta, zeta square and zeta cubed is equal to one. So one is a third root of unity. I want to show you here that if you take the real cube root of three and you times it by any power of zeta three here, and then you cube that, what's going to happen as well by multiplicative properties here, you're going to get the cube root of three cubed, and then you're going to get zeta cubed to the nth power for which the cube root of three cube gives you three, and then zeta cubed is going to give you one, one to the nth power is still just one, this gives you three. So basically what I'm saying here is that the cube, a cube root of three is going to have the form, you take the real cube root of three times any third root of unity, any complex third root of unity. This will give you all the cube roots of three. Why is that? Well, clearly we just saw a moment ago that the real cube root of three times a complex root of unity does give you a cube root of three. Now the cube roots of three are in fact the solutions to the polynomial x cubed minus three. That polynomial has at most three solutions because its degree is three. And as there are three complex third roots of unity, you times each of those by the real cube root of three, we get all of the possibilities right there. So that tells us that if we take the field, q adjoin the real cube root of three times a principal complex third root of unity, that will in fact give us the splitting field. Why is it the splitting field? Well, because this field does contain the three roots to our polynomial cube. Alright, so the polynomial splits inside of that field. Well, to be a splitting field, it has to be a minimal extension, right? The field, the polynomial splits yes, but is there any smaller field that does it? Well, notice that this field has degree six over the rational numbers. How do I know that? That's because I can factor this thing as q adjoin the cube root of three zeta over q adjoin the cube root of three. Oh boy, let me clean up this a little bit. Give myself a little bit more space. And then we also have that it factors as that. And then we have the cube root of three over q, like so, for which this first one we already mentioned is going to be three, because the minimal polynomial of the cube root of three is q right here. And so that's a degree that's a degree three polynomial, which gives us a degree three extension. And this right here is going to be to degree two. And the basic reason for that is, since the cube root of three, I chose it to be the real cube root of three. And this zeta three is a non real number. The extension, the extension of q adjoin zeta over q, is it going to be two? Because this polynomial, excuse me, this root zeta does satisfy the polynomial x squared plus x plus one. You can show that's the minimal polynomial for zeta right here. This gives us a degree two extension over q. Now throwing the cube root of three into there doesn't do a look of anything, because these are all real numbers. This is a real field and so to adjoin a non real number is going to be just like you adjoin it over q, like here. So we're going to get that this is two, this is three, so the product is in fact six. And so if you adjoin one root, you get a degree three extension, but it doesn't split over there. And so divided by divisibility, we need to have an, we need to have a subfield of this field, which is divisible by three, but it's not three, which only leaves six here. So we do have in fact this is a splitting field. It's the smallest field extension of q for which the polynomial q of x splits over. Now we arrive to a very important property of field extensions, that field extensions are unique up to algebraic closure. And so suppose we have a field f, and suppose we have a set of polynomials s, then the splitting field of s exists and is unique inside of f bar there. So I do have to put a little bit of clarification here. When we talk about it's unique inside of f bar, given two algebraic closures of a field, you know, say something like f bar versus f hat. Okay, we have previously shown that these two algebraic closures are necessarily isomorphic to each other. And in fact, if we have some isomorphism fee between them, this map fee, when restricted to f, the base field, this is just going to be the identity map. Okay, fee restricted to f is just the identity on the set f right there. So we prove that such a nice isomorphism is going to exist between algebraic closures. Now, if you do have the situation where a field e is the subset of f bar and e prime is a subset of f hat, if these are two different splitting fields over f for the same set of polynomials s. So you know that these are the splitting fields for the polynomial s right here. So e is the splitting field of s inside of f bar and e prime is the splitting field of s inside of f hat. Then it turns out that this field extension, excuse me, this field isomorphism fee right here, if you take fee of e, this gives you a subfield of f bar. Okay, well, so this gives this will give us a splitting field of s because of the isomorphic property, the polynomials s will factor in fee of e and it will fact be minimal. So we're going to have to get that since fee of e is a splitting field for s, the uniqueness of splitting fields in f hat is going to guarantee that fee of e is actually equal to e prime. Therefore two splitting fields for the same polynomials are always isomorphic, even if they're contained in different algebraic closures. So in this theorem that we're going to do right now, we're going to say for a fixed algebraic closure, the splitting fields are isomorphic and so what we just mentioned a moment ago is if you have distinct algebraic closures since they are since they are isomorphic, that isomorphism will restrict to the splitting fields as well. Alright, so let's proceed to prove this. Note that a specific polynomial splits over a field if and only if all of its roots are contained and this is a consequence of the factor theorem. We alluded to this fact earlier in this video here. So suppose we have some subset of x inside of the algebraic closure of f, f bar and but more specifically suppose that this set f is the set of all roots to the polynomials s because the set of polynomials s it splits over f bar because it's the algebraic closure and so in f bar we have all the roots of s called that set x. Alright, so then in particular what's going to happen is the set s is going to split over any field extension k over f if and only if k contains the set of roots x because they like you said by the factor theorem if you split into linear factors you contain all the roots and if you contain all the roots it'll it'll split. So we look at the fields that contain the roots of the polynomials and s those subfields of the algebraic closure and so that's what we then defined the set e to be. We're going to take the intersection of all fields k for which what do we require of k well clearly it's contained inside this fixed algebraic closure but it contains the set x the set of roots of s right here now the intersection of field is a field so e is going to be a subfield of f bar now since each of the fields k contains the field f that means f is contained inside the intersection of the k's therefore f is a subfield of e so e is an extension field of f but similarly since every field in this intersection contains x we see that x will be a subset of this intersection so x is a subset of e so e is a field extension of f that contains x so by the above observations all of the polynomials in s must split over e. I claim that then e is the splitting field for s because by construction since we're using these intersections e is necessarily the smallest field extension of f for which s splits over because since e is a field for which s splits over because remember s is going to split over k if and only if k contains x so we define this set to be the intersection of all fields that contain x but we alternatively could define this set to be the intersection of all fields that extend f that s splits over okay since s splits over e e is actually one of the fields in this intersection um and therefore this shows the minimality of e right here that in fact e is the splitting field of s inside of f bar because it's the smallest field for which um s splits over because if there was a smaller field it would be in this intersection and therefore the intersection would have been smaller than e giving us that contradiction there so we do get that splitting fields exist okay well why are they unique up to algebraic closure here well by construction like we saw before this field e contains the field f it contains the set of roots x from the algebraic closure so in particular the set the field f the field e i should say contains the subfield f adjoint x so what is this right here f adjoint x here this means the set uh that that is the smallest field that extends f that contains all the elements x well this is a field that contains f that contains x so all the polynomials in s would split over the pollen over the field f adjoint x so by construction f adjoint x would be inside of this intersection would be one of those fields k so therefore e is a subset of that but also e contains it so in particular this splitting field e is equal to f adjoint x and so the splitting field is exactly that the splitting field is the base field adjoint all of the roots of every polynomial inside of s we alluded to this fact earlier when we talked about our examples but then this proof solidifies it and so as this set x is fixed um with regard to the algebraic closure that then determines uniquely the set e the splitting field the splitting field is exactly just the base field adjoint all of the roots of the polynomials which are contained inside of the algebraic closure so splitting fields are unique uh inside of an algebraic closure they exist which then finishes this video uh on splitting fields