 In this video, we're going to prove what one might call the analog of Lagrange's theorem for field theory. Because remember what Lagrange's theorem tells us is that every subgroup of a finite group divides the order of the group. And then from that, you can infer this statement about indices of a group. Where if the index of a subgroup measures the number of cosets, then you can factor indices, right? So if you have some subgroup K, the number of cosets of K inside of G is equal to the number of cosets of some other group H inside of G times the number of cosets of K inside of G, where here we have K as a subgroup of H, which is a subgroup of G. So when it comes to counting cosets, we get this property, the index factors, okay, between some intermediate subgroup. Now, the same is also true for field extensions when we look at their degree. Some people call it the index of the field extension. We generally call it the degree here. So imagine we have three fields. We have the field F, which sits inside the field E, which sits inside the field K. So K is a field extension of F. It's also a field extension of E. E is this intermediate field that sits in between F and K, excuse me. And so we can then talk about the degrees of these extensions in all the cases. And so what we claim is that the degree of the extension K over F is equal to the degree of the extension K over E times the degree of the extension E over F. So you can factor degrees of field extensions analogous to how you could with coset, the number of cosets for groups. And so that's why we call this the field theory version of Lagrange's theorem. And this will be very important because we're going to see as we study Galois theory that these degrees of field extensions actually have something to do with indices of subgroups when we look at the Galois group. But we get a little ahead of ourselves there. So let's prove this identity here. This is a counting argument here for which let A be a basis of E when we view E as an F vector space. And let B be a basis of K when we view K as an E vector space. Now, it makes sense to view K as both an E vector space and an F vector space. The basis, when we look at it as an F vector space, will be much larger than when we look at it as an E vector space. So be aware that this basis B is viewing K as an E vector space right now. Now, in particular, since E is a subset, it's a subfield of K, the elements of A, which belong to E, can also be viewed as elements of K. And so in particular, we can add to track, multiply, and divide these things. And so consider the set C, which is going to be the set of products between elements from A and B. In particular, C is the set of all possible products where all the possible products alpha times beta where alpha is an arbitrary element of the basis A and beta is an arbitrary element of the basis B. And so we claim that this set C forms a basis for K viewed as an F vector space. Okay, so let's first make sure we know how many elements are in C in the first place. So for the sake of simplicity, well, not not simplicity, just for the sake of clarification, right? Thus, cardinality of the set alpha is the, since it's a basis of E over F, it's going to be the dimension of E as an F vector space. That by definition is the degree E over F. And likewise, the cardinality of B is the dimension of K as an E vector space. And that by definition is the degree of the extension K over E. So is the set C, does it, is its cardinality equal to the cardinality of A times the cardinality of B? That seems likely. And that's what we want to prove here. But what we have to, in order to do that, we have to prove that each of these possible products alpha times beta are distinct, because if any of them overlap with each other, we might not get the full, the full amount of elements there. And so let's consider the possibility that alpha beta equals alpha prime beta prime, where alpha and alpha prime are potentially different elements of A and beta beta prime are potentially different elements of B. Now, when you look at that equation, alpha beta equals alpha prime beta prime, I'm going to move, I'm going to set the right hand side equal to zero, so we're going to get alpha beta minus alpha prime beta prime. This is equal to zero. Now, the significance here is that when you look at that, it's like, I have some linear combination of elements from a field that adds up to be zero. Could I make some argument about linear independence or something? That's exactly what we're going to try to make the argument right here. Because notice the elements of alpha, excuse me, the elements of A, including here alpha and alpha prime, these are elements of E. And when we look at B here, B is a basis of K as an E vector space. So B is this E basis of K. So what we claim here, what does it mean to be an E basis? We mean that every element of K can be written as a linear combination of things from B, but your coefficients can range over elements of E. Well, the things inside of A come from E. So when you look at this equation right here, you can think of the alphas, alpha and alpha prime here. These are our coefficients in a linear combination because they belong to the set E, like so. But beta and beta prime, they came from B. And B is a linear independent set. So the only way that this linear combination can add up to be zero is that the coefficients have to all be zero. So alpha and alpha prime have to all be zero in that situation. So the only way you get agreement that alpha, beta equals alpha prime, beta prime was that the alphas, alpha and alpha prime had to be zero. But wait a second, alpha was, excuse me, A was also a basis for E over F. And alpha prime came from A. If alpha was equal to zero, that would mean zero belongs to the basis, but no basis can contain zero because it's linearly dependent. That gives us a contradiction. So it turns out that there is no possibility that alpha, beta equals alpha prime, beta prime. The only other possibility is that these elements were in fact the same thing. That is, you would have to have that beta equals beta prime. And then in that situation, you can divide beta from both sides. So since you got alpha, beta equals alpha prime beta, you can divide beta from both sides inside of K. You'd get that alpha equals alpha prime. So the only way you can get those things to be the same thing, the same product, is if you have the same factors. So this tells us that we can identify in a natural way C with the set A cross B. There's this bijection between the two sets for which honestly the map makes sense to go the other way around. If you have an ordered pair of alpha, beta, you can identify without the product alpha, beta, like so. This map will be one to one and clearly onto by the definition of set C. So there's a bijection between those. So this gives us that the cardinality of the set C is equal to the cardinality of A times B. For which this is the Cartesian product. As sets, this is equal to the cardinality of A times the cardinality of B. And as we observed earlier, the cardinality of A is the degree of K over E. And the cardinality of B is the degree E over F. Therefore, because of this statement right here, the cardinality of C equals this factor degree. To prove our equation here, we have to prove that C is a basis for K over F. Because if C was a basis, then that's the dimension of the vector space K over F. And that's exactly what this degree is measuring. So we now will proceed to show that C is in fact a basis for K over F. So take an arbitrary element gamma inside of K. Well, by what we've observed earlier, since B is an E basis for K over E. That means every element gamma can be written as a linear combination of the betas. So you have some beta 0, beta 1, beta 2, beta 3, whatever. But the coefficients in that linear combination are going to come from E. But A, the set A is a basis for E over F. So each of those AIs can be written uniquely as a linear combination of some C's times the alphas, like so. So the element AI that you say right here, it can be written as some combination of using alpha ij's and coefficients Cij's, like so. Where now the coefficient Cij comes from the base field F. And the alpha ij's of course belong to, they belong to the set A, like so. Now taking M to be the maximum here of all of these MIs, for which MI here, this is some finite number in this combination, right? Like how many of the alphas do you need? You potentially can use all of them if you have to, right? Setting some of these coefficients to be 0 if they're not already defined is the idea here. We can then come up with a single M and a single N that takes care of everything here. So look at the combination of gamma again. If we look at it as an element in an E vector space, we have this coordinate vector right there. But each of those AIs can be written in the following sum. You have a sum where j ranges from 0 to M of Cij alpha ij, where again we might have to throw a couple of zero coefficients in there so that all of the sums are consistent in this situation. But then this is a combination inside the field K. We could distribute the beta i's throughout this. So we get the double sum where i ranges from 0 to N, j ranges from 0 to M, and you get Cij times alpha ij beta i. For which as the alpha ij's range over the elements of A, the beta i's range over the elements of B, you have all these possible products. These are what the elements of C looks like. So this shows that gamma can be expressed as a linear combination of things from C. That is, gamma belongs to the span of C. And since gamma was an arbitrary element, this indicates that C is a spanning set for K viewed as a F vector space. Alright, so that's a good start. We have a spanning set. Is it a basis? Is it linearly independent? So consider that we have some combination of the alphas and betas that adds up to be zero in this situation here. Alright, so can this combination give you a non-trivial combination of zero? Now since B is a basis for K over E, then it has to be the only way, because you can factor this sum like we did above, the only way that a sum of the betas can add up to be zero, that happens if and only if each and every one of the coefficients is zero. So the coefficients of an individual beta i is the sum of the Cij alpha ij's. So each of those combinations has to equal zero, which remember alpha ij comes from capital A and Cij comes from F. But A itself is a basis for E over F. So the only way that you can take this combination to equal zero is that if each of those coefficients were equal to zero for each of the alpha ij's. And so each of the Cij's has to equal zero, but the Cij's are the coefficients of this original linear combination. This shows in fact that C is a linearly independent set and therefore is then a basis for our field extension K over F. This then proves Lagrange's theorem for fields that finite extensions can be factored in the way that we've seen here. And in fact the proof that we just went through never actually requires that things be finite dimensional. This would also be true for infinite dimensional field extensions. Now cardinalities, these infinite cardinals, multiplication might not mean as much as you might think. And for our lecture series we're really good at both to be concerned with finite extensions. But be aware that our theorem actually is perfectly applicable for infinite extensions well that the commonatorics we made in this argument applies to infinite cardinals as well. And so that the theorem doesn't need to be modified for the infinite case as well. Let's look at an important corollary of this result here. So imagine we have the field extension E over F and suppose this is a finite extension and as we observed earlier if it's a finite extension it must necessarily be an algebraic extension. Take some element beta inside of the extension field E. Then we know by what we've just observed that the degree of the minimal polynomial of beta over F has to divide the degree of E over F and why is that? Well that's because if we have the field F and we have the extension field E then the simple extension F a joint beta would have to be an intermediate field between those two. So applying the previous statement right there we then get the degree of the extension E over F. It has to factor as E over F a joint beta times F a joint beta over F for which this second one right here is equal to the degree of this minimal polynomial of beta over F like so. And so that gives us the divisibility condition we want. In particular if E itself is a simple algebraic extension this tells us that the degree of the minimal polynomial beta must actually divide the degree of the minimal polynomial of alpha here. In fact we even get better that the polynomials in fact divide one another. Okay let's look at an example of these principles in practice here. Let's consider the element alpha which looks like the square root of 3 plus the square root of 5. Now this is an element which we can produce by radicals and so as we saw earlier in this lecture we will be able to construct a minimal polynomial utilizing the fact that we have radicals here. So if you move one of these square roots to the other side I'm going to take the square root of 3. We get x minus the square root of 3 that's equal to the square root of 5. Squaring both sides we see the following. x minus the square root of 3 squared that will give you x squared minus 2x times the square root of 3 plus 3. When you square root the right hand side the square root of 5 squared is 5. So attracting 5 from both sides we can move that over here. You can end up with x squared minus 2x times the square root of 3 plus minus 2 in that situation. So we get that. Now what we can do next is we can move this square root to the right hand side so we get the following equation. x squared minus 2 equals 2x times the square root of 3. If we square both sides again so we square this side we square this side. On the right hand side we're going to have to foil it again. x squared minus 2 squared that will give you x to the fourth minus 4x squared plus 4. On the right hand side when you square things you're going to get x squared clearly. You'll get 2 squared which is 4, square root of 3 squared which is 3, 4 times 3 is 12. You end up with 12x squared for which then when you move that to the other side you'll get the following polynomial. p of x equals x to the fourth minus 16x squared plus 4. This would then be the minimal polynomial of alpha over the rational number. So you get this degree 4 polynomial. This will be our minimal polynomial. You might recognize this polynomial if you've been paying attention in this lecture series. It's like I feel like I've seen this somewhere before. Actually the very first video of this lecture 26 we talked about this exact same polynomial. x to the fourth minus 16x squared plus 4. Now you might recall in that situation it's like, huh, I thought you said the roots in that situation were plus or minus the square root of, oh boy, what was it, 8 plus the 2 times the square root of 15. But now you produce this other number, the square root of 2 plus the square root of 5. No, that wasn't what it was. Sorry, the square root of 3 plus the square root of 5. So do we have a fifth root of this polynomial? It's degree 4, it should only have 4. Let's investigate this thing a little bit further. Now this is an irreducible polynomial. So if you take Q adjoin alpha, look at that field over Q, the degree of the extension will in fact be the degree of this, if it's irreducible minimal polynomial. In which case that gives you that this field is degree 4, okay? And clearly on the other hand though, alpha belongs to the field Q adjoin the square root of 3, square root of 5. Clearly these two elements would add together, give you alpha. So alpha belongs to that. And so we look at this field extension. This field extension would factor, okay? It's going to factor as Q adjoin the square root of 3, square root of 5, over Q adjoin the square root of 3, and then Q adjoin the square root of 3 over Q here. And so let's compute these things. This one, Q adjoin the square root of 3 over Q is clearly going to be degree 2 because this is a simple extension and your irreducible polynomial there would be X squared minus 3, okay? What about this one? Q adjoin the square root of 3, join the square root of 5 over Q adjoin the square root of 3. This is also a simple extension because that field Q adjoin the square root of 3 times the square root of 5. You can think of that as Q adjoin the square root of 3, adjoin the square root of 5, like so. You've added one new element. I mean, I know of its algebraic consequences, of course, but you've added a new element to Q adjoin the square root of 3. So over this base field, this is a simple extension right here. And so the degree of this extension is going to come from the degree of the minimal polynomial, which the minimal polynomial here will be X squared minus 5. You can take the exact same rational polynomial, which is still a polynomial over Q adjoin the square root of 3. That would still be irreducible. And its root is going to be the square root of 5, negative square root of 5 as well. So this is also a degree 2 extension. So you get degree 2 times degree 2, which is equal to 4. So look at that. Because alpha belongs to Q adjoin the square root of 3 times the square root of 5, it has to be the case that Q adjoin the square root of alpha is actually a subfield of Q adjoin the square root of 2, excuse me, square root of 3, square root of 5, like so. But this is a degree 4 extension over Q. This is a degree 4 extension over Q. And so because these degrees divide each other, or even better than that, if you have a four-dimensional subspace of a four-dimensional vector space, they actually have to be one and the same thing. So these vector spaces are equal to each other, which means these fields are actually one and the same thing. Interesting enough, this field Q adjoin the square root of 3, square root of 5 is actually the same thing here as just adjoining a simple element. This happens a lot. It turns out that adjoining two elements, two algebraic elements, you can often replace it with a single element that takes care of it. So a double extension can actually be turned into a simple extension. We'll explore more of this in the future. Now, before we go to the next example, I'm sure many of us are curious about what's happening here, right? How do you have this, right? Is this a new root of this polynomial? The answer is no, it's not. I do want you to consider what happens when we square this element here. If you take the square root of 3 plus the square root of 5 and you square it, well, you square the square root of 3, you're going to get 3, then you're going to get 2 times the square root of 3 times the square root of 5, like so, and then you'll get the square root of 5 squared, which gives you 5, which this gives you 8 plus 2 times, well, the square root of 3 times the square root of 5, that's the same thing as the square root of 15. So I want you to be aware that the square root of 3 plus the square root of 5, when you square it, it gives you 8 plus 2 root 15. 15. So the square root of three plus the square root of five is a square root of eight plus two root 15, for which there's two of these of course, you have the positive one and the negative one. So I want you to be aware that we've just argued that the square root of three plus the square of five is actually equal to the square root of eight plus two root 15. So we are calling these two different elements alpha, they actually are the one and the same element. This is what can get a little bit tricky about radical elements is that different radicals can actually be equal to each other. They look different, but they're actually one of the same alpha, one of the same. Yeah, in this case, they're both this element alpha, these two different radicals gave you the same thing. One way that you could try to determine whether these things are equal to each other, well, you could try some type of numerical analysis. But of course, that only gives you that they're probably equal. We can actually look at their minimal polynomials, that if these two different radical elements have the same minimal polynomial, so at the very least they have the same minimal polynomial, you can make an argument from there that they're actually one of the same thing. Because notice that both of these elements, the square root of three plus the square root of five and the square root of eight plus two root square root of 15, they are both square roots of eight plus two root 15, which is forcing them to be the same thing, because there's the positive and negative case. All right, let's look at another example here. Let's look at this one. Take the field where we're going to adjoin the cube root of five and the square root of negative five. Consider this field extension. Well, look at the field F, which will define to be the field Q adjoin the cube root of five. Notice that this F is a subfield of the real numbers, because the cube root of five is by definition a real number. And so taking rational numbers and throwing in a real number, you can only produce real numbers in that situation. Since F is a subfield of the real numbers, clearly the square root of negative five is an imaginary number, purely imaginary number. It doesn't belong to F. And so since the square root of negative five doesn't belong to, excuse me, the square root of negative five doesn't belong to the real numbers because it's purely imaginary. And since F is a real subfield, that means the square root of four i, excuse me, square root of five i doesn't belong to F. This is a trick that we're going to use a lot, that if you have a real subfield, you can use that to eliminate imaginary numbers or complex numbers that have non-trivial imaginary components. All right, so that's an important thing to see here. The square root of negative five does not belong to F. On the other hand, the minimum polynomial of the square root of negative five is going to be x squared plus five. This is a polynomial over the rational field, but it's also a polynomial, a irreducible polynomial in F a joint x. Because if this polynomial was reducible, we'd have to have a root, which means the square root of negative five would be an element of F, which is not. So we get that the degree of E over F is degree two. So that's a nice calculation there. And again, that's because the minimum polynomial is x squared plus five. But then if we look at the cube root of five, its minimal polynomial as a rational polynomial is x cubed minus five. So this tells us that the degree of the extension F over Q is equal to three. Now, because of the factorization, we have E over F, we have F over Q, we then can compute the degree of E over Q, which by factorization, it'll be the degree of E over F and F over Q. So you get two times three, which is six. And so this factorization principle can be very helpful in computing degrees of extensions, field extensions. And if we know the degree of a field extension, we can often utilize that to show that certain fields are equal to each other, certain algebraic elements are one and the same thing. So this is a very, very important property. Let me end this lecture with a theorem that I'm actually going to leave as an exercise to the viewer here. Given a field extension E over F, then the following are equivalent. E is a finite extension of F, that's condition one. There exist algebraic elements alpha one, alpha two, up to alpha N that belong to E such that E is equal to the extension F, a joint alpha one, alpha two, alpha three, up to alpha N. So you do have this finite number of elements you can join to F to create this extension E. And that condition two is equivalent to the last condition here. There exist elements alpha one, alpha two, alpha N, that belong to the field E for some number N, of course, such that E equals the field extension F of joint alpha one all the way through alpha N. And that then contains the proper subfield F of joint alpha one all the way to alpha N minus one, which that contains the proper subfield F of joint alpha one all the way up to alpha N minus two, all the way down to F of joint alpha one, all the way down to F here. And in each of these situations, these are proper containments that adjoining the next element makes the field get bigger. And each of these elements were algebraic along the way. That is, this field is algebraic over that field. This field is algebraic over that field. And then all the way down to this is algebraic over that. And so like I said, I leave this as an exercise to the viewer here. It's a consequence of the properties we've already proven already. That's where we're going to end lecture 26 for now. Thanks for watching. If you learned anything about algebraic or finite extensions, please like these videos, subscribe to the channel to see more videos like this in the future. And if you have any questions, please post them in the comments below. And I'll be glad to answer them as soon as I can.