 In this video, I wanted to give us an example of a construction of a finite field using Kroniker's Theorem. We have, of course, looked at the finite field z-adjoint p, where p is a prime, in that situation. These, as a ring, is just the integer's mod p. It turns out that you can construct a finite field for any order p to the n, where p is a prime number. And I want to, in this video, construct the field of four elements. I say the field, because up to isomorphism, they're all equivalent, they're all isomorphic. We'll prove this in the future, that there's really only one. But I just want to give you the construction right now, because Kroniker's Theorem is the tool to do so. By Kroniker's Theorem, you can construct a simple extension of a field by using an irreducible polynomial. So look at the finite binary field z-2. So z-2 only has two elements, right? You have 1 and 0 here. And you're going to follow the usual rules of addition, addition and multiplication mod 2, right? So when you're working mod 2 here, we'll just write addition, you know it's mod 2. 0 plus 0 is 0, 0 plus 1 is 1, 1 plus 1 is 1, and 1 plus 1 is 0. So that's just the cyclic group of order 2. If you think of it in terms of multiplication, like so, 0 times anything is 0, and 1 times 1 is 1. So we get our Cayley tables for the operations in that situation. This is our field of order 2. We can, from this, construct a field of order 4. To do that, we need to find an irreducible polynomial over z-adjoin 2. For which there is 1 and only 1 such polynomial, but 1 is enough. The polynomial x squared plus x plus 1 is an irreducible polynomial over z-adjoin, over z-2 adjoin x. Now how do I know that? Since it's a quadratic polynomial, degree 2, right? It'll be irreducible if and only if it has a root. Now z-2 has only two numbers, and those are the only two possible roots. We can check those. We can plug in 0, in which case you get 0 plus 0 plus 1, which is 1 mod 2, so that's not a root. You can plug in 1 and 1 here, so 1 plus 1 plus 1, which is 3, which is 1 mod 2. That doesn't quite work either. So p of x is irreducible because we brute force tried every possible number. There's only two. It didn't take that long. It's irreducible because it doesn't have a root. Quadratic polynomials have to have a linear factor if they reduce, and they have to have a root to have a linear factor. Therefore, this is an irreducible polynomial. So then mimicking the construction from Kroniker's theorem, we can construct a simple extension of z-2 by adjoining a root of p of x onto z-2. This is going to form a field, which we're going to call that field f4, because it's a field with four elements, and I'll explain what that is in a second. So f4 by construction is z-2 adjoined alpha, where alpha is a root of the polynomial p of x. And I claim that everything in f4 can be written as a plus b alpha, where a and b are binary scalars, 0, 1, and then alpha is that root of the polynomial p of x. Alright, so how do we know that? So first of all, since by construction alpha is a root to the polynomial p of x, that means that 1 plus alpha plus alpha squared is equal to 0. It's equal to 0 because it's a root. We created a root, we added it to our field, and so it must satisfy this algebraic relationship that 1 plus alpha plus alpha squared is equal to 0 here. Now, if I minus alpha from both sides of this equation and I minus 1 from both sides, you'll end up with alpha squared is equal to negative alpha minus 1. These are our coefficients. These are coefficients in z2. But in z2, there is no such thing as negative 1, or better yet, negative 1 is really just positive 1. So negative alpha minus 1 is really just alpha plus 1. So because of this relationship, 1 plus alpha plus alpha squared equals 0, that means that alpha squared is equal to 1 plus alpha. And then by induction, we can deal with other powers as well. If you have something like alpha cubed, this would be alpha squared times alpha. Alpha squared is 1 plus alpha. You have alpha here. You can distribute the alpha here. You're going to end up with alpha plus alpha squared, which alpha squared is, like I said before, 1 plus alpha. And then you're going to end up with a 2 alpha plus 1. 2 alpha is just 0. You just end up with 1 right here. Alpha cubed is equal to 1 in this situation. Alpha is, in fact, a third root of unity for this field, interesting enough. And so by reduction techniques like this, we can actually construct the Cayley table for any possible combination here. Okay? Oh, before we do that, let me, of course, go back to this statement above here, right? How do I know that everything can be written like this? Well, if you are a field with binary coefficients and you join alpha, clearly you have to have this. But you might also need c times alpha squared because alpha squared is inside that set. You would need d times alpha cubed because alpha cubed is inside that set. You need e times alpha to the fourth because alpha to the fourth is inside that set. Because if this set is closed under multiplication, you have to have alpha times alpha times alpha times alpha and everything else, right? But because of these reduction principles, I don't need alpha squared because I can write that as 1 plus alpha. And by induction, all these higher powers of alpha can be written as a combination of a scalar and some multiple of alpha. So this statement right here is, in fact, correct. That 1 and alpha form a basis for this vector space here. The reason why this is the field of four elements is because this vector space as a z2 vector space has dimension 2. Now, as the vector space itself has size 2, the cardality here is going to be 2 squared, which is equal to 4. So this does give us four elements. And one of the four elements here, you get 0, you get 1, you get alpha and you get 1 plus alpha. So think about the possibilities here. A plus B times alpha, you have two options for A, two options for B, two times two is four, like I was saying them before. And so you get these options here. With respect to addition, if you look at just this sector, this is just addition on z2. But then if you throw an alpha here, because this new element alpha is linear independent with 1, when you take, well, clearly add 0 to everything, you're just going to get back that element, it's the additive identity. If you add 1 to it, well, this part's the same. Alpha plus 1 is just equal to alpha plus 1. That's what it is, because they're independent. You can't simplify it any better than that. But on the other hand, if you take alpha plus 1 and you add 1 to that, that's the same thing as alpha plus 2. And since we're working mod 2, the 2 reduces and you just get back alpha, alpha plus 0 in that situation. Addition is going to be commutative. It's an abelian group with respect to addition here. Alpha plus 0 is alpha plus 1 is what it is. Alpha plus alpha is equal to 2 times alpha, which 2 and 0 are the same number since we're in characteristic 2 field. So that gives you 0. And the same thing here, alpha plus 1 plus alpha is going to be 1 plus 2 alpha, which when you reduce mod 2, you just get back 1. So alpha plus 1 plus alpha is 1. The remaining calculations are similar. If you take 1 plus alpha plus 1 plus alpha, you're going to get 2 plus 2 alpha that reduces to 0. So addition here, notice that the addition, if you look carefully at this, this is just the client 4 group. This is Z2 cross Z2 with respect to the usual addition. That's always going to happen with these finite fields. If you look at just the additive structure, you're going to get an elementary abelian group. Multiplication is where things get really interesting. How does multiplication work? Well, if you look at just this part of our finite field, this is just going to look like multiplication in Z2. But then 1 is going to act like the multiplicative identity here. So 1 times alpha is alpha, 1 times 1 plus alpha is 1 plus alpha. Multiplication is commutative, so see that. So this is the sector where things get interesting. So like we saw earlier, alpha times alpha, which is alpha squared, this equals 1 plus alpha. So we get that. By the calculation we did up here, you're going to get that alpha times 1 plus alpha is 1, which of course means 1 plus alpha times alpha is also equal to 1. Alpha and 1 plus alpha are multiplicative inverses. 1 over alpha is equal to 1 plus alpha in this ring. That's why it's a field. Every element has a multiplicative inverse. And then finally, what is 1 plus alpha times 1 plus alpha? Well, since it's a ring, we can do the FOIL method here. So you can end up with 1 plus alpha plus alpha plus alpha squared. So this is going to give you 1 plus 2 alpha plus alpha squared, 2 alpha is 0, of course. Alpha squared is the same thing as, of course, 1 plus alpha. So you end up with a 2 plus alpha, but 2 is 0, so you end up with alpha. Each and every one of these possible products is computed by the relation that alpha squared equals 1 plus alpha. This actually happens. That is, we can derive every relationship. We can derive the whole Cayley table from the irreducible polynomial itself, because alpha squared equals 1 plus alpha came from the observation that 1 plus alpha plus alpha squared equals 0. So the fact that alpha was the root of this irreducible polynomial led to the construction of a field extension, a simple extension. And the case of a simple extension of a finite field, that's going to give you another finite field. And we will see in the future that you can create finite fields by generalizing this concept. For the moment, though, we've now seen the field, the finite field of order 4.