 In the previous video of lecture 30, I had made an allegory that finite fields are two rings like finite cyclic groups are two group theory, okay? And I can actually solidify that analog in this theorem right here. Suppose we have a field, okay? And suppose, since f is a field, that means the non-zero elements of that field form a group, okay? If G is a finite subgroup of the group of units of a field, then I claim that G is actually a cyclic group, which necessarily is a finite cyclic group because we're assuming that G is a finite subgroup. This is an important example for finite fields because if f is a finite field, then f star is a finite group, and therefore every subgroup of f star, including f star itself, is a finite subgroup. So for finite fields, the group of units is in fact cyclic. So when you look at a finite field multiplicatively, it's basically just a cyclic group plus this zero element, all right? Of course, additively, it's an elementary beelding group. And so it's kind of interesting how you glue those things together. You take an elementary beelding group with respect to addition, you take a finite cyclic group multiplicatively plus a zero, you make a field, and that's every finite field. But we're going to prove a slightly general argument. So f could be an infinite field, but if you take a finite subgroup of the group of units, that finite subgroup has to be cyclic, okay? So since f is finite, it has some finite cardinality called the order of that thing in. Now, if G were not cyclic, first of all, I should mention that since f is a field, f star, it's going to be an abelian group because multiplication is commutative inside of a field. So G is a finite abelian group. So therefore the fundamental theorem of finite abelian groups applies to G here. And for which case we can factor G into a product of cyclic groups. Now if G is itself not cyclic, that means it has a non-trivial exponent. We know by Lagrange's theorem, if you take any element of your group, alpha, and you raise it to the order of the group, so alpha raised to n, you're going to get back one. That's, again, a consequence of Lagrange's theorem. Now for a finite abelian group, we often care about this concept of an exponent. The exponent is the smallest number, the smallest positive integer m, such that if you raise any element in the finite abelian group to m, you get back one. Now if you take something like the Klein-4 group, z cross, z2 cross z2, sure the order of the group is equal to four, but its exponent is equal to two. If you square any element in the Klein-4 group, you're going to get back the identity. And so if you have a finite abelian group which is not cyclic, then its exponent will be strictly smaller than the order. So if it's not cyclic, we're going to get that its exponent m is strictly less than its order n. So I want you to consider the polynomial x to the m minus one, which then can be viewed as a polynomial over this field. I mean, it only uses the coefficients one and negative one. That really has nothing to do with any specific field, because every field has that. Now if you take your element alpha and plug it into this polynomial, you're going to get alpha to the m minus one, but by observation since m is the exponent of this group, you're going to get alpha to the m is equal to one, one minus one is equal to zero. So alpha, which belongs to g, is a root of this polynomial. But since, and this is true for every element of g, but since m is less than n, that means this polynomial has more roots than its degree, which is a contradiction. That then shows that the exponent and the order of the finite group must be equal, which then forces it to be cyclic, and therefore that proves our theorem g is cyclic. And I just love this proof because it's like, wait, wait, wait, wait, wait, wait, wait, because polynomials can't have more roots than their degrees, subgroups of a, multiplicative subgroups of a field have to be cyclic. It's just so cool how those two seemingly unrelated elements come together for such an elegant, elegant proof right here. So yes, for a finite field, the unit group of that field is necessarily going to be cyclic. So for finite fields, we really are interested in this situation because since the whole set f star is cyclic as a multiplicative group, that means it has a generator. In fact, it has multiple generators. That is, there exists some nonzero element of f such that every nonzero element of f can be expressed as an exponent, as a power of that element called alpha or something like that, right? Such a thing is called a primitive root. And there's an analog for this, of course, in number theory, because in number theory, we're very much interested in the ring zp, where p is, of course, a prime in that situation. In which case, number theory talks about the primitive root. It's just a nonzero element such that every other nonzero element is a power of that element. We call it a primitive root, okay? Number theory is really just a special case. Elementary number theory in that regard with primitive roots is just a special case of what we're talking about right now. Zp is a finite field. This is true for every finite field, including zp as well. They have primitive roots. Now, of course, in number theory, we also talk about the ring zp to the k. They also have primitive roots. This shows a skew to what we're talking about right now, because this is not a field, because it does contain no potent elements, which in a field, the only no potent element is zero. But in particular, for a finite field, every finite field has a primitive root. Let's call that element alpha, okay? If you take fp, a joint alpha, this primitive root, that gives you f, because every element of f, other than zero, is a power of alpha. Subsequently, we can produce every element just by taking powers of alpha right here. We really can. And so for finite fields, every finite field is a simple extension of the prime field fp. Remember, a simple extension is you would join one element to a field, and you get a bigger field, okay? So every finite field is a simple extension of the base field. We will see a similar result for fields of characteristics zero and such, but that one's a lot more complicated. For finite fields, it's just so much simpler. Finite fields are just so well-behaved. Let's do an example of this. Consider the field Z2. So we're just going to work mod 2 here, binary coefficients zero and one. And take the degree 4 polynomial 1 plus x plus x to the fourth. I claim that this polynomial is irreducible. Now, if it has a root, then it has a linear factor. Let's show it has no roots. If you plug in zero, you end up with one. If you plug in one, you end up with one as well. So this polynomial has no linear factors because it has no roots. But it is a degree 4 polynomial. It could factor into quadratic terms. Could that be possible? Well, over Z2, we proved this in a previous video when we talked about the field of order 4. There's only one polynomial that's degree 2 that's irreducible over Z2, and that's exactly x squared plus x plus 1. So as there's no linear factors 2f, if it factors at all, it has to factor into two quadratics, but there's only one irreducible quadratic. So it's got to be x squared plus x plus 1 quantity squared. But if you actually take x squared plus x plus 1 squared and you factor it mod 2, you're going to see the following. First of all, since you're squaring it and we're working mod 2, we can do freshman exponentiation. The exponent 2 distributes over rings of characteristic 2. So x squared plus x plus 1 squared is the same thing as x squared squared plus x squared plus 1, which that gives us x squared, excuse me, x to the fourth plus x squared plus 1, which is not the same thing as 1 plus x plus x to the fourth, all right? And therefore, this is not the same polynomial. And therefore, that's exhausted every possibility. I mean, because we're over a finite field, there's only a finite number of factorizations one can do for f. We've exhausted all of them. So f has to be irreducible in that situation. So since this is an irreducible polynomial, we can take a root of this polynomial, call it alpha, and we can join it to z2. That will then give us the field of order, excuse me, 2 to the fourth. That is, this would then give us the field of order 16. So f2, I'll put it as z2 here. As z2 joined this element alpha, this then gives us f16. Now, the important thing we should know about alpha here is it's the root of this polynomial. So we get that alpha to the fourth plus alpha plus 1 is equal to 0. Moving alpha 1 to the other side, we get alpha to the fourth equals alpha plus 1, although this is negative, but working mod 2, negative doesn't mean anything. So you get alpha to the fourth is equal to alpha plus 1. This will help us with our reductions here. So be aware that f16 as a vector space is just the same thing as z2 to the fourth. So we could think of everything as just a binary column vector with four entries. So we have two options. We have two options for the first one, two options for the second, two options for the third, two options for the fourth. We can make that identification in that situation. Although we're not going to think of it that way, we're going to think of instead, not as a coordinate vector, but we think of it more as we have something here, some constant coefficient, then a1 alpha plus a2 alpha squared plus some a3 alpha cubed, like so. Because once you get up to alpha to the fourth, you can reduce it based upon this algebraic relationship on alpha. So we never need a linear combination with a higher power of alpha than alpha cubed. But also by our previous result, I should mention that the group of units for f16 is in fact z15. So this is a cyclic group of order 15. So that means every non-zero element inside of this ring should be written as a power of alpha. And so what we're going to do is go through every single one of them. There are 15 of them, but it's really not that exhaustive. It's exhaustive, but it's not exhausting. Clearly alpha to the first gives us alpha. Alpha squared gives us alpha squared, alpha cubed gives us alpha cubed. Because remember, our goal is to write everything as a linear combination of 1 alpha, alpha squared, alpha cubed. That is we want to write everything as a linear combination of the first four powers of alpha. So because of our relationship, alpha to the fourth is equal to 1 plus alpha. That gives us the first one. Now alpha to the fifth is the same thing as alpha times alpha to the fourth. Alpha to the fourth is 1 plus alpha. So you distribute that and you're going to get alpha plus alpha squared. That gives us alpha to the fifth. And we work through this recursively, right? Alpha to the fifth, excuse me, alpha to the sixth is alpha times alpha to the fifth. So we times this by alpha. We're going to get alpha squared plus alpha cubed. Then when we do alpha to the seventh, this is going to equal alpha times alpha squared plus alpha cubed. This is going to give you alpha cubed plus alpha to the fourth. We then reduce the alpha to the fourth as 1 plus alpha. So we get 1 plus alpha and alpha cubed. So we do that again, right? For alpha to the eighth, we take the previous line. We times that by alpha. So that, if you times the previous line by alpha, you're going to get alpha plus alpha squared plus alpha to the fourth. Alpha to the fourth is the same thing as 1 plus alpha. The alpha is canceled because we're working mod 2 and we're left with 1 plus alpha squared. Alpha to the ninth, we times the top by alpha. We get alpha plus alpha cubed, no reduction there. If you times this one by alpha, you'll get alpha to the tenth. Alpha is then going to become alpha squared. Alpha cubed will come in alpha to the fourth, which gives us 1 plus alpha. Alpha to the eleventh is formed by taking alpha to the tenth and times it by alpha. So you're going to get one becomes alpha. Alpha becomes alpha squared. Alpha squared becomes alpha cubed, no reduction necessary there. Alpha to the twelfth times everything above by alpha. You're going to get alpha squared, alpha cubed. Then you'll get an alpha to the fourth, which comes 1 plus alpha. So now we got everything here represented. 1 plus alpha plus alpha squared plus alpha cubed. Then if you take this times alpha to the twelfth by alpha, that gives you alpha to the thirteenth. The above line, if we times it by alpha, we're going to end up with alpha plus alpha squared plus alpha cubed plus alpha to the fourth. Alpha to the fourth becomes 1 plus alpha. So the alphas cancel. We're left with 1 plus alpha squared plus alpha cubed. We're almost there. Let's see, alpha to the thirteenth. If you times it by alpha, we'll give us alpha to the fourteenth. Take this times it by alpha. You get alpha plus alpha cubed plus alpha to the fourth, which is 1 plus alpha. The alphas cancel. You get 1 plus alpha cubed. And then if we do this one more time, because we claim we're at the end, alpha to the fourteenth times alpha to alpha to the fifteenth, then we claim that alpha was a primitive root here. It should be 1 here. If we do that, you're going to get 1 plus alpha to the cube times that by alpha. You're going to get alpha plus alpha to the fourth. Alpha to the fourth is 1 plus alpha. The alphas cancel. We get back 1. And so when you look at this list, we get 15 different elements. The only thing who's missing, of course, is 0, but we throw that into the field there. We get 15 different elements. There was no repetition. And exactly at the fifteenth moment, this thing repeats itself. And so we do see that alpha, in fact, is this primitive root for the field f16 that we observed right here. And this can be done for essentially every finite field. Clearly, the exact calculations will differ, but this is one of the cool things about finite fields. Multiplicatively, they are cyclic. They always have these primitive roots. One element will rule them all and end the darkness by them.