 So, let's continue our construction of finite field and we have been talking about the minimum polynomials. So, I wrote down several interesting properties, let me just write down the definition here. Before somebody give you an element of finite field with t power m elements, what is the minimum polynomial? There is a polynomial belonging to the finite field with just p elements. So, I have two things that are needed and beta should be 0 and then what degree is minimum. So, we wrote down a few properties last time, I think I didn't highlight probably the most important property and I did not even mention a few other important properties. So, let me do that a little bit more quickly. So, this is the first thing. The first property which is quite important which I think we missed out last time, I am not sure if I mentioned the property. The first property is that m beta of x has to be reduced. It is so important to write it down as a property. Why should it be reducible? How do you prove this? Yeah, exactly. So, that is the thing. So, you always assume the opposite statement and come up with the contradiction. Suppose you say it is reducible. Then what should happen? m beta of x should be able to be, you should be able to write it as some a of x times b of x. Both of them have degrees quickly less than the degree of m beta of x. Now, you put x equals beta and then on the left-hand side you will get 0. On the right-hand side it should also, but surely one of those polynomials should have beta as the root and that has to be lesser degree and that gives you a contradiction. So, that is the idea here. So, again prove this by contradiction. So, you assume m beta of x equals some a of x times b of x where degree of a of x and degree of b of x are quickly lesser than degree of m beta of x. Now, you see beta has to be a root of either a of x and b of x and that gives a contradiction. 0 equals beta times b of beta implies beta of 0 or b of beta of 0 which is a contradiction. What does it contradict? It contradicts the minimality of the degree of m beta that we assume. So, that is the end of the proof. So, that is the first thing. And another property that we saw, I mentioned that the original side completes the property. So, I said if you have an f of x times zpx and f of beta of 0, then what happens? m beta of x should divide f of x. So, what is important about this property is the probability. And this probability is very important. So, we already know a polynomial in zpx for which every beta and fpm will be a root. What is that polynomial? So, you can take f of x to be x power p power m minus x. What do we know about this polynomial? f of beta, we know f of beta is 0. So, that implies the minimal polynomial of any element of this finite field with p power m element has to divide x power p power m minus x. So, that is another, this probability is quite important. We use it quite extensively. So, let me ask you a couple of interesting questions based on these results that we have mentioned. So, we see that if you have an element of finite field and if you have a minimal polynomial for it, the coefficients from zp, then it is the minimal polynomial is irreducible. What about kind of a converse for this statement? Suppose I say there is an irreducible polynomial over zbx. And for that polynomial, I know the beta of a root, beta and fpm at p power m. I know the root for some irreducible polynomial. Will that mean that be the minimal polynomial for beta? Can I immediately say that will be the minimal polynomial? Sorry? Okay. So, suppose I say there is some irreducible polynomial in zpx for which beta is a root. Does it mean that that polynomial is the minimal polynomial for beta? It has to be true. So, that converse also has to be true. It is kind of invariant because beta is the root for this irreducible polynomial. If you had a minimal polynomial for beta, what should happen? That should derive this. The only way that polynomial can derive this is if it is itself. That is the definition of irreducible. So, it is also true. So, converse for here is also true. If at any point in time, you come across an irreducible polynomial for which you know that beta is the root, then you know that that has to be the minimal polynomial. Yes, that makes sense also. So, likewise here, there is kind of a converse for this. You know that if you have an element in your field and if you have a minimal polynomial for it, that minimal polynomial divides x bar p power m minus x. Remember, this is an irreducible polynomial. Okay. Suppose now I give you another irreducible polynomial which is a factor of x bar p power m minus x. Okay. Suppose there is an irreducible polynomial which is a factor of x bar p power m minus 1. I am doing kind of the reverse of this. So, I found the minimal polynomial and now I know that it divides this. But let us do the reverse. So, let us start with this x bar p power m minus x. Find an irreducible polynomial which is a factor of that. Okay. Is it true that there is an element in the field for which that is a minimal polynomial? Okay. So, these converse are quite important in understanding the structure of the field. Okay. So, that is why I am kind of trying to beat this out. So, think about it. Suppose I know factor is x bar p power m minus x. Since I am crazy here, somebody gives me a factoring algorithm. It may be a factor. I find an irreducible polynomial with coefficients from z p. Okay. That also is there. Now, do I know that there will be an element in the finite field f p power m for which this will be the irreducible. If this will be the minimal polynomial, can I say that? Okay. Will there be an element in this finite field which will do a root of this irreducible polynomial that I found? Okay. It has to be true. Remember why? What happens to x bar p power m minus x? It factors into linear factors over f p power m. Over f p power m, we can write it as x minus something times x minus something times x minus something times x minus something. Somebody else gives you an algorithm which factors it and gives you a factor over z p x. Now, that also has to respect this factorization, which means this irreducible polynomial has to have some element of the finite field as a root. Okay. And that tells you that there has to be a minimal polynomial. Okay. So, there is some kind of a link here between irreducible polynomials x bar p power m minus x and elements of the finite field. Okay. So, that link is very strong. Okay. It is not a very loose link. And you can go from one to the other back and forth very easily. Okay. So, that really ties down the field and its constructions and polynomials in some very nice way. Okay. So, that is what we are going to explore. We will start the examples and then we will slowly see some interesting nice final constructions which will give you a lot of thought. Okay. So, think about that for a while. It is important to know that the strong relationship between irreducible polynomials and finite field elements factors of x bar p power m minus x and all that. Okay. All right. So, let us see a couple of examples. We will begin with really simple examples. We will begin with, say, S2. It is a very example. Zero one. Okay. It is the first example I am going to give you. So, what do I want to check? If it is S2, I want to see how x bar m is x. So, x bar minus x and S2 is what? Same as x bar plus x, right? So, that is going to factor very easily, right? So, x times x plus 1 and that is what you are expected. Okay. So, in S2, it is 5 trivial. So, factorization is nothing much to look at in terms of minimal polynomials and all. So, this is a minimal polynomial of zero and this is a minimal polynomial of one. Okay. Right? Both are irreducible. So, all these things are very trivial. There is nothing much to look at. Okay. So, here I have to be careful about minus. Okay. So, minus is not the same as plus. I am looking at x bar 3 minus x. Okay. I want to factor this. I know it is going to factor as x times x minus 1 times x minus 2. Okay. It is a thing. It will seem nice. So, as the x minus 1 times x minus 2, you might think there should be an x term where it will not have any x term. Well, it is minus 3x, right? So, that is zero. Okay. So, you simply have x bar plus 2 or x bar minus 1, which comes out like this. Okay. Right? So, when I write down an abstract statement that x bar p power of minus x is factors into linear factors and f p power n, which might seem very trivial to you, but when you actually see it, it is very surprising. Okay. So, x bar 3 minus x factors like this. Okay. So, x bar normal of zero. This is the minimal power normal of 1 and this is the minimal power normal of 2. So, okay. Check this factorization. I mean, you have to be careful when you check this factorization. Remember, you are working in f3. So, it is a bit more trivial. It is not trivial. So, let us look at x bar in terms of completing the story. Okay. 0, 1, 2, 3, 4. Then, what should happen to x bar 5 minus 1 minus x. It has to be x times. So, let us try to factor it in a little bit differently. Okay. So, x times x bar 4 minus 1. So, what is x bar 4 minus 1? x bar minus 1 times x bar plus 1. Am I right? Okay. In any field, this is true. Right? x bar minus p square. We will learn all these formulas from young age. It is true in any field. It is a polynomial identity. Okay. So, x bar 4 minus 1 is x bar minus 1 times x bar plus 1. And then, what do you do? You have to factor these things further. So, what about, which are the elements in the square that give you 1? Okay. 4 and 1. Right? So, this will show the factor as x minus 1 times x minus 4. And this has to be the other two. x minus 2 times x minus 3. Okay? So, you can mention the factor that way. And it gives you another way of thinking about how this polynomial has to factor. So, you know the final answer has to be that way. But then, how to get to it is an interesting, interesting kind of. Okay? So, in F5, some crazy things like this happen. Okay? So, let's go to F4, which is of more interest to us. And I will use my polynomial construction for F4, which is 0, 1, alpha, 1, 4, 5, 4. And I'll just follow this. Okay? So, remember, it's mod 2. Right? Mod 2 addition to 0. Okay? So, now let's do x bar 4 minus x. Okay? So, the factoring will become a little bit more interesting. Okay? So, let's try what I did before. Let me pull x out. So, that's the first thing you can do. It's very easy. x bar 3 minus 1. Now, what do you do with 8 bar 3 minus b bar 3? Okay? It's also plus, right? So, the plus and minus are the same. And so, it's 2 is 0. Okay? So, 8 bar 3 plus b bar 3, we can again do. 8 plus b. And then, right? So, I already identified m0 of x. I already identified m1 of x. So, what has to happen with x square plus x plus 1? It has to be x plus alpha times x plus 1 plus alpha. So, remember, this is also alpha squared. It is just an easy notation for me. So, we have to work out x plus 1. And this has to be x plus alpha times alpha squared. So, this is basically the minimal polynomial of alpha. It's also the minimal polynomial of alpha squared. Okay? Which is not surprising. We saw all the properties. Then, f4 alpha and alpha squared has to have the same minimal polynomial. Right? So, we saw that this property, right? If you also copy, you don't get a different minimal polynomial. You get the same minimal polynomial. Okay? So, you've observed one interesting thing, right? We constructed this f4 with the reducible polynomial x square plus x plus 1. Okay? And what happens in f4? x square plus x plus 1 nicely factors into linear terms. Okay? So, x square plus x plus 1 has no roots in f2, but it has roots in f4. Not only that, it has all its roots in f4. Okay? So, all the roots you might want are in f4. Okay? So, such nice properties are true with polynomials in general over finite states. Okay? So, any polynomial you can find a lot in a field in which factor into linear terms. Okay? So, that's a nice result to have. Okay? So, that's the f4. And let's do a few more examples like those. Maybe a little bit differently just to give you a feel for how this might work. Okay? So, let me take f8. Okay? So, it might be a little bit uncomfortable with this. I'll just pull out the final answer very, very quickly and we'll see how it goes. Okay? So, let's say that I don't think that's true here is alpha plus 3 is not going to matter. Alpha plus 7 is 1. And we'll take alpha plus 3 to be alpha square plus 1. Alpha plus 1. Okay? So, this is a finite state. So, remember how am I constricting this? I'm using basically power of x to be. Power of alpha to be alpha plus 3 plus alpha plus 1. Okay? All right? And I know, so remember what is f8? f8 is set of all polynomials in alpha of degree less than or equal to 2. Okay? So, I can write it out that way. But I know that there is a primitive element in this field. And I know that in that field, alpha is a primitive element. Okay? So, I can state like that. Alpha itself will be a primitive element. Okay? So, I know that also. So, I know that this will be 2. Okay? So, if you don't believe me, you can try this. Okay? 0, 1, alpha, alpha square, what is alpha square 3? Alpha square plus 1. What is alpha square 4? Alpha square plus alpha. What is alpha square 4? Alpha square plus alpha plus 1. What is alpha square 6? Alpha square plus 1. And then alpha square plus 7 will be 1. Okay? And that gives you all the fields. All the field elements. So, alpha will be a primitive element if I do this. Okay? So, this is something that's true in a field. Okay? So, now let's try to look at x star 8 plus x and see how it works out. Okay? So, factoring is a little bit more painful here. So, usually you have to do it in some other way. Okay? So, let's see. Okay? So, some properties, I know the minimum problems are easy to guess now. So, it's a 0. What is n0 of x? x itself. What is n1x? x plus 1. So, these two things are the easiest to write down. Now, I have all the balsas to deal with. Right? And what do you know? I know alpha, alpha square and alpha per square. They're all roots of the minimal problem of alpha. Right? So, I have a minimal problem of alpha. Right? I'll have some minimal problem of alpha. I know that alpha, alpha square and alpha per square are all roots of the minimal problem of alpha. So, let's try this exercise. Okay? So, this seems like a full body exercise to you. You'll see at the end of it you will get something interesting. Let's try this. Okay? I know that this product has to divide the minimal problem of alpha. Right? I know that. So, let's try this. Let's say alpha times x plus alpha square times x plus alpha per square. Okay? So, let me see. Okay? So, you have remember alpha per. So, we have the table in the corner. Let me just go back there. Copy it and paste it in the space. Okay? So, that's the table. So, we do what we want to explore plus alpha plus alpha square. What's alpha plus alpha square? Alpha per square. Right? And then alpha per three times x plus alpha per four. Okay? And then let's multiply this out. We have x per three plus then what? Alpha per four plus alpha per four. So, that will go away. Right? And then the x term is going to be alpha per eight plus alpha per three. What is alpha per eight? Same as alpha. So, alpha plus alpha per three. What is alpha plus alpha per three? One. Okay? So, it just directs. And then what's the constant term? Alpha per three times alpha per four which is alpha per seven which is one. Okay? So, that's, is that great or not? That's a great result. Why is that a great result? I found the minimal polynomial fault. Alpha. Okay? See, x plus alpha times x plus alpha square times x plus alpha per four has to divide the minimal polynomial as alpha. I know that. Okay? So, it has to be like that. But then what happens to the product itself? That itself belongs to, belongs to what? Z two x. Okay? That coefficients only zero and one. And then I know that that is irreducible also. So, clearly this has to be equal to the minimal polynomial of alpha which is also equal to minimal polynomial of alpha square. Which is also equal to the minimal polynomial of alpha per four. Okay? So, likewise if you look at the other elements. Okay, what are the other elements? That's remaining here. Alpha, alpha square and alpha per three are done. So, if you look at alpha per three, alpha per three, alpha per six and then alpha per four, alpha per five, they're all conjugates. Right? They all have the same minimal polynomial. Okay? So, if you do that multiplication also. Okay? So, it's a bit of work. So, I'm not going to do it for you. But if you do that, you'll see you'll get x per three plus x per plus one. Okay? And that will be the minimal polynomial of alpha per three. It will also be the minimal polynomial of alpha per six. It will also be the minimal polynomial of alpha. Okay? So, that's the nice thing about the pink. So, that's the thing. So, what will happen is if you take x per eight plus x and then you factor it. You will get, you'll get what? Expands x plus one times of alpha. So, one plus alpha per six. So, now, if I want to factor this, if I want to get through the alphas, so to speak. Okay? So, this factoring is nice, but then there is alpha. Okay? So, maybe I don't want alpha. How do I do that? Okay? So, I have to do the grouping, right? So, I group the x plus alpha per one, x plus alpha per four per one. What will I get? What will I get? X per three plus x plus one. Okay? And then I group the remaining terms. I group X per three plus X per four. Okay? So, that gives me the complete factoring of x per eight plus x in over z two itself. Okay? So, I mean, there's lots of results which we didn't prove. We just used here. I mean, these things are generally true also in general. But we'll come to that later. But for now, this is a good recipe we have for finding. Okay? So, if you go to the larger field, doing this is a bit more messy. So, I'm going to stop. Maybe I'll give you an example with f16 and then we'll stop. Okay? So, I think this is example five. So, if you see f16, I'll give you an answer. It will allow you to calculate it. Okay? So, here, if you see x per 16 affects the factor like this. Okay? So, at times, it's just going to give you the final answer for now. And then later on, we'll see how the food is done. Okay? X plus one will have x per four plus x per one. And then you will have x per four plus x per one. You'll have x per four plus x per one. Okay? So, this is how we factor. Okay? Okay? So, that's an astonishing result which is true. You might have seen the previous two, previous two factors. There is one common connection between the two, which is truly an astonishing result if you haven't seen it before. Hence out, when you factor x per p per m plus x, you will get every irreducible polynomial in zpx whose degree divides them. You will get that as the factor. Okay? So, that is the general rule here. We will prove that. It's not too hard to prove. But it's, it's quite difficult to prove. Okay? So, when you go, when you see if it's a part of this polynomial, that's why these things are all quite strongly linked. Okay? So, when you do x per p per m plus x in general, when you factor it into terms like this, k into terms in zpx, turns out every irreducible polynomial whose degree divides them will show up as a factor. So, you see here, x per p per m plus x per p per m plus x per p per m plus x per p per m. So, you see here, this is x per 2 per 4 plus x, right? Okay? So, every irreducible polynomial, every binary irreducible polynomial whose degree divides 4, what does it mean to say degree divides 4? Degree can be 1, 2 or 4. Every irreducible polynomial will show up as a factor. And those will be exactly the factors. There will be nothing. Okay? It's quite a surprise in this sense. So, you go back and see for x per 8 plus x. This is what? x per 2 per 3 plus x, right? Every irreducible polynomial whose degree divides 3, what is that? 1 and 3 has shown up here, binary irreducible polynomial. Okay? So, it's a powerful result which ties up these polynomials together with the finite fields. And the way the finite field comes in nicely to tie up everything together. Okay? Okay. So, we're going to try and prove some of these things. And some of the ideas may be a little abstract, but I think it's quite important to get to the bottom of it to see how all these things are related. It's a very nice way of seeing it. So, let's do that. So, the first thing we're going to see is the following result. Okay? So, let me say that by each other, a few abstract results is what we're going to see next. And these may not help you directly in terms of other temporal curves, but I think it's good to see them once just to know what they are and to get a feel for how these things work in algebra. Okay? So, the first result we'll see is this interesting guy. Suppose f is the extension of zp. Okay? Some finite extension. Okay? So, f is a finite field which is an extension of zp. What do you mean by extension of zp? It contains zp as a subfield, the proper subfield. So, that's an extension of zp. Okay? So, let's just say it that way. Okay? So, let's say it's an extension of zp that contains all roots of x-par p-par m. Okay? So, it contains all the roots of this. Okay? So, what do you mean by all the roots of this? How many roots does it have? That's p-par m roots. Okay? Counting multiplicity. So, that's p-par m roots. Okay? So, then it turns out then the roots of x-par p-par m and x-par m, the field, basically some x-par p-par m. Okay? Basically, the form of field would be p-par m elements. Okay? The roots form a field with p-par m elements. And, of course, obviously, the root of p-par m is contained in it. But this is really true. I don't have to really say it. So, basically, it contains all the roots of it. So, all the roots are in that. It's contained in the root of p. Okay? So, there are a lot of things to prove if you want to claim this result. The first thing you have to show is that no two roots of x-par p-par m minus x can repeat. So, x-par p-par m minus x cannot have repeated roots. Okay? So, there is a way to show this. Okay? So, that's a proof that we'll assume. So, the way you show it is, you have to formally define the derivative. Okay? So, you can define the derivative of polynomials. Whenever you have, and there will be a product rule for the derivative, right? So, f of x times g of x has a product rule. And the same thing will be holding for polynomials. I know there is no calculus here, but you can formally define the derivatives. What is the formal definition? So, say x-par m derivative is n times x-par n minus n. That is clearly well defined. That's no problem. Define formally. Define itself if the polynomial has repeated roots, then that root is also a root of the derivative. Okay? So, you use that idea and show that this polynomial and its derivative are relatively prime in any field. You can show that. It's not very high. So, once you show that, it cannot have repeated roots. So, once this f has all the roots of x-par p-par m minus x, it has all the roots of x-par p-par m minus x. Okay? So, that's the first step and we'll assume it. Okay? So, this is the root of x-par p-par m from a field with p-par m elements. Okay? So, this we have to prove by the field axioms. Okay? So, this means the first step we assume. So, we show roots of x-par p-par m minus x and x are basically 0 of the root. You know what is the root. That's all these things you know already. And then, maybe there are other roots. So, this will be, or maybe I don't need to even qualify as. Let me just say that alpha 1, alpha 2, 4 and 2, alpha p-par m minus 2. Okay? So, you can argue that the 0 of s will be a root of this. You can also argue that the 1 of s will be a root of that. Okay? So, other than that, it will have p-par m minus 2 and they'll all be distant. Okay? So, now we have to just show axiom after axiom for the field. Okay? So, what do you need for the fields? So, we have any two roots here. Alpha i plus alpha j is also a root. Okay? So, maybe just a better notation is to say alpha 1 is 0. Alpha 2 is 1. And then, go from 3 to 4 on this. Okay? So, I think that's a good notation. All the way to the top. Okay? So, now I have to show alpha i plus alpha j is also a root. So, that's easy to show. Alpha i plus alpha j goes to the top p-par m. Okay? Minus alpha i plus alpha j. Right? What will this be? So, I know this is a field that's characteristic p. Right? It contains zp. So, it's a characteristic p field. So, that will be alpha i plus alpha j raised to the top p-par m. If you alpha i top p-par m plus alpha j top p-par m. And then, we are subtracting this and this is clearly 0. Okay? So, the addition is satisfied. Okay? So, you take any two roots of this guy and then add them up. You also get the root of x-par p-par m minus x-in-f. Okay? So, you are not going to go outside of this thing. So, the closure for addition is satisfied. So, maybe the closure for multiplication is also to be satisfied. It's not very hard to show. Okay? Once again, you just use the same idea. You raise to the top p, each of them raise to the top p. You can do it when you replace the alpha i, alpha j. Okay? So, in the closure, you can show in the same way. All these things are not very hard. So, you just show closure for addition, closure for multiplication and inverse and so forth. Okay? Just based on these properties, you can show. All right? Think about it. I mean, inverse might be a little bit more twisted. You'll have to show for each element alpha i, alpha i raised to the top, alpha i squared, alpha i par 3. All of those things will also be roots. Okay? And then eventually it has to repeat. And then there it repeats, there will be an inverse. Okay? So, that's how you show the inverse. Okay? So, this is closure for addition. You show similarly closure for multiplication, inverse and so forth. Okay? So, there is some work involved there, but it's not very hard. You can show that when this can be done. Okay? So, that's the first result we have, a more abstract result. In any extension field, if you have roots for x par p par m minus x, okay? And then that f p m itself will be inside that. So, this is the first result that we're going to use. We'll keep that aside. Okay? So, we'll just remember this, x par p par m minus x. So, this is kind of like a converse, you know? So, the way to think about it is, if you have a field with p par m elements, the linear factors of x par p par m minus x are there in that field. The converse kind of thing is also true. In any field, if x par p par m minus x factors, those factors, those linear terms, those roots form the field at p par. Okay? So, that is also true. Both roots. All right. So, the next result is, so the next result is the following. The thing about how I want to first this, so that it's easy to do well. Okay. So, let me do a more definite result. So, this is kind of like a property of primitive polynomials, minimal polynomials. Okay? So, this is an interesting property of minimal polynomials. And here we'll use a result, we'll use a proof technique which will give you an idea of how many of these proofs are going to work. Okay? So, let me use the property. Okay? So, the property is as follows. Suppose beta belonging to f p par m is primitive. Okay? So, what do I mean by primitive? It generates the entire f p par m in multiplicative fashion. It's all there is p par m minus 1. Okay? So, it's primitive. Degree of m beta of x equals, that's the result. It's, so suppose we're going to use a technique which will be a little bit surprising and a bit confusing. You can see it's the first time but it's a very powerful technique. Okay? So, it's as follows. Okay? So, beta, okay. So, beta is an element of f p par m. You will have, suppose, suppose degree of m beta of x equals p. It's sum b. So, m beta of x is a minimal polynomial for beta, n f p par m. And it has some degree b. Okay? I know that this guy is actually a irreducible polynomial over, over, over z p par m. You know it's an irreducible polynomial over z p. Okay? So, what do we want to say now? Okay? So, now, if I, okay, so the, so the wedding can be done real slick fast. I just want to make sure that I don't get over something. Okay? So, now what I can do is I can construct a field with power steps, power equals m beta of x. So, what I'm going to take, is I'm going to take, I'm going to take the polynomial. So, I have a field construction method, right? So, the problem is, okay. So, how will you say d divides m? Yeah, yeah. So, I haven't proved that. I'm going to prove it. I'll prove it. Eventually that is also proved. Yeah. So, but I have to show z equal to m. It's a bit more twisted here. We'll come to it. There are multiple ways of proving it. The reason I'm proving it this way is it gives you a flavor of these proofs and it's a nice way of thinking about it. Okay? So, you're going to construct a field with power steps equals m beta of x. So, some other field, let's say. Okay? So, it seems like it's some other field. Let's say we construct this field. Okay? So, how many elements? Okay? So, the first thing we know is d is less than or equal to m. Okay? So, that's true. So, now I want to show that d has to be greater than or equal to m. And how do I do that? So, that's the thing which I'm, that's the thing I'm struggling with here. It's already naturally quickly finished this. It looks like the... So, you construct a field with power steps equals m beta of x. And that's going to have how many elements? Okay? So, let's say this field is f. Okay? Size of f is going to be equal to p power b. Right? So, it's going to have p power b elements. Okay? And there's a way to argue that p power b will have to be at least greater than or equal to p power m because beta is primitive. So, I'm just getting confused by that. I don't know why. Okay? So, here is... Okay? So, this field... Okay? So, beta is the root of this m beta of x. So, you have to say that this field will contain beta. Okay? So, that's the main idea. So, this field will be a subfield of this original field. And it will contain beta. Okay? So, that's the main notion here. So, you can think of f as a subfield of f beta m. And it contains beta. Okay? So, that's a crucial idea. I think it's... I hope it's clear. I don't know if it's very clear or if it's confusing to you. So, you start with the minimal polynomial of beta. Okay? And you construct the... Okay? So, I guess, I mean, we can just... We can do it in so many different ways, but I think this is what's... So, I think it would be an interesting group where it looks like it's running into trouble and putting it down very clearly. So, basically, you take this minimal polynomial and you construct a field. You're going to get a field with beta d elements. And this field you can think of as being contained in f p param. And it will also have beta in it. Okay? So, from there, you have to argue that it will have p power d has to be greater than i equal to p power m. Okay? So, that's the crucial idea, but I'm thinking why I'm struggling so hard with this. So, here... So, size of f is p power d. I know that beta power d power d minus 1 is going to be equal to 1. Okay? And from there, since beta is primitive, you should be able to... d has to be greater than i equal to m. Okay? So, in that, it shows the proof that hopefully this logic was maybe not very confusing. Okay? So, I don't know. I mean, this is not very clear. I'll come back and try to fix it a little bit later. This is how the proof goes. Okay? So, you have to think about this a little bit. So, if you have a primitive element in this field, it will have a minimal polynomial. And then you construct a field which has that minimal polynomial as the reducible polynomial power x. Okay? So, you have to argue that every element in this new field will also... can be also thought of as being an fp param. Okay? So, if you think about it, it looks the same thing. So, you take... you take a root of this minimal polynomial associated with the beta that you have in your fp param. And then you look at all the polynomials in beta of degree less than or equal to d. It will have the same... if you can map it one-to-one to an element in fp param. Okay? So, it's clear. So, think about this. So, you have... So, you will have... you will have the same roots as an fp param. Okay? So, you can map this field as... into this fp param as a sub-field. Okay? And then you say beta has to belong to the sub-field in fp param. And then it has to be p power d minus 1. And that beta is primitive in that ordinal way. So, d has to be at least as greater than or greater than or greater than or greater than. Okay? So, that's how we prove it. It's a bit abstract. And I know if you haven't seen it, it may not be very condensed here. And then you go through the steps one-to-one. That's the idea. Okay? So, similar to the way in which we showed that zp is contained in this field as... Okay? So, we started with one and then we showed that there's an isomorphic copy of zp inside this fp. So, you do the same thing here. So, you construct the field with the minimal polynomial of beta. And then you do a one-to-one mapping from this field to the polynomials involving beta in the ordinal s. Okay? And then that has to have at least p power n elements. Here you have p power d elements. So, d has to be greater than or greater than. Okay? Is that good enough? Or do you want me to go into the dirty details here? Okay? Okay. Okay. So, when I say contains, it should be an isomorphic copy. So, let me maybe go through it. Okay? So, let's say power fx is said to be mb power fx. Okay? So, you got beta from fp power n. Okay? And you have power fx to be mb power fx. So, maybe power fx is something like, let's say 0 plus 1x plus 1 a d minus 1 x power d minus 1. So, that's our degree d, no? So, a d x d. Okay? Let's see. We got power fx. Okay? So, if I construct a field, if I construct a field as the first, it would be, okay? So, maybe I should use some other notation here. Sorry for that. Okay? So, if I construct a field, okay? So, maybe I should use alpha here. Right? And then I'm going to say pi of alpha of 0. Okay? So, this is the field that I constructed here. Right? So, then I have to kind of argue that if I take, see, alpha as such that, pi of alpha is 0. Right? Okay? So, I have to kind of map this alpha in s to the beta in fp power n. Okay? So, this is my kind of isomorphism. Okay? So, let's say this is isomorphism. I'm talking about. Okay? So, you take the alpha in s and map it to the beta in fp power n. So, what does that mean? So, if I do alpha squared in s, what will I map it to? Beta squared in fp power n. Okay? So, in general, every polynomial here is 0 plus o1 alpha plus o1 2. So, a b minus 1 alpha b minus 1 will be mapped throughout the same a0 plus 1, but then every alpha will become beta. Okay? Plus 1 to a b minus 1 beta power b minus 1. Okay? So, this will be in fp power n. So, this way you will have an isomorphic copy of s in fp power n. Okay? Is that okay? All right? So, that beta specific a is in fp power n. Alpha is also in f. Okay? But I know beta is a primitive element of fp power n. Okay? So, in this sub c in fp power n, in the isomorphic copy in fp power n, I have at least 3 power n elements. Right? So, likewise, in this side also I should have 3 power n elements. Okay? So, it has to be at least as because that is the idea. Okay? So, first you make this isomorphism. It's a bit confusing. It's very formal. But please, please make it clear. So, the question is clear. So, you're saying beta is contained in f. The reason why I say it's not contained, it's just there's an isomorphic copy of s inside fp power n and beta will be there. That's the idea. The reason, the crucial idea is that f also has 0. Right? So, you have f beta versus 0. So, this thing is addition will also, multiplication will also properly okay. Okay? So, if I take two elements here and multiply, I'm going to set pi of also to be 0. But if I do here in fp power n, what will happen? You'll get some term that I know pi of beta is 0 in fp power n also. So, I can do the modular reduction. I will get something inside here. So, it's a perfect isomorphism. It's a nice isomorphism that way. Okay? So, what do you know about this guy? Size is greater than or equal to p power n. Okay? But at least p power n elements of this form. Okay? Because beta is there. So, beta square is there, beta power 3 is there. So, until beta power p power n minus 1 is there and also clearly 0 is also there. Okay? So, p power n, there is a variable in this. Okay? So, that means this size also is greater than or equal to p power n. In fact, the size you can say is equal to. That doesn't matter. Okay? So, it's greater than or equal to p power n. Here also, it's p power n. So, what is the size of s? p power d. So, p power d has to be greater than or equal to p power n and d is greater than or equal to p power n. Okay? So, it's a bit of an abstract proof and the reason why I did it is to just show you an abstract proof. Okay? So, how these abstract proofs work, you see? I was seeing these things before. It seems that the crazy, the next makes a perfect sense. You think about it very carefully. So, when people use words like contains beta, you should be very careful. So, usually people don't use that word very loosely in algebra. They'll be very careful about what they mean. But since this is true, we can just loosely say it contains beta. Okay? All right? So, that's the other powerful statements in this proof which you should remember. Okay? So, we said elements of s p power n are roots of x power p power n minus x. Okay? Now, you also have minimal polynomials and so that's also some strange results that prove. If you have an irreducible polynomial, let's say binary irreducible polynomial and if you have a root for it in some field, okay? The field that you construct with this irreducible polynomial is also fully contained in that field. Okay? So, that's the other powerful statement here. Okay? And that's used again and again in various abstract proofs in finite fields. Okay? So, if you have a minimal polynomial, okay? So, some element in some finite field, okay? Some big field. Then you construct a field with that minimal polynomial as your irreducible polynomial. That entire field is contained inside this abstract field. And that is a powerful notion that is used in many abstract proofs. Okay? So, we use that over and over again to prove a lot of simple statements. And, I mean, usually in my Erecent R-coding class, I keep such abstractness to a minimum because it's mostly not needed and strictly needed in Erecent R-coding. But anyway, for many of you, this might be your first real course in algebra. So, it might be a good idea to see some of these ideas and see how some structure which is very abstract is broken down into something very nice and specific. Okay? So, that's an opportunity you don't get too much. Okay? So, we'll stop here for today. It's been a very abstract notion. Okay? So, what I'm going to do real quick maybe in the next class is to take a specific example of this and show you how it works. Okay? So, I'm going to take a very simple example, maybe like F8 or something, and we'll construct a field, find a minimal polynomial, and then construct some other field with it, then I'll show you how both of them are really pretty much the same. We'll do that later. Yeah. So, I guess here it is more like homomorphism or something here. Okay? Yeah?