 So, we have been looking at minimum polynomials and we saw this interesting little property that if you have an element of finite field would be 3 power m elements. So, we had a formula for the minimum polynomial basically for product over say gamma n c beta x minus gamma and what is c beta basically the set of conjugates of beta and that is the beta beta power p beta power p squared. So, I will say finally it has to end somewhere so it probably ends up p power d and then you should have the beta power maybe d minus 1. So, you should have the beta power p power d equals p beta should be the smallest. So, this is the set of conjugates of beta and so you will see so this is a powerful formula. So, it gives you an explicit formula for the minimum polynomial in terms of the elements of the finite field. If you have beta you have all you have to do is just keep exponentiating it to 3 you get all the conjugates and then take the polynomial with those conjugates as roots it is the minimal polynomial. So, in particular this k will belong to z p x which is not a very obvious maybe these two fans can be turned off it is about to help I am not really communicating from these fans I have to think about it. So, this is a kind of a non-trader statement and that is something which is useful for us. So, let us look at this set of conjugates and see what we can do if we have the fans on maybe this is on. So, the first thing first question you might want to ask is if you have an element beta and f p power m how many conjugates will it have? So, the answer lies in this little equation here. So, what does this imply? Beta power p power d minus 1 equals 1. So, this is something that has to happen for any element. So, for what kind of d can this happen is the question that is at one point. The other thing is you know that f p power m also has a primitive element. So, beta if it is an arbitrary element of f p power m it can be written as the primitive element power something. So, those two together somehow tie down these conjugates very nicely for us and we will look at that a little bit more in detail. So, it is interesting to list out all the conjugates quickly, list out all the minimal polynomials quickly and that is something that we will do. So, let us do that next. So, if you start with f p power m in 0, 1, alpha, alpha squared, 1 to alpha power p power 1 minus 2 is alpha of a primitive element. Any beta is going to be some alpha power i. So, we will ask the question what are the conjugates? So, beta so it is going to be alpha power i then what? Alpha power i times p then what? Alpha power i times p squared. So, 1 to some alpha power i times p power. In our notation we have been doing d minus 1. And what is the property? The final property that we have is alpha power i p power b equals alpha power i. So, this becomes basically the same as alpha power i times p power b minus 1 equals 1. And this should be and then d should be the smallest integer for which this can happen. So, there is also something like that which needs to be true. So, if this has to happen then what do we know? p power m minus 1 has to be right. Thanks p power b minus 1. So, that has to happen. If you have alpha power some element being some power being equal to 1 then the order of alpha which is actually p power m minus 1 has to divide i times p power d minus 1. So, this will give you some control over it. So, in fact it also turns out that d will have to divide m. So, that seems like a slightly counterintuitive kind of answer here. So, this is also another way in which you can view this. So, let us come here you can slowly conclude that d has to divide m. So, think about how you might want to prove it. I am not going to skip that here for now. So, this has to happen. Anyway this is not a very real fact. Think about it for a while. Yes. No, no, no. There is an i here. So, it won't happen. So, d will have to divide m. It is the previous class. I am sorry. I have beta power p power d minus 1. Yeah. And beta power p power m minus 1 is also 1 plus beta belongs to b power m. Okay. Yeah, you can prove it in various ways. But anyway, so this has to happen. There is this various other way. So, this equation also is important to remember. So, you should see that d is the smallest integer for which p power m minus 1 divides i times p power d minus 1. This will help you in the computation. But another way to look at it if you go through all our results and think about it carefully, you can show that this can happen only if d divides m. So, that is another result that you can look at. Anyway, so that is not too critical for us. So, this is a useful factor. So, the number of conjugates of any element of a field will be what? It is controlled by m. Okay. So, depending on m, only the divisors of m can address the number of conjugates. So, that is an important factor to remember. Okay. So, you can prove this. All right. So, let's take a few examples and see how these conjugates work out. And then I will tell you some general facts. Okay. So, let's take some very simple examples. The simplest example is what for you. Okay. So, here we have 0, 1 out of the square root of the weight of alpha plus 6. Okay. And alpha plus 7 is 1. It turns out the other equation was not too relevant. So, maybe one of you can take alpha plus 3 if you want to look at it. Okay. So, let's try to look at C0. Okay. What are the conjugates? Zero. Okay. Just one. Okay. Zero. So, you raise zero to any power. You just keep getting zero. Okay. So, what are the conjugates? One. Just one. Okay. What are the conjugates of alpha? Okay. So, here you have to do some work. You can have alpha. And what is P in this case? 2, right? So, we have eight elements. So, P is 2. So, we will do alpha square. And then, alpha plus 4. And then, that's it. Alpha plus 8 is the same as alpha. So, you start. Okay. And see, so now the next conjugates, you must think maybe after I don't see alpha squared. So, what will happen if you write the conjugates of C alpha squared? You will get the same thing as C alpha. Okay. So, that's something that's nice about these things. So, in fact, all these three are the same. Okay. So, once you have a set of conjugates, they're all conjugates of each other. You can't go. So, they're repeatedly raising to the power P. Okay. And then, you're doing the same thing. So, that's to work out that. Okay. So, the next thing, next element to look at is C alpha power 3. And that would have alpha power 3, alpha power 6, and then alpha power 12. What is alpha power 12? It's alpha power 5. Okay. And then, this is the same as. Okay. So, that's it. All right. So, now, if you want to find minimal polynomials, to each set of conjugates, you can associate a minimal polynomial. Right. So, the conjugates of C0 to C0, you associate polynomial X for C1, X plus 1. Okay. You have to do the computation. And if your assume alpha power 3 is 1 plus alpha, you're going to get X power 3 plus X plus 1. Okay. So, that's what's happening here. You're going to get X power 3 plus X plus 1. Okay. What will you get for C alpha power 3? Okay. So, now, you can use all the sum total of the previous results that we used. You know, it is going to be a minimal, the irreducible polynomial of degree 3. There are only two of them. One of them we already have. The other one has to appear as the minimal polynomial corresponding to the other element. So, that would be X power 3 plus X power 3. So, now, instead of this rule, we said, say, alpha power 3 is 1 plus alpha squared. What would have happened? C alpha will correspond to X power 3 plus X power plus 1. And then, the other one, C alpha power 3 will correspond to X power 3 plus X plus 1. So, anyway, both fields are isomorphic. We know that. It's not a big deal. But then, just the structure and the form will slightly differ in terms of the isomorphism. Okay. So, you'll have to think of any other thing. All right. So, what people do, see, if you look at the list of conjugates here, this alpha has just really carries no information. You know, it's a primitive element. All the information is only in the powers. Okay. So, when people write conjugates, they'll usually drop the alpha. Okay. So, when they write conjugates, this element is kind of irrelevant. Okay. We don't really care about what it's the conjugate of. Because the power there is complicated, right? How do you write alpha power something as 0? You have to say minus infinity and all that. It's just not very nice. So, that's usually dropped. Nobody cares about the conjugates as 0. For 1, you write alpha power 0. Okay. So, instead of 1, you write alpha power 0. Okay. So, it's very common to write the conjugates as fellows. You write the conjugates as 1 as 0. The conjugates of C alpha. Okay. So, let me write 1, 1, 1. C 1, 2, 4. And then, for alpha power 3, you write 3, 5, 6. Okay. So, drop the alpha. Okay. And, in fact, even this 1, right, is the C alpha power 0. Even this alpha is not really carrying any information on the C alpha, C alpha power 0. So, instead of this, people usually write C 0. And then, this would be C 1 and this would be C 3. Okay. So, these are all notations. And there are some conflict as such, right. So, from one notation to the other. C 1 here is very different from the C 1 here. Okay. So, remember that confusion in notation. Instead of alpha power 1, it's simply dropping the alpha and saying C 1. Okay. So, this is very, very common. And it's done all the time. Okay. So, let's look at another example of F16. For a blue 1, alpha 4, all the way to alpha power 14. And now, alpha power 15 is gone. And then, let's say alpha power 4 or something like that. Okay. So, let's just say that. If we do that, then, what would be C 0? C 0, we're always going to do 0. There's no problem. Okay. Remember, C 0, C alpha power 0 is actually the conjugates of 1, which is just 1 itself, which is alpha power 0. Okay. So, you realize it like that. What about C 1? Okay. So, instead of thinking of this 1 as actually being C alpha and raising alpha to the power 2 and all that, I can simply take this 1 and multiply that 1 by 2 and take modulo 16. Okay. It's the same as taking alpha power that and raising it to the power 2 and then reducing it modulo. Okay. Simply take the 1 and multiply by 2. Okay. So, I have 1. Then, I multiply 1 by 2. Okay. 2. So, what about this bottom thing? Any ideas? Is there like a view? H bar? Okay. So, 1, 2 and then you multiply 2 by 2 again, you get 4. You multiply 4 by 2 again. What do you get? 8. Okay. And then you multiply 8 by 2 again. 1. You get 16. That modulo 15, it goes back to 1. Okay. So, the way to read this is basically this is C alpha, conjugates of alpha. Alpha power 1 and these are alpha power 1, alpha power 2, this is alpha power 4, this is alpha power 8. Okay. And then the next one goes back to alpha again, which is the same. So, you write this in this way. Okay. The next thing to look at is C3. C3 would be 3, 6, 12, 9, 18, but 18 is the same as 3. So, modulo 15. Okay. And you basically think of this as multiplication by 2 modulo 15. Okay. Okay. All right. So, you multiply by 2, do modulo. Okay. Okay. So, when what is left? So, what is left? 5 is left. Right. So, C5. 5, then multiply by 2, you get 10. Multiply by 2, you get 20, which is back to 5 again. Okay. And when you go C7, I'll write it down. You can check it. No, it won't come in this order. There will be this. Okay. So, after 7, you'll have 14. After 14, you'll have 13. So, 13 you'll have 11. After 11, you'll have 7 again. So, you stop. Okay. So, set sets have a name and they are called cyclotomic cossets. Okay. So, that's the name for it. If you think of them just in terms of numbers, these are called cyclotomic cossets under multiplication by 2, modulo 15. Okay. So, that's the way to talk of these sets. Okay. So, there are how many sets? There are five sets sets. They partition the entire set 0 to 14. Okay. What do we mean by partition? The union of all these things makes the set 0 to 14. No two of these sets will have anything in common. So, the cyclotomic cossets under multiplication by 2, modulo 15, partition the set 0 to 14. Okay. Right. So, this is what happens. There is a general version of this. Okay. So, before that, let's just think of this set once again. Okay. So, what about minimal parameters? C 0 is minimal parameter for 1 which is just x plus 1. Okay. What about C 1? Minimal parameter for alpha, alpha square, alpha 4 and alpha 8. The minimal parameter corresponding to C 1 would be x power 4 plus x plus 1. So, let's do that. You could do x power 4 plus x plus 1. Okay. What about 510? x power 4 plus x plus 1. It has to be, it has to be an irreducible polynomial of degree 2. Right. There's nothing you can do. It has to be x power 4 plus x plus 1. Okay. See, remember the number of elements in the cyclotomic cosset tells you something about the degree of the minimal polynomial. Okay. And you know the degree will divide m, the degree s, there also will divide. So, there is a showing d divides m, the way I wrote it down. Okay. So, what about 7, 11, 13, 13? Okay. So, a useful trick to remember for small cases is if you have x power 4 plus x plus 1 as the minimal polynomial, there is a reflection, right? What is the reflection? x power 4 plus x power 3 plus 1. What will be its roots? It will be 1 by the roots of these things. So, what is 1 by 1? It has to be 14, right? See, alpha is 1 by alpha of alpha power minus 1. Model of 15, it becomes alpha power 14. Okay. So, the inverse of this will be this guy. So, we used to have a relationship. They will give you mirror image polynomials as the minimal polynomials. Okay. So, you know immediately that this will correspond to x power 4 plus x power 3 plus 1. Okay. What about the c3? Yeah. So, the only meaning missing minimal polynomial of degree 4 was this one. And notice here, what is its reflection? It is itself. Okay. So, you won't get anything new. Okay. So, it will be nicely in a very proper way. I mean, of course, we knew that it has to work out like this. Nothing else can happen. Okay. All right. So, let me once again read this definition of cyclic atomic process. Okay. So, the idea is you take the fact 0, 1. So, let me say this here. You have to say under multiplication by prime p modulo p power m minus 1. Okay. So, that's, that always goes along with cyclic atomic process. The parameters of cyclic atomic process are p and m. p is a prime. Then modulo p power m minus 1. Okay. So, the idea is you take any element i from the set. The cyclic atomic process of i is basically i, ip, ip square, so on. But eventually it has to stop somewhere, right? So, to stop somewhere, let's say i p power p minus 1. So, i p power p equal by modulo p is minus 1 step. So, i p power p equals i modulo p power minus 1. Okay. So, it has a lot of properties. This b will be such that b divides m. All those things you can show. Okay. So, it's not very hard to show these things. All right. So, an interesting thing is the CIs are also related to conjugates. Right? C i is basically set of conjugates of alpha, par i. Okay. So, you can show this being conjugate is what's called an equivalence relation. Okay. So, that also will come very easily. So, two elements belonging to the same cyclic atomic process will be something called an equivalence relation. So, what's nice about an equivalence relation is you can show it has to partition your original space. If you have a set and you can demand an equivalence relation on it, then you'll also get a partition of the original set. Okay. So, this is something I'm not going to prove, but I'll state this as a fact. So, it's easy to see by construction that there's interior also. What is this fact? The cyclic atomic process from a partition of the set is zero running to zero. Okay. So, this is the fact which comes through this equivalence relation stuff. It's not very hard to show. Okay. So, what do I mean by partitioning? If you take union over i, C i, what will you get? Okay. So, let me call this set as s. You'll get s. Okay. So, that's very probably very easy to show. It's not very hard. The elements are long there. And then what is the other thing? C i intersect C j is what? So, it's null set. Whenever i is not equal to C j. Oh, no, no, no. Okay. So, let me write it this way. So, this is dangerous. It's already a null set. Okay. If C i is not equal to C j. What is the best way of doing it? What is the best way of doing it? Okay. So, you either have C i equal C j for i and j or C i intersect C j is a null set. You can't have anything in between. You cannot have C i and C j overlapping partially. That's what this statement is supposed to do. Okay. I didn't write it carefully. So, let me write it carefully. Okay. So, i j either C i equal C j or what? C i intersect C j is the null set. Okay. One of these two things is true. Okay. So, if I reduce the number of i's to form a proper partition, then I can make the statement. But anyway, so, they form a partition. And then you either have C i equal C j or C i intersect C j is a null set. It's clear, right? Okay. So, this is in general. So, let's do maybe a couple of more examples. And then we'll try to form this point. It's quite important to see that. Okay. So, see those facts. I think I didn't state it. So, if alpha is a primitive element. Okay. So, let me just also write a few more things. A few more facts about this primitive elements and all that. Okay. So, let's just get rid of these other facts. Okay. If you have alpha belonging to F t, then being primitive. Okay. So, what i is alpha per i also primitive. Okay. So, that's an important question. So, it's good to know that. Okay. So, if you have alpha being primitive. So, alpha per i for i equals 0 to p power m minus 2 generates the entire non-zero sets. That's fine. For what i will you have alpha per i also being primitive. So, the question basically is to figure out all of us. Alpha per i. Okay. So, what's the formula for it? What is the formula? It's very easy. It doesn't be equal to i comma. Is that correct? Does that make sense? Okay. Right. So, i and p power m minus 1 will have some common factor. So, say for instance, if it has some common factor. Then you have to divide p power m minus 1. That's a common factor to get the order of alpha per i. Okay. So, this is a fact which is in general true. It's alpha. So, in fact, instead of p power m minus 1, you can replace with order of alpha. Okay. So, you can replace with order of alpha. Okay. So, this is a general fact. Alpha has order some m. And you raise alpha to the power i. The order of alpha power i is going to be that order divided by the gcd of gcd. Okay. Think about it for a while. You see, it has to be this way. You can prove this. It's not hard to prove. If you show this, we'll divide that. That will divide this. And you get the answer. Okay. So, it's not very hard. Okay. To prove this, you have to show that LHS will divide the RHS. And RHS will also divide that. Okay. So, once you show that, it's done. That's the way to prove it. But it's a fact which is easy to see. Okay. The smallest power that will take alpha power i to 1 is what? p power m minus 1. Right. So, you need p power m minus 1 power. And then if you already have i, what is common between i and p power m minus 1? You can divide that. Okay. So, whatever is not common, you have to have that. I mean, otherwise, it will not work. So, that's the... It's kind of intuitive in a way. But you can prove it rigorously if you like. Okay. So, that's the idea. So, given this fact, how do you answer that question? When is alpha power i also primitive? GCD of i and p power m minus 1 has to be 1. Which means it has to be relatively prime with p power m minus 1. Okay. So, this is true whenever... Okay. So, this is true. Alpha power i is primitive. This one only is. But i and p power m minus 1 are relatively prime or GCD of i, p power m minus 1. Okay. So, this is the fact which is good to remember. Okay. So, we go back to this example. We go back to this example here. So, what i will alpha power i be primitive? Okay. So, everything in C1, right? What else? C7. Okay. So, it seems to be going in... In... According to cyclodynamic process, right? According to conjugate. So, if an element is primitive, its conjugate also will be primitive, right? So, because... How do you get from one element to its conjugate? You take this i and multiply by p. p and p power m minus 1 can have no common factor. Right? It's not possible for p and p power m minus 1 to have common factor. So, clearly if i and GCD of i and p power m minus 1 is 1, then i p, p power m minus 1 will also be 1. Okay. So, it will go on and on like that. Okay. So, primitive elements occur as conjugates. If one element is primitive, all its conjugates are also primitive. Okay. So, that's another nice fact to remember. Okay. All conjugates... So, this implies conjugates of alpha power i are also primitive. Okay. So, if alpha power i is primitive, then the cyclotomic concept of i will have exactly how many elements? If alpha power i is primitive, C i, the cyclotomic concept of i has to have how many elements? It has to have m elements. Right? So, that is for that. Right? So, that also is another thing that is employed. Okay? So, alpha power i is primitive. Okay? This is only in one direction. Okay? Remember that C i has equal to m. The other direction is not true. Okay? You can try to write it down and try to actually prove it rigorously. You will see the other way you can never show. In fact, there is a counter example. What is the counter example? It's right here, right? Right here. The C 3 and X 16 is a counter example. Okay? Alpha power i is primitive. Okay? I order this five. Right? GCD of 3 and 15. Okay? So, you can divide that by... So, you immediately get the order of the five. But then, it has four elements. Basically, the opposite is not true. Okay? So, that is something here. Okay? So, another thing that will be true is... I'll conjugate the alpha power i is primitive. Another fact which you can quickly show. Alpha power i is primitive. It can only go to alpha power minus i is also primitive. So, that's another thing. Does that make sense? Alpha power i is primitive. Minus i. It should have the same order. You can use the same formula as you did before. It has to have the same order. You can show that. Alpha power i is primitive. Minus i. So, you can also use the polynomial approach, if you like. Okay? So, alpha power minus i. The root minimal polynomial will be there. The middle image of the minimal polynomial for alpha power i. And that is primitive. That gives you a primitive element. It's also primitive. It cannot be primitive. So, those are various ways of doing it. So, I think that's... That should be all the facts we need about this stuff here. Okay? So, let's take one complicated looking example. Okay? And we'll try to figure it out. I'm going to say it's... If you turn out finally that the example is quite easy. When it starts out, since it might not seem like a starting example to you. So, let's look at f128. Okay? So, let's try to form cyclatomic cosets. Okay? We'll try to do that. It's not a good deal. C0 is going to be... C0 is nothing much to do that. What will be C1? C2, 4, 8, 16, 42, 64, right? Okay? So, let's try to form C2. What will be C2? Yeah. C2 is not so interesting. C3. Okay? 3, 6, 12, 24, 48, 96. And then what do you do? It's 192. You have to subtract 128. I'm going to get 60. 65, right? So, 127. Multiplication by 2, modular. 127. Okay? So, what is 65? That takes you back to 3. Okay? So, you can do this. Keep on doing. Then I'm going to make an argument for it. Okay? So, what kind of a number is 127? No. What kind of number is 7? I'm sorry. See, you know, this is 2.7, right? 7 is the prime number. Okay? So, which means what? Every CI should have size over 1 or 7. It can't have anything else. Okay? So, you only have the one gate which is 1. You can't have more than one gate which has the same size 1. Why? Because there's only one reduce of the terminal of degree 1. Okay? So, you can't have the same gate repeating. Okay? So, what should happen? Everything else should have size exactly equal to 7. Okay? Which means what? If I ask you, what is the degree of the minimal polynomial of alpha par i at x? What is your answer? So, if i is 0, then it is 1. If i is not 0, it's 7. If 7 is, if i is not 0. Okay? So, that's an interesting observation. Okay? The other observation is I think 127 is also prime. Okay? So, what happens if 127 is prime? So, go back to the formulas of alpha par i. Okay? So, when is alpha par i? What is the order of alpha par i? Beta r minus 1, 127. Data by GCD of i comma 127. What will be GCD of i comma 127? Always d1. Okay? So, every element in f128 is primitive except for 0 and 1. Okay? So, that's another thing which is nice about this kind of thing. So, f128 that way, even though it looks a bit scary, it's very, very easy. Okay? So, it's not too hard to think of what will happen in f128. Okay? So, this kind of a question is important for us. Okay? So, you will see that in code construction, that shows up a lot. Okay? Oftentimes, we may not care about the exact minimal polynomial. It might be needed in some steps, but it may not be too crucial. But what you will care about is the degree of the minimal polynomial. You might want to know that after that time for whatever reason. So, to find the degree of the polynomial, do you really need to know the structure of the finite field? That's the question that's important. Do you need to know the structure of the finite field? No. What we need is only the cyclotomic cross-section and that is simply multiplication by 2 modulo 127. So, you only need arithmetic, which is very simple. And you can find the degree of the minimal polynomial in that. Okay? So, an interesting exercise is to repeat the same thing for us, 256. So, this is a field which shows up often in practice. Can you just write what's so nice about f256? Eight bits. Eight bits, right? So, every element can be represented with eight bits. What is so nice about eight bits? These guys who program cannot think beyond eight bits being one byte and that's why it's so easy to think. So, people can have illustrious real estate too. They can only think in terms of bytes. They limited that. So, since they like that, they like f256 a lot. So, you'll see a lot of codes. There will be codes over f256 that are used in practice. So, they have their own limitations. They like real bytes. Okay. So, let's start here for today. And we'll pick up from here. Okay. So, this will be kind of the last lecture for your quiz. Your quiz will come next Tuesday. And from Wednesday onwards, we'll be doing problems on these slides.