 Okay, so we're looking at some abstract properties of finite fields. I took the last class that I was talking about, some kind of isomorphism and all that. So let's spend some time with that and see if we can clarify. I mean, I want to clarify some of those motions. So let's see a few examples. The first one I wanted to talk about was F4. Remember the form in which we constructed F4 so far looks like this, right? So alpha squared is on this alpha and then what alpha by 3 would be 1, right? So this is the form in which we've been constructing F4, right, polynomials and modulo x squared plus x plus 1, okay? So someone may come along and say he has some other field, okay, some other field. It's exactly four elements, right? And maybe he calls it A, B, C and D, okay? And then he might say all these things. He might say, so here, excuse me. So here you can define between 0, 1, 1 and alpha squared for plus operation. That's the plus operation, right? So this is how the plus operation gets defined, right? So that's how you define the plus operation. You also have the time separation, right? So you define it. So this is how the time separation works in F4 that we defined, okay? So someone might come along and say he has some other field, A, B, C, D, and then he will define... So now in a field, you define the plus and the time separation, right? So you might define plus and not times like this. So you might say plus and they might put A right here and then D here and then D here, C here, okay? And then you might define the time separation also. You might do the A, A, A, and then you might say this is D and this is B, okay? So someone might come along and say I have another C and this is the plus and this is the times and you can check all the field axioms with these operations. You can check, for instance, that if the identity would be A, the multiplicative identity will be B. Every element will have an identity inverse. Every element will have a multiplicative inverse. You can check all those things, okay? So you can do that and you'll see that this is a valid field, right? But then you have to think about whether these two fields are really different, okay? Except for the fact that I have called 0 as A, 1 as B, alpha as C and alpha square as D. There is really no difference. If you do C squared here in this field, what do you get? C times C is D, okay? So it's exactly the same thing. It's also true that D equals D plus C, okay? So all those things are true. So in a way, these two fields are not really different. They are the same in some other terms, okay? So how do you write that very formally is the following thing. And that's where these ideas of isomorphism will enter the picture. Okay? So you can have an isomorphism between G and F4. Okay? So what is an isomorphism? It's rescue a one-to-one and on-to-map between two sets that are equal in size. Okay? So it's a one-to-one map which respects the plus and the dot. Okay? It should be consistent with the addition and multiplication. So let me define this function here. So in G, I have A, B, C, D. And let's say this function S maps A to 0, B to 1, C to alpha, and B to alpha squared. Okay? So this is my one-to-one map of the invertible. And this function S has to respect addition and multiplication. What do I mean by that? So I have alpha, beta, and G and alpha plus beta, right? Should map for F of alpha plus beta should be alpha of alpha plus alpha of beta. And then F of alpha times beta should be alpha of alpha times alpha of beta for alpha beta and G. Okay? So this should happen. So let's say a one-to-one map from G to F4 of, let's say, what is this map called? A proper map from G into F4 which respects the addition and multiplication. Creates some kind of a homomorphism. Or in this case, it's an isomorphism because both of them are the same. Okay? Which respects the addition and the multiplication. Okay? So if you do addition first in G and then invoke S, it is the same as invoking F first and then doing the addition in F4. Okay? Remember, this plus is in F4, right? What about this plus? It's in G, okay? So it's a different kind of object. So such things are isomorphic and clearly they don't add to a set of possible fields. Nobody can say that G is not only a new field. The thing you do in G, you can also do in F4. So it's not anything new. Okay? Fundamentally, it's not new. Okay? So that is something to remember when you think of, oh my goodness. I think of this isomorphism idea. Okay? So now what we did in the last class was something a little bit more refined than this. A little bit more abstract maybe. Okay? So let me describe what we did in last class maybe with the picture. Okay? So we first started with a field of, okay, some kind of field. Okay? And that's why there was an element beta on it. Okay? Right? So here, from the beta, we constructed this from beta x. Okay? Interestingly, we see that then beta x has beta as a root. Okay? Remember this is, this is all zp. Okay? So this was part of plastic p. Okay? This is all zpx. It has beta as a root and it is also irreducible. Okay? So what we can do with this guy is we can use it to construct a field, construct a field g. Okay? Use it, power effect sequence and beta effect. We can do this. Right? So this picture is a nice picture to keep in mind. Okay? So if you have a field x of characteristic p and then you have some element beta and s, you go to its minimal polynomial. You know it's a polynomial in zpx. You know this irreducible. Okay? Right? Once you know, once you have an irreducible polynomial in zpx, you can always use it as your power effects in the field construction and come up with a field. Okay? What is this field? This field is called polynomials in some alpha. Okay? And what does that alpha do? Right? So the coefficients are from zp, right? For the polynomials. Additional is modulo p, like in zp. And for multiplication, you have to use power of alpha equal to 0. Okay? So that is the roots for constructing this field. Okay? So please remember that. One can show is there is a homomorphism from here to here. Okay? From this field, you can take every element of this field and map it to some element of this field s in such a way that addition and multiplication are present. Okay? You can do that. Okay? So if you take two elements here and add them here and then invert this homomorphism, we will go to some element. That will be the same element that you would get if you first invert f on each of these things and then these additions. Okay? And that comes very easily. I mean it's not so hard to come up with that. So that field here is basically the field that is generated by beta. Right? Think about it. You take beta here and then look at all the polynomials in beta with coefficients from zp. Okay? That is clearly inside f. Right? So what is that part which is isomorphic to this gate here? So that gate is, you look at all a0 plus a1 beta plus a1 whole am. So a, d minus y, d minus 1, beta dot d minus 1. What is d? d is the degree of this type. Right? So if you look at this set for all oi in zp. Right? This is a set which completely is contained in f. Right? There is no problem. And every element of this set is isomorphic to, I mean this, not every element, this set actually is isomorphic to g. Okay? So that's an isomorphism. Right? Why do you get the isomorphism? The alpha that you used in constricting this field, you map it to the beta that you had here. Okay? And then every element will, there will map every element here. And you know this is inside that field f and that base in g. Clearly, there's an isomorphism. Okay? So this is the idea which kind of connects fields and polynomials in a very, very tight way. Okay? And the connection is also quite much tighter because you know that x star p power m minus x factors in, into linear factors in a field of size p power m. Okay? So every element will have a irreducible factor of, a lot of mineral polynomial which is an irreducible factor of x star p power m minus x. Okay? So there's a very nice connection among these polynomials with x star p power m minus x. And then there's also this field which you can construct if, and then that is inside this. Okay? At this point this s, s is a general finite field. Somebody tells me that there's a finite field with p power m elements and I can nail it down like this. Okay? So I know it has an isomorphic field like this. Okay? So if every finite field has that. All right? So that's the, that's the idea here. So let me depict this with one explicit example and then maybe you'll see what I mean. Okay? So the example you're going to take is about 16. Because that's where the most, the non-trivial example comes. So let's say we do 16 in the following way. Okay? I'll take alpha dot 4 with 1 plus alpha. And at 16 means characteristic is 2. 2 is equal to 0. Okay? So it's all modular 2. Minus is the same as plus. And you have alpha dot 4 is 1 plus alpha. Alpha is the primitive element. Okay? So I'm going to say alpha is the primitive element. Alpha dot 14 is 1. Okay? So let's try and construct a few of these things. So the elements of this field will be 0, 1, alpha, alpha squared, alpha 3. Okay? When alpha 4 will be 1 plus alpha, what about alpha 4 plus 5? Alpha plus. Alpha square I'm interested in alpha 10 for some reason. Okay? So let's say I'll define alpha 10 real quick. If you take alpha 4 plus 5, it's where it will be. So that will be alpha square plus alpha 4. So what is alpha 4? Alpha plus 1. So it's going to be alpha square plus alpha plus 1. Okay? So these are quick computations. Okay? All right? So if you now try and compute the minimal polynomial of alpha part 5. Okay? So I won't try and compute the minimal polynomial of alpha part 5. Okay? So it turns out, so I'm going to give you the answer directly. I mean it's not too easy to, difficult to see. So you see alpha part 10 has only alpha square, alpha and 1. And then alpha part 5 has alpha plus alpha squared. Then you can see, x squared plus x plus 1 will be a minimal polynomial for alpha part 5. Okay? I know this result from some other way, but anyway. So just take it from me. This will be the minimal polynomial for alpha part 5. You can quickly check it. It's an irreducible polynomial. You put in alpha part 5, what happens? The alpha part 10 plus alpha part 4, 5 plus 1, which is 0. Okay? So clearly this is a valid answer. I mean it's not a wrong answer. So that's the minimal polynomial. Okay? So now, if I construct a field with this minimal polynomial, what field will I get? So use it to construct a field. What field will I get? My construction x squared plus x plus 1. If I use it as a power fx. I'll get f4. Okay? So that's it. So I'll get f4 if I construct it. So I'll use this as power fx. I'm going to write f4, which may be I'll denote as gamma, 1 plus gamma. The same as gamma squared. And then gamma part 3 is 1. Okay? Right? So what this means is there is an isomorphic copy of f4 sitting inside my f16. And what is that isomorphic copy? Okay? So from here you can go into this and that will be 0, 1, alpha part 5 and alpha part 10. Okay? So this guy is actually isomorphic to f4. Okay? You can quickly check that. I mean anything you do with 0, 1, alpha part 5, alpha part 10, you will never leave that set. That's the first thing you have to check to see that it's a valid subfield, right? You will never leave that set. You multiply it with a square or add whatever you do, it will be inside that set. Okay? That is the first thing you can check. And you can also check that it's a field. It's not very hard. Okay? And it's isomorphic to f4. All right? So this is what I mean by this thing of going from an element to its minimal polynomial and constructing the field with that minimal polynomial as the aggregation of the polynomial. And that gives you an isomorphic copy of whatever you construct inside. Okay? So this is an abstract idea which is used in a lot of abstract results which frankly are not really needed in error control coding. But I guess it might be of interest to you. It's good to know anyway. Okay? So that's the point. All right. Any questions? Clarifications are okay. All right? So one place where I use this idea was to show that if the property that we are inputting is data belonging to fprm is perimeter, then what happens? That implies the key of m beta bytes equals mk. Okay? So there's also another way to show this which is just to minimize the isomorphism and all that which is probably much quicker. Any ideas that anyone think about it? Is there any other way to show it which doesn't require isomorphism and all this fancy stuff? No? Yes? Maybe? Hmm. Yeah, but you're using the isomorphism again. I don't want anything else. So you can use the fact that beta and beta part p have the same minimal polynomial. Okay? So what should happen? The minimal polynomial of beta should have beta as its root. Should have beta part p as its root. Beta part p square as its root. So on, then it can never repeat. Because beta is perimeter, it can never repeat value other than pprm minus 1. So it will have m different roots in next ppr. So that is a quick way of saying that this m beta should have degree equal to m. In fact, that's another tool you can use. Okay? Which is maybe a quick step that I wanted to do the isomorphism first to do this. So one more result for which you definitely need the isomorphism. Okay? So that's very good. All right? So perimeter, you have degree of m beta that's being equal to m. Okay? And there are much stronger results which are true. In fact, you can show any minimal polynomial has to have degree that divides them. Okay? Not just less than or equal to m. It should divide them. We will prove that. But before that, we'll see a quick result about isomorphism of fields. Okay? So this is a very important result. Let me write it down. Two fields for any two fields, ppr m governments are isomorphism. Okay? So remember, so far the only thing which has been a sticking point for us in the whole area is we have never shown that for every m and every p, there is a field with ppr m. Never shown that explicitly. Whenever we have a reducible polynomial of degree m over zp, we know there is a field of ppr f. That's the point. We will come to it soon enough. But let's say somebody comes up with two dozen fields, both of size ppr m. Both of number of elements ppr m. And then both of them have to be isomorphic region. Yes, that is something that we can show. Okay? And that goes through this xp ppr m minus 1 and this ppr m minus x factorization and the isomorphism that I did just now. Okay? So let me try and do that. Not too hard. Okay? So I could call these things as theorems if you like them. I'm just avoiding the word. Just to keep you a little bit more happy. Okay? So theorems sounds funny. Sounds like a good thing. Okay? All right. So two fields. But here we have the status of a theorem. It's a fairly big statement. It's not a very small statement. Okay? So two fields with ppr m elements are isomorphic tools. Okay? So let's say f and g are two fields with ppr m elements. Okay? So of course. Okay? So the crucial notion is xpr ppr m minus x will factor into linear factors over both all the elements f and all the elements of g. That is also true. And nevertheless, the factors of xpr ppr m minus x over zpx, they will be unique. They will be the same. So if you put together, you should get the same factorization. Okay? So that's a strong limitation. You cannot really have too many different things. And that is the crux of this proof. Okay? So let's say, let's start with the, let's start with the field f. Let's say beta is primitive in f. Okay? So there's a primitive element. Let's say that's beta. Okay? Now look at m beta of x. Okay? This is an irreducible factor of, irreducible degree m factor of, factor of what? xpr ppr m minus x. We know that. And that's, it's, it's over zpx. Right? m beta of x is over zpx. Right? Okay? So what that means is there will be an element of g for which the same m beta of x is a minimal problem. It has to happen. So let's ppr m minus x factors into the same factors over zpx. And it factors into linear factors over g. So there should be some elements which combine together and give you the same m beta of x in that factorization. Okay? So clearly there should be an element of g for which m beta of x is an minimal problem. Okay? So that is the link. Once you have that, everything is done. Okay? So this implies, so all that I said just now, which probably I didn't, I'm not going to write that, write down those things very precisely. So what I said, xppr m minus x equals the element of g of minus gamma. Okay? And m beta of x divides this one. Right? Which equals this. So that implies that I close, let's say some gamma prime in g such that, such that m gamma prime of x equals m beta of x. Okay? Right? So what can possibly be the next step? Yeah. So you take this m gamma prime of x and then create a field with that as your irreducible polynomial. Okay? That is clearly in inside g and it is equal to all of g. Now that's also isomorphic to s. Okay? Or let's say enomorphic to s, but then the sides are the same. So that's to be isomorphic. Okay? So it's also isomorphic to s. So through that field which we constructed, s and g have to be isomorphic. Okay? So that's the idea. So you construct a field with power x equal to m gamma prime of x equal to m beta of x. This k on the one side is isomorphic to s and the other side is isomorphic to g. So from there, that's the idea. Okay? So that's the idea. So it's kind of a slightly abstract idea, but you see the central idea, central link that connects these two things is the x part p power m minus x and how it has to factor into linear factors over any field with p power n elements. Okay? So once you have that, everything gets tied on the terminal problem that's complemented. Everything comes from it. Every unique factorization, all those things, they all flow into this. Essentially this works. Okay? So what is so nice about it is once you have a degree m with reducible polynomial over z p x, the field where a field with p power m elements is bounded. So there cannot be any other field. Somebody says there's some other field. Anything he can do, he or she can do with that field. You can also do with this field that you have. Because both of them are asymptomatic. There's nothing new that will come out of any field that is unbounded. Okay? So any two fields with p power m elements are the same. All right? So that's the one yourself that comes out of this. So the last thing we have to show is the existence of a field with p power m elements. Okay? So that's the next thing we're going to show. And I'm going to show it in a rather quick way, skipping through most of the spots. Okay? I'll do only maybe one proof very clearly. And then after that, the next part I will skip through real quick. Is there a question? What is the question? In which one is there p-reducible polynomial that exists? Yeah, yesterday I spoke, in my last class I spoke briefly about it. It has to be true, right? Every reducible polynomial which is a factor of x star p power m elements. Okay? For that, there has to be an element of every field for which it will be a minimal polynomial. Because that's pretty simple. I mentioned it very briefly in the beginning of last class. That has to be true. Okay? In fact, it's also true that for every reducible polynomial there will be some element. The easiest thing is to just construct a field with that reducible polynomial as pi of x. Right? And for that alpha that you used, this has to be the reducible polynomial. Right? So for every reducible polynomial, there is a field element with that polynomial as its minimal polynomial also. But also, all these things are true. So for every reducible polynomial divides x power m elements? Yeah. Yeah. So that's obvious from that, right? So you can do it that way. So you take, so for instance, the point is making, if you have a reducible polynomial of degree m, it has to divide x power p power m minus x. How do you show that? So you start with that reducible polynomial, construct a field with p power m elements. Okay? You know that the reducible polynomial is the minimal polynomial of the element alpha that you assume for that. Right? In the construction, you take an alpha and that is the root s and beta of x are the reducible polynomial. And that this has to be the minimal polynomial. And that we have to divide x power p power m minus x. All these results are nicely proved because of that. Okay? So I'm going to formally kind of state a couple of, make a statement which will kind of include what you're saying also. Okay? So this next fact is also quite big. That's the status of the theorem, but I need to try and just put down as a fact. X power p power m minus x equals the product of all monic reducible polynomials of degree b, that the right side. Okay? So this kind of covers what you're trying to say also. There are various ways. I mean, what you said will be one of the proofs, one of the part of the proof of this case. Okay? So x power p power m minus x is the product of all monic reducible polynomials of degree d, that the right side. Okay? So if you take for instance, p equals 2. So let's just take a couple of examples to illustrate what this means. If you take p equals 2, so if you take p equals 2, m equals 4. Okay? What it means is x power 16 minus x is the product of all monic reducible binary polynomials of degree 1, 2 and so. Okay? What are the polynomials of degree? I can do plus here if p is 2. What are the degree 1 reducible polynomials x and plus 1? What are the degree 2 reducible polynomials? Only the degree 2 reducible polynomials. Okay? And then degree 4 turns out that I forgot one. x power 4 plus x plus 1, x power 4 plus x power 3 plus x power plus x plus 1. Okay? So that's the idea. Okay? So we're going to need a lot of elements in this proof. I may not be able to go through all the things. And like I said, it's not too crucial that we know the truth of this result. More important is to know the statement precisely. Okay? So if you want one more example, you can do p equals 2, 1 equals 3. So that would mean x power 8 plus x. x times x plus 1 times x power 3 plus x plus 1 times x. Okay? So that's the kind of statement that this is made. So it's nice to know that. Okay? So another important place where such a result is used, you may wonder why this result is used, is to figure out if you have a big field, what are the small fields that are contained inside? Okay? So as I showed you some results where I said, you have a big field, you take an element, find its minimal polynomial, and then construct the field with that minimal polynomial. That field is contained inside this bigger field. So now, how do I find out what fields are contained inside this bigger field? I'll go through and look at Ld that divides M, and every fp power d has to be now inside fp power n, another d divides it. So that's one of the kind of corollaries of the statement, if you think about it. Okay? So because you have a minimal polynomial of degree d, right? And that's going to come up somewhere in the factorization. So that's how we improve this. Anyway, we'll come to that later. So all these things are nice to know. So if you have a very large field, you're naturally interested in figuring out what are the small fields that are contained in it, and this gives you clues as to what can happen. Okay? So let's try to prove it. So maybe I should just keep the proof. The proof will repeatedly use this idea of isomorphism. She will go to an element theta in fp power m, minimal polynomial, and exercise the factorization. Okay? So I don't know if it's needed to show this, so let me see if I can... Okay, so let me show maybe a couple of quick facts, at least a couple of steps, and then we'll see. Okay? So there are two things to be proved here. First thing is, if there is a reducible polynomial of degree d, which is like d divides m, you have to show that that reducible polynomial divides x bar p power m minus 1. That's one thing we have to show. And then there's also another thing we have to show. If there is a reducible factor of a polynomial which is a factor of x bar p power m minus 1, then you have to show d divides. So if I show those two, then the statement is proved. Right? Did you have a question? Yes, you can. No, it won't help well. So actually, in fact, one of them will not... its root will not be primitive. See, I said primitive element has degree m minimal polynomial. The opposite is not true. Degree m minimal polynomial does not mean the element is primitive. Primateive comes from some other element. Okay? So we'll see that also. Okay. Okay, so let's try to prove at least one part of it. Okay? Yes? Degree m polynomial will have at least one polynomial. Degree m polynomial will have... Oh, no, no, no, no. Either all the roots are primitive or no root is primitive. That should be the result. For every irreducible polynomial, either all the roots will be primitive or none of them will be primitive. In fact, all the roots will have the same order. It's easy to show that. We'll come to it. Let me just quickly get rid of this. Because that means that we can find at least m integral. Yes, yes. What is exactly m? At least m. Most cases are a little bit more than m. Okay? All right. So let's move on and finish this proof if there are no more questions. Okay, so the first thing is... Let's say... Let's say some polynomial. Let's say private facts. Is irreducible? Degree d... Stop d-development. Okay? So if you take d equals 1, the statement is quite trivial. In fact, if private facts is x or x plus 1, x minus 1, then it is very... So let's say... Okay, let me not say that. If private facts is equal to x, this is done, right? If private facts is equal to x, then what happens? x clearly divides x part d power m minus x. There's nothing much to do. Okay, so this is a one-deal. It's no big deal. The only way to keep things is to... x divides x part d power m minus x. And we are done. So we can assume private facts is not equal to... So if we do that, private facts says degree d. So one thing we know is... So by Wigner, she was asking this question. So if private facts is irreducible and has degree d, private facts divides what? Private facts divides... See, private facts... So one thing to remember is private facts is not x and it has degree d. So private facts divides x times x part d power d minus 1 minus 1. Am I right? Which implies private facts is not x. So it implies private facts divides x part d power d minus 1. This has to happen. So how do we show this step? So you take private facts as the reducible polynomial with which you are constructing a finite field. You know that private facts itself will be the reducible... will be a valid minimal polynomial for that. And that has to divide the x part d power d minus 1. You know it's not x. So it has to divide x part d power d minus 1 minus 1. So now I have to show that when d divides m, x part d power d minus 1 minus 1 will derive x part d power m minus 1. That can be shown. So we can show. This side will skip. It's not very hard. I'll tell you the steps. It's not very hard. x part d power d minus 1 minus 1 divides x part d power m minus 1 minus 1. So the crucial thing here is whenever d divides m. The crucial step to show here is that in fact p power d minus 1 divides p power m minus 1 whenever. This is definitely as d divides m. So this is the step which you can show. It's not very hard. So if d divides m, it's very easy to show that it divides. You can also show that it's definitely as d does not divide m. p power d minus 1 will not divide p power m minus 1. So think about how you might want to show it. You have to divide m by d. Then use that quotient and remind that to write this. It requires a minor algebraic modification or manipulation that this can be shown. So once you show this, whenever d divides m, what happens? This guy divides that guy. And that's enough to show that this polynomial will also divide that guy. So if you have x part a minus 1, then with x part b minus 1, you have to show that if a divides b, then clearly the polynomial will also divide that guy. So you can easily show that it's not very hard to do that. So that's the idea. So that proves in one direction that if you have p f x doing an irreducible polynomial of degree d and d divides m, then p f x will divide the x part p power m minus x. So the opposite direction is also in a very similar way. So let me see. If you have this power f x divides x part p power m minus x, then you have to show d divides m. So that's what we're going to do. So here what you do is the following. So it's not very hard. The first assumption you get rid of is that p f x equals x. So if p f x equals x, you can get rid of. So once you get rid of that, p f x is not equal to x, p f x divides x part p power m minus 1 minus 1. So you get rid of. So assume p f x is not equal to x. Equal to x is quite easy. And that implies p f x divides x part p power m minus 1 minus 1. So if you do this, then what will happen is you can show that p power d minus 1 will divide p power m minus 1. So if you have p f x degree equals d, then from here you can show that this implies p power d minus 1. They'll have to divide p power d minus 1 by x. But they have to divide p power m minus 1. The reason is you can show that there has to be an element of this field which has order p power d minus 1. You can show that. In fact, you take this p f x with p f x to construct a field. It will have p power d element. So there will be an element in that field whose order is exactly equal to p power d minus 1. And that field will be inside a field of p power m because p f x divides x power p power m minus 1. So this f power f p power m has an element of order p power d minus 1. That can happen only if p power d minus 1 divides p power m minus 1. That also we know. So once from here you can go to d divide. Because we know that result already. So p power d minus 1 divides p power m minus 1. So this one only is d divide. So you use this result and you get that. In fact, this result is true for any n. So I don't even need to pre-print here. So for any number n to have n power d minus 1, it divides n power m minus 1. It's only if d divides n. So that is true. So once you have such results, you can prove it. So for this step you need to show that there exists some alpha n f p power m sub square order of alpha equals p power d minus 1. How do you prove it? You start with power f x and construct the field. It has an element of order p power d minus 1. But we know that that field is isomorphic to a subfield of f p power m precisely because that is true. So that is true. Oh my goodness. What happened? So that has an element of order p power d minus 1. That means p power d minus 1 divides p power m minus 1. And that means d divided by 1. So that's the proof for this result. I know it went a little bit quick, but like I said, the main thing is to remember that this is true. So from here, you can show using this result an interesting property. So first of all, equate the degrees on both sides. What happens when you equate the degrees on both sides? On the left-hand side you have p power m. On the right-hand side you will have what? You will have an expression like this. Suppose you write down so m of m is the number of irreducible polynomials m over z p. What is the degree of the right-hand side there? On the left-hand side you have p power m. On the right-hand side you will have summation over d, dividing m, what? d times m of d. So every d that divides m, you have m of d such polynomials. You multiply them all together. We are going to get d times m of d, add it all up. p power m has to be equal to this. So such equations are, I don't know if you have taken a course on number theory, you will see that such equations have enormous potential for exploitation. You can show some very interesting properties with such equations. So this formula involving d dividing m, there is a very standard way to invert it. You can invert this. It is called Mobius inversion. You can use some properties of very interesting results in number theory. It is both standard and reasonable number theory. So you can use something to invert it. So you will get a formula for n of m which is reasonably explicit. And then you bound that formula and you can show n of m is strictly greater than 1. Greater than or equal to 1, strictly greater than 0. So from here you can show, I am skipping it, but you can show using the inversion which is called Mobius inversion. So you use Mobius inversion and you get a formula for, even if I am going to go into the details here, I can give you references if you like. You can show n of m is greater than or equal to 1. So what does that mean? There was a result that I have after from day one, right? That there exists at least one monic irreversible polynomial for every m and a q. So this is how you can go through that result. There is also another way to show the existence of f p power m directly. I am skipping that, but this is a nice proof. You already know, also know that there is an irreversible polynomial for every m and a q. So the step is a bit more involved number theory. You don't need to do that. So I think this kind of slowly is bringing us to an end of what I wanted to do in planetary, except for this last and most important property of minimal parameters. So this is probably the most important property. That okay, I don't know, I don't know about the number. This is the most critical one and this is what is most useful in coding. So the final property that we are going to see is most useful for communications and marathon coding. So this is the real place where you have to pay real attention. So if you have beta n f p power m, so it takes some time to develop this property. I want to do it carefully. First we will define conjugates of beta. Basically conjugates of beta are from the set beta, beta is power p, beta power p squared, so on. So this cannot be an infinite set. So it will repeat somewhere. It might repeat somewhere depending on beta. So we can't say where it will repeat very easily when there is a way to precisely say how it will repeat. But anyway, so for arbitrary beta and f p power m, it will eventually repeat. So this will be a finite set. Don't think of this as an infinite set or something. In the finite set. That's the first thing. One thing we know already is the minimal polynomial of beta will have all the elements of this set as roots. The conjugates of beta will be roots of the minimal polynomial. So this will be the conjugates of beta will be this thing puts us m beta x. So let's say this thing kind of ends. Let's say beta has got beta power. So let me leave it here. So I have to figure out where it ends. So let's say this will have a notation for it. I call this c beta. Let us say we assume size of c beta equals some d. So let this be some d. It has d elements. So this will be different roots of m beta of x. So one thing we can show, we can be sure is if you look at this polynomial, product of alpha is an alpha n c beta x minus alpha. What can I say about this polynomial? This will have to divide m beta of x. So in fact it turns out this is equal to m beta x. So we can show this. So let's say we draw on the left side this and this half. Let's say m beta of x is equal to this. So we are going to be able to get some derivative of this. So let's say we draw on the left side this. Let's say you draw what is also equal to m beta of x. So it also becomes equal to m beta of x. Let's say this is equal to m beta of x, that is equal to m beta of x. So we will get some derivative of x that is equal to this. So let's say we draw the product of alpha. We will get some derivative of x. Let's see if it goes through. Let's look at this polynomial, f of x, which is product plus. If all does this look, it looks x minus theta times x minus theta dot b times x minus theta dot b square polynomial. So where will it go? Theta b, you have d minus 1. So all the way to d. So x minus theta dot, theta I think it's d minus 1. So 1 is basically theta 0. So that's how it will go. Let's say we multiply result and we get certain coefficients. So that's going to be x0 plus f1 x orbital fb x power b. And so that will add details. So let's say we multiply it out and we get these positions. So we need to show, we need to show what fi belongs to zp. If we do that, then we are done. So how do we show that? So for that, we'll use a few results. So the first thing I'm going to show is f i power p equals f i. So that's what we will show. So this is equivalent to showing f i power p equals f i. So you might wonder why this is true, but see for elements of zp, clearly f i power p will be equal to zp. But it also turns out in a larger field, if you have any element which satisfies this equation, it also belongs to zp. That is also true. It's not too hard to try to prove this. It's something you can try and prove. So that's another fact. So let me write down. f p power m is where x is alpha such that alpha power p power b equals alpha. Then alpha belongs to f p power b, which is contained in. So this can only happen if b divides m. This cannot happen if b doesn't divide m, which is contained in f p power m. So that is also true. So we're going to flag this fact, which is quite easy to show. Then we'll just show this case. So f i power b equals f power p. So the first step in that is to raise this equation we have on both sides to the power p. If you do that, if you do that, you'll look at even case or odd case carefully, but you'll see that the right expression does not change. The reason is you raise it to the power p where it will happen. You'll get here x power p minus beta power p. Assuming b p is odd, for instance, just for starting. You'll get x power p minus beta power p. Here you'll get x power p minus beta power p squared. This will become the next term so on. What will be the last term? x power p minus beta power p power d, but that has to be equal to only then you'll have exactly d elements in your conjugate list. So that will become like the first element. Except that instead of x you have x power p. So what happens if you come this side and raise the left hand side to the power p? You raise this guy to the power p. What happens? Again it will become f 0 power p plus f 1 power p x power p plus f 2 power p x power p. Now this is like an identity in x power p. You simply replace x power p to y and equate coefficients on both sides. You should get f i power p equals f 1. So that's the idea. How do you prove this? You raise both sides to power p. So you might wonder what if p is even? The only even is q and for even the 2 it doesn't matter whether it's minus or plus. Everything works out the same. So that's the way you prove this system. You can show that f i power p equals f i and that gets liberal. So this is quite crucial for coding theory. You'll see that later. So this is probably the most important result that you have to remember about minimal polynomials. How do you find minimal polynomial of element? Find all its conjugates first. Take the linear terms x minus conjugate, multiply them out, you'll get the minimum polynomial. It's important. So let me conclude this with one final couple of minutes. I'm going to take 10 more minutes on the first case. So hopefully it's okay. Don't mind that. We'll start at 2, 5 more minutes left. So let's quickly see one sub three. I'll just write it down real quick and then we go on to examples of polynomial. So the statement that I wrote down here, I'm going to write it down like a result. f p power d is contained at f p power m. This one only is d divided by z. So this is a result which we kind of proved in the previous statements. I'm not going to write down a special proof for it again that this is true. So f p power b is contained in f p power m and this statement also, you'll remember, this is also important. If any element satisfies alpha power p power b equals alpha, then it belongs to the subfield also. So this is a fairly important statement. So if you take for instance, f p power some large number, let's say f p power, let's say 20. And if you want to go from f p to f z, you can think of various directions. So you have f 4 and then what? f 4 is what? f 2 squared. Then you have f 2 power 3. You can think of some kind of hierarchy here. Then you have f 2 power 6. And then you also have f 2 power 4. So I'm putting errors to indicate the containment. This is the kind of picture that you can draw to show how this containment usually works. So even though there are so many of these finite fields, they're all kind of contained within each other. So it's not really, it's totally interconnected. It's not that each of these fields is all distinct and they're not connected to each other. Because of this subfield result, this whole number of finite fields, they're all kind of connected to each other and all these minimal polynomials will keep showing up. So if you factor x power 2 power 12 plus x, you'll get all the minimal polynomials that can ever be there everywhere. So they're all connected in somewhere and they'll keep appearing again and again and again in different forms. In an isomorphic sense, they're all the same. So that's one point I wanted to make. Another point I wanted to make is the following. So we will never use it in the course but it's important if you're doing this for some other reason. So I've been talking about z p and then extending z p to f p power m to a polynomial power x irreducible degree m over z p. This is what we've been talking about. And this is our construction. And this is a unique construction. As in there's no other field of size p power m, we know all that. So these two are we've been talking about. So there is an extension possible where instead of z p, you consider somewhere f p power something. So for instance what I can do is I can start with some q which is first say some p power s and then look at f q. Then we take f q and we can extend f q to f q power m. How do you do it? You do it in the exact same way. Except that you need a power x which is irreducible over f q x and degree m. So you can think of like a chain of fields. You can start with f p. Then use let's say this is pi 1 x. You can use pi 0 x which is irreducible over f p x degree s. And then you can extend it for that in the exact same method that we used. The a 0 plus a 1 x. Except that the a i will come from f q and you'll have a degree m and you'll do everything in the model of pi 1 x. So you can do this double extension. In fact it will also be true that there is a degree, there exists a degree m s pi of x in irreducible power section f p x which will give you the direct extension also. It is also possible. In fact that seems to be it right. These are two different ways of constructing the same thing. You can do this. And every statement I made with p can be replaced with q which is a power of prime. For instance this statement. This statement. This statement. You can replace p with q where q is a power of prime. q equals p power x. You can replace it. So you'll get even this statement can be replaced that way. This statement also can be replaced that way. Instead of p you can have a power of prime. Which means n of m is greater than or equal to 1 for every q not necessarily p. For every power of prime also you have an irreducible polynomial of degree m for every m and every q. So all these things are possible. So for every m and every q you have an irreducible polynomial that is also true. All the factoring is true. All the conjugates instead of raising it to the power p you'll be raising it to the power q when you do the conjugates. All those things you can do in this higher fields also. But we'll hardly ever use it. For us the most interesting case in coding is to pick p equals 2. And then m equals 2, 3, 4. So this is the most interesting case in coding. But in general the fields are just always. So I'm going to stop here today. Next week we'll start doing some tutorials. When is the quiz? What day? Tuesday of next week. Then we'll have class on Monday next week. Monday, Tuesday, Wednesday, Thursday and then another Monday. So Monday I'll do some things and then from Tuesday onwards we'll start looking at problems. All of you have access to the problem set. You know where the problem set is. You go to my web page, you navigate towards the course. You'll get two assignments. One will be on linear block codes. Another assignment will be on finite fields etc. So both those assignments are included. There are also solutions I believe for some of you.