 lecture number what lecture eight okay all right so I gave you a brief introduction into fields so we're going to start looking at finite fields that's going to be the main topic we'll begin with I am not sure how far we get into this okay so the first definition we need is for a group and let me define what a group is a set of elements a set with one one binary operation right okay let me say let me be very precise I said G sorry to define write it down later I said G with one binary operation I'll denote it plus okay one binary operation plus is a group if a few conditions are satisfied so what's a binary operation if you want a technical definition binary operation is a mapping from what to what plus there's a mapping from G cross G to G itself okay so what do I mean by G cross G it's an ordered pair of elements from G okay take two elements as G crossed okay so you need a bunch of conditions the first condition which will which will mostly be very clear is that plus should be associative okay what do I mean by associative a plus b plus c should be the same as a plus b plus c there should be no ambiguity if you were to add three or more elements should be very clear anyway you add you should get the same answer it's associative in most cases this is a little bit difficult to prove but based on our intuition it'll be very clear so I will never prove it I'll just simply say it's associative and it's obvious okay the other two things will usually need careful proof there exists an element I'll call it zero belonging to G such that such that what a plus zero equals a for all a and G okay so that's the zero element the identity okay so this is the identity case it's associative this is identity zero is called the identity element and the next condition is what facilitates subtraction okay so far you can think of this plus as the addition defined on this group and how do you do subtraction you define what's called an additive inverse for every a and G there exists okay I'll call it minus a in G such that what a plus minus a will give you this zero which is the identity okay so remember what is this zero again it was this identity okay so this facilitates subtraction suppose if you want to do subtract B from a then you add minus B to a okay so that's the that's the notion for this inverse okay all these three as I said the second in property two and three are typically proved to show it's a group and the first associativity is usually usually be very clear from the definition and our intuitions okay no no no I think for identity that will come out to be true okay but anyway so commutativity is not required okay in a group okay but most groups that we will be seeing in this class at least will be commutative okay so what's commutativity commutative group plus a plus B is the same as B plus a for all AB in G okay there's another word for commutative it's abelian named after a very famous mathematician Abel okay so these these are also this this is as I said it's not necessary for us but since most of the groups almost all the groups will be seeing in this class will be abelian or commutative well when I say group I'll naturally assume it's commutative if it's not commutative I'll explicitly say so okay there's also another so this notation plus is usually called addition okay so so you can have any other notation in place of in place of plus right I could have any anything else I put want want to do here and depending on your artistic skills you can draw any symbol you like there and call that your addition okay notice that symbol has nothing to do with any of the properties okay what is that plus it's just a mapping from G cross G to G that's the only thing I can call that plus anything else I want okay I simply called it addition right this is the additive notion okay in future we'll think of states still sets where there are two different operations addition and multiplication and we will require different different slightly different kind of properties combining the two okay but you should be comfortable with this okay so you can have the same group definition without plus with some other symbol okay maybe dot or maybe times or maybe something else okay so one can give a lot of examples I think people here we mostly familiar with groups I'm not going to spend too much time giving an example but I just want to point out one operation which is usually done with groups which is one notation with this plus okay for instance if I have some natural number n okay okay so this is something which is important we'll use it also you should suppose I have a natural number n I can take an a belonging to the group and then add a to itself n times okay remember I can do this n times only when n is a natural number okay doesn't make sense to add it 1.5 times or anything okay so this will work only when n is a natural number so instead of writing this down painfully a plus a plus a n times I'll have a short hand notation for it I'll simply say some natural number n times a okay okay so remember this makes a lot of sense with the plus so when I write n times a it means I'm adding a to itself n times it's just short hand notation it's useful frequently when we when we write down long long operations okay so when instead of plus you use times or dot okay typically this would become what a dot a dot a dot a and that notation will be a power n okay in the multiplicative notation in the additive notation it is n times a in the multiplicative notation it could become a power n. Similarly this inverse okay in the multiplicative notation will become a power minus 1 and the identity in the multiplicative notation is typically written as 1 okay you don't think of the multiplicative identity if you refuse a multiplicative notation the identity you would write as 1 as opposed to 0. But it is all equivalent, it is the same as far as a group is concerned, there is no difference, just change in notation. So maybe at this point we will see an example which will be particularly useful to us. I am not going to give you the standard simple examples, I will give you the one example which I think is slightly more interesting. So I will use this example Zn, what is Zn, some of you might be familiar, it is the integer 0, 1 through n minus 1. And then I am going to say addition and multiplication are divided, defined modulo n. I am defining two binary operations on this set, the two binary operations are denoted plus and dot, I will call it addition and multiplication. Both of them are supposed to be done modulo n, why? Because I want my binary operations to result in some value in this set, I cannot jump out of my set. So I am just saying modulo n, both of them are valid binary operations. So I am going to make one statement first, I am going to say Zn is a group with respect to what? With respect to plus it is a group, it is very clear that it is a group with respect to plus. So what all should I show? It shows the group, first thing I should show associativity, but you know associativity is valid for addition over the general integers, whether you do modulo n or not, associativity is valid, it is not a big deal, I do not have to prove, I am not going to prove it here. Next thing I have to show is what? Identity, what is the additive identity? 0, 0 is the same 0 here, the additive inverse is what? For 1 it is n minus 1, for 2 it is n minus 2, for n minus 2 it is what? 2, again you can show all these things, if one element is an inverse of the other, the inverse of that is the element itself, all these things you can show. And you can show for instance that additive identity is unique, you cannot have two different zeros. So all these things you can show. So it is not a big deal. What about the other operation dot? Can I say Zn is a group with respect to dot? So there is some problem there. So maybe I need to look at Zn without 0, this notation is what? I have removed the element 0 from Zn. So I will have a notation for this, I will simply denoted Zn star, this is also a standard notation. If any time you have a group, additive group, you remove that 0 element, you will get something else. Can I say Zn star is a group with respect to dot, it appears to be associative, it is associative, it appears to have an identity, what is the identity? One is the identity, but what it may not have is the inverse all the time. So you will need some further conditions. You cannot say this very clearly with respect to group. So it turns out the Zn star is a group with respect to dot, I can say if and only if n is prime. So I can prove this very easily but let us see an example. It is very easy to show the one wave. If n is not prime, it is very easy to show Zn star will not be a group. All you need is one example. You can take say for instance Z4, if you can take Z4 star, what is Z4 star? 1, 2, 3. What element do you think won't have an inverse? So you see what is the, one can think of the multiplicative inverse of 1, what will be the multiplicative inverse of 1? 1 itself, that is the identity. For 3 also you will have an inverse, what is the 3? What is the inverse of 3? 3 itself. So 3 times 3 will be what? Do you see this? It is going a little bit too fast. So you look at this multiplication. If you do 3 dot 3 mod 4, what do you get? 1, which means the inverse of 3 is 3 itself. It can happen. All these things can happen in these small groups that we are looking at. The element can have, there can be its own inverse. It does not have to be 1 by whatever. So 3, 3, what about 2? 2 does not clearly have an inverse. One way of quickly checking it is try each of the elements and if none of them is the inverse, clearly there is no inverse. You can also prove this in a general way. You can show if n is composite, any number that divides n can never have an inverse. So it is not very difficult to show it. So 2 does not have an inverse. So this is not a group with respect to dot. Why 2 has no inverse? As I said this can be generalized. If you look at z n star, a dividing n will not have an inverse. Any a that divides n will not have an inverse. So you can think of that. But now we have to show the other case. I am going to show z p star will be a group with respect to multiplication. So I need to show that also. If you look at z p star, what will it be? 1, 2, p minus 1 where p is prime. Suppose I look at this guy. You can show this will be a group. So only thing that you have to show which is non-trivial is that every element has an inverse. You can prove it in several ways. I do not know if you have seen proofs of these statements. You can prove it in several ways. You take any element a. Look at all these guys. a dot 1, a dot 2, a dot 3, so on till a dot p minus 1. Can any of them be the same? Yeah, why cannot you have to argue that none of these guys modulo p will be the same. Why will they not be the same? Suppose you have a dot i being equal to a dot j, then what happens? A times i minus j has to be 0 mod n, mod p and p has to divide either a or i minus j. That is the property of prime numbers. Prime number divides the product of two numbers. Then it has to either divide one of them or the other. It cannot separate into two and one dividing this and another dividing the other. That cannot happen with prime numbers. So, if you look at this guy, a dot i is not equal to a dot j mod p. So, you consider these guys. No two of them are the same. So, how many distinct guys you have here? p minus 1 distinct numbers. One of these things has to be equal to 1 and that one will be the inverse. So, that implies that exists k such that a dot k equals 1 mod p. So, that is the proof. So, you see the only property I used for p is what? When p divides the product of two numbers, it will have to divide any one of the two. So, that is the construction that gave me the inverse. So, that shows that z p star will be a field with respect to, will be a group with respect to the multiplication, modulo p multiplication. You can see several examples and convince yourself. So, let us see an example and try to compute this inverse. It is an interesting thing to compute. Let us take for instance, a slightly non-trivial case. We will take z 5, 0, 1, 2, 3, 4, 5. If you take z p star, oops, you are right. I am going to define addition and multiplication modulo 5. What is z 5 star? 1, 2, 3, 4. So, I want you to spend a few minutes, no, you probably do not need a few minutes. Try to find the inverse of each of these elements. What is 1 inverse? 1, 2 inverse, 3. So, that should give you 3 inverse is 2 and then 4 inverse will be 4 itself. So, that is how you go about computing. Is that clear? So, another thing I want to illustrate is what is this power notation. So, basically in the z 5 star, so if I write for instance, the powers of 2, 2 would be 2. What would be 2 squared? 4. 4. What is 2 power 3? 3. What is 2 power 4? 1. After this, it would repeat. So, this is the notation for powers. So, multiplicative notation, just short hand for saying I am multiplying 2 with itself 3 times. I simply did not notice 2 power 3. So, you notice you can use all your regular knowledge of numbers when you do these things. Nothing is wrong with that. Suppose I take 3. 3 squared is what? No, it is mod 5. So, it is 4. 3 power 3 is 2 and then 3 power 4 is 1 and it would repeat. You do not have to do 3 power 4. You can simply multiply this 2 with 3. Do you see that? You do 3 power 3 already. It is 2. Simply multiply this with this. You can do modulo individually also. That is how the modulo operation works. Suppose I take 4. What is 4 squared? 1. 1. And it repeats. So, one can intuitively see in any group with the finite number of elements, if you keep taking successive powers, eventually it has to reach 1 and repeat. So, it has to happen. So, you can use maybe if you want fancy pigeonhole principle to prove such things. But if it is finite number of elements, if it does not repeat, then it will have infinite. It is a very simple contradiction there. So, it has to repeat. So, this is a useful property for elements of a group. So, we will study that very closely. So, let us go back to Z5 now and look at it as a group with respect to plus. So, as a group with respect to plus. So, I am going to look at just this plus. And I want you to do these things once again. What is the inverse? What is minus 1? 4. 4. What is minus 2? 3. 3. What is minus 3? 2. 2. What is minus 4? 1. What is minus 0? 0. 0. It is all these things you can readily guess. And now instead of powers, I would have what? I would have sums. So, I will have this other notation. So, if I take 1 and add it, I can get 1 or 2 times 1 which would be 2, 3 times 1 which would be 3, 4 times 1 which would be 4, and then 5 times 1 which would be 0. So, you see that. And then after this, it will repeat. 6 times 1 would be 1 again. It will repeat. I am not going to really just try very hard to distinguish. It should be clear to you. So, I know I have to but it is a little bit more painful to keep. Maybe I will put that dot operator as a slightly bigger dot. It should be clear. So, suppose you want to do this with 2. 2 times 2 would be 4, 3 times 2 would be 1. Am I right? 4 times 2 would be 3, 5 times 2 would be 0. And then once you get to the identity, you keep repeating. Is that clear? So, you can do this for addition also. So, these are examples. So, that is the notation. Z 5 times. Yes. 5 plus 5 times 1. No, it is 5 times 1. Now, I am adding 1 5 times 2 itself. There is no 5. I mean this number. See, when I use this notation, I guess that is the problem. See, this notation is different. No, I said NA. So, NA, that notation N. So, this can be any natural number. So, if you want, you can generalize this and write neat results if you like, but it is the idea. The next notion that is useful from a group for us is one of order. Suppose I have a group G and the operation dot. So, I will use the multiplicative notation for its operation. So, I will say G is finite. Order is valid whether G is finite or not. So, maybe I will say G is finite later. So, the order of an element is something that is important. So, suppose I have an element A belonging to G. So, here is the definition. Order of A, maybe I will write down G here just to illustrate that it is G and then maybe actually I should write the dot also, but I guess it is clear. Order of A is the, write down what the definition is. Smallest N belonging to N such that A power N is 1. So, you saw why this has to be valid. Of course, if G is not finite, you could have an infinite number of, infinite order for an element it could possibly have, but for finite groups, it is very clear why this thing has to be finite. So, there is a very interesting result if G is finite and we will use this quite often. If G is finite, then order of A divides order of G for any A inch. I will prove this for an abelian group. If I have to, it is true for any group. I never said anything about abelian group. It is true for any group. It is easiest to prove for an abelian group just because I am lazy. So, I do not know or maybe I can prove it for a general group. I do not know. I will start with a proof. Maybe if I see that I am showing it for a general case, it will be nice. Otherwise, I will just say I will prove it for the abelian case. Yes, very much. I am using the multiplicative notation. So, I said one belonging to G once time. The point at which the successive powers of A start repeating. I showed you how you keep on taking successive powers. Eventually you have to repeat. The first point where it repeats is the order. It is reasonably important for us in some cases. So, for instance, we go back to the previous example that I had. In Z phi star, what is order of 2? 4. Order of 3? 4. Order of 4? 2. And you see the previous results. What is the size of Z phi star? What is the size of Z phi star? 4. And you see all of these orders divide 4. So, possible orders are only 2 and 4. So, it is not this thing. Even here you can verify that. Here Z phi has order what? The size 5. And you see the order of each of these elements will divide 5. It is very easy to check. So, I am thinking if I should prove this or not. So, let me not prove it. I will just take it as a fact. So, it is likely it is not very difficult proof. I can easily do it. I can have to use some division arguments and all that. So, I will assume that this is true. It is a little bit intuitive to see why it should also be true. There are lots of interesting results in the meanwhile. For instance, one of the results that you can use is that A power mod G would be equal to 1. So, one of the corollaries from this which is very important for us is that A power mod G equals 1. Why is this true? Because order divides the group. So, if you raise it to the mod G, this I will prove. So, it is very easy. I know order of A n G is going to divide the size of G. That implies A power size G is going to be some constant K times order of A n G. Now, I can rewrite this as what? A power order of G raised to the power K and this becomes 1. So, that is the way to prove such results. This is an interesting result. If you have any group finite group size mod G, any element raised to the power G is going to be equal to 1 just by this result. So, if you want me to look at this example, if I look at Z p star, what are the elements of Z p star 1, 2, 2, p minus 1? So, what is the size of Z p star? p minus 1. If I take any element A in Z p star, what is going to happen? Using this result A power p minus 1 is going to be equal to 1 mod p. This is a very famous result. Anyone knows what this result is called? Yeah, Fermat's little theorem or something like that. You can call it, I think it is called Fermat's little theorem. Anyway, it does not matter what it is called. This is a very famous result in number theory. If you are in the habit of writing competitive exams in mathematics, this is the result that is used all the time to simplify your calculation in the modulo p. They last for some crazy number raised to the power usually the year 2007 or 1993 or something, modulo or something and you use all these results to simplify your calculation. So, you see it follows from a very nice result in group theory, which is more general. So, remember I want to emphasize once again that this result is true whether or not G is a billion or anything. This result is always true. Okay, so I think that is pretty much all we need from group theory. We will jump ahead and define fields. So, here is the definition. Set f, I will denote fields usually with f with two operations plus and dot set to 0. So, this is the to be a field if a bunch of conditions are satisfied. The first condition is f with plus is an abelian group. Second, f. So, before I write the second condition, I think I can write the second condition. It is not a big deal. f star, what is f star? F minus the additive identity. So, suppose 0 is the additive identity. f star with dot is again another abelian group. So, as I said most groups we see will be abelian. So, it is not a big deal. What is f star? f star is f without the additive identity. So, those are conditions that each of these operations have to satisfy independent of the other. f with plus has to be a abelian group. f star with dot has to be an abelian group. So, there is nothing that relates the two. So, then you would have distributive relationship. a dot b plus c should be equal to a dot b plus a dot c for all abc. Is that clear? I think I have covered everything. There might be some inter relating thing that I might have missed. I think that is about it. So, I do not think I need anything more. I believe that should be enough. Maybe there is something that is missing there. So, that is the definition of the field. And what is the finite field? If there is a set f which satisfies this condition and the size of f is finite, it is going to become a finite field. So, those are the conditions. So, this is the distributive law. So, it is usual to denote the multiplicative identity as 1. So, 0 and 1. So, using the distributive property and all these things and the abelian nature and all that, you can show some very interesting results. So, for instance, you can show some things you can show just based on this definition. 0 dot with. So, you can also extend the definition of c. Technically, you do not need dot for 0. But using the distributive property and all that, you can show 0 dot with any a will be what? And again be the additive identity. So, you can extend those definitions also to dot. So, all these things you can show. So, there will be inverse and there will be a lot of interesting interplay between the two things just based on these properties. The point, the reason why I am not going to go into detail and prove all of that is most of them are intuitive and obvious to you, will be obvious to you. Most of the properties that you expect based on what you know from real numbers and rational numbers will hold in general for fields. So, it is not very non-trivial. So, all these things you can show. So, as I said, we are interested in finite fields, a field F with size of F finite. First of all, are there any that should be the first question, is there any it is possible to have all these conditions satisfied for the field, have operations like this and still have finite, still have it be a field and the point I can I can simply justify by showing you this example. So, if you look at this, this set 0, 1, 2, 3, 4 with addition and multiplication modulo 5, what is this? This you can show will satisfy all these properties. We already checked that this set is an additive group modulo 5 and then without the 0, we already checked it is a multiplicative group and how do we know distribution holes? We know distribution holes, right? It is just integer addition multiplication. I can prove it, but it is going to take too much time and based on your intuition, based on your knowledge of addition and multiplication of integers, you know distribution holes. So, this in fact is a field. So, since we know it is a field, we will call it F5. So, this 5 will indicate always the size of the field. So, F5 is this field. There are very surprising properties you can show. For instance, you can show there is only essentially one field of size 5 and this is the F5. So, one can say the F5. So, it is not true for groups that are several groups of a given order, right? But for fields, finite fields, there is essentially only one of a given size. So, that you can show. But it is not clear if this is going to extend or are there any other fields you might ask. So, the first example is F5, this we already saw, but this can be generalized now. What can I generalize it to? Fp, right? Anytime I have a prime number of elements plus and dot modulo p, just by the previous proof that I gave, this is going to be a finite field, okay? It is going to be a field. I can add, I can subtract, I can multiply, I can divide, except with 0. Sir, given a size you are playing, whatever this is, that is why you want to be If a field of success exists, it is a unique case, up to isomorphism model. We will maybe prove it, I do not know. So, it is not, there is not a lot of variety in finite fields, okay? So, it is just all, it is all essentially the same. You do not have to worry too much about structure and all that. So, it is enough if we see construction of fields. And once you know one way of constructing it, you know, essentially all the ways of constructing it. So, you do not have to worry too much about it. So, that already said that, but it requires proof obviously. Okay, so that is the first example and it is quite non-trivial and it is very, very useful to get comfortable with this FP, okay? So, in coding mostly in practice, what P will we choose? Okay, what field do you think we will be interested in error control coding? P equals 2, right? You only want 0 and 1, it is all binary, right? So far, we have only seen codes over F2, right? I have seen vector spaces over the binary field, okay? So, we are only seen codes over F2. You can define codes over general fields, but in general, we are always interested in the case P equals 2, okay? So, F2 is 0, 1 is of great interest to us, okay? Addition and multiplication, modulo 2, okay? It is a very simple field, you have studied it in various ways, okay? All right. So, the next thing I am going to do, okay? It is going to look at some properties concerning polynomials over fields, okay? So, we will need that, okay? So, the general thing is what? So, where are we going from here, okay? So, the final destination would be, can we characterize all finite fields or can we find all finite fields, okay? Some strange things are true here. You cannot have finite fields of any length, okay? Any size, okay? They have to be of certain specific sizes, okay? So, all those things will prove, they are not very difficult to prove as well. But to go through and fully do the construction, you need some notion of polynomials over fields, okay? So, some ideas of what to do with polynomials, what is it that you can do with polynomials, how to think of polynomials, what is multiplication, what is division, etc., okay? So, those kind of things are useful to have, okay? So, since we just now saw Fp, I am going to talk about polynomials over Fp, just give you some ideas, okay? So, this will not be very rigorous, just I hope most of it is preview for you, polynomials over, let us say a general field F, for a field F, okay? So, as long as you have a field and you have your polynomials with coefficients from the field, most of the things you naturally expect for polynomials with real coefficients will carry over, okay? So, what do we mean by a polynomial, okay? In general, a polynomial is denoted a0 plus a1x plus a2x squared plus so on till what? Some degree, right? I will say an x to the power m, okay? So, this is a polynomial, okay? This is a classic polynomial and these coefficients, if they come from F, this is supposed to be a polynomial over F, okay? So, there is a nice notation for this. So, instead of saying all this, I will say F and then I put square brackets and say x, okay? That means this is what? This is the set of all polynomials over F, okay? The set is very clear, okay? So, and you have different polynomials of different degrees. What is the degree of a polynomial? Largest non-zero term, right? Which is the exponent of the largest non-zero coefficient. So, all those things will carry over here as well. See, there are a couple of things you would be very careful about, okay? So, I think I do not want to get into such great detail about polynomials. Polynomials and evaluations are two different things, okay? So, when you say polynomial, it has got a certain x power, all these things and you can divide them, you can multiply them, they will give you different numbers. Other thing is where you evaluate them over, okay? So, F of x, always you think of functions as F of x, okay? So, when you say function F of x, there is an ambiguity that are you referring to the function evaluated at x or the function itself, okay? The function itself is some kind of a mapping and all those things. I assume you are a little bit familiar with that, but you are raising a point which is highly technical, I do not want to go into it. If it is not disturbing you, then it is fine, okay? So, polynomial and its evaluation over some x or a value x can be two different things, okay? You should always keep that in mind. I will give you an example, okay? Maybe at that point it will be more clear, okay? So, these are polynomials. So, let us look at the set very closely, okay? So, I have to define a few operations. I can define a few operations over F, okay? Plus over F, okay? I think most of you know what this is, okay? I will add two polynomials over F, add the corresponding coefficients, right? When you do not have the coefficient, what do you do? That is 0, okay? So, when a coefficient is not there, you assume it is 0 times whatever power and then you add with 0 and you get it, okay? So, this is easy, okay? So, this extends from your notion of polynomial addition, okay? I can also do, okay? I am sorry, Fx. I can also define a multiplication over Fx, okay? I will multiply two polynomials. I do the exact same thing, right? You multiply term by term, collect the common terms, do the addition over F, okay? Except that, all your additions will now be over F, multiplications will be over F for the coefficients, okay? The place holder X will follow the same rules as you had before. X square times X bar 3 will be what? X bar 5, do not go around reducing it modulo anything, okay? That does not make any sense. It is just a place holder for you in the polynomial, okay? That is just something. We do not want to think of it as anything else, okay? So, only the coefficients have to be reduced modulo p if you are working over Fp for instance or according to the field operation, okay? Whatever the field operation is telling, okay? So, this also extends, okay? So, this extends from a regular polynomial multiplication, okay? What about groups? Is F of X a group over plus? Yes, yeah? What is the zero element? Zero. What is the inverse of a polynomial? All the coefficients you make minus, okay? Do not always think in terms of Fp. In general, you make it, make all your coefficients minus of what they are and if you add those two, what will happen? You will get zero, the zero polynomial, okay? So, both of them, okay? What about multiplication? Is F of X a group? Okay? So, what will be missing? Yeah, inverse will not be there, right? So, you cannot expect the inverse of a polynomial to exist. It would not be there, okay? For instance, X is a polynomial over any F, right? What is its inverse? You cannot have another polynomial multiplying X and giving you one. It is not possible, right? So, the degree will always be a problem. So, dot is not a group. So, in fact, F of X technically is what is called a ring, okay? So, it has its own intriguing properties. There are a lot of things about dividing and multiplying in rings, which is very, very intriguing, okay? So, all that is fine. One property that will be very, very crucial for us is division, okay? Dividing. So, remember, dot is not a, F of X over dot is not a group. So, I cannot define inverse for a polynomial, but I can still divide two polynomials, okay? As long as I maintain a quotient and a reminder, okay? It is the same thing as dividing in numbers, right? When you divide two integers, if they do not divide each other properly, you will have both a quotient and a reminder and you have to keep both. I can do the same thing with polynomials. I am sure you have learnt long division in your, at some point in school and I am sure you have forgotten it because you never had to use it, okay? So, if you have forgotten it, please go and remind yourself what long division is, the way of writing down the, what you divide inside and then you put one after the other, right? You remember how to do that, okay? So, long division is very important. So, division is an important thing. So, I want to spend some time doing that, okay? Suppose we are given two polynomials, A of X and B of X, okay? So, what do I mean when I say divide A of X by B of X? What do I mean by that? Okay? What do I mean by that? Okay, exactly. So, this, the technical way of divide, defining division, okay? So, one way you would do it is what? Put A of X here, right? And then you put B of X here and then you keep finding things, filling it out till you get what? Till you get something whose degree is less than B of X, okay? A proper way of clearly writing it down is the following, okay? So, like you said, find Q of X, R of X, belonging to F of X such that, such that what? A of X equals Q of X times B of X plus R of X and what should be true now? Degree of R of X should be strictly less than degree of B of X, okay? So, this is what you were doing when you did this long division, okay? You were trying to find a Q of X and R of X such that you can write A of X in this form. This is division, okay? Yeah, there are some uniqueness properties and all that, I guess that is very clear, okay? If you follow this long division process, if you have an A of X and B of X with the field, you will get only one answer, you cannot get multiple answers, right? So, you will stop at one unique way and if you can prove all those things, it is not a very difficult thing, okay? So, so far, you might have been used to dividing over what? Real numbers, right? So, real numbers as coefficients. Now, if you have any other coefficient from some other field, will the same division carry over? I have claimed it does, okay? But what about long division? Can you do long division with the same thing? What should disturb you? There should be something, but it should work, no? Mostly, what do you do? How do you, how do you do long division? You take this coefficient, then you put something there so that this times this will be equal to that coefficient, right? So, let us try something over, say, F5. Let us try a long division over F5 and then see how it works out. I am quickly running out of time. I will just do this and then we will pick up from here as we go along, okay? So, let us take, let us take F5x, okay? I will take a of x to be, let us take some polynomial, x power 3, no, x power 4 plus 2x squared plus x plus 1, maybe b of x is x squared plus x plus 1, okay? Suppose I have to divide these two over F5, okay? So, maybe I will put a 2 here just to complicate matters, 2x squared plus x plus 1. Oh, no, this makes things very easy. So, maybe I should make it x squared plus x plus 1. I will just leave it like this, okay? F5 of x. How will you go about dividing it? Just follow the same method, okay? You do x power 4 plus 2x squared plus x plus 1 and then you put x squared plus x plus 1 here. What will be your first term? x squared, right? You put x squared and then you do x power 4 plus x power 3 plus x squared and then you do what? You add these, subtract these two, okay? So, these two will cancel. You will get minus x power 3, okay? Plus x squared plus x plus 1. If you have this minus, you see that it is actually what? Minus 1, right? So, it is always good to go back and replace it with the field development that you are familiar with, okay? So, you will simply write minus 1 as 4, okay? So, now what do you do? You have to find something here which went divided. It will give you this. So, you see you can actually do it. There is nothing that prevents you from doing this. 4x power 3 plus 4x squared plus 4x, okay? So, you divide these two. What do you get? You subtract these two. What do you get? These two are going to cancel. What will happen to these two things? 2x squared, do you agree? Okay? Minus 3 is the same. Plus 2x plus 1, okay? So, then you again do the same thing. You get a plus 2 here. You get 2x squared plus 2x plus 2. What do you get? You get 4, okay? So, at that point I can stop. Why? I got a number which I got a polynomial here which had degree less than my b of x. So, what can I write? I can write a of x equals x squared plus 4x plus 2 times b of x plus 4, okay? And as he pointed out, there are some uniqueness properties about this. And you can see the uniqueness is clear. If you go through this process, you will definitely get only 1q of x and 1r of x, okay? For the same a of x and b of x, you cannot keep getting different things, okay? So, it is not a big deal. So, this is how you do division, okay? And division can be very different. Okay? You can check that this multiplication will also work out. You multiply and reduce everything, modulo of i, reduce all the coefficients, modulo of i, you will get the same answer as a of x, okay? So, this is long division. I think we will stop here. I will pick up from here in the next class and you will see the way we will proceed from here is, I think I will probably remind you of a couple of properties of polynomials and then we will jump to the construction of a general field, okay? You will see it is very easy and simple. It is not very difficult, okay?