 So, this is lecture 9 right, okay so we have been looking at polynomials over a general field right, fx, okay so this is the set of polynomials, polynomials with coefficients from a field f, okay so we have been looking at these polynomials and the last thing I did was show you, show you an example of division, how do you divide a polynomial a of x by b of x to get a quotient and a reminder and we saw how, I think I gave an example in f5 right, just to show you that it carries over from your familiar real and rational fields as well, if you have a finite field also, you could think of polynomials over that finite field, okay, alright so let's keep moving ahead with that, so the division, how did the division work, let me once again remind you real quick, the division worked as follows, so if you were to divide a of x by b of x, you would get a quotient q of x and a reminder r of x, all of these case will lie in fx as well and the degree of r of x would be less than the degree of b of x, okay, so you should be reminded of division with integers, okay when you see this, so you notice this polynomials over fx seem to be very similar like integers, similar to integers, right, so you have two integers, you can always divide one by the other and get a quotient and a reminder, okay, so you'll see the similarity is valid in more than one way, okay, so you can always think of polynomials in a similar way as you think of integers, okay, so at least as far as division is concerned, okay, many of the properties based on division carry over to polynomials as well, okay, so for instance you say if r of x is 0, what would you say, the reminder is 0, what do you say, say b of x divides a of x, okay, so a shorthand for that is just this, okay, so basically a of x will then be a multiple of b of x, right, it will be some polynomial times b of x, okay, b of x, so the way to read this is to say b of x divides a of x, okay, so that's the way to think about, okay, so there's also a notion of devices, okay, if you can find some polynomial b of x that divides a of x then b of x is supposed to be a divisor of a of x and based on that you can conclude whether or not devices exist for any polynomial, okay, so if you take any integer, the first thing people are interested in maybe interested in finding out is whether it's prime or not, does it have any devices, right, so one of those things which you learn a lot, I mean can you factorize an integer, so same way since division works, one can also imagine factorizing a polynomial, okay, but it'll be slightly different because see there's always, so there's always a trivial factorization of an integer, right, any integer can be written as 1 times n, okay, and you have to rule that out, you can't consider that as a proper factorization, similarly for polynomials there'll be more trivial factorizations, for instance if you want to see an example, okay, suppose I take a polynomial say a of x, let's say x squared plus 2x plus 1 and let's say it's in f5x, okay, suppose I say this, okay, suppose you want to think about factoring a of x, okay, so you want to write a of x as a product of two other polynomials, okay, there are various trivial ways of doing it, for instance I could write it as what, 1 times this, okay, that's too trivial, maybe 1 is not really something that you want, but you can also write it in a different way, I can write it as 2 times some other polynomial, what will I put here, I have to write it as 2 times, yeah, so you can divide this by 2, you can divide each coefficient by 2 or multiply each coefficient by 2 inverse, what's 2 inverse in f5, 3, right, so you multiply each coefficient by 3, you will get a polynomial there, so you would write 3x squared plus x plus 3, right, so is this a proper factorization, you have to rule out all these things, this is also trivial, okay, so this is also a trivial factorization, okay, so this is one thing I think it's true for any field, but I want to remind you that these factorizations are not considered factorizations, okay, so only what, which is considered a non-trivial factorization, if you can write it as, see each of these things should be polynomials of degree at least 1 and degree less than the total degree of a of x, okay, so that's only then it's a non-trivial factorization, right, each factor should have degree at least 1 and its degree of course will be less than the degree of a of x, okay, so that's non-trivial, okay, so can you find a non-trivial factorization for a of x here, x squared plus 2x plus 1, okay, so maybe from your knowledge of formulas for x plus 1 whole square you might be able to write it, but remember this is f5, will anything change now, won't change, you have to just reduce everything model of 5 and I've chosen numbers so that you're comfortably safe within that limit, okay, so you can see this will factor nicely as x plus 1 into x plus 1, okay, okay, so one thing you might have learned finding linear factors is very easy, okay, and for polynomials finding linear factors is typically very easy, okay, so why is that true, so this is how you find linear factors, finding linear factors of a of x is, okay, so how do you do that, if you want, if you have to look at that, there is a way of, there's a way of doing it, suppose, okay, suppose, suppose you have, so suppose let me say a of x belongs to fx, okay, so here's this result which will help you find linear factors, okay, a of alpha equals 0, okay, for some alpha in f, it'll take alpha in f, okay, if and only if, what, based on the division algorithm, x minus alpha divides a of x, okay, so a very familiar root based linear factor approach, you can easily, easily prove it using the division algorithm, right, right, so you put in a of alpha there, you'll see r of alpha will also be 0, okay, just work out very, very easily, okay, so you can show these things very, you know, very straightforward fashion, okay, so you can use this result quickly to find linear factors, okay, so even in a case where it may not be very obvious, the factorization may not be very obvious, right, the linear factorization, you can use this result to quickly find linear factors, so I'll throw a few polynomials at you now and ask you to find and ask you to try to factorize and the first attempted factorizing is to try to find linear factors, okay, finding higher degree factors is not as trivial as this, okay, so it can be a little bit more complicated, but linear factors can be very, very easily found, okay, so let's see a few examples to just try home this point, okay, so the first thing we'll see is the polynomial x square plus 1 belonging to f2x, okay, we'll see a very, very simple case, okay, how do you find linear factors, okay, you have to find roots for x square plus 1 in, let's say in f2, okay, I'm only trying to factorize it in f2x, okay, so you want to find factors for this in f2, okay, so f2 contains what, only two elements, 0 and 1, okay, the roots can be either 0 or 1, just substitute each of those values and see which one is the root, okay, so is x equals 0 a root, obviously not, okay, x equals 1 would be a root, so you know what, x minus 1 but x minus 1 is the same as what, x plus 1 in f2x, so x plus 1 has to divide x square plus 1, okay, so that is one thing we have found, okay, so you find that, okay, so if there is one linear factor for x square plus 1 what will be the other factor, it will also be a linear factor, right, in this case it will actually also be x plus 1, in fact x square plus 1 will be x plus 1, the whole square in f2x, okay, it's a very strange factorization, so you should immediately be reminded of, you should immediately convince yourself of one thing, okay, the same polynomial, if I change the field, the factorization will change, okay, so for instance instead of f2x if I had said this is qx, okay, will you have been able to factor it? No, in qx it doesn't factor, what about rx? Real number field rx, even there it doesn't factor, where will it factor? If you go to complex fields you can factor this, okay, so you can do that, so there is something there which is very interesting, so a polynomial that doesn't factor in a field can possibly factor in a larger field which contains this field, okay, so that can happen, yeah, okay, so that's something to watch out for, okay, just to drive home the point, let's try to see x square plus 1 the same polynomial, but I'll say now this is in f3x, so first thing you should convince yourself is x square plus 1 in f3x is a completely different polynomial from x square plus 1 in f2x, okay, the 1 is totally different, okay, the 1 that you had in f2x, if you add 1 to 1 what do you get? 0, in f3x the 1 you have, if you add 1 to 1 what do you get? 2, okay, it's not 0, and you know it's definitely not the additive identity in f3, okay, it's completely different, okay, so these two polynomials while I've written them down exactly the same way, they have the same only by looks, okay, by behavior they are completely different and you should not say these two are equal, okay, remember that, okay, but try to factor this, let's see, okay, you don't have any factors, okay, so it's fact is it is, the correct term is it's irreducible in f3x, okay, so if a polynomial has no factors you say it is irreducible, okay, so this factor is irreducible, this polynomial is irreducible in f3x, okay, so a polynomial which has no factors called irreducible and there is some distinction between irreducible and prime, in this case we can say it is prime, okay, that's not beat the point too much, but technically there is a distinction between irreducible things and prime things in theory, but we will just simply say it's irreducible, we just stick to irreducible, we won't say prime, okay, we just simply say irreducible, that's a convenient thing, we can think of it as prime numbers if it's confusing you too much, okay, this is a prime in that, okay, so you can keep looking at examples like this, it's quite intriguing to see the same polynomial factor in different ways in different fields, okay, so let's see one more example, okay, I went from 1 to 3, I think it's because I've read it f3 here, I'm sorry about that, I'll come back to my third, let's look at x squared plus x plus 1, I want you to look at this in f2x and also x squared plus x plus 1 and f3x, okay, let's see, try to do the factorization, in f2 it's irreducible, right, so you're able to quickly find it's irreducible, in f3 what happens? Yeah, it factors as x minus 1 times, see I have to be careful about minus in f3, right, because minus 1 is not the same as plus 1, okay, x plus, is that also x minus 1? Yeah, so it's actually x minus 1 squared, okay, so remind yourself that, okay, this is another reminder that strange things can happen when you change the field, okay, a polynomial that looks exactly the same will behave in a completely different, in one case it is irreducible, another case it has, no, don't, okay, so you can write it as x plus 2, but it's the same as x minus 1, it makes no difference, okay, so the integers, when you see them modulo something, you can write one integer modulo m in any ways you want, suppose you're looking at this modulo 3, modulo 3, 2 is the same as minus 1, it's the same as 5, it's the same as 8, right, I mean all these numbers are the same, once you say modulo 3, okay, so I like writing it as x minus 1 squared, okay, it's not the same as anything else, okay, so let's go to slightly more complicated examples, but now I'll just stick to f2x, okay, I'll just stick to f2x, okay, so let me start with x bar 3, I'm sorry, I wanted to write x bar 3 here for some reason, okay, from now on we'll stick to f2x, you can go to fpx, it's more confusing, just stick to f2x, okay, but we have to be slightly more careful when you look at higher degree polynomials, right, so for cubic polynomials, if it is reducible you will have a linear factor, I may write it wrong, you have to have a linear factor, right, you cannot escape a linear factor, if you have a factor of degree 2 what will be the other factor, it will be linear, okay, so anyway you should have a linear factor, okay, without having a linear factor this will not, this will not reduce, okay, so it's enough if you check for roots, okay, so you put in x equals 0, you don't get a root, x equals 1 you don't get a root, you can conclude this is irreducible, it's not a problem, okay, so let's try something more, something like say x bar 4 plus x plus 1, okay, so you can only say no linear factor now, you have to rule out degree 2 factors, I'm sorry, yeah, suppose it factors into 2, you can write it as a of x into b of x, what can be the degrees of a of x and b of x, yeah, so anyone has to have one degree, cannot have a linear factor, but this polynomial it's not true, it could have 2 factors both of degree 2, okay, so you have to rule out degree 2 factors as well, linear factors you can immediately rule out, right, you know immediately there's no linear factor, okay, so you put x equals 0 you don't get anything, x equals 1 you don't get anything, it's okay, but first for degree 2 factors you have to do some more work, okay, it turns out actually it's enough if you eliminate irreducible factors of degree 2, okay, right, so when you check for, check for whether a number is composite or not, integer is composite or not, what do you do, you only divide by primes, you don't have to divide by other composite numbers, right, if it's divisible by a composite number obviously it's divisible by a prime also, so it's enough if you divide by irreducible polynomials, okay, so you have to check for irreducible polynomials of degree 2, if you do that search you'll see this guy is the only irreducible polynomial of degree 2 and f2x, okay, look at all other degree 2 polynomials, all of them will be reducible, okay, if you want you can take a minute and enumerate all the degree 2 polynomials of in f2x, how many degree 2 polynomials will you have in f2x? Eight, right, eight, am I right? Okay, degree 2, man, degree 2 means one of the coefficients is 1, okay, the coefficient of x squared is 1, that's what I mean by degree 2, degree less than or equal to 2 is 8, you're right, if you say degree equal to 2, how many? 4, okay, that's correct, okay, you can write down all the 4 and you'll see this is the only irreducible polynomial, okay, all the others will have linear factors, okay, it's very easy to show that, okay, so it's enough if you check if x power 4 plus x plus 1 has x squared plus x plus 1 as a factor, please check that, how will you check it? You have to divide, that's all, you have to use long division, there's no other way here, you can't, right, you have to divide x power 4 plus x plus 1 by x squared plus x plus 1 in where, where should you divide that? In f2x, okay, do that long division and check if the reminder is 0, how do we know irreducible polynomials? Yeah, basically you have to list all the polynomials and check if it's irreducible, well, there are smarter ways and it's actually, okay, I'll comment about it as we go along, yeah, but for now, for starters, you can say I list all the polynomials and check whether each is irreducible, yeah, so it's, it's not a factor, right, you'll have a non-zero reminder, so this, this will end up being irreducible, okay, so you can do these checks if you want, okay, so it turns out there are quite a few irreducible polynomials, okay, it's not surprising and just like I said, there's an analogy between these polynomials and integers, right, these polynomials over a field behave very much like integers and there are quite a few primes as well, right, okay, how many primes are there roughly? There's some count, no, how do they grow in n? If you look at all the numbers from 0 to n, how many primes roughly can you expect for large n, log n, right, so that's the result, so log n is quite a large, when it grows with n, it doesn't, doesn't go off down to 0, right, so there are quite a few, so likewise here also, there are quite a few irreducible polynomials, so the typical way of finding irreducible polynomials, he asked a question is to randomly generate polynomials of a given degree, okay, and keep checking, you'll quickly find an irreducible polynomial, okay, so it's not very difficult in practice, it turns out one can show some very nice results for every m in n, okay, positive integer m, there is a, there is at least one, in fact, typically more than one, at least one irreducible polynomial of degree m in, I'll say fpx now, this is what we need, p prime, any p prime, this is the result that we need now, in fact, you can generalize this a little bit and show any finite field also, it will be true, okay, so we know there is for every degree m in every characteristic p, okay, in every fpx, I'm sorry, I haven't said anything about characteristic items, sorry for that, in fpx, there will be at least one irreducible polynomial, okay, so that's the result we'll accept without proof, okay, if I have to prove it, it's quite unnecessary, it's a tangent, so we'll accept this without proof, okay, so that is just like we accept that our prime numbers, we'll accept this also, okay, so this is true, okay, any questions about irreducible polynomials, okay, these will play a very important role in the construction, so if there's something disturbing me, disturbing you about this, you should ask me right now, is there any question, okay, all right, good, all right, so we've seen enough facts, I think about polynomials and this is pretty much what you should need, okay, so I don't think we'll need anything more, so we'll, let's try to jump into construction of finite fields, okay, so we'll do it, we'll do this in a very simple fashion and in many ways this, the way I do it will not be very rigorous as well, okay, so if you're interested you should read books and a good book, very good book is Algebra by Michael Aten, a cheap edition is available, strongly suggest, very strongly suggest that you are interested in these things, please buy this book for yourself and go ahead and read it, it's a very good book and if you want to really understand all the technical details you can look at it, okay, so I'll do it in a way which is simple, simple and intuitive and appeals to all, okay, it's not a very rigorous way, okay, so first thing we'll see is we'll assume, okay, so far I've not shown you anything, only finite fields we've seen so far are what, fp, okay, so we've seen fp, p for p prime and hopefully you're reasonably familiar with this, okay, so one of the results that I showed for groups which will be very crucial later on also as we go along is order of an element, okay, so order of an element in a finite group, what does it do, it divides something, what does it divide, divides the size of the group, okay, so if I take for instance fp, fp star is what, it's a multiplicative group, multiplicative order of any element in fp, okay, we'll divide what, p minus 1, okay, so be very careful, if it has to divide p then you'll be happy, because very few things divide p, okay, so it has to divide p minus 1, okay, so just a reminder so that you don't forget that result, okay, so now we will try to prove some properties assuming fq exists for some q, okay, I have not said anything about existence of fields, I'll say, suppose, no, I didn't say such things, I said any two fields of the same size, there'll be isomorphic, that is the result, I never said the size is only p, and I'll come to it, there is a close relationship between fp for a prime p and any field that you can possibly have, I'll come to it, there's a very close relationship but it's not definitely not isomorphic, there are other fields than fp, fields other than fp, okay, suppose fq exists for some q, okay, some very powerful results are possible for q, okay, just based on simple arguments, you can show some really, really powerful results for what q can be, okay, some of those surprising results, if you haven't seen it before it'll be very surprising for you, okay, so suppose fq exists, okay, I don't know anything about fq, I'm just saying fq exists, okay, what all should I define for defining fq? I should define a set with q elements, okay, I should define, I define it such that fq with an addition operator becomes a group, okay, and then I should also define it so that the fq star which is the additive's identity thrown away should be a multiplicative group, okay, I'm not saying anything about all those things, okay, I'm just saying it, I assume it exists, okay, I don't know about the elements of fq but I know two elements of fq should always be there, what are the two elements that I know should be there? The additive identity of the additive group fq and the multiplicative identity of the multiplicative group fq star should be there, okay, 0 and 1 belong to fq, I know these two will be there, okay, so one might start by saying what is 0 equals 1, okay, so if you multiplicate the identity and additive entity are the same, you don't get anything interesting, okay, so it's just everything is 0, okay, you don't get anything, okay, so we'll obviously assume 0 and 1 are not the same, okay, so what's the only way of coming up with other elements, okay, I know there are operations that are plus and dot, those two operations have not defined but they exist in this field, if the field exists they have to be there, using those operations, can I create more elements of fq just using 0 and 1, okay, right, you see that's just about the only way I can create, the only way I can create more elements is if I add 1 to itself, right, if I do anything else I won't get new elements, okay, right, so I could think of 1 plus 1, okay, this will belong to fq, okay, remember I don't know what plus is, I'm just saying there should be a plus and I can always add 1 to itself and get another element in fq, there's nothing that stops me from repeating this, I can add 1 plus, 1 plus 1 and I'll still be in fq, okay, so now since I can keep on doing it I'll need a shortcut, shorthand, okay, so instead of writing 1 plus 1 plus 1 n times, okay, what will be my shorthand, okay, remember I'll write this simply as n, okay, I'll write this simply as n itself, okay, let's assume that it's n times 1, okay or n, n, 1 added to itself n times, it's what it's assumed to be, okay, n, okay, now I know q is finite, okay, so remember q is finite, right, okay, so but this n can keep on going, right, it can keep on adding and obviously since fq itself is finite, since all these things have to belong to fq, eventually what should happen? Yeah, you should go to 0 itself, okay, in fact it has to repeat but from that argument you can easily show that there will be a smallest n for which n will be equal to this 0, the additive identity is 0, okay, do you notice that, okay, so since q is finite, there exists smallest n belonging to n such that n equals 0 in what, in fq, okay, this n that I defined in fq, how did I define n, how did I define n in fq, I took the 1 and added to itself n times, that's how I got the n in fq, okay, this n in natural number is little different, right, that's just the number, okay, there will exist a smallest n such that n is 0 in fq, okay, the next surprising result is, okay, this smallest n has to be prime, has to be a prime number, okay, okay, so how will you argue that it has to be? Okay, that all those things need a lot of proofs, okay, I haven't come to that, eventually I'll show some isomorphism like that but let's not go to that but why should, what's the simple argument for why? Exactly, see, so why are you going to fp now? There's no going to fp, there's no field, you know, I mean just let's not jump into something else, eventually we'll get that, the simple argument is suppose n were not prime, I can write it as a times b and even in fq that will hold, right, okay, so how do you prove it? If n is not prime, okay, if n is not prime, if n equals ab in n then n will be a times b in fq as well, okay, why will that be true? You can, you have to write that and prove it by the distribution properties, you can prove these things, okay, you can take the distribution property of the field, you can take the distributive property of the field and you can show a times b will be n in fq, okay, it's very easy. You remember, how did I define a in fq? What is a in fq? 1 plus 1 plus 1, a times, what is b in fq? 1 plus 1 plus 1, b times, if you multiply those two, will you get 1 plus 1 plus 1 n times? You can use the distributive property of the field, it's just one, just keep on doing it, you'll have same one added to itself n times, so we'll get the n, okay, now if n is 0, what have I shown? a or b has to be 0 in the field, okay, so then what will be violated? n will not be the smallest entry anymore, so obviously there should have been a smaller number, okay, so therefore n has to be prime, okay, and 0 equals a b implies a is 0 or b is 0 and you get a contradiction, okay, so therefore n has to be prime, okay, so this prime number which is the smallest number for which 1 plus 1 plus 1 added that prime number of times equals 0 in fq is called the characteristic of fq, okay, so this is the definition of characteristic, smallest prime p for which p equals 0 in fq, okay, remember what is p in fq? 1 plus 1 plus 1 added b times, okay, so remember this is a very powerful result, okay, so I never knew, I don't even know if fq exists, okay, I only know, so if somebody tells me that fq exists, then there has to be a characteristic for this fq which is p, right, it's a prime number, which is a prime number p and that p will be 0 in fq or 1 added to itself, p times will be 0 in fq, okay, so what have I shown right now? So just based on this, I've shown 0, 1, 2, 3, all these till p minus 1 will be actually a subset of fq, this is what I've shown, okay, for some p prime, okay, p is prime, okay, I will call this set as fp, okay, remember I have not shown that this is the field fp, okay, I'm going to call this set as fp, right now this is some subset of fq, okay, and all the operations among these elements are defined in fq in some field which I don't even know whether it exists or not, okay, but the only thing I know is it's got p elements and each of the elements are 1 added to itself that many times, if I say 2, it's 1 added to itself twice, if I say 3, it is 1 added to itself 3 times in fq, okay, of course since I've called it fq, what am I going to show? I'm going to show next that this set, this fp will be exactly the fp that we saw before, okay, it'll be isomorphic technically but we'll simply take it as equality, okay, okay, so that's the next result we'll see, see fp contained in fq is actually the finite field fp, okay, so how do you show these things? How do you show that this set will be the same as the finite field fp? You have to show, yeah, exactly, you have to show every operation here will result in the exact same operation in the original fp that you knew, it has to, right, you take any two things and add, what are you doing? You're taking one and adding to itself, so it'll be the same thing as the sum and why do you have to do modulo p? Where will the modulo p come? Because p is 0 in my fq, I've shown that, okay, so modulo p will naturally come because p is 0, whether you multiply or add, multiplication is also easy because then use the distributive property, okay, so it's all just sums of ones, so you can show a times b is same as a times b, where will the modulo p come? Because p is 0 in fq, okay, so the exact same thing will come, okay, so you can write down the proof, this requires proof, okay, so based on your familiarity with mathematics, it might be easy or difficult for you but it's very simple proof, there is no fancy concept here, okay, so imagine what we have shown now, it's quite powerful, I've said fq is some field, I don't know if it exists or not, but just because it exists, what should it contain? It should contain some fb, that's the only thing that we have shown so far, okay, is that clear? Any questions? Any thoughts on what is it that I've actually shown and what is it that I'm claiming? Oh, it's okay, it's clear enough, okay, okay, so the next statement we'll show will be slightly more, yeah one of those to see because if one of them are non-zero, suppose a is non-zero, it'll have an inverse, multiplicative inverse, so multiply both sides, you'll get b to be 0, so any one has to be 0, but that can be shown, okay, so there are some the next result is even more interesting, okay, so now any, so you have a bigger set fq, which contains the field fp, okay, and what can you do in fq? You can add two elements of fq, and it contains a field fp and you can also multiply two elements of fq, okay, so naturally you'll get addition and scalar multiplication and fq will become a vector space over fq, over fp, okay, so I don't know if you're not familiar with abstract vector spaces, this might be a very surprising fact to you, but what is how do you define a vector space? What's the definition of a vector space? It's a set of vectors over some scalar field, okay, so you need a scalar field and then you need a set, what are the, what are the, what else do you need for a vector space? You need to be able to add any two vectors and result in another vector, can you do that in fq? Can you add two elements of fq and result in another element of fq? Yeah, that's what the addition operator will give you, okay, and now my base field is fp, can I multiply an element of fp with an element of fq? Yeah, I just used the multiplication operation that was defined in fq, I can multiply, I'll still be in fq, so both of those hold and fp is a field, I know it's a field, so fq will have to be a vector space over fp. Sir, can we tell that any field is a vector space over itself, yeah, if you take a field, yeah, it will do, right, so you notice the argument, okay, so it's a slightly abstract argument, but basically the way you argue is if you have a and b belonging to fq, a plus b belongs to fq, that's my vector addition now, this is addition in fq, which I will call as vector addition if you want, okay, what about multiplication? If you have a belonging to fp and b belonging to fq, how can I do a, a dot b? Yeah, it will also begin in fq, this will be multiplication in in fq, I will call it my scalar multiplication, okay, I can happily do this, okay, so fq will have to be a vector space over fp, okay, now what do I know about vector spaces? Once I come to vector spaces, I mean very familiar domain, okay, fq is in fact finite, it's only got a finite number of elements, so it's very easy to see, fq will in fact be a finite dimensional vector space over fq, fp, okay, so that's the next step, so more is true, fq will be a finite dimensional vector space over fp, okay, what else is true? Suppose I say, suppose dimension is m, okay, it has to be some finite number m, I will say dimension is m, okay, now what else do I know about vector spaces? What else do vector spaces have? How do I describe all the vectors in a vector space? Use a basis, okay, so that implies now there exists a basis, let me call it b1, b2, how many vectors will be there in the basis? bm for fq over fp, what does it mean now? An arbitrary vector in fq, let me call it alpha, I'm just sorry, x, x, what else? Alpha and fq can be written as alpha equals a1, b1 plus a2, b2, where will all these gates belong? ba will belong to what? Fq, okay, it's the basis but there will be elements of fq, right, all basis vectors are belong to the vector space, right, so we all belong to a2, b2 plus am, bm, so where those basis, where do ai belong? fp, okay, for every alpha and fq, I can uniquely write it in this form and it's also vice versa, for every set of ai that I come up with, I will have an element in fq, so how many elements will there be in fq? How many choices do I have for each ai? p power m, so each ai can be an arbitrary element of fp, right, so that would be p possibilities, I have m of them to choose, so q has to be equal to p power m, okay, so I wanted to step back and imagine the power of abstract mathematics at work, okay, so started with fq, you don't even know if it exists or not, okay, said somebody tells you fq exists, the two pages of work, you can come to the fact that this q has to be equal to power of a prime, it cannot be 10 for instance, okay, so that's an amazing thing that we have shown, okay, so it's quite non-trivial, okay, you should be surprised by it if you're not surprised, okay, so just because somebody said a finite field exists for q, you can show q has to be equal to p power m and the proof is very, I mean it doesn't involve any magic, right, it's some very simple step-by-step argument, there's nothing more to it, I don't think I've really missed out on any important step in the proof, I've shown pretty much everything, okay, so it's quite simple to see, q will be p power m, okay, so now this analogy is more powerful than just saying q, just giving you an idea of what q can be, okay, what else have I done, how did I go to the vector space, what was my addition in the vector space, it's the exact same addition as in Fq, so what will be addition in Fq, it will be addition of vectors over Fp, so how do I add two vectors now, okay, addition in Fq, suppose I ask you to add two vectors, alpha 1 and alpha 2 in Fq, if you have to add it, what will you do first, first write it as a vector over Fp in some basis, okay, we'll choose some basis and write it over Fp, okay, so you write alpha 1 as a11 b1 plus a12 b2, so until a1 m, bm, and then you write alpha 2 as a21 b1, a22 b2 plus a2 mbm, then what is alpha 1 plus alpha 2, it's very easy, this is simple vector addition, a11 plus a21, why do I know how to add a11 and a21, both of them are in Fp, simply add mod p, there is nothing more to do, okay, b1 plus so on till a1 m plus a2 mbm, in fact, we have a representation for each element of Fq, okay, we don't know it, we didn't, do you see that it, Fq is just an m dimensional vector space over Fp, how will you represent vectors, m dimensional vector space over a field, you can use coordinate vectors, okay, suppose, okay, only thing I don't know is I don't know the basis, right, I haven't told you what the basis is, I don't know the basis, but I can represent every vector as simply a coordinate vector, I just won't be able to translate it into the actual vector, but I will know what the coordinates are, right, just list out all the coordinates, there are p by m possibilities, right, simply write it down, okay, and you can even add them, okay, only thing is you won't know what they actually are, okay, you will only know their coordinates, okay, so in your, in the vector spaces you are used to, like R3 and all, coordinate is everything, right, you don't care beyond that, but this vector space coordinate is not everything, because you don't, you don't know the real elements, the b1 and bm, you don't know it, you don't know it, actually it will turn out also very simple, we can find that very easily, but we don't know them yet, but we can represent each element of Fq as coordinates, and in fact add them also, there's no problem, just add the coordinates, individual coordinates are added, modulo p, okay, so let's see that example right away, okay, suppose I say F16, okay, I haven't told you if it exists or not, suppose somebody tells you F16 exists, okay, then I can write F16 as what, I can write it as 00000010010 0011, so on till what, 1111, okay, and I even know how to add, how will I add 0011 and 1111, simple vector addition modulo 2, that's all, for instance 0011 plus 1111 will be what, 1100, I know it belongs there, what is it that I cannot do in a field, what else should I be able to do, should be able to multiply and I have no clue how to multiply, because vector space will not take you anywhere as far as multiplication is concerned, right, multiplication, how did I argue for multiplication I said, I have to use the multiplication in Fq, I mean I haven't defined what multiplication is, that's the only other trick we have to add to this construction, okay, so we already know what F2 power m or Fp power m will look like, it'll be all coordinate vectors over Fp and I also know how to add them, okay, the addition is also going to be the same as vector addition, the only thing I don't know is how to multiply two elements of Fq, okay, you don't think of multiplying two vectors, but I mean it's the same, how do you multiply two elements of Fq, how do you multiply these two, to any two vectors in this, okay, so hopefully you're convinced by now that finite fields are the simplest things in the world, yeah F16 will have to be over F2, no, right, what will be the characteristic of F16, has to be 2, okay, so right, why does it have to be 2, q has to be p power m where p is the characteristic, okay, so maybe I didn't write it down, Fq exists implies q is p power m where p is characteristic, okay, any q can be uniquely written as p power m, if it has only one prime factor it'll be very easily figured out, so once I give you q you can find out what the characteristic will be also, okay, so there's no F16 over F3, it's nothing like that, only for vector spaces you have, since I've said it's already a field that is a more complicated multiplication, once you define that you'll see, I mean it cannot be over arbitrary fields, it doesn't work, okay, so any questions on how this worked out, okay, all right, okay, so good, so the next thing we have to show is multiplication, okay, and for multiplication I will use, I will basically what I'll do is I'll present the construction for Fq for any q which is p power m and then I'll argue why the multiplication works, okay, but I want you to think back over Fp, okay, if you had worried only about addition, Fn should have worked for any n, okay, multiplication was made possible why, we're doing modulo p and p was prime, okay, so we'll use a similar idea here, we'll construct a set of objects, modulo some other thing for which will make multiplication work, okay, so basically that's the next thing we'll do, I'm dangerously close to the end of the period, I think we have 10 more minutes, I think I have time for doing this, okay, so let's do construction of Fq, okay, I think I already wrote down, construction of Fq for q equals p power m, p prime, okay, I know all these things now, previously I didn't know, I know all these things have to be true, Fq cannot exist, just by the same proof you can show that Fq cannot exist, okay, I assumed, I started with the assumption that Fq exists and then I showed q is p power m, so if q is not p power m obviously that cannot exist now, anytime it exists you have to come to this q equals p power, okay, all right, so this is how you construct it, I'll simply present the construction and then I'll go back and appeal to this as well, okay, so you can show Fq which is defined as a of x in Fpx such that set of all polynomials in Fpx such that degree of a of x is less than or equal to m minus 1 can be made into a, can be made into a field, okay, okay, first of all some sanity check, okay, first of all does the set have p power m elements, if I say set of all polynomials in Fpx degree is less than or equal to m minus 1, it will have p power m elements, okay, that's the first check, based on what I know about addition in Fp, Fq, does that same thing work, what if I add two polynomials, what will I get, is it the same as vector space addition, yeah, it's the same as vector space addition over the coefficient because these x powers are just placeholders, they're just telling you what to add and it's exactly like vector addition, so both of those work, okay, so whatever I know about Fq is not contradicted by the set it and the polynomial addition operation, so I can make the addition a simple polynomial addition, okay, okay, so for multiplication we need we need something prime, right, how did multiplication work the last time, you did it modulo, some prime number, here we will need an irreducible polynomial of degree m, okay, so I'll start with pi of x being an irreducible polynomial, irreducible in Fpx degree equal to m, okay, why do I know such a polynomial will exist, yeah, I just gave you the result, okay, so it's true that for any m, any degree m I'll have an irreducible polynomial pi of x, I will define multiplication modulo, polynomial multiplication modulo p of x and pi of x, okay, so what do I mean by modulo in polynomial things, when I multiply two polynomials and if the degree is less than or equal to m minus 1, obviously the degree can be greater than, right, so if I take two polynomials from this field and multiply I can go outside, how do I reduce it, I'm going to say modulo pi of x, what should I do when I do modulo pi of x, I should divide by pi of x and take the reminder and the reminder will definitely have degree less than or equal to m minus 1 and it will belong here, okay, so this multiplication satisfies the closure property, okay, you multiply any two, you will definitely come back to this field, there's no problem, okay, but what's the crucial thing you have to check in finite field, in field multiplication, inverse and inverse will exist by the same argument that I gave before, okay, you simply take a polynomial and multiply by all other polynomials modulo pi of x, no two of them will be the same, why, if any of them are the same pi of x will have to be reducible, the same argument is before, okay, since that cannot happen no two of them have to be the same, since all of them are distinct and you have exactly q of them, one of them will have to be equal to 1 and that will be the inverse, okay, so there's no change in the proof, okay, so that's why I kept saying these polynomials are the same as integers in many ways, okay, whatever you did with integers you can do with these polynomials also, you'll have these prime numbers as far as factorization is concerned, they're very very similar, okay, so that's the proof, I don't even have to write it down, right, all of you are convinced you're all shaking your head very vigorously, okay, right, so it's a very easy proof, the same proof will carry over as an Fp, okay, just pay the same proof this will work out, okay, so that's it, that's the end of the construction, these four lines, okay, who were told you finite fields are difficult, I don't know what they were saying, right, they're the simplest things in the world, okay, how do you construct any finite fields, simply take all polynomials of degree less than or equal to m minus 1, how do you add simple polynomial addition, how do you multiply, you find the reducible, find some irreducible polynomial of degree m, you do model work, that's it, exactly, that can be a basis, so now we can go back and find basis and anything else if you want, okay, so I know what the objects are, I know what the vectors are, okay, so I can easily go back and find basis elements and all, I'll come back to that more, okay, all right, so we'll stop here, do more