 Okay, so let's begin, so now we've been looking at, in yesterday's class, we were looking at my god, we are now not looking at everything, so let's look at this, let's put serious problem, it's not coming back again, okay, so this is fine now you can see everything, so we've been looking at polynomials for the finite fields and I was trying to point out that there is some connection between, this is a very important connection between irreducible polynomials and fields, the danger of keeping it on this side, it seems very loose and I'm not able to tighten it See anything that, maybe I'm doing it the wrong direction, alright, so I learned to look at that So we've been looking so far at polynomials or finite fields, specifically I think they gave me some examples of f2x and then f3x and we saw that there are quite a different variety of polynomials Some of them are reducible, there are many many which are reducible, right, so I think and I was commenting how in the infinite fields like the rational field and the real field and complex field, we take reducible polynomials and get bigger fields from them So similar thing is true here and we'll now first see some constructions of other finite fields Some simple examples, very quick, just to give you a feel for how it works and then we'll do it a bit more formally, okay, so the first field we're going to see is what's called f4 It's also denoted gf4, I'll come back to that later f4 is a finite field with four elements So I'll show you the construction now, it's a very simple construction I'll see how easy it is to think about this construction and we can add to it later So the crucial idea is polynomials over f2, so f4 can be thought of as constructed using polynomials over f2 So this is the construction, I'll provide the construction first and then we go back and interpret it carefully So f4 can be constructed as follows So any field will have the additive identity which we'll call 0 and then outside of the additive identity you'll have the multiplicative group and that will have a multiplicative identity which we'll call 1 and then f4 should have just two other elements So let's see what we can do with 0 and 1, can you do anything with 0 and 1 to create new elements? The only thing you can possibly do is 1 plus 1 But I'm going to say 1 plus 1 is 0 So I'm going to put that constraint here in f4, it turns out if you don't have 1 plus 1 is 0, you can never have an f4 So we'll come back to that later on, but anyways, f4 will have 1 plus 1 being 0 So basically addition will be modulo 2 So it will continue to be modulo 2 addition like we had before So plus 0, 1 and then we need two other elements So the way to think of them, one way to construct them is to say I'll have the two elements as alpha and 1 plus alpha So this is one description, we'll come back and generalize this but I'll tell you how to make this into a field So first of all this is alpha and 1 plus alpha, we don't know what they are But let's say this is the field, but I have to now define multiplication But as far as addition is concerned, check real quick that addition is okay You can check here now real quick that addition is okay If I add 1 with alpha that will happen, I get 1 plus alpha, that's there If I add alpha with itself that will happen, alpha plus alpha is going to be 0 Because I'm going to say that it's 2 alpha and then addition is modulo 2, so it's 0 So you have to treat alpha like an indeterminate unknown variable Treat it like an indeterminate Or if you really don't like x, you can treat it like your famous i i is square root of minus 1, so you know how comfortably you are using i Every time we deal with complex numbers, think of this alpha like the i But instead of the complex numbers and real numbers, we have this over just f2 So f2 and alpha is some indeterminate variable which we are using to define f4 So the best way to think about it is like it's i So i is square root of minus 1, we have alpha Now only thing that's undefined now is multiplication So for multiplication we need some rule, so for that it turns out the following rule is good enough You will say alpha square equals alpha plus 1 So I can do that So if you remember in the complex multiplication Oh my god, did I make some mistake with that We use that or not? What is that analysis? Excuse me, I'm telling you Anyway, so alpha square is alpha plus 1, so for instance in the complex numbers Thank you Anyway, so for the complex numbers you also have a similar rule for multiplication, what is that rule? i square is minus 1 So like this I'm going to say alpha square is alpha plus 1 So now I have to check that this is a proper field So even for the complex numbers you would have checked this I'm not sure if you remember, but you would have checked that a plus b i had a multiplicative inverse Right, 1 by a plus b i, how do you do 1 by a plus b i? A minus b i by a square plus b square So you know that, that is the inverse of 1 by a, a plus b i So if you do a plus b i times a minus b i by a square plus b square you would get 1 As long as a and b are either of them it's not 0 You know you can always find a 1 by a plus b i Except that it's only finite, it's much easier to check For 1 it is trivial, multiplicative inverse of 1 is 1 itself, what about for alpha? It has to be a minus b i by a element, right? And I already told you that it's a free, so you know it exists So what has to be the multiplicative inverse of alpha? S b 1 plus alpha, right? So you have to work on that So you do alpha times n plus alpha, what will you get? You get alpha plus alpha squared, then alpha squared is alpha plus 1 So it cancels and you will get 1 So everything is satisfied and this becomes a b So the first thing to get used to in finite fields is There will be an indeterminate alpha In every finite field you will have that This is similar to the indeterminate i that you have in the complexes And it turns out there are these whole bunch of wonderful objects called algebraic number fields Which you would have never learnt I think in the engineering mathematics that you saw Where you can extend the rational numbers in a similar way You can add an indeterminate alpha or beta and extend the rational numbers So unfortunately in your math you have never seen it, it's a nice thing If you have seen it then this finite field will be easier for you But if you haven't seen it, if the only thing you have seen is complex numbers Then the i is good enough, okay? i square is minus 1, like that In every finite field we will see we will have an alpha which is an indeterminate And it will have satisfy some rule, okay? That rule will be useful for multiplication For addition you won't need anything, it will be very simple So this will be a general form of the construction So I will show you one more example and then we will plan into the general theory Where we will see how to think of these things in general In general there is a very different way to think about it There is also a very strong nice structure in it But for now we will just look at one more example So we should erase this You know this monitor is not smart enough to distinguish between the pen and the finger Maybe somebody should design a better monitor Alright, so let's move on to another example The example I want to give is F9 It's a slightly more complicated example So here it turns out in F9 also You will have 0 and 1, is that a question? Is that a question? What is the question? Is that a problem? How do you justify? I will come to it slowly enough So the way you think about it, we will think about it in some other way That's my choice, I mean this is a valid field now, I have an F4 That there is more to it But I know I am not justifying all my choices, I will do it later This is just for you to get used to how these things will look Suddenly if you don't have a field for these things I will use this in a lot of examples also So it's just good to have a couple of examples for this So F9 is once again a field with 9 elements So I am going to have a 0 and a 1 So here I will have 1 and then I will have 1 plus 1 I won't force 1 plus 1 to be 0, I will call 1 plus 1 as 2 Just using our familiar number system, 1 plus 1 I will call as 2 But 1 plus 1 plus 1 I will force 2 as 0 So it turns out for F9 you have to have something like this Otherwise it will not work So if you want a field with 9 elements, you have to have 0, 1 And you should have here 1 plus 1 doing something And you should have 1 plus 1 plus 1 is 0 So these are obviously not trivial facts Like in non-trivial, we will prove it later on But for now I will just accept it So we should be 0, 1 and 2 And then we will have nothing else So now we need 6 other elements to complete this F9 And once again I will use some indeterminate alpha And now I will make elements for both So we will have So let's say 2 alpha 2 alpha 2 alpha plus 1 2 alpha plus 2 We have 2 alpha plus 1 2 alpha So what I have done here is basically So if you think about it There are 9 elements here They are all polynomials of degree Less than or equal to 1 in alpha Do you agree on that? Those are the 9 elements that I have done here Even if you go back here These are also polynomials of degree 1 in alpha With coefficients 0 and 1 Binary coefficients All polynomials of degree less than or equal Same thing I am doing here also I am going to take all polynomials in alpha Which is some indeterminate Of degree less than or equal to 1 But then the coefficients are Coefficients are ternary So I have 0, 1 and 2 I am allowing 3 coefficients So that gives me 9 polynomials So addition is taken care of I told you 3 is equal to 0 So addition is taken care of Add any 2 I will get Another element from this field There is no problem Both of them are guaranteed Addition is very easy So multiplication I need a rule I use this rule So alpha squared So I can use several rules One rule I can use is alpha squared plus 1 to 0 So you can remember Alpha squared plus 1 And in ternary in f3 it was irreducible So x squared plus 1 was irreducible So I will always have to pick an irreducible guy Anyway alpha squared plus 1 equal to 0 Is a fine enough rule Or you can choose some other rule So this is one rule you can use And you can now check that this will be a field So every element will have an inverse It's a bit painful to go through and check So let me see who can give me the inverse of alpha 2 alpha 2 alpha is fine 2 alpha is the inverse of alpha Alpha times 2 alpha is 1 2 alpha squared Alpha squared is 2 2 times 2 is 4 So inverse of alpha is 2 alpha So it looks like There is some scary computation involved With this alpha squared plus 1 equal to 0 It's very easy You have to do some polynomial multiplication You have to use it using this alpha squared Plus 1 equal to 0 You can always do it It's not very hard So every non-zero element will have a valid inverse But all these things are not very obvious So you can keep doing the constructions with this But there is a general way in which to think of these things Which is what we are going to see next So it's very easy The general way is also not too hard But if you don't have these examples And mine will be a bit confused So we are going to use these two examples later on Every time I introduce something I will use examples from these two fields F4 and F9 just to keep things very simple Maybe we will see a fade also So now let's go over the construction Of finite fields Before we finally begin construction So let's think of How do we want to think of a finite field Somebody will see I have a finite field Somebody says I have a field with A finite number of elements So slowly let's see If we can extract some more information about this field So we will ask this person some questions And then slowly try to figure out What is there inside this field So the first thing you know that has to be there in any field is The additive identity So we can agree with this person That additive identity will call us 0 If he has to agree and we will also agree So it levels 0 Outside of this 0 you have a multiplicative group And that multiplicative group has to be an identity And that has to be 1 So that is two things we can agree on There is no problem with that And now with just these two What else can I do I can try to construct more fields Now I can ask this person who has this finite field You tell me what is 1 plus 1 Because you know you agreed just It was multiplicative identity 1 Now I can take this and add it to itself You tell me what is 1 plus 1 Plus 1 plus 1 You might tell you 3, so on and so forth Now you can do adding 1 to itself How many times As many times as you want But you know that the field is finite So eventually what has to happen 1 plus 1 plus 1 several times Has to go to 0 You might say has to repeat But then if it repeats you can also say You can know everything to one side And say eventually the first repetition has to happen at 0 It cannot happen anywhere else Think about it So when you do 1 plus 1 plus 1 plus 1 You keep getting 1, 2, 3, 4 Eventually it has to repeat back at 0 It cannot repeat somewhere in the middle If you are repeating then what do you have 1 plus 1 plus 1 some x times equals 1 plus 1 plus 1 from y times You know all the y to this side You might add with the attitude inverse of 1 So you would get 1 plus 1 plus 1 x minus y times being 0 So you must have encountered 0 before it So it will repeat definitely at 0 So there is a number of times A minimum number of times You have to add 1 with itself To get 0 So you can ask that information From this person who has this finite field How many times should I add 1 to itself To give me 0 I know 0 has to happen Because it's a finite field It cannot keep on going on and on forever It has to stop somewhere The repetition has to happen So let's say where it happens So let's find the characteristic Of the field So that's the idea of the characteristic 1 plus 1 1 plus 1 plus 1 That's the problem Has to repeat somewhere Has to repeat So this implies There exists minimum P Such that 1 plus 1 plus 1 So on Equals 0 in F So this needs a bit of query But I kind of verbally prove it for you I think you can think about it It's not too hard to imagine this So when you add 1 plus 1 plus 1 to itself 7 times, there should be a number P Which for which when you add a P Times you should get the 0 on the finite field So now we can ask from this person Who has this finite field what that number P is Now I'm going to claim something stronger There is a reason why we call this guy P This number has to be correct It cannot be a composite number It cannot have any factors So that means we'll try to prove it But before that this number P is called characteristic So this P is called characteristic of X And any finite field Has to have a finite characteristic The infinite field What about them? Like the rational numbers They don't have a finite characteristic In fact for those fields Characteristic is 0 It's the crazy definition So when somebody says field of characteristic 0 They mean an infinite field Like the rational, reals Or the algebraic number fields Or the complexes, the function fields So many other fields are there One of those fields is what they mean by saying Characteristic 0 So they don't have a finite characteristic But any finite field has to have a finite characteristic And moreover the characteristic of a finite field Has to be prime So let's try to quickly prove that Characteristic Of a finite field Is prime So standard way to prove many of these results Is to assume the opposite and show that there is a contradiction So P I know Is the smallest number of times So I have to add 1 to itself So we need 0 in the field Suppose P is r times s Suppose P equals r times s Then what will happen You can do this identity 1 plus 1 plus 1 p times You can move the distribution property of the field And show that it will be the same as 1 plus 1 plus 1 r times multiplied by 1 plus 1 plus 1 s times So this is not too bad to prove Sure 1 plus 1 plus 1 p times On the left hand side you have 1 plus 1 plus 1 p times On the right hand side also you will have 1 plus 1 plus 1 p times So you use the distribution Take the first one meant to play on the right You will have s ones The second one meant to play on the right You will have s more ones So r times this is p So you will have p times 1 P, p1 is adding on the right hand side also So now I know this guy is 0 So now suppose 1 plus 1 plus 1 r times is 0 If it is 0 then I already have a contradiction What is the contradiction? r is less than p And 1 plus 1 plus 1 r times is 0 That is not supposed to happen P was supposed to be the meaning Now suppose it is non-zero I am going to claim that 1 plus 1 plus 1 s times has to be 0 That is that true Because non-zero I know this guy has a multiplicative inverse in f I will multiply by the multiplicative inverse On both sides What will happen? This guy will become 1 itself So you will have 0 times that So you will have 1 plus 1 plus 1 s times is 0 So in any field One property that is true is If a b is 0 Either a has to be 0 or b has to be 0 That is true in any field So that is a general fact So a b is 0 A is not 0 then you multiply the inverse So if a b is 0 Either a is 0 or b is 0 That is true in any field So from here you will get a contradiction It is easy to write down I am not going to write it down for you So you can say from here Either 1 plus 1 plus 1 r times has to be 0 Or 1 plus 1 plus 1 s times has to be 0 And that contradicts the minimality of p So p has to be front Without implies So that is the end of the proof What is QED Some Latin stuff I think so hence prove So that is the end of the proof So that is the end of the proof So that is nice So that is nice We already made some pretty big problems So we started with Some finite field somebody said We have so much more information about the field There is a 0 and a 1 We also know that it has a characteristic p Which is a prank So now we can now Find p elements of the field What are the p elements I know that f contains 0 It has 1 It also has 1 plus 1 2 So on full it has to p minus 1 It might have more stuff But it has definitely up to p minus 1 So that is p minus 1 So up to that I know it is not 0 If I add it p times I am going to get 0 So we can just stop there In fact the next statement is interesting It turns out It is not a finite field Now I can say something Very very definite So this is true This guy 0, 1, 2, p minus 1 Which is a subset of f Is also more effective to up to z Are very very familiar More p integer So it is exactly z p in this case The 0 is the 0 of z p 1 is the 1 of z p 2 is the 2 of z p So you can try proving this It is not very hard How many other times you want So when you multiply it is the same as doing modulo p in m in p It is the exact same thing It does not change anything So if you are adding 1 plus 1 plus 1 Some r times inside this And 1 plus 1 plus 1 s times inside The same as multiplying the integers r and s Modulo p How many times you have to add 1 Others would be p times And 3 times when you add something It goes to 0 So this guy is isomorphic The way I am constructing 0 to p Minus 1 Will determine everything about the p Not everything A significant part of f Yes Completely So the question is So the finite field Is very limited in some sense Cannot do all kinds of arbitration That is how the p is So that is why when I told you In f9 for instance 1 plus 1 plus 1 has to be 0 So there are some statements about that We will prove that So these are the next facts that we are saying So there are 2 facts which are very easy to build down And then that gives you A fantastic finite field So in any finite field There will be an isomorphic copy of zp So z2, z3, z5, z7 Things you know very well The integers modulate So let me give you an example If you say for instance p is 7 And then You have to multiply 1 plus 1 plus 1 By 1 plus 1 plus 1 plus 1 You know these 2 guys belong to F I am going to say this is equivalent to Multiplying 3 times 4 mod 7 Which is what 12 mod 7 which is 5 Which is 1 plus 1 plus 1 Plus 1 in f That is also one more This is in z7 It is very easy to prove it You multiply this out how many 1s you will have 1 plus 1 plus 1 12 1s you will have But you know 7 of them will add to 0 Because my p was 7 So you will have, you will be left with 5 Which is 1 plus 1 plus 1 So it is kind of a trivial statement That it is very important So the fact that you have also p inside Any finite field So this p is a characteristic Which is an important Particle of the finite field You should first figure out What are the mistakes The next statement Is even more interesting It really fits It fits very well F Is what The finite dimensional vector space Works this way Which was Not only Does F contain an isomorphic property of zp It is also a finite dimensional Vector space over zp The fact that it is finite dimension is not too critical The whole field itself is finite Of course the dimension cannot be infinite Fact that it is a vector space is critical So here there is an abstract vector space So this is what I said we will use the abstract definition So you have an object s Which is unknown to you You know that there is a field inside it I want to show that this object Whatever it is has to be a Finite dimensional vector space over So it is very easy to show It is not very hard So I mean I don't know You can think about Two elements of the field F If you add them You should get another element of the field That is the property of the field body So addition has nothing to do You might have to check the scalar multiplication And that is also true because the scalar is again part of the field You multiply some scalar With another element of the field You will get another element of the field There is no problem So it is kind of a trivial thing to check What is slightly non-trivial is Suppose I tell you the dimension is m Non-trivial Non-trivial factor If I tell you the dimension is m What is the size of F? It has to be p power So the proof is quite easy So I will leave it You have to check the axioms It is not very hard If p is a field You just check the axioms It will work out One of the factors Suppose m is the dimension of Zp Then Size of F is p power So it will come along with Just a finite field Somebody told me it is a finite field I asked for a characteristic I know it is a prime number And then the next thing I ask is As a vector space where there is p What will be the dimension Then I know that it has p power m elements Remember Points are special numbers If you have 1 prime and you have p power m You cannot have another prime for which the same number will be p power n or something It is not possible So now you can see why When I said F9 1 plus 1 plus 1 has to be 0 Why is that proof? If I have a finite field with 9 elements 9 is 3 squared That is the only prime number for which It is a power So the characteristic has to be 3 So 1 plus 1 plus 1 has to be 0 Otherwise I will not have a field The same thing with F4 F4 can be written as a power of prime In the unique way 2 squared So 2 has to be the characteristic of F4 So 1 plus 1 has to be 0 in it So these characteristic 2 fields Are quite crucial in coding They are very nicely represented With bits so they are used a lot in coding So characteristic 2 is quite important Characteristic 2 is the simplest When you add 2 things you do not do anything with this Very easy Point 2 is the same as XOR But other characteristics are also At least they are very nice So we will deal with them in another way So this fact is very important So one thing we have proven now is Any finite field Has to have what kind of number of elements 9 power m If I have a number like 10 I cannot have a finite field With 10 elements So that is a highly non-trivial fact Before we started if I just told you There is no finite field size 10 elements You may not have believed me But that is true So you cannot have a finite field With number of elements Which is not a power of m Power of a prime So it has to be prime or power of a prime So it has to be That you prove So you can prove it just based on the same argument It is not a prime and the characteristic Any finite field has to have Prime power number of elements So this is an important fact So that is what is interesting So it turns out This is not the end of finite field structure There are lots of very interesting structures For finite fields And so far I have not mentioned anything about polynomials So polynomials over finite fields Play a very important role in Understanding the property, understanding the structures And all that we will come to it slowly And the both interpand very very Inseparably polynomials over finite fields And the finite field constructions So we will slowly come to it Even this is a non-trivial fact And we have been able to slowly prove Without any non-elementary step in any of these Basic elementary prime numbers I think great about it We are able to come to a statement Where the size of the finite field has to be power of a prime Any questions I am sorry Connection between Yeah, we will come to it So from this way So now how do we describe all the Elements of the finite field? That is the next question So far we have defined only zp Which is a small part of it presumably Can I p power m? They met the other elements How do you define the other elements? So you have to know So you can do it several ways now So I know f is the Let us preserve that zp What can I ask next? I can ask for what? For the basis of This vector space over there That will give me a complete description of The finite field So that is the next thing you can ask Basis of f over zp So let us say we have Alpha 1, Alpha 2 It is very common to use alpha for elements of finite field So it is not My convention everybody uses alpha So you have m of them Some m elements Of the finite field Which form a basis for it over zp I do not know what these guys are But I know that should be a basis I can find out Once I have this What are all the elements of f? Then I know f is A1, Alpha 1 Plus A2, Alpha 2 So until m, Alpha m Then what are these guys? Or elements of f? That is it So not only do I have basis elements I also have one description for Every element of the finite field I have this Alpha 1, Alpha m being some basis elements Then I can do Okay So this finite field This vector space kind of description of the finite field Is very very good for one thing One of the operations in the field Can be very easily done with this vector space description What is that operation? Look at it See I mean ultimately I am interested in doing Computation of this finite field I am not doing This for some abstract pleasure or something like that I want some maybe also that I want to be able to do computations for this I can represent every element of this Finite field in a computer program very easily I have a vector of Then m each m is an entry From zp I can nicely represent it How will I add any two of these guys? Adding is very easy Very trivial all I have to remember is Modular p for the coefficient That is all Addition of two elements is very easy in this But this personation What do I do for multiplication? It is a bit more complicated It is messy in this notation What all information do you need I mean if you ask this guy who gave you the finite field I mean you need to know a lot of things You need to know what happens when you multiply Alpha 1 with Alpha 2 What element do you get What happens when you multiply Alpha 1 with Alpha 3 You need so many descriptions Of multiplications It does not seem like a very Efficient way to do multiplication And that is true Multiplication still remains a challenge You can do it But you have to ask first for the information So you have to say What is Alpha 1 times Alpha 2 You give me that as An element of this vector Alpha 1 times Alpha 3 All this information you have to give Then you can multiply But you need a big lookup table For all these things Maybe it is not a very easy thing to do Now it turns out The multiplicative structure of finite fields Is also very easy And it has a very nice thing See one thing we are not exploiting here is At least not directly is The multiplicative group We have not tried to understand much about the multiplicative group We have the additive group Which is this nice vector space But the multiplicative group is not clear We don't have a handle on It turns out that group is really really simple I would be very surprised at how easy That multiplicative group is So it turns out to be generated by just one element It is very very nice So we will see that next The next thing we are going to see Is how to understand the multiplicative group better So So before that, before we go and understand The multiplicative group I am going to give you a general construction For f p power m So I will give you a general construction So this construction So What we have to show till now If there is a field f with finite number of elements Size of f has to be equal to p power m But we don't know for every p and every m A field of size p power m exists We don't know that There is a field It has to have some prime power s The size We showed you examples for f4 and f9 What about f16? What about f8? What about f121? I don't know The other fields of that size That's a very valid question We don't know if there are fields of That size for any prime power So this construction will answer that question It will tell you At least partially, if you believe one fact That it is a There exists always for p and m There always exists at least one finite field Of size f p power m So the first thing we will start with this A polynomial which I call pi x Which is Irreducible Degree m In fpx So fp What it is? So fp is the same as fp And in fpx I want a degree m polynomial Which is irreducible So you might wonder What if there is no degree m polynomial Which is irreducible So it turns out that cannot be proved That is the fact we will prove later on But for now For every degree In every p There will be always at least one In fpx So that is the fact You have to believe that If this is the fact it exists If such a polynomial exists There is no problem But it is acceptable For a few examples For every degree you saw At least one irreducible polynomial The example that we saw It turns out you can show in general That it is proved for any degree We will maybe come back and prove it later So once you have this It turns out The following construction works fp power m You can make it as a0 plus a1 alpha So until am minus 1 alpha power m minus 1 Remember alpha is the indeterminate The indeterminate that we have So these a's will come from zp And then The most important thing is pi of alpha equals So this is the generalization Of your simple construction I gave you for a 4n f9 I gave you two example construction And this is the most general version of it For any p and any n You can do this construction You look at You take your indeterminate alpha And look at all polynomials Of degrees less than or equal to degrees 2 less than m With coefficients from zp This is similar to what I did for a 4n f9 You can go back and check The exact same thing I did And then I need a condition On alpha for the multiplication See for addition this is good enough For multiplication I need to know How to simplify Multiplication So that I will use pi of alpha as 0 Remember pi of alpha is the Degree m So when you do pi of alpha as 0 It gives you an expression For alpha power m In terms of Lower powers And that is good enough Once I have alpha power m in terms of Lower powers that's good enough I can also find alpha power m plus 1 M plus 2 and so on Again repeatedly I can do So this basically gives you Alpha power m In terms of And that's good enough So when you multiply two polynomials You simply use this rule repeatedly And simplify it and we will get back With another expression And since pi of alpha is irreducible It turns out this is a thing The irreducibility is crucial It's not irreducible it will not work Since pi of alpha is irreducible This guy is a field So we can prove that it's not very hard We will prove it soon enough So how do you prove it The only non trivial thing is Existence of multiplicative inverse Do you agree or not Addition and multiplication are being done Modulo Power of alpha Another way to think of multiplication If you are confused about this pi of alpha rule You take two of these polynomials You multiply them Just like polynomial multiplications You will get a bigger polynomial You divide the pi of alpha You will get a reminder It's the exact same thing As using this alpha power m rule Over and over again So if you have a of alpha And b of alpha m p power m a of alpha times b of alpha Is the same Same as what Times n of p power m Same as a of alpha b of alpha Mod power alpha So how do I do mod I multiply and then I do that long division With pi of alpha, whatever the reminder I get I take that, so it's the same as that So that is your Multiplication So how do I show that everything has to have a multiplicative inverse Every element has to have a multiplicative inverse You remember For zp we showed that Remember the proof we did for zp What is the proof we did for zp I wrote down all the elements And then I multiplied i with Each of the elements And I argued p times i And p times j mod p Can never be the same What is that proof If it is same then p times i minus j No, not p times I am sorry Would I get that right A times i and A times j A times i minus j cannot be equal to 0 mod p Because p has to divide A times i minus j And that's not going to happen The exact same proof is true For all the elements But I know the p is reducible If I have an a is alpha I am multiplied by all these elements But a is alpha times d is alpha Cannot be equal to A is alpha times c is alpha If it is equal Then what will happen? p is alpha has to divide A is alpha times d is alpha minus c is alpha And that is not going to happen Because p is alpha is a reducible The exact same proof is Same proof as As for zp So you will never have a repetition When you take your particular polynomial And multiply by all the other elements Of fp para Which means in particular there should be One polynomial Which will multiply and give me one So that's the simple proof That you have for this result The same proof as we did for zp Except that now we have the polynomials And you know p is also very reducible So it cannot be product of two factors On that side So as you see So nothing major about it And at this point we will start For this class And tomorrow we will pick up from here So the only thing we haven't seen Is the multiplicative structure And we have understood it well enough The multiplicative structure the only thing That seems to be in this construction is Product of two polynomials Any reducible polynomial Like I said there is much more Stronger multiplicative structure Which we will see in general class And then we will specialize to this construction So in terms of there is more to this construction Than meets the eye for instance There is more of a finite field around this You can essentially show that Any other finite field of size p para Will be isomorphic to this So it's not very surprising You think of the finite dimension vector space Any two are the same The multiplicative structure you cannot have anything So it's the same Any two finite fields will be isomorphic So this one construction is good enough You don't have to worry about anything else So we will see those facts in the next class