 Okay, so we're going to start looking at, we're going to look once again at cyclic codes a little bit differently from a, maybe slightly different perspective. Okay, so let me first, so first thing I'll do is I'll just take measure to binary cyclic codes. Then maybe let one will go non-binary outside. Let's give some part of the presentation simple. I'll do this. So what's the cyclic code? If you look at the definition, a linear code, a cyclic code. The following condition is true. If you have C1, C2, Cn belonging to the code, that implies. So you can do it in so many different ways. I'll do it when I'm so glad to look at it. Okay, so this is the definition of a cyclic code. So basically this is from here to here you go by a cyclic. Okay, so that makes the code cyclic. So I have presented that there's only one cyclic shift. But even if I take this code word and do a cyclic shift several times, say some 10 times or 20 times, I'll still get another code word. Because I can do it one step at a time and then I'll get. Okay, also I can do a cyclic left shift. Okay, what happens if I do a cyclic left shift? A cyclic left shift is the same as n minus 1 right shift. So that is also covered. So everything is covered in this definition. So one very easy way to construct cyclic codes is the following idea. So let's say something of a naive construction. So this is not the way most people picture cyclic codes. But one way of doing the construction is to simply say, I'll take a generator matrix. I'll take a first row, let's say the first row is G1. Okay, second row will be, so I need notation here. So let me just write R of G1. Okay, so I'll use this notation here. So C is this, I'll call this as R of C. Okay, just to show that this is a cyclic right shift. So R of G1, R squared G1, what is R squared G1? I rotated cyclic codes twice. And then all the way down to n minus 1. Okay, so I could take my generator matrix to be at this top. No, no, no, I want to do n minus 1. I know there's some k and all in all. I'll come to it later. Okay, so I'll take an arbitrary resource for G1. I'm not set any property for it. Some arbitrary resource for G1. And then keep cyclically shifting it to the right. Still I exhaust all the possible cyclic shifts. Okay, so I'll put n here. But sometimes you may not need all n. If you don't need all n to exhaust all the possible cyclic shifts, you can stop some other point. But otherwise you'll have to do all the n. Okay, so the first claim is this will always generate the cyclic code. Okay, so that is not very hard to prove. Okay, so the code generated by this will be cyclic. So if you have one code word, it would have been generated by some message multiplying G. If you take a cyclic shift of that, right, all you have to do is cyclically shift the message also and you'll get the code word. Okay, so this is obviously a cyclic code. Okay, so think about that a little bit if you're confused. Basically you have some message, both set of messages multiplying this. Okay, so if you rotate a code word, if you click shift by 1, all I have to do is shift my message bits by 1 and I'll get the corresponding shift in every place. Okay, so this is obviously cyclic. It's also obviously linear. So what is the difficulty here? What do you think is going to be a bit more painful to compute? Sorry, that is fine. But even before that, what are the parameters of the code that you're interested in? Lock length is fine. Okay, so let's say then what is the next parameter interested in? Dimension, right? So how do you do the dimension? How to find the rank. Okay, and the rank might be something. You do something less than that. Usually these things will be, rank will be definitely less than that. Okay, I'm going to do cyclic shift. There will be some linear dependency somewhere. Okay, so what about k? It's not clear that you have a good handle on k. Okay, and it's also a very painful construction. If you want some guaranteed minimum distance or anything like that, you don't know how to do it. There are so many other things which you don't know about this code. You just know it's cyclic. But then, any advantage you hope for because of the cyclic nature is not really coming through in this construction. It's like, there's a construction, a generator matrix. You might as well do a non-cyclic one. What's the point in doing a cyclic if you're not using any major structure in it? Okay, so this is definitely a valid construction, but it doesn't tell you anything more about dimension, minimum distance, etc. Okay, so what's nicer is to view these cyclic codes in this ring of polynomials, modulates, parent plus one like we did in class, and then think of ideals and polynomials and all that, and that gives you much nicer construction. It also tells you a lot more about k. Okay, so it turns out if you fix an n, you don't have cyclic codes for all possible k. So there are only some k for which cyclic codes are possible. So these are highly non-trivial results. If you just start with this construction, how do you go about proving something like that? You may not even be able to convince anybody of that. Okay, on the other hand, the way we do it, we'll be able to quickly find out for what k will n, k, cyclic code exist. Okay, so how do you think about constructing it? What are the easy ways of doing it, etc. All that will be clear. Okay, so while this is a very intuitive and simple construction, it's not very informative. What we will do is a little bit more complicated, but it will give you all the information you want. Okay, so what is this idea of ring? So the first idea is to use this ring. So I'll denote our Rn. Okay, so basically Rn contains n-ary polynomials of degree less than or equal to n minus 1. Okay, so this is the binary polynomial of degree less than or equal to n minus 1. Addition and multiplication. I'm going to do modulo x bar n plus 1. Okay, so that's the, that's this ring Rn. Okay, so addition being modulo x bar n plus 1 is okay. I mean it doesn't really affect anything. But multiplication modulo x bar n plus 1 really will affect the multiplication. Okay, so it's not the normal polynomial multiplication. A lot of things will change. Anything that crosses x bar n will fold back. Okay, so one nice thing about this ring is, a cyclic right shift is represented by multiplication by, so that is the nicest thing about this ring. And that's because this ring also represents the binary vector space, n-dimensional vector space. There's nothing, there's no problem there. It definitely represents the n-dimensional binary vector space. How do I think of it as the n-dimensional binary vector space? Yeah, simply take the coefficients as 0 or 1 that becomes n-dimensional binary vector space. I don't have to worry about polynomial. One nice thing about moving to polynomials is that, you have a nice kind of mathematical description for what a cyclic right shift is. Okay, so it's basically a multiplication by x in Rn. Okay, so that's the first idea. The next idea that we saw, which is again very useful, is what happens to a cyclic code. A cyclic code becomes one ideal one other. So I defined the ideals for you, so let me define that once again here. An ideal of Rn of any ring, okay, so ideal of ring R, okay? So a subset, okay, so first of all it has to be a subspace. Let me say a subspace said to be an ideal. Okay, so let me be careful about subspace here. So a subset over a path is an ideal for following conditions as satisfied. First condition i is an additive subgroup. Okay, what do we mean by additive subgroup? Okay, ring with respect to its addition operation is a group. Okay, it's an abelian group and this ideal that we look at, the subset that we look at should be a subgroup of that additive subgroup. Okay, so in particular in Rn, in Rn the binary vector space itself is the additive group. Okay, the additive group is the binary vector space. So I can think of the subgroup as a subspace. Okay, in Rn I can say i has to be a subspace. So basically if you want an ideal in Rn, it should first of all be a linear code. Only linear codes are candidates for ideas. Okay, so after that there is another definition which is again very important which will bring in the multiplication. Okay, so if you have some o-effects on ab and b-effects on i, then other clouds, o-effects times b-effects along square. Okay, so if you take any polynomial in the ideal and any other polynomial anywhere in the ring, multiply the two together, it should still be a guide. Okay, so what's the definition of an ideal? And it doesn't seem very obvious why an ideal should be a cyclic code. And in Rn it's quite easy. Okay, so in Rn, so if you go to Rn, one will imply i is a linear code, two is what? Okay, so if you take any polynomial in Rn, okay, so I have to a-effects here, so I think I should not put a-effects here, you know. I don't know what these rings are. Okay, it's just ab. Just ab, right? So in two, everything is a polynomial. And in Rn everything is a polynomial. Okay, so what does two mean? Okay, so you have a polynomial in the ideal. Okay, and you can take any polynomial in R. Okay, so in particular I'll choose a basis set of polynomials. Once I choose a basis set of polynomials, since I know already i is a linear block code, I can add up all of them, right? It's closed under addition also. So I can add up all of them and replace this arbitrary element of R with only the basis elements. Okay, so if I have the basis polynomials, satisfying this property, that's enough. Okay, so two basically will reduce you. dfx and i implies x power i times dfx is an i for, well, i equals 1 to n minus 1. Okay, i equals 0 is kind of trivial, dfx is again written, right? So it's enough if you have this property, h power i times dfx should also be in the ideal. That's enough. So once you have that, the entire prove is satisfied. Yes, because any arbitrary polynomial in R, Rn can be written as summation ai x power i. So that's good enough. Okay, so it works out. So that's the main idea. So you notice that this is basically nothing but cyclic code. Okay, so this makes i equal to 64. Okay, so it's quite easy that way. You want to spell too hard to show these properties. So an ideal in Rn is basically a cyclic code that will help you. Okay, so let's take a couple of examples. Okay, so I'm going to start this. I'm sorry. Yeah, actually xbx is also enough. Okay, so, yeah, I mean, I think it's also easy to see why xbx is enough. So if xbx is an i, then you use the same property over and over again. You get x power i dfx. Right? I don't have to. I put this out explicitly just to say xx times dfx is also enough. Because when you use the same property over and over again, you'll get x power i dfx. Okay, so let's look at some, let's go up 4. Okay, so up 4 is going to be 0, 1, x and plus x. Okay, so we have to start writing out all these parameters. It's a little bit, not a very boring answer. Okay, and then you'll have x power 3, x power 3 plus 1. I'm sorry, x power plus x. x power 3 plus x power 3 plus 1. Okay, so we'll have 16 polynomials in R4 and there they are. Okay, so suppose I ask you to identify ideals. Yes, I've said, okay, fine, ideals in Rn are cyclic codes. And I was promising you a simple way of coming up with cyclic codes. But then, where is R4? Can you identify the ideals? Can you stop any ideal very quickly? Okay, so it's, I mean, the entire R4 will be an ideal. Okay, fine. Okay, so R4 itself is an ideal. Anything else? For instance, 0 is an ideal. Okay, these two things are trivial. Okay, so we will not consider these two as non-trivial ideals. It's just trivial ideals. Anything else? Yeah, so you have, you need some method to generate the idea. I guess you already know the answer, so maybe you're giving the answer to me. But you would ideally need somewhere coming up with the ideas. Okay, so one, one very easy method to come up with ideals is to say, set up all multiples of one element of R4. Okay, so that's called the ideal generated by an element. Okay, so that's one way of generating an ideal. Ideal generated by some AFX4. Okay, so let me just mark the R4 in Rn. Okay, so this is basically, we will denote this with this notation. Okay, that is set up all AFX, BFX, such that BFX is one. Yeah, yeah, we'll come to that. I think there are some of these results last time, but anyway, I want to emphasize some of them here. Okay, so this is one way to generate an ideal. And you can show quickly that this will be an ideal. Okay, you have to prove that this is an ideal. How do you prove it? You have to verify the two properties. First thing you have to verify that it's an additive subgroup. You add any two multiples of AFX like this. You'll again get AFX times B1 of X to B2 of X. And that again belongs to R, so it's closed under addition. That's not a problem. What about multiplication? So multiply any element of this AFX times this guy with another element of R, what do I get? Once again, get another multiple of AFX, right? So it becomes AFX into BFX. You see that? Okay, so you can do other multiplication first. B1 of X times B2 of X, reduce it, model X bar n plus 1, you'll get some other polynomial. So both of them are the same. Okay, so eventually you get the element of this idea. Okay, so this is one very nice way of generating ideals in a ring. So in general, these ideals are called principal ideals. Okay, so these are called principal ideals. Okay, and let's see some examples of principal ideals in R4. So take the idea generated by one, what will I get? Yeah, again R4, right? Is it clear? Does it seem okay? Ideal generated by one multiple of R4. What about any other element? Can I get ideal generated by any other element to be R4? Can any other element? I think for all, let me use the polynomial. The polynomial that will? Ideal has to be like, what do you want the ideal? I don't want the ideal. Okay, think about this for a little while more. What else can give me R4? So what do I need? So what does it mean if I say AFX is R4? In particular, there's one element in R4 that needs to be a multiple of AFX. What is that element? One, right? So if I have one being a multiple of AFX, that's enough. If I show that one belongs to this ideal, then entire R4 will be in the ideal. Okay, that's the crucial element. That's an argument that's often used when you want to show an ideal is the entire ring. All you have to show is that some unit or one in this case belongs to the ideal. So when can I have AFX times BFX equals one multiple of AFX plus one? Is it possible? Sorry? Sorry? AFX is x bar i. It's definitely not the way it is. It's not the way it is. It's not the way it is what they need or what. Think about it. It seems like a lot of interesting discussions are happening, so maybe I should wait for a while and go solve this. There's no GCC. No GCC. The GCC is one. Okay. So that seems like an interesting place to start. So he's suggesting GCD of AFX comma x bar n plus one should be equal to one. So that's a nice property that the GCD satisfies. GCD can also be written as a linear combination of the two polynomials. So it means if I have something like this, it implies that exist polynomials BFX and CFX, what's that? What's that? AFX BFX plus CFX. X bar n plus one equals one. So if you do mod X bar n plus one, it seems like you're going to one. So it seems like a very valid idea. Can you give me an example of an AFX in R4? X. So what multiplied by X will give me one? X bar n minus one. So you take X times X bar four X bar three. You will go to X bar four which is actually one modulo X bar four plus one. So those things will come. So there are also other cases. So if you have something like... See, you cannot have any other GCD. So if it is X bar plus X, for instance. So for example, X equals four. What about this way? X times X plus one. This will not be... So it will be something contained in R4. So these are... You need a strict GCD. That's the thing. There are also other ways. I think you can take any other power here. You will get that. And there are also other things. But you will be careful if you say irreducible. If it's irreducible, it should not be a divisor of R4. X bar n four plus one. So that might work out. I mean, you might have something like X bar three plus X plus one which might work out. So there are several cases which will give you the entire R4. That's some of this general interest. It's not a great interest to us. Just ask me. Yes? Can you say that any element will be equal to the element that is written by the GCD of that element? Yeah, we will come to that. Okay. So the next question kind of based on what he's asking is to say... When can I say... This is the question that you're asking, right? When can I say that ASX and BFX is the same idea? So for that... To answer all these questions, you need something to understand something more about the idea. So all these questions can be very easily answered if you understand something very basic about the idea. So that's what we'll see next. Next result we'll see, we'll talk about something known as the generator polynomial of an ideal or some kind of a unique generator for an ideal which will fix all these things. So the problem here is the same idea can be generated by multiple elements. And you have to find that element which is in some way unique to that ideal which will not generate something else that you don't want to generate. So then you can answer all these questions very, very easily. So let's try and answer that question that comes from the generator polynomial idea. So this motion of the generator polynomial is important. If you have an ideal in Rn, we'll denote this by GFX. GFX is there. So I'm dealing with binary cases here so it's quite easy. GFX is the least degree polynomial in I. So that's called... That is something special, it's called the generator. So the least degree polynomial in I is the generator polynomial GFX. So it turns out this will be in some ways unique. Just so you can show, they'll do only one least degree. So let's go and prove some properties for it. There are a lot of properties that the generator polynomial has and you'll see some very interesting results will come out of it. So this will basically fix... Once you understand the role of a generator polynomial it will fix all possible ideals in R4 or any Rn. You can quickly generate all the ideals in Rn. You'll know exactly what they are, what they can be. So all these repetitive stuff with different things generating the same ideal and all all the confusion will go away if you always fix to the generator polynomial. So there are many properties. The first one is it's unique. GFX is unique. So all these proofs will go back to this least degree assumption and the fact that you have an idea. So how do you prove uniqueness? Uniqueness is typically proved by contradiction. Suppose you have GFX and G prime effects. G prime effects and G double prime effects being both two least degree polynomials in the same ideal I. So when you say two least degree both of them should have equal degree. One is lesser than the other then both of them cannot be least. So both of them have equal degree and then both of them are in the ideal. So when you say equal degree in time, what does it mean? The last term, the last term is some expiry. Same thing for last term is expiry. Now what happens if I do G prime plus G double prime? Sorry? So if you do G prime effects plus G double prime effects I know it's still in the ideal and it has degrees strictly less than ideal. So the only way it will work out is there has to be zero in which case G prime and G double prime are equal. So you can quickly show that. So this implies this equals zero and that implies G prime. So that's the way to show that GFX is unique. So if somebody gives you a generator polynomial and says somebody gives you an arbitrary element of Rn and says find out something about the ideal generated by that. The first thing you should try to find out is the generator polynomial. Once you find the generator polynomial it will be okay. So suppose somebody asks you a question, ideal generated by AFX same as the ideal generated by BFX, how do you answer that? Find the generator polynomial for ideal generated by AFX and the generator polynomial for ideal generated by BFX. If both of them are the same then they are the same idea. Both of them are different and they cannot be the same idea. So it's unique to the ideal. The second property is that BFX will divide by Rn. So it means that it will come before this. So I think this is fine. So this really nails down every single ideal out there. Every ideal has a generator polynomial. And I know that it has to divide X bar n plus 1. So how do I nail down all the ideals? I start with X bar n plus 1. Find all its devices. Take one, each one will generate an ideal. To be a generator polynomial for an ideal. And those are the only ideals there. Nothing else is there. Any other ideal I come up with should be equal to one of these. So I will show this GFX divides at bar n plus 1. So basically the proof is to divide X bar n plus 1 by GFX. So what will happen when we do that? We will get a quotient and we will get a remainder. The important argument here is this RFX has degree of 50 less than degree of GFX. And RFX belongs to the ideal. Why does RFX belong to the ideal? If you reduce this equation modulo X bar n plus 1, you get RFX is QFX times GFX. So if you do QFX times GFX in Rn, what will you get? Small RFX. So it belongs to I. It has degrees strictly less than the least degree polynomial in I. So it should only be 0. It cannot be anything else. So that implies and RFX is in I. These things will imply RFX equals 0 and QFX divides at bar n plus 1. So there is an even more slightly more interesting property. Which is quite useful. If you have AFX in I implies that it exists. MFX such that AFX equals MFX times GFX in S2X. So if I don't add this thing, this statement is obvious. I know my ideal is generated right. Let me step back a little bit. So let me do this more carefully. This basically implies what I wanted to say. So once this result is shown, this implies that I equals GFX. Every element of the ideal is actually a multiple of GFX. Not only in Rn, it's a multiple of S2X itself. Without the model X bar n plus 1, it's a multiple. You see that? So that's a little bit more special property. All our multiplications have been modulo X plus 1, X bar n plus 1. So what I'm saying here is every polynomial in the ideal, which is in Rn, remember it's a polynomial of degree less than or equal to n minus 1. You can write it as some polynomial times GFX and you won't have to be any modular. Itself, it will be a degree less than or equal to n minus 1 polynomial so that no modular will be necessary. This is not very hard to show. Once again, the proof is... So this is what is important. Generator polynomial actually generates the entire ideal. Once I prove that, this is also a linear. So how do I prove this result? Once again, you have to divide AFX by GFX. You know AFX has a degree less than or equal to n minus 1 and GFX also has a degree less than or equal to n minus 1. And when you divide, if AFX has a degree greater, the remainder will have a certain degree and you can upper bound the degree and you'll see really, RFX again has to belong to I. There's nothing more you can do to me. So you can again show a very similar argument that this has to be true. Is it okay? Think about this for a while. Maybe there is some subtlety there. Only the least degree polynomial in the ideal divides X bar n plus 1. AFX may not be the least degree polynomial. So it may not necessarily divide. So only GFX divides. So first of all, I started with an abstract ideal. I didn't say anything about what generates it. So it's some ideal in Rn. So what I've shown after these things is any ideal in Rn is generated by one polynomial which belongs to it. So previously, we were starting with the definition of principle ideal as in all the multiples of a particular polynomial. Here I've shown that an arbitrary ideal is also principle and it is in fact generated by its generator polynomial which is a unique element in it. And that's the least degree polynomial. Okay, do you have a question? We know that C of X is the degree polynomial in the ideal. And we also know that every element in the ideal generates the ideal. No, no, we don't know. No such statement has been told. Every element in the ideal does not generate the entire ideal. If you want to count an example for it, you take the entire R4. You take the entire R4. It's an ideal, right? Not all elements generate the entire R4. You take X plus 1, it generates only a subset of R4. Every element in the ideal will not generate the entire ideal. All that is not true. The effects will generate the entire ideal, yes. I'm sorry? Okay. Step back a little bit. Wait, wait, wait. A effects is an arbitrary element of the ideal. It might divide X bar n plus 1. It may not divide X bar n plus 1. I don't know. If it divides, then? A effects will generate the entire ideal. Okay. Why should that be true? See, okay, so I think people are mixing up so many different things. See, in an ideal, if I have an element in an ideal, it does not mean that GCD of A effects, comma, X bar n plus 1 should be equal to 1. There's no such condition, okay? So maybe I should step back a little bit and give you some explicit examples of ideals in R4, and then maybe I should come to this. I just went a little bit faster, I think. Maybe I should step back a little bit. See. So this is not a property satisfied by every element. Okay? Only some elements satisfy. It's because satisfied when the ideal generated by it becomes equal to R3. All that is at the completely different propositions. Okay? Now, I'm starting with an arbitrary ideal in RF. If I did all that, the reason I did all that was just to point out that answering questions about ideas requires you to think a little bit if you don't know anything about the generated problem. You just say, okay, what is going on? Do they affect BFX? If you ask questions about the ideal generated by A effects, the same as ideal generated by BFX, you have to do a lot of work. I mean, you have to think about what is happening and all that. Once you understand the generated problem and stuff, everything will become easy, is what I said. Okay? So then I started with an arbitrary ideal in RF. Okay? Now, in this ideal, there are polynomials. It is not necessary that all of these polynomials should either divide X bar n plus 1 or not divide X bar n plus 1. I don't know. They might divide. They may not divide. I have no problems with that. Okay? It is not necessary that GCD of the elements in this ideal with X bar n plus 1 should be one. That's also not needed. It's just an arbitrary ideal. The only thing that's needed here is any two polynomials in the ideal, if I add them, I should get another guy in the ideal. Any polynomial in this ideal, if I multiply it with any polynomial from RL, I should get another elemental ideal itself. Those are the only two conditions that this ideal satisfies. Forget about all those example problems that I asked you with all these. Those things don't apply anymore. So just to get the thinking going. Okay? So let's see maybe some explicit examples. Sorry? I know my g of X divides every A of X to nothing point. Ask for this proof. Yes. No, you don't want to ask. Because if you divide otherwise, I divide by A. Otherwise, I'll have a dependent on the RL of X. Yeah. In RL, it's okay. Yeah. How do you divide the A of X? L. I'm saying it's true in F2 itself. Yeah. How did you get to RL of X? I mean, RL of X, if it's not true, then I always have, I can find the remainder of what's in RL of X. Why will you divide? Why will you divide? Dividing in F2 of X or RL of X? What is division in RL of X? See, in RL, this ideal has really no description. I haven't told you how to, how the ideal is generated or anything like that. No. What are these g of X being the least easy problem in time to automatically make division in RL of X equal to the remainder of 2 of X for whenever I divide the g of X? That's what I'm saying in the last statement. That's exactly what I said here. That's what I said in the proof books. So, since g of X is least big deal, A of X will have a degree greater. A of X also has a degree less than or equal to n minus 1. So, when you do a division, you're not going to do a star n plus 1 anywhere. So, it will not enter the picture at all. So, everything will be in F2 of X itself. That's why I said this multiplication property is true in G of X. So, usually when I write I equals generated by G of X, I don't need that this is true. I don't need that the multiplication should be in F2 of X. See, that's if the multiplication is in RL. But I'm saying here the generated polynomial is so special that not only does it have this property in RL, it also has the property in X2 of X. So, I think we really need to look at a few examples. So, let's look at a few examples. So, I'm going to take in R4, let's say the ideal is generated done. Let's try X2 plus 1. So, this will be what? So, how do you come up with these elements? So, every single multiple X, X2 plus 1. There are 16% guys there. So, let's see how many different. 0 of course will be what? Then you have 1. Then you have X. So, X multiplied by this is X power 3 plus X. Then you have X plus 1. So, what happens to multiply the X plus 1? X power 3 plus X plus 1. And then what else will you have? What about a squared? Multiply the squared what will happen? You won't get anything more. So, now you can check everything else. I think you won't get anything else. You'll get only these four guys. I'm going to come to it. So, here we got four guys. Let me ask a few more questions. Just so that it's working out. So, what's the GCD? So, if you have AXX and I, what's GCD of AX comma X power 4 plus 1? No, my X power 4 plus 1. So, for the same AX, if AX is 0, then GCD is not a proper definition. Forget about it. The AX is X power plus 1. What is the GCD? X power plus 1. X power 3 plus X. X power plus 1. So, X power 3 plus X power plus X plus 1. What about GCD of this? It should be this guy, right? X power 3 plus X power plus X plus 1. Because X power plus 1 also factors as X plus 1 times this guy. So, you'll get this GCD. So, let's just answer the question that GCD of AX with X power plus 1 can be anything. It's really very hard to control. So, that's one. Then, is there any other element of ideal which will generate the same idea? So, first thing I want to point out is, if you take AX in this ideal, what will be this ideal generated by AX? It will be some other ideal, but can I say anything more precise about it? It will be contained in I itself. It cannot go outside of I. So, if I take X power plus 1, I get the entire ideal. If I take X power 3 plus X, what will I get? You'll get the entire one. What will happen if I take X power 3 plus X power plus X plus 1? You'll get only two guys. You'll get zero and that. You won't get anything else. So, you can check these things. All these things you have to check. So, you can check if the ideal generated by X power 3 plus X is also equal to the same I. I'm using this I. I think the confusion is sometimes I use I inside the example. Sometimes I use a general ideal I outside. So, you should remember that when I say an ideal outside of the example, it's a general ideal. It's not something generated by X power plus 1 is R4. So, remember that. This is I. What will be X power 3 plus X power plus X plus 1? This will just have two guys. Zero and X power 3 plus X power plus X plus 1. You can test it out. It's a bit more of work, but you can figure out what. Okay? All right? Okay. So, let me make one more statement now. So, maybe I should see, should we look at one more example? Maybe you should look at one more example. Okay? So, once again in R4, let's try and look at the ideal generated by, let me take a simple example. Let me do X power 3 plus X power 3. Okay? So, let's try that. Okay? We'll do this. Let's try. Okay? So, let's see what happens. So, you'll have zero X power 3 plus 1. What happens when I multiply this with X? X plus 1, right? We've got X plus 1. Okay? And what happens if I multiply by X plus 1? Okay? Okay? Okay? And then X squared. X squared plus X. X squared plus 1. Okay? And then you'll have X squared plus X plus 1. What will you get for that? Yeah, it will be X squared plus X. No, no, you'll get X squared plus X. Okay? So, you won't get anything more. Okay? You'll get X squared plus X again. Okay? And then if you multiply by X power 3, here what will you get? X power 3 plus X squared. Okay? And then multiply by X power 3 plus 1. What will you get? I'm sorry? X power plus 1, right? Okay? And that's it. I think that is it. You won't have anything more. Okay? Inject that you'll have 8 elements. You won't have anything else. Right? See, the first thing is an ideal in Rn is a linear block code, which already means that the number of elements in it will be a power of 2. Okay? So, if you have 16 elements in the ring, it doesn't matter. So, multiply by 16. The ideal generated by anything will either have size 1 or 2 or 4 or 8 or 16. It cannot be anything else in between. Okay? So, once you cross 4, you know it has to stop at 8. It cannot be the entire thing, because X power 3 plus 1 has a common divisor with X power 4. So, it has to stop at 8. And once it comes to 8, you can stop. You don't have to do any more. Okay? Anything else you'd like to repeat? Okay? So, now you can try to ask the same question. Which of these other elements generate the same ideal? Which generates a sub-ideal? You can try to list out everything. So, you'll get all kinds of different ideals depending on which polynomial you get. Okay? So, next question I'm going to... So, now we're done with the examples. We're going to go through one more general idea. Suppose I look at an ideal generated by some A-sites. And say A-sites divides X power into some. So, let's look at this situation. Different situation. Initially, I said I have an ideal. I don't know anything about it. We know that we can define a generator polynomial generated by the minimal degree polynomial there and all that is fine. So, now suppose I say I start with a divisor of X power n plus 1 and look at the ideal generated by it. Okay? What can I say? Can I say anything about the generator? What will be the generator polynomial? Can I say it's equal to A-sites? Or maybe it's some divisor of A-sites. How do you prove a result like this? I don't know anything about the A-sites. What do you do by the way? X power n plus 1, no? And if not, come back and become a... F2, that's not what I mean when I say this. When I say this, I don't mean multiple and F2 alone. Of course, if I multiply only in F2, then there's no case to be made. The procedure is under the polynomial. But I'm not saying anything about F2. I'm going to multiply this with arbitrary degree polynomials. As you have to determine something else. Okay. Then that will be J of X. And it will have a larger length of confidence. Which means J of X would not generate the entire ideal. Could be fine. Here, I suppose to be generated by A-sites. Yes. So if I have a generated J of X, which divides A-sites, then the ideal J of X will be larger than the ideal J of X. That's okay. As long as J of X is a multiple of A-sites and R-n. That's okay now. See what I have to show is... No. Please, please, please. Not easily. A-sites is not least degree. J of X is least degree. J of X is not least degree. J of X is not least degree. That's okay now. I never said A-fix is the least degree polynomial. Why should both of them be right? M-fix is not the right one. Yes, generating I. Because J of X divides J, then J of X generates a bigger I. No, no, no. Be careful. Be careful. How can you say you'll be generating a bigger I? Don't just assume that very quickly. You have to prove that. You're not wrong. That means proof. That's my only point. What you have to show is... You cannot get a divisive of X power n plus 1, which is lesser degree than A-fix, as a multiple of A-fix and R-n. You have to show that. Think about it. You want to show that or not. Anyway, I think we're running out of time. We'll stop here. But that's the property that we have to show. This is true. We generally are equal to A-fix. You have to show that. That requires proving. So it's not an immediate fact. That would be very obvious to you. So used to the integers. We have a great deal of disarmament on X. Here it shows, you have to express... You have to explain things about it. n plus 1. Okay. You have to express the degree less than A-fix. You have to explain things about it. Yeah. You have to explain. You have to say I'm the second. I'm the second. Yeah. So you have to show like that. This is a way to show that it cannot happen. Okay? So you have to... I mean, so you have to go from R-n to F-2. And there you have to use arguments like what he said just now. Okay? So you have to argue that it has to be that way. And then from there you can argue. All right? So we'll come to it. Will A-fix derive the experiment percent of the problem? I'll prove this in next class. Okay? So explicitly. And then we'll pick up something.