 Alright, so we are going to do some problems for today and these are some additional problems. You will find them in the usual page. You go to my page. You don't know the exact full address but you have to click on courses. This will be a tab on the left. And then you click on e if I want to. We take you to a page which has the things and these are additional questions. So that's the information on how to get one. So we are going to start with the first one. So here is the first question. So almost all these questions showed up in some quiz or finals or something like that. If you indicate with that. So you have variable x which belongs to p of 32. So 2 power 5, characteristic 2p. And then you have some function or an operator that's defined on this. So it's called... Everything that's tr, you can expand it as stress if you like. It's called tr, so it's a standard variable. And this is p of x plus x squared plus x plus 4 plus x plus 8 plus x plus 16. So that's the formula. So give me any x and g of 32. I'll apply this and give you place of x. So the first chart asks you to compute, basically asks you to show the problem. Place of x squared, what will it be? So if I square it, what's going to happen? I have to take this whole thing and square it. It will do a square plus x plus 4 plus x plus 8. x plus 32 and that's x itself. So see immediately what it is equal to. Place of x. So place of x is something which once squared gives place of x again. So you can pull this to the side. Now you will see place of x times. Place of x plus 1 equals 0 which means place of x equals 0 or 1. Minus 1 which is the same as 1 because we know we are enough. Characteristic 2 feet. So the range of place of x is only 0 or 1. So place of x is a special operator that way. It takes any element of g of 32 and sends it to binary, 0 or 1. And the part B asks you to show the next interesting property. Place of x plus y will be what? So it's not very hard. Instead of x you put x plus y. x will be x plus y and everything is just a power of 2. So you will eventually get x power 2 plus x, y power 2, then x power 4 plus y power 4, etc. So it's very easy to show that this will be the same as place of x plus y. So we've seen a very interesting property for place. It's also an operator which preserves summation. It behaves well with summation. If you take place of x plus y, you will get place of x plus place of y. It's not quite, you can say it's linear and roughly, but if you multiply x with some alpha, place of alpha x will not be the same as alpha tends to be. So it doesn't work out that way. So it's not linear with respect to multiplicative constants. But that's too much to expect. But as far as summation is concerned, it is linear. So linearity is usually with respect to this bit. So if you take 0 or 1, then it is linear. And then the next question asks you to find, find all x such that place of x equals 0. How do you do that? Sorry? Yeah, I mean, so what we're doing here is to just say, take all the 32 elements and then, of course, 0 is a quick answer. Take all the 32 elements and then, do what? We'll just compute one after the other and we'll get the answer. So there are some shortcuts you can take. So for instance, if you take an element which is a square, for instance, if I evaluate place of x squared, what will I get? Same as place of x. So you can use such tricks. So this is also true. Am I right or am I wrong? Place of x squared is the same as place of x. So x squared plus x squared plus x squared plus x squared plus x plus x So that means your clue as to how to shorten your computation. What is this clue? How does this help you? For each trigonatomic process, you just say an F and it's enough if you compute for 1. So then you can quickly go and see. So you'll see that 16 of the elements of Gf 32, will have trace 0. The remaining 16 will have trace 1. So, that is another property that you can show for the trace function in general also. So, trace is a special linear operator from 3R32 to 3R4. So, part 2 basically expands on it, we will ask you to show the second question is similar to the first question, similar to one I am going to skip that you write on your own. So, what is the part except that you do something slightly more general than going to binary. So, you can take a look at it, so it is a bit interesting. So, let us go to the third question. So, the third question is about solving quadratic equations over finite fields. So, you have a quadratic set equation is basically f of x equals x square plus x plus 0 and k belongs to g of 42, some constant which is in g of 32. x is a variable and x also belongs to g of 32. We will try to get to a formula for solving for x. So, if you have this for a real equation with real coefficients, you know the formula already. So, it is going to be minus 2 plus or minus square root of b square minus 4 a c by 2. So, finite fields of characteristic 2, it is not going to work. Otherwise, the formula will work. If you do not have characteristic 2, the formula will be very happily work. So, you can happily use it. There is no problem. So, that is a very, that is a very just first on the farm. If you go back and look at how to derive it, how many of you know how to derive it? What do you do to derive it? You have to complete the square. So, here you cannot complete the square because there is no 2. So, that will work in any field except for characteristic 2. So, in characteristic 2, in some cases you can use this trick. Just do the trick that you use. So, path a shows you how it works. Path a says, suppose you have a solution. Suppose that there is an x in g of 32 which will solve this. So, let us look at it. So, what are the various possibilities? Suppose x plus k could be irreducible over g of 32. In which case there will be no solution. Or there should be 2 solutions. There cannot be just one solution over any field. Why is that? So, if you have one solution, you know it is attacked. The other thing is linear and that will definitely have a solution over any field. So, because it is a quadratic, you can say quickly that it has to either have 2 roots in g of 32 or no roots at all. So, let us say first of all that there is a root. So, there is a root. Which means that I put x of that. I put x plus k equals 0. So, what it can do is, what the question asks you to do is to show certain properties of k. So, it shows. So, you can show the following. The question asks you to show it. I am just saying you can show. k plus k squared plus k power 4 plus k power 8 plus k power 16. What is this? We know this as a function. What is this? k plus k. So, k plus k has to be equal to 2. You can show this. How do you do that? k is equal to x plus x squared. So, k is equal to x plus x squared. And then you do a trace. That is the same as trace of x plus trace of x squared. But you know trace of x plus x squared is the same as trace of x squared. So, you can do the manipulation if you like. You will get 0. So, what does this immediately mean? The previous question, you listed all the elements of G of 32 which had trace 0. So, there are 16 such elements. So, whenever k takes one of those 16 values, you can possibly have a root. If it takes something else, then the trace is 1 and that violates the requirement that you have. So, there cannot be a solution in 15. So, that is the first nice application of the trace. So, if you have trace as a kind of a trivial result, it is a nice application to show that only if the trace is 0 for k, you will have roots in G of 32. I am sorry? The trace of x can be 1. Yeah, trace of k should be 0. x is not irrelevant to this. So, x squared plus x plus k, the question that I am asking is for what k do you have a solution? And there is also a reason why you don't have to look at a x squared plus b x plus k. You can manipulate. You do a very simple substitution to get it to x squared plus x plus 1. So, that is not what you have. You just put one, let's say equal to the big substitution. You will get to this. Yeah, so this is a very general form of a quadratic equation. k is not the variable. k is some constant that I am picking ahead of time. k is not a constant. k is a constant that I have chosen ahead of time. k is a constant that has been chosen. Suppose x squared plus x plus k has a solution. There is some x that solves that equation. Then you can show trace of k has to be 0. Which means what? You use what is called the contract positive or whatever. So, trace of k is 1, then there cannot be any x for which that is 2. So, a implies b means what? Not a. Yeah, not b implies not a. Not a does not imply not b. So, that is the converse of the statement. So, hopefully you will read this logic before. But remember that a implies b does not mean that not a implies not b. Only not b implies not a. That is called the contract positive. So, what does x have in the coefficient? x has a? What do you mean by x having a coefficient? Alpha x plus k. Alpha x plus k. See then what you should do is you should do the substitution. Y equals alpha x by alpha. One of those things. So, if you do the substitution you will see, I think it is. You can go back to a y squared plus y plus some other constant equal to 0. So, it is not a very hard thing to do. Something very simple to think about. Given an alpha and g of 32, can I always find other limited g of 32 values? You are asking does every element in g of 32 have a square root? What do you think? Why? Yeah. So, there will be always a square root in g of 32. The reason is if alpha power i, if i is even then clearly there is a square root. Simply take i by 2. If i is odd what will you do? You add 31 to i. So, you know alpha for 31 is? 1. So, you add 31 to i. So, 31 plus i will be even. So, you take that by 2. We will get a square root. So, every element has a square root. So, what is the connection in asking you that question? Yeah. Yeah. So, a squared plus bh plus c is not necessarily the most general form of the quadratic equation. You can go from there to x squared plus x plus something. Or maybe y squared plus y plus something. I think it is bc by a in general. I forget this. It is a very standard substitution. These are tricks you can do even for cubic equations. You do not have to necessarily look at a x property plus b x property. There are ways to move it around so that you get one least like x cubed plus a x plus c. These are standard things that you can do. Yes. It is a bit non-trivial. So, the question is how do you get to this if you did not know anything else? It takes some clever tricks. I mean it is not a very obvious answer. I know the answers I am doing it, but otherwise it is not very obvious. Is there a similar popular thing that you can do? Yeah, in every field you can do it. So, here you can see what I have done. I have gone up to 16 and stopped. So, you can start with x and keep on stirring it till you go to a non-trivial stir. So, when you come back, you just stop it. That is the price and that is how to store all these conditions. All these nice properties it is how to store. It has some very powerful results. So, I will just take it for a few days. Any linear functional from this value of field to the lower field is in general trace of some constant times x. There is nothing else that is linear. It has a very generic property. It is not very relevant right now, but it has very strong properties. Alright. So, what we asked you to do the following. Find i and j, subscribe, x of k power i plus k power j equals 0. So, basically you know this is 0. So, you know k plus k square plus k power 4 plus k power 8 plus k power 16 is 0. So, you just look at this equation. You can identify an x and an x square. So, from here. So, you can write it as k plus k square plus k power 8. Plus k square plus k power 8 power square is 0. So, the trace of k equal to 0 already gives you a solution for x. So, now what you do? You simply set one of the rules to be this. So, k square plus k power 8 is one rule. Is it okay? Alright. So, you can identify that. So, what is the other rule? Square of this. Square of this will not be the answer. What is the other rule? 1 plus this. So, 1 plus this will be the other rule. Think about it. See, any quadratic equation, the middle term has to be equal to the sum of the three rules. So, the middle sum is 1 already. 1 by b minus b by a if you like. b and a are 1 here and minus is the same. So, sum has to be 1. So, one rule is given. The other rule is obvious. 1 plus k square plus k power 8. You can also do the product. You get the same answer. So, what is the problem? How much property model equals? Square of this. Square of this. Square of this. Square of this. Square of this. So, these are tricks you can use to solve quadratic equations in a finite field. So, I did it in GF 32. If you try the same trick in GF 16, it will not work. The reason is, you notice there will be only four terms there. k plus k square plus k power 4 plus k plus 8 equal to 0. You won't have this split nicely. So, but in any odd m, GF 2 power m, you can use the same trick. You have to have trace of k equal to 0 and then you split it, group it. You will get the answer. It's very easy to solve quadratic equations in 2 power m. For m even, I will let you look at them. I mean, there is a standard method. It's not that it's very hard. It can be done, but that's some other modification. It's not a very simple method like this. Is that okay? Okay. So that's the third question. I'm going to go to the fourth question. Okay, so even though I've written it, I'll urge you to once again go back and work out all these questions one after the other, like you didn't know the answer. It's very important. Like I said, I mean, this is... there are a million things like this in finite states, right? So, I mean, every equation will have a method like this. So, you have to be comfortable enough with these kind of manipulations. Okay. All right. Okay, so the fourth question is also a bit of a previous question. Okay, so let me try and go through that. You have alpha belonging to 0 to 2 power m. It's just primitive. I think it's partly printed in your... in the sheets again. You don't have to print both sides. I'm sorry about that. Well, I have the whole question in here. Okay. Okay. So this is primitive. And then, primitive and trutus unity. Okay, so I'll do the primitive and trutus unity. Okay, so this is the first information that's given to you. The next information given to you is n is such that 1 plus x plus x squared plus... x power n minus 2 plus x power n minus 1 which you can write in short as i equals 0 to n minus 1 x power i is irreducible. Of course, I have to say which field it is. It's irreducible over this rate. Okay, so g of 2x. In parameters with coefficients, with binary coefficients, we can't factor this into a product of... If it's not non-dynamic, of course, you can factor it. You know, you can factor it. So I'll give you the factors. So that works. But for binary, it's known that it's irreducible. So this is what's given to you and you have to find the fellow. Here, the first thing you have to find is the minimal polynomial of alpha over g of 2. It's not mentioned. You have to find n alpha over g of 2. So what's the polynomial of alpha over g of 3? That's the first thing you have to find. Sorry? Yeah, I mean, usually you have a binary field. It's okay. I mean, if you like, I can say 3. So this is a polynomial with binary coefficients which will have alpha as a root. And among those, pick the smallest one. Which one? Will it be this? How do you know alpha as a root? So that's the first part in proving the... So it's the only non-trivial thing in proving that. Okay? So you know that x bar n plus 1 equals... So x bar n plus 1 has alpha as a root. Okay? So if you take some polynomial, f of x, which is f of 1, then you know f of alpha equals 0. Okay? So now I'm going to factor x bar n plus 1. x plus 1 times whatever I wrote there. Okay? This has alpha as a root. Okay? So now x bar n clearly does not have alpha as a root. So alpha has to be a root of this other thing. Okay? So alpha has a root of this guy. And then I also know that is irreducible. Okay? So clearly that has to be the minimal polynomial of alpha over 3. Okay? It has binary coefficients, right? That's also good. Part b is a little bit more tedious. I'm going to leave you to it. If you really get stuck, ask me again on Thursday. We'll go through this. Okay? Find the smallest positive integer J such that 2 power J equals 1 model. Is that okay? Okay? Find the smallest positive integer J such that 2 power J equal to 1 model. It's not very hard. It's very simple. But you may not. So question sounds like it's totally different from the first part, right? So okay, we'll use the first part. Okay? So how do you use the first part? How do you think about the secretomic process? Because 2 power J equal to 1 affects the length of the secretomic process. And the fact that the minimal polynomial of alpha has degree n minus 1 says a lot about the secretomic process. So you have to use that idea and then come to the answer. Okay? Sorry? Yeah. Yeah, I think about it. It'll be something very simple. Okay? So it has to be... So you have to use that idea and get to the answer. Okay? So otherwise it just sounds like, man, how do you think about that? So it doesn't seem to be unrelated. But the relation is just secretomic process. Okay? So you know how do you find the minimal polynomial of alpha over gs2? To take all its secretomic lambda, alpha, alpha squared, so on, right? What should be the size of the secretomic process? It has to be equal to n minus 1. Only then the reducible polynomial will have n minus 1. Okay? So our controls the size of the secretomic process with the smallest j such that 2 power J becomes equal to 1. Okay? And that has to happen only with n or n minus 1. And so that's the idea. Okay? Okay? So think about it. So if you really start to come back and talk to me, I'll discuss this in more detail. But that's the idea. Okay? So you have to connect the size of the secretomic process to the degree of the reducible polynomial. And size of the secretomic process is controlled by this k. Smallest j such that 2 power j equal to 1 mod n. Right? Because we're doing multiplication by 2, modular n. Okay? So you start with 1. Okay? So how do you do secretomic process for alpha? Okay? So you start with 1. And then you do 2. 2 squared. 2 power 3. So all we keep doing with those modular n till you get a repetition. I know that up to 2 power n minus 2, there is no repetition. Maybe it's n minus 1. Okay? So think about it. So 2 power n minus 2, there is no repetition. Although I know that there's no repetition. And otherwise we did knew the minimal polynomial will not be n minus 1. Okay? So of course in 2 power n minus 1, it has to be equal to 1 mod n. Okay? So that's how it will work out. Okay? So that's the idea behind this relationship. So as you can imagine, there are many more complicated relations between polynomials, reproducibility and how the number 2 behaves when successive powers are raised. Okay? So this is 2 squared important. This k, right? Okay? This j, smallest j such that 2 power j equal to 1 mod n will be the size of the secretomic process for alpha. Okay? And that will be the degree of the minimal polynomial of alpha. Both of these will be modulo n, of course. Okay? Now if I want to... So here it's very easy. So suppose in general, I want the minimal polynomial of alpha part 3. Okay? So I have to do 3, 3 times 2, 3 times 2 squared, so on. So the smallest j for which 3 times 2 power j equals 1 mod n will give me the degree of that. Only under the conception that you can throw 3 away, because you keep 3 there. There's no even how 3 will go. Okay? 3 times 2 power j equals 1, smallest j. Okay? So the same result is true here. We've just used it in a special case here and we know the answer. Okay? So I'm going to move on to the first question. So if you try it again on your own. Okay? So the reason why I'm not writing it down and I write it down, it will be probably clear to you, but think about it and try it on your own. If you get stuck, come back and talk to me. I'll explain it more. Okay? So the first question is again interesting. It plays with multiple constructions of gf2 and f2. Okay? So I'm going to define gf16 in the following way. F of alpha and gf2 alpha. What do you mean by gf2 alpha? Paranormal with primary coefficients in the variable alpha. Okay? Such that degree of f of alpha is less than or equal to 3. Okay? So I'm going to define gf16 if I do addition and multiplication model of f of alpha equal to something irreducible polynomial. I'll pick up to be alpha power 4 plus alpha power 3 plus alpha power 2 plus alpha plus 1. Okay? So I've chosen a dv's middle polynomial. So if I do this, what happens? So this polynomial is irreducible, but it is not a primitive polynomial. Right? So alpha will not be a primitive element of this field and this construction. Okay? Something else will be the primitive element. What's the twist here? We have to pay attention to that side. Okay? If you don't do that, the answers will go down. Okay? On part a, I've asked to find f of alpha in this gf16 in the before constructed gf16 with minimal polynomial x power 4 plus x power 3. Okay? Seems to make sense. So think about it. So how would you go about doing this? Sorry? Okay. So how would you? I mean, instead of more systematic way of doing it, that's my question. Hmm. Yeah. So how would you find a solution to do this? Because that's a bit painful, right? I have to try each and every possibility. Is there a way to simplify that work? You know, you know, something about disomorphism. Maybe you can use that. Using this as a polynomial. And then? Yeah. Yeah, exactly. So that's the idea. That will simplify your work considerably. Okay? So what should we do? You construct a gf16 with x power 4 plus x power 3 plus 1 as your pi of beta, so to speak. Okay? And then find the element in that whose minimal polynomial is x power 4 plus x power 3 plus x power plus x power 1. That you can use some standard constructions. Right? You know what it is, right? So you, based on your experience, you can quickly find that. Okay? So you know, like, for instance, alpha power 3. So beta power 3 would form that. So beta power 3 is not a good thing to use. We would use its vector notation. Okay? And then you map, when you map beta to alpha, that element will go to the element corresponding to this. Okay? Is that okay? So we'll be back to that. So you use that mapping and then invert that mapping, you'll get to get to this. And then you can quickly check that that also is the root of this. Once you find one root, other roots are easy. You just repeatedly square it. We'll get to that. Is that okay? Is that clear? Okay? So maybe we should try this. And then people are a little bit confused. Okay? So let's say I construct a copy of an isomorphic copy of gf16, that's the same thing, but it's a little bit different. So it is of the form 0, 0, plus 0, 1, beta, plus 0, 2, beta, plus 1, 0, 3, beta, plus 4. You know, there is a binary. And then what will I do? I will insert, but beta, plus beta, plus 1 equals 0. Okay? So addition and multiplication will be modular this. Okay? I'm sorry? Oh, sorry. Okay? So I'll do that. Okay? So here, it's a little bit easier because I know beta is a primitive element. Okay? So because this is a primitive polynomial. Okay? So I know beta is a primitive element of gf16. Okay? I'm sorry, gf16 dot primitive. So x bar 4 plus x bar 3 plus 1 is a little bit different from what we are used to at standard. But it only does, so it only does flipping, right? So it's not too bad to deal with. Okay? So we know that beta bar 3 will have minimal polynomial this bit. Okay? We're all right. So now, what will happen when I map? So what should I do next? That is a bit confusing to me. So this is that. So now I'm going to map, I'm going to do isomorphism, right? So I'm going to take 1 to itself and then I'm going to map beta to alpha, beta square to alpha square, beta bar 3 to alpha bar 3, right? So what should happen if I do that? Can I do that? It's not to alpha. It's not to alpha. Okay? So that's what I should do, right? We're clear? So I have to map beta bar 3 to alpha, is that correct? Yeah. So I should do that. So I have to map beta bar 3 to alpha and that should hopefully give me the right thing. Okay? So I will go from, I will assign this beta bar in that other field. Okay? So that's what's confusing me. So yeah, I can do that, right? So the beta bar 4 will also have the same minimal polynomial as beta. So beta bar 4 I know is 1 plus beta bar 3. So once I find any one element, I'm good enough. Okay? So it looks like the answer suggests that 1 plus alpha should be a root of x bar 3 plus x bar 3. Okay? So let's do that. So the isomorphism has to be beta bar 3 mapping to alpha. Okay? And then 1 plus beta bar 3 equals beta bar 4 here. Okay? So that means 1 plus alpha should give you an element, should have. Okay? And then n beta bar 4 as well. Alpha bar 4 plus alpha 3 plus 1. So 1 plus alpha should have 1 plus x bar 3 plus x bar 4 as minimal polynomial. Seems to be okay. But let's check that. Okay? So it could be wrong. It would be wrong here. Okay? So let's check that. So let's check this. 1 plus alpha bar 3 plus alpha bar 4 plus alpha bar 3 is the same as 1 in this field. So it's alpha plus plus alpha bar 3. Is that correct? Okay? And then you have 1 plus alpha bar 4. Okay? So I get that. Okay? So it becomes 1 plus alpha plus alpha bar 3 plus alpha bar 4 and that goes to 0. Okay? So 1 plus alpha is an element which has 1 plus x bar 3 plus x bar 4 as the minimal polynomial. Okay? So what are the other elements? 1 plus alpha squared, 1 plus alpha bar 4 which will become 1 plus. So you have to keep doing the model of alpha. You can get all that. Okay? Now we repeat the same thing with x bar 4 plus x plus 1. Okay? So there you have to be a little bit more careful to go to this element. So here we were a little bit lucky. We got this very quickly. There you may not get it very quickly but you have to work on it and then get to the suitable polynomial. I think it will be something else. Anyway. So you should be able to get that. Okay? So you can try and repeat the same thing for this. Is it okay? Okay? So we were a little bit twisted though. We used alpha somehow. Somehow you can always use the tried and trusted method. Okay? Just try a few cases. It's enough if you find one root. Okay? Then the other roots can be quickly found. In fact I think in the exam if I remember correctly most people took that route. Okay? So you just try a few cases. 1 plus alpha and that's a prime candidate. Right? You say 1 plus alpha. You know it has to be a prime candidate. You try that. Try a few cases and you'll succeed. Okay? So let's try the trial and error method for X power 4 plus X plus 1. What can you try? So if I want a root of X power 4 plus X plus 1. Okay? So that's part b. Maybe I didn't tell you what part b is. Okay? So part b is repeating the same things for X power 4 plus X plus 1. Okay? Find all solutions for X power 4 plus X plus 1. So we know the roots of X power 4 plus XQ plus 1. We know the roots of X power 4 plus XQ plus X plus 1. Yeah. So you can also do. Yeah. So you can do various things but for instance X power 4 plus X plus 1 will have 1 by the other thing as the root. But then when you have to do that 1 by 1 plus alpha. If it comes to be doing that modulo alpha power 4 then it's okay. It's not very hard. Something like alpha plus alpha square does not divide anything. It's very tough to get. I don't know. So he's suggesting let's try alpha plus alpha square. Okay? Is that what you're trying? Is what you're suggesting? Let's try it. Okay? So let's try alpha plus alpha square. Okay? So what will you do for that? Alpha square plus alpha power 8 plus alpha plus alpha square plus alpha square. Okay? So this is what we get. Alpha power 8. Sorry? Alpha power alpha 4. Okay? Where is this 0? So, that seems fine. So, actually alpha power 5 will be 1, right? Is that okay? So, you use the fact that this is equal to alpha power 3, because you use alpha power 5 equals 1. So, then you can quickly see that this also goes to 0. So, alpha power alpha star is a root of x star, 4 plus x star and then you keep repeatedly sparing it till you exhaust it. So, this is also a equally nice method, but if you want to sound like you are being very smart and using isomorphisms and all, we can use the isomorphism and get to the answer also. But in general, trying out one after the other may be very inefficient. So, because the number of roots are quite small, using the isomorphism is a clean answer in general for bigger fields that you work very nicely. For small fields like this, you can simply try some random things and then you will get it. So, we are going to use the arithmetic category. Notice that in this construction, alpha has other sides, not an other sustained element. So, different set up. Okay. So, I think we will stop with this. We have done five questions. The sixth one is not too bad, you can try it. Okay. So, I will urge you to go and try these questions and the other questions in the website. Once again, the website is here. If you go to my page, click on courses 512. I have added a new assignment, additional questions. Okay. So, there are usual additional questions that are quite long. It is 21 questions set. Okay. So, we will try and go through all of them by this week for fluency. But tomorrow is going to be a presentation class. Joshi and Swarna and who else? And Shailesh and Shailesh. Okay. Thank you.