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Currently I am working on matlab in order to find Cyclotomic Cosets for 45. As 45 in not in the format of 2^m-1, matlab give me an error. I am trying to write algorithm in matlab/octave for my research work. My desired result i-e Generating polynomial for (45,33) is X^12 + x^3 + 1. I am considering this as my primitive polynomial, off course it is non primitive. And on this base I want to generate generating polynomial for (45,29), (45,23,) and (45,7), (45,1).

Thanks alot.

user153842
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  • I have no idea what you are asking about. – mercio May 28 '14 at 14:40
  • I want to know the relationship between C(15) and C(45). – user153842 May 28 '14 at 14:52
  • It would help us all, if you gave us more context in the question body. Like confirm my guess of this being in characteristic two? – Jyrki Lahtonen May 29 '14 at 05:13
  • Thanks alot.... @Jyrki Lahtonen .. I am trying to write algo in matlab/octave for my research work. My desired result i-e Generating polynomial for (45,33) is X^12 + x^3 + 1. I am considering this as my primitive polynomial, off course it is non primitive. And on this base I want to generate generating polynomial for (45,29) etc – user153842 May 29 '14 at 14:01
  • In other words if I have alpha = x^12 + x^3 + 1, what should be its gf(x,y)?? – user153842 May 29 '14 at 14:46
  • Generating polynomial for what? A binary cyclic code? I'm afraid I don't remember Matlab functions related to finite fields (other than that I found them very kludgy! If $x^{12}+x^3+1$ is the minimal polynomial of a given 45th root of unity $\alpha$, then $x^4+x+1$ is the minimal polynomial of $\alpha^3$. So if you are looking for a generator polynomial of a length $45$ BCH-code of designed distance $5$ and dimension $29$, then that is $$(x^{12}+x^3+1)(x^4+x+1)=x^{16}+x^{13}+x^{12}+x^7+x^3+x+1.$$ – Jyrki Lahtonen May 29 '14 at 16:17
  • But you should edit these extras into the main question body. For example, if somebody who is knowledgable about Matlab shows up, they may miss these comments. – Jyrki Lahtonen May 29 '14 at 16:18
  • Thanks a lot ... as matlab has impose restrictions on calculating fields and generation polynomial.... for (45,29) with d = 5 the given result is correct is there any way to calculate generator polynomial for other dimensions like 23,7 ... etc i-e (45,23) and (45,7)... – user153842 Jun 01 '14 at 09:43
  • If $j\in C(5,45)$, then $\alpha^j$ is a ninth root of unity, so... – Jyrki Lahtonen Jun 04 '14 at 21:07

1 Answers1

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Presumably this is binary? At least the remark about $2^m-1$ points that way.

Anyway, over a field of characteristic two cyclotomic cosets modulo $n$ are the sets of the form $C(j,n)=\{k\in\Bbb{Z}_{n}\mid k\equiv 2^i j\pmod{n}\ \text{for some natural number $i$}\}$. So when $n=45$ we get $$ \begin{aligned} C(0,45)&=\{0\}\\ C(1,45)&=\{1,2,4,8,16,32,19,38,31,17,34,23\}\\ C(3,45)&=\{3,6,12,24\}\\ C(5,45)&=\{5,10,20,40,35,25\}\\ C(7,45)&=\{7,14,28,11,22,44,43,41,37,29,13,26\}=-C(1)\\ C(9,45)&=\{9,18,36,27\}\\ C(15,45)&=\{15,30\}\\ C(21,45)&=\{21,42,39,33\}=-C(3) \end{aligned} $$ So you see that the minimal polynomial of a root of unity $\zeta$ of order $45$ is of degree twelve. The cyclotomic polynomial (over $\Bbb{Q}$) $$ \Phi_{45}(x)=\frac{(x^{45}-1)(x^3-1)}{(x^{15}-1)(x^9-1)}=x^{24}-x^{21}+x^{15}-x^{12}+x^9-x^3+1. $$ is of degree 24, so it splits into a product of two irreducible factors - both of degree 12. The zeros of these two factors are $\zeta^k$ with $k\in C(1,45)$ and $k\in C(7,45)$ respectively. As $C(7,45)\equiv-C(1,45)\pmod{45}$, the two factors will be reciprocals of each other. I guess we could take advantage of this bit in figuring them out. But it is more efficient to take advantage of the fact that $\zeta^3$ is of order fifteen. Because 15 is still divisible by three, we also get the converse. If $\omega$ is of order fifteen, and $\alpha^3=\omega$, then $\alpha$ will necessarily be of order $45$. As $15$ is of the form $2^m-1$ for $m=4$ we can utilize the factorization of $\Phi_{15}(x)$ over the binary field. Even Matlab should be able to give you that. I happen to have it memorized, so let's just state that $$ \Phi_{15}(x)=x^8-x^7+x^5-x^4+x^3-x+1\equiv(x^4+x+1)(x^4+x^3+1)\pmod 2. $$ Let the two factors here be $f_1(x)=x^4+x+1$ and $f_2(x)=x^4+x^3+1$. In view of the above the factors of $\Phi_{45}(x)$ in $\Bbb{F}_2[x]$ must be $$ f_1(x^3)=x^{12}+x^3+1\qquad\text{and}\qquad f_2(x^3)=x^{12}+x^9+1. $$

Jyrki Lahtonen
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  • The cyclotomic cosets modulo 15 (resp. 45) are, indeed, connected. By taking a coset modulo 45, and reducing everything modulo 15 gives you a coset modulo 15. Doing this exercise for the listed cosets tells you quite a bit extra. Occasionally you get repetitions. Here $C(1,45)$ covers $C(1,15)$ thrice. The same with $C(7)$. Also $C(9,45)$ and $C(3,45)$ both give you $C(3,15)$. I guess it is relatively easy to analyze when any of this happens. – Jyrki Lahtonen May 28 '14 at 14:42