Number of (binary) cyclic codes of length 21

Question

what are all the binary cyclic codes of length $21$? Is it possible to find all values of k for which $[21,k]$ is a binary cyclic code? How do i go about this problem, does finding the cyclotomic cosets have something to do with the solution? I looked at Describe all the cyclic codes of length $7$. but it really does not help much. thanks alot.

Answer 1

Hm... If you did not find that answer helpful (it tells you exactly how to find all such codes if you replace 7 with 21) I am not sure what could possibly help. Maybe repetition? Let's try.

The number of cyclic binary codes of length n corresponds to ideals of $F_2[x]/(x^n-1)$, and those correspond to divisors of $x^n-1$.

A factorization into irreducibles is $x^{21}-1=(1+x)(1+x+x^2)(1+x^2+x^3)(1+x+x^3)(1+x^2+x^4+x^5+x^6)(1+x+x^2+x^4+x^6),$ and that gives you all 64 divisors.

You did not mention what $k$ is supposed to be, but I guess it is the dimension of the code. This is simply $21$ minus the degree of the chosen generator polynomial (more generally, the generator polynomial is of degree $n-k$ where $n$ is the block length), and I'll let you puzzle out what the possibilities are.

Cyclotomic cosets would be helpful for factoring by hand. Including the details of that would exceed the scope of this post.