Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

I need to compute a generator polynomial for a binary cyclic code of length 12 and dimension 5. I know that factorization of $(x^{12}+1)$ over $GF(2)$ is $(x+1)^4(x^2+x+1)^4$. What will be next step?

Thanks for any advice.

share|cite|improve this question
Is it possible this process? I know that factorization of $(x^{12}+1)$ is $(x+1)^4(x^2+x+1)$ and I need to find the product of factor of degree $n-k$ (in this case 7). So the generator polynomial can be $x^7+x^5+x^4+x^3+x^2+x+1$. Is it true? Can I get a parity check polynomial from this generator polynomial? – James Oct 16 '12 at 6:25

Linear binary cyclic codes of length 12 are ideals of the quotient ring $\mathbb{F}_2[x]/(x^{12}-1)$. Since this ring is a principal ideal ring, each ideal has the form $(f(x))$ where $f(x)$ divides $x^{12}-1$. In order to have a 5 dimensional code, you will need an $f(x)$ with degree $12-5=7$.

You have already factored $x^{12}-1=x^{12}+1$. Now you just have to find all possible ways to get a divisor of $x^{12}+1$ with degree 7.

Let $p(x)=x+1$ and $q(x)=x^2+x+1$. Check that $p(x)^3q(x)^2$ and $p(x)q(x)^3$ are the only possibilities.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.