Let $C$ be the cyclic code of length $63$ over $\mathbb{F}_2$ that has generator polynomial $$g(x)=(x+1)(x^6+x+1)(x^6+x^5+1)$$
We are asked to comment on the minimum distance of $C$. However I don't understand how to find the minimum distance. Can anyone explain how to do this? Thanks in advance!
Best Answer
You can always first check what the BCH-bound gives you.
It is known (and easy to check) that the irreducible polynomial $p(x)=x^6+x+1$ is primitive. In other words $n=63$ is the smallest integer such that $p(x)\mid x^n-1$. If $\alpha\in\Bbb{F}_{64}$ is a zero of $p(x)$ then by Galois theory of finite fields so is $\alpha^2$. Of course, $\alpha^0$ is a zero of the factor $x+1$.
IMO the key to this exercise is that the third factor is the reciprocal polynomial of $p(x)$, that is $$ \tilde{p}(x)=x^6+x^5+1=x^6p(\frac1x). $$ From this we see immediately $\alpha^{-1}$ and $\alpha^{-2}$ are zeros of the factor $\tilde{p}(x)$.
So we have shown that $\alpha^j$ is a zero of the generator polynomial polynomial for five consecutive values of $j=-2,-1,0,1,2$. The BCH-bound then tells us that the minimum distance of the code is at least $5+1=6$. I leave it to you to verify that the generator polynomial is itself of weight six. We can thus conclude that the minimum distance of this code is $d_{min}=6$.
It may be worth observing that because $x+1$ is a factor of the generator, all the codewords have $x=1$ as a zero. Therefore they must have an even weight.