Suppose we have a binary cyclic code $C$ with length $N=2^d-1$ defined by a primitive root $N^{th}$ root of unity (so generating polynomial is the minimal polynomial for the root).
I want to show two things:
(1) If $g(X)\in\mathbb F_2[X]$, then $g(X)^2=g(X^2)$. I think this is clear because $g(X)\in\mathbb F_2[X]$ but how to write it down in a formal way?
(2) $C$ is a BCH code with design distance $3$, rank $d$ and length $N$.
The BCH code is with design distance $x$ is the cyclic code of length $N$ over $\mathbb F$ defined by the ideal $J=\{P(X)\in\mathbb F[X]: P(\alpha)=P(\alpha ^2)=...=P(\alpha^{x-1})=0\}$ where $1\le x\le N$ and $\alpha$ a primitive root of $X^N-1$ in field $K$
Same problem as in Part (1), it seems clear from the definitin of the BCH code but how to write it down?
Extended hints:
(1) Any polynomial of $\Bbb{F}_2[X]$ is of the form $$ g(X)=\sum_{i=0}^n g_iX^i $$ for some natural number $n$ and coefficients $g_i\in \Bbb{F}_2$. You are expected to just check it out that the identity holds. A shortcut is to observe that $\Bbb{F}_2[X]$ is a commutative ring of characteristic two. In such rings squaring a binomial is easier than usual, because $$ (a+b)^2=(a+b)(a+b)=a^2+ab+ba+b^2=a^2+2ab+b^2=a^2+b^2. $$ This result is widely known as freshman's dream for some reason (rolls eyes).
(2) Elements of $C$ are multiples of the minimal (=generator) polynomial $f(X)$ of $\alpha$. So you are given that $f(\alpha)=0$. What can you say about $f(\alpha^2)$ in light of part (1)? Your task is basically to check that a polynomial is divisible $f(X)$, if and only if both $\alpha$ and $\alpha^2$ are its zeros.