Finding minimum distance of a cyclic code

Let $q$ be a prime and let $\alpha$ be a non-zero element of $\mathbb{F}_q$. Assume that $n$ is the smallest natural number such that $\alpha^n=1$ and $g(x)$ is a divisor of $x^n-1\in\mathbb{F}[x]$ with $\alpha,\alpha^2,\alpha^3$ as roots.

Let $C$ be the $q$-ary cyclic code with $g(x)$ as the generator polynomial. Show that the minimum distance of $C$ is $4$ if $n>4$.

I am not sure where to start for this question. Any hints will be greatly appreciated. Thank you!

• What's the weight of $g(x)=(x-\alpha)(x-\alpha^2)(x-\alpha^3)$? May 3 '18 at 19:27

Suppose $f(x)$ is a nonzero codeword in $C$ with weight at most 3, say $$f=ax^i+bx^j+cx^k,$$ where $0 \le i <j<k < n$. Since $g(x)|f(x)$, then $$f(\alpha)=f(\alpha^2)=f(\alpha^3)=0,$$ so $$a\alpha^{il}+b\alpha^{jl}+c\alpha^{kl}=0,l=1,2,3.$$ Or in matrix form, $$[v_1,v_2,v_3][a,b,c]^T=0,$$ where $v_1=[\alpha^i,\alpha^{2i},\alpha^{3i}]^T$.
The matirx $[v_1,v_2,v_3]$ is a vandermonde type matrix, and we know $\alpha^{i},\alpha^{j},\alpha^{k}$ are pairwisely distinct, so the matrix is invertible. This implies that $a=b=c=0$, a contradiction!
So weight of $f$ is at least 4.