I know that Hamming codes can be arranged in cyclic form. But my question is how can I prove this.
My idea was to find a generator/primitive polynomial $p(x)$? For example I want to show that the $[15,11]$ Hamming code can be written in a cyclic form. Then the generator polynomial $p(x)$ must divide $x^{15}+1$. The factorization of this is: $x^{15}+1=(x+1)(x^2+1)(x^4+x+1)(x^4+x^3+1)(x^4+x^3+x^2+x+1)$. But what is now my $p(x)$ and why?
And what is the next step in my proof?
Thanks in advance.
Before answering your question, not every Hamming code is equivalent to some cyclic code. For instance, the ternary $[4,2,3]_3$ Hamming code (aka the tetracode) is not equivalent to any cyclic code.
But it is true that binary Hamming codes can all be seen as cyclic codes. In fact, the $[2^r-1,2^r-r,3]$ Hamming code is the primitive narrow-sense BCH code of length $n = 2^r-1$. More generally, the $q$-ary $[\frac{q^r-1}{q-1},\frac{q^r-1}{q-1}-r,3]_q$ Hamming code with $\gcd(r,q-1)=1$ is the narrow-sense BCH code with its defining set being the $q$-cyclotomic coset $C_1$ of $1$ modulo $n$.
So, if you would like a generator polynomial of the $[15,11,3]$ Hamming code, take the $2$-cyclotomic coset $C_1=\{1,2,4,8\}$ of $1$ modulo $15$. The generator polynomial $p(x)$ you want is the corresponding minimal polynomial
$$\prod_{i\in C_1}(x-\alpha^i),$$
where $\alpha$ is a primitive $15$th root of unity in $\mathbb{F}_{2^4}$.
For instance, take the irreducible polynomial $x^4+x+1$ over $\mathbb{F}_2$. If you regard $\mathbb{F}_{16}$ as $\mathbb{F}_2[x]/(x^4+x+1)$, the element $x$ is a primitive $15$th root of unity. Writing this element as $\alpha$, your $p(x)$ is
$$\begin{align*}p(x) &= \prod_{i\in \{1,2,4,8\}}(x-\alpha^i)\\ &= (x-\alpha)(x-\alpha^2)(x-\alpha^4)(x-\alpha^8)\\ &= x^4+x+1.\end{align*}$$
If you would like the proof that the $[\frac{q^r-1}{q-1},\frac{q^r-1}{q-1}-r,3]_q$ Hamming code with $\gcd(r,q-1)=1$ is the narrow-sense BCH code with defining set $C_1$, assume that $\gcd(r,q-1)=1$. Let $\alpha$ be a primitive element of $\mathbb{F}_{q^r}$, where $\alpha^{q-1}$ is a primitive $n$th root of unity. Construct a cyclic code (actually the narrow-sense BCH code) of length $n=\frac{q^r-1}{q-1}$ by taking the minimal polynomial $M_{\alpha^{q-1}}(x)$ of $\alpha^{q-1}$ as its generator polynomial. The nonzero elements of the subfield $\mathbb{F}_q$ in $\mathbb{F}_{q^r}$ are powers of $\alpha$, where the power is a divisor of $n$. Since $\gcd(r,q-1)=1$, the only element that can be a power of $\alpha^{q-1}$ is the identity. Hence, writing the elements of $\mathbb{F}_{q^r}$ as $r$-tuples in $\mathbb{F}_{q}^r$, no $r$-tuples corresponding to $\alpha^0, \alpha^{q-1}, \dots, \alpha^{(q-1)(n-1)}$ are multiples of another using only elements of $\mathbb{F}_q$, which means that these are simply the points on projective geometry $\operatorname{PG}(r-1,q)$. Hence, the $r \times n$ matrix $H$ in which the columns are the $r$-tuples corresponding to $\alpha^0, \alpha^{q-1}, \dots, \alpha^{(q-1)(n-1)}$ is the parity-check matrix of the $[\frac{q^r-1}{q-1},\frac{q^r-1}{q-1}-r,3]_q$ Hamming code.
