Construct minimal polynomials using cyclotomic cosets 
Decompose $f(x)=x^{18}-x^3$ in irreducible factors over $\mathbb{F}_2$

$f=x^3(x^{15}-1)$ and since the multiplicative order of $15$ mod $2$ is $4$ , the splitting field is $\mathbb{F}_{2^4}=\mathbb{F}_{16}=\mathbb{F}_{2}[x]/g(x)$, with $g\in\mathbb{F}_2[x]$ irreducible and of degree $4$.
The cyclotomic cosets are:


*

*$C_0=\{0\}$

*$C_1=\{1,2,4,8\}$

*$C_3=\{3,6,9,12\}$

*$C_5=\{5,10\}$

*$C_7=\{7,11,13,14\}$


For each of them there is a minimal polynomial:


*

*$m_0=x-\alpha^0=x-1$

*$m_1=(x-\alpha^1)(x-\alpha^2)(x-\alpha^4)(x-\alpha^8)$

*$m_3=(x-\alpha^3)(x-\alpha^6)(x-\alpha^9)(x-\alpha^{12})$


and so on, so that: $x^{15}-1=m_0m_1m_3m_5m_7$.
But how to find explicitly the $m_i$'s without doing all the multiplications ?
I read on M.Tomlinson book Error-Correction Coding and Decoding that
$m_1=x^4+(α+α^2+α^4+α^8)x^3+(α^3+α^6+α^{12}+α^9+α^5+α^{10})x^2+(α^7+α^{14}+α^{13}+α^{11})x+α^{15}$
and that the sums in the brakets are either $1$ or $0$, but did not understand neither how to construct $m_1$ nor how to decide the values.
 A: To construct $m_1$, just do the basic algebra starting with  $m_1=(x-\alpha^1)(x-\alpha^2)(x-\alpha^4)(x-\alpha^8)$ (if needed (not in this case) use the fact that $\alpha ^{15}=1$). The coefficients will either be $0$ or $1$ since you are working over $GF(2)$. If you add the coefficients of $x^3$, $x^2$ and $x$ you get $\alpha^{14}+ \ldots + \alpha$. Now, $\alpha$ generates (by choice) the multiplicative group of zero's of $x^{15} -1$, hence $\alpha \neq 1$. However, $x^{15}-1=(x-1)(x^{14}+ x^{13} \ldots + 1)$ , so $\alpha^{14}+ \ldots + \alpha = 1$ (remember that you are working over characteristic $2$), which yields $m_1= x^4+ \rho x^3 +\delta x^2 +(1+\rho + \delta)x +1$. Now you should notice that the roots of the minimum polynomial corresponding to coset $C_7$ are multiplicative inverses of the ones corresponding to $C_1$. This means that $m_7$ will be the reciprocal polynomial of $m_1$. Suppose now that $\rho = 0$ and $\delta = 1$, then the reciprocal of $m_1= x^4 + x^2 + 1$ is equal to $m_1$ which would mean that $m_1$ has $8$ different roots, which cannot be the case. Suppose that $\rho =1$ and $\delta = 1$, then $m_1=x^4 + x^3 + x^2 + x + 1 $, impossible for the exact same reason. So either $\rho = 1$ and $\delta = 0$ or both $\rho$ and $\delta $ are $0$, depending on the choice of $\alpha$. One gets $m_1 =x^4 + x^3 +1$ and $m_7= x^4 + x + 1$ or the other way round.
