Minimal polynomials Can someone explain to me how the minimal polynomials in page 4 of this document are obtained? Please help me.
http://web.ntpu.edu.tw/~yshan/BCH_code.pdf
It should be something standard about minimal polynomials you should only check the Field in the example.. I get $a^2$, but I am not sure about the other two. 
Alternatively can someone show me how to do exercise 5 in http://www.ms.uky.edu/~corso/teaching/math362/Coding-Theory-3.pdf page 465 ??
 A: Those polynomials can be easily explained. The minimal polynomial of $\alpha$ is
$$\phi_1(x)=x^4+x+1,$$
because we defined $\alpha$ to be one of the zeros of this polynomial.
Because $\alpha$ is a primitive element, it is of order $15$. Therefore $\beta=\alpha^3$ is of order five - in other words $\beta^5=1$. This implies that
$$
0=\beta^5-1=(\beta-1)(\beta^4+\beta^3+\beta^2+\beta+1).
$$
Because $\beta\neq1$ we see that $\beta$ is a zero of
$$
\phi_2(x)=x^4+x^3+x^2+x+1.
$$
On the other hand $\beta$ generates the field $GF(16)$, so its minimal polynomial has to be of degree four. Therefore $\phi_2(x)$ is the minimal polynomial. Alternatively we can see that $\beta$ has as its conjugates (act on it by the Frobenius automorphism)
$\beta^2=\alpha^6$, $\beta^4=\alpha^{12}$ and $\beta^8=\beta^3=\alpha^9$ four distinct elements of $GF(16)$.
On the other hand $\gamma=\alpha^5$ is of order three ($=15/5$) only. Therefore it belongs to the subfield $GF(4)$ (that contains precisely the roots of unity of order $4-1$). Hence its minimal polynomial is the only irreducible quadratic polynomial over $GF(2)$, namely
$$
\phi_3(x)=x^2+x+1.
$$
We could also argue as above starting from 
$$
0=\gamma^3-1=(\gamma-1)(\gamma^2+\gamma+1).
$$

If we want to continue and find the minimal polynomial of $\alpha^7$ it is simplest to observe that $\alpha^7$ is a conjugate of $\alpha^{2\cdot7}=\alpha^{14}=\alpha^{-1}$. But obviously $\alpha^{-1}$ is a zero of the polynomial
$$
x^4\phi_1(\frac1x)=x^4(\frac1{x^4}+\frac1x+1)=1+x^3+x^4.
$$
Admittedly that is not much use for the purposes of constructing larger minimum distance BCH-codes because once we get to $\alpha^7$ we have exhausted all the elements of $GF(16)\setminus GF(2)$ as zeros of the generator polynomial. Thus the resulting code is kinda tiny. It is the repetion code of length 15.
