Let $\alpha$ be a root of the polynomial $^2++3$ in $GF(49)$.
Question: I want to express every element of $GF(49)^*$ both as a polynomial in $\alpha$ of degree less than $2$ and as a power of $\alpha$.
Would love to know if my approach is correct:
The elements of the finite field $GF(p^n)$ is defined as:
$$GF(p^n) = (0,1,2,3,4,5,...,p-1) \cup(p,p+1,p+2,...,p+p-1)\cup...\cup(p^{n-1},p^{n-1} +1,p^{n-1}+2,...,p^{n-1}+p-1)$$
The order of the field is given by $p^n$ while $p$ is called the characteristic of the field.
Thus,
$GF(49) = (0,1,2,3,4,5,6...,46,47,48)$
Where the order of the field (i.e. number of elements) is $49$ and $7$ is called the characteristic of the field.
Given that $\alpha$ is the root of the polynomial $x^2 + x + 3$, the elements of $GF(7^2)$ can be represented as polynomials in $\alpha$ with degree less than $3$, using modulus $(\alpha^2 + \alpha +3)$.
Then I did this:
$$ \begin{array}{|c|c|} \hline \text { Powers } & \text { polynomials in } \alpha \text { with degree less than 3 } \\ \hline 0 & 0 \\ \hline 1 & 1 \\ \hline \alpha & \alpha \\ \hline \alpha^{2} & \alpha^{2} \\ \hline \alpha^{3} & \alpha+3 \\ \hline \alpha^{4} & \alpha^{2}+3 \alpha \\ \hline \alpha^{5} & 3 \alpha^{2}+\alpha+3 \\ \hline \alpha^{6} & \alpha^{2}+6 \alpha+9 \\ \hline \alpha^{7} & 6 \alpha^{2}+8 \alpha+3 \\ \hline \alpha^{8} & 8 \alpha^{2}+9 \alpha+18 \\ \hline \alpha^{9} & 9 \alpha^{2}+26 \alpha+24 \\ \hline \alpha^{10} & 26 \alpha^{2}+15 \alpha+27 \\ \hline \alpha^{11} & 15 \alpha^{2}+53 \alpha+78 \\ \hline \alpha^{12} & 53 \alpha^{2}+93 \alpha+45 \\ \hline \alpha^{13} & 93 \alpha^{2}+8 \alpha+159 \\ \hline \alpha^{14} & 8 \alpha^{2}+252 \alpha+279 \\ \hline \alpha^{15} & 252 \alpha^{2}+271 \alpha+24 \\ \hline \alpha^{16} & 271 \alpha^{2}+271 \alpha+756 \\ \hline \alpha^{17} & 276 \alpha^{2}+1027 \alpha+813 \\ \hline \alpha^{18} & 1027 \alpha^{2}+537 \alpha+828 \\ \hline \alpha^{19} & 537 \alpha^{2}+1855 \alpha+3081 \\ \hline \alpha^{20} & 1855 \alpha^{2}+3618 \alpha+1611 \\ \hline \alpha^{21} & 3618 \alpha^{2}+244 \alpha+5565 \\ \hline \alpha^{22} & 244 \alpha^{2}+9183 \alpha+10854 \\ \hline \alpha^{23} & 9183 \alpha^{2}+10610 \alpha+732 \\ \hline \alpha^{24} & 10610 \alpha^{2}+9915 \alpha+27549 \\ \hline \alpha^{25} & 9915 \alpha^{2}+16939 \alpha+31830 \\ \hline \alpha^{2 .} & 16939 \alpha^{2}+41745 \alpha+29745 \\ \hline \alpha^{21} & 41745 \alpha^{2}+46684 \alpha+50817 \\ \hline \alpha^{21} & 46684 \alpha^{2}+9072 \alpha+125235 \\ \hline \alpha^{29} & 9072 \alpha^{2}+171919 \alpha+140052 \\ \hline \alpha^{30} & 171919 \alpha^{2}+130980 \alpha+27216 \\ \hline \alpha^{31} & 130980 \alpha^{2}+199135 \alpha+515757 \\ \hline \alpha^{32} & 199135 \alpha^{2}+646737 \alpha+392940 \\ \hline \alpha^{13} & 646737 \alpha^{2}+193805 \alpha+597405 \\ \hline \alpha^{34} & 193805 \alpha^{2}+1244142 \alpha+1940211 \\ \hline \alpha^{31} & 1244142 \alpha^{2}+2134016 \alpha+581415 \\ \hline \alpha^{36} & 2134016 \alpha^{2}+662727 \alpha+3732426 \\ \hline \alpha^{37} & 662727 \alpha^{2}+5866442 \alpha+6402048 \\ \hline \alpha^{31} & 5866442 \alpha^{2}+5739321 \alpha+1988181 \\ \hline \alpha^{38} & 5739321 \alpha^{2}+7854623 \alpha+17599326 \\ \hline \alpha^{40} & 7854623 \alpha^{2}+23338647 \alpha+17217963 \\ \hline \alpha^{41} & 23338647 \alpha^{2}+9363340 \alpha+23563869 \\ \hline \alpha^{42} & 9363340 \alpha^{2}+46902516 \alpha+70015941 \\ \hline \alpha^{41} & 46902516 \alpha^{2}+79379281 \alpha+28090020 \\ \hline \alpha^{44} & 79379281 \alpha^{2}+18812496 \alpha+140707548 \\ \hline \alpha^{45} & 18812496 \alpha^{2}+220086829 \alpha+238137843 \\ \hline \alpha^{48} & 220086829 \alpha^{2}+219325347 \alpha+56437488 \\ \hline \alpha^{47} & 219325347 \alpha^{2}+276524317 \alpha+660260487 \\ \hline \end{array} $$
There are no solutions in $GF(7)$ to $x^2+x+3=0$, so you should think of the elements of $GF(49)$
as $n_1+n_2\alpha$ where $\alpha$ is a root of $x^2+x+3$ and $n_1$ and $n_2$ are integers between $0$ and $6$
-- not as $(0,1,2,3,4,5,6,...,46,47,48).\;$ For example, $\alpha^2=-\alpha-3$.