Find a generator polynomial of a 16-ary Reed-Solomon code of dimension 10.
My work so far:
First, $$k=q-1-\left(\delta -1\right)$$
$$10=15-1-\left(\delta -1\right)$$
$$\implies \delta =6 $$
and then noting that the generator polynomial $g(x)$ of a Reed-Solomon code will have degree $\delta-1$, which means $g(x)$ will have degree 5.
I then calculated $g(x)$ by noting that any $q$-ary Reed-Solomon code of length $q-1$ will be generated by a $g(x)$ such that
$$g(x) = \prod^{a+\delta -1}_{i=a+1}\left(x-\alpha^{i}\right), a \geq0, 2\leq\delta\leq q-1 $$
Setting $a=0$ and $\delta=6$, and letting $\alpha$ be a root of the the minimal polynomial $x^4+x+1$,
$$\prod^{5}_{i=1}\left(x-\alpha^i\right)=\left(x-\alpha\right)\left(x-\alpha^2\right)\left(x-\alpha^3\right)\left(x-\alpha^4\right)\left(x-\alpha^5\right)$$
expanding this I have
$$x^5-x^4\left(\alpha^5+\alpha^4+\alpha^3+\alpha^2+\alpha\right)-x^3\left(\alpha^9+\alpha^8+2\alpha^7+2\alpha^6+2\alpha^5+\alpha^4+\alpha^3\right)-x^2\left(\alpha^{12}+\alpha^{11}+2\alpha^{10}+2\alpha^9+2\alpha^8+\alpha^7+\alpha^6\right)=x\left(\alpha^{14}+\alpha^{13}+\alpha^{12}+\alpha^{11}+\alpha^{10}\right)-\alpha^{15}$$
and finally after using a log table for the primitive elements of GF$\left(2^4\right)$, and reducing everything modulo 2, I have $$g(x)=x^5+x^4\left(\alpha^3+\alpha+1\right)+x^3\left(\alpha^2\right)+x^2\left(\alpha^2+\alpha\right)+x\left(\alpha\right)+1$$
So my question is, does this look like the correct generator for this code?
Update: I used the comments below to find $g(x)$ to be $$g(x)=x^5+x^4\left(\alpha^3+\alpha+1\right)+x^3\left(\alpha^2\right)+x^2\left(\alpha^2+\alpha\right)+x\left(\alpha\right)+1$$
Is the final step to find a value $\alpha$ satisfying $\alpha^{3}= \alpha +1$?
Best Answer
After working through all of the comments a generator polynomial for this [15, 10, 6] Reed-Solomon code, where $\alpha$ is a primitive element of $x^4+x+1$, is $$g(x)=x^5+x^4\left(\alpha^3+\alpha+1\right)+x^3\left(\alpha^2\right)+x^2\left(\alpha^2+\alpha\right)+x\left(\alpha\right)+1$$