Construct a Reed Solomon code: find the parity check matrix I am trying to solve the following exercise, but I need a check/opinion on how to solve it.

Construct a Reed-Solomon code with dimensions $[12,7]$ over $\mathbb{F}_{13}$ and find a parity check matrix for the code $C$. Hint: $2$ is a primitive element of $\mathbb{F}_{13}$.


First thing: I have $\delta=12-7+1=6$, so the minimum distance is exactly $6$. Also, I choose to build a narrow-sense code, so the defining set is $T = \mathcal{C}_1 \cup \ldots \cup \mathcal{C}_{5}$.
As $12=n=13-1$, then $\mathcal{C}_i=\{ i \}$, so the generator polynomial is $$g(x)=(x-2)(x-2^2)(x-2^3)(x-2^4)(x-2^5)=(x-2)(x-4)(x-8)(x-3)(x-6)$$
Now, I can work out the computations and find $h(x)$,check polynomial, dividing $x^{12}-1$ by $g(x)$, but it seems a bit heavy to me. Is there any other possibility to compute the check polynomial faster?  And so also the parity check matrix.

As an alternative to @rschweib's answer and possibly requiring little computation you have good look-up tables, a cyclic Reed-Solomon code whose generator polynomial has $2, 2^2, 2^3, 2^4, 2^5$ as roots has parity check matrix
$$H = \left[\begin{matrix}
1&2&2^2&2^3&\quad \cdots&2^{11}\\
1&2^2&(2^2)^2&(2^2)^3&\quad \cdots&(2^2)^{11}\\
1&2^3&(2^3)^2&(2^3)^3&\quad \cdots&(2^3)^{11}\\
1&2^4&(2^4)^2&(2^4)^3&\quad \cdots&(2^4)^{11}\\
1&2^5&(2^5)^2&(2^5)^3&\quad \cdots&(2^5)^{11}
\end{matrix}\right]$$
