Reed Solomon Code Working in GF(32). Polynomial is $x^5+x^2+1$. $\alpha$ is primitive element. $t = 3$. RS code. $n = 31$, $k = 25$. I have obtained generator polynomial $x^6+\alpha^{10}x^5+\alpha^{30}x^4+\dots$
How do I obtain generator matrix? I believe I write down coefficients in increasing order of $x$ over $31$ columns and $25$ rows and keep shifting since it is cyclic. Is this correct?

The answer given by Sudarsan is one possible generator matrix. If
$${\bf d} = (d_0, d_1, \ldots , d_{k-1}) \longleftrightarrow 
d(x) = d_0 + d_1x + \cdots + d_{k-1}x^{k-1}$$
is the data polynomial, then the codeword polynomial corresponding
to the codeword ${\bf c} = {\bf d}G$ (where $G$ is the generator
matrix in Sudarsan's answer) is $d(x)g(x)$.  Thus, using this generator
matrix gives us a nonsystematic Reed-Solomon code meaning that
$\bf d$ is not a subvector of ${\bf c}$.
The generator matrix $\hat{G}$ of a systematic cyclic code, in which the codewords
are of the form 
$${\bf c} = {\bf d}\hat{G} = ({\bf r}, {\bf d}) = (r_0, r_1, \ldots, r_{n-k-1}, d_0, d_1, \ldots, d_{k-1}),$$
is obtained as follows.  We use the fact that $g_{n-k} = 1$.


*

*The first row is the same as in Sudarsan's answer.

*The second row is obtained by first shifting the first row to the right
by one place (as in Sudarsan's answer) but then subtracting $g_{n-k-1}$
times the first row from the shifted second row.   What this does
is modify the second row to put a $0$ below the $g_{n-k}$ in the first
row.  The first two rows thus look like
$$\left[ \begin{array}{} g_0 & g_1  & \dots & g_{n-k-1} & 1 &0 & \dots & 0 & 0\\ -g_0g_{n-k-1} & g_0-g_1g_{n-k-1}  & \dots & g_{n-k-2}-g_{n-k-1}^2& 0 & 1 &\dots & 0 & 0  \end{array}\right]\\
{\Large \Downarrow}\\
\left[ \begin{array}{} p_{0,0} & p_{0,1}  & \dots & p_{0,n-k-1} & 1 &0 & \dots & 0 & 0\\ p_{1,0} & p_{1,1}  & \dots & p_{1, n-k-1} & 0 & 1 &\dots & 0 & 0  \end{array}\right]$$

*The third row is obtained first shifting the newly constructed second
row to the right by one place but then subtracting $p_{1,n-k-1}$ times
the first row from so as to put a $0$ below the $1$ in the first
row. Note that this forms a $3\times 3$ identity matrix on the right.

*Lather, rinse, repeat, till you get $\hat{G} = [P \mid I]$ where $I$
is the $k\times k$ identity matrix and $P$ is the $k\times (n-k)$ matrix
formed on the left as the shifting and subtracting is done. The first row
of $\hat{G}$ is, of course, the same as the first row of $G$ in Sudarsan's
answer. The rows of $\hat{G}$ are, of course, codewords, and the corresponding
codeword polynomials are 
$$g(x),\\x^{n-k+1} - \left(x^{n-k+1} \mod g(x)\right),
\\ x^{n-k+2} - \left(x^{n-k+2} \mod g(x)\right),\\ \vdots
\\x^{n-1} - \left(x^{n-1} \mod g(x)\right),$$
The codeword polynomial corresponding to ${\bf d}\hat{G}$
is $x^{n-k}d(x) - \left(x^{n-k}d(x) \mod g(x)\right)$ where the second term on the right is the residue of $x^{n-k}d(x)$, a polynomial of degree $n-1$,
modulo the generator polynomial $g(x)$. This residue is of degree $n-k-1$
or less, while $x^{n-k}d(x)$ has no nonzero coefficients of degree
smaller than $n-k$, reflecting the fact that the codeword polynomial
$x^{n-k}d(x) - \left(x^{n-k}d(x) \mod g(x)\right)$ has all the data symbols 
"in the clear" in the
high-order coefficients followed by the parity symbols; that is,
we have a systematic code.
