Generator matrices of codes I need a little help please.
The Klein curve $X\subset \mathbb{P}^{2} $ given by homogeneous equation:
$x^{3}y+y^{3}z+z^{3}x=0$.


*

*Show that the space $L(m(z))$ is generated by elements of the form
$\left \{x^{i}y^{j}|2i+3j\leq 3m, i-2j \leq  m \right \}$.
My first question: How to prove this?

*From this generator system, select a basis of $L(m(z))$ and write out generator matrices of codes$(X,P,m(z))_{L}$ for $m=2,3,5$ where $P=X(F_{8})\setminus  \left \{ Q_{1},Q_{2} \right \}  $ $\left ( Q_{1}=(1:0:0);Q_{2}=(0:1:0) \right )$
I have found basis for $m=2,3,5$ . For instance, for $m=2$ I got a basis $\left \{ 1,x,x^{2},y,yx,y^{2} \right \}$. And my second question: What is its generator matrix?
Thanks a lot!
Added: 24 rational points that i have founded:
$P_{\infty }= (0,1,0);P_{1}=(1,0,0);P_{2}=(0,0,1)$;
$P_{3}=(1,\alpha^{4} ,\alpha^{2});P_{4}= (\alpha^{4},1,\alpha^{2});P_{5}=(\alpha^{2}
\alpha^{4},1)$;
$P_{6}=(1,\alpha^{5},\alpha^{5});P_{7}=(\alpha^{5},1 ,\alpha^{5})
P_{8}= (\alpha^{5},\alpha^{5},1)$;
$P_{9}=(\alpha,\alpha^{2},1);P_{10}=(1,\alpha,\alpha^{2});P_{11}=(\alpha^{2},1 ,\alpha)$;
$P_{12}= (\alpha^{3},1,\alpha^{3}); P_{13}=(\alpha^{3},\alpha^{3},1);P_{14}=(1,\alpha^{3},\alpha^{3});$
$P_{15}=(1,1 ,\alpha); P_{16}=(\alpha,1,1);P_{17}=(1,\alpha,1);$
$P_{18}=(\alpha^{5},\alpha ,1); P_{19}=(1,\alpha^{5},\alpha);P_{20}=(\alpha,1,\alpha^{5});$
$P_{21}=(\alpha^{4},\alpha^{3} ,1); P_{22}=(1,\alpha^{4},\alpha^{3});P_{23}=(\alpha^{3},1,\alpha^{4});$
 A: As you are interested in $F_8$, we can assume that we are working over fields of characteristic two.
The notation of the question seems to be a bit inconsistent?! For $x$ and $y$ to be functions on (an affine part of) the curve, they need to be ratios of forms of same degree - more often than not $x=X/Z$ and $y=Y/Z$. I assume that the equation of the curve is actually given in terms of the homogeneous
coordinates:
$$
X^3Y+Y^3Z+Z^3X=0.
$$
Sorry about mangling your question in this way, but I honestly cannot understand it otherwise.
Dehomogenization (divide by $Z^4$) turns this into
$$
F(x,y)=x^3y+y^3+x=0.\tag{*}
$$
The curve intersects the line at infinity $Z=0$ at the points $Q_1=(1:0:0)$ and $Q_2=(0:1:0)$. Here $Q_1$ is in the affine piece $X=1$, so it is the origin $u=v=0$
in the coordinate system $u=Y/X, v=Z/X$. That dehomogenization looks like
$$
P(u,v)=u+u^3v+v^3=0.
$$
The partial derivative $\partial P/\partial u =1+u^2v$ is non-zero at $Q_1$, so $v$ is a local parameter at $Q_1$. From a study of the terms of $P(u,v)$ we see that $u$ must have a zero of multiplicity three at $Q_1$. Therefore $x=1/v$ has a simple pole at $Q_1$ and
$y=u/v$ has a zero of multiplicity $2$ there. Thus we can conclude that the monomial
$\nu_{Q_1}(x^iy^j)=2j-i$. In other words, if $i-2j>0$, then $x^iy^j$ has a pole of order $i-2j$ at $Q_1$, and otherwise it does not have a pole there.
Similarly $Q_2$ is the origin of the affine piece $R(s,t)=s^3+t+st^3=0$, where $s=X/Y$ and $t=Z/Y$. Here $\partial R/\partial t(Q_2)\neq0$, so $s$ is a local parameter at $Q_2$, and $t$ has a zero of order $3$. So $x=s/t$ has a double pole at $Q_2$, and $y=1/t$ a pole of order three there. Thus the monomial $x^iy^j$ has a pole of order $2i+3j$ at $Q_2$.
AFAICT your exercise #1 asks us to identify the space of functions that have a pole of order at most $3m$ at $Q_2$, a pole of order at most $m$ at $Q_1$, and no other poles.
The affine curve on the $xy$-plane determined by $(*)$ seems to be non-singular. If $\partial F/\partial x=x^2y+1$ vanishes at some point, we must have $y=1/x^2$ there. But 
$$F(x,1/x^2)=x+x^{-6}+x=x^{-6}$$
never vanishes on an affine point of the curve, so we have the claimed non-singularity.
From this it follows that for a function to have no poles on the affine $xy$-part of the curve, it must be a polynomial on $x$ and $y$. The claim of your question #1 follows from this, our pole number calculations, and the answer (once we figure it out) to your other question.
In order to write down a generator matrix for the resulting Goppa code, we need an enumeration of the $F_8$-rational points of the Klein curve. Presumably you have been given this in class, but I had to dig for it. You may need to translate my numbering of points to match with that of your source.
Let us first find the points $(x,y)\in X(F_8)\setminus\{Q_1,Q_2\}$ with $x\neq0$.
We need an auxiliary variable $w=x^2y$. Multiplying $(*)$ by $x^6$ gives
$$
x^9y+x^6y^3+x^7=0\Leftrightarrow x^7(1+w)=w^3.
$$
This implies that we must have 
$$
x^7=\frac{w^3}{1+w}.
$$
Here we have phenomenal luck! Any non-zero element of $F_8$ satisfies the equation $x^7=1$,
so the above equation implies $w^3=1+w$. Let $\alpha$ be a root of $\alpha^3=\alpha+1$. It is well known that such an element $\alpha$ is a primitive element of $F_8$. The  solutions of $w^3=1+w$ are then $w=\alpha$ and its conjugates $w=\alpha^2$ and $w=\alpha^4=\alpha+\alpha^2$. If you have trouble with the arithmetic of $F_8$, please take a look
at my other answer. Here we can pick any non-zero value for $x$, and then solve for $y=w/x^2$ for any of the three possible values of $w$. With $w=\alpha$ we get seven $F_8$-rational points:
$$
P_i=(\alpha^i,\alpha^{1-2i}),\qquad\text{for $i=1,2,\ldots,7$.}
$$
With $w=\alpha^2$ we get seven other $F_8$-rational points:
$$
P_{7+i}=(\alpha^i,\alpha^{2-2i}),\qquad\text{for $i=1,2,\ldots,7$.}
$$
And with $w=\alpha^4$ we get seven more:
$$
P_{14+i}=(\alpha^i,\alpha^{4-2i}),\qquad\text{for $i=1,2,\ldots,7$.}
$$
Finally we observe that if $x=0$, then $(*)$ implies that $y=0$ also, and we have our
last point $P_{22}=(0,0)$. Thus the curve has 22 $F_8$-rational points in the affine $xy$-plane.
Remember that we can reduce the exponents of $\alpha$ modulo 7.
Below I will list the values of the functions $1,x,y$ at the 22 points on the three rows of the following matrix. You get the values of the other monomial with the obvious operations.
$$
\pmatrix{1&1&1&1&1&1&1&1&1&1&1&1&1&1&1&1&1&1&1&1&1&1\cr
\alpha&\alpha^2&\alpha^3&\alpha^4&\alpha^5&\alpha^6&1&\alpha&\alpha^2&\alpha^3&\alpha^4&\alpha^5&\alpha^6&1&\alpha&\alpha^2&\alpha^3&\alpha^4&\alpha^5&\alpha^6&1&0\cr
\alpha^6&\alpha^4&\alpha^2&1&\alpha^5&\alpha^3&\alpha&1&\alpha^5&\alpha^3&\alpha&\alpha^6&\alpha^4&\alpha^2&\alpha^2&1&\alpha^5&\alpha^3&\alpha&\alpha^6&\alpha^4&0\cr}
$$
Hopefully you can make some sense out of this. Too bad I don't have your textbook so that I could help you more.
