Algebraic aspect of (Riemann) surface Let $X$ be a Riemann surface of genus $1$ and let $x \in X$. Set $P = x^3$ considered as a divisor on $X$. Then let $$V_j = \{f: M \rightarrow \mathbb{C} \cup \{\infty\} : f \text{ is a meromorphic function on } X \text{ with } \operatorname{ord}_x(f) \geq -j\}$$ for $j = 2, 3, 6$ and $\operatorname{ord}_x(f) \geq -j$ means the order of poles of $f$ at $x$ is never greater than $j$.
This $V_j$ is a $\mathbb{C}-$vector space, and by the Riemann–Roch theorem $$\dim_\mathbb{C}(V_j) = j$$ for $j = 2, 3, 6.$
In fact, $V_j \cong \mathbb{C}^j$ as vector spaces which gives that $V_2$ is a proper subspace of $V_3.$ In fact, $V2, V3 \leq V_6$. So the we can extend a basis $\{1, a\}$ of the isomorphic copy of $V_2$ to a basis of isomorphic copy of $V_3$ in the form $\{1, a, b\}$. 
Here $a$ is a non-constant meromorphic function in $V_2$ and $b$ is a non-constant meromorphic function in $V_3$ such that $b \notin \operatorname{span}_\mathbb{C}(1, a)$.
By this construction, I would like to verify that $a ,b$ considered as elements of $V_6$ will satisfy a polyomial with coefficient in $\mathbb{C}.$ Then this should give me a way to think (or embed) $X$ in to a projective space $\mathbb{CP}^2.$
Can anyone help of how the polynomial part and embed part can be done ?

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*Find $C$ such that $b^2-Ca^3$ has a pole of order $5$ at $x$, 

*find $C_2$ such that $b^2-Ca^3-C_2 ab$ has a pole of order $4$,

*find $C_3$ such that $b^2-Ca^3-C_2 ab-C_3 a^2$ has a pole of order $3$, 

*find $C_4$ such that $b^2-Ca^3-C_2 ab-C_3 a^2-C_4 b$ has a pole of order $2$, 

*find $C_5$ such that $b^2-Ca^3-C_2 ab-C_3 a^2-C_4 b-C_5a$ has a pole of order $1$

No rational function has only one pole thus it is constant $$b^2-Ca^3-C_2 ab-C_3 a^2-C_4 b-C_5a-C_6=0$$

Then $z\to [a(z):b(z):1]$ is a rational map and a morphism $X\to \Bbb{P}^2$.
It is injective because $a-a(z_0),b-b(z_0)$ have respectively 2 and 3 poles thus the same number of zeros. If $z_1\ne z_0,a(z_1)=a(z_0),b(z_1)=b(z_0)$ then $(b-b(z_0))/(a-a(z_0))$ has no pole thus it is constant which is a contradiction. $z\to [a(z):b(z):1]$ is an embedding because for the same reason $a-a(z_0),b-b(z_0)$ never have a common double zero at $z_0$.
