I have been reading Rick Miranda's Book on Algebraic Curves and Riemann Surfaces and there is a proposition whose proof I don't complete understand.
The proposition states that curves of genus one are cubic plane curves.
Is proof is the following : If $X$ is an algebraic curve and we have a divisor with degree $3$ this divisor is very ample and so $dimL(D)=3$ if $deg(D)=3$, using Riemann-Roch, we see that $\phi_D$ will map $X$ to $\mathbb{P}^2$. Since $deg(D)=3$ the hyperplane divisor will have degree $3$ and so the image is a cubic curve.
Now I get that the degree of the smooth projective curve $Y=\phi(X)$ will be $3$, but how does he know that this a plane curve? We just know that $\phi_D$ gives an embedding, we don't know any more additional structure to the Riemann Surface. What is the part that I am missing? Thanks in advance.
As you say above, $\phi(X)$ is embedded in $\mathbb{P}^2$, which makes it a smooth plane curve. I should mention that a smooth projective curve in $\mathbb{P}^2$ is called a smooth plane curve, as a matter of terminology. I believe that this is all that Miranda is claiming, but citing some theorems we can say a bit more.
We have an (a priori) analytic subvariety of $\mathbb{P}^2$, given by $\phi(X)=Y$. Using Chow's Theorem we can conclude that $Y$ is actually algebraic. Hence, $Y=Z(f_1,\ldots, f_r)$ for $f_1,\ldots, f_r\in \mathbb{C}[x,y,z]$ homogeneous polynomials. However, a version of the Hauptidealsatz from commutative algebra now implies that $Y=Z(f)$, i.e. it is cut out by a single homogeneous polynomial. By a result on the degree (found in Miranda), we know that if $Y=Z(f)$, and $Y$ is of degree $3$, then $\deg(f)=3$. So, it follows that $Y$ is cut out by a single degree $3$ equation, $f(x,y,z)\in \mathbb{C}[x,y,z]$.