Degree and fiber of scheme morphisms in $\mathbb P^1$ induced by linear system of divisors on an elliptic curve Consider curves (in the sense of schemes) over $\mathbb C$. Hartshorne, Algebraic Geometry says that, given an elliptic curve $X$, and a closed point $P_0$, the linear system $|2P_0|$ induces a morphism in $\mathbb P^1$ of degree 2.
Now, I know that the induced morphism exists, but why has it degree 2? And, if I took $|3P_0|$, would it bee of degree $3$.
Moreover, I'd like to investigate the fibers of such (a bit abstractly described) morphism. Case $d=2$ involves hyperellipticity and I can deal with it. But Case $d=3$? Is it true, for example, that there are fibers made of at least one rational point and at least one (closed) non rational point? Or are they somehow "separated"?
Thank you in advance.

For any projective curve $X$ and an effective Cartier divisor $D$, if there exists a non-constant $f\in |D|$, $f$ induces a rational map $f:X\to\mathbb{P}^1$ and if $X$ is smooth, this is a morphism. Further the inverse image of a point in $\mathbb{P}^1$ is completely contained in $D$ and thus $\deg f\leq\deg D$. So, in your case, one gets the degree of the map given by $2P_0$ has degree at most 2. But, it can not be degree one, since degree one maps are isomorphisms and the two have different genus. If you took $3P_0$, it has now three linearly independent sections and the natural map is not to the line, but the plane and then it will give you an embedding. So, you have to be more specific as to what you mean. 
The latter part is unclear. What does rational point mean to you? You are working over the complex numbers.
