Finding ramification points of a map to $\mathbb{P}^1$ given by a divisor Let $X$ be a elliptic curve over $k$ ("curve" in the sense of Hartshorne Chapter IV). A closed point $P_0\in X$ gives rise to a base-point free linear system $|2P_0|$ of dimension 1, which in turn gives rise to a morphism $X\rightarrow \mathbb{P}^1$ of degree 2. This morphism is given by selecting a basis $f_0,f_1$ of $\Gamma(X,\mathcal{L}(2P_0))$ and setting $P\mapsto [f_0(P):f_1(P)]$. How do I spot the ramification points of this map? It seems obvious that $P_0$ should be a ramification point, but I don't quite see why. I believe answering this question will help me understand a more fundamental question: why is this map degree 2? We can think of $\Gamma(X,\mathcal{L}(2P_0))$ as rational functions on $X$ with a pole of order $\leq 2$ at $P_0$, so is there then some canonical choice of basis $f_0,f_1$ that makes the answers to the above questions clear?
Edit: I found this thread that seems to answer the question. I mostly understand the solution here, although the very last step is still giving me trouble. We choose $x\in \Gamma(X,\mathcal{L}(2P_o))$ is such that $\{1,x\}$ form a basis of this vector space. Since $x$ has a pole of order 2 at $P_0$ (and no other poles), we conclude that $x$ has exactly two zeros (by Riemann-Roch). We can say the same thing for $x-\alpha$ for any $\alpha\in \mathbb{C}$. The claim is that for generic $\alpha\in \mathbb{C}$, these two zeros are distinct. I don't yet understand this claim.
 A: (I mostly have the case $k = \Bbb C$ in mind, but it should work more or less for other field.)
Any elliptic curve is a smooth cubic curve $X \subset \Bbb P^2$ (by Riemann-Roch), and there is a group law such that $Q_1 +Q_2+Q_3 = 0$ if there is $L \subset \Bbb P^2$ such that $L \cap X = Q_1 + Q_2 + Q_3$ (as divisor). In particular, let $Q_0 = -P/2$. By definition, $Q_0$ is the point such that $L_{P_0}X \cap X = 2P_0 + Q_0$.
Lines through $Q_0$ are parametrised by $\Bbb P^1$, call this family $\{\ell_t\}_{t \in \Bbb P^1}$. Consider the morphism $X \to \Bbb P^1, P \mapsto t$ if $P \in \ell_t$. You can check that this is simply the morphism induced by $|2P_0|$. It follows that ramification points of the map are geometrically points $P$ such that $Q_0 \in T_{P}X$. Algebraically, there are element of order $2$ in the group law.
By Riemann-Hurwitz, you can check that there will be $4$ ramification points. It is a fun exercise to check it using this algebraic or geometric description of the ramification point.
