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Let $X$ be the Riemann surface of genus $g$ defined by $y^2=\prod_{i=1}^{2g+2}(x-e_i)$. Let $\pi:X\longrightarrow S$, be a degree two branched covering of the Riemann sphere $S$. Let $\pi(p_i)=e_i$, hence the ramification points are $p_1,\ldots, p_{2g+2}$.

I'm trying to show that the ramification divisor $R_\pi=p_1+\cdots+p_{2g+2}$ is linearly equivalent to the divisor $(2g+2).p_j$ for some fixed $j, 1\le j\le2g+2$.

I already know that $2.p_i\sim 2.p_j, \forall p_i$.

What about $f:X\longrightarrow S, f(x,y)=y/(x-e_j)^{g+1}=\frac{\sqrt{\prod_{i=1}^{2g+2}(x-e_i)}}{(x-e_j)^{g+1}}$ ? Is $f$ meromorphic on $X$? I'm confused whether $f$ is meromorphic or not because of the square root.

If $f$ is meromorphic then I can conclude that $div(f)=p_1+\cdots+p_{2g+2}-(2g+2).p_j$, and hence the linear equivalence $R_\pi=p_1+\cdots+p_{2g+2} \sim (2g+2).p_j$.

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    I'm slightly confused. Are you saying that you have been given $2[p_i]\sim 2[p_j] \ \forall i$? I am skeptical if what you are trying to show is true in general. Take $g=1$ for example, so we have $p,q,r,s$. To say that $q+r+s \sim 3p$ is equivalent to saying that the points $q,r,s$ are collinear via the embedding given by the linear system $|3p|$. – 2017-02-10
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    Take $h(x,y)=\frac{x-e_i}{x-e_j}$, then $(h)=2p_i-2p_j$. I'm trying to show e.g. $p_1+p_2+p_3+p_4\sim 4.p_1$. Notice $X$ is hyperelliptic. – 2017-02-10
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    A similar equivalence relation is in this book https://www.math.ucsd.edu/~eizadi/207A-14/Dolgachev-topics.pdf page 122, formula 5.12. – 2017-02-10

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