Basis for $L(D)$ over the hyperelliptic curve $y^2=x^6-1$ Let $X$ be the Riemann Surface associated to the curve $y^2=x^6-1$ and let $\pi$ be the x-projection. Let $p+q=\pi^{-1}(\infty)$ and $r_{k}=(e^{ki\pi/3},0)$. I need to find a basis for the Riemann-Roch space $L(D)$ with $D=p+q+r_0+r_3$.I have found out that $dimL(D)=3$ and I think that the function $x$ is in $L(D)$. How could I complete the basis? I mean, how can I find another function in $L(D)$?
 A: You've already found that $x \in L(D)$. Since $1 \in L(D)$, too, it suffices to find one more linearly independent function.
Let $\DeclareMathOperator{\div}{div} \infty_+ = [1:1:0]$ and $\infty_- = [1:-1:0]$ be the points of $C$ at infinity, and let $\newcommand{\iinfty}{\underline{\infty}} \iinfty = \infty_+ + \infty_-$. Let $\zeta = e^{2\pi i/6}$ be a primitive $6^\text{th}$ root of $1$. Then
\begin{align*}
\div(y) &= r_0 + \cdots + r_5 - 3 \iinfty \\\
\div(x - \zeta^k) &= 2 r_k - \iinfty \, .
\end{align*}
So to construct a function with simple poles at $r_0$ and $r_3$, we should take a ratio of $y$, $x - 1$, and $x - \zeta^3 = x + 1$. So we compute
\begin{align*}
\div\left(\frac{y}{x^2-1}\right) &= \div\left(\frac{y}{(x-1)(x+1)}\right) = r_0 + \cdots + r_5 - 3 \iinfty - \left(2r_0 - \iinfty + 2r_3 - \iinfty\right)\\
&= r_1 + r_2 + r_4 + r_5 - r_0 - r_3 - \iinfty \, .
\end{align*}
Thus $\frac{y}{x^2 - 1} \in L(D)$, and by examining the orders of poles at $\infty_+$ and $r_0$, we find that $\left\{1, x, \frac{y}{x^2 - 1}\right\}$ is linearly independent.
