Hyperelliptic Riemann Surface satisfying Hausdorff property

48 Views Asked by At

Let $X$ and $Y$ be smooth affine plane curves defined respectively by $$y^2 = h(x)$$ $$ \text{and} \space w^2 = k(z)$$

where $h(x)$ is of degree $2g + 1 + \epsilon$ and $k(x) = z^{2g + 2}h(1/z)$.

Let $U=\{(x, y) \in X\mid x \ne 0\}$ and $V = \{(z, w) \in Y\mid z \ne 0\}$.

Define an isomorphism $\Phi : U \rightarrow V$ by $$\Phi(x,y) = (z, w) = (1/x, y/x^{g+1})$$

Let $Z$ be Riemann surface obtained by glueing $X$ and $Y$ together along $U$ and $V$ via $\Phi$.

How do I show that $Z$ is Hausdorff?


Further Detail: RICK MIRANDA'S ALGEBRAIC CURVES AND RIEMANN SURFACES, PAGE -60.