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Let $C = V(y^2 - x^4 - 1) \subset \Bbb A^2$ and set $C_0 = C \times \{0\}, C_1 = C \times \{1\} \subset \Bbb A^3$.

Consider the space $X$ obtained by quotienting $C_0 \sqcup C_1$ but the relation $((x, y) ; 0) \sim (1/x, y/x^2)$ for all $y$ and all $x \neq 0$. According Miranda III.1.7, $X$ is a smooth projective curve (and hyperelliptic). Thus $X = Proj(S)$ for some graded ring $S$.

My question: can you describe explicitly the ring $S$ (up to isomorphism of $k$-algebras)?

Typically, the comments in this question tells that if we replace $C$ by $\Bbb A^1$ and $\sim$ by $(x;0) \simeq (1/x, 1)$, we get $S = k[x_0,x_1]$, yielding $X = \Bbb P^1_k$. But what is $S$ in our case?

Alphonse
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  • Maybe one could try to find a graded ring $S$ and two distinct elements $z_0, z_1 \in S$ such that the rings $S_{(z_j)}$ of elements of degree zero in the localizations $S_{z_j}$ are isomorphic to $k[x,y]/(y^2 - x^4 - 1)$ ($j = 0,1$) and this isomorphism commutes with the map $S_{(z_0)} \to S_{(z_1)}$ corresponding to $(x,y) \mapsto (1/x, y/x^2)$. – Alphonse Oct 23 '18 at 09:30
  • See the case when $C$ is replaced by $\Bbb A^1$, we can put $z_j=x_j$, and $$S_{(z_j)} = { s(z_0,z_1)/z_j^m, m \geq 0, s \text{ homogenous of degree m} } \cong k[x_{1-j} / x_j]$$ with $k[x_{1-j} / x_j] \to k[x_j / x_{1-j}]$ given by the inversion map. – Alphonse Oct 23 '18 at 09:31

1 Answers1

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Let $S = \frac{k[X,Y,Z]}{(Y^2 - (X^4 + Z^4))}$, where we give $Y$ weight $2$, and $X$ and $Z$ weight $1$. Then the polynomial $F = Y^2 - (X^4 + Z^4)$ is weighted homogeneous of degree $4$, so the quotient $S$ is a graded ring. Thus the completed curve $\overline{C}$ is given by the equation $Y^2 = X^4 + Z^4$ in the weighted projective space $\mathbb{P}(1,2,1)$. (In general, one usually considers a hyperelliptic curve of genus $g$ as living in $\mathbb{P}(1, g+1, 1)$.)

We can see that this is the ring obtained from the gluing construction Miranda gives by looking at affine open subsets of $\operatorname{Proj}(S)$. On the affine open $U_2$ where $Z \neq 0$, we have affine coordinates $$ x = \frac{X}{Z} \quad \text{and} \quad y = \frac{Y}{Z^2} $$ and the dehomogenization of $F$ with respect to $x$ and $y$ is $$ y^2 = \frac{Y^2}{Z^4} = \frac{X^4}{Z^4} + 1 = x^4 + 1 \, . $$ On the affine open $U_0$ where $X \neq 0$, we have affine coordinates $$ z = \frac{Z}{X} \quad \text{and} \quad w = \frac{Y}{X^2} $$ and the dehomogenization of $F$ with respect to $z$ and $w$ is $$ w^2 = \frac{Y^2}{X^4} = 1 + \frac{Z^4}{X^4} = 1 + z^4 \, . $$ The change of coordinates map on $U_0 \cap U_2$ is given by $$ z = \frac{Z}{X} = \frac{1}{x} \qquad \qquad w = \frac{Y}{X^2} = \frac{Y/Z^2}{X^2/Z^2} = \frac{y}{x^2} \, , $$ which shows that we recover Miranda's gluing construction.

Viktor Vaughn
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