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While I think it is very difficult to solve the question using only the machinery presented in Miranda at this point, the Hurwitz formula comes up in the next chapter and is probably the most elementary tool to use. In case of the first curve, you can consider restricting a projection $\mathbb{P}^3 \setminus \{x_1=x_2=0\} \to \mathbb{P^1}$ given by $[x_0,x_1,x_2,x_3]\mapsto [x_1,x_2]$ to the curve C.

This map has degree 4, and there are 4 fibers of the map with cardinality 2 instead of 4. We can calculate this explicitly from the defining equations. For if $[1,c]$ is a point in $\mathbb{P}^1$, we must solve the system of equations

$x_0x_3 = 2c$

$x_0^2+x_3^2+1+c^2 = 0.$

Setting $x_3= 2c/x_0$ yields

$x_0^2+\frac{4c^2}{x_0^2} + 1 + c^2 = 0.$

This equation has four solutions $x_0$ unless $c$ is one of the four roots of $c^4-14c^2+1$, and in those cases there are two solutions.

It follows from the Hurwitz formula that

$2g(C)-2 = 4(2g(\mathbb{P}^1)-2)+4\cdot 2,$

and thus $g(C)=1$.

While I think it is very difficult to solve the question using only the machinery presented in Miranda at this point, the Hurwitz formula comes up in the next chapter and is probably the most elementary tool to use. In case of the first curve, you can consider restricting a projection $\mathbb{P}^3 \setminus \{x_1=x_2=0\} \to \mathbb{P^1}$ given by $[x_0,x_1,x_2,x_3]\mapsto [x_1,x_2]$ to the curve C.

This map has degree 4, and there are 4 fibers of the map with cardinality 2 instead of 4. We can calculate this explicitly from the defining equations. For if $[1,c]$ is a point in $\mathbb{P}^1$, we must solve the system of equations

$x_0x_3 = 2c$

$x_0^2+x_3^2+1+c^2 = 0.$

Setting $x_3= 2c/x_0$ yields

$x_0^2+\frac{4c^2}{x_0^2} + 1 + c^2 = 0.$

This equation has four solutions $x_0$ unless $c$ is one of the four roots of $c^4-14c^2+1=0$, and in those cases there are two solutions.

It follows from the Hurwitz formula that

$2g(C)-2 = 4(2g(\mathbb{P}^1)-2)+4\cdot 2,$

and thus $g(C)=1$.

While both the other solutions are technically correct, they are far, far more complex than intended by the text.

While I think it is very difficult to solve the question using only the machinery presented in Miranda at this point, the Hurwitz formula comes up in the next chapter and is probably the most elementary tool to use. In case of the first curve, you can consider restricting a projection $\mathbb{P}^3 \setminus \{x_1=x_2=0\} \to \mathbb{P^1}$ given by $[x_0,x_1,x_2,x_3]\mapsto [x_1,x_2]$ to the curve C.

This map has degree 4, and there are 4 fibers of the map with cardinality 2 instead of 4. We can calculate this explicitly from the defining equations. For if $[1,c]$ is a point in $\mathbb{P}^1$, we must solve the system of equations

$x_0x_3 = 2c$

$x_0^2+x_3^2+1+c^2 = 0.$

Setting $x_3= 2c/x_0$ yields

$x_0^2+\frac{4c^2}{x_0^2} + 1 + c^2 = 0.$

This equation has four solutions $x_0$ unless $c$ is one of the four roots of $c^4-14c^2+1$, and in those cases there are two solutions.

It follows from the Hurwitz formula that

$2g(C)-2 = 4(2g(\mathbb{P}^1)-2)+4\cdot 2,$

and thus $g(C)=1$.

While I think it is very difficult to solve the question using only the machinery presented in Miranda at this point, the Hurwitz formula comes up in the next chapter and is probably the most elementary tool to use. In case of the first curve, you can consider restricting a projection $\mathbb{P}^3 \setminus \{x_1=x_2=0\} \to \mathbb{P^1}$ given by $[x_0,x_1,x_2,x_3]\mapsto [x_1,x_2]$ to the curve C.

This map has degree 4, and there are 4 fibers of the map with cardinality 2 instead of 4. We can calculate this explicitly from the defining equations. For if $[1,c]$ is a point in $\mathbb{P}^1$, we must solve the system of equations

$x_0x_3 = 2c$

$x_0^2+x_3^2+1+c^2 = 0.$

Setting $x_3= 2c/x_0$ yields

$x_0^2+\frac{4c^2}{x_0^2} + 1 + c^2 = 0.$

This equation has four solutions $x_0$ unless $c$ is one of the four roots of $c^4-14c^2+1$, and in those cases there are two solutions.

It follows from the Hurwitz formula that

$2g(C)-2 = 4(2g(\mathbb{P}^1)-2)+4\cdot 2,$

and thus $g(C)=1$.

While both the other solutions are technically correct, they are far, far more complex than intended by the text.

While I think it is very difficult to solve the question using only the machinery presented in Miranda at this point, the Hurwitz formula comes up in the next chapter and is probably the most elementary tool to use. In case of the first curve, you can consider restricting a projection $\mathbb{P}^3 \setminus \{x_1=x_2=0\} \to \mathbb{P^1}$ given by $[x_0,x_1,x_2,x_3]\mapsto [x_1,x_2]$ to the curve C. This map has degree 4, and there are 4 fibers of the map with cardinality 2 instead of 4. It follows from the Hurwitz formula that

$2g(C)-2 = 4(2g(\mathbb{P}^1)-2)+4\cdot 2,$

and thus $g(C)=1$.

While both the other solutions are technically correct, they are far, far more complex than intended by the text.

While I think it is very difficult to solve the question using only the machinery presented in Miranda at this point, the Hurwitz formula comes up in the next chapter and is probably the most elementary tool to use. In case of the first curve, you can consider restricting a projection $\mathbb{P}^3 \setminus \{x_1=x_2=0\} \to \mathbb{P^1}$ given by $[x_0,x_1,x_2,x_3]\mapsto [x_1,x_2]$ to the curve C.

This map has degree 4, and there are 4 fibers of the map with cardinality 2 instead of 4. We can calculate this explicitly from the defining equations. For if $[1,c]$ is a point in $\mathbb{P}^1$, we must solve the system of equations

$x_0x_3 = 2c$

$x_0^2+x_3^2+1+c^2 = 0.$

Setting $x_3= 2c/x_0$ yields

$x_0^2+\frac{4c^2}{x_0^2} + 1 + c^2 = 0.$

This equation has four solutions $x_0$ unless $c$ is one of the four roots of $c^4-14c^2+1$, and in those cases there are two solutions.

It follows from the Hurwitz formula that

$2g(C)-2 = 4(2g(\mathbb{P}^1)-2)+4\cdot 2,$

and thus $g(C)=1$.