7 added 5 characters in body

The intersection of the two quadrics in $\mathbb{P}^3$ is a complete intersection and defines an elliptic curve, so the genus should be is 1. A way to see it this is to pick a point $p$ on $C$ and project from $p$ onto a general hyperplane. The image curve $C'$ is of degree one less than the original curve, hence $C'$ is a plane curve of degree 3. Since cubics have genus 1, we are done.

Another way to see this that $g(C)=1$ is by computing cohomology of the sequence $$0 \to O_{P^3}(-4) \to O_{P^3}(-2)\oplus O_{P^3}(-2) \to O_{P^3}\to O_C \to 0$$(This is the resolution of $O_C$ as an $O_{P^3}$ module, which is easy to write down for complete intersections). Using this and the standard formulae for cohomology on $P^n$, we get $g=h^1(O_C)=1$.

Yet another way to see it is by looking at the curve as a divisor of type $(2,2)$ on the quartic surface $X_0X_3-2X_1X_2$. In general, by the adjunction formula, divisors of type $(a,b)$ have arithmetic genus $(a-1)(b-1)$, so again we get g=1.

The twised cubic $C$ is the (isomorphic) image of $P^1$ under the 3-uple embedding $f_3:(u,v)\to (u^3,u^2v,uv^2,v^3)$, so since $P^1$ has genus 0, C has genus 0. Of course, this computation could be carried out using a projection, and $C'$ would be a plane curve of degree 2.

6 deleted 19 characters in body

The intersection of the two quadrics in $\mathbb{P}^3$ is a complete intersection and defines an elliptic curve, so the genus should be 1. One A way to see it is to pick a point $p$ on $C$ and project from $p$ onto a general hyperplane. The image curve $C'$ is of degree one less than the original curve, hence $C'$ is a plane curve of degree 3. Since cubics have genus 1, we are done.

Another way to see this is by computing cohomology of the sequence $$0 \to O_{P^3}(-4) \to O_{P^3}(-2)\oplus O_{P^3}(-2) \to O_{P^3}\to O_C \to 0$$(This is the resolution of $O_C$ as an $O_{P^3}$ module, which is easy to write down for complete intersections). Using this and the standard formulae for cohomology on $P^n$, we get $g=h^1(O_C)=1$.

Another

Yet another way to see it is by looking at the curve as a divisor of type $(2,2)$ on the quartic surface $X_0X_3-2X_1X_2$. In general, by the adjunction formula, divisors of type $(a,b)$ have arithmetic genus $(a-1)(b-1)$, so again we get g=1.

EDIT: Yet another way to see it is to pick a point $p$ on $C$ and project from $p$ onto a general hyperplane. The image curve $C'$ is of degree one less than the original curve, hence $C'$ is a plane curve of degree 3. Since cubics have arithmetic genus 1, we are done.

The twised cubic $C$ is the (isomorphic) image of $P^1$ under the 3-uple embedding $f_3:(u,v)\to (u^3,u^2v,uv^2,v^3)$, so since $P^1$ has genus 0, C has genus 0. Of course, this computation could be carried out using a projection, and $C'$ would be a plane curve of degree 2.

5 deleted 14 characters in body

The intersection of the two quadrics in $\mathbb{P}^3$ is a complete intersection and defines an elliptic curve, so the genus should be 1. One way to see this is by computing cohomology of the sequence $$0 \to O_{P^3}(-4) \to O_{P^3}(-2)\oplus O_{P^3}(-2) \to O_{P^3}\to O_C \to 0$$(This is the resolution of $O_C$ as an $O_{P^3}$ module, which is easy to write down for complete intersections). Using this and the standard formulae for cohomology on $P^n$, we get $g=h^1(O_C)=1$.

Another way to see it is by looking at the curve as a divisor of type $(2,2)$ on the quartic surface $X_0X_3-2X_1X_2$. In general, by the adjunction formula, divisors of type $(a,b)$ have arithmetic genus $(a-1)(b-1)$, so again we get g=1.

EDIT: Yet another way to see it is to pick a general point $p$ on $C$ and project from $p$ onto the a general hyperplaneat infinity. The image curve $C'$ is of degree one less than the original curve, hence $C'$ is a plane curve of degree 3. Since cubics have arithmetic genus 1, we are done.

The twised cubic $C$ is the (isomorphic) image of $P^1$ under the 3-uple embedding $f_3:(u,v)\to (u^3,u^2v,uv^2,v^3)$, so since $P^1$ has genus 0, C has genus 0. Of course, this computation could be carried out using a projection, and $C'$ would be a plane curve of degree 2.

4 added 270 characters in body
3 added 92 characters in body; deleted 1 characters in body
2 added 139 characters in body
1