I'm working on this problem: Show that a rational normal curve $C$ of degree $3$ cannot be an intersection of two quadrics.
Here is my solution. Let
$$J=\{\text{$f$ is homogeneous of degree $2$ and $f(x)=0$ for all $x$ in $C$}\}.$$
Then clearly $J$ is a $k$-vector space with dimension $\geq3$. If $C$ is an intersection of two quadrics $p,q$, then $J$ is a collection of homogeneous polynomials of degree $2$ in the radical of the homogeneous ideal generated by $p$ and $q$. However, it is possible that there exists a homogeneous polynomial $h$ of degree $2$ such that $h$ is $k$-linearly independent of $p,q$ and $h^n=pu+qv$ for some $u,v$. However it seems to be true that there does not exist such $h$ and it leads to a contradiction. How can I show that?