Let $K$ be an algebraically closed field and $\{P_1, \dots, P_k\}\subseteq\mathbb{P}_K^2$. Can we always find two algebraic curves $F$ and $G$, i.e. homogeneous polynomials $F, G\in K[X, Y, Z]$, such that $V(F, G)=\{P_1, \dots, P_k\}$?
This problem was given to me as a homework problem for $K=\mathbb{R}$, where the answer is obviously yes: If $P_i=(x_i:y_i:z_i)$, where we can WLOG assume that $z_i\neq 0$ for all $i$ after a suitable coordinate transformation, then just take $$ F = G = \prod_{i=1}^k ((X-x_i)^2+(Y-y_i)^2), $$ which obviously yields $V(F) = V(G) = \{P_1, \dots, P_k\}$ over $\mathbb{R}$.
A similar construction can be used to show that we can always find curves $F$ and $G$ as desired if $K$ is not algebraically closed (just take some polynomial in $X$ without a root in $K$ and homogenise to get a homogeneous polynomial in $K[X, Y]$; the curve it defines will only have $(0:0:1)$ as a point).
But what if $K$ is algebraically closed? I tried using Riemann-Roch, but did not succeed as one would have to control the infinitely many points in $\mathbb{P}_K^2\setminus\{P_1, \dots, P_k\}$, which seems impossible with this approach.
