Factoring a polynomial which shares the same zeros as another

Question

I have this problem:

Let $V=V(Y^2-X^2(X+1))\subset \mathbb{A}^2$, the zero set of the polynomial $G(X,Y)=Y^2-X^2(X+1)\in K[X,Y]$ for a field $K$.

In my notes it says: If $F(X,Y)\in I(V)$ then $F(X,Y)=H(X,Y)(Y^2-X^2(X+1))$ for some $H\in K[X,Y]$.

This seems obvious but I'm not sure if we can definitely do this when $K$ is not algebraically closed. I want to use the following result:

"If $F\in K[X,Y]$ is irreducible and $V(F)$ is infinite, then $I(V(F))=(F)$"

If $K$ is algebraically closed then the above holds, so no problem there, but in my notes we don't assume this.

Answer 1

The result is true by the Nullstellensatz if $K$ is algebraically closed, as you correctly noted.
It may be however be false for non algebraically closed $K$. Here is a counterexample:

For the field $K=\mathbb F_2=\mathbb Z/2\mathbb Z$ the zero set of $G$ is $$V=V(G)=\{(0,0),(1,0)\}\subset \mathbb A^2( \mathbb F_2)=\mathbb F_2^2$$
The polynomial $F(X,Y)=Y\in \mathbb F_2[X,Y]$ vanishes on $V$, i.e. $Y\in I(V)$, but there is clearly no polynomial $H(X,Y)\in \mathbb F_2[X,Y]$ for which $Y=H(X,Y)(Y^2-X^2(X+1))$

Answer 2

If $F,G\in K[X,Y]$ have no common factors, then $V(F,G)$ is a finite set (possibly empty) even over the algebraic closure of $K$. In your situation, $H=Y^2-X^2(X+1)$ is irreducible and so if $F\in I(V)$ where $V=V(H)$ and $F$ is not a multiple of $H$, then $F,H$ have no common factors and thus they have only finitley many common zeroes over $\overline{K}$, the algebraic closure. But, clearly all zeroes of $H$ are zeroes of $F$ over $\overline{K}$ and this set is infinite unless empty. This is empty means $H$ is a unit and so $V=\emptyset$ and of course $F$ is a multiple of $H$.