<s>[INST] $g(Z^3,Z^2) = 0$ implies $g(X,Y) = (X^2 - Y^3)h(X,Y)$ Let $F$ be a field and $g(X,Y) \in F[X,Y]$. Suppose that $g(Z^3,Z^2) = 0$ where $g(Z^3, Z^2) \in F[Z]$. I want to show that $g(X,Y) = (X^2 - Y^3) h(X,Y)$ for some $h(X,Y) \in F[X,Y]$. 
My first idea is that $g(X,Y)$ can be expressed as $g(X,Y) = X^2 p(X,Y) +Y^3 q(X,Y) + a_1XY^2 + a_2XY + a_3Y^2 + a_4X + a_5Y + a_6$. Using the condition, we obtain $$0 = Z^6 (p(Z^3, Z^2) + q(Z^3, Z^2) + a_1Z) + a_2Z^5 + a_3 Z^4 + a_4 Z^3 + a_5 Z^2 + a_6.$$
It follows that all $a_k$ are $0$. So $p(Z^3, Z^2) + q(Z^3, Z^2) = 0$. Moreover, $1, Z^2, Z^3, Z^4, Z^5$ in these two polyomials can only be obtained be substituting for $1, Y, X, Y^2, XY$ respectively. Thus the coefficients of these monomials in $p(X,Y)$ and $q(X,Y)$ must match. Let $\tilde{p}(X,Y)$ and $\tilde{q}(X,Y)$ be the sum of the monomials which have either $X^2$ or $Y^3$ as a factor. It remains to show that $p(X,Y) = q(X,Y) \mod (X^2 - Y^3)$. 
However, from here onwards I'm stuck. I would like to solve it with basic algebra, no geometry. 
 [/INST] It is better to start with (by division algorithm)
$$g(X,Y)=(X^2-Y^2)h_1(X,Y)+X\cdot h_2(Y)+h_3(Y)$$
Plugging $X=Z^3, Y=Z^2$ gives:
$$g(Z^3,Z^2)=(Z^6-Z^6)h_1(Z^3,Z^2)+Z^3\cdot h_2(Z^2)+h_3(Z^2)$$
which implies
$$Z^3\cdot h_2(Z^2)+h_3(Z^2)=0\quad\quad \quad \text{for all }Z $$
The first component contains odd order terms only, the second component contains even order terms only. So they must both be zeros.
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