common zeroes of multivariable polynomials

Question

Let $P_1(X,Y),\cdots,P_n(X,Y)$ be polynomials of $\mathbb C[X,Y]$ not all zero and $S$ be an infinite subset of $\mathbb C^2$ such that $P_1,\cdots,P_n$ vanish on $S$. My question: do there exist a polynomial $G\in\mathbb C[X,Y]$ and an infinite subset $T$ of $S$ such that for every integer $j$ ($1\le j\le n$) $P_j(X,Y)=G(X,Y)Q_j(X,Y)$ (with $Q_j(X,Y)\in\mathbb C[X,Y]$) and an index $i$ ($1\le i\le n$) with $Q_i$ not vanishing on $T$?

I tried to apply the Nullstellensatz, but I did not manage to prove this assertion.

Thanks in advance for any answer.

Answer 1

For each $1 \leq j \leq n$, we write

$$\displaystyle P_j(x,y) = \prod_{\ell=1}^{k_j} P_{\ell,j}(x,y)$$

where each $P_{\ell,j}$ is geometrically irreducible. Put $S(\ell_1, j_1; \ell_2, j_2)$ for the common zero locus of $P_{\ell_1, j_1}$ and $P_{\ell_2, j_2}$. Then

$$\displaystyle S \subseteq \bigcup_{j_1 \ne j_2} \bigcup_{\ell_1, \ell_2} S(\ell_1, j_1; \ell_2, j_2).$$

By Bezout's theorem, $S(\ell_1, j_1; \ell_2, j_2)$ is finite unless $P_{\ell_1, j_1}, P_{\ell_2, j_2}$ are proportional (since they are geometrically irreducible). Thus, the hypothesis that $S$ is not finite implies that there exist $(\ell_1, j_1), (\ell_2, j_2)$ such that $P_{\ell_1, j_1}$ and $P_{\ell_2, j_2}$ are proportional. But by the definition of $S$, each $P_j$ vanishes on $S$, which implies that for each $1 \leq j \leq n$ there exists $\ell \leq k_j$ such that $P_{\ell,j}$ vanishes on $S$. In particular, the vanishing locus of $P_{\ell,j}$ must be contained in $S(\ell_1, j_1; \ell_2, j_2)$. By the same argument $P_{\ell,j}$ is proportional to $P_{\ell_1, j_1}$ and $P_{\ell_2, j_2}$. Setting $G$ to be this common polynomial, we see that $G$ divides $P_j$ for $1 \leq j \leq n$. This proves your claim.