Zeroes of prime polynomials in the algebraic torus (A Hilbert's Nullstellensatz for Laurent polynomials?)

Question

1. Let $Q\in\mathbb C[z_1,\dots,z_D]$ be a prime polynomial and let $Z(Q)$ be the algebraic hypersurface of its zeroes. Assume that $P\in\mathbb C[z_1,\dots,z_D]$ is a polynomial which has zeroes at least in $Z(Q)\cap (\mathbb C^*)^D$, where $\mathbb C^*:=\mathbb C\setminus \{0\}$. Hence, it vanishes at least in the intersection of the algebraic hypersurface of the polynomial $Q$ with the algebraic torus. The question is: Does $P$ belongs to the principal ideal of $Q$, $P\in (Q)$?

2. Moreover, let $Q_1,\dots,Q_N\in\mathbb C[z_1,\dots,z_D]$ be different prime polynomials and consider its product $Q:=Q_1\cdots Q_N$. Suppose a polynomial $P\in\mathbb C[z_1,\dots,z_D]$ which has zeroes at least in $Z(Q)\cap (\mathbb C^*)^D$. Does $P$ belong to the principal ideal of $Q$, $P\in(Q)$?

Answer 1

If $P$ vanishes on a nonempty open subset $U$ of $Z(Q)$, then it vanishes on all of $Z(Q)$. This is simply because $Z(P)$ is closed, and if $U\subseteq Z(P)$, then $Z(Q)=\overline{U}\subseteq Z(P)$. The closure of any nonempty $U$ is equal to $Z(Q)$ because $Z(Q)$ is irreducible. Hence, $P\in \sqrt{\langle Q\rangle}=\langle Q\rangle$ as required. We choose $U=Z(Q)\cap(\mathbb C^\ast)^D$, which is open and it is nonempty unless you have $Z(Q)\subseteq Z(z_1\cdots z_D)$. Since $Q$ is irreducible, $U$ is nonempty if and only if $Q\notin\{ z_1,\ldots, z_D \}$.

For your second question, the answer remains the same because we still have $\sqrt{\langle Q\rangle} = \langle Q\rangle$ and the same argument works, as long as none of the $Q_i$ is a coordinate function: $Z(Q)$ is just the union of the $Z(Q_i)$ and you need to make sure that each of those irreducible varieties intersects $U$ in a nonempty open subset.

However note, that if $Q_1=Q_2$, say, then the ideal generated by $Q$ is not radical.