Let $U\subset \mathbb{A}^n_{\mathbb{C}}$ be any Zariski open affine subvarity. Let $M$ be an $\mathcal{O}(U)$-module. Suppose $M$ satisfies $M\overset{L}{\otimes}\mathbb{C}_{\mathfrak{M}}\cong 0$ for every maximal ideal $\mathfrak{M}\subset \mathcal{O}(U)$, where $\mathbb{C}_{\mathfrak{M}}=\mathcal{O}(U)/\mathfrak{M}$.

My questions are:

(1) If $M$ is countably generated over $\mathcal{O}(U)$, then must $M$ be $0$? This is true for $n=1$, because using the Koszul resolution of $\mathbb{C}_{\mathfrak{M}}$, we see that $M$ is a module over the fraction field $\mathbb{C}(z)$, so it cannot be countably generated unless it is $0$.

(2) Intuitively, one wanted to prove by induction on $n$. For this we need to take $M\overset{L}{\otimes}(\mathcal{O}(U)/I)$ for different ideals $I$. A priori, the derived tensor is not necessarily concentrated in degree $0$. So does (1) hold with $M$ replaced by a bounded complex of modules $M^\bullet$, where each cohomology module is countably generated? If so, how does the induction work?

Let $U\subset \mathbb{A}^n_{\mathbb{C}}$ be any Zariski open affine subvarity. Let $M$ be an $\mathcal{O}(U)$-module. Suppose $M$ satisfies $M\overset{L}{\otimes}\mathbb{C}_{\mathfrak{M}}\cong 0$ for every maximal ideal $\mathfrak{M}\subset \mathcal{O}(U)$, where $\mathbb{C}_{\mathfrak{M}}=\mathcal{O}(U)/\mathfrak{M}$.

My questions are:

(1) If $M$ is countably generated over $\mathcal{O}(U)$, then must $M$ be $0$? This is true for $n=1$, because using the Koszul resolution of $\mathbb{C}_{\mathfrak{M}}$, we see that $M$ is a module over the fraction field $\mathbb{C}(z)$, so it cannot be countably generated unless it is $0$.

(2) Intuitively, one wanted to prove by induction on $n$. For this we need to take $M\overset{L}{\otimes}(\mathcal{O}(U)/I)$ for different ideals $I$. A priori, the derived tensor is not necessarily concentrated in degree $0$. So does (1) hold with $M$ replaced by a bounded complex of modules $M^\bullet$, where each cohomology module is countably generated?