[proofplan] We construct a [holomorphic function](/page/Holomorphic%20Function) $f \in \mathcal{O}(\mathbb{C}^n)$ with $\operatorname{div}(f) = D$ for a given effective divisor $D = \sum_{j=1}^N n_j V_j$ on $\mathbb{C}^n$. The proof proceeds by encoding $D$ as a Cousin II datum on a polydisc cover, verifying the cocycle condition, and then invoking [Oka's theorem on Cousin II](/theorems/3412) to solve the datum. The topological obstruction vanishes because $\mathbb{C}^n$ is contractible. [/proofplan] [step:Cover $\mathbb{C}^n$ by polydiscs and construct local defining functions for $D$] Let $D = \sum_{j=1}^N n_j V_j$ be an effective divisor on $\mathbb{C}^n$, where each $V_j$ is an irreducible analytic hypersurface in $\mathbb{C}^n$ and each $n_j$ is a positive integer. The condition that $D$ is locally finite means that every compact subset of $\mathbb{C}^n$ meets only finitely many of the $V_j$. Choose an open cover $\{U_\alpha\}_{\alpha \in A}$ of $\mathbb{C}^n$ by polydiscs. Since $D$ is locally finite, each $U_\alpha$ meets only finitely many of the hypersurfaces $V_j$, say $V_{j_1}, \dots, V_{j_m}$ (where $m$ depends on $\alpha$). On each $U_\alpha$, we construct a [holomorphic function](/page/Holomorphic%20Function) $h_\alpha \in \mathcal{O}(U_\alpha)$ with $\operatorname{div}(h_\alpha) = D|_{U_\alpha}$. For each irreducible hypersurface $V_{j_\ell}$ intersecting $U_\alpha$, the [Weierstrass Preparation Theorem](/theorems/3381) applied after a generic linear change of coordinates provides a Weierstrass polynomial \begin{align*} W_{j_\ell,\alpha}: U_\alpha \to \mathbb{C} \end{align*} that is a monic polynomial in one of the variables (after the coordinate change) with holomorphic coefficients in the remaining variables, such that $V_{j_\ell} \cap U_\alpha = \{z \in U_\alpha : W_{j_\ell,\alpha}(z) = 0\}$ and $W_{j_\ell,\alpha}$ vanishes to order $1$ along $V_{j_\ell}$. Define \begin{align*} h_\alpha := \prod_{\ell=1}^m W_{j_\ell,\alpha}^{\,n_{j_\ell}} \in \mathcal{O}(U_\alpha). \end{align*} By construction, $h_\alpha$ vanishes exactly along the hypersurfaces $V_{j_\ell}$ that meet $U_\alpha$, and vanishes to order $n_{j_\ell}$ along $V_{j_\ell}$, so $\operatorname{div}(h_\alpha) = D|_{U_\alpha}$. [guided] We are given an effective divisor $D = \sum_{j=1}^N n_j V_j$ on $\mathbb{C}^n$, where each $V_j$ is an irreducible analytic hypersurface and each $n_j$ is a positive integer. The local finiteness condition means that every compact subset of $\mathbb{C}^n$ meets only finitely many of the $V_j$. We choose an open cover $\{U_\alpha\}_{\alpha \in A}$ of $\mathbb{C}^n$ by polydiscs. Since $D$ is locally finite, each $U_\alpha$ meets only finitely many hypersurfaces $V_{j_1}, \dots, V_{j_m}$ (where $m$ depends on $\alpha$). Why do we need the [Weierstrass Preparation Theorem](/theorems/3381)? An irreducible analytic hypersurface $V_j$ is locally defined as the zero set of a [holomorphic function](/page/Holomorphic%20Function), but we need a canonical choice that correctly tracks the multiplicity. The [Weierstrass Preparation Theorem](/theorems/3381) provides this: after a generic linear change of coordinates (chosen so that $V_j$ is not contained in any hyperplane $\{z_n = \text{const}\}$), the local defining equation can be taken to be a Weierstrass polynomial \begin{align*} W_{j_\ell,\alpha}: U_\alpha \to \mathbb{C} \end{align*} that is monic in one of the variables (after the coordinate change) with holomorphic coefficients in the remaining variables, such that $V_{j_\ell} \cap U_\alpha = \{z \in U_\alpha : W_{j_\ell,\alpha}(z) = 0\}$ and $W_{j_\ell,\alpha}$ vanishes to order exactly $1$ along $V_{j_\ell}$. Why does this matter for multiplicities? Since each $W_{j_\ell,\alpha}$ vanishes to order exactly $1$ along $V_{j_\ell}$, raising it to the power $n_{j_\ell}$ produces a [holomorphic function](/page/Holomorphic%20Function) vanishing to order precisely $n_{j_\ell}$ along $V_{j_\ell}$. We then define \begin{align*} h_\alpha := \prod_{\ell=1}^m W_{j_\ell,\alpha}^{\,n_{j_\ell}} \in \mathcal{O}(U_\alpha). \end{align*} The local finiteness of $D$ ensures that this product is a finite product on each $U_\alpha$, hence holomorphic. By construction, $h_\alpha$ vanishes exactly along the hypersurfaces $V_{j_\ell}$ that meet $U_\alpha$, and vanishes to order $n_{j_\ell}$ along $V_{j_\ell}$, so $\operatorname{div}(h_\alpha) = D|_{U_\alpha}$. [/guided] [/step] [step:Verify that the local defining functions form a Cousin II datum] On any non-empty overlap $U_\alpha \cap U_\beta$, both $h_\alpha$ and $h_\beta$ have the same divisor $D|_{U_\alpha \cap U_\beta}$. Therefore their ratio \begin{align*} g_{\alpha\beta} := \frac{h_\alpha}{h_\beta}: U_\alpha \cap U_\beta \to \mathbb{C} \end{align*} is a [holomorphic function](/page/Holomorphic%20Function) whose divisor is $\operatorname{div}(h_\alpha) - \operatorname{div}(h_\beta) = D|_{U_\alpha \cap U_\beta} - D|_{U_\alpha \cap U_\beta} = 0$. A [holomorphic function](/page/Holomorphic%20Function) with vanishing divisor has no zeros, so $g_{\alpha\beta} \in \mathcal{O}^*(U_\alpha \cap U_\beta)$ is nowhere-vanishing and holomorphic. The collection $\{g_{\alpha\beta}\}_{\alpha, \beta \in A}$ satisfies the multiplicative cocycle condition on each triple overlap $U_\alpha \cap U_\beta \cap U_\gamma$: \begin{align*} g_{\alpha\beta} \cdot g_{\beta\gamma} = \frac{h_\alpha}{h_\beta} \cdot \frac{h_\beta}{h_\gamma} = \frac{h_\alpha}{h_\gamma} = g_{\alpha\gamma}. \end{align*} Therefore $\{(U_\alpha, h_\alpha)\}$ is a Cousin II datum with transition cocycle $\{g_{\alpha\beta}\} \in \check{Z}^1(\{U_\alpha\}, \mathcal{O}^*)$. [/step] [step:Verify the hypotheses of Oka's theorem on Cousin II] To apply [Oka's theorem on Cousin II](/theorems/3412), we must verify two conditions: 1. **$\mathbb{C}^n$ is a domain of holomorphy.** The space $\mathbb{C}^n$ is itself a domain of holomorphy: any [holomorphic function](/page/Holomorphic%20Function) $f \in \mathcal{O}(\mathbb{C}^n)$ is entire, and there is no larger connected [open set](/page/Open%20Set) in $\mathbb{C}^n$ to which all entire functions extend (since $\mathbb{C}^n$ is already all of $\mathbb{C}^n$). 2. **$H^2(\mathbb{C}^n, \mathbb{Z}) = 0$.** The space $\mathbb{C}^n$ is contractible: the homotopy \begin{align*} H: \mathbb{C}^n \times [0,1] &\to \mathbb{C}^n, \\ (z, t) &\mapsto tz \end{align*} contracts $\mathbb{C}^n$ to the origin. A contractible space has vanishing singular cohomology in all positive degrees: $H^q(\mathbb{C}^n, \mathbb{Z}) = 0$ for all $q \geq 1$. In particular, $H^2(\mathbb{C}^n, \mathbb{Z}) = 0$. Both hypotheses are satisfied. [/step] [step:Apply Oka's theorem to solve the Cousin II datum and obtain the global holomorphic function] By [Oka's theorem on Cousin II](/theorems/3412), since $\mathbb{C}^n$ is a domain of holomorphy and $H^2(\mathbb{C}^n, \mathbb{Z}) = 0$, the Cousin II datum $\{(U_\alpha, h_\alpha)\}$ is solvable. This means there exists a [meromorphic function](/page/Meromorphic%20Function) $f \in \mathcal{M}(\mathbb{C}^n)$ such that for every $\alpha \in A$: \begin{align*} \frac{f}{h_\alpha} \in \mathcal{O}^*(U_\alpha), \end{align*} i.e., $f/h_\alpha$ is holomorphic and nowhere-vanishing on $U_\alpha$. [guided] What does solvability of the Cousin II datum mean concretely? It means we can find a single global function $f$ that, on each $U_\alpha$, differs from the local function $h_\alpha$ by a nowhere-vanishing holomorphic factor. Equivalently, $f$ and $h_\alpha$ have the same zero set with the same multiplicities on $U_\alpha$. The cocycle $\{g_{\alpha\beta}\}$ is the obstruction to patching the local functions $h_\alpha$ into a global function. Oka's theorem says this obstruction vanishes when $H^1(\Omega, \mathcal{O}^*) = 0$, which holds when both $H^1(\Omega, \mathcal{O}) = 0$ (Cousin I solvability) and $H^2(\Omega, \mathbb{Z}) = 0$ (no topological obstruction). [/guided] [/step] [step:Verify that $f$ is holomorphic and $\operatorname{div}(f) = D$] Since $f/h_\alpha \in \mathcal{O}^*(U_\alpha)$ and $h_\alpha \in \mathcal{O}(U_\alpha)$, we have $f = h_\alpha \cdot (f/h_\alpha)$ on $U_\alpha$. Since $h_\alpha$ is holomorphic and $f/h_\alpha$ is holomorphic and nowhere-vanishing, the product $f$ is holomorphic on $U_\alpha$. Since $\{U_\alpha\}$ covers $\mathbb{C}^n$, the function $f$ is holomorphic on all of $\mathbb{C}^n$: $f \in \mathcal{O}(\mathbb{C}^n)$. The divisor of $f$ on each $U_\alpha$ is \begin{align*} \operatorname{div}(f)|_{U_\alpha} = \operatorname{div}(h_\alpha) + \operatorname{div}(f/h_\alpha) = D|_{U_\alpha} + 0 = D|_{U_\alpha}, \end{align*} where $\operatorname{div}(f/h_\alpha) = 0$ because $f/h_\alpha$ is nowhere-vanishing. Since $\{U_\alpha\}$ covers $\mathbb{C}^n$, the divisor identity $\operatorname{div}(f) = D$ holds globally. Since $D$ is effective (all multiplicities $n_j > 0$), the local defining functions $h_\alpha$ are holomorphic (not merely meromorphic), confirming that $f$ is holomorphic rather than meromorphic. [guided] We need to verify two things: that $f$ is holomorphic (not merely meromorphic), and that $\operatorname{div}(f) = D$. **Holomorphicity.** On each $U_\alpha$, we have $f/h_\alpha \in \mathcal{O}^*(U_\alpha)$ from the Cousin II solution, so $f = h_\alpha \cdot (f/h_\alpha)$ on $U_\alpha$. Since $h_\alpha \in \mathcal{O}(U_\alpha)$ is holomorphic and $f/h_\alpha$ is holomorphic and nowhere-vanishing, their product $f$ is holomorphic on $U_\alpha$. Since the polydiscs $\{U_\alpha\}$ cover $\mathbb{C}^n$, the function $f$ is holomorphic on all of $\mathbb{C}^n$: $f \in \mathcal{O}(\mathbb{C}^n)$. Why does effectiveness matter here? The divisor $D$ is effective, meaning all multiplicities $n_j > 0$. This ensures the local defining functions $h_\alpha = \prod_{\ell=1}^m W_{j_\ell,\alpha}^{\,n_{j_\ell}}$ are holomorphic (not meromorphic) -- they have zeros but no poles. If $D$ had negative multiplicities (prescribing poles), the $h_\alpha$ would be meromorphic, and the resulting $f$ would also be meromorphic. **Divisor computation.** On each $U_\alpha$, the divisor of $f$ is \begin{align*} \operatorname{div}(f)|_{U_\alpha} = \operatorname{div}(h_\alpha) + \operatorname{div}(f/h_\alpha) = D|_{U_\alpha} + 0 = D|_{U_\alpha}, \end{align*} where $\operatorname{div}(f/h_\alpha) = 0$ because $f/h_\alpha$ is nowhere-vanishing. Since $\{U_\alpha\}$ covers $\mathbb{C}^n$, the divisor identity $\operatorname{div}(f) = D$ holds globally. **Non-uniqueness.** The function $f$ is determined up to multiplication by a nowhere-vanishing entire function: if $\tilde{f}$ is another solution, then $\tilde{f}/f$ is an entire function with no zeros, hence of the form $e^g$ for some $g \in \mathcal{O}(\mathbb{C}^n)$. This non-uniqueness is inherent -- the construction prescribes the zero set and multiplicities, but not the "non-vanishing part" of $f$. [/guided] [/step]