[proofplan] We identify the two sides by proving both inclusions inside the ambient space $A(\Omega) \subset A^2(\Omega)$. The analytic input is the previously established [Skoda $A^2$ Division Theorem on Strictly Pseudoconvex Domains](/page/Skoda%20A2%20Division%20Theorem): on a bounded smoothly bounded strictly pseudoconvex domain, generators $g_1,\dots,g_p \in A(\Omega)$ divide every $f \in A(\Omega)$ that belongs locally to the holomorphic ideal they generate, with global coefficients in $A^2(\Omega)$. After verifying that this external theorem applies to the present $\Omega$, $g$, and $f$, the reverse inclusion is formal for elements already lying in $A(\Omega)$, because global $A^2$ coefficients are holomorphic on $\Omega$ and therefore give local holomorphic coefficients. [/proofplan]

[step:Define the two ideal membership conditions inside $A^2(\Omega)$] Define the tuple map \begin{align*} g: \Omega &\to \mathbb{C}^p \\ z &\mapsto (g_1(z),\dots,g_p(z)), \end{align*} and define its Euclidean length by \begin{align*} |g|(z) := \left(\sum_{j=1}^p |g_j(z)|^2\right)^{1/2}, \qquad z \in \Omega. \end{align*} Here $\overline{\Omega}$ denotes the closure of $\Omega$ in $\mathbb{C}^m$, $\partial\Omega := \overline{\Omega} \setminus \Omega$ denotes its boundary, and $\mathcal{L}^{2m}$ denotes $2m$-dimensional Lebesgue measure under the real identification $\mathbb{C}^m \cong \mathbb{R}^{2m}$. The space $A(\Omega)$ denotes the [algebra](/page/Algebra) of [holomorphic functions](/page/Holomorphic%20Function) on $\Omega$ that are continuous on $\overline{\Omega}$, and $A^2(\Omega)$ denotes the [Bergman space](/page/Bergman%20Space) of holomorphic functions $u: \Omega \to \mathbb{C}$ satisfying \begin{align*} \int_\Omega |u(z)|^2\,d\mathcal{L}^{2m}(z) < \infty. \end{align*} Since $\Omega$ is bounded and every function in $A(\Omega)$ is continuous on the compact set $\overline{\Omega}$, we have $A(\Omega) \subset A^2(\Omega)$. For $f \in A(\Omega)$, membership $f \in A(\Omega) \cap I_{\mathrm{loc}}(g)$ means local holomorphic membership in the [ideal](/page/Ideal) generated by $g_1,\dots,g_p$: for every point $a \in \Omega$ there are an open neighbourhood $U_a \subset \Omega$ of $a$ and holomorphic maps $b_{a,j}: U_a \to \mathbb{C}$, for $1 \leq j \leq p$, such that \begin{align*} f(z) = \sum_{j=1}^p g_j(z)b_{a,j}(z), \qquad z \in U_a. \end{align*} Membership $f \in I_{A^2}(g)$ means that $f \in A^2(\Omega)$ and there exist functions $h_1,\dots,h_p \in A^2(\Omega)$ such that \begin{align*} f(z) = \sum_{j=1}^p g_j(z)h_j(z), \qquad z \in \Omega. \end{align*} Inside the theorem we compare $A(\Omega) \cap I_{\mathrm{loc}}(g)$ with $A(\Omega) \cap I_{A^2}(g)$ as subsets of $A^2(\Omega)$. [/step]

[step:Apply Skoda's $A^2$ division theorem after verifying its hypotheses]Let $f \in A(\Omega) \cap I_{\mathrm{loc}}(g)$. We invoke the previously established [Skoda $A^2$ Division Theorem on Strictly Pseudoconvex Domains](/page/Skoda%20A2%20Division%20Theorem), whose relevant form is the following: if $D \subset \mathbb{C}^m$ is bounded, strictly pseudoconvex, and has smooth boundary, if $G_1,\dots,G_p \in A(D)$, and if $F \in A(D)$ belongs at every point of $D$ to the local holomorphic ideal generated by $G_1,\dots,G_p$, then there exist $H_1,\dots,H_p \in A^2(D)$ such that $F = \sum_{j=1}^p G_jH_j$ on $D$. We verify these hypotheses with $D := \Omega$, $G_j := g_j$ for $1 \leq j \leq p$, and $F := f$. The domain hypotheses hold because the theorem statement assumes that $\Omega \subset \mathbb{C}^m$ is bounded, strictly pseudoconvex, and has smooth boundary. The generator hypotheses hold because $g_j \in A(\Omega)$ for every $1 \leq j \leq p$. The function hypothesis holds because $f \in A(\Omega)$, and the local ideal hypothesis is exactly the definition of $f \in I_{\mathrm{loc}}(g)$ adopted in the previous step. Therefore there exist holomorphic functions $h_1,\dots,h_p: \Omega \to \mathbb{C}$ such that \begin{align*} f(z) = \sum_{j=1}^p g_j(z)h_j(z), \qquad z \in \Omega, \end{align*} and, for each $1 \leq j \leq p$, \begin{align*} \int_\Omega |h_j(z)|^2\,d\mathcal{L}^{2m}(z) < \infty. \end{align*} Thus $h_j \in A^2(\Omega)$ for every $j$, and hence $f \in I_{A^2}(g)$. Since the original choice of $f$ already lies in $A(\Omega)$, this proves $f \in A(\Omega) \cap I_{A^2}(g)$.[/step]

[guided]We start with $f \in A(\Omega) \cap I_{\mathrm{loc}}(g)$. By the convention fixed in the first step, this means two things: first, $f: \Omega \to \mathbb{C}$ is holomorphic and continuous up to $\partial\Omega$; second, for every point $a \in \Omega$, the germ of $f$ at $a$ belongs to the holomorphic ideal generated by the germs of $g_1,\dots,g_p$ at $a$. The point of Skoda's theorem is to turn these local holomorphic divisions into one global division with square-integrable holomorphic coefficients. We use the previously established [Skoda $A^2$ Division Theorem on Strictly Pseudoconvex Domains](/page/Skoda%20A2%20Division%20Theorem). In the form needed here, it says: if $D \subset \mathbb{C}^m$ is bounded, strictly pseudoconvex, and has smooth boundary, if $G_1,\dots,G_p \in A(D)$, and if $F \in A(D)$ belongs locally on $D$ to the holomorphic ideal generated by $G_1,\dots,G_p$, then there are functions $H_1,\dots,H_p \in A^2(D)$ satisfying \begin{align*} F(w) = \sum_{j=1}^p G_j(w)H_j(w), \qquad w \in D. \end{align*} This is a separate analytic division theorem; the present argument uses it only to prove one inclusion in the ideal equality. We now check the hypotheses one by one. Take $D := \Omega$, $G_j := g_j$ for $1 \leq j \leq p$, and $F := f$. The domain condition is satisfied because $\Omega$ is assumed bounded, strictly pseudoconvex, and smoothly bounded. The generator condition is satisfied because each $g_j$ belongs to $A(\Omega)$. The function condition is satisfied because $f \in A(\Omega)$. Finally, the local ideal condition is exactly the statement $f \in I_{\mathrm{loc}}(g)$ under the convention that $I_{\mathrm{loc}}(g)$ is interpreted through local holomorphic membership. The theorem therefore gives holomorphic maps $h_j: \Omega \to \mathbb{C}$, for $1 \leq j \leq p$, with \begin{align*} f(z) = \sum_{j=1}^p g_j(z)h_j(z), \qquad z \in \Omega, \end{align*} and with \begin{align*} \int_\Omega |h_j(z)|^2\,d\mathcal{L}^{2m}(z) < \infty, \qquad 1 \leq j \leq p. \end{align*} The holomorphicity together with the displayed integrability condition is precisely $h_j \in A^2(\Omega)$. Hence $f$ has a global $A^2$ division by $g_1,\dots,g_p$, so $f \in I_{A^2}(g)$. Because this same $f$ was chosen from $A(\Omega)$, we have proved $f \in A(\Omega) \cap I_{A^2}(g)$.[/guided]

[step:Convert global $A^2$ division into local holomorphic membership] Let $f \in A(\Omega) \cap I_{A^2}(g)$. To prove the reverse inclusion into the left side of the theorem, we use the $A(\Omega)$ condition for membership in $A(\Omega) \cap I_{\mathrm{loc}}(g)$. By definition of $I_{A^2}(g)$, there are $h_1,\dots,h_p \in A^2(\Omega)$ such that \begin{align*} f(z) = \sum_{j=1}^p g_j(z)h_j(z), \qquad z \in \Omega. \end{align*} Each $h_j \in A^2(\Omega)$ is holomorphic on $\Omega$. Hence, for every $a \in \Omega$, taking $U_a := \Omega$ and $b_{a,j} := h_j|_{U_a}$ gives holomorphic maps $b_{a,j}: U_a \to \mathbb{C}$ satisfying \begin{align*} f(z) = \sum_{j=1}^p g_j(z)b_{a,j}(z), \qquad z \in U_a. \end{align*} Thus $f$ satisfies the local holomorphic membership condition, so $f \in A(\Omega) \cap I_{\mathrm{loc}}(g)$. [/step]

[step:Conclude equality of the two ideals as subsets of $A^2(\Omega)$] The previous two steps prove \begin{align*} A(\Omega) \cap I_{\mathrm{loc}}(g) \subset A(\Omega) \cap I_{A^2}(g) \end{align*} and \begin{align*} A(\Omega) \cap I_{A^2}(g) \subset A(\Omega) \cap I_{\mathrm{loc}}(g). \end{align*} The $A(\Omega)$ factor on the right-hand side is essential: the ideal $I_{A^2}(g)$ may contain holomorphic square-integrable functions that do not extend continuously to $\overline{\Omega}$. Thus the two membership conditions agree inside $A^2(\Omega)$ only after imposing the common ambient condition $f \in A(\Omega)$. Therefore \begin{align*} A(\Omega)\cap I_{\mathrm{loc}}(g)=A(\Omega)\cap I_{A^2}(g). \end{align*} [/step]