[proofplan] Part (1) is proved by showing that the associated graded ring $G_{\hat{\mathfrak{a}}}(\hat{R})$ is isomorphic to $G_{\mathfrak{a}}(R)$, which is Noetherian by the [Associated Graded Ring Is Noetherian](/theorems/2948) theorem. A lemma on graded rings then lifts Noetherianness from the associated graded to $\hat{R}$. Part (2) follows from (1) together with a flatness argument: $\hat{R}$ is the inverse limit $\varprojlim R/\mathfrak{a}^n$, each $R/\mathfrak{a}^n$ is a flat $R/\mathfrak{a}^n$-module (being the identity), and we show $\hat{R}$ is flat over $R$ using the [Artin-Rees Lemma](/theorems/2950) to control the interaction of the $\mathfrak{a}$-adic filtration with submodules. Part (3) reduces to the case $M = R^n$ via a finite presentation and the five lemma. [/proofplan] [step:Prove (1): $\hat{R}$ is Noetherian via the associated graded ring] The $\mathfrak{a}$-adic completion is $\hat{R} = \varprojlim_{n} R/\mathfrak{a}^n$, equipped with the ideal $\hat{\mathfrak{a}} = \ker(\hat{R} \to R/\mathfrak{a})$. The $\hat{\mathfrak{a}}$-adic filtration on $\hat{R}$ has quotients \begin{align*} \hat{\mathfrak{a}}^n / \hat{\mathfrak{a}}^{n+1} \cong \mathfrak{a}^n / \mathfrak{a}^{n+1} \end{align*} for all $n \geq 0$. To verify this isomorphism: the natural map $R \to \hat{R}$ sends $\mathfrak{a}^n$ into $\hat{\mathfrak{a}}^n$, inducing a map $\mathfrak{a}^n/\mathfrak{a}^{n+1} \to \hat{\mathfrak{a}}^n/\hat{\mathfrak{a}}^{n+1}$. Surjectivity follows from the fact that $\hat{\mathfrak{a}}^n$ is generated by products of $n$ elements from $\hat{\mathfrak{a}}$, and each such element is a limit of elements from $\mathfrak{a}$. Injectivity follows from $\mathfrak{a}^n \cap \ker(R \to \hat{R}/\hat{\mathfrak{a}}^{n+1}) = \mathfrak{a}^{n+1}$, which holds because $R/\mathfrak{a}^{n+1} \hookrightarrow \hat{R}/\hat{\mathfrak{a}}^{n+1}$ is an isomorphism (a standard property of the $\mathfrak{a}$-adic completion). Therefore the associated graded rings are isomorphic: \begin{align*} G_{\hat{\mathfrak{a}}}(\hat{R}) = \bigoplus_{n \geq 0} \hat{\mathfrak{a}}^n/\hat{\mathfrak{a}}^{n+1} \cong \bigoplus_{n \geq 0} \mathfrak{a}^n/\mathfrak{a}^{n+1} = G_{\mathfrak{a}}(R). \end{align*} Since $R$ is Noetherian, the [Associated Graded Ring Is Noetherian](/theorems/2948) theorem gives that $G_{\mathfrak{a}}(R)$ is Noetherian. It remains to lift Noetherianness from $G_{\hat{\mathfrak{a}}}(\hat{R})$ to $\hat{R}$. The $\mathfrak{a}$-adic completion $\hat{R}$ is complete with respect to $\hat{\mathfrak{a}}$ (i.e., $\hat{R} \cong \varprojlim \hat{R}/\hat{\mathfrak{a}}^n$), and $\bigcap_{n \geq 0} \hat{\mathfrak{a}}^n = 0$. Under these conditions, if $G_{\hat{\mathfrak{a}}}(\hat{R})$ is Noetherian, then $\hat{R}$ is Noetherian. The argument: let $J$ be an ideal of $\hat{R}$. The associated graded module $G(J) = \bigoplus_{n \geq 0} (J \cap \hat{\mathfrak{a}}^n)/(J \cap \hat{\mathfrak{a}}^{n+1})$ is a graded ideal of $G_{\hat{\mathfrak{a}}}(\hat{R})$, hence finitely generated. Choose homogeneous generators $\bar{x}_1, \ldots, \bar{x}_s$ with $\bar{x}_k \in (J \cap \hat{\mathfrak{a}}^{d_k})/(J \cap \hat{\mathfrak{a}}^{d_k+1})$, and lift to $x_k \in J \cap \hat{\mathfrak{a}}^{d_k}$. The ideal $J' = (x_1, \ldots, x_s) \subseteq J$ satisfies $G(J') = G(J)$. One then shows $J' = J$ by a completeness argument: for any $y \in J$, successive approximation modulo $\hat{\mathfrak{a}}^N$ gives $y \in J' + (J \cap \hat{\mathfrak{a}}^N)$ for all $N$, and $\bigcap_N (J' + J \cap \hat{\mathfrak{a}}^N) = J'$ by completeness and $\bigcap_N \hat{\mathfrak{a}}^N = 0$. [guided] The strategy for part (1) is to use the associated graded ring as a "bridge" to transfer Noetherianness. The chain of reasoning is: $R$ Noetherian $\implies$ $G_{\mathfrak{a}}(R)$ Noetherian (by [Associated Graded Ring Is Noetherian](/theorems/2948)) $\implies$ $G_{\hat{\mathfrak{a}}}(\hat{R})$ Noetherian (by the isomorphism of graded rings) $\implies$ $\hat{R}$ Noetherian (by lifting from the graded ring to the complete ring). The first implication is the content of [theorem 2948](/theorems/2948): since $R$ is Noetherian and $\mathfrak{a}$ is an ideal, the graded ring $G_{\mathfrak{a}}(R) = \bigoplus \mathfrak{a}^n/\mathfrak{a}^{n+1}$ is Noetherian. The second step is the isomorphism $G_{\hat{\mathfrak{a}}}(\hat{R}) \cong G_{\mathfrak{a}}(R)$. This comes from the fact that completion does not change the successive quotients: $\hat{\mathfrak{a}}^n/\hat{\mathfrak{a}}^{n+1} \cong \mathfrak{a}^n/\mathfrak{a}^{n+1}$. The key ingredient is the isomorphism $R/\mathfrak{a}^n \cong \hat{R}/\hat{\mathfrak{a}}^n$ for each $n$, which is a standard property of $\mathfrak{a}$-adic completions. The third step lifts Noetherianness. The argument is an "approximation" technique: if $J \trianglelefteq \hat{R}$, its graded version $G(J)$ is finitely generated over $G_{\hat{\mathfrak{a}}}(\hat{R})$. Lifting generators and using completeness (specifically $\bigcap_n \hat{\mathfrak{a}}^n = 0$), one constructs a finite generating set for $J$. Why is completeness essential? Without it, we could only say $J' + J \cap \hat{\mathfrak{a}}^N = J$ for all $N$, but could not conclude $J' = J$. The completeness condition $\hat{R} = \varprojlim \hat{R}/\hat{\mathfrak{a}}^n$ ensures that an element in $\bigcap_N (J' + \hat{\mathfrak{a}}^N)$ already lies in $J'$. [/guided] [/step] [step:Prove (2): $\hat{R}$ is flat over $R$, so $\hat{R} \otimes_R (-)$ is exact] By the [Characterization of Flat Modules](/theorems/2839), it suffices to show that for every injective $R$-linear map $f: N \hookrightarrow M$ between finitely generated $R$-modules, the induced map $\operatorname{id}_{\hat{R}} \otimes f: \hat{R} \otimes_R N \to \hat{R} \otimes_R M$ is injective. We use the $\mathfrak{a}$-adic filtration. For any finitely generated $R$-module $M$, the natural map \begin{align*} \hat{R} \otimes_R M \to \hat{M} = \varprojlim_{n} M/\mathfrak{a}^n M \end{align*} is an isomorphism when $M$ is free of finite rank (since $\hat{R} \otimes_R R^k \cong \hat{R}^k$ and $\widehat{R^k} \cong \hat{R}^k$). For the general case, we use this in part (3). For flatness directly: consider the inverse system of short exact sequences \begin{align*} 0 \to \mathfrak{a}^n M / \mathfrak{a}^n N' \to M/\mathfrak{a}^n N' \to M/\mathfrak{a}^n M \to 0, \end{align*} where $N' = f(N)$. The [Artin-Rees Lemma](/theorems/2950) gives a stable $\mathfrak{a}$-filtration on $N'$: there exists $c \geq 0$ such that $\mathfrak{a}^n M \cap N' = \mathfrak{a}^{n-c}(\mathfrak{a}^c M \cap N')$ for all $n \geq c$. In particular, $\mathfrak{a}^{n-c} N' \subseteq \mathfrak{a}^n M \cap N' \subseteq \mathfrak{a}^{n-c} N'$ up to a shift, so the $\mathfrak{a}$-adic topology on $N'$ coincides with the subspace topology induced from the $\mathfrak{a}$-adic topology on $M$. This means the sequence \begin{align*} 0 \to N'/(\mathfrak{a}^n M \cap N') \to M/\mathfrak{a}^n M \end{align*} is exact for each $n$, and by Artin-Rees, the inverse system $\{N'/(\mathfrak{a}^n M \cap N')\}_n$ has surjective transition maps. Taking inverse limits of exact sequences with surjective transition maps preserves exactness: \begin{align*} 0 \to \hat{N'} \to \hat{M} \end{align*} is exact. Since $N' \cong N$ via $f$ (which is injective), $\hat{N} \to \hat{M}$ is injective. By part (3) (proved below for free modules and then extended), $\hat{R} \otimes_R N \cong \hat{N}$ and $\hat{R} \otimes_R M \cong \hat{M}$ for finitely generated modules, so $\hat{R} \otimes_R N \to \hat{R} \otimes_R M$ is injective. Since flatness can be tested on finitely generated modules (by the [Characterization of Flat Modules](/theorems/2839)), $\hat{R}$ is flat over $R$, and therefore $\hat{R} \otimes_R (-)$ is exact. [/step] [step:Prove (3): $\hat{R} \otimes_R M \cong \hat{M}$ for finitely generated $M$] **Free case.** If $M = R^k$ for some $k \geq 1$, then \begin{align*} \hat{R} \otimes_R R^k \cong \hat{R}^k \quad \text{and} \quad \widehat{R^k} = \varprojlim_{n} R^k/\mathfrak{a}^n R^k \cong \left(\varprojlim_{n} R/\mathfrak{a}^n\right)^k = \hat{R}^k. \end{align*} The natural map sends $x \otimes (r_1, \ldots, r_k) \mapsto (xr_1, \ldots, xr_k)$, which is manifestly an isomorphism. **General case.** Let $M$ be a finitely generated $R$-module. Choose a finite presentation: \begin{align*} R^s \xrightarrow{\;\alpha\;} R^t \xrightarrow{\;\beta\;} M \to 0. \end{align*} This is exact (such a presentation exists because $M$ is finitely generated over the Noetherian ring $R$, so $\ker \beta$ is finitely generated, admitting a surjection from $R^s$). Apply the functor $\hat{R} \otimes_R (-)$. By the [Right Exactness of Tensoring](/theorems/2838): \begin{align*} \hat{R} \otimes_R R^s \xrightarrow{\;\operatorname{id} \otimes \alpha\;} \hat{R} \otimes_R R^t \xrightarrow{\;\operatorname{id} \otimes \beta\;} \hat{R} \otimes_R M \to 0 \end{align*} is exact. Apply the completion functor $\widehat{(-)}$ to the same presentation. Completion is right exact on finitely generated modules over Noetherian rings (the transition maps in the inverse system $\{M/\mathfrak{a}^n M\}$ are surjective, and the kernel system satisfies the Mittag-Leffler condition by Artin-Rees), so: \begin{align*} \widehat{R^s} \xrightarrow{\;\hat{\alpha}\;} \widehat{R^t} \xrightarrow{\;\hat{\beta}\;} \hat{M} \to 0 \end{align*} is exact. Consider the commutative diagram: \begin{align*} \hat{R} \otimes_R R^s &\xrightarrow{\;\operatorname{id} \otimes \alpha\;} \hat{R} \otimes_R R^t \xrightarrow{\;\operatorname{id} \otimes \beta\;} \hat{R} \otimes_R M \to 0 \\ \downarrow \varphi_s &\qquad\qquad \downarrow \varphi_t \qquad\qquad\qquad \downarrow \varphi_M \\ \widehat{R^s} &\xrightarrow{\;\hat{\alpha}\;} \widehat{R^t} \xrightarrow{\;\hat{\beta}\;} \hat{M} \to 0 \end{align*} where $\varphi_s, \varphi_t, \varphi_M$ are the natural maps $\hat{R} \otimes_R (-) \to \widehat{(-)}$. The diagram commutes by naturality of these maps. By the free case, $\varphi_s$ and $\varphi_t$ are isomorphisms. Both rows are exact. By the five lemma (applied to the two rightmost squares, using that $\varphi_s$ is surjective and $\varphi_t$ is an isomorphism), $\varphi_M: \hat{R} \otimes_R M \to \hat{M}$ is an isomorphism. [guided] The strategy for part (3) is a standard "resolve and compare" argument. We know the result for free modules by direct computation, and we extend it to all finitely generated modules using a presentation. For the free case $M = R^k$: the tensor product $\hat{R} \otimes_R R^k$ distributes as $\hat{R}^k$ (tensor product commutes with finite direct sums). The completion $\widehat{R^k}$ is computed as $\varprojlim R^k/\mathfrak{a}^n R^k$. Since $\mathfrak{a}^n R^k = (\mathfrak{a}^n)^k$ (the $\mathfrak{a}$-adic filtration acts component-wise), we get $\varprojlim (R/\mathfrak{a}^n)^k = (\varprojlim R/\mathfrak{a}^n)^k = \hat{R}^k$. The natural map is the identity on $\hat{R}^k$, hence an isomorphism. For the general case: since $R$ is Noetherian and $M$ is finitely generated, $M$ admits a finite presentation $R^s \to R^t \to M \to 0$. We apply both functors to this presentation and use the five lemma. The five lemma requires: (i) both rows are exact, and (ii) the vertical maps for $R^s$ and $R^t$ are isomorphisms. Condition (ii) is the free case. Condition (i) for the top row follows from [Right Exactness of Tensoring](/theorems/2838). Condition (i) for the bottom row uses right exactness of completion, which relies on the Artin-Rees lemma to ensure that the inverse limit of the kernel system is well-behaved. The five lemma then forces $\varphi_M$ to be an isomorphism: surjectivity follows from surjectivity of $\varphi_t$ and the diagram chase; injectivity follows from injectivity of $\varphi_t$ and surjectivity of $\varphi_s$. [/guided] [/step]