Let $(A,m)$ be a Noetherian local ring and $\tilde{A}$ its completion by the maximal ideal $m$.
Are the following three statements true?
(i)
If $\tilde{A}$ is free over $A$, then $A\cong \tilde{A}$
(ii)
If $\tilde{A}$ is finitely generated as $A$ module, then $A\cong \tilde{A}$
(iii)
If $\tilde{A}$ is finitely generated as $A$ algebra, then $A\cong \tilde{A}$
Here are short proofs of (i) and (ii).
(i) If $A$ is a Noetherian local ring with maximal ideal $\mathfrak{m}$ then the completion $\hat A$ is a local ring with maximal ideal $\hat{\mathfrak{m}}=\mathfrak{m}\hat A$. The map $A\to \hat A$ is flat, and induces an isomorphism $A/\mathfrak{m}\to \hat A/\hat{\mathfrak{m}}=\hat A/\mathfrak{m}\hat A$. If $\hat A$ is free over $A$ then its rank as a free module is the dimension of $\hat A/\mathfrak{m}\hat A$ as an $A/\mathfrak{m}$-module. Since this is one, the free module has rank one, so $A\to \hat A$ is an isomorphism.
(ii) Since $A$ is Noetherian, a finitely generated module is finitely presented. In general, a finitely presented flat module is projective, and projective modules over a Noetherian local ring are free. So if $\hat A$ is finitely generated as an $A$-module then it is free. Now use (i).
Please let me know if this is wrong. I believe the answer to all three questions is yes.
Let $ (A, \mathfrak{m}) $ be a Noetherian, local ring and let $ \widehat{A} = \varprojlim_{i \in \mathbb{N}} A/\mathfrak{m}^{i} $ be its completion.
Let us assume that $ \widehat{A} $ is free over $ A $ as an $ A $-module. This means that there is an isomorphism $ \phi: \widehat{A} \to \oplus_{i \in \Lambda} A $.
If $ q_{j}: A \to \oplus_{i \in \Lambda} A $ is the natural injection, then let $ f_{j} $ equal $ q_{j}(1) $, and let $ e_{j} $ equal $ \phi^{-1}(f_{j}) $. Let $ \psi_{j}: \oplus_{i \in \Lambda} A \to A $ map $ \{a_{i} f_{i}\}_{i \in \Lambda} $ to $ a_{j}f_{j} $ and let $ \pi: A \to A/\mathfrak{m} $ be the natural quotient morphism.
Suppose that the cardinality of $ \Lambda $ is at least two. If this is the case, then the morphism $ \pi \circ \psi_{j} \circ \phi $ maps $ \widehat{A} $ surjectively onto $ A/\mathfrak{m} $. Since $ \widehat{A} $ is a local ring, the kernel is the unique, maximal ideal $ \mathfrak{m} \widehat{A} $. However, explicit computation shows that the kernel $ \mathfrak{a}_{j} $ of $ \pi \circ \psi_{j} $ is equal to $ (\oplus_{i \in \Lambda \setminus \{j\}} A) \oplus \mathfrak{m} $. Since $ f_{i} $ is in $ \mathfrak{a}_{j} $ for $ i \ne j $, \begin{align*} e_{i} &=\phi^{-1}(f_{i}) \\ &\in \phi^{-1}(\mathfrak{a}_{j}) \\ &= \mathfrak{m} \widehat{A}. \end{align*} As a result, if $ 1 $ is equal to $ \sum_{i \in \Lambda} a_{i} e_{i} $, then \begin{align*} \mathfrak{m} \widehat{A} & \not \ni 1 \\ &= \sum_{i \in \Lambda} a_{i} e_{i} \\ & \in \mathfrak{m}\widehat{A} \end{align*} This is a contradiction, so $ \Lambda $ is a set containing one element. As a result, $ \widehat{A} \cong A $. This hopefully proves i).
It is well known that a finitely generated module $ N $ over a Noetherian, local ring $ (A,\mathfrak{m}) $ is flat if and only if it is free.
The ring $ \widehat{A} $ is a flat $ A $-module. Therefore, if it is finitely generated, it is free. Now the work done in part i) proves that $ \widehat{A} \cong A $.
The dimension of $ \widehat{A} $ is equal to the dimension of $ A $. So if $ \widehat{A} $ is a finitely generated $ A $-algebra, then $ \widehat{A} \cong A[f_{1},\dots,f_{\ell}] $ for some $ f_{1},\dots,f_{\ell} \in \widehat{A} $. Since the dimension of $ \widehat{A} $ is equal to that of $ A $,
\begin{align*}
\widehat{A} & \cong A[f_{1},\dots,f_{\ell}] \\
& \cong A[x_{1},\dots,x_{\ell}]/\mathfrak{a}
\end{align*}
for some ideal $ \mathfrak{a} \subseteq A[x_{1},\dots,x_{\ell}] $ such that the height of $ \mathfrak{a} $ is equal to $ \ell $ and $ \mathfrak{a} $ is $ A $-flat.
So $ \operatorname{Frac}(A)[X]/\mathfrak{a} $ is dimension zero over $ \operatorname{Frac}(A) $. Also $ (A/\mathfrak{m})[X]/\mathfrak{a}) $ is dimension zero over $ A/\mathfrak{m} $.
As a result, the length of $ A[X]/\mathfrak{a} $ is finite as an $ A $-module. In Chapter 3, Chain Conditions, page 15, of Commutative Ring Theory by Hideyuki Matsumura, Matsumura says that this is equivalent to $ A[X]/\mathfrak{a} $ satisfying both the a.c.c. and d.c.c. conditions. So $ A[X]/\mathfrak{a} $ is an Artinian $ A $-module. Let $ y_{i} $ equal $ x_{i}+\mathfrak{a} $. The chain $ \langle y_{i} \rangle \supsetneq \langle y_{i}^{2} \rangle \supsetneq \cdots \subsetneq \langle y_{i}^{n} \rangle \supsetneq \cdots $ is a descending chain. So there is an $ n $ such that $ \langle y_{i}^{n+1} \rangle =0 $. As a result, $ A[y_{i}] $ is a finitely generated $ A $-module. This is equivalent to the statement that $ A[X]/\mathfrak{a} $ is a finitely generated $ A $-module. Now our work in part ii) shows that $ \widehat{A} \cong A $.
So I also think (iii) is true.
Claim: if $A$ is a Noetherian local ring and its $\mathfrak m$-adic completion $A^\wedge$ is essentially of finite type over $A$, then $A$ is a complete local ring.
Sketch of the proof. Say $A^\wedge$ is the localization of a finite type $A$-algebra $B$ at some prime $\mathfrak p$ lying over $\mathfrak m$. Since $A^\wedge \to (B_\mathfrak p)^\wedge$ is an isomorphism, we see from Tag 02HT that $A \to B$ is etale at $\mathfrak p$. Thus we may assume $A \to B$ is etale. We may even assume $B$ is standard etale over $A$ by Tag 00UE. Then by Tag 00UF we can find a finite faithfully flat ring map $A \to C$ such that for every prime $\mathfrak q \subset C$ lying over $\mathfrak m$ there is an $A$-algebra map $B_\mathfrak p \to C_\mathfrak q$. Since $B_\mathfrak p = A^\wedge$ is complete we conclude that $C_\mathfrak q$ is complete (hints: use Tag 04GH and the fact that complete local rings are henselian). We finish by the following lemma.
Lemma: If $A$ is a Noetherian local ring and there exists a finite faithfully flat ring map $A \to C$ such that $C_\mathfrak q$ is complete for all $\mathfrak q \subset C$ lying over $\mathfrak m$, then $A$ is complete.
Proof. Consider the map $C \to C^\wedge$ where $C^\wedge$ is the $\mathfrak m$-adic completion of $C$. Then $C^\wedge$ is the (finite) product of the completions of the local rings of $C$ at the prime ideals $\mathfrak q$ of $C$ lying over $\mathfrak m$, see Tag 07N9. Of course this is exactly the set of maximal ideals of $C$. It follows from the assumption that $C \to C^\wedge$ is a map of $C$-modules which is an isomorphism after localization at all maximal ideals. Hence it is an isomorphism. Since $C$ is isomorphic to $A^{\oplus n}$ as an $A$-module for some $n > 0$, we conclude $A$ is complete.
