Question
I'm reading this answer. Let $A$ be a Noetherian ring, $I$, $J$ ideals of $A$. Let $\hat{A}$, $\hat{J}$ be $I$-adic completions of $A$ and $J$, respectively. Then, $\hat{J} \simeq J\hat{A}$ ?
What I know
$\hat{A} = \varprojlim(A/I^n)$, and $\hat{J} = \varprojlim(J/I^nJ)$. I have the homomorphism \[ \psi \colon J\hat{A} \ni j (a_n \bmod I^n)_n \mapsto (ja_n \bmod I^nJ)_n \in \hat{J}. \] It is easy to show that this is injective, but I have trouble proving the surjectivity of $\psi$.
What I've tried
I can express $(j_n \bmod I^nJ)_n \in \hat{J}$ (let's denote this as $([j_1], [j_2], \dots)$) as \begin{align*} ([j_1], [j_2], [j_3], \dots) & = ([j_1], [j_1], [j_1], \dots) \\ & \quad + (0, [j_2 - j_1], [j_2 - j_1], \dots) \\ & \quad + (0, 0, [j_3 - j_2], \dots) \\ & \qquad\qquad \vdots \\ & = ([j_1], [j_1], [j_1], \dots) \\ & \quad + ([j_2 - j_1], [j_2 - j_1], [j_2 - j_1], \dots) \\ & \quad + ([j_3 - j_2], [j_3 - j_2], [j_3 - j_2], \dots) \\ & \qquad\qquad \vdots \\ \end{align*} Since $j_{n+1} - j_n \in I^nJ$, there exist $a_n \in I^n$ and $j'_n \in J$ such that $j_{n+1} - j_n = a_nj_n'$. Then, \[ j_1 ([1], [1], \dots) + j_1' ([a_1], [a_1], \dots) + j_2' ([a_2], [a_2], \dots) + \dots \in J\hat{A} \] is mapped to $([j_1], [j_2], \dots) \in \hat{J}$ by $\psi$. Hence $\psi$ is surjective.
Is this proof correct ? I didn't use the fact that $I$, $J$ are finitely generated. Does $\hat{J} \simeq J\hat{A}$ hold when $A$ is not Noetherian ?
Is there another proof ? I will be grateful for any help !
