Let $M$ be a finitely generated module over a Noetherian ring $A$. Let $\mathfrak{a}\subset A$ be an ideal and denote $\hat{M}$ to the $\mathfrak{a}$-adic completion of $M$. Then how can I show that $\widehat{\mathfrak{a}M}=\hat{\mathfrak{a}} \hat{M}$?
I know that $\hat{A}$ is a flat $A$-module. I need some help. Thanks.
$\newcommand\ideal{\mathfrak}$The completion of a finitely generated module $M$ over a Noetherian ring $A$ can be obtained by extension of scalars: $\hat M\cong\hat A\otimes_AM$. By associativity of tensor product: \begin{align} \hat{\ideal a}\otimes_{\hat A}\hat M &\cong(\hat A\otimes_A\ideal a)\otimes_{\hat A}(\hat A\otimes_AM)\\ &\cong((\hat A\otimes_A\ideal a)\otimes_{\hat A}\hat A)\otimes_AM\\ &\cong(\hat A\otimes_A\ideal a)\otimes_AM\\ &\cong\hat A\otimes_A(\ideal a\otimes_AM) \end{align} From the canonical epimorphism $\ideal a\otimes_AM\twoheadrightarrow\ideal aM$, we get the commutative diagram below, where the top row is surjective:$\require{AMScd}$ \begin{CD} \hat A\otimes_A(\ideal a\otimes_AM)@>>>\hat A\otimes_A(\ideal aM)\\ @V\sim VV@VV\sim V\\ \hat{\ideal a}\otimes_{\hat A}\hat M@>>>\widehat{\ideal aM} \end{CD} Consequently, the bottom row is surjective as well, but since its image is $\hat{\ideal a}\hat M$, this concludes the proof.
$\def\a{\mathfrak{a}}$There's no need for Noetherianity assumptions. The result is still true for an arbitrary ring $A$ and an arbitrary $A$-module $M$, as long as $\mathfrak{a}$ is a finitely generated ideal. Indeed, one has inclusions of $\hat{A}$-submodules of $\hat{M}$ $$ \a\hat{M}\subset\hat{\a}\hat{M}\subset\widehat{\a M}, $$ so it suffices to see that the left module equals the right one. Let $f_1,\dots,f_r\in A$ be generators of $\a$. We have a surjection $A^{\oplus r}\xrightarrow{(f_1,\dots,f_r)}\a M$. By 0315 (2), we have a surjection $\smash{\hat{A}}^{\oplus r}\xrightarrow{(f_1,\dots,f_r)}\widehat{\a M}$. But the image of $\smash{\hat{A}}^{\oplus r}\xrightarrow{(f_1,\dots,f_r)}\hat{M}$ is $\a \hat{M}$.
