If $M$ be a finitely generated module over a Noetherian ring $A,$ then $\widehat{aM}=\hat{a} \hat{M}.$

Question

Let $M$ be a finitely generated module over a Noetherian ring $A.$ Let $\hat{M}$ be the $a$-adic completion of $M.$ Then how can I show that $\widehat{aM}=\hat{a} \hat{M}.$

I know that $\hat{A}$ is flat $A$ module. I need some help. Thanks.

Answer 1

$\newcommand\ideal{\mathfrak}$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.