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

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.

$$\newcommand\ideal{\mathfrak}$$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.