If $A^{\wedge}$ is the completion of a local ring, do we always have $\mathfrak{m}^{\wedge}=\mathfrak{m}A^{\wedge}$?

Let $A$ be a local ring with maximal ideal $\mathfrak{m}$, and let $A^{\wedge}$ be the $\mathfrak{m}$-adic completion of $A$. Then $A^{\wedge}$ is a local ring with maximal ideal $\mathfrak{m}^{\wedge}=\ker(A^{\wedge}\to A/\mathfrak{m})$. In the case of the $p$-adic numbers, we have $$\mathfrak{m}^{\wedge}=\mathfrak{m}A^{\wedge}.$$ Does this equality always hold?

• Are you sure you typed that equation right? – Sam Cassidy Apr 28 '18 at 11:35
• @SamCassidy Thanks, fixed. – user501746 Apr 28 '18 at 12:03
• Anyway it seems like your question is answered on page 109 of Atiyah-Macdonald. It holds if A is Noetherian. – Sam Cassidy Apr 28 '18 at 15:15