1

Let $(R,m)$ be a Noetherian local ring and $\hat{R}$ its $m$-adic completion. Then there is a one to one correspondence between $m$-primary ideals in $R$ and $\hat{m}$-primary ideals in $\hat{R}$.

Suppose $I\in \hat{R}$ is an $\hat{m}$-primary ideal. It is enough to show $I=(I\cap R)\hat{R}$. I tried to show $I/(I\cap R)\hat{R}=0$. If not, then $\hat{m}$ is its associated prime ideal. We thus have $\hat{R}/\hat{m}\hookrightarrow I/(I\cap R)\hat{R}$. Is it enough to show the contradiction? Other approaches are also appreciate. Thanks.

Display name
  • 453

1 Answers1

0

Choose $n$ such that $I\supset \hat m^n$. The canonical isomorphism $R/m^n\to \hat R/ \hat m^n$ induces $$(I\cap R) / m^n \overset{\sim}\longrightarrow I/\hat m^n$$

so that

$$I=I\cap R +\hat m^n=I\cap R +m^n\hat R =(I\cap R)\hat R.$$

Acrobatic
  • 1,281
  • 1
  • 3
  • 12