Let $(R,m)$ be an (integral,excellent,regular...) local ring, and $(\widehat{R},\widehat{m})$ be its completion. Let ${a_{\lambda}}$ be a decreasing net of ideals, and $\widehat{a_{\lambda}}$ be the completion in $\widehat{R}$. If we denote $a=\bigcap a_{\lambda}$, such that $V(a)=\bigcup V(a_{\lambda})$ in $\mathrm{Spec}R$, do we have $\widehat{a}=\bigcap \widehat{a_{\lambda}}$?
I have some ideas on this question. First consider an exact sequence $0\longrightarrow a \longrightarrow R\longrightarrow \underleftarrow{\lim}R/a_{\lambda}$, taking completion we have an exact sequence $0\longrightarrow \widehat{a} \longrightarrow \widehat{R}\longrightarrow (\underleftarrow{\lim}R/a_{\lambda})^{\wedge}$.
Since $R/ a_{\lambda}\hookrightarrow \widehat{R}/\widehat{a_{\lambda}}$ is injective, we have an injection $\underleftarrow{\lim}R/a_{\lambda} \hookrightarrow \underleftarrow{\lim}\widehat{R}/\widehat{a_{\lambda}}$. Thus, if we can embedd $(\underleftarrow{\lim}R/a_{\lambda})^{\wedge}$ into $\underleftarrow{\lim}\widehat{R}/\widehat{a_{\lambda}}$, we have the conslusion. But I don't know under which assumptions this statement would be true...
