TITLE: If $P$ is a prime ideal in a commutative Noetherian local ring $R$, is $P\hat{R}$ a prime ideal in $\hat{R}$?
QUESTION [8 upvotes]: Do prime ideals expand to prime ideals in the completion? 

I believe this is the case since I think $R/P\equiv \hat{R}/P\hat{R}$, although Atiyah-Macdonald explicitly mentions the preservation of quotients only with respect to powers of maximal ideal (I am assuming completion with respect to the maximal ideal here).
EDIT: I guess there is more going on here. If I take $S=k[x]$ with $k$ a field, then $(x+1)S$ is prime in $S$ but $x+1$ is a unit $k[[x]]$. So the modified question is: 

When do prime ideals expand to prime ideals in the completion?

REPLY [4 votes]: No, this is not true: prime ideals do not extend to prime ideals after completion. 
Consider $S=\mathbb C[X,Y]$ and its localization $R=S_M$at $M=(X,Y)\subset S$, so that $R=\mathbb C[X,Y]_{(X,Y)}$ is a local ring with maximal ideal  $\mathfrak m=(X,Y)R$.
The completion of $R$ along $\mathfrak m$ is the ring of formal power series $\hat R=\mathbb C[[X,Y]]$.
 The principal prime ideal $\mathfrak p=(Y^2-X^2-X^3)\subset R$ has as extension the principal ideal $\hat {\mathfrak p}=\mathfrak p\hat R=(Y^2-X^2-X^3)\hat R$ which is no longer prime:
Indeed $Y^2-X^2-X^3=(Y+X\sqrt {1+X})(Y-X\sqrt {1+X})\in \hat {\mathfrak p}\;$  although $Y+ X\sqrt {1+X},Y-X\sqrt {1+X}\notin \hat{\mathfrak p}$