# If $P$ is a prime ideal in a commutative Noetherian local ring $R$, is $P\hat{R}$ a prime ideal in $\hat{R}$?

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?

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For the original question, the answer is no. Consider the ideal generated by $y^2-x^2(x+1)$ in $\mathbb C[x,y]_m$ where $m$ is the maximal ideal $(x,y)$ of $\mathbb C[x,y]$. In your edit, $S$ is no longer a local ring. The modified question is too general. – user18119 Mar 14 '12 at 21:09
@QiL: Thanks for pointing out that mistake and for the counterexample. – Jake Voigt Mar 14 '12 at 21:21
Dear @QiL, believe it or not, I hadn't seen your comment when I posted my answer. I suppose I had started to write and then came back to my answer after some interruption. Anyway you were first but the similarity of our examples is amusing. – Georges Elencwajg Mar 15 '12 at 1:21
@GeorgesElencwajg, don't worry about that. I understood as such. – user18119 Mar 15 '12 at 12:21

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 $Y+ X\sqrt {1+X},Y-X\sqrt {1+X}\notin \hat{\mathfrak p}$