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$(R,m)\subseteq (S,n)$ is a local extension of rings and $S$ is a finitely generated $R$-module. If $P$ is a prime ideal of $R$ such that $P\subset m^2$ and $P'$ is a prime ideal in $S$ such that $P'\cap R=P$, is $P' \subset n^2$?

Following the counterexample by Georges, I was wondering if this works under additional assumptions. Such as if $S$ is a domain or if $R,S$ are both complete?

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In the example of Georges $R, S$ are both complete.

Here is an example with $R, S$ complete local domains: let $k$ be a field, $R=k[[x,y]]$, $S=k[[x, y, z]]$ with $z^2=y^3$. Let $P'=(x^2+z)S$. As $(x,y,z)^2=(x^2,xy, y^2, xz, yz)$, $P'$ is not contained in $(x,y,z)^2={\mathfrak n}^2$.

I claim that $P'\cap R=(y^3-x^4)R\subset (x,y)^2={\mathfrak m}^2$ which is then a counterexample.

Proof of the claim: Notice that $S=R\oplus zR$. Let $$f=(x^2+z)(h_0(x,y)+zh_1(x,y))\in P'\cap R.$$ Then $f=x^2h_0+y^3h_1+z(h_0+x^2h_1)$. Hence $h_0=-x^2h_1$ and $f=(y^3-x^4)h_1$. On the other hand $$y^3-x^4=z^2-x^4=(z-x^2)(z+x^2)\in P'\cap R$$ and the claim is proved.

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No.

Take $R=k=$ a field, $P=m=m^2=(0)$ and $S=k[X]/(X^2)=k[\epsilon ],\; P'=n=(\epsilon)\nsubseteq n^2=(0) $

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  • $\begingroup$ Thanks, does it work under additional assumptions. Such as if $S$ is a domain or if $R,S$ are both complete? $\endgroup$ – Jake Voigt Mar 15 '12 at 21:43

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