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Let $R\subseteq S$ be an integral extension of commutative rings with identity. Let $P$ be a prime ideal in $R$ and $Q$ a prime ideal in $S$. If $Q=PS$ and $P=Q\cap R$ what can we say about $Q^n\cap R$? This ideal always contains $P^n$, but when does equality hold?

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What have you tried? Do you know an example where equality does not hold? –  Georges Elencwajg Sep 12 '11 at 8:08

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For example if $R\subseteq S$ is flat: the extension then actually is faithfully flat and thus for every ideal $I\subseteq R$ we have $IS\cap R=I$. Applying this to $I=P^n$ yields $$P^n = SP^n\cap R = (SP)^n \cap R = Q^n \cap R.$$

However flatness is not a frequent property of integral extensions ...

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Flatness is more common then you think, due to the "miracle flatness theorem": Is $R$ is regular, $S$ is Cohen-Macaualy, and $R \subseteq S$ is finite, then $S$ is flat over $R$. –  David Speyer Sep 12 '11 at 17:29
    
But for $\dim (R)>2$ the condition of being Cohen-Macaulay strongly restricts the admissible singularities of $S$. Moreover even in the $1$-dimensional case one (almost) completely excludes the case of non-normal domains $R$. –  Hagen Sep 13 '11 at 7:20

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