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Let $\Bbb Q \neq K' \subset K$ be number fields such that $K/\Bbb Q$ and $K'/\Bbb Q$ are Galois. Let $P$ be a prime ideal of $\mathcal O_K$.

Is it true that $$ P^r \cap K' = (P \cap K')^r $$ for any $r \geq 1$?

I know that $\supset$ hold very generally. I tried to factor $ P^r \cap K'$ into a product of prime ideals in $K'$, without success.

Thank you!

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    Related: https://math.stackexchange.com/questions/63828 – 2017-01-10
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    Another answer would be: If $A$ is a Dedekind domain and $B$ is a domain such that $A \subset B$ is integral, then $B$ is a torsion-free $A$-module, so it is a flat $A$-module, since $A$ is a Dedekind domain (see [here](http://math.stackexchange.com/q/158427)). Since $A \subset B$ is flat and integral, [it is](http://math.stackexchange.com/q/385402) faithfully flat, therefore $I^{ec}=I$ for any ideal $I$ of $A$. [In particular](https://math.stackexchange.com/questions/63828), if $I=(P \cap K')^r$, then we have $$(P \cap K')^r = B(P \cap K')^r \cap A = P^r \cap A.$$ – 2017-02-10
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    In my above comment, I still need to check that $B(P \cap K')^r = B(P \cap A)^r = P^r$, which is moreless saying $((P^c)^r)^e = (P^r)^{ce} = P^r$, which is not the case! We have $I^{ec} = I$ for all ideals $I$ of $A$, but not $J^{ce} = J$ for all ideals $J$ of $B$. – 2017-02-25

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