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Let $o$ be a Dedekind domain, $K$ be its field of fractions, $L\mid K$ is a finite separable field extension, and $O$ be the integral closure of $o$ in $L$. Suppose $p$ is a prime ideal of $o$ and $I$ is an ideal of $O$ such that $pO+I=O$ (here, $pO$ is the ideal of $O$ generated by $p$). Then we must have $p+I \cap o=o$ ?

Thanks in advance.

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I think so. Let me know if any of my steps are incorrect.

Let $p \mathcal O = P_1^{e_1} \cdots P_s^{e_s}$, and let $I = Q_{1}^{f_1} \cdots Q_{t}^{f_t}$ for primes $P_i, Q_i$ and integers $e_i, f_i \geq 1$. If $R^n$ is a power of a prime $R$ of $\mathcal O$ which lies over the prime $r$ of $\mathcal o$, then $$R^n \cap \mathcal o \supseteq (R \cap \mathcal o)^n = r^n.$$ Now $p \mathcal O + I = \mathcal O$, so $p \mathcal O$ and $I$ are relatively prime, i.e. the collections of primes $\{P_i\}, \{Q_i\}$ have nothing in common (if you add two of ideals, the order at each prime is the minimum of the two orders). Let $Q_i$ lie over the prime $q_i$ (you could have $q_i = q_j$ for $i \neq j$, but it doesn't matter). Then $$I \cap \mathcal o = (Q_1^{f_1} \cdots Q_t^{f_t}) \cap \mathcal o = Q_1^{f_1} \cap \cdots \cap Q_t^{f_t} \cap \mathcal o $$ $$ = \bigcap\limits_{i=1}^t (Q_i^{f_i} \cap \mathcal o) \supseteq \bigcap\limits_{i=1}^t q_i^{f_i} = \prod\limits_{i=1}^t q_i^{f_i}$$ Since $P$ is relatively prime to all the $Q_i$, also $p$ is relatively prime to each of the (not necessarily distinct) primes $q_1, ... , q_t$. So $p$ and $I \cap \mathcal o$ are comaximal.

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