Are the contractions of relatively prime ideals relatively prime? 
Let $R\subset S$ be Dedekind domains such that $S$ is a finite extension of $R$. Let $P$ be a non-zero prime ideal of $R$. Let $PS$ be the extension of $P$ in $S$. Let $I$ be an ideal of $S$ such that $PS+I=S$. Then, is $P+I\cap R = R$?

Here is my attempt. By lying-over, there exists a prime ideal $Q$ of $S$ such that $Q\cap R=P$. Thus, $PS\subset Q$ so that $Q+I=S$. Suppose $P+I\cap R\neq R$. Since $R$ is of Krull-dimension $1$, $I\cap P\subset P+ I\cap P=P$. I suspect $I\subset \sqrt{Q}$ in this situation. Pick $\alpha\in I$ such that $\alpha\notin Q$.
Let $n$ be the least positive integer such that $\alpha^n + a_{n-1}\alpha^{n-1}+\cdots + a_0\in Q$ where $a_i\in R$ for each $i$ and $a_0\in P$.
(Note that such $n$ exists. Since $\alpha$ is integral over $R$, there exists $n$ such that $\alpha^n + a_{n-1}\alpha^{n-1}+\cdots + a_0=0$ where $a_i\in R$. Hence $\alpha(\alpha^{n-1}+\cdots a_1)= -a_0$. The right hand side is in $R$ and the left hand side is in $I$. Hence $a_0\in I\cap R\subset P$)
Now with this minimal $n$, we apply the same argument. To be specific, since $\alpha(\alpha^{n-1}+\cdots a_1)= -a_0\in P\subset Q$ and $\alpha\notin Q$, $\alpha^{n-1}+\cdots a_1 \in Q$ because $Q$ is a prime.
I am stuck here. If we can prove that $a_1\in P$, by the minimality of $n$, we must have $n=1$. Thus, $\alpha\in \sqrt{Q}$. Hence, we can conclude that $I+PS\subset\sqrt{Q}\neq S$, which is a contradiction.
How do I prove this?

You should use the prime decomposition for Dedekind rings and the behavior of prime ideals in extensions ( we say prime ideals, but since they we consider only non-zero ideals, they are in fact maximal ideals)
Not first that if $I_1$, $I_2$ are ideals of $R$, then $I_1 + I_2= R$ implies $I_1 S + I_2 S = S$. However, if $J_1$, $J_2$ are ideals of $S$ and $J_1 + J_2 = S$, it may happen that $(J_1 \cap R) + (J_2 \cap R )\ne R$.
So we have to show that $J S + I = S$ implies $J + (I \cap R) = R$. 
Consider the prime decomposition of $J$ in $R$ and of $I$ in $S$. 
$$J = P_1 \cdot \ldots P_k \\
I = Q_1 \cdot \ldots Q_l$$
We have $P_i \ne Q_j \cap R$ for all $i$, $j$. Indeed, if we had that then  $J \subset P_i \subset Q_j $, and $I\subset Q_j$ so $J S + I \subset Q_j$, contradiction. 
Denote now by $P'_j= Q_j \cap R$. They are maximal ideals of $R$ ( the extension $R\subset S$ is integral. Moreover, we have $P'_j \ne P_i$ for all $i\ne j$. Now we get 
$$I\cap R \supset P'_1 \cdot \ldots P'_l$$ ( elementary thing). It's enough now to show that 
$$P_1 \cdot\ldots  P_k + P'_1 \cdot \ldots P'_l = R$$
This is true, since there is no maximal ideal containing both of the terms of the sum ( since $P_i \ne P'_j$ for all $i \ne j$).
