I was a little bit puzzled with the following problem that I have recently come across:
Let $R$ be a Dedekind domain and let $P$ be a prime ideal in $R$. Is it true that $P^k$ is an irreducible ideal? (Here, an irreducible ideal $I$ means that if $I=J \cap K$ for some ideals $J$ and $K$, then either $I=J$ or $I=K$.)
Here is how I have progressed: We know that in a Dedekind domain, ideals can be factored uniquely into product of prime ideals. So, here if $P^k$ is not irreducible then $P^k=I\cap J$ for some ideals $I$ and $J$ of $R$ such that $I\supset P^k$ and $J\supset P^k$, where the inclusions are proper. Now, we factorize the ideals $I$ and $J$ into product of prime ideals viz. $I=\wp_{\alpha_1}^{\beta_1}\cdots\wp_{\alpha_k}^{\beta_k}$ and $J=\wp_{\gamma_1}^{\delta_1}\cdots\wp_{\gamma_k}^{\delta_k}$. Now, in a Dedekind domain, contains means divides. So, $I=\wp_{\alpha_1}^{\beta_1}\cdots\wp_{\alpha_k}^{\beta_k}$ divides $P^k$ and $J=\wp_{\gamma_1}^{\delta_1}\cdots\wp_{\gamma_k}^{\delta_k}$ divides $P^k$. Now, unique factorization implies that $I=P^a$ and $J=P^b$ for some integers $a,b \leq k$. Without loss of any generality, suppose that $ a < b$. Then $I \cap J = P^b$. So, $b$ must be equal to $k$ and hence $P^k$ is irreducible.
I am not sure if I'm missing out something...
Yes, it's true and your argument is fine.
Note that in the integers the statement could be restated as: if $\operatorname{lcm} (m,n)=p^k$ then $m$ or $n$ is equal to $p^k$.
And, indeed also for Dedekind domains the intersection of the ideals is their LCM in the semigroup of ideals, which has unique factorization.