Dedekind domain with a finite number of prime ideals is principal I am reading a proof of this result that uses the Chinese Remainder Theorem on (the finite number of) prime ideals $P_i$. In order to apply CRT we should assume that the prime ideals are coprime, i.e. the ring is equal to $P_h + P_k$ for $h \neq k$, but I can't see it. How does it follow?

Here's one proof. Let $R$ be a Dedekind ring and assume that the prime ideals are $\mathfrak{p}_1,\ldots,\mathfrak{p}_n$. Then $\mathfrak{p}_1^2,\mathfrak{p}_2,\ldots,\mathfrak{p}_n$ are coprime. Pick an element $\pi \in \mathfrak{p}_1\setminus \mathfrak{p}_1^2$ and by CRT you can find an $x\in R$ s.t.
$$ x\equiv \pi\,(\textrm{mod } \mathfrak{p}_1^2),\;\; x\equiv 1\,(\textrm{mod } \mathfrak{p}_k),\; k=2,\ldots,n $$
Factoring we must have $(x)=\mathfrak{p}_1$. It follows that all prime ideals are principal, so all ideals are principal and $R$ is a PID.
EDIT: The definition of a Dedekind domain is a Noetherian integrally closed, integral domain of dimension 1. The last condition means precisely that every nonzero prime ideal is maximal, so maximality of nonzero primes is tautological. Maximal ideals are always coprime.
