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?
 A: Hint $\ $ Nonzero prime ideals are maximal, hence comaximal $\, P + Q\ =\ 1\, $ if $\, P\ne Q.$
Another (perhaps more natural) way to deduce that semi-local Dedekind domains are PIDs is to exploit the local characterization of invertibility of ideals. This yields a simpler yet more general result, see the theorem below from Kaplansky, Commutative Rings. A couple theorems later is the fundamental result that a finitely generated ideal in a domain is invertible iff it is locally principal. Therefore, in Noetherian domains, invertible ideals are global generalizations of principal ideals. To best conceptually comprehend such results it is essential to understand the local-global perspective.


A: 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.
A: Different prime ideals are coprime because in a Dedekind domain every nonzero prime ideal is maximal. If $P$ and $Q$ are nonzero prime ideals then $P+Q$ is an ideal containing both $P$ and $Q$ and so must be the whole ring if $P\ne Q$.
