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.

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    $\begingroup$ Sorry, I really mean why are the $P$ 's coprime? (In the sense that the sum of two of them gives the entire ring.) $\endgroup$ – user14174 Jan 2 '12 at 12:36
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    $\begingroup$ If $M_1$ and $M_2$ are distinct maximal ideals, then $M_1+M_2=R$ because $M_1+M_2$ is an ideal (sum of two ideals is an ideal) that properly contains the maximal ideal $M_1$ (as it contains both $M_1$ and $M_2$, and $M_2\not\subset M_1$). $\endgroup$ – Arturo Magidin Jan 2 '12 at 18:19

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 $P\ne Q$.

  • $\begingroup$ I'd like to make a comment that probably would make my question less trivial. I note that all of your answers are taking as granted that the ideals are maximal. From the definition I read it is not immediate to me why those ideals are maximal. $\endgroup$ – user14174 Jan 2 '12 at 20:42
  • $\begingroup$ Btw, the definition says: a Dedekind ring is an integral domain with unit such that every proper ideal is (in a unique way) a product of prime ideals. $\endgroup$ – user14174 Jan 2 '12 at 20:53
  • $\begingroup$ @Lmn6: That's OK, because remember that saying an ideal $\mathfrak{p}$ divides an ideal $\mathfrak{q}$ is the same as saying $\mathfrak{q}\subset \mathfrak{p}$. $\endgroup$ – user641 Jan 2 '12 at 22:23
  • $\begingroup$ @Lmn6, there are several equivalent definitions of Dedekind domains. One of them is "integrally closed Noetherian domain with Krull dimension one (i.e., every nonzero prime ideal is maximal)." $\endgroup$ – lhf Jan 2 '12 at 22:40
  • $\begingroup$ @SteveD: Yes, but your argument should work assumed that the prime ideals we are talking about are principal. But I don't know if that is the case; does it follow straight from the definition? $\endgroup$ – user14174 Jan 9 '12 at 19:30

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.

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