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I just wondering if the following statement is true.

If $R$ is a Dedekind domain and $P$ is a prime ideal of $R$, then $R_P$ is a PID.

$R_P$ means $R$ localize at $P$.


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What definition of Dedekind domain are you working with? (One definition is "$R$ is a Dedekind domain if and only if $R_P$ is a PID for all primes $P$.") – Qiaochu Yuan Mar 13 '12 at 3:13
up vote 2 down vote accepted

Yes. Every Dedekind domain is a Krull domain and every nonzero prime ideal of a Dedekind domain is maximal. Therefore if $R$ is Dedekind and $P$ is a nonzero prime ideal of $R$, then $R_P$ is a PID--in fact, it's a discrete valuation ring.

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The answer depends crucially on which of many equivalent definitions of Dedekind domain that you employ. Here's an answer that covers all bases. Find your Prufer analog $(n)$ in the following list, then trace the path from $(n)$ to $(4)$ in the proofs in [1]

THEOREM $\ \ $ Let $\rm\:D\:$ be a domain. The following are equivalent:

(1) $\rm\:D\:$ is a Prufer domain, i.e. every nonzero f.g. (finitely generated) ideal is invertible.
(2) Every nonzero two-generated ideal of $\rm\:D\:$ is invertible.
(3) $\rm\:D_P\:$ is a Prufer domain for every prime ideal $\rm\:P\:$ of $\rm\:D.\:$
(4) $\rm\:D_P\:$ is a valuation domain for every prime ideal $\rm\:P\:$ of $\rm\:D.\:$
(5) $\rm\:D_P\:$ is a valuation domain for every maximal ideal $\rm\:P\:$ of $\rm\:D.\:$
(6) Every nonzero f.g. ideal $\rm\:I\:$ of $\rm\:D\:$ is cancellable, i.e. $\rm\:I\:J = I\:K\ \Rightarrow\ J = K\:$
(7) $\: $ (6) restricted to f.g. $\rm\:J,K.$
(8) $\rm\:D\:$ is integrally closed and there is an $\rm\:n > 1\:$ such that for all $\rm\: a,b \in D,\ (a,b)^n = (a^n,b^n).$
(9) $\rm\:D\:$ is integrally closed and there is an $\rm\: n > 1\:$ such that for all $\rm\:a,b \in D,\ a^{n-1} b \ \in\ (a^n, b^n).$
(10) Each ideal $\rm\:I\:$ of $\rm\:D\:$ is complete, i.e. $\rm\:I = \cap\ I\: V_j\:$ as $\rm\:V_j\:$ run over all the valuation overrings of $\rm\:D.\:$
(11) Each f.g. ideal of $\rm\:D\:$ is an intersection of valuation ideals.
(12) If $\rm\:I,J,K\:$ are nonzero ideals of $\rm\:D,\:$ then $\rm\:I \cap (J + K) = I\cap J + I\cap K.$
(13) If $\rm\:I,J,K\:$ are nonzero ideals of $\rm\:D,\:$ then $\rm\:I\ (J \cap K) = I\:J\cap I\:K.$
(14) If $\rm\:I,J\:$ are nonzero ideals of $\rm\:D,\:$ then $\rm\:(I + J)\ (I \cap J) = I\:J.\ $ ($\rm LCM\times GCD$ law)
(15) If $\rm\:I,J,K\:$ are nonzero ideals of $\rm\:D,\:$ with $\rm\:K\:$ f.g. then $\rm\:(I + J):K = I:K + J:K.$
(16) For any two elements $\rm\:a,b \in D,\ (a:b) + (b:a) = D.$
(17) If $\rm\:I,J,K\:$ are nonzero ideals of $\rm\:D\:$ with $\rm\:I,J\:$ f.g. then $\rm\:K:(I \cap J) = K:I + K:J.$
(18) $\rm\:D\:$ is integrally closed and each overring of $\rm\:D\:$ is the intersection of localizations of $\rm\:D.\:$
(19) $\rm\:D\:$ is integrally closed and each overring of $\rm\:D\:$ is the intersection of quotient rings of $\rm\:D.\:$
(20) Each overring of $\rm\:D\:$ is integrally closed.
(21) Each overring of $\rm\:D\:$ is flat over $\rm\:D.\:$
(22) $\rm\:D\:$ is integrally closed and prime ideals of overrings of are extensions of prime ideals of $\rm\:D.$
(23) $\rm\:D\:$ is integrally closed and for each prime ideal $\rm\:P\:$ of $\rm\:D,\:$ and each overring $\rm\:S\:$ of $\rm\:D,\:$ there is at most one prime ideal of $\rm\:S\:$ lying over $\rm\:P.\:$
(24) For polynomials $\rm\:f,g \in D[x],\ c(fg) = c(f)\: c(g)\:$ where for a polynomial $\rm\:h \in D[x],\ c(h)\:$ denotes the "content" ideal of $\rm\:D\:$ generated by the coefficients of $\rm\:h.\:$ (Gauss' Lemma)
(25) Ideals in $\rm\:D\:$ are integrally closed.
(26) If $\rm\:I,J\:$ are ideals with $\rm\:I\:$ f.g. then $\rm\: I\supset J\ \Rightarrow\ I|J.$ (contains $\:\Rightarrow\:$ divides)
(27) the Chinese Remainder Theorem $\rm(CRT)$ holds true in $\rm\:D\:,\:$ i.e. a system of congruences $\rm\:x\equiv x_j\ (mod\ I_j)\:$ is solvable iff $\rm\:x_j\equiv x_k\ (mod\ I_j + I_k).$

[1] Bazzoni and S. Glaz, Prüfer rings, Multiplicative Ideal Theory in Commutative Algebra, Springer-Verlag (2006), pp. 55–72.

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Well, only 27 equivalences? I read (but not checked) that Gilmer gave 40 is his book. – user26857 Dec 12 '12 at 11:13

This is the case, and as Jack says there are even better adjectives for the $R_P$. Perhaps the easiest way to see that these are principal is the following. Note that $R_P$ is still a Dedekind domain, so we have unique factorization of ideals; however, from the basics of localization we know that there is only one prime ideal $PR_P$. If we can show that this ideal is principal, then we are done.

For this, take an element $\pi$ (it's always called this) in $P - P^2$ and show that this is a generator for $PR_P$—what factorizations can $\pi R_P$ have?

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I'm assuming in this that your definition of Dedekind is "integrally closed Noetherian domain in which all non-zero prime ideals are maximal". As Qiaochu says, the $R_P$ being principal can be part of the definition. Working with the condition "ideals factor into primes" it seems less clear to me how to prove this directly. Equivalences between all of these are probably proved (in the cleverest way possible) in Serre's Local Fields. – Dylan Moreland Mar 13 '12 at 3:37

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