Theorem
An integral domain R is a Dedekind domain if and only if R is a hereditary ring.
First, let’s recall the definitions:
- Integral Domain: A commutative ring with unity (1=0) and no zero divisors.
- Dedekind Domain: An integral domain R which satisfies the following three conditions:
- R is Noetherian.
- R is integrally closed in its field of fractions K.
- Every non-zero prime ideal of R is maximal.
- Hereditary Ring: A ring R (not necessarily commutative) such that every (left/right) ideal is a projective module. Since we are in the context of commutative integral domains, this means every ideal I⊆R is projective as an R-module.
- Projective Module: An R-module P is projective if for any surjective R-module homomorphism f:M→N and any R-module homomorphism g:P→N, there exists an R-module homomorphism h:P→M such that f∘h=g.
- Invertible Ideal: For an integral domain R with field of fractions K, a non-zero fractional ideal I is invertible if there exists a fractional ideal J such that IJ=R. A crucial fact is that for an ideal I⊆R in an integral domain R, I is projective if and only if it is invertible (as long as I={0}). The zero ideal {0}is always projective.
We need to prove both directions of the equivalence.
Proof: Dedekind Domain ⟹ Hereditary Ring
Assume R is a Dedekind domain. We want to show that R is a hereditary ring, meaning every ideal I⊆Ris projective.
- The zero ideal I={0} is always projective (it’s a direct summand of the free module R0={0}).
- Let I be a non-zero ideal of R.
- A fundamental property of Dedekind domains is that every non-zero fractional ideal is invertible.
- Since I is a non-zero ideal contained in R, it is a non-zero fractional ideal.
- Therefore, by the property of Dedekind domains, I is invertible.
- As mentioned in the definitions, for a non-zero ideal I in an integral domain R, I is projective if and only if it is invertible.
- Since I is invertible, it must be projective.
- Since this holds for any non-zero ideal I, and the zero ideal is also projective, every ideal of R is projective.
- Therefore, by definition, R is a hereditary ring.
Proof: Hereditary Ring ⟹ Dedekind Domain
Assume R is an integral domain that is also a hereditary ring. We want to show that R is a Dedekind domain by verifying the three conditions.
Since R is hereditary, every ideal I⊆R is projective. For non-zero ideals, this means they are invertible.
- R is Noetherian:
-
An ideal I is invertible only if it is finitely generated.
-
Proof: If I is invertible, II−1=R. This means 1∈II−1, so 1=∑i=1naibi for some ai∈I and bi∈I−1. For any x∈I, we have x=x⋅1=x∑aibi=∑(xbi)ai. Since x∈I and bi∈I−1, each xbi∈II−1=R. Thus, x is an R-linear combination of a1,…,an. This shows that I is generated by {a1,…,an}, so I is finitely generated.
-
Since R is hereditary, every ideal (including non-zero ones) is projective, and thus every non-zero ideal is invertible.
-
Therefore, every non-zero ideal is finitely generated. The zero ideal is also finitely generated (by 0).
-
A ring in which every ideal is finitely generated is, by definition, Noetherian.
- Every non-zero prime ideal P of R is maximal:
-
Let P be a non-zero prime ideal of R.
-
Suppose there exists an ideal M such that P⊊M⊆R. We want to show M=R.
-
Since R is hereditary, both P and M are projective (as they are ideals). Since P={0}, P is invertible. If M=R, it’s trivially invertible. If M=R and M={0}, then M is also invertible.
-
Because ideals in R are invertible (if non-zero), we have the property that for ideals A,B, A⊆B implies A=BC for some integral ideal C. (Specifically, C=B−1A).
-
Applying this to P⊂M, we get P=MI for some integral ideal I=M−1P.
-
Since P is a prime ideal and P=MI, by the definition of prime ideal, we must have M⊆P or I⊆P.
-
We assumed P⊊M, so M⊆P.
-
Therefore, we must have I⊆P, which means M−1P⊆P.
-
Multiply by M: M(M−1P)⊆MP, which simplifies to P⊆MP.
-
Since P is an ideal in R, MP⊆P always holds.
-
Thus, we have P=MP.
-
We already established that R is Noetherian, so P is finitely generated.
-
By Nakayama’s Lemma (applied to the finitely generated R-module P), the condition P=MP implies there exists an element m∈M such that (1−m)P={0}.
-
Since R is an integral domain and P is a non-zero ideal, we must have 1−m=0.
-
So, m=1. Since m∈M, this means 1∈M.
-
An ideal containing 1 must be the entire ring, so M=R.
-
This shows that there is no ideal strictly between P and R. Therefore, P is a maximal ideal.
- R is integrally closed:
-
We will use the property that an integral domain R is a Dedekind domain if and only if its localization RM at every maximal ideal M is a Discrete Valuation Ring (DVR). We already know R is Noetherian and its non-zero primes are maximal.
-
Let M be any maximal ideal of R. Consider the localization RM.
-
RM is a local Noetherian integral domain (localization preserves these properties). Its unique maximal ideal is MRM.
-
Since R is hereditary, the ideal M is projective in R. Projectivity is preserved under localization, so MRM is a projective RM-module.
-
In a local ring (A,m), an ideal I is projective if and only if it is principal (and non-zero).
-
Proof Sketch: If I is projective, it’s invertible. Let II−1=A. Then 1=∑xiyi with xi∈I,yi∈I−1. Since A is local, not all xiyi can be in the maximal ideal m, so at least one, say x1y1, must be a unit in A. Then y1=ux1−1 for some unit u. For any x∈I, xy1∈II−1=A. So x(ux1−1)∈A, which implies x∈(x1)A. Thus I⊆(x1)A. Since x1∈I, we have I=(x1)A, so I is principal. The converse (principal ⟹ projective in a domain) is also true.
-
Applying this to RM, the maximal ideal MRM is projective, hence it must be principal.
-
A local Noetherian integral domain whose maximal ideal is principal is a DVR.
-
Thus, RM is a DVR for every maximal ideal M of R.
-
An integral domain R is integrally closed if (and only if) RM is integrally closed for all maximal ideals M. Since every DVR is integrally closed, RM is integrally closed for all M.
-
Alternatively, it’s a known result that if RM is a DVR for all maximal M, then R itself is integrally closed (because R=⋂M maximalRM).
-
Therefore, R is integrally closed.
Conclusion: We have shown that if R is a hereditary integral domain, it is Noetherian, every non-zero prime ideal is maximal, and it is integrally closed. Therefore, R is a Dedekind domain.
Combining both directions, we have proven that an integral domain R is a Dedekind domain if and only if it is a hereditary ring.