Dedekind domain

A Dedekind domain admits UFI

Let 𝑅 be an integral domain. Then 𝑅 admits unique factorization of ideals if¹ 𝑅 is a Dedekind domain ring

Dedekind implies UFI

Let 𝐼 ◃𝑅 be a nonzero proper ideal. First we show that if a prime factorization exists, it is necessarily unique. Suppose

𝐼=𝔭1⋯𝔭𝑟=𝔮1⋯𝔮𝑟

whence 𝔮1⋯𝔮𝑟 ⊆𝔭1, so without loss of generality 𝔮1 ⊆𝔭1 by ^D2. Since dim⁡𝑅 =1, 𝔮1 is maximal, whence 𝔮1 =𝔭1. Multiplying both sides by 𝔭−11 =𝔮−11 as a Product ideal gives

𝔭−11𝐼=𝔭2⋯𝔭𝑟=𝔮2⋯𝔮𝑟

since Prime ideals are invertible in a Dedekind domain, so we can induce that the factorization is unique.

To prove existence, we use the Noetherian property and Prime ideals are invertible in a Dedekind domain. Let I be the set of all ideals of 𝑅 for which there exists no prime factorization, and assume towards I ≠∅, whence there exists a maximal element 𝐼 ∈I. Now 𝐼 must be contained in a maximal ideal 𝔭 (which is prime), and since 𝑅 ⊆𝔭−1 we have

𝐼⊆𝐼𝔭−1⊆𝔭𝔭−1=𝑅

Since Prime ideals are invertible in a Dedekind domain guarantees 𝐼𝔭−1 ≠𝐼, it follows from the maximality of 𝐼 in I that 𝐼𝔭−1 has a prime factorization

𝐼𝔭−1=𝔭1⋯𝔭𝑟

whence

𝐼=𝐼𝔭−1𝔭=𝔭1⋯𝔭𝑟𝔭

is a prime factorization of 𝐼, i.e. 𝐼 ∉I, a contradiction.

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Footnotes

1. It seems to be possible to strengthen this to an iff. ↩