A Dedekind domain admits UFI

Let $𝑅$ be an integral domain. Then $𝑅$ admits unique factorization of ideals if¹ $𝑅$ is a Dedekind domain #m/thm/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 = 𝔭 1 \dots 𝔭 𝑟 = 𝔮 1 \dots 𝔮 𝑟

whence $𝔮 1 \dots 𝔮 𝑟 \subseteq 𝔭 1$ , so without loss of generality $𝔮 1 \subseteq 𝔭 1$ by ^D2. Since $d i m 𝑅 = 1$ , $𝔮 1$ is maximal, whence $𝔮 1 = 𝔭 1$ . Multiplying both sides by $𝔭 - 1 1 = 𝔮 - 1 1$ as a Product ideal gives

𝔭 - 1 1 𝐼 = 𝔭 2 \dots 𝔭 𝑟 = 𝔮 2 \dots 𝔮 𝑟

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 $I \neq \emptyset$ , whence there exists a maximal element $𝐼 \in I$ . Now $𝐼$ must be contained in a maximal ideal $𝔭$ (which is prime), and since $𝑅 \subseteq 𝔭 - 1$ we have

𝐼 \subseteq 𝐼 𝔭 - 1 \subseteq 𝔭 𝔭 - 1 = 𝑅

Since Prime ideals are invertible in a Dedekind domain guarantees $𝐼 𝔭 - 1 \neq 𝐼$ , it follows from the maximality of $𝐼$ in $I$ that $𝐼 𝔭 - 1$ has a prime factorization

𝐼 𝔭 - 1 = 𝔭 1 \dots 𝔭 𝑟

whence

𝐼 = 𝐼 𝔭 - 1 𝔭 = 𝔭 1 \dots 𝔭 𝑟 𝔭

is a prime factorization of $𝐼$ , i.e. $𝐼 \notin I$ , a contradiction.

#state/tidy | #lang/en | #SemBr

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

A Dedekind domain admits UFI

Footnotes