Ideals of a Dedekind domain need at most two generators

Let $𝑅$ be a Dedekind domain and $𝐼 ⊴ 𝑅$ be an ideal. Then for any $𝛼 \in 𝐼$ , there exists a $𝛽 \in 𝐼$ such that $𝐼 = ⟨ 𝛼, 𝛽 ⟩$ . #m/thm/ring

Proof

It suffices to prove the case where $𝐼$ is not principal. Since A Dedekind domain admits UFI, we have

𝐼 = \prod 𝑖 𝔭 𝑎 𝑖 𝑖 .

for distinct prime ideals $𝔭 𝑖$ and exponents $𝑎 𝑖 \in ℕ$ . Choose some $𝛼 \in 𝐼$ and consider $⟨ 𝛼 ⟩ \subseteq 𝐼$ . Since A Dedekind domain is a CDR, we have

⟨ 𝛼 ⟩ = (\prod 𝑖 𝔭 𝑏 𝑖 𝑖) (\prod 𝑗 𝔮 𝑐 𝑗 𝑗)

where $𝔮 𝑖$ are distinct from each other and the $𝔭 𝑖$ and $𝑏 𝑖, 𝑐 𝑖 \in ℕ$ with $𝑎 𝑖 \leq 𝑏 𝑖$ .

We seek

𝛽 \in ⋂ 𝑖 (𝔭 𝑎 𝑖 𝑖 ∖ 𝔭 𝑎 𝑖 + 1 𝑖) ∖ ⋂ 𝑗 𝔮 𝑗 .

For each $𝔭 𝑖$ , let $𝑥 𝑖 \in 𝔭 𝑎 𝑖 𝑖 ∖ 𝔭 𝑎 𝑖 + 1 𝑖$ , and for each $𝔮 𝑖$ , let $𝑦 𝑖 \notin 𝔮 𝑖$ . By the Chinese remainder theorem for rings, we have a surjective homomorphism

𝜑 : 𝑅 ↠ ⨁ 𝑖 𝑅 𝔭 𝑎 𝑖 + 1 𝑖 \oplus ⨁ 𝑗 𝑅 𝔮 𝑖 .

Then any

𝛽 \in 𝜑 - 1 {(𝑥 1 + 𝔭 𝑎 1 + 1 1, \dots, 𝑦 1 + 𝔮 1, \dots)}

will do the trick.

We claim $𝐼 = ⟨ 𝛼, 𝛽 ⟩$ . Note

𝐼 = \prod 𝑖 𝔭 𝑎 𝑖 𝑖 ⊋ ⟨ 𝛼, 𝛽 ⟩ ⊋ (\prod 𝑖 𝔭 𝑏 𝑖 𝑖) (\prod 𝑗 𝔮 𝑐 𝑗 𝑗) = ⟨ 𝛼 ⟩ .

and also $⟨ 𝛼, 𝛽 ⟩ ⊋ ⟨ 𝛽 ⟩$ where $⟨ 𝛽 ⟩$ is not divisible by any $𝔭 𝑎 𝑖 + 1 𝑖$ or $𝔮 𝑗$ . Hence $⟨ 𝛼, 𝛽 ⟩ = 𝐼$ .

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