A Dedekind domain with finitely many prime ideals is a UFD

Let $𝑅$ be a Dedekind domain with finitely many prime ideals. Then every prime ideal is principal, whence $𝑅$ is a PID and in particular a UFD. #m/thm/ring

Proof

Let ${𝔭 𝑖} 𝑛 𝑖 = 1$ enumerate all prime ideals in $𝑅$ . By a similar construction to that in the proof of Ideals of a Dedekind domain need at most two generators, we can choose for each $𝑖$ a $𝛽 𝑖 \in 𝔭 𝑖$ which is not in $𝔭 2 𝑖$ or $𝔭 𝑗$ for $𝑗 \neq 𝑖$ via the Chinese remainder theorem for rings. It follows that $⟨ 𝛽 𝑖 ⟩ = 𝔭 𝑖$ .

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