proof of Dedekind domains with finitely many primes are PIDs

Proof.

Let $\\mathfrak{p}_{1},\\ldots,\\mathfrak{p}_{k}$ be all the primes of a Dedekind domain $R$ . If $I$ is any ideal of $R$ , then by the Weak Approximation Theorem we can choose $x\\in R$ such that $\u_{\\mathfrak{p}_{i}}((x))=\u_{\\mathfrak{p}_{i}}(I)$ for all $i$ (where $\u_{\\mathfrak{p}}$ is the $\\mathfrak{p}$ -adic valuation). But since $R$ is Dedekind, ideals have unique factorization; since $(x)$ and $I$ have identical factorizations, we must have $(x)=I$ and $I$ is principal.∎

单词	ProofOfDedekindDomainsWithFinitelyManyPrimesArePIDs
释义	proof of Dedekind domains with finitely many primes are PIDs Proof. Let $\\mathfrak{p}_{1},\\ldots,\\mathfrak{p}_{k}$ be all the primes of a Dedekind domain $R$ . If $I$ is any ideal of $R$ , then by the Weak Approximation Theorem we can choose $x\\in R$ such that $\u_{\\mathfrak{p}_{i}}((x))=\u_{\\mathfrak{p}_{i}}(I)$ for all $i$ (where $\u_{\\mathfrak{p}}$ is the $\\mathfrak{p}$ -adic valuation). But since $R$ is Dedekind, ideals have unique factorization; since $(x)$ and $I$ have identical factorizations, we must have $(x)=I$ and $I$ is principal.∎
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