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For Dedekind ideals, divisibility reverses inclusion

Statement

Assume the Axiom of Choice.

Let $R$ be a Dedekind domain and let $I, J \subseteq R$ be nonzero integral ideals. Then $I \supseteq J$ if and only if $I$ divides $J$ .

Facts & Assumptions

Given: The Axiom of Choice, a Dedekind domain $R$ , and nonzero integral ideals $I, J$ .

[L1]

Nonzero fractional ideals factor uniquely into prime powers, and every integral ideal has nonnegative exponents (Unique factorization of nonzero fractional ideals into prime powers).

Proof

technique · direct

1.1L1givenalgebra

Write $I = \prod_{p} p^{a_{p}}$ and $J = \prod_{p} p^{b_{p}}$ with $a_{p}, b_{p} \geq 0$ by [L1]. Then $I \supseteq J$ exactly when $a_{p} \leq b_{p}$ for every $p$ , and that coordinatewise inequality is exactly the condition that $J = I \prod_{p} p^{b_{p} - a_{p}}$ .

2.1step 1.1∎

Therefore $I \supseteq J$ if and only if $I$ divides $J$ .

Depends on

Unique factorization of nonzero fractional ideals into prime powers

Used by

Every nonzero ideal in a Dedekind domain is generated by two elements Corollary

Dependency tree · two levels

5 results within two dependency steps of this one, each drawn at its shortest distance from it. An arrow runs from a result to what uses it, so the chart reads left to right and ends at this result, which carries a heavier outline. Every node is a link to that result. Click elsewhere on the chart to enlarge it.

Sources

J. P. May, Notes on Dedekind Rings (standard reference, not scraped)
J. S. Milne, A Primer of Commutative Algebra, §20 (standard reference, not scraped)