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A minimal prime over a principal nonzerodivisor has height one
Statement
Let be a Noetherian commutative ring, let be a nonzerodivisor, and let be a prime ideal minimal over . Then .
Facts & Assumptions
Given: A Noetherian commutative ring , a nonzerodivisor , and a prime ideal minimal over .
Every prime minimal over a principal ideal has height at most (Krull's principal ideal theorem).
The principal-ideal reduction passes to a Noetherian local domain whose maximal ideal is minimal over the image of (Reduce the principal ideal theorem to a Noetherian local domain).
A Noetherian local domain has dimension zero exactly when it is a field (A Noetherian local domain has dimension zero exactly when it is a field).
Proof
By [L1], .
Apply [L2] to a minimal prime of . Because is a nonzerodivisor, , so . In the reduced local domain , the image of lies in the maximal ideal. If that maximal ideal had height , then [L3] would make a field, forcing to be a unit, contradiction. Hence the maximal ideal of has height , so has height at least .
Steps 1.1 and 2.1 give .
Depends on
Used by
Dependency tree · two levels
16 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
- Allen B. Altman and Steven L. Kleiman, A Term of Commutative Algebra, 13th ed., §21 (standard reference, not scraped)
- J. S. Milne, A Primer of Commutative Algebra, v4.03, §21 (standard reference, not scraped)
- Melvin Hochster, Dimension theory and systems of parameters (standard reference, not scraped)