ExampleConstruction: AI-adaptedVerification: Literature-sourcedverified 2026-09-24 (gpt-6-sol)

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A principal ideal generated by a zero divisor can have a minimal prime of height zero

Example

In the ring $R = k [x, y] / (x y)$ the element $\overset{ˉ}{x}$ is a zero divisor, and the prime ideal $(\overset{ˉ}{x})$ is minimal over the principal ideal $(\overset{ˉ}{x})$ while having height $0$ .

Facts & Assumptions

Given: A field $k$ and the ring $R = k [x, y] / (x y)$ .

[L1]

Minimal primes are exactly the primes of height zero (Minimal primes are exactly the primes of height zero).

Verification

technique · direct computation

1.1givenalgebra

In $R$ , one has $\overset{ˉ}{x} \overset{ˉ}{y} = 0$ with both factors nonzero, so $\overset{ˉ}{x}$ is a zero divisor. Also $R / (\overset{ˉ}{x}) ≅ k [y]$ is a domain, hence $(\overset{ˉ}{x})$ is a prime ideal minimal over the principal ideal $(\overset{ˉ}{x})$ .

2.1L1step 1.1

The ideal $(\overset{ˉ}{x})$ is one of the two minimal primes of $k [x, y] / (x y)$ , so [L1] gives $ht ((\overset{ˉ}{x})) = 0$ . Therefore the exact-height-one conclusion fails for a zerodivisor generator.

3.1step 1.1step 2.1∎

So a principal ideal generated by a zero divisor can indeed have a minimal prime of height zero.

Depends on

Minimal primes are exactly the primes of height zero

Used by

Nothing in the library uses this result yet.

Dependency tree · two levels

2 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. S. Milne, A Primer of Commutative Algebra, v4.03, §§18, 21 (standard reference, not scraped)
Melvin Hochster, Dimension theory and systems of parameters (standard reference, not scraped)