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A minimal prime over a principal nonzerodivisor has height one

Statement

Let R be a Noetherian commutative ring, let xR be a nonzerodivisor, and let p be a prime ideal minimal over (x). Then ht(p)=1.

Facts & Assumptions

Given: A Noetherian commutative ring R, a nonzerodivisor xR, and a prime ideal p minimal over (x).

[L1]

Every prime minimal over a principal ideal has height at most 1 (Krull's principal ideal theorem).

[L2]

The principal-ideal reduction passes to a Noetherian local domain whose maximal ideal is minimal over the image of x (Reduce the principal ideal theorem to a Noetherian local domain).

[L3]

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

technique · direct
1.1

By [L1], ht(p)1.

L1given
2.1

Apply [L2] to a minimal prime qp of R. Because x is a nonzerodivisor, xq, so qp. In the reduced local domain A=(R/q)p/q, the image of x lies in the maximal ideal. If that maximal ideal had height 0, then [L3] would make A a field, forcing x/1 to be a unit, contradiction. Hence the maximal ideal of A has height 1, so p has height at least 1.

L2L3step 1.1given
3.1

Steps 1.1 and 2.1 give ht(p)=1.

step 1.1step 2.1

Depends on

Used by

Dependency tree · two levels

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Sources