Alphabeta Math
CorollaryStatement: Literature-sourcedProof: AI-adaptedprecheck passaudited 2026-08-31
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Prime ideals and dimension of a DVR

Statement

Let V be a discrete valuation ring with maximal ideal m. Then the only prime ideals of V are (0) and m. In particular, dim⁡V=1.

Facts & Assumptions

Given: A discrete valuation ring V with maximal ideal m=(π).

[L1]

Every nonzero ideal of V is (πn)=mn for a unique n≥0 (Ideals in a DVR are powers of the maximal ideal).

[F1]

The Krull dimension of a nonzero ring is the supremum of the lengths of its strict chains of prime ideals (Krull dimension of a nonzero ring).

[A1]

A discrete valuation ring is a domain, so (0) is prime.

Proof

technique · direct
1.1L1givenalgebra

Let p be a nonzero prime ideal of V. Choose 0≠x∈p. By [L1], x=uπn for some unit u and some n≥1, so πn∈p. Since p is prime, π∈p. Therefore m=(π)⊆p, and maximality of m forces p=m.

2.1F1A1step 1.1∎

By [A1], (0) is prime, and step 1.1 shows there are no other nonzero primes besides m. Therefore the prime spectrum has exactly the strict chain (0)⊊m. Its length is 1, and no longer chain exists. Hence [F1] gives dim⁡V=1.

Depends on

Used by

Dependency tree · two levels

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Sources