One-dimensional Noetherian domain implies Cohen-Macaulay: Difference between revisions
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===Verbal statement=== | ===Verbal statement=== | ||
Any [[one-dimensional domain]] (i.e. a [[Noetherian domain]] whose [[Krull dimension]] is exactly one) is [[Cohen-Macaulay ring|Cohen-Macaulay]]. | Any [[one-dimensional Noetherian domain]] (i.e. a [[Noetherian domain]] whose [[Krull dimension]] is exactly one) is [[Cohen-Macaulay ring|Cohen-Macaulay]]. | ||
==Proof== | ==Proof== | ||
By definition, every maximal ideal has codimension one. | |||
Because any nonzero element is a nonzerodivisor, we can clearly, for every maximal ideal, pick a regular sequence of length one. Thus, the depth of any maximal ideal is one. | |||
Hence, for any maximal ideal, the depth equals the codimension, so the ring is Cohen-Macaulay. |
Latest revision as of 16:28, 12 May 2008
This article gives the statement and possibly, proof, of an implication relation between two commutative unital ring properties. That is, it states that every commutative unital ring satisfying the first commutative unital ring property must also satisfy the second commutative unital ring property
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Statement
Property-theoretic statement
The property of commutative unital rings of being a one-dimensional Noetherian domain is stronger than the property of being a Cohen-Macaulay ring.
Verbal statement
Any one-dimensional Noetherian domain (i.e. a Noetherian domain whose Krull dimension is exactly one) is Cohen-Macaulay.
Proof
By definition, every maximal ideal has codimension one.
Because any nonzero element is a nonzerodivisor, we can clearly, for every maximal ideal, pick a regular sequence of length one. Thus, the depth of any maximal ideal is one.
Hence, for any maximal ideal, the depth equals the codimension, so the ring is Cohen-Macaulay.