The Stacks project

Lemma 15.105.5. Let $A$ be a ring. The following are equivalent

  1. $A$ has weak dimension $\leq 0$,

  2. $A$ is absolutely flat,

  3. $A$ is reduced and every prime is maximal, and

  4. every local ring of $A$ is a field.

Proof. The equivalence of (1) and (2) is immediate.

Proof of (1) $\Rightarrow $ (3). Assume $A$ is absolutely flat. Then every ideal of $A$ is pure, see Algebra, Definition 10.108.1. Hence every finitely generated ideal is generated by an idempotent by Algebra, Lemma 10.108.5. If $f \in A$, then $(f) = (e)$ for some idempotent $e \in A$ and $D(f) = D(e)$ is open and closed (Algebra, Lemma 10.21.1). This implies every prime ideal of $A$ is maximal by Algebra, Lemma 10.26.5. Moreover, if $f$ is nilpotent, then $e = 0$ hence $f = 0$. Thus $A$ is reduced.

Proof of (3) $\Rightarrow $ (4). If $A$ is reduced and every prime is maximal, then every local ring is a field by Algebra, Lemma 10.25.1.

Proof of (4) $\Rightarrow $ (2). If every local ring of $A$ is a field, then every $A$-module is flat by Algebra, Lemma 10.39.18. $\square$

Comments (4)

Comment #3575 by shanbei on

In the fourth to the last line,

Instead of "Since A every local ring of A...", it should be "Since every local ring of A".

In the second to the last line,

Instead of "... closed under specialization", it should be "... closed under generalization".

Comment #9737 by Olivier Benoist on

Isn't the consequence that all local rings of A are fields equivalent to the other assertions? (It seems to me that it implies (3)). In this case, it might be useful to state it as a fourth equivalent condition.

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