
Lemma 15.90.5. Let $A$ be a ring. The following are equivalent

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

2. $A$ is absolutely flat, and

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

In this case every local ring of $A$ is a field.

Proof. The equivalence of (1) and (2) is immediate. Assume $A$ is absolutely flat. This implies every ideal of $A$ is pure, see Algebra, Definition 10.107.1. Hence every finitely generated ideal is generated by an idempotent by Algebra, Lemma 10.107.5. If $f \in A$, then $(f) = (e)$ for some idempotent $e \in A$ and $D(f) = D(e)$ is open and closed (Algebra, Lemma 10.20.1). This already implies every ideal of $A$ is maximal for example by Algebra, Lemma 10.25.5. Moreover, if $f$ is nilpotent, then $e = 0$ hence $f = 0$. Thus $A$ is reduced.

Assume $A$ is reduced and every prime of $A$ is maximal. Let $M$ be an $A$-module. Our goal is to show that $M$ is flat. We may write $M$ as a filtered colimit of finite $A$-modules, hence we may assume $M$ is finite (Algebra, Lemma 10.38.3). There is a finite filtration of $M$ by modules of the form $A/I$ (Algebra, Lemma 10.5.4), hence we may assume that $M = A/I$ (Algebra, Lemma 10.38.13). Thus it suffices to show every ideal of $A$ is pure. Since $A$ every local ring of $A$ is a field (by Algebra, Lemma 10.24.1 and the fact that every prime of $A$ is minimal), we see that every ideal $I \subset A$ is radical. Note that every closed subset of $\mathop{\mathrm{Spec}}(A)$ is closed under specialization. Thus every (radical) ideal of $A$ is pure by Algebra, Lemma 10.107.4. $\square$

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".

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