Lemma 15.104.5. Let $A$ be a ring. The following are equivalent

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

$A$ is absolutely flat, and

$A$ is reduced and every prime is maximal.

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

Lemma 15.104.5. Let $A$ be a ring. The following are equivalent

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

$A$ is absolutely flat, and

$A$ 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 10.108.1. 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 already implies every ideal of $A$ is maximal for example by Algebra, Lemma 10.26.5. 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 10.39.3). 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 10.39.13). Thus it suffices to show every ideal of $A$ is pure. Since every local ring of $A$ is a field (by Algebra, Lemma 10.25.1 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 generalization. Thus every (radical) ideal of $A$ is pure by Algebra, Lemma 10.108.4. $\square$

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## Comments (2)

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