Lemma 15.91.5. 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 15.91.5. 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 10.107.1. 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 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 generalization. Thus every (radical) ideal of $A$ is pure by Algebra, Lemma 10.107.4. $\square$

Your email address will not be published. Required fields are marked.

In your comment you can use Markdown and LaTeX style mathematics (enclose it like `$\pi$`

). A preview option is available if you wish to see how it works out (just click on the eye in the toolbar).

Unfortunately JavaScript is disabled in your browser, so the comment preview function will not work.

All contributions are licensed under the GNU Free Documentation License.

## Comments (2)

Comment #3575 by shanbei on

Comment #3699 by Johan on