Lemma 15.104.19. Let $J$ be a set. For each $j \in J$ let $A_ j$ be a valuation ring with fraction field $K_ j$. Set $A = \prod A_ j$ and $K = \prod K_ j$. Then $A$ has weak dimension at most $1$ and $A \to K$ is a localization.

Proof. Let $I \subset A$ be a finitely generated ideal. By Lemma 15.104.18 it suffices to show that $I$ is a flat $A$-module. Let $I_ j \subset A_ j$ be the image of $I$. Observe that $I_ j = I \otimes _ A A_ j$, hence $I \to \prod I_ j$ is surjective by Algebra, Proposition 10.89.2. Thus $I = \prod I_ j$. Since $A_ j$ is a valuation ring, the ideal $I_ j$ is generated by a single element (Algebra, Lemma 10.50.15). Say $I_ j = (f_ j)$. Then $I$ is generated by the element $f = (f_ j)$. Let $e \in A$ be the idempotent which has a $0$ or $1$ in $A_ j$ depending on whether $f_ j$ is $0$ or not. Then $f = g e$ for some nonzerodivisor $g \in A$: take $g = (g_ j)$ with $g_ j = 1$ if $f_ j = 0$ and $g_ j = f_ j$ else. Thus $I \cong (e)$ as a module. We conclude $I$ is flat as $(e)$ is a direct summand of $A$. The final statement is true because $K = S^{-1}A$ where $S = \prod (A_ j \setminus \{ 0\} )$. $\square$

Comment #700 by Kestutis Cesnavicius on

'is has' --> 'has'

Comment #3576 by shanbei on

Fifth~sixth line:

"depending on whether f_j is 0 or 1" should be "depending on whether f_j is 0 or not"

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