The Stacks project

Lemma 38.19.5. Let $A \to B$ be a local ring map of local rings which is essentially of finite type. Let $N$ be a finite $B$-module which is flat as an $A$-module. If $A$ is henselian, then $N$ is a filtered colimit

\[ N = \mathop{\mathrm{colim}}\nolimits _ i F_ i \]

of free $A$-modules $F_ i$ such that all transition maps $u_ i : F_ i \to F_{i'}$ of the system induce injective maps $\overline{u}_ i : F_ i/\mathfrak m_ AF_ i \to F_{i'}/\mathfrak m_ AF_{i'}$. Also, $N$ is a Mittag-Leffler $A$-module.

Proof. We can find a morphism of finite type $X \to S = \mathop{\mathrm{Spec}}(A)$ and a point $x \in X$ lying over the closed point $s$ of $S$ and a finite type quasi-coherent $\mathcal{O}_ X$-module $\mathcal{F}$ such that $\mathcal{F}_ x \cong N$ as an $A$-module. After shrinking $X$ we may assume that each point of $\text{Ass}_{X_ s}(\mathcal{F}_ s)$ specializes to $x$. By Lemma 38.19.4 we see that there exists a fundamental system of affine open neighbourhoods $U_ i \subset X$ of $x$ such that $\Gamma (U_ i, \mathcal{F})$ is a free $A$-module $F_ i$. Note that if $U_{i'} \subset U_ i$, then

\[ F_ i/\mathfrak m_ AF_ i = \Gamma (U_{i, s}, \mathcal{F}_ s) \longrightarrow \Gamma (U_{i', s}, \mathcal{F}_ s) = F_{i'}/\mathfrak m_ AF_{i'} \]

is injective because a section of the kernel would be supported at a closed subset of $X_ s$ not meeting $x$ which is a contradiction to our choice of $X$ above. Since the maps $F_ i \to F_{i'}$ are $A$-universally injective (Lemma 38.7.5) it follows that $N$ is Mittag-Leffler by Algebra, Lemma 10.89.9. $\square$

Comments (0)

Post a comment

Your email address will not be published.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.

In order to prevent bots from posting comments, we would like you to prove that you are human. You can do this by filling in the name of the current tag in the following input field. As a reminder, this is tag 05U7. Beware of the difference between the letter 'O' and the digit '0'.