Lemma 10.126.13. Suppose $R \to S$ is a local homomorphism of local rings. Assume that $S$ is essentially of finite presentation over $R$. Let $M$ be a finitely presented $S$-module. Then there exists a directed set $(\Lambda , \leq )$, and a system of local homomorphisms $R_\lambda \to S_\lambda $ of local rings together with $S_\lambda $-modules $M_\lambda $, such that

The colimit of the system $R_\lambda \to S_\lambda $ is equal to $R \to S$. The colimit of the system $M_\lambda $ is $M$.

Each $R_\lambda $ is essentially of finite type over $\mathbf{Z}$.

Each $S_\lambda $ is essentially of finite type over $R_\lambda $.

Each $M_\lambda $ is finite over $S_\lambda $.

For each $\lambda \leq \mu $ the map $S_\lambda \otimes _{R_\lambda } R_\mu \to S_\mu $ presents $S_\mu $ as the localization of $S_\lambda \otimes _{R_\lambda } R_\mu $ at a prime ideal.

For each $\lambda \leq \mu $ the map $M_\lambda \otimes _{S_\lambda } S_\mu \to M_\mu $ is an isomorphism.

**Proof.**
As in the proof of Lemma 10.126.11 we may first write $R = \mathop{\mathrm{colim}}\nolimits R_\lambda $ as a directed colimit of local $\mathbf{Z}$-algebras which are essentially of finite type. Next, we may assume that for some $\lambda _1 \in \Lambda $ there exist $f_{j, \lambda _1} \in R_{\lambda _1}[x_1, \ldots , x_ n]$ such that

\[ S = \mathop{\mathrm{colim}}\nolimits _{\lambda \geq \lambda _1} S_\lambda , \text{ with } S_\lambda = (R_\lambda [x_1, \ldots , x_ n]/ (f_{1, \lambda }, \ldots , f_{m, \lambda }))_{\mathfrak q_\lambda } \]

Choose a presentation

\[ S^{\oplus s} \to S^{\oplus t} \to M \to 0 \]

of $M$ over $S$. Let $A \in \text{Mat}(t \times s, S)$ be the matrix of the presentation. For some $\lambda _2 \in \Lambda $, $\lambda _2 \geq \lambda _1$ we can find a matrix $A_{\lambda _2} \in \text{Mat}(t \times s, S_{\lambda _2})$ which maps to $A$. For all $\lambda \geq \lambda _2$ we let $M_\lambda = \mathop{\mathrm{Coker}}(S_\lambda ^{\oplus s} \xrightarrow {A_\lambda } S_\lambda ^{\oplus t})$. We leave it to the reader to see that this works.
$\square$

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