Lemma 15.13.3. Let $A = \mathop{\mathrm{lim}}\nolimits A_ n$ be a limit of an inverse system $(A_ n)$ of rings. Suppose given $A_ n$-modules $M_ n$ and $A_{n + 1}$-module maps $M_{n + 1} \to M_ n$. Assume
the transition maps $A_{n + 1} \to A_ n$ are surjective with locally nilpotent kernels,
$M_1$ is a finite projective $A_1$-module,
$M_ n$ is a finite flat $A_ n$-module, and
the maps induce isomorphisms $M_{n + 1} \otimes _{A_{n + 1}} A_ n \to M_ n$.
Then $M = \mathop{\mathrm{lim}}\nolimits M_ n$ is a finite projective $A$-module and $M \otimes _ A A_ n \to M_ n$ is an isomorphism for all $n$.
Proof.
By Lemma 15.11.3 the pair $(A, \mathop{\mathrm{Ker}}(A \to A_1))$ is henselian. By Lemma 15.13.1 we can choose a finite projective $A$-module $P$ and an isomorphism $P \otimes _ A A_1 \to M_1$. Since $P$ is projective, we can successively lift the $A$-module map $P \to M_1$ to $A$-module maps $P \to M_2$, $P \to M_3$, and so on. Thus we obtain a map
\[ P \longrightarrow M \]
Since $P$ is finite projective, we can write $A^{\oplus m} = P \oplus Q$ for some $m \geq 0$ and $A$-module $Q$. Since $A = \mathop{\mathrm{lim}}\nolimits A_ n$ we conclude that $P = \mathop{\mathrm{lim}}\nolimits P \otimes _ A A_ n$. Hence, in order to show that the displayed $A$-module map is an isomorphism, it suffices to show that the maps $P \otimes _ A A_ n \to M_ n$ are isomorphisms. From Lemma 15.3.4 we see that $M_ n$ is a finite projective module. By Lemma 15.3.5 the maps $P \otimes _ A A_ n \to M_ n$ are isomorphisms.
$\square$
Comments (0)
There are also: