Lemma 18.42.4. Let $\mathcal{C}$ be a site. Let $\Lambda $ be a Noetherian ring. Let $I \subset \Lambda $ be an ideal. The sheaf $\underline{\Lambda }^\wedge = \mathop{\mathrm{lim}}\nolimits \underline{\Lambda /I^ n}$ is a flat $\underline{\Lambda }$-algebra. Moreover we have canonical identifications

\[ \underline{\Lambda }/I\underline{\Lambda } = \underline{\Lambda }/\underline{I} = \underline{\Lambda }^\wedge /I\underline{\Lambda }^\wedge = \underline{\Lambda }^\wedge /\underline{I} \cdot \underline{\Lambda }^\wedge = \underline{\Lambda }^\wedge /\underline{I}^\wedge = \underline{\Lambda /I} \]

where $\underline{I}^\wedge = \mathop{\mathrm{lim}}\nolimits \underline{I/I^ n}$.

**Proof.**
To prove $\underline{\Lambda }^\wedge $ is flat, it suffices to show that $\underline{\Lambda }^\wedge (U)$ is flat as a $\Lambda $-module for each $U \in \mathop{\mathrm{Ob}}\nolimits (\mathcal{C})$, see Lemmas 18.28.2 and 18.28.3. By Lemma 18.42.3 we see that

\[ \underline{\Lambda }^\wedge (U) = \mathop{\mathrm{lim}}\nolimits \underline{\Lambda /I^ n}(U) \]

is a limit of a system of flat $\Lambda /I^ n$-modules. By Lemma 18.42.1 we see that the transition maps are surjective. We conclude by More on Algebra, Lemma 15.27.4.

To see the equalities, note that $\underline{\Lambda }(U)/I\underline{\Lambda }(U) = \underline{\Lambda /I}(U)$ by Lemma 18.42.2. It follows that $\underline{\Lambda }/I\underline{\Lambda } = \underline{\Lambda }/\underline{I} = \underline{\Lambda /I}$. The system of short exact sequences

\[ 0 \to \underline{I/I^ n}(U) \to \underline{\Lambda /I^ n}(U) \to \underline{\Lambda /I}(U) \to 0 \]

has surjective transition maps, hence gives a short exact sequence

\[ 0 \to \mathop{\mathrm{lim}}\nolimits \underline{I/I^ n}(U) \to \mathop{\mathrm{lim}}\nolimits \underline{\Lambda /I^ n}(U) \to \mathop{\mathrm{lim}}\nolimits \underline{\Lambda /I}(U) \to 0 \]

see Homology, Lemma 12.31.3. Thus we see that $\underline{\Lambda }^\wedge /\underline{I}^\wedge = \underline{\Lambda /I}$. Since

\[ I \underline{\Lambda }^\wedge \subset \underline{I} \cdot \underline{\Lambda }^\wedge \subset \underline{I}^\wedge \]

it suffices to show that $I \underline{\Lambda }^\wedge (U) = \underline{I}^\wedge (U)$ for all $U$. Choose generators $I = (f_1, \ldots , f_ r)$. For every $n$ we obtain a short exact sequence

\[ 0 \to K_ n/(I^ n)^{\oplus r} \to (\Lambda /I^ n)^{\oplus r} \xrightarrow {(f_1, \ldots , f_ r)} I/I^{n + 1} \to 0 \]

where $K_ n = \{ (x_1, \ldots , x_ r) \in \Lambda ^{\oplus r} \mid \sum x_ i f_ i \in I^{n + 1}\} $. We obtain short exact sequences

\[ 0 \to \underline{K_ n/(I^ n)^{\oplus r}}(U) \to \underline{(\Lambda /I^ n)^{\oplus r}}(U) \to \underline{I/I^{n + 1}}(U) \to 0 \]

A calculation shows $K_ n = K + (I^ n)^{\oplus r}$, hence the transition maps $K_{n + 1}/(I^{n + 1})^{\oplus r} \to K_ n/(I^ n)^{\oplus r}$ are surjective. Hence the system of modules on the left hand side has surjective transition maps and a fortiori has ML. Thus we see that $(f_1, \ldots , f_ r) : (\underline{\Lambda }^\wedge )^{\oplus r}(U) \to \underline{I}^\wedge (U)$ is surjective by Homology, Lemma 12.31.3 which is what we wanted to show.
$\square$

## Comments (2)

Comment #6266 by Owen on

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