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 $I = (f_1, \ldots , f_ r)$. This gives a short exact sequence $0 \to K \to \Lambda ^{\oplus r} \to I \to 0$. We obtain short exact sequences

$0 \to \underline{(K \cap I^ n)/I^ nK}(U) \to \underline{(\Lambda /I^ n)^{\oplus r}}(U) \to \underline{I/I^ n}(U) \to 0$

By Artin-Rees (Algebra, Lemma 10.51.2) the system of modules on the left hand side has ML. (It is zero as a pro-object.) Thus we see that $(\underline{\Lambda }^\wedge )^{\oplus r}(U) \to \underline{I}^\wedge (U)$ is surjective by Homology, Lemma 12.31.3 which is what we wanted to show. $\square$

## Comments (1)

Comment #6266 by Owen on

If the last sequence were short exact, it would imply $(\Lambda^\wedge)^{\oplus r}\to I^\wedge$ is an isomorphism. The sequence $0\to K/(K\cap (I^n)^{\oplus r})\to(\Lambda/I^n)^{\oplus r}\to I/I^n\to0$ does appear to be exact, and the first nonzero term does have ML.

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