The Stacks project

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

If the last sequence were short exact, it would imply is an isomorphism. The sequence does appear to be exact, and the first nonzero term does have ML.

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