The Stacks project

Proof. Since sheafification is exact it is clear that $0 \to \underline{A} \to \underline{B} \to \underline{C} \to 0$ is an exact sequence of abelian sheaves. Thus $0 \to \underline{A} \to \underline{B} \to \underline{C}$ is an exact sequence of abelian presheaves. To see that $\underline{B} \to \underline{C}$ is surjective, pick a set theoretical section $s : C \to B$. This induces a section $\underline{s} : \underline{C} \to \underline{B}$ of sheaves of sets left inverse to the surjection $\underline{B} \to \underline{C}$. $\square$

Proof. Choose a presentation $\Lambda ^{\oplus m} \to \Lambda ^{\oplus n} \to Q \to 0$. This gives an exact sequence $M^{\oplus m} \to M^{\oplus n} \to M \otimes Q \to 0$. By Lemma 18.42.1 we obtain an exact sequence

which proves the lemma. (Note that taking sections over $U$ always commutes with finite direct sums, but not arbitrary direct sums.) $\square$

Proof. Let $I \subset \Lambda$ be a finitely generated ideal. By Algebra, Lemma 10.39.5 it suffices to show that $\underline{M}(U) \otimes _\Lambda I \to \underline{M}(U)$ is injective. As $\Lambda$ is coherent $I$ is finitely presented as a $\Lambda$-module. By Lemma 18.42.2 we see that $\underline{M}(U) \otimes I = \underline{M \otimes I}$. Since $M$ is flat the map $M \otimes I \to M$ is injective, whence $\underline{M \otimes I} \to \underline{M}$ is injective. $\square$

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

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

has surjective transition maps, hence gives a short exact sequence

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

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

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

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. $\square$

Proof. Proof of (1). If $M$ is generated by $x_1, \ldots , x_ r$ then $x_1, \ldots , x_ r$ define global sections of $\underline{M}$ which generate it, hence $\underline{M}$ is of finite type. Conversely, assume $\underline{M}$ is of finite type. Let $U \in \mathcal{C}$ be an object which is not sheaf theoretically empty (Sites, Definition 7.42.1). Such an object exists as we assumed $\mathop{\mathit{Sh}}\nolimits (\mathcal{C})$ is not the empty topos. Then there exists a covering $\{ U_ i \to U\}$ and finitely many sections $s_{ij} \in \underline{M}(U_ i)$ generating $\underline{M}|_{U_ i}$. After refining the covering we may assume that $s_{ij}$ come from elements $x_{ij}$ of $M$. Then $x_{ij}$ define global sections of $\underline{M}$ whose restriction to $U$ generate $\underline{M}$.

Assume there exist elements $x_1, \ldots , x_ r$ of $M$ which define global sections of $\underline{M}$ generating $\underline{M}$ as a sheaf of $\underline{\Lambda }$-modules. We will show that $x_1, \ldots , x_ r$ generate $M$ as a $\Lambda$-module. Let $x \in M$. We can find a covering $\{ U_ i \to U\} _{i \in I}$ and $f_{i, j} \in \underline{\Lambda }(U_ i)$ such that $x|_{U_ i} = \sum f_{i, j} x_ j|_{U_ i}$. After refining the covering we may assume $f_{i, j} \in \Lambda$. Since $U$ is not sheaf theoretically empty, there is at least one $i \in I$ such that $U_ i$ is not sheaf theoretically empty. Then the map $M \to \underline{M}(U_ i)$ is injective (details omitted). We conclude that $x = \sum f_{i, j}x_ j$ in $M$ as desired.

Proof of (2). Assume $\underline{M}$ is a $\underline{\Lambda }$-module of finite presentation. By (1) we see that $M$ is of finite type. Choose generators $x_1, \ldots , x_ r$ of $M$ as a $\Lambda$-module. This determines a short exact sequence $0 \to K \to \Lambda ^{\oplus r} \to M \to 0$ which turns into a short exact sequence

by Lemma 18.42.1. By Lemma 18.24.1 we see that $\underline{K}$ is of finite type. Hence $K$ is a finite $\Lambda$-module by (1). Thus $M$ is a $\Lambda$-module of finite presentation. $\square$