The Stacks project

Proof. Let $\text{Mod}^{fg}_{A, I}$ be the category of inverse systems $(M_ n)$ of finite $A$-modules satisfying: (1) $M_ n$ is annihilated by $I^ n$ and (2) $M_{n + 1}/I^ nM_{n + 1} = M_ n$. By the correspondence between coherent sheaves on $X$ and finite $A$-modules (Lemma 30.9.1) it suffices to show $\text{Mod}^{fg}_{A, I}$ is equivalent to the category of finite $A^\wedge $-modules. To see this it suffices to prove that given an object $(M_ n)$ of $\text{Mod}^{fg}_{A, I}$ the module

is a finite $A^\wedge $-module and that $M/I^ nM = M_ n$. As the transition maps are surjective, we see that $M \to M_1$ is surjective. Pick $x_1, \ldots , x_ t \in M$ which map to generators of $M_1$. This induces a map of systems $(A/I^ n)^{\oplus t} \to M_ n$. By Nakayama's lemma (Algebra, Lemma 10.20.1) these maps are surjective. Let $K_ n \subset (A/I^ n)^{\oplus t}$ be the kernel. Property (2) implies that $K_{n + 1} \to K_ n$ is surjective, in particular the system $(K_ n)$ satisfies the Mittag-Leffler condition. By Homology, Lemma 12.31.3 we obtain an exact sequence $0 \to K \to (A^\wedge )^{\oplus t} \to M \to 0$ with $K = \mathop{\mathrm{lim}}\nolimits K_ n$. Hence $M$ is a finite $A^\wedge $-module. As $K \to K_ n$ is surjective it follows that

as desired. $\square$

Proof. Let $\alpha =(\alpha _ n) : (\mathcal{F}_ n) \to (\mathcal{G}_ n)$ be a morphism of $\textit{Coh}(X, \mathcal{I})$. The cokernel of $\alpha $ is the inverse system $(\mathop{\mathrm{Coker}}(\alpha _ n))$ (details omitted). To describe the kernel let

for $l \geq m$. We claim:

1. the inverse system $(\mathcal{K}'_{l, m})_{l \geq m}$ is eventually constant, say with value $\mathcal{K}'_ m$,

2. the system $(\mathcal{K}'_ m/\mathcal{I}^ n\mathcal{K}'_ m)_{m \geq n}$ is eventually constant, say with value $\mathcal{K}_ n$,

3. the system $(\mathcal{K}_ n)$ forms an object of $\textit{Coh}(X, \mathcal{I})$, and

4. this object is the kernel of $\alpha $.

To see (a), (b), and (c) we may work affine locally, say $X = \mathop{\mathrm{Spec}}(A)$ and $\mathcal{I}$ corresponds to the ideal $I \subset A$. By Lemma 30.23.1 $\alpha $ corresponds to a map $f : M \to N$ of finite $A^\wedge $-modules. Denote $K = \mathop{\mathrm{Ker}}(f)$. Note that $A^\wedge $ is a Noetherian ring (Algebra, Lemma 10.97.6). Choose an integer $c \geq 0$ such that $K \cap I^ n M \subset I^{n - c}K$ for $n \geq c$ (Algebra, Lemma 10.51.2) and which satisfies Algebra, Lemma 10.51.3 for the map $f$ and the ideal $I^\wedge = IA^\wedge $. Then $\mathcal{K}'_{l, m}$ corresponds to the $A$-module

where the last equality holds if $l \geq m + c$. So $\mathcal{K}'_ m$ corresponds to the $A$-module $K/K \cap I^ mM$ and $\mathcal{K}'_ m/\mathcal{I}^ n\mathcal{K}'_ m$ corresponds to

for $m \geq n + c$ by our choice of $c$ above. Hence $\mathcal{K}_ n$ corresponds to $K/I^ nK$.

We prove (d). It is clear from the description on affines above that the composition $(\mathcal{K}_ n) \to (\mathcal{F}_ n) \to (\mathcal{G}_ n)$ is zero. Let $\beta : (\mathcal{H}_ n) \to (\mathcal{F}_ n)$ be a morphism such that $\alpha \circ \beta = 0$. Then $\mathcal{H}_ l \to \mathcal{F}_ l$ maps into $\mathop{\mathrm{Ker}}(\alpha _ l)$. Since $\mathcal{H}_ m = \mathcal{H}_ l/\mathcal{I}^ m\mathcal{H}_ l$ for $l \geq m$ we obtain a system of maps $\mathcal{H}_ m \to \mathcal{K}'_{l, m}$. Thus a map $\mathcal{H}_ m \to \mathcal{K}_ m'$. Since $\mathcal{H}_ n = \mathcal{H}_ m/\mathcal{I}^ n\mathcal{H}_ m$ we obtain a system of maps $\mathcal{H}_ n \to \mathcal{K}'_ m/\mathcal{I}^ n\mathcal{K}'_ m$ and hence a map $\mathcal{H}_ n \to \mathcal{K}_ n$ as desired.

To finish the proof of (1) we still have to show that $\mathop{\mathrm{Coim}}= \mathop{\mathrm{Im}}$ in $\textit{Coh}(X, \mathcal{I})$. We have seen above that taking kernels and cokernels commutes, over affines, with the description of $\textit{Coh}(X, \mathcal{I})$ as a category of modules. Since $\mathop{\mathrm{Im}}= \mathop{\mathrm{Coim}}$ holds in the category of modules this gives $\mathop{\mathrm{Coim}}= \mathop{\mathrm{Im}}$ in $\textit{Coh}(X, \mathcal{I})$. Parts (2) and (3) of the lemma are immediate from our construction of kernels and cokernels. $\square$

Proof. Omitted. Hint: Look on affine opens, use Lemma 30.23.1, and use Algebra, Lemma 10.20.1. $\square$

Proof. It suffices to check this locally on $X$. Hence we may assume $X$ is affine, i.e., we have a situation as in Lemma 30.23.1. The functor is the functor $\text{Mod}^{fg}_ A \to \text{Mod}^{fg}_{A^\wedge }$ which associates to a finite $A$-module $M$ the completion $M^\wedge $. Thus the result follows from Algebra, Lemma 10.97.2. $\square$

Proof. To prove this we may work affine locally on $X$. Hence we may assume $X = \mathop{\mathrm{Spec}}(A)$ and $\mathcal{F}$, $\mathcal{G}$ given by finite $A$-module $M$ and $N$. Then $\mathcal{H}$ corresponds to the finite $A$-module $H = \mathop{\mathrm{Hom}}\nolimits _ A(M, N)$. The statement of the lemma becomes the statement

via the equivalence of Lemma 30.23.1. By Algebra, Lemma 10.97.2 (used 3 times) we have

where the second equality uses that $A^\wedge $ is flat over $A$ (see More on Algebra, Lemma 15.65.4). The lemma follows. $\square$

Proof. Proof of (1). The uniqueness implies it suffices to construct $(\mathcal{F}, a, \beta )$ Zariski locally on $X$. Thus we may assume $X = \mathop{\mathrm{Spec}}(A)$ and $\mathcal{I}$ corresponds to the ideal $I \subset A$. In this situation Lemma 30.23.1 applies. Let $M'$ be the finite $A^\wedge $-module corresponding to $(\mathcal{F}_ n)$. Let $N$ be the finite $A$-module corresponding to $\mathcal{G}$. Then $\alpha $ corresponds to a map

whose kernel and cokernel are annihilated by $I^ t$ for some $t$. Recall that $N^\wedge = N \otimes _ A A^\wedge $ (Algebra, Lemma 10.97.1). By More on Algebra, Lemma 15.89.17 there is an $A$-module map $\psi : M \to N$ whose kernel and cokernel are $I$-power torsion and an isomorphism $M \otimes _ A A^\wedge = M'$ compatible with $\varphi $. As $N$ and $M'$ are finite modules, we conclude that $M$ is a finite $A$-module, see More on Algebra, Remark 15.89.20. Hence $M \otimes _ A A^\wedge = M^\wedge $. We omit the verification that the triple $(M, N \to M, M^\wedge \to M')$ so obtained is unique up to unique isomorphism.

The proof of (2) is exactly the same and we omit it. $\square$

Proof. Suppose $(\mathcal{F}_ n)$ is an object of $\textit{Coh}(X, \mathcal{I})$ which is annihilated by $\mathcal{I}^ c$ for some $c \geq 1$. Then $\mathcal{F}_ n \to \mathcal{F}_ c$ is an isomorphism for $n \geq c$. Hence if we set $\mathcal{F} = \mathcal{F}_ c$, then we see that $\mathcal{F}^\wedge \cong (\mathcal{F}_ n)$. This proves the first assertion.

Let $\mathcal{F}$, $\mathcal{G}$ be objects of $\textit{Coh}(\mathcal{O}_ X)$ such that either $\mathcal{F}$ or $\mathcal{G}$ is annihilated by $\mathcal{I}^ c$ for some $c \geq 1$. Then $\mathcal{H} = \mathop{\mathcal{H}\! \mathit{om}}\nolimits _{\mathcal{O}_ X}(\mathcal{G}, \mathcal{F})$ is a coherent $\mathcal{O}_ X$-module annihilated by $\mathcal{I}^ c$. Hence we see that

see Lemma 30.23.5. This proves the statement on homomorphisms.

The notation $\mathop{\mathrm{Ext}}\nolimits $ refers to extensions as defined in Homology, Section 12.6. The injectivity of the map on $\mathop{\mathrm{Ext}}\nolimits $'s follows immediately from the bijectivity of the map on $\mathop{\mathrm{Hom}}\nolimits $'s. For surjectivity, assume $\mathcal{F}$ is annihilated by a power of $I$. Then part (1) of Lemma 30.23.6 shows that given an extension

in $\textit{Coh}(U, I\mathcal{O}_ U)$ the morphism $\mathcal{G}^\wedge \to (\mathcal{E}_ n)$ is isomorphic to $\mathcal{G} \to \mathcal{E}^\wedge $ for some $\mathcal{G} \to \mathcal{E}$ in $\textit{Coh}(\mathcal{O}_ U)$. Similarly in the other case. $\square$

Proof. The question is local on $X$ hence we may assume $X$ is affine, i.e., we have a situation as in Lemma 30.23.1. In this case, if $(\mathcal{F}_ n)$ corresponds to the finite $A^\wedge $ module $M$, then $\bigoplus \mathop{\mathrm{Ker}}(\mathcal{F}_{n + 1} \to \mathcal{F}_ n)$ corresponds to $\bigoplus I^ nM/I^{n + 1}M$ which is clearly a finite module over $\bigoplus I^ n/I^{n + 1}$. $\square$

Proof. Since $f^* : \textit{Coh}(\mathcal{O}_ Y) \to \textit{Coh}(\mathcal{O}_ X)$ is right exact we have

hence the pullback of a system is a system. The construction of cokernels in the proof of Lemma 30.23.2 shows that $f^* : \textit{Coh}(Y, \mathcal{J}) \to \textit{Coh}(X, \mathcal{I})$ is always right exact. If $f$ is flat, then $f^* : \textit{Coh}(\mathcal{O}_ Y) \to \textit{Coh}(\mathcal{O}_ X)$ is an exact functor. It follows from the construction of kernels in the proof of Lemma 30.23.2 that in this case $f^* : \textit{Coh}(Y, \mathcal{J}) \to \textit{Coh}(X, \mathcal{I})$ also transforms kernels into kernels. $\square$

Proof. If $X$ and $X'$ are affine, then this follows immediately from More on Algebra, Lemma 15.89.3. To prove it in general we let $Z_ n \subset X$, $Z'_ n \subset X'$ be the $n$th infinitesimal neighbourhoods of $Z$, $Z'$. The induced morphism $Z_ n \to Z'_ n$ is a homeomorphism on underlying topological spaces. On the other hand, if $z' \in Z'$ maps to $z \in Z$, then the ring map $\mathcal{O}_{X, z} \to \mathcal{O}_{X', z'}$ is flat and induces an isomorphism $\mathcal{O}_{X, z}/\mathcal{I}_ z \to \mathcal{O}_{X', z'}/\mathcal{I}'_{z'}$. Hence it induces an isomorphism $\mathcal{O}_{X, z}/\mathcal{I}_ z^ n \to \mathcal{O}_{X', z'}/(\mathcal{I}'_{z'})^ n$ for all $n \geq 1$ for example by More on Algebra, Lemma 15.89.2. Thus $Z'_ n \to Z_ n$ is an isomorphism of schemes. Thus $f^*$ induces an equivalence between the category of coherent $\mathcal{O}_ X$-modules annihilated by $\mathcal{I}^ n$ and the category of coherent $\mathcal{O}_{X'}$-modules annihilated by $(\mathcal{I}')^ n$, see Lemma 30.9.8. This clearly implies the lemma. $\square$

Proof. First, assume $X = \mathop{\mathrm{Spec}}(A)$ is affine. Let $I, J \subset A$ be the ideals corresponding to $\mathcal{I}, \mathcal{J}$. Then $V(I) = V(J)$ implies we have $I^ c \subset J$ and $J^ d \subset I$ for some $c, d \geq 1$ by elementary properties of the Zariski topology (see Algebra, Section 10.17 and Lemma 10.32.5). Hence the $I$-adic and $J$-adic completions of $A$ agree, see Algebra, Lemma 10.96.9. Thus the equivalence follows from Lemma 30.23.1 in this case.

In general, using what we said above and the fact that $X$ is quasi-compact, to choose $c, d \geq 1$ such that $\mathcal{I}^ c \subset \mathcal{J}$ and $\mathcal{J}^ d \subset \mathcal{I}$. Then given an object $(\mathcal{F}_ n)$ in $\textit{Coh}(X, \mathcal{I})$ we claim that the inverse system

is in $\textit{Coh}(X, \mathcal{J})$. This may be checked on the members of an affine covering; we omit the details. In the same manner we can construct an object of $\textit{Coh}(X, \mathcal{I})$ starting with an object of $\textit{Coh}(X, \mathcal{J})$. We omit the verification that these constructions define mutually quasi-inverse functors. $\square$