The Stacks project

Proof. Let $\mathcal{B} = \bigoplus I^ n\mathcal{O}_ X = f^*\widetilde{B}$. Then $\bigoplus I^ n\mathcal{F}$ is a finite type graded $\mathcal{B}$-module. Hence the result follows from Lemma 30.19.3 part (1). $\square$

Proof. The statement is local on $Y$, hence this reduces to the case where $Y$ is affine. In the affine case the result follows from Lemma 30.20.1. Details omitted. $\square$

Proof. Write $M_ n = H^ p(X, I^ n\mathcal{F})$ for $n \geq 1$ and $M_0 = H^ p(X, \mathcal{F})$ so that we have maps $\ldots \to M_3 \to M_2 \to M_1 \to M_0$. Setting $B = \bigoplus _{n \geq 0} I^ n$, then $M = \bigoplus _{n \geq 0} M_ n$ is a finite graded $B$-module, see Lemma 30.20.1. Observe that the products $B_ n \otimes M_ m \to M_{m + n}$, $a \otimes m \mapsto a \cdot m$ are compatible with the maps in our inverse system in the sense that the diagrams

commute for $n, m' \geq 0$ and $m \geq m'$.

Proof of (1). Choose $d_1, \ldots , d_ t \geq 0$ and $x_ i \in M_{d_ i}$ such that $M$ is generated by $x_1, \ldots , x_ t$ over $B$. For any $c \geq \max \{ d_ i\} $ we conclude that $B_{n - c} \cdot M_ c = M_ n$ for $n \geq c$ and we conclude (1) is true.

Proof of (2). Let $c$ be as in the proof of (1). Let $n + m \geq c$. We have $M_{n + m} = B_{n + m - c} \cdot M_ c$. If $c > n$ then we use $M_ c \to M_ n$ and the compatibility of products with transition maps pointed out above to conclude that the image of $M_{n + m} \to M_ n$ is contained in $I^{n + m - c}M_ n$. If $c \leq n$, then we write $M_{n + m} = B_ m \cdot B_{n - c} \cdot M_ c = B_ m \cdot M_ n$ to see that the image is contained in $I^ m M_ n$. This proves (2).

Let $K_ n \subset M_ n$ be the kernel of the map $M_ n \to M_0$. The compatibility of products with transition maps pointed out above shows that $K = \bigoplus K_ n \subset M$ is a graded $B$-submodule. As $B$ is Noetherian and $M$ is a finitely generated graded $B$-module, this shows that $K$ is a finitely generated graded $B$-module. Choose $d'_1, \ldots , d'_{t'} \geq 0$ and $y_ i \in K_{d'_ i}$ such that $K$ is generated by $y_1, \ldots , y_{t'}$ over $B$. Set $c = \max (d'_ i, d'_ j)$. Since $y_ i \in \mathop{\mathrm{Ker}}(M_{d'_ i} \to M_0)$ we see that $B_ n \cdot y_ i \subset \mathop{\mathrm{Ker}}(M_{n + d'_ i} \to M_ n)$. In this way we see that $K_ n = \mathop{\mathrm{Ker}}(M_ n \to M_{n - c})$ for $n \geq c$. This proves (3).

Consider the following commutative solid diagram

Since the kernel of the surjective arrow $I^ n \otimes _ A M_0 \to I^ nM_0$ maps into $K_ n$ by the above we obtain the dotted arrow and the composition $I^ nM_0 \to M_{n - c} \to M_0$ is the canonical map. Then clearly the composition $I^ nM_0 \to M_{n - c} \to I^{n - 2c}M_0$ is the canonical map for $n \geq 2c$. Consider the composition $M_ n \to I^{n - c}M_0 \to M_{n - 2c}$. The first map sends an element of the form $a \cdot m$ with $a \in I^{n - c}$ and $m \in M_ c$ to $a m'$ where $m'$ is the image of $m$ in $M_0$. Then the second map sends this to $a \cdot m'$ in $M_{n - 2c}$ and we see (4) is true.

Part (5) is an immediate consequence of (4) and the definition of morphisms of pro-objects. $\square$

Proof. Let $c = \max \{ c_ p, c_{p + 1}\} $, where $c_ p, c_{p + 1}$ are the integers found in Lemma 30.20.3 for $H^ p$ and $H^{p + 1}$.

Let us prove part (1). Consider the short exact sequence

From the long exact cohomology sequence we see that

Hence by Lemma 30.20.3 part (2) we see that this is contained in $I^{n - c}H^ p(X, \mathcal{F})$ for $n \geq c$.

Note that part (3) implies part (2) by definition of the Mittag-Leffler systems.

Let us prove part (3). Fix an $n$. Consider the commutative diagram

This gives rise to the following commutative diagram

with exact rows. By Lemma 30.20.3 part (4) the kernel of $\beta $ is equal to the kernel of $\alpha $ for $m \geq c$. By a diagram chase this shows that the image of $\gamma $ is contained in the kernel of $\delta $ which shows that part (3) is true (set $k = n + m$ to get it). $\square$

Proof. This follows from Lemma 30.20.4 as follows. Set $M = H^ p(X, \mathcal{F})$, $M_ n = H^ p(X, \mathcal{F}/I^ n\mathcal{F})$, and denote $N_ n = \mathop{\mathrm{Im}}(M \to M_ n)$. By Lemma 30.20.4 parts (2) and (3) we see that $(M_ n)$ is a Mittag-Leffler system with $N_ n \subset M_ n$ equal to the image of $M_ k$ for all $k \gg n$. It follows that $\mathop{\mathrm{lim}}\nolimits M_ n = \mathop{\mathrm{lim}}\nolimits N_ n$ as topological modules (with limit topologies). On the other hand, the $N_ n$ form an inverse system of quotients of the module $M$ and hence $\mathop{\mathrm{lim}}\nolimits N_ n$ is the completion of $M$ with respect to the topology given by the kernels $K_ n = \mathop{\mathrm{Ker}}(M \to N_ n)$. By Lemma 30.20.4 part (1) we have $K_ n \subset I^{n - c}M$ and since $N_ n \subset M_ n$ is annihilated by $I^ n$ we have $I^ n M \subset K_ n$. Thus the topology defined using the submodules $K_ n$ as a fundamental system of open neighbourhoods of $0$ is the same as the $I$-adic topology and we find that the induced map $M^\wedge = \mathop{\mathrm{lim}}\nolimits M/I^ nM \to \mathop{\mathrm{lim}}\nolimits N_ n = \mathop{\mathrm{lim}}\nolimits M_ n$ is an isomorphism of topological modules¹. $\square$

Proof. This is a reformulation of the theorem on formal functions (Theorem 30.20.5) in the case of a complete Noetherian base ring. Namely, in this case the $A$-module $H^ p(X, \mathcal{F})$ is finite (Lemma 30.19.2) hence $I$-adically complete (Algebra, Lemma 10.97.1) and we see that completion on the left hand side is not necessary. $\square$

Proof. This is just a reformulation of a special case of the theorem on formal functions, Theorem 30.20.5. Let us spell it out. Note that $\mathcal{O}_{Y, y}$ is a Noetherian local ring. Consider the canonical morphism $c : \mathop{\mathrm{Spec}}(\mathcal{O}_{Y, y}) \to Y$, see Schemes, Equation (26.13.1.1). This is a flat morphism as it identifies local rings. Denote momentarily $f' : X' \to \mathop{\mathrm{Spec}}(\mathcal{O}_{Y, y})$ the base change of $f$ to this local ring. We see that $c^*R^ pf_*\mathcal{F} = R^ pf'_*\mathcal{F}'$ by Lemma 30.5.2. Moreover, the infinitesimal neighbourhoods of the fibre $X_ y$ and $X'_ y$ are identified (verification omitted; hint: the morphisms $c_ n$ factor through $c$).

Hence we may assume that $Y = \mathop{\mathrm{Spec}}(A)$ is the spectrum of a Noetherian local ring $A$ with maximal ideal $\mathfrak m$ and that $y \in Y$ corresponds to the closed point (i.e., to $\mathfrak m$). In particular it follows that

In this case also, the morphisms $c_ n$ are each closed immersions. Hence their base changes $i_ n$ are closed immersions as well. Note that $i_{n, *}\mathcal{F}_ n = i_{n, *}i_ n^*\mathcal{F} = \mathcal{F}/\mathfrak m^ n\mathcal{F}$. By the Leray spectral sequence for $i_ n$, and Lemma 30.9.9 we see that

Hence we may indeed apply the theorem on formal functions to compute the limit in the statement of the lemma and we win. $\square$

Proof. The fibre $X_ y$ is finite, and by Morphisms, Lemma 29.20.7 it is a finite discrete space. Moreover, the underlying topological space of each infinitesimal neighbourhood $X_ n$ is the same. Hence each of the schemes $X_ n$ is affine according to Schemes, Lemma 26.11.8. Hence it follows that $H^ p(X_ n, \mathcal{F}_ n) = 0$ for all $p > 0$. Hence we see that $(R^ pf_*\mathcal{F})_ y^\wedge = 0$ by Lemma 30.20.7. Note that $R^ pf_*\mathcal{F}$ is coherent by Proposition 30.19.1 and hence $R^ pf_*\mathcal{F}_ y$ is a finite $\mathcal{O}_{Y, y}$-module. By Nakayama's lemma (Algebra, Lemma 10.20.1) if the completion of a finite module over a local ring is zero, then the module is zero. Whence $(R^ pf_*\mathcal{F})_ y = 0$. $\square$

Proof. The fibre $X_ y$ is of finite type over $\mathop{\mathrm{Spec}}(\kappa (y))$. Hence $X_ y$ is a Noetherian scheme by Morphisms, Lemma 29.15.6. Hence the underlying topological space of $X_ y$ is Noetherian, see Properties, Lemma 28.5.5. Moreover, the underlying topological space of each infinitesimal neighbourhood $X_ n$ is the same as that of $X_ y$. Hence $H^ p(X_ n, \mathcal{F}_ n) = 0$ for all $p > d$ by Cohomology, Proposition 20.20.7. Hence we see that $(R^ pf_*\mathcal{F})_ y^\wedge = 0$ by Lemma 30.20.7 for $p > d$. Note that $R^ pf_*\mathcal{F}$ is coherent by Proposition 30.19.1 and hence $R^ pf_*\mathcal{F}_ y$ is a finite $\mathcal{O}_{Y, y}$-module. By Nakayama's lemma (Algebra, Lemma 10.20.1) if the completion of a finite module over a local ring is zero, then the module is zero. Whence $(R^ pf_*\mathcal{F})_ y = 0$. $\square$