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Notes on prismatic cohomology

Kiran S. Kedlaya

Section 6 Derived completeness

Reference.

[117], tag 091N, 0BKF. See also [117], tag 0BKH for the case of a noetherian ring, where some simplifications occur.
We fill in a missing detail from Section 5, namely the distinction between classical and derived completeness of a module with respect to an ideal. As you might imagine, the latter is best treated within the framework of derived categories; we will do as much as we can in classical language, and end with a few statements which one may want to postpone until after reading Section 10.

Subsection 6.1 The trouble with classical completion

Definition 6.1.1.

For \(A \in \Ring\) and \(I\) a finitely generated ideal of \(A\text{,}\) an \(A\)-module \(M\) is classically \(I\)-complete if the natural map \(M \to \lim_n M/I^n M\) is an isomorphism. In particular, this means that \(M\) is \(I\)-adically separated: \(\bigcap_n I^n M = 0\text{.}\)

Remark 6.1.2. Completions behaving badly.

Much of our intuition about completion of modules with respect to an ideal is derived from the case of finitely generated modules over a noetherian ring. A few pitfalls to keep in mind include the following.
  • Classically \(I\)-complete modules do not form an abelian subcategory of the category of all \(A\)-modules. For example, it is possible for the quotient of \(A\) by a principal ideal to be noncomplete; see [117], tag 05JD. This is remedied by using derived \(I\)-complete modules instead; see Proposition 6.3.1.
  • The completion functor \(M \mapsto \lim_n M/I^n M\) preserves surjections, but it is not even right exact even on finitely presented modules. (This is arguably not surprising because completion is the composition of a right exact functor with a left exact functor.) See [117], tag 05JF and also Example 6.1.4.
  • The completion of a flat module (or even a flat \(A\)-algebra) need not be flat; see Example 6.1.3. This phenomenon will force us to adopt Definition 6.5.1.
If we drop the restriction that \(I\) be finitely generated, then things get even stranger. For example, completion is no longer an idempotent operation; see [117], tag 05JA.

Example 6.1.3.

For \(A \in \Ring\text{,}\) \(A\llbracket x \rrbracket\) is the \(x\)-adic completion of the flat \(A\)-algebra \(A[x]\text{,}\) but is flat over \(A\) if and only if \(A\) is coherent (every finitely generated ideal is finitely presented); see [117], tag 0ALB. See [117], tag 0AL8 for a concrete example.
On a related note, one can construct a ring \(A\) and an element \(f \in A\) such that \(A_f \llbracket x \rrbracket\text{,}\) which is the \(x\)-adic completion of the flat \(A \llbracket x \rrbracket\)-algebra \(A \llbracket x \rrbracket_f\text{,}\) is not itself flat over \(A \llbracket x \rrbracket\text{.}\) See again [117], tag 0AL8.
One can see some additional issues with classical completion from the following basic example (adapted from [129], Example 3.20; see also [117], tag 0G3F).

Example 6.1.4.

Take \(A = \ZZ_p, I = pA\text{.}\) Let \(M_0\) be the set of sequences \((x_n)_{n=0}^\infty\) over \(\ZZ_p\) with \(\lim_{n \to \infty} x_n = 0\text{.}\) Let \(M_1 \subset M_0\) be the set of sequences \((x_n)_{n=0}^\infty\) with \(x_n \equiv 0 \pmod{p^n}\text{.}\) Let \(M_2 \subset M_1\) be the set of sequences \((x_n)_{n=0}^\infty\) with \(\lim_{n \to \infty} x_n/p^n = 0\text{.}\)
The modules \(M_0, M_1, M_2, M_0/M_1, M_1/M_2\) are all classically \(I\)-complete (note that \(M_0\) is isomorphic to \(M_2\) via the map \((x_n) \mapsto (x_n p^n)\)). However, \(M_0/M_2\) is not \(I\)-adically separated: we have \(\bigcap_n p^n (M_0/M_2) = M_1/M_2\text{.}\) (In other words, the closure of \(M_2\) in \(M_0\) is equal to \(M_1\text{.}\)) Consequently, applying the completion functor to the exact sequence
\begin{equation*} 0 \to M_2 \to M_0 \to M_0/M_2 \to 0 \end{equation*}
yields the sequence
\begin{equation*} 0 \to M_2 \to M_0 \to M_0/M_1 \to 0 \end{equation*}
which is not exact in the middle.

Subsection 6.2 Derived completeness

Definition 6.2.1.

For \(A \in \Ring\) and \(I\) a finitely generated ideal of \(A\text{,}\) an \(A\)-module \(M\) is derived \(I\)-complete if for each \(f \in I\text{,}\)
\begin{equation} \Hom_A (A_f, M) = 0 \mbox{ and } \Ext^1_A(A_f, M) = 0.\tag{6.1} \end{equation}
By Lemma 6.2.3, it will suffice to check this condition for \(f\) running over a generating set of \(I\) (or of any ideal with the same radical as \(I\)).

Remark 6.2.2.

When working with (6.1), note that \(A_f\) admits the following free resolution as an \(A\)-module:
\begin{equation*} 0 \to A[T] \stackrel{\times (1-Tf)}{\to} A[T] \stackrel{T \mapsto f^{-1}}{\to} A_f \to 0 \end{equation*}
Consequently, for any \(A\)-module \(M\text{,}\) \(\Ext^n_A(A_f, M) = 0\) for \(n \geq 2\text{.}\) Since \(\Hom_A = \Ext^0_A\text{,}\) this means that (6.1) can be reformulated as
\begin{equation} \Ext^n_A(A_f, M) = 0 \qquad (n \geq 0).\tag{6.2} \end{equation}
For example, this makes it clear that if any two terms in a short exact sequence are derived \(I\)-complete, then so is the third.

Proof.

Let \(I\) be the set in question. For \(f \in I, g \in A\text{,}\) the functor \(\Hom_A(A_{fg}, \bullet)\) factors as
\begin{equation*} M \mapsto \Hom_A(A_f, M) \mapsto \Hom_A(A_g, \Hom_A(A_f, M)). \end{equation*}
From the spectral sequence for a composition of functors (or more elementary considerations), we see that if \(\Ext^n_A(A_f, M) = 0\) for all \(n \geq 0\text{,}\) then \(\Ext^n_A(A_{gf}, M) = 0\) for all \(n \geq 0\text{;}\) hence \(fg \in I\text{.}\)
For \(f,g \in I\text{,}\) the sequence
\begin{equation*} 0 \to A_{f+g} \to A_{f(f+g)} \oplus A_{g(f+g)} \to A_{fg(f+g)} \to 0 \end{equation*}
is exact because \(f,g\) generate the unit ideal in \(A_{f+g}\text{.}\) (As an aside, see [81], Lemma 1.6.12 for another application of this observation.) Since \(f(f+g), g(f+g), fg(f+g) \in I\) by the previous paragraph, using the snake lemma we obtain \(f+g \in I\text{.}\) Consequently, \(I\) is an ideal of \(A\text{.}\)
For any \(f \in I\) and any positive integer \(n\text{,}\) \(A_f = A_{f^n}\text{.}\) Hence \(I\) is a radical ideal of \(A\text{.}\) (Compare [117], tag 091Q.)

Proof.

If \(M\) is classically \(I\)-complete, we have
\begin{equation*} \Hom_A(A_f, M) = \Hom_A(A_f, \lim_n M/I^n M) = 0. \end{equation*}
To show that \(\Ext^1_A(A_f, M) = 0\text{,}\) consider an extension
\begin{equation} 0 \to M \to E \to A_f \to 0.\tag{6.3} \end{equation}
For each \(n \geq 0\text{,}\) pick \(e_n \in E\) mapping to \(f^{-n} \in A_f\) and set \(\delta_n = fe_{n+1} - e_n \in M\text{.}\) Since \(M\) is complete, we may define the elements
\begin{equation*} e_n' = e_n + \delta_n + f \delta_{n+1} + f^2 \delta_{n+2} + \cdots \end{equation*}
which satisfy \(f e_{n+1}' = e_n'\text{;}\) we thus obtain a map \(A_f \to E\) splitting the sequence by mapping \(f^{-n}\) to \(e_n'\text{.}\)
In the converse direction, by an elementary argument (Exercise 6.7.1) we can reduce to the case where \(I = (f)\text{.}\) That, we must show that if (6.1) holds, then for any \(x_0, x_1, \ldots \in M\) there exists \(x \in M\) such that \(x \equiv x_0 + fx_1 + \cdots + f^{n-1} x_{n-1} \pmod{f^nM}\) for each \(n\text{.}\) To this end, form an extension as in (6.3) by taking
\begin{equation*} E = M \oplus \bigoplus_n Ae_n/(x_n - fe_{n+1} + e_n) \end{equation*}
with \(e_n\) mapping to \(f^{-n}\) in \(A_f\text{;}\) note that by the snake lemma, \(M/f^n M = E/f^n E\) for all \(n\text{.}\) Since the extension splits by hypothesis, there is an element \(x + e_0 \in E\) which generates a copy of \(A_f\) in \(E\text{.}\) We then have
\begin{equation*} x + e_0 = x - x_0 + fe_1 = x - x_0 + x_1 + f^2 e_2 = \cdots; \end{equation*}
this yields the desired result. (Compare [117], tag 091R.)

Proof.

The following can be viewed as an algebraic version of the open mapping theorem from functional analysis.

Proof.

Subsection 6.3 The category of derived-complete modules

We next introduce the category of derived \(I\)-complete modules and its basic properties. This leads naturally to the operation of derived completion.

Proof.

See [117], tags 0G1U, 091V, 091U.

Proof.

To prove (1), choose any \(u \in 1+I\text{,}\) then apply derived Nakayama (Proposition 6.3.1 to \(M = A/(u)\) to deduce that \(u \in A^\times\text{.}\)
To prove (2), apply part (3) of Proposition 6.3.1.
For (3), see [117], tag 0G3H.

Remark 6.3.3.

With notation as in Example 6.1.4, Proposition 6.3.1 implies that the \(A\)-module \(M_0/M_2\text{,}\) which is not \(I\)-adically separated, is nonetheless derived \(I\)-complete.

Remark 6.3.4.

The category of derived \(I\)-complete \(A\)-modules does not have the property that filtered colimits are exact ([117], tag 0ARC). In particular, it is not a Grothendieck abelian category.
On the other hand, a countably filtered colimit of derived \(I\)-complete \(A\)-modules is again derived \(I\)-complete. The point is that any potential witness to the failure of completeness can be expressed using only countably many module elements.

Remark 6.3.5.

For \(I = (f_1,\dots, f_n)\text{,}\) the derived \(I\)-completion functor from Proposition 6.3.1 can be described as the composition of the derived \(f_i\)-completions for \(i=1,\dots,n\) (in any order). These individual functors can be described concretely using Lemma 6.4.1. For an alternate description in the language of derived categories, see Proposition 6.6.2.

Definition 6.3.6.

Let \(I\) be a finitely generated ideal of \(A \in \Ring\text{.}\) We may then promote the derived \(I\)-completion functor from \(A\)-modules to \(A\)-algebras as follows.
For any \(A\)-algebra \(B\text{,}\) the multiplication map on \(B\) defines a morphism \(B \otimes_A B \to B\) of \(A\)-modules. Let \(\widehat{B}\) denote the derived \(I\)-completion of \(B\text{.}\) Now consider the composition
\begin{equation*} \widehat{B} \otimes_A \widehat{B} \to \widehat{B \otimes_A B} \to \widehat{B} \end{equation*}
where the first map is induced by the individual maps \(B \to B \otimes_A B\) and the second map is induced by the multiplication morphism. This gives us a multiplication map on \(\widehat{B}\) which gives it the structure of an \(A\)-algebra. In particular, the derived \(I\)-completion \(\widehat{A}\) of \(A\) is itself an \(A\)-algebra, and the ring \(\widehat{B}\) is also an \(\widehat{A}\)-algebra.

Subsection 6.4 Derived \(f\)-completion

The following lemma gives an explicit recipe for derived completion with respect to a principal ideal.

Proof.

From the exact sequence
\begin{equation*} 0 \to A \to A_f \to A_f/A \to 0 \end{equation*}
we obtain a morphism \(M \to \Ext^1_A(A_f/A, M)\) which is an isomorphism whenever \(M\) is derived \(I\)-complete; we claim that the target of this map is in fact \(\widehat{M}\text{.}\) Namely, if \(M \to N\) is another morphism with \(N\) derived \(I\)-complete, then by functoriality of \(\Ext^1_A\) in the second argument we obtain a unique morphism \(\Ext^1_A(A_f/A, M) \to \Ext^1_A(A_f/A, N) \cong N\) through which \(M \to N\) factors. This yields (1); by writing \(A_f/A = \colim_n f^{-n}A/A\text{,}\) we may then deduce (2).
As an application of Lemma 6.4.1, we obtain a criterion that lets us forget about the difference between classical and derived completions in many cases of interest.

Proof.

The derived completion \(\widehat{M}\) fits into an exact sequence given in (6.4). By the assumption about bounded torsion, the projective system formed by the \(M[f^n]\text{,}\) in which the transition maps are multiplicaton by \(f\text{,}\) is essentially zero (that is, any sufficiently long composition is the zero map). Hence the \(R^1\) term vanishes and we obtain the desired conclusion. (Compare [18], Lecture III, Lemma 2.4.)
A key application of Lemma 6.4.2 is the following.

Proof.

For \(x \in R[f^\infty]\text{,}\) we have \(f^{p^n} x = 0\) for some nonnegative integer \(n\text{.}\) We then have \(f^{p^n} x^{p^n} = 0\text{,}\) and then \(fx = 0\) because \(R\) is perfect.

Remark 6.4.4.

One way to make a ring which is derived \(f\)-complete but not classically \(f\)-complete is to start with a ring \(A\text{,}\) an element \(f \in A\text{,}\) and a module \(M\) which is derived \(f\)-complete but not classically \(f\)-complete (e.g., see Remark 6.3.3), and then form the square-zero extension \(A \oplus M\text{.}\)

Subsection 6.5 Flatness and smoothness

Definition 6.5.1.

Let \(I\) be a finitely generated ideal of \(A \in \Ring\text{.}\) We say that \(M \in \Mod_A\) is \(I\)-completely flat (resp. \(I\)-completely faithfully flat) if \(M/IM\) is a flat (resp. faithfully flat) \(A/I\)-module and \(\Tor_i^A(A/I, M) = 0\) for \(i \gt 0\text{.}\) Equivalently, for any \(N \in \Mod_{A/I}\text{,}\) \(\Tor_i^A(N,M) = 0\) for \(i \gt 0\text{.}\) If \(M\) is (faithfully) flat, then it is \(I\)-completely (faithfully) flat (Exercise 6.7.6), but not conversely (see Remark 6.1.2).
More generally, we say that \(M\) has finite \(I\)-complete Tor amplitude if there exists some \(c \gt 0\) such that \(\Tor_i^A(A/I, M) = 0\) for \(i \geq -c\text{.}\) Equivalently, for any \(N \in \Mod_{A/I}\text{,}\) \(\Tor_i^A(N,M) = 0\) for \(i \geq c\text{.}\)

Definition 6.5.2.

Let \(I\) be a finitely generated ideal of \(A \in \Ring\text{.}\) A derived \(I\)-complete \(A\)-algebra \(R\) is \(I\)-completely étale (resp. \(I\)-completely smooth, \(I\)-completely ind-smooth) if \(R \otimes_A^L A/I\) is concentrated in degree 0 where it is an étale (resp. smooth, ind-smooth) \(A/I\)-algebra. That is, \(R \otimes_A^L A/I\) is an étale (resp. smooth, ind-smooth) \(A/I\)-algebra and \(\Tor_i^A(R, A/I) = 0\) for \(i \gt 0\text{.}\)

Proof.

This follows from an algebraization theorem originally due to Elkik [45]. For a more modern treatment, see [9].

Remark 6.5.4.

An \(I\)-completely smooth morphism is sometimes called a “formally smooth” morphism, but this is not entirely accurate: the latter simply means that the infinitesimal lifting criterion is satisfied ([117], tag 00TI). An \(I\)-completely smooth morphism is formally smooth, but formal smoothness is a meaningful condition without any reference to derived completeness.
By contrast, a smooth morphism of rings is one which is formally smooth and of finite presentation (see [117], tag 02H6). In general an \(I\)-completely smooth morphism is not of finite type and hence not smooth; for instance, \(\ZZ_p \to \ZZ_p[x]^\wedge_{(p)}\) is \(p\)-completely smooth but not of finite type.

Subsection 6.6 Derived completeness in the derived category

If you don’t yet know what derived categories are, skip this discussion for now and return after you have read Section 10.

Definition 6.6.1.

Let \(I\) be a finitely generated ideal of \(A \in \Ring\text{.}\) An object \(K\) in the derived category \(D(A)\) of \(A\)-modules is derived \(I\)-complete if for each \(f \in I\text{,}\)
\begin{equation*} R\Hom_A(A_f, K) = 0. \end{equation*}
The argument of Lemma 6.2.3 carries over to show that it suffices to check this condition at a generating set of \(I\text{,}\) or of any other ideal with the same radical.
One immediate consequence of this definition is that if any two terms of a distinguished triangle are derived \(I\)-complete, then so is the third. In particular, the mapping cone of a morphism between derived \(I\)-complete complexes is derived \(I\)-complete; see Remark 6.6.6.

Proof.

See [117], tags 091V, 091Z, 0G1U.

Remark 6.6.3.

In general, an \(A\)-module admits derived \(I\)-completions “as a module”, as an object in \(\Mod_A\) as per Proposition 6.3.1; and also “as a complex”, i.e., by identifying the module with a singleton complex in \(D(A)\) concentrated in degree 0 and then applying Proposition 6.6.2. The former is \(H^0\) of the latter, but the other cohomology groups of the latter carries some extra information that is invisible at the module-theoretic level; see Example 6.6.5 for a typical example.
However, if an \(A\)-module \(M\) is derived \(I\)-complete “as a module”, it is also derived \(I\)-complete “as a complex”: Proposition 6.6.2 says that the latter condition applied to \(M[0]\) can be tested at the level of cohomology groups.

Remark 6.6.4.

In the case where \(I = (f)\) is a principal ideal, we may adapt the proof of Lemma 6.4.1 to show that the derived \(f\)-completion of an \(A\)-module \(M\) as a complex is concentrated in degrees -1 and 0, with the cohomology in degree -1 being \(\lim_n M[f^n]\text{.}\) Consequently, Lemma 6.4.2 can be upgraded to assert that the derived \(f\)-completion as a complex is also isomorphic to the classical \(f\)-completion.

Example 6.6.5.

Take \(A = \ZZ, I = (p), M = \QQ/\ZZ\text{.}\) Then the ordinary completion of \(M\) vanishes, as does the derived completion of \(M\) as a module, but the derived completion of \(M\) as a complex is the group \(\ZZ_p\) concentrated in degree \(-1\text{.}\)

Remark 6.6.6.

We will frequently used the derived Nakayama’s lemma (Proposition 6.6.2) in the following form: for \(f\colon K^\bullet \to L^\bullet\) a morphism of derived \(I\)-complete complexes, \(f\) is a quasi-isomorphism if and only if the induced morphism \(K^\bullet \otimes^L_A A/I \to L^\bullet \otimes^L_A A/I\) of complexes over \(A/I\) is a quasi-isomorphism. To be precise, this will follow from applying Proposition 6.6.2 to the mapping cone of \(f\) (Definition 10.2.2).

Definition 6.6.7.

Let \(I\) be a finitely generated ideal of \(A \in \Ring\text{.}\) We say that a complex \(K\) of \(A\)-modules is \(I\)-completely flat if for any \(I\)-torsion \(A\)-module \(N\text{,}\) the derived tensor product \(K \otimes_A^L N\) is concentrated in degree \(0\text{.}\) Equivalently, \(K \otimes_A^L A/I\) is concentrated in degree \(0\) where it is a flat \(A/I\)-module. If in addition \(K \otimes_A^L A/I\) is faithfully flat as an \(A/I\)-module, we say that \(K\) is \(I\)-completely faithfully flat. Note that these definitions agree with the corresponding notions for modules (Definition 6.5.1).
Similarly, we say that \(K\) has finite \(I\)-complete Tor amplitude if there exists some \(c \geq 0\) such that for any \(I\)-torsion \(A\)-module \(N\text{,}\) \(K \otimes_A^L N\) is concentrated in degrees \(\geq -c\text{.}\)

Exercises 6.7 Exercises

1.

Let \(I\) be a finitely generated ideal of \(A \in \Ring\) and choose \(M \in \Mod_A\text{.}\) Suppose that for each \(f \in I\text{,}\) the map \(M \to \lim_n M/f^n M\) is surjective. Prove that the map \(M \to \lim_n M/I^n M\) is surjective.
Hint.
See [117], tag 090S.

2.

Let \(I\) be a finitely generated ideal of \(A \in \Ring\text{.}\) Let \(M \to N\) be a morphism of derived \(I\)-complete \(A\)-modules such that \(M/IM \to N/IN\) is surjective. Prove that \(M \to N\) is surjective.

3.

Let \(A\) be a ring which is derived \(f\)-complete, but not classically \(f\)-complete, for some \(f \in A\) (e.g., see Remark 6.4.4). Let \(I\) be the kernel of the surjective (by Lemma 6.2.4) map from \(A\) to its classical \(f\)-completion \(\lim_n A/f^n\text{.}\)
  1. Show that \(I^2 = 0\text{.}\)
  2. Show that any classically \(f\)-complete \(A\)-module is also a module over \(\lim_n A/f^n\) (that is, it is annihilated by \(I\)).
  3. Deduce that \(A\text{,}\) as a module over itself, cannot be written as the quotient of a classically \(f\)-complete \(A\)-module.
Hint.
See [117], tag 0G3G.

4.

Let \(I\) be a finitely generated ideal of \(A \in \Ring\text{.}\)
  1. Show that the kernel of any morphism between classically \(I\)-complete \(A\)-modules is classically \(I\)-complete.
  2. Let \(M\) be an \(A\)-module which can be written as the cokernel of a morphism between classically \(I\)-complete \(A\)-modules. Show that for any classically \(I\)-complete \(A\)-module \(N\) and any surjective morphism \(f\colon N \to M\) of \(A\)-modules, \(\ker(f)\) is classically \(I\)-complete.
  3. Deduce that the abelian closure of the category of classically \(I\)-complete \(A\)-modules consists of the cokernels of morphisms between classically \(I\)-complete \(A\)-modules. By Exercise 6.7.3, this can be strictly smaller than the category of derived \(I\)-complete \(A\)-modules. (See however Exercise 6.7.5.)

5.

Let \(f\) be a non-zerodivisor in some \(A \in \Ring\text{.}\) Show that \(M \in \Mod_A\) is derived \(f\)-complete if and only if there exists a short exact sequence
\begin{equation*} 0 \to K \to L \to M \to 0 \end{equation*}
in which \(K,L \in \Mod_A\) are \(f\)-torsion-free and \(f\)-adically complete.
Hint.
[117], tag 09AT.

6.

Let \(I\) be a finitely generated ideal of \(A \in \Ring\text{.}\) Show that for any flat \(A\)-module \(M\text{,}\) the derived \(I\)-completion of \(M\) as a complex is \(I\)-completely flat.
Hint.
Use the adjunction property of derived \(I\)-completion.

7.

Let \(A \in \Ring\) be noetherian and let \(I\) be an ideal of \(A\text{.}\) Suppose that \(M \in D(R)\) is \(I\)-completely flat. Then the derived \(I\)-completion of \(M\) is concentrated in degree \(0\text{,}\) where it is a flat \(A\)-module.
Hint.
See [17], Lemma 5.15.

8.

Suppose that \(A \in \Ring\) is derived \(I\)-complete for some finitely generated ideal \(I\text{.}\) Prove that a map \(M_1 \to M_2\) between finite projective \(A\)-modules is surjective (resp. bijective) if and only if the induced map \(M_1/IM_1 \to M_2/IM_2\) is surjective (resp. bijective). In particular, a finite projective \(A\)-module \(M\) is free if and only if \(M/IM\) is a free \(A/I\)-module.
Hint.
For surjectivity, apply derived Nakayama (Proposition 6.3.1) to the cokernel. For bijectivity, first verify that \(M_1\) and \(M_2\) have the same rank everywhere (using the fact that \(I \subseteq \Rad(A)\)), then recall that a surjective map between projective modules of the same finite rank is a bijection.

9.

Suppose that \(A \in \Ring\) is derived \(I\)-complete for some finitely generated ideal \(I\text{.}\) Prove that the base extension functor from finite projective \(A\)-modules to finite projective \(A/I\)-modules is essentially surjective.
Hint.
Start with a finite projective \(A/I\)-module \(\overline{M}\text{.}\) Choose a finite free \(A/I\)-module \(\overline{F}\) and a projector \(\overline{U}\colon \overline{F} \to \overline{F}\) whose image is isomorphic to \(\overline{M}\text{.}\) View \(\overline{F}\) as the reduction of a finite free \(A\)-module \(F\) and choose an endomorphism \(U_0\colon F \to F\) lifting \(\overline{U}\text{.}\) Show by induction that for each positive integer \(n\text{,}\) \(U_0\) is congruent modulo \(I\) to an endomorphism \(U_n\) such that \(U_n^2 \equiv U_n \pmod{I^{n+1}}\text{.}\) Take the limit to lift to the classical \(I\)-completion of \(A\text{,}\) then use the fact that the kernel of the map from \(A\) to the classical completion has square zero (Exercise 6.7.3) to lift to \(A\text{.}\) Conclude that we have produced a projector on \(F\) whose image has base extension isomorphic to \(\overline{M}\text{.}\)

10.

Suppose that \(A \in \Ring\) is derived \(I\)-complete for some finitely generated ideal \(I\text{.}\) Prove that the natural map \(\Pic(A) \to \Pic(A/I)\) is an isomorphism.
Hint.

11.

Let \(I\) be a finitely generated ideal of \(A \in \Ring_\delta\) containing \(p\text{.}\) Let \(A \to B\) be a morphism in \(\Ring\) such that \(B\) is derived \(I\)-complete and \(A \to B\) is \(I\)-completely etale. Then \(A \to B\) promotes uniquely to a morphism in \(\Ring_\delta\text{.}\)
Hint.
Using Remark 2.3.3, it suffices to lift \(A \to B\) to a morphism \(W_2(A) \to W_2(B)\text{.}\) Achieve this by combining Proposition 6.5.3 with Proposition 3.5.1. (Compare [25], Lemma 2.18.)

12.

Let \(I\) be a finitely generated ideal of \(A \in \Ring\) containing \(p\text{.}\) Prove that the derived \(I\)-completion of \(A\) (as a module, viewed as an \(A\)-algebra using Definition 6.3.6) admits a unique \(\delta\)-ring structure compatible with \(A\text{.}\) (That is, Exercise 2.5.8 remains true when the classical completion is replaced by the derived completion.)
Hint.
Use the characterization of \(\delta\)-structures on \(A\) in terms of \(W_2(A)\) (Remark 2.3.3).

13.

Let \((A, I)\) be a bounded prism. Show that for any flat \(A\)-module \(M\text{,}\) the derived \((p, I)\)-completion of \(M\) as a complex is concentrated in degree 0 and is both classically \((p,I)\)-complete and \((p,I)\)-completely flat.

14.

Let \((A, I)\) be a prism. Prove that the class of \(I\) in \(\Pic(A)\) is \(p\)-torsion.
Hint.
First use Exercise 6.7.10 to argue that \(\Pic(A) \to \Pic(A/p)\) is an isomorphism. Then show that \(\phi\) induces the \(p\)-power map on \(\Pic(A/p)\text{,}\) and apply Lemma 5.3.6 to conclude. See also [25], Lemma 3.6.
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