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crystalline.tex
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\input{preamble}
% OK, start here.
%
\begin{document}
\title{Crystalline Cohomology}
\maketitle
\phantomsection
\label{section-phantom}
\tableofcontents
\section{Introduction}
\label{section-introduction}
\noindent
This chapter is based on a lecture series given by Johan de Jong
held in 2012 at Columbia University.
The goals of this chapter are to give a quick introduction to
crystalline cohomology. A reference is the book \cite{Berthelot}.
\medskip\noindent
We have moved the more elementary purely algebraic discussion of divided
power rings to a preliminary chapter as it is also useful
in discussing Tate resolutions in commutative algebra.
Please see Divided Power Algebra, Section \ref{dpa-section-introduction}.
\section{Divided power envelope}
\label{section-divided-power-envelope}
\noindent
The construction of the following lemma will be dubbed the
divided power envelope. It will play an important role later.
\begin{lemma}
\label{lemma-divided-power-envelope}
Let $(A, I, \gamma)$ be a divided power ring.
Let $A \to B$ be a ring map. Let $J \subset B$ be an ideal
with $IB \subset J$. There exists a homomorphism of
divided power rings
$$
(A, I, \gamma) \longrightarrow (D, \bar J, \bar \gamma)
$$
such that
$$
\Hom_{(A, I, \gamma)}((D, \bar J, \bar \gamma), (C, K, \delta)) =
\Hom_{(A, I)}((B, J), (C, K))
$$
functorially in the divided power algebra $(C, K, \delta)$ over
$(A, I, \gamma)$. Here the LHS is morphisms of divided
power rings over $(A, I, \gamma)$ and the RHS is morphisms of
(ring, ideal) pairs over $(A, I)$.
\end{lemma}
\begin{proof}
Denote $\mathcal{C}$ the category of divided power rings
$(C, K, \delta)$. Consider the functor
$F : \mathcal{C} \longrightarrow \textit{Sets}$ defined by
$$
F(C, K, \delta) =
\left\{
(\varphi, \psi)
\middle|
\begin{matrix}
\varphi : (A, I, \gamma) \to (C, K, \delta)
\text{ homomorphism of divided power rings} \\
\psi : (B, J) \to (C, K)\text{ an }
A\text{-algebra homomorphism with }\psi(J) \subset K
\end{matrix}
\right\}
$$
We will show that
Divided Power Algebra, Lemma \ref{dpa-lemma-a-version-of-brown}
applies to this functor which will
prove the lemma. Suppose that $(\varphi, \psi) \in F(C, K, \delta)$.
Let $C' \subset C$ be the subring generated by $\varphi(A)$,
$\psi(B)$, and $\delta_n(\psi(f))$ for all $f \in J$.
Let $K' \subset K \cap C'$ be the ideal of $C'$ generated by
$\varphi(I)$ and $\delta_n(\psi(f))$ for $f \in J$.
Then $(C', K', \delta|_{K'})$ is a divided power ring and
$C'$ has cardinality bounded by the cardinal
$\kappa = |A| \otimes |B|^{\aleph_0}$.
Moreover, $\varphi$ factors as $A \to C' \to C$ and $\psi$ factors
as $B \to C' \to C$. This proves assumption (1) of
Divided Power Algebra, Lemma \ref{dpa-lemma-a-version-of-brown}
holds. Assumption (2) is clear
as limits in the category of divided power rings commute with the
forgetful functor $(C, K, \delta) \mapsto (C, K)$, see
Divided Power Algebra, Lemma \ref{dpa-lemma-limits} and its proof.
\end{proof}
\begin{definition}
\label{definition-divided-power-envelope}
Let $(A, I, \gamma)$ be a divided power ring.
Let $A \to B$ be a ring map. Let $J \subset B$ be an ideal
with $IB \subset J$. The divided power algebra $(D, \bar J, \bar\gamma)$
constructed in Lemma \ref{lemma-divided-power-envelope}
is called the {\it divided power envelope of $J$ in $B$
relative to $(A, I, \gamma)$} and is denoted $D_B(J)$ or $D_{B, \gamma}(J)$.
\end{definition}
\noindent
Let $(A, I, \gamma) \to (C, K, \delta)$ be a homomorphism of divided
power rings. The universal property of
$D_{B, \gamma}(J) = (D, \bar J, \bar \gamma)$ is
$$
\begin{matrix}
\text{ring maps }B \to C \\
\text{ which map }J\text{ into }K
\end{matrix}
\longleftrightarrow
\begin{matrix}
\text{divided power homomorphisms} \\
(D, \bar J, \bar \gamma) \to (C, K, \delta)
\end{matrix}
$$
and the correspondence is given by precomposing with the map $B \to D$
which corresponds to $\text{id}_D$. Here are some properties of
$(D, \bar J, \bar \gamma)$ which follow directly from the universal
property. There are $A$-algebra maps
\begin{equation}
\label{equation-divided-power-envelope}
B \longrightarrow D \longrightarrow B/J
\end{equation}
The first arrow maps $J$ into $\bar J$ and $\bar J$ is the kernel
of the second arrow. The elements $\bar\gamma_n(x)$ where $n > 0$
and $x$ is an element in the image of $J \to D$ generate $\bar J$
as an ideal in $D$ and generate $D$ as a $B$-algebra.
\begin{lemma}
\label{lemma-divided-power-envelop-quotient}
Let $(A, I, \gamma)$ be a divided power ring.
Let $\varphi : B' \to B$ be a surjection of $A$-algebras with kernel $K$.
Let $IB \subset J \subset B$ be an ideal. Let $J' \subset B'$
be the inverse image of $J$. Write
$D_{B', \gamma}(J') = (D', \bar J', \bar\gamma)$.
Then $D_{B, \gamma}(J) = (D'/K', \bar J'/K', \bar\gamma)$
where $K'$ is the ideal generated by the elements $\bar\gamma_n(k)$
for $n \geq 1$ and $k \in K$.
\end{lemma}
\begin{proof}
Write $D_{B, \gamma}(J) = (D, \bar J, \bar \gamma)$.
The universal property of $D'$ gives us a homomorphism $D' \to D$
of divided power algebras. As $B' \to B$ and $J' \to J$ are surjective, we
see that $D' \to D$ is surjective (see remarks above). It is clear that
$\bar\gamma_n(k)$ is in the kernel for $n \geq 1$ and $k \in K$, i.e.,
we obtain a homomorphism $D'/K' \to D$. Conversely, there exists a divided
power structure on $\bar J'/K' \subset D'/K'$, see
Divided Power Algebra, Lemma \ref{dpa-lemma-kernel}.
Hence the universal property of $D$ gives an inverse $D \to D'/K'$ and we win.
\end{proof}
\noindent
In the situation of Definition \ref{definition-divided-power-envelope}
we can choose a surjection $P \to B$ where $P$ is a polynomial
algebra over $A$ and let $J' \subset P$ be the inverse image of $J$.
The previous lemma describes $D_{B, \gamma}(J)$ in terms of
$D_{P, \gamma}(J')$. Note that $\gamma$ extends to a divided power
structure $\gamma'$ on $IP$ by
Divided Power Algebra, Lemma \ref{dpa-lemma-gamma-extends}. Hence
$D_{P, \gamma}(J') = D_{P, \gamma'}(J')$ is an example of a special
case of divided power envelopes we describe in the following lemma.
\begin{lemma}
\label{lemma-describe-divided-power-envelope}
Let $(B, I, \gamma)$ be a divided power algebra. Let $I \subset J \subset B$
be an ideal. Let $(D, \bar J, \bar \gamma)$ be the divided power envelope
of $J$ relative to $\gamma$. Choose elements $f_t \in J$, $t \in T$ such
that $J = I + (f_t)$. Then there exists a surjection
$$
\Psi : B\langle x_t \rangle \longrightarrow D
$$
of divided power rings mapping $x_t$ to the image of $f_t$ in $D$.
The kernel of $\Psi$ is generated by the elements $x_t - f_t$ and
all
$$
\delta_n\left(\sum r_t x_t - r_0\right)
$$
whenever $\sum r_t f_t = r_0$ in $B$ for some $r_t \in B$, $r_0 \in I$.
\end{lemma}
\begin{proof}
In the statement of the lemma we think of $B\langle x_t \rangle$
as a divided power ring with ideal
$J' = IB\langle x_t \rangle + B\langle x_t \rangle_{+}$, see
Divided Power Algebra, Remark \ref{dpa-remark-divided-power-polynomial-algebra}.
The existence of $\Psi$ follows from the universal property of
divided power polynomial rings. Surjectivity of $\Psi$ follows from
the fact that its image is a divided power subring of $D$, hence equal to $D$
by the universal property of $D$. It is clear that
$x_t - f_t$ is in the kernel. Set
$$
\mathcal{R} = \{(r_0, r_t) \in I \oplus \bigoplus\nolimits_{t \in T} B
\mid \sum r_t f_t = r_0 \text{ in }B\}
$$
If $(r_0, r_t) \in \mathcal{R}$ then it is clear that
$\sum r_t x_t - r_0$ is in the kernel.
As $\Psi$ is a homomorphism of divided power rings
and $\sum r_tx_t = r_0 \in J'$
it follows that $\delta_n(\sum r_t x_t - r_0)$ is in the kernel as well.
Let $K \subset B\langle x_t \rangle$ be the ideal generated by
$x_t - f_t$ and the elements $\delta_n(\sum r_t x_t - r_0)$ for
$(r_0, r_t) \in \mathcal{R}$.
To show that $K = \Ker(\Psi)$ it suffices to show that
$\delta$ extends to $B\langle x_t \rangle/K$. Namely, if so the universal
property of $D$ gives a map $D \to B\langle x_t \rangle/K$
inverse to $\Psi$. Hence we have to show that $K \cap J'$ is
preserved by $\delta_n$, see
Divided Power Algebra, Lemma \ref{dpa-lemma-kernel}.
Let $K' \subset B\langle x_t \rangle$ be the ideal
generated by the elements
\begin{enumerate}
\item $\delta_m(\sum r_t x_t - r_0)$ where $m > 0$ and
$(r_0, r_t) \in \mathcal{R}$,
\item $x_{t'}^{[m]}(x_t - f_t)$ where $m > 0$ and $t', t \in I$.
\end{enumerate}
We claim that $K' = K \cap J'$. The claim proves that $K \cap J'$
is preserved by $\delta_n$, $n > 0$ by the criterion of
Divided Power Algebra, Lemma \ref{dpa-lemma-kernel} (2)(c)
and a computation of $\delta_n$
of the elements listed which we leave to the reader.
To prove the claim note that $K' \subset K \cap J'$.
Conversely, if $h \in K \cap J'$ then, modulo $K'$ we can write
$$
h = \sum r_t (x_t - f_t)
$$
for some $r_t \in B$. As $h \in K \cap J' \subset J'$
we see that $r_0 = \sum r_t f_t \in I$. Hence $(r_0, r_t) \in \mathcal{R}$
and we see that
$$
h = \sum r_t x_t - r_0
$$
is in $K'$ as desired.
\end{proof}
\begin{lemma}
\label{lemma-divided-power-envelope-add-variables}
Let $(A, I, \gamma)$ be a divided power ring.
Let $B$ be an $A$-algebra and $IB \subset J \subset B$ an ideal.
Let $x_i$ be a set of variables. Then
$$
D_{B[x_i], \gamma}(JB[x_i] + (x_i)) = D_{B, \gamma}(J) \langle x_i \rangle
$$
\end{lemma}
\begin{proof}
One possible proof is to deduce this from
Lemma \ref{lemma-describe-divided-power-envelope}
as any relation between $x_i$ in $B[x_i]$ is trivial.
On the other hand, the lemma follows from the universal property
of the divided power polynomial algebra and the universal property of
divided power envelopes.
\end{proof}
\noindent
Conditions (1) and (2) of the following lemma hold if $B \to B'$ is flat
at all primes of $V(IB') \subset \Spec(B')$ and is very closely related
to that condition, see
Algebra, Lemma \ref{algebra-lemma-what-does-it-mean}.
It in particular says that taking the divided power
envelope commutes with localization.
\begin{lemma}
\label{lemma-flat-base-change-divided-power-envelope}
Let $(A, I, \gamma)$ be a divided power ring.
Let $B \to B'$ be a homomorphism of $A$-algebras.
Assume that
\begin{enumerate}
\item $B/IB \to B'/IB'$ is flat, and
\item $\text{Tor}_1^B(B', B/IB) = 0$.
\end{enumerate}
Then for any ideal $IB \subset J \subset B$ the canonical map
$$
D_B(J) \otimes_B B' \longrightarrow D_{B'}(JB')
$$
is an isomorphism.
\end{lemma}
\begin{proof}
Set $D = D_B(J)$ and denote $\bar J \subset D$ its divided power ideal
with divided power structure $\bar\gamma$. The universal property of
$D$ produces a $B$-algebra map $D \to D_{B'}(JB')$, whence a map as in
the lemma. It suffices to show that
the divided powers $\bar\gamma$ extend to $D \otimes_B B'$ since then
the universal property of $D_{B'}(JB')$ will produce a map
$D_{B'}(JB') \to D \otimes_B B'$ inverse to the one in the lemma.
\medskip\noindent
Choose a surjection $P \to B'$ where $P$ is a polynomial algebra over $B$.
In particular $B \to P$ is flat, hence $D \to D \otimes_B P$ is flat by
Algebra, Lemma \ref{algebra-lemma-flat-base-change}.
Then $\bar\gamma$ extends to $D \otimes_B P$ by
Divided Power Algebra, Lemma \ref{dpa-lemma-gamma-extends}; we
will denote this extension
$\bar\gamma$ also. Set $\mathfrak a = \Ker(P \to B')$ so that
we have the short exact sequence
$$
0 \to \mathfrak a \to P \to B' \to 0
$$
Thus $\text{Tor}_1^B(B', B/IB) = 0$ implies that
$\mathfrak a \cap IP = I\mathfrak a$.
Now we have the following commutative diagram
$$
\xymatrix{
B/J \otimes_B \mathfrak a \ar[r]_\beta &
B/J \otimes_B P \ar[r] &
B/J \otimes_B B' \\
D \otimes_B \mathfrak a \ar[r]^\alpha \ar[u] &
D \otimes_B P \ar[r] \ar[u] &
D \otimes_B B' \ar[u] \\
\bar J \otimes_B \mathfrak a \ar[r] \ar[u] &
\bar J \otimes_B P \ar[r] \ar[u] &
\bar J \otimes_B B' \ar[u]
}
$$
This diagram is exact even with $0$'s added at the top and the right.
We have to show the divided powers on the ideal
$\bar J \otimes_B P$ preserve the ideal
$\Im(\alpha) \cap \bar J \otimes_B P$, see
Divided Power Algebra, Lemma \ref{dpa-lemma-kernel}.
Consider the exact sequence
$$
0 \to \mathfrak a/I\mathfrak a \to P/IP \to B'/IB' \to 0
$$
(which uses that $\mathfrak a \cap IP = I\mathfrak a$ as seen above).
As $B'/IB'$ is flat over $B/IB$ this sequence remains exact after
applying $B/J \otimes_{B/IB} -$, see
Algebra, Lemma \ref{algebra-lemma-flat-tor-zero}. Hence
$$
\Ker(B/J \otimes_{B/IB} \mathfrak a/I\mathfrak a \to
B/J \otimes_{B/IB} P/IP) =
\Ker(\mathfrak a/J\mathfrak a \to P/JP)
$$
is zero. Thus $\beta$ is injective. It follows that
$\Im(\alpha) \cap \bar J \otimes_B P$ is the
image of $\bar J \otimes \mathfrak a$. Now if
$f \in \bar J$ and $a \in \mathfrak a$, then
$\bar\gamma_n(f \otimes a) = \bar\gamma_n(f) \otimes a^n$
hence the result is clear.
\end{proof}
\noindent
The following lemma is a special case of
\cite[Proposition 2.1.7]{dJ-crystalline} which in turn is a
generalization of \cite[Proposition 2.8.2]{Berthelot}.
\begin{lemma}
\label{lemma-flat-extension-divided-power-envelope}
Let $(B, I, \gamma) \to (B', I', \gamma')$ be a homomorphism of
divided power rings. Let $I \subset J \subset B$ and
$I' \subset J' \subset B'$ be ideals. Assume
\begin{enumerate}
\item $B/I \to B'/I'$ is flat, and
\item $J' = JB' + I'$.
\end{enumerate}
Then the canonical map
$$
D_{B, \gamma}(J) \otimes_B B' \longrightarrow D_{B', \gamma'}(J')
$$
is an isomorphism.
\end{lemma}
\begin{proof}
Set $D = D_B(J)$ and denote $\bar J \subset D$ its divided power ideal
with divided power structure $\bar\gamma$. The universal property of
$D$ produces a homomorphism of divided power rings $D \to D_{B'}(J')$,
whence a map as in the lemma. It suffices to show that
there exist divided powers on the image of
$D \otimes_B I' + \bar J \otimes_B B' \to D \otimes_B B'$
compatible with $\bar \gamma$ and $\gamma'$ since then
the universal property of $D_{B'}(J')$ will produce a map
$D_{B'}(J') \to D \otimes_B B'$ inverse to the one in the lemma.
\medskip\noindent
Choose elements $f_t \in J$ which generate $J/I$. Set
$\mathcal{R} = \{(r_0, r_t) \in I \oplus \bigoplus\nolimits_{t \in T} B
\mid \sum r_t f_t = r_0 \text{ in }B\}$ as in the proof of
Lemma \ref{lemma-describe-divided-power-envelope}. This lemma shows that
$$
D = B\langle x_t \rangle/ K
$$
where $K$ is generated by the elements $x_t - f_t$ and
$\delta_n(\sum r_t x_t - r_0)$ for $(r_0, r_t) \in \mathcal{R}$.
Thus we see that
\begin{equation}
\label{equation-base-change}
D \otimes_B B' = B'\langle x_t \rangle/K'
\end{equation}
where $K'$ is generated by the images in $B'\langle x_t \rangle$
of the generators of $K$ listed above. Let $f'_t \in B'$ be the image
of $f_t$. By assumption (1) we see that the elements $f'_t \in J'$
generate $J'/I'$ and we see that $x_t - f'_t \in K'$. Set
$$
\mathcal{R}' =
\{(r'_0, r'_t) \in I' \oplus \bigoplus\nolimits_{t \in T} B'
\mid \sum r'_t f'_t = r'_0 \text{ in }B'\}
$$
To finish the proof we have to show that
$\delta'_n(\sum r'_t x_t - r'_0) \in K'$ for
$(r'_0, r'_t) \in \mathcal{R}'$, because then the presentation
(\ref{equation-base-change}) of $D \otimes_B B'$ is identical
to the presentation of $D_{B', \gamma'}(J')$ obtain in
Lemma \ref{lemma-describe-divided-power-envelope} from the generators $f'_t$.
Suppose that $(r'_0, r'_t) \in \mathcal{R}'$. Then
$\sum r'_t f'_t = 0$ in $B'/I'$. As $B/I \to B'/I'$ is flat by
assumption (1) we can apply the equational criterion of flatness
(Algebra, Lemma \ref{algebra-lemma-flat-eq}) to see
that there exist an $m > 0$ and
$r_{jt} \in B$ and $c_j \in B'$, $j = 1, \ldots, m$ such
that
$$
r_{j0} = \sum r_{jt} f_t \in I \text{ for } j = 1, \ldots, m,
\quad\text{and}\quad
r'_t = \sum c_j r_{jt}.
$$
Note that this also implies that $r'_0 = \sum c_j r_{j0}$.
Then we have
\begin{align*}
\delta'_n(\sum r'_t x_t - r'_0)
& =
\delta'_n(\sum c_j (\sum r_{jt} x_t - r_{j0})) \\
& =
\sum c_1^{n_1} \ldots c_m^{n_m}
\delta_{n_1}(\sum r_{1t} x_t - r_{10}) \ldots
\delta_{n_m}(\sum r_{mt} x_t - r_{m0})
\end{align*}
where the sum is over $n_1 + \ldots + n_m = n$. This proves what we want.
\end{proof}
\section{Some explicit divided power thickenings}
\label{section-explicit-thickenings}
\noindent
The constructions in this section will help us to define the connection
on a crystal in modules on the crystalline site.
\begin{lemma}
\label{lemma-divided-power-first-order-thickening}
Let $(A, I, \gamma)$ be a divided power ring. Let $M$ be an $A$-module.
Let $B = A \oplus M$ as an $A$-algebra where $M$ is an ideal of square zero
and set $J = I \oplus M$. Set
$$
\delta_n(x + z) = \gamma_n(x) + \gamma_{n - 1}(x)z
$$
for $x \in I$ and $z \in M$.
Then $\delta$ is a divided power structure and
$A \to B$ is a homomorphism of divided power rings from
$(A, I, \gamma)$ to $(B, J, \delta)$.
\end{lemma}
\begin{proof}
We have to check conditions (1) -- (5) of
Divided Power Algebra, Definition \ref{dpa-definition-divided-powers}.
We will prove this directly for this case, but please see the proof of
the next lemma for a method which avoids calculations.
Conditions (1) and (3) are clear. Condition (2) follows from
\begin{align*}
\delta_n(x + z)\delta_m(x + z)
& =
(\gamma_n(x) + \gamma_{n - 1}(x)z)(\gamma_m(x) + \gamma_{m - 1}(x)z) \\
& = \gamma_n(x)\gamma_m(x) + \gamma_n(x)\gamma_{m - 1}(x)z +
\gamma_{n - 1}(x)\gamma_m(x)z \\
& =
\frac{(n + m)!}{n!m!} \gamma_{n + m}(x) +
\left(\frac{(n + m - 1)!}{n!(m - 1)!} +
\frac{(n + m - 1)!}{(n - 1)!m!}\right)
\gamma_{n + m - 1}(x) z \\
& =
\frac{(n + m)!}{n!m!}\delta_{n + m}(x + z)
\end{align*}
Condition (5) follows from
\begin{align*}
\delta_n(\delta_m(x + z))
& =
\delta_n(\gamma_m(x) + \gamma_{m - 1}(x)z) \\
& =
\gamma_n(\gamma_m(x)) + \gamma_{n - 1}(\gamma_m(x))\gamma_{m - 1}(x)z \\
& =
\frac{(nm)!}{n! (m!)^n} \gamma_{nm}(x) +
\frac{((n - 1)m)!}{(n - 1)! (m!)^{n - 1}}
\gamma_{(n - 1)m}(x) \gamma_{m - 1}(x) z \\
& = \frac{(nm)!}{n! (m!)^n}(\gamma_{nm}(x) + \gamma_{nm - 1}(x) z)
\end{align*}
by elementary number theory. To prove (4) we have to see that
$$
\delta_n(x + x' + z + z')
=
\gamma_n(x + x') + \gamma_{n - 1}(x + x')(z + z')
$$
is equal to
$$
\sum\nolimits_{i = 0}^n
(\gamma_i(x) + \gamma_{i - 1}(x)z)
(\gamma_{n - i}(x') + \gamma_{n - i - 1}(x')z')
$$
This follows easily on collecting the coefficients of
$1$, $z$, and $z'$ and using condition (4) for $\gamma$.
\end{proof}
\begin{lemma}
\label{lemma-divided-power-second-order-thickening}
Let $(A, I, \gamma)$ be a divided power ring. Let $M$, $N$ be $A$-modules.
Let $q : M \times M \to N$ be an $A$-bilinear map.
Let $B = A \oplus M \oplus N$ as an $A$-algebra with multiplication
$$
(x, z, w)\cdot (x', z', w') = (xx', xz' + x'z, xw' + x'w + q(z, z') + q(z', z))
$$
and set $J = I \oplus M \oplus N$. Set
$$
\delta_n(x, z, w) = (\gamma_n(x), \gamma_{n - 1}(x)z,
\gamma_{n - 1}(z)w + \gamma_{n - 2}(x)q(z, z))
$$
for $(x, z, w) \in J$.
Then $\delta$ is a divided power structure and
$A \to B$ is a homomorphism of divided power rings from
$(A, I, \gamma)$ to $(B, J, \delta)$.
\end{lemma}
\begin{proof}
Suppose we want to prove that property (4) of
Divided Power Algebra, Definition \ref{dpa-definition-divided-powers}
is satisfied. Pick $(x, z, w)$ and $(x', z', w')$ in $J$.
Pick a map
$$
A_0 = \mathbf{Z}\langle s, s'\rangle \longrightarrow A,\quad
s \longmapsto x,
s' \longmapsto x'
$$
which is possible by the universal property of divided power
polynomial rings. Set $M_0 = A_0 \oplus A_0$ and
$N_0 = A_0 \oplus A_0 \oplus M_0 \otimes_{A_0} M_0$.
Let $q_0 : M_0 \times M_0 \to N_0$ be the obvious map.
Define $M_0 \to M$ as the $A_0$-linear map which sends
the basis vectors of $M_0$ to $z$ and $z'$. Define $N_0 \to N$
as the $A_0$ linear map which sends the first two basis vectors
of $N_0$ to $w$ and $w'$ and uses
$M_0 \otimes_{A_0} M_0 \to M \otimes_A M \xrightarrow{q} N$
on the last summand. Then we see that it suffices to prove the
identity (4) for the situation $(A_0, M_0, N_0, q_0)$.
Similarly for the other identities. This reduces us to the case
of a $\mathbf{Z}$-torsion free ring $A$ and $A$-torsion free modules.
In this case all we have to do is show that
$$
n! \delta_n(x, z, w) = (x, z, w)^n
$$
in the ring $A$, see Divided Power Algebra, Lemma \ref{dpa-lemma-silly}.
To see this note that
$$
(x, z, w)^2 = (x^2, 2xz, 2xw + 2q(z, z))
$$
and by induction
$$
(x, z, w)^n = (x^n, nx^{n - 1}z, nx^{n - 1}w + n(n - 1)x^{n - 2}q(z, z))
$$
On the other hand,
$$
n! \delta_n(x, z, w) = (n!\gamma_n(x), n!\gamma_{n - 1}(x)z,
n!\gamma_{n - 1}(x)w + n!\gamma_{n - 2}(x) q(z, z))
$$
which matches. This finishes the proof.
\end{proof}
\section{Compatibility}
\label{section-compatibility}
\noindent
This section isn't required reading; it explains how our discussion
fits with that of \cite{Berthelot}.
Consider the following technical notion.
\begin{definition}
\label{definition-compatible}
Let $(A, I, \gamma)$ and $(B, J, \delta)$ be divided power rings.
Let $A \to B$ be a ring map. We say
{\it $\delta$ is compatible with $\gamma$}
if there exists a divided power structure $\bar\gamma$ on
$J + IB$ such that
$$
(A, I, \gamma) \to (B, J + IB, \bar \gamma)\quad\text{and}\quad
(B, J, \delta) \to (B, J + IB, \bar \gamma)
$$
are homomorphisms of divided power rings.
\end{definition}
\noindent
Let $p$ be a prime number. Let $(A, I, \gamma)$ be a divided power ring.
Let $A \to C$ be a ring map with $p$ nilpotent in $C$.
Assume that $\gamma$ extends to $IC$ (see
Divided Power Algebra, Lemma \ref{dpa-lemma-gamma-extends}).
In this situation, the (big affine) crystalline site of
$\Spec(C)$ over $\Spec(A)$
as defined in \cite{Berthelot}
is the opposite of the category of systems
$$
(B, J, \delta, A \to B, C \to B/J)
$$
where
\begin{enumerate}
\item $(B, J, \delta)$ is a divided power ring with $p$ nilpotent in $B$,
\item $\delta$ is compatible with $\gamma$, and
\item the diagram
$$
\xymatrix{
B \ar[r] & B/J \\
A \ar[u] \ar[r] & C \ar[u]
}
$$
is commutative.
\end{enumerate}
The conditions
``$\gamma$ extends to $C$ and $\delta$ compatible with $\gamma$''
are used in \cite{Berthelot} to insure that
the crystalline cohomology of $\Spec(C)$ is the same as the crystalline
cohomology of $\Spec(C/IC)$. We will avoid this issue
by working exclusively with $C$ such that $IC = 0$\footnote{Of course there
will be a price to pay.}. In this case,
for a system $(B, J, \delta, A \to B, C \to B/J)$ as above,
the commutativity of the displayed diagram above implies $IB \subset J$ and
compatibility is equivalent to the condition that
$(A, I, \gamma) \to (B, J, \delta)$ is a homomorphism of divided
power rings.
\section{Affine crystalline site}
\label{section-affine-site}
\noindent
In this section we discuss the algebraic variant of the crystalline site.
Our basic situation in which we discuss this material will be as
follows.
\begin{situation}
\label{situation-affine}
Here $p$ is a prime number, $(A, I, \gamma)$ is a divided power
ring such that $A$ is a $\mathbf{Z}_{(p)}$-algebra, and $A \to C$ is a
ring map such that $IC = 0$ and such that $p$ is nilpotent in $C$.
\end{situation}
\noindent
Usually the prime number $p$ will be contained in the
divided power ideal $I$.
\begin{definition}
\label{definition-affine-thickening}
In Situation \ref{situation-affine}.
\begin{enumerate}
\item A {\it divided power thickening} of $C$ over $(A, I, \gamma)$
is a homomorphism of divided power algebras $(A, I, \gamma) \to (B, J, \delta)$
such that $p$ is nilpotent in $B$ and a ring map $C \to B/J$ such that
$$
\xymatrix{
B \ar[r] & B/J \\
& C \ar[u] \\
A \ar[uu] \ar[r] & A/I \ar[u]
}
$$
is commutative.
\item A {\it homomorphism of divided power thickenings}
$$
(B, J, \delta, C \to B/J) \longrightarrow (B', J', \delta', C \to B'/J')
$$
is a homomorphism $\varphi : B \to B'$ of divided power $A$-algebras such
that $C \to B/J \to B'/J'$ is the given map $C \to B'/J'$.
\item We denote $\text{CRIS}(C/A, I, \gamma)$ or simply $\text{CRIS}(C/A)$
the category of divided power thickenings of $C$ over $(A, I, \gamma)$.
\item We denote $\text{Cris}(C/A, I, \gamma)$ or simply $\text{Cris}(C/A)$
the full subcategory consisting of $(B, J, \delta, C \to B/J)$ such that
$C \to B/J$ is an isomorphism. We often denote such an object
$(B \to C, \delta)$ with $J = \Ker(B \to C)$ being understood.
\end{enumerate}
\end{definition}
\noindent
Note that for a divided power thickening $(B, J, \delta)$ as above
the ideal $J$ is locally nilpotent, see
Divided Power Algebra, Lemma \ref{dpa-lemma-nil}.
There is a canonical functor
\begin{equation}
\label{equation-forget-affine}
\text{CRIS}(C/A) \longrightarrow C\text{-algebras},\quad
(B, J, \delta) \longmapsto B/J
\end{equation}
This category does not have equalizers or fibre products in general.
It also doesn't have an initial object ($=$ empty colimit) in general.
\begin{lemma}
\label{lemma-affine-thickenings-colimits}
In Situation \ref{situation-affine}.
\begin{enumerate}
\item $\text{CRIS}(C/A)$ has products,
\item $\text{CRIS}(C/A)$ has all finite nonempty colimits and
(\ref{equation-forget-affine}) commutes with these, and
\item $\text{Cris}(C/A)$ has all finite nonempty colimits and
$\text{Cris}(C/A) \to \text{CRIS}(C/A)$ commutes with them.
\end{enumerate}
\end{lemma}
\begin{proof}
The empty product is $(C, 0, \emptyset)$. If $(B_t, J_t, \delta_t)$ is a
family of objects of $\text{CRIS}(C/A)$ then we can form the product
$(\prod B_t, \prod J_t, \prod \delta_t)$ as in
Divided Power Algebra, Lemma \ref{dpa-lemma-colimits}.
The map $C \to \prod B_t/\prod J_t = \prod B_t/J_t$ is clear.
\medskip\noindent
Given two objects $(B, J, \gamma)$ and $(B', J', \gamma')$ of
$\text{CRIS}(C/A)$ we can form a cocartesian diagram
$$
\xymatrix{
(B, J, \delta) \ar[r] & (B'', J'', \delta'') \\
(A, I, \gamma) \ar[r] \ar[u] & (B', J', \delta') \ar[u]
}
$$
in the category of divided power rings. Then we see that we have
$$
B''/J'' = B/J \otimes_{A/I} B'/J' \longleftarrow C \otimes_{A/I} C
$$
see Divided Power Algebra, Remark \ref{dpa-remark-forgetful}.
Denote $J'' \subset K \subset B''$
the ideal such that
$$
\xymatrix{
B''/J'' \ar[r] & B''/K \\
C \otimes_{A/I} C \ar[r] \ar[u] & C \ar[u]
}
$$
is a pushout, i.e., $B''/K \cong B/J \otimes_C B'/J'$.
Let $D_{B''}(K) = (D, \bar K, \bar \delta)$
be the divided power envelope of $K$ in $B''$ relative to
$(B'', J'', \delta'')$. Then it is easily verified that
$(D, \bar K, \bar \delta)$ is a coproduct of $(B, J, \delta)$ and
$(B', J', \delta')$ in $\text{CRIS}(C/A)$.
\medskip\noindent
Next, we come to coequalizers. Let
$\alpha, \beta : (B, J, \delta) \to (B', J', \delta')$ be morphisms of
$\text{CRIS}(C/A)$. Consider $B'' = B'/ (\alpha(b) - \beta(b))$. Let
$J'' \subset B''$ be the image of $J'$. Let
$D_{B''}(J'') = (D, \bar J, \bar\delta)$ be the divided power envelope of
$J''$ in $B''$ relative to $(B', J', \delta')$. Then it is easily verified
that $(D, \bar J, \bar \delta)$ is the coequalizer of $(B, J, \delta)$ and
$(B', J', \delta')$ in $\text{CRIS}(C/A)$.
\medskip\noindent
By Categories, Lemma \ref{categories-lemma-almost-finite-colimits-exist}
we have all finite nonempty colimits in $\text{CRIS}(C/A)$. The constructions
above shows that (\ref{equation-forget-affine}) commutes with them.
This formally implies part (3) as $\text{Cris}(C/A)$ is the fibre category
of (\ref{equation-forget-affine}) over $C$.
\end{proof}
\begin{remark}
\label{remark-completed-affine-site}
In Situation \ref{situation-affine} we denote
$\text{Cris}^\wedge(C/A)$ the category whose objects are
pairs $(B \to C, \delta)$ such that
\begin{enumerate}
\item $B$ is a $p$-adically complete $A$-algebra,
\item $B \to C$ is a surjection of $A$-algebras,
\item $\delta$ is a divided power structure on $\Ker(B \to C)$,
\item $A \to B$ is a homomorphism of divided power rings.
\end{enumerate}
Morphisms are defined as in Definition \ref{definition-affine-thickening}.
Then $\text{Cris}(C/A) \subset \text{Cris}^\wedge(C/A)$ is the full
subcategory consisting of those $B$ such that $p$ is nilpotent in $B$.
Conversely, any object $(B \to C, \delta)$ of $\text{Cris}^\wedge(C/A)$
is equal to the limit
$$
(B \to C, \delta) = \lim_e (B/p^eB \to C, \delta)
$$
where for $e \gg 0$ the object $(B/p^eB \to C, \delta)$ lies
in $\text{Cris}(C/A)$, see
Divided Power Algebra, Lemma \ref{dpa-lemma-extend-to-completion}.
In particular, we see that $\text{Cris}^\wedge(C/A)$ is a full subcategory
of the category of pro-objects of $\text{Cris}(C/A)$, see
Categories, Remark \ref{categories-remark-pro-category}.
\end{remark}
\begin{lemma}
\label{lemma-list-properties}
In Situation \ref{situation-affine}.
Let $P \to C$ be a surjection of $A$-algebras with kernel $J$.
Write $D_{P, \gamma}(J) = (D, \bar J, \bar\gamma)$.
Let $(D^\wedge, J^\wedge, \bar\gamma^\wedge)$ be the
$p$-adic completion of $D$, see
Divided Power Algebra, Lemma \ref{dpa-lemma-extend-to-completion}.
For every $e \geq 1$ set $P_e = P/p^eP$ and $J_e \subset P_e$
the image of $J$ and write
$D_{P_e, \gamma}(J_e) = (D_e, \bar J_e, \bar\gamma)$.
Then for all $e$ large enough we have
\begin{enumerate}
\item $p^eD \subset \bar J$ and $p^eD^\wedge \subset \bar J^\wedge$
are preserved by divided powers,
\item $D^\wedge/p^eD^\wedge = D/p^eD = D_e$ as divided power rings,
\item $(D_e, \bar J_e, \bar\gamma)$ is an object of $\text{Cris}(C/A)$,
\item $(D^\wedge, \bar J^\wedge, \bar\gamma^\wedge)$ is equal to
$\lim_e (D_e, \bar J_e, \bar\gamma)$, and
\item $(D^\wedge, \bar J^\wedge, \bar\gamma^\wedge)$ is an object of
$\text{Cris}^\wedge(C/A)$.
\end{enumerate}
\end{lemma}
\begin{proof}
Part (1) follows from
Divided Power Algebra, Lemma \ref{dpa-lemma-extend-to-completion}.
It is a general property of $p$-adic completion that
$D/p^eD = D^\wedge/p^eD^\wedge$. Since $D/p^eD$ is a divided power ring
and since $P \to D/p^eD$ factors through $P_e$, the universal property of
$D_e$ produces a map $D_e \to D/p^eD$. Conversely, the universal property
of $D$ produces a map $D \to D_e$ which factors through $D/p^eD$. We omit
the verification that these maps are mutually inverse. This proves (2).
If $e$ is large enough, then $p^eC = 0$, hence we see (3) holds.
Part (4) follows from
Divided Power Algebra, Lemma \ref{dpa-lemma-extend-to-completion}.
Part (5) is clear from the definitions.
\end{proof}
\begin{lemma}
\label{lemma-set-generators}
In Situation \ref{situation-affine}.
Let $P$ be a polynomial algebra over $A$ and let
$P \to C$ be a surjection of $A$-algebras with kernel $J$.
With $(D_e, \bar J_e, \bar\gamma)$ as in Lemma \ref{lemma-list-properties}:
for every object $(B, J_B, \delta)$ of $\text{CRIS}(C/A)$ there
exists an $e$ and a morphism $D_e \to B$ of $\text{CRIS}(C/A)$.
\end{lemma}
\begin{proof}
We can find an $A$-algebra homomorphism $P \to B$
lifting the map $C \to B/J_B$. By our definition of
$\text{CRIS}(C/A)$ we see that $p^eB = 0$ for
some $e$ hence $P \to B$ factors as $P \to P_e \to B$.
By the universal property of the divided power envelope we
conclude that $P_e \to B$ factors through $D_e$.
\end{proof}
\begin{lemma}
\label{lemma-generator-completion}
In Situation \ref{situation-affine}.
Let $P$ be a polynomial algebra over $A$ and let
$P \to C$ be a surjection of $A$-algebras with kernel $J$.
Let $(D, \bar J, \bar\gamma)$ be the $p$-adic completion of
$D_{P, \gamma}(J)$. For every object $(B \to C, \delta)$ of
$\text{Cris}^\wedge(C/A)$ there
exists a morphism $D \to B$ of $\text{Cris}^\wedge(C/A)$.
\end{lemma}
\begin{proof}
We can find an $A$-algebra homomorphism $P \to B$ compatible
with maps to $C$. By our definition of
$\text{Cris}(C/A)$ we see that $P \to B$ factors as
$P \to D_{P, \gamma}(J) \to B$. As $B$ is $p$-adically complete
we can factor this map through $D$.
\end{proof}
\section{Module of differentials}
\label{section-differentials}
\noindent
In this section we develop a theory of modules of differentials
for divided power rings.
\begin{definition}
\label{definition-derivation}
Let $A$ be a ring. Let $(B, J, \delta)$ be a divided power ring.
Let $A \to B$ be a ring map. Let $M$ be an $B$-module.
A {\it divided power $A$-derivation} into $M$ is a map
$\theta : B \to M$ which is additive, annihilates the elements
of $A$, satisfies the Leibniz rule
$\theta(bb') = b\theta(b') + b'\theta(b)$ and satisfies
$$
\theta(\delta_n(x)) = \delta_{n - 1}(x)\theta(x)
$$
for all $n \geq 1$ and all $x \in J$.
\end{definition}
\noindent
In the situation of the definition, just as in the case of usual
derivations, there exists a {\it universal divided power $A$-derivation}
$$
\text{d}_{B/A, \delta} : B \to \Omega_{B/A, \delta}
$$
such that any divided power $A$-derivation $\theta : B \to M$ is equal to
$\theta = \xi \circ d_{B/A, \delta}$ for some $B$-linear map
$\Omega_{B/A, \delta} \to M$. If $(A, I, \gamma) \to (B, J, \delta)$
is a homomorphism of divided power rings, then we can forget the
divided powers on $A$ and consider the divided power derivations of
$B$ over $A$. Here are some basic properties of the divided power
module of differentials.
\begin{lemma}
\label{lemma-omega}
Let $A$ be a ring. Let $(B, J, \delta)$ be a divided power ring and
$A \to B$ a ring map.
\begin{enumerate}
\item Consider $B[x]$ with divided power ideal $(JB[x], \delta')$
where $\delta'$ is the extension of $\delta$ to $B[x]$. Then
$$
\Omega_{B[x]/A, \delta'} =
\Omega_{B/A, \delta} \otimes_B B[x] \oplus B[x]\text{d}x.
$$
\item Consider $B\langle x \rangle$ with divided power ideal
$(JB\langle x \rangle + B\langle x \rangle_{+}, \delta')$. Then
$$
\Omega_{B\langle x\rangle/A, \delta'} =
\Omega_{B/A, \delta} \otimes_B B\langle x \rangle \oplus
B\langle x\rangle \text{d}x.
$$
\item Let $K \subset J$ be an ideal preserved by $\delta_n$ for
all $n > 0$. Set $B' = B/K$ and denote $\delta'$ the induced
divided power on $J/K$. Then $\Omega_{B'/A, \delta'}$ is the quotient
of $\Omega_{B/A, \delta} \otimes_B B'$ by the $B'$-submodule generated
by $\text{d}k$ for $k \in K$.
\end{enumerate}
\end{lemma}
\begin{proof}
These are proved directly from the construction of $\Omega_{B/A, \delta}$
as the free $B$-module on the elements $\text{d}b$ modulo the relations
\begin{enumerate}
\item $\text{d}(b + b') = \text{d}b + \text{d}b'$, $b, b' \in B$,
\item $\text{d}a = 0$, $a \in A$,
\item $\text{d}(bb') = b \text{d}b' + b' \text{d}b$, $b, b' \in B$,
\item $\text{d}\delta_n(f) = \delta_{n - 1}(f)\text{d}f$, $f \in J$, $n > 1$.
\end{enumerate}
Note that the last relation explains why we get ``the same'' answer for
the divided power polynomial algebra and the usual polynomial algebra:
in the first case $x$ is an element of the divided power ideal and hence
$\text{d}x^{[n]} = x^{[n - 1]}\text{d}x$.
\end{proof}
\noindent
Let $(A, I, \gamma)$ be a divided power ring. In this setting the
correct version of the powers of $I$ is given by the divided powers
$$
I^{[n]} = \text{ideal generated by }
\gamma_{e_1}(x_1) \ldots \gamma_{e_t}(x_t)
\text{ with }\sum e_j \geq n\text{ and }x_j \in I.
$$
Of course we have $I^n \subset I^{[n]}$. Note that $I^{[1]} = I$.
Sometimes we also set $I^{[0]} = A$.
\begin{lemma}
\label{lemma-diagonal-and-differentials}
Let $(A, I, \gamma) \to (B, J, \delta)$ be a homomorphism
of divided power rings. Let $(B(1), J(1), \delta(1))$ be the coproduct
of $(B, J, \delta)$ with itself over $(A, I, \gamma)$, i.e.,
such that
$$
\xymatrix{
(B, J, \delta) \ar[r] & (B(1), J(1), \delta(1)) \\
(A, I, \gamma) \ar[r] \ar[u] & (B, J, \delta) \ar[u]
}
$$
is cocartesian. Denote $K = \Ker(B(1) \to B)$.
Then $K \cap J(1) \subset J(1)$ is preserved by the divided power
structure and
$$
\Omega_{B/A, \delta} = K/ \left(K^2 + (K \cap J(1))^{[2]}\right)
$$
canonically.
\end{lemma}
\begin{proof}
The fact that $K \cap J(1) \subset J(1)$ is preserved by the divided power
structure follows from the fact that $B(1) \to B$ is a homomorphism of
divided power rings.
\medskip\noindent
Recall that $K/K^2$ has a canonical $B$-module structure.