## 60.3 Some explicit divided power thickenings

The constructions in this section will help us to define the connection on a crystal in modules on the crystalline site.

Lemma 60.3.1. 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 $(B, J, \delta )$.

Proof. We have to check conditions (1) – (5) of Divided Power Algebra, Definition 23.2.1. 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 $\gamma$. $\square$

Lemma 60.3.2. 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

$\delta _ n(x, z, w) = (\gamma _ n(x), \gamma _{n - 1}(x)z, \gamma _{n - 1}(x)w + \gamma _{n - 2}(x)q(z, z))$

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 $(B, J, \delta )$.

Proof. Suppose we want to prove that property (4) of Divided Power Algebra, Definition 23.2.1 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 23.2.2. 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. $\square$

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