Lemma 60.6.3. 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 $K = \mathop{\mathrm{Ker}}(B(1) \to B)$. Then $K \cap J(1) \subset J(1)$ is preserved by the divided power structure and

$\Omega _{B/A, \delta } = K/ \left(K^2 + (K \cap J(1))^{}\right)$

canonically.

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.

Recall that $K/K^2$ has a canonical $B$-module structure. Denote $s_0, s_1 : B \to B(1)$ the two coprojections and consider the map $\text{d} : B \to K/K^2 +(K \cap J(1))^{}$ given by $b \mapsto s_1(b) - s_0(b)$. It is clear that $\text{d}$ is additive, annihilates $A$, and satisfies the Leibniz rule. We claim that $\text{d}$ is a divided power $A$-derivation. Let $x \in J$. Set $y = s_1(x)$ and $z = s_0(x)$. Denote $\delta$ the divided power structure on $J(1)$. We have to show that $\delta _ n(y) - \delta _ n(z) = \delta _{n - 1}(y)(y - z)$ modulo $K^2 +(K \cap J(1))^{}$ for $n \geq 1$. The equality holds for $n = 1$. Assume $n > 1$. Note that $\delta _ i(y - z)$ lies in $(K \cap J(1))^{}$ for $i > 1$. Calculating modulo $K^2 + (K \cap J(1))^{}$ we have

$\delta _ n(z) = \delta _ n(z - y + y) = \sum \nolimits _{i = 0}^ n \delta _ i(z - y)\delta _{n - i}(y) = \delta _{n - 1}(y) \delta _1(z - y) + \delta _ n(y)$

This proves the desired equality.

Let $M$ be a $B$-module. Let $\theta : B \to M$ be a divided power $A$-derivation. Set $D = B \oplus M$ where $M$ is an ideal of square zero. Define a divided power structure on $J \oplus M \subset D$ by setting $\delta _ n(x + m) = \delta _ n(x) + \delta _{n - 1}(x)m$ for $n > 1$, see Lemma 60.3.1. There are two divided power algebra homomorphisms $B \to D$: the first is given by the inclusion and the second by the map $b \mapsto b + \theta (b)$. Hence we get a canonical homomorphism $B(1) \to D$ of divided power algebras over $(A, I, \gamma )$. This induces a map $K \to M$ which annihilates $K^2$ (as $M$ is an ideal of square zero) and $(K \cap J(1))^{}$ as $M^{} = 0$. The composition $B \to K/K^2 + (K \cap J(1))^{} \to M$ equals $\theta$ by construction. It follows that $\text{d}$ is a universal divided power $A$-derivation and we win. $\square$

Comment #4222 by Dario Weißmann on

Concerning the second paragraph of the proof. I suggest replacing "We will show this by induction ...." (till the end of the paragraph) by the shorter (and easier) direct calculation: The claim is clear for $n=1$. Assume $n>1$. Note that $\delta_i(z-y)$ lies in $(K\cap J(1))^{}$ for $i>1$. Calculating modulo $K^2 + (K\cap J(1))^{}$ we have The claim follows.

Comment #4402 by Dario Weißmann on

Noticed two typos in the fix: "Calculating module..." and "This prove ..."

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