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 $.
$\square$

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