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 $\gamma$. $\square$

In your comment you can use Markdown and LaTeX style mathematics (enclose it like $\pi$). A preview option is available if you wish to see how it works out (just click on the eye in the toolbar).

Unfortunately JavaScript is disabled in your browser, so the comment preview function will not work.

In order to prevent bots from posting comments, we would like you to prove that you are human. You can do this by filling in the name of the current tag in the following input field. As a reminder, this is tag 07HH. Beware of the difference between the letter 'O' and the digit '0'.