Lemma 23.5.3. Let $p$ be a prime number. Let $A$ be a ring such that every integer $n$ not divisible by $p$ is invertible, i.e., $A$ is a $\mathbf{Z}_{(p)}$-algebra. Let $I \subset A$ be an ideal. Two divided power structures $\gamma , \gamma '$ on $I$ are equal if and only if $\gamma _ p = \gamma '_ p$. Moreover, given a map $\delta : I \to I$ such that

$p!\delta (x) = x^ p$ for all $x \in I$,

$\delta (ax) = a^ p\delta (x)$ for all $a \in A$, $x \in I$, and

$\delta (x + y) = \delta (x) + \sum \nolimits _{i + j = p, i,j \geq 1} \frac{1}{i!j!} x^ i y^ j + \delta (y)$ for all $x, y \in I$,

then there exists a unique divided power structure $\gamma $ on $I$ such that $\gamma _ p = \delta $.

**Proof.**
If $n$ is not divisible by $p$, then $\gamma _ n(x) = c x \gamma _{n - 1}(x)$ where $c$ is a unit in $\mathbf{Z}_{(p)}$. Moreover,

\[ \gamma _{pm}(x) = c \gamma _ m(\gamma _ p(x)) \]

where $c$ is a unit in $\mathbf{Z}_{(p)}$. Thus the first assertion is clear. For the second assertion, we can, working backwards, use these equalities to define all $\gamma _ n$. More precisely, if $n = a_0 + a_1p + \ldots + a_ e p^ e$ with $a_ i \in \{ 0, \ldots , p - 1\} $ then we set

\[ \gamma _ n(x) = c_ n x^{a_0} \delta (x)^{a_1} \ldots \delta ^ e(x)^{a_ e} \]

for $c_ n \in \mathbf{Z}_{(p)}$ defined by

\[ c_ n = {(p!)^{a_1 + a_2(1 + p) + \ldots + a_ e(1 + \ldots + p^{e - 1})}}/{n!}. \]

Now we have to show the axioms (1) – (5) of a divided power structure, see Definition 23.2.1. We observe that (1) and (3) are immediate. Verification of (2) and (5) is by a direct calculation which we omit. Let $x, y \in I$. We claim there is a ring map

\[ \varphi : \mathbf{Z}_{(p)}\langle u, v \rangle \longrightarrow A \]

which maps $u^{[n]}$ to $\gamma _ n(x)$ and $v^{[n]}$ to $\gamma _ n(y)$. By construction of $\mathbf{Z}_{(p)}\langle u, v \rangle $ this means we have to check that

\[ \gamma _ n(x)\gamma _ m(x) = \frac{(n + m)!}{n!m!} \gamma _{n + m}(x) \]

in $A$ and similarly for $y$. This is true because (2) holds for $\gamma $. Let $\epsilon $ denote the divided power structure on the ideal $\mathbf{Z}_{(p)}\langle u, v\rangle _{+}$ of $\mathbf{Z}_{(p)}\langle u, v\rangle $. Next, we claim that $\varphi (\epsilon _ n(f)) = \gamma _ n(\varphi (f))$ for $f \in \mathbf{Z}_{(p)}\langle u, v\rangle _{+}$ and all $n$. This is clear for $n = 0, 1, \ldots , p - 1$. For $n = p$ it suffices to prove it for a set of generators of the ideal $\mathbf{Z}_{(p)}\langle u, v\rangle _{+}$ because both $\epsilon _ p$ and $\gamma _ p = \delta $ satisfy properties (1) and (3) of the lemma. Hence it suffices to prove that $\gamma _ p(\gamma _ n(x)) = \frac{(pn)!}{p!(n!)^ p}\gamma _{pn}(x)$ and similarly for $y$, which follows as (5) holds for $\gamma $. Now, if $n = a_0 + a_1p + \ldots + a_ e p^ e$ is an arbitrary integer written in $p$-adic expansion as above, then

\[ \epsilon _ n(f) = c_ n f^{a_0} \gamma _ p(f)^{a_1} \ldots \gamma _ p^ e(f)^{a_ e} \]

because $\epsilon $ is a divided power structure. Hence we see that $\varphi (\epsilon _ n(f)) = \gamma _ n(\varphi (f))$ holds for all $n$. Applying this for $f = u + v$ we see that axiom (4) for $\gamma $ follows from the fact that $\epsilon $ is a divided power structure.
$\square$

## Comments (1)

Comment #9550 by Ryo Suzuki on