**Proof.**
Any element of $IB$ can be written as a finite sum $\sum \nolimits _{i=1}^ t b_ ix_ i$ with $b_ i \in B$ and $x_ i \in I$. If $\gamma $ extends to $\bar\gamma $ on $IB$ then $\bar\gamma _ n(x_ i) = \gamma _ n(x_ i)$. Thus, conditions (3) and (4) in Definition 23.2.1 imply that

\[ \bar\gamma _ n(\sum \nolimits _{i=1}^ t b_ ix_ i) = \sum \nolimits _{n_1 + \ldots + n_ t = n} \prod \nolimits _{i = 1}^ t b_ i^{n_ i}\gamma _{n_ i}(x_ i) \]

Thus we see that $\bar\gamma $ is unique if it exists.

If $IB = 0$ then setting $\bar\gamma _ n(0) = 0$ works. If $I = (x)$ then we define $\bar\gamma _ n(bx) = b^ n\gamma _ n(x)$. This is well defined: if $b'x = bx$, i.e., $(b - b')x = 0$ then

\begin{align*} b^ n\gamma _ n(x) - (b')^ n\gamma _ n(x) & = (b^ n - (b')^ n)\gamma _ n(x) \\ & = (b^{n - 1} + \ldots + (b')^{n - 1})(b - b')\gamma _ n(x) = 0 \end{align*}

because $\gamma _ n(x)$ is divisible by $x$ (since $\gamma _ n(I) \subset I$) and hence annihilated by $b - b'$. Next, we prove conditions (1) – (5) of Definition 23.2.1. Parts (1), (2), (3), (5) are obvious from the construction. For (4) suppose that $y, z \in IB$, say $y = bx$ and $z = cx$. Then $y + z = (b + c)x$ hence

\begin{align*} \bar\gamma _ n(y + z) & = (b + c)^ n\gamma _ n(x) \\ & = \sum \frac{n!}{i!(n - i)!}b^ ic^{n -i}\gamma _ n(x) \\ & = \sum b^ ic^{n - i}\gamma _ i(x)\gamma _{n - i}(x) \\ & = \sum \bar\gamma _ i(y)\bar\gamma _{n -i}(z) \end{align*}

as desired.

Assume $A \to B$ is flat. Suppose that $b_1, \ldots , b_ r \in B$ and $x_1, \ldots , x_ r \in I$. Then

\[ \bar\gamma _ n(\sum b_ ix_ i) = \sum b_1^{e_1} \ldots b_ r^{e_ r} \gamma _{e_1}(x_1) \ldots \gamma _{e_ r}(x_ r) \]

where the sum is over $e_1 + \ldots + e_ r = n$ if $\bar\gamma _ n$ exists. Next suppose that we have $c_1, \ldots , c_ s \in B$ and $a_{ij} \in A$ such that $b_ i = \sum a_{ij}c_ j$. Setting $y_ j = \sum a_{ij}x_ i$ we claim that

\[ \sum b_1^{e_1} \ldots b_ r^{e_ r} \gamma _{e_1}(x_1) \ldots \gamma _{e_ r}(x_ r) = \sum c_1^{d_1} \ldots c_ s^{d_ s} \gamma _{d_1}(y_1) \ldots \gamma _{d_ s}(y_ s) \]

in $B$ where on the right hand side we are summing over $d_1 + \ldots + d_ s = n$. Namely, using the axioms of a divided power structure we can expand both sides into a sum with coefficients in $\mathbf{Z}[a_{ij}]$ of terms of the form $c_1^{d_1} \ldots c_ s^{d_ s}\gamma _{e_1}(x_1) \ldots \gamma _{e_ r}(x_ r)$. To see that the coefficients agree we note that the result is true in $\mathbf{Q}[x_1, \ldots , x_ r, c_1, \ldots , c_ s, a_{ij}]$ with $\gamma $ the unique divided power structure on $(x_1, \ldots , x_ r)$. By Lazard's theorem (Algebra, Theorem 10.81.4) we can write $B$ as a directed colimit of finite free $A$-modules. In particular, if $z \in IB$ is written as $z = \sum x_ ib_ i$ and $z = \sum x'_{i'}b'_{i'}$, then we can find $c_1, \ldots , c_ s \in B$ and $a_{ij}, a'_{i'j} \in A$ such that $b_ i = \sum a_{ij}c_ j$ and $b'_{i'} = \sum a'_{i'j}c_ j$ such that $y_ j = \sum x_ ia_{ij} = \sum x'_{i'}a'_{i'j}$ holds^{1}. Hence the procedure above gives a well defined map $\bar\gamma _ n$ on $IB$. By construction $\bar\gamma $ satisfies conditions (1), (3), and (4). Moreover, for $x \in I$ we have $\bar\gamma _ n(x) = \gamma _ n(x)$. Hence it follows from Lemma 23.2.4 that $\bar\gamma $ is a divided power structure on $IB$.
$\square$

## Comments (0)