# The Stacks Project

## Tag 07H1

Lemma 23.4.2. Let $(A, I, \gamma)$ be a divided power ring. Let $A \to B$ be a ring map. If $\gamma$ extends to $B$ then it extends uniquely. Assume (at least) one of the following conditions holds

1. $IB = 0$,
2. $I$ is principal, or
3. $A \to B$ is flat.

Then $\gamma$ extends to $B$.

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 10.80.4) 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 $y_j = \sum x_ia_{ij} = \sum x'_{i'}a'_{i'j}$ holds1. 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 $IB$. $\square$

1. This can also be proven without recourse to Algebra, Theorem 10.80.4. Indeed, if $z = \sum x_ib_i$ and $z = \sum x'_{i'}b'_{i'}$, then $\sum x_ib_i - \sum x'_{i'}b'_{i'} = 0$ is a relation in the $A$-module $B$. Thus, Algebra, Lemma 10.38.11 (applied to the $x_i$ and $x'_{i'}$ taking the place of the $f_i$, and the $b_i$ and $b'_{i'}$ taking the role of the $x_i$) yields the existence of the $c_1, \ldots, c_s \in B$ and $a_{ij}, a'_{i'j} \in A$ as required.

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\begin{lemma}
\label{lemma-gamma-extends}
Let $(A, I, \gamma)$ be a divided power ring.
Let $A \to B$ be a ring map.
If $\gamma$ extends to $B$ then it extends uniquely.
Assume (at least) one of the following conditions holds
\begin{enumerate}
\item $IB = 0$,
\item $I$ is principal, or
\item $A \to B$ is flat.
\end{enumerate}
Then $\gamma$ extends to $B$.
\end{lemma}

\begin{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 \ref{definition-divided-powers} 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.

\medskip\noindent
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 \ref{definition-divided-powers}.
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.

\medskip\noindent
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 \ref{algebra-theorem-lazard})
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\footnote{This
can also be proven without recourse to
Algebra, Theorem \ref{algebra-theorem-lazard}. Indeed, if
$z = \sum x_ib_i$ and $z = \sum x'_{i'}b'_{i'}$, then
$\sum x_ib_i - \sum x'_{i'}b'_{i'} = 0$ is a relation in the
$A$-module $B$. Thus, Algebra, Lemma \ref{algebra-lemma-flat-eq}
(applied to the $x_i$ and $x'_{i'}$ taking the place of the $f_i$,
and the $b_i$ and $b'_{i'}$ taking the role of the $x_i$) yields
the existence of the $c_1, \ldots, c_s \in B$
and $a_{ij}, a'_{i'j} \in A$ as required.}.
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 \ref{lemma-check-on-generators} that $\bar\gamma$
is a divided power structure on $IB$.
\end{proof}

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