The Stacks project

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 10.81.4) 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 $y_ j = \sum x_ ia_{ij} = \sum x'_{i'}a'_{i'j}$ 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 10.81.4. 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 10.39.11 (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.

Comments (0)


Post a comment

Your email address will not be published. Required fields are marked.

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.

All contributions are licensed under the GNU Free Documentation License.




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 07H1. Beware of the difference between the letter 'O' and the digit '0'.