The Stacks project

23.4 Extending divided powers

Here is the definition.

Definition 23.4.1. Given a divided power ring $(A, I, \gamma )$ and a ring map $A \to B$ we say $\gamma $ extends to $B$ if there exists a divided power structure $\bar\gamma $ on $IB$ such that $(A, I, \gamma ) \to (B, IB, \bar\gamma )$ is a homomorphism of divided power rings.

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 $IB$. $\square$

Lemma 23.4.3. Let $(A, I, \gamma )$ be a divided power ring.

  1. If $\varphi : (A, I, \gamma ) \to (B, J, \delta )$ is a homomorphism of divided power rings, then $\mathop{\mathrm{Ker}}(\varphi ) \cap I$ is preserved by $\gamma _ n$ for all $n \geq 1$.

  2. Let $\mathfrak a \subset A$ be an ideal and set $I' = I \cap \mathfrak a$. The following are equivalent

    1. $I'$ is preserved by $\gamma _ n$ for all $n > 0$,

    2. $\gamma $ extends to $A/\mathfrak a$, and

    3. there exist a set of generators $x_ i$ of $I'$ as an ideal such that $\gamma _ n(x_ i) \in I'$ for all $n > 0$.

Proof. Proof of (1). This is clear. Assume (2)(a). Define $\bar\gamma _ n(x \bmod I') = \gamma _ n(x) \bmod I'$ for $x \in I$. This is well defined since $\gamma _ n(x + y) = \gamma _ n(x) \bmod I'$ for $y \in I'$ by Definition 23.2.1 (4) and the fact that $\gamma _ j(y) \in I'$ by assumption. It is clear that $\bar\gamma $ is a divided power structure as $\gamma $ is one. Hence (2)(b) holds. Also, (2)(b) implies (2)(a) by part (1). It is clear that (2)(a) implies (2)(c). Assume (2)(c). Note that $\gamma _ n(x) = a^ n\gamma _ n(x_ i) \in I'$ for $x = ax_ i$. Hence we see that $\gamma _ n(x) \in I'$ for a set of generators of $I'$ as an abelian group. By induction on the length of an expression in terms of these, it suffices to prove $\forall n : \gamma _ n(x + y) \in I'$ if $\forall n : \gamma _ n(x), \gamma _ n(y) \in I'$. This follows immediately from the fourth axiom of a divided power structure. $\square$

Lemma 23.4.4. Let $(A, I, \gamma )$ be a divided power ring. Let $E \subset I$ be a subset. Then the smallest ideal $J \subset I$ preserved by $\gamma $ and containing all $f \in E$ is the ideal $J$ generated by $\gamma _ n(f)$, $n \geq 1$, $f \in E$.

Proof. Follows immediately from Lemma 23.4.3. $\square$

Lemma 23.4.5. Let $(A, I, \gamma )$ be a divided power ring. Let $p$ be a prime. If $p$ is nilpotent in $A/I$, then

  1. the $p$-adic completion $A^\wedge = \mathop{\mathrm{lim}}\nolimits _ e A/p^ eA$ surjects onto $A/I$,

  2. the kernel of this map is the $p$-adic completion $I^\wedge $ of $I$, and

  3. each $\gamma _ n$ is continuous for the $p$-adic topology and extends to $\gamma _ n^\wedge : I^\wedge \to I^\wedge $ defining a divided power structure on $I^\wedge $.

If moreover $A$ is a $\mathbf{Z}_{(p)}$-algebra, then

  1. for $e$ large enough the ideal $p^ eA \subset I$ is preserved by the divided power structure $\gamma $ and

    \[ (A^\wedge , I^\wedge , \gamma ^\wedge ) = \mathop{\mathrm{lim}}\nolimits _ e (A/p^ eA, I/p^ eA, \bar\gamma ) \]

    in the category of divided power rings.

Proof. Let $t \geq 1$ be an integer such that $p^ tA/I = 0$, i.e., $p^ tA \subset I$. The map $A^\wedge \to A/I$ is the composition $A^\wedge \to A/p^ tA \to A/I$ which is surjective (for example by Algebra, Lemma 10.96.1). As $p^ eI \subset p^ eA \cap I \subset p^{e - t}I$ for $e \geq t$ we see that the kernel of the composition $A^\wedge \to A/I$ is the $p$-adic completion of $I$. The map $\gamma _ n$ is continuous because

\[ \gamma _ n(x + p^ ey) = \sum \nolimits _{i + j = n} p^{je}\gamma _ i(x)\gamma _ j(y) = \gamma _ n(x) \bmod p^ eI \]

by the axioms of a divided power structure. It is clear that the axioms for divided power structures are inherited by the maps $\gamma _ n^\wedge $ from the maps $\gamma _ n$. Finally, to see the last statement say $e > t$. Then $p^ eA \subset I$ and $\gamma _1(p^ eA) \subset p^ eA$ and for $n > 1$ we have

\[ \gamma _ n(p^ ea) = p^ n \gamma _ n(p^{e - 1}a) = \frac{p^ n}{n!} p^{n(e - 1)}a^ n \in p^ e A \]

as $p^ n/n! \in \mathbf{Z}_{(p)}$ and as $n \geq 2$ and $e \geq 2$ so $n(e - 1) \geq e$. This proves that $\gamma $ extends to $A/p^ eA$, see Lemma 23.4.3. The statement on limits is clear from the construction of limits in the proof of Lemma 23.3.2. $\square$

[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.

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