## 23.2 Divided powers

In this section we collect some results on divided power rings. We will use the convention $0! = 1$ (as empty products should give $1$).

Definition 23.2.1. Let $A$ be a ring. Let $I$ be an ideal of $A$. A collection of maps $\gamma _ n : I \to I$, $n > 0$ is called a *divided power structure* on $I$ if for all $n \geq 0$, $m > 0$, $x, y \in I$, and $a \in A$ we have

$\gamma _1(x) = x$, we also set $\gamma _0(x) = 1$,

$\gamma _ n(x)\gamma _ m(x) = \frac{(n + m)!}{n! m!} \gamma _{n + m}(x)$,

$\gamma _ n(ax) = a^ n \gamma _ n(x)$,

$\gamma _ n(x + y) = \sum _{i = 0, \ldots , n} \gamma _ i(x)\gamma _{n - i}(y)$,

$\gamma _ n(\gamma _ m(x)) = \frac{(nm)!}{n! (m!)^ n} \gamma _{nm}(x)$.

Note that the rational numbers $\frac{(n + m)!}{n! m!}$ and $\frac{(nm)!}{n! (m!)^ n}$ occurring in the definition are in fact integers; the first is the number of ways to choose $n$ out of $n + m$ and the second counts the number of ways to divide a group of $nm$ objects into $n$ groups of $m$. We make some remarks about the definition which show that $\gamma _ n(x)$ is a replacement for $x^ n/n!$ in $I$.

Lemma 23.2.2. Let $A$ be a ring. Let $I$ be an ideal of $A$.

If $\gamma $ is a divided power structure^{1} on $I$, then $n! \gamma _ n(x) = x^ n$ for $n \geq 1$, $x \in I$.

Assume $A$ is torsion free as a $\mathbf{Z}$-module.

A divided power structure on $I$, if it exists, is unique.

If $\gamma _ n : I \to I$ are maps then

\[ \gamma \text{ is a divided power structure} \Leftrightarrow n! \gamma _ n(x) = x^ n\ \forall x \in I, n \geq 1. \]

The ideal $I$ has a divided power structure if and only if there exists a set of generators $x_ i$ of $I$ as an ideal such that for all $n \geq 1$ we have $x_ i^ n \in (n!)I$.

**Proof.**
Proof of (1). If $\gamma $ is a divided power structure, then condition (2) (applied to $1$ and $n-1$ instead of $n$ and $m$) implies that $n \gamma _ n(x) = \gamma _1(x)\gamma _{n - 1}(x)$. Hence by induction and condition (1) we get $n! \gamma _ n(x) = x^ n$.

Assume $A$ is torsion free as a $\mathbf{Z}$-module. Proof of (2). This is clear from (1).

Proof of (3). Assume that $n! \gamma _ n(x) = x^ n$ for all $x \in I$ and $n \geq 1$. Since $A \subset A \otimes _{\mathbf{Z}} \mathbf{Q}$ it suffices to prove the axioms (1) – (5) of Definition 23.2.1 in case $A$ is a $\mathbf{Q}$-algebra. In this case $\gamma _ n(x) = x^ n/n!$ and it is straightforward to verify (1) – (5); for example, (4) corresponds to the binomial formula

\[ (x + y)^ n = \sum _{i = 0, \ldots , n} \frac{n!}{i!(n - i)!} x^ iy^{n - i} \]

We encourage the reader to do the verifications to make sure that we have the coefficients correct.

Proof of (4). Assume we have generators $x_ i$ of $I$ as an ideal such that $x_ i^ n \in (n!)I$ for all $n \geq 1$. We claim that for all $x \in I$ we have $x^ n \in (n!)I$. If the claim holds then we can set $\gamma _ n(x) = x^ n/n!$ which is a divided power structure by (3). To prove the claim we note that it holds for $x = ax_ i$. Hence we see that the claim holds 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 the claim for $x + y$ if it holds for $x$ and $y$. This follows immediately from the binomial theorem.
$\square$

Example 23.2.3. 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. Then $I = pA$ has a canonical divided power structure. Namely, given $x = pa \in I$ we set

\[ \gamma _ n(x) = \frac{p^ n}{n!} a^ n \]

The reader verifies immediately that $p^ n/n! \in p\mathbf{Z}_{(p)}$ for $n \geq 1$ (for instance, this can be derived from the fact that the exponent of $p$ in the prime factorization of $n!$ is $\left\lfloor n/p \right\rfloor + \left\lfloor n/p^2 \right\rfloor + \left\lfloor n/p^3 \right\rfloor + \ldots $), so that the definition makes sense and gives us a sequence of maps $\gamma _ n : I \to I$. It is a straightforward exercise to verify that conditions (1) – (5) of Definition 23.2.1 are satisfied. Alternatively, it is clear that the definition works for $A_0 = \mathbf{Z}_{(p)}$ and then the result follows from Lemma 23.4.2.

We notice that $\gamma _ n\left(0\right) = 0$ for any ideal $I$ of $A$ and any divided power structure $\gamma $ on $I$. (This follows from axiom (3) in Definition 23.2.1, applied to $a=0$.)

Lemma 23.2.4. Let $A$ be a ring. Let $I$ be an ideal of $A$. Let $\gamma _ n : I \to I$, $n \geq 1$ be a sequence of maps. Assume

(1), (3), and (4) of Definition 23.2.1 hold for all $x, y \in I$, and

properties (2) and (5) hold for $x$ in some set of generators of $I$ as an ideal.

Then $\gamma $ is a divided power structure on $I$.

**Proof.**
The numbers (1), (2), (3), (4), (5) in this proof refer to the conditions listed in Definition 23.2.1. Applying (3) we see that if (2) and (5) hold for $x$ then (2) and (5) hold for $ax$ for all $a \in A$. Hence we see (b) implies (2) and (5) hold for a set of generators of $I$ as an abelian group. Hence, by induction of the length of an expression in terms of these it suffices to prove that, given $x, y \in I$ such that (2) and (5) hold for $x$ and $y$, then (2) and (5) hold for $x + y$.

Proof of (2) for $x + y$. By (4) we have

\[ \gamma _ n(x + y)\gamma _ m(x + y) = \sum \nolimits _{i + j = n,\ k + l = m} \gamma _ i(x)\gamma _ k(x)\gamma _ j(y)\gamma _ l(y) \]

Using (2) for $x$ and $y$ this equals

\[ \sum \frac{(i + k)!}{i!k!}\frac{(j + l)!}{j!l!} \gamma _{i + k}(x)\gamma _{j + l}(y) \]

Comparing this with the expansion

\[ \gamma _{n + m}(x + y) = \sum \gamma _ a(x)\gamma _ b(y) \]

we see that we have to prove that given $a + b = n + m$ we have

\[ \sum \nolimits _{i + k = a,\ j + l = b,\ i + j = n,\ k + l = m} \frac{(i + k)!}{i!k!}\frac{(j + l)!}{j!l!} = \frac{(n + m)!}{n!m!}. \]

Instead of arguing this directly, we note that the result is true for the ideal $I = (x, y)$ in the polynomial ring $\mathbf{Q}[x, y]$ because $\gamma _ n(f) = f^ n/n!$, $f \in I$ defines a divided power structure on $I$. Hence the equality of rational numbers above is true.

Proof of (5) for $x + y$ given that (1) – (4) hold and that (5) holds for $x$ and $y$. We will again reduce the proof to an equality of rational numbers. Namely, using (4) we can write $\gamma _ n(\gamma _ m(x + y)) = \gamma _ n(\sum \gamma _ i(x)\gamma _ j(y))$. Using (4) we can write $\gamma _ n(\gamma _ m(x + y))$ as a sum of terms which are products of factors of the form $\gamma _ k(\gamma _ i(x)\gamma _ j(y))$. If $i > 0$ then

\begin{align*} \gamma _ k(\gamma _ i(x)\gamma _ j(y)) & = \gamma _ j(y)^ k\gamma _ k(\gamma _ i(x)) \\ & = \frac{(ki)!}{k!(i!)^ k} \gamma _ j(y)^ k \gamma _{ki}(x) \\ & = \frac{(ki)!}{k!(i!)^ k} \frac{(kj)!}{(j!)^ k} \gamma _{ki}(x) \gamma _{kj}(y) \end{align*}

using (3) in the first equality, (5) for $x$ in the second, and (2) exactly $k$ times in the third. Using (5) for $y$ we see the same equality holds when $i = 0$. Continuing like this using all axioms but (5) we see that we can write

\[ \gamma _ n(\gamma _ m(x + y)) = \sum \nolimits _{i + j = nm} c_{ij}\gamma _ i(x)\gamma _ j(y) \]

for certain universal constants $c_{ij} \in \mathbf{Z}$. Again the fact that the equality is valid in the polynomial ring $\mathbf{Q}[x, y]$ implies that the coefficients $c_{ij}$ are all equal to $(nm)!/n!(m!)^ n$ as desired.
$\square$

Lemma 23.2.5. Let $A$ be a ring with two ideals $I, J \subset A$. Let $\gamma $ be a divided power structure on $I$ and let $\delta $ be a divided power structure on $J$. Then

$\gamma $ and $\delta $ agree on $IJ$,

if $\gamma $ and $\delta $ agree on $I \cap J$ then they are the restriction of a unique divided power structure $\epsilon $ on $I + J$.

**Proof.**
Let $x \in I$ and $y \in J$. Then

\[ \gamma _ n(xy) = y^ n\gamma _ n(x) = n! \delta _ n(y) \gamma _ n(x) = \delta _ n(y) x^ n = \delta _ n(xy). \]

Hence $\gamma $ and $\delta $ agree on a set of (additive) generators of $IJ$. By property (4) of Definition 23.2.1 it follows that they agree on all of $IJ$.

Assume $\gamma $ and $\delta $ agree on $I \cap J$. Let $z \in I + J$. Write $z = x + y$ with $x \in I$ and $y \in J$. Then we set

\[ \epsilon _ n(z) = \sum \gamma _ i(x)\delta _{n - i}(y) \]

for all $n \geq 1$. To see that this is well defined, suppose that $z = x' + y'$ is another representation with $x' \in I$ and $y' \in J$. Then $w = x - x' = y' - y \in I \cap J$. Hence

\begin{align*} \sum \nolimits _{i + j = n} \gamma _ i(x)\delta _ j(y) & = \sum \nolimits _{i + j = n} \gamma _ i(x' + w)\delta _ j(y) \\ & = \sum \nolimits _{i' + l + j = n} \gamma _{i'}(x')\gamma _ l(w)\delta _ j(y) \\ & = \sum \nolimits _{i' + l + j = n} \gamma _{i'}(x')\delta _ l(w)\delta _ j(y) \\ & = \sum \nolimits _{i' + j' = n} \gamma _{i'}(x')\delta _{j'}(y + w) \\ & = \sum \nolimits _{i' + j' = n} \gamma _{i'}(x')\delta _{j'}(y') \end{align*}

as desired. Hence, we have defined maps $\epsilon _ n : I + J \to I + J$ for all $n \geq 1$; it is easy to see that $\epsilon _ n \mid _{I} = \gamma _ n$ and $\epsilon _ n \mid _{J} = \delta _ n$. Next, we prove conditions (1) – (5) of Definition 23.2.1 for the collection of maps $\epsilon _ n$. Properties (1) and (3) are clear. To see (4), suppose that $z = x + y$ and $z' = x' + y'$ with $x, x' \in I$ and $y, y' \in J$ and compute

\begin{align*} \epsilon _ n(z + z') & = \sum \nolimits _{a + b = n} \gamma _ a(x + x')\delta _ b(y + y') \\ & = \sum \nolimits _{i + i' + j + j' = n} \gamma _ i(x) \gamma _{i'}(x')\delta _ j(y)\delta _{j'}(y') \\ & = \sum \nolimits _{k = 0, \ldots , n} \sum \nolimits _{i+j=k} \gamma _ i(x)\delta _ j(y) \sum \nolimits _{i'+j'=n-k} \gamma _{i'}(x')\delta _{j'}(y') \\ & = \sum \nolimits _{k = 0, \ldots , n}\epsilon _ k(z)\epsilon _{n-k}(z') \end{align*}

as desired. Now we see that it suffices to prove (2) and (5) for elements of $I$ or $J$, see Lemma 23.2.4. This is clear because $\gamma $ and $\delta $ are divided power structures.

The existence of a divided power structure $\epsilon $ on $I+J$ whose restrictions to $I$ and $J$ are $\gamma $ and $\delta $ is thus proven; its uniqueness is rather clear.
$\square$

Lemma 23.2.6. Let $p$ be a prime number. Let $A$ be a ring, let $I \subset A$ be an ideal, and let $\gamma $ be a divided power structure on $I$. Assume $p$ is nilpotent in $A/I$. Then $I$ is locally nilpotent if and only if $p$ is nilpotent in $A$.

**Proof.**
If $p^ N = 0$ in $A$, then for $x \in I$ we have $x^{pN} = (pN)!\gamma _{pN}(x) = 0$ because $(pN)!$ is divisible by $p^ N$. Conversely, assume $I$ is locally nilpotent. We've also assumed that $p$ is nilpotent in $A/I$, hence $p^ r \in I$ for some $r$, hence $p^ r$ nilpotent, hence $p$ nilpotent.
$\square$

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