Lemma 23.2.4 (07GP)—The Stacks project

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. $\square$

Comments (2)

Comment #1331 by Shishir Agrawal on February 28, 2015 at 00:02

It doesn't really affect the proof, but when verifying axiom (5), at the point when axiom (2) is applied $k$ times, it appears to me that there is a small typo: I believe it should be that $\gamma_j(y)^k = \frac{(kj)!}{(j!)^k}\gamma_{kj}(y)$ , rather than $\gamma_j(y)^k = \frac{(kj)!}{k!(j!)^k}\gamma_{kj}(y)$ as is asserted.

Comment #1351 by Johan on March 12, 2015 at 19:40

Indeed! Thanks very much. See here for corresponding change.

Comments (2)

Post a comment