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

  2. 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

It doesn't really affect the proof, but when verifying axiom (5), at the point when axiom (2) is applied times, it appears to me that there is a small typo: I believe it should be that , rather than as is asserted.

Comment #1351 by on

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

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