The Stacks project

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

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

Comparing this with the expansion

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

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

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

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. $\square$