**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.
$\square$

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