Lemma 60.3.2. Let $(A, I, \gamma )$ be a divided power ring. Let $M$, $N$ be $A$-modules. Let $q : M \times M \to N$ be an $A$-bilinear map. Let $B = A \oplus M \oplus N$ as an $A$-algebra with multiplication

\[ (x, z, w)\cdot (x', z', w') = (xx', xz' + x'z, xw' + x'w + q(z, z') + q(z', z)) \]

and set $J = I \oplus M \oplus N$. Set

\[ \delta _ n(x, z, w) = (\gamma _ n(x), \gamma _{n - 1}(x)z, \gamma _{n - 1}(x)w + \gamma _{n - 2}(x)q(z, z)) \]

for $(x, z, w) \in J$. Then $\delta $ is a divided power structure and $A \to B$ is a homomorphism of divided power rings from $(A, I, \gamma )$ to $(B, J, \delta )$.

**Proof.**
Suppose we want to prove that property (4) of Divided Power Algebra, Definition 23.2.1 is satisfied. Pick $(x, z, w)$ and $(x', z', w')$ in $J$. Pick a map

\[ A_0 = \mathbf{Z}\langle s, s'\rangle \longrightarrow A,\quad s \longmapsto x, s' \longmapsto x' \]

which is possible by the universal property of divided power polynomial rings. Set $M_0 = A_0 \oplus A_0$ and $N_0 = A_0 \oplus A_0 \oplus M_0 \otimes _{A_0} M_0$. Let $q_0 : M_0 \times M_0 \to N_0$ be the obvious map. Define $M_0 \to M$ as the $A_0$-linear map which sends the basis vectors of $M_0$ to $z$ and $z'$. Define $N_0 \to N$ as the $A_0$ linear map which sends the first two basis vectors of $N_0$ to $w$ and $w'$ and uses $M_0 \otimes _{A_0} M_0 \to M \otimes _ A M \xrightarrow {q} N$ on the last summand. Then we see that it suffices to prove the identity (4) for the situation $(A_0, M_0, N_0, q_0)$. Similarly for the other identities. This reduces us to the case of a $\mathbf{Z}$-torsion free ring $A$ and $A$-torsion free modules. In this case all we have to do is show that

\[ n! \delta _ n(x, z, w) = (x, z, w)^ n \]

in the ring $A$, see Divided Power Algebra, Lemma 23.2.2. To see this note that

\[ (x, z, w)^2 = (x^2, 2xz, 2xw + 2q(z, z)) \]

and by induction

\[ (x, z, w)^ n = (x^ n, nx^{n - 1}z, nx^{n - 1}w + n(n - 1)x^{n - 2}q(z, z)) \]

On the other hand,

\[ n! \delta _ n(x, z, w) = (n!\gamma _ n(x), n!\gamma _{n - 1}(x)z, n!\gamma _{n - 1}(x)w + n!\gamma _{n - 2}(x) q(z, z)) \]

which matches. This finishes the proof.
$\square$

## Comments (4)

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