Lemma 60.2.4. Let $(B, I, \gamma )$ be a divided power algebra. Let $I \subset J \subset B$ be an ideal. Let $(D, \bar J, \bar\gamma )$ be the divided power envelope of $J$ relative to $\gamma$. Choose elements $f_ t \in J$, $t \in T$ such that $J = I + (f_ t)$. Then there exists a surjection

$\Psi : B\langle x_ t \rangle \longrightarrow D$

of divided power rings mapping $x_ t$ to the image of $f_ t$ in $D$. The kernel of $\Psi$ is generated by the elements $x_ t - f_ t$ and all

$\delta _ n\left(\sum r_ t x_ t - r_0\right)$

whenever $\sum r_ t f_ t = r_0$ in $B$ for some $r_ t \in B$, $r_0 \in I$.

Proof. In the statement of the lemma we think of $B\langle x_ t \rangle$ as a divided power ring with ideal $J' = IB\langle x_ t \rangle + B\langle x_ t \rangle _{+}$, see Divided Power Algebra, Remark 23.5.2. The existence of $\Psi$ follows from the universal property of divided power polynomial rings. Surjectivity of $\Psi$ follows from the fact that its image is a divided power subring of $D$, hence equal to $D$ by the universal property of $D$. It is clear that $x_ t - f_ t$ is in the kernel. Set

$\mathcal{R} = \{ (r_0, r_ t) \in I \oplus \bigoplus \nolimits _{t \in T} B \mid \sum r_ t f_ t = r_0 \text{ in }B\}$

If $(r_0, r_ t) \in \mathcal{R}$ then it is clear that $\sum r_ t x_ t - r_0$ is in the kernel. As $\Psi$ is a homomorphism of divided power rings and $\sum r_ tx_ t - r_0 \in J'$ it follows that $\delta _ n(\sum r_ t x_ t - r_0)$ is in the kernel as well. Let $K \subset B\langle x_ t \rangle$ be the ideal generated by $x_ t - f_ t$ and the elements $\delta _ n(\sum r_ t x_ t - r_0)$ for $(r_0, r_ t) \in \mathcal{R}$. To show that $K = \mathop{\mathrm{Ker}}(\Psi )$ it suffices to show that $\delta$ extends to $B\langle x_ t \rangle /K$. Namely, if so the universal property of $D$ gives a map $D \to B\langle x_ t \rangle /K$ inverse to $\Psi$. Hence we have to show that $K \cap J'$ is preserved by $\delta _ n$, see Divided Power Algebra, Lemma 23.4.3. Let $K' \subset B\langle x_ t \rangle$ be the ideal generated by the elements

1. $\delta _ m(\sum r_ t x_ t - r_0)$ where $m > 0$ and $(r_0, r_ t) \in \mathcal{R}$,

2. $x_{t'}^{[m]}(x_ t - f_ t)$ where $m > 0$ and $t', t \in I$.

We claim that $K' = K \cap J'$. The claim proves that $K \cap J'$ is preserved by $\delta _ n$, $n > 0$ by the criterion of Divided Power Algebra, Lemma 23.4.3 (2)(c) and a computation of $\delta _ n$ of the elements listed which we leave to the reader. To prove the claim note that $K' \subset K \cap J'$. Conversely, if $h \in K \cap J'$ then, modulo $K'$ we can write

$h = \sum r_ t (x_ t - f_ t)$

for some $r_ t \in B$. As $h \in K \cap J' \subset J'$ we see that $r_0 = \sum r_ t f_ t \in I$. Hence $(r_0, r_ t) \in \mathcal{R}$ and we see that

$h = \sum r_ t x_ t - r_0$

is in $K'$ as desired. $\square$

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