## Tag `07HC`

Chapter 51: Crystalline Cohomology > Section 51.2: Divided power envelope

Lemma 51.2.4. 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 $K = \mathop{\rm Ker}(\Psi)$ 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

- $\delta_m(\sum r_t x_t - r_0)$ where $m > 0$ and $(r_0, r_t) \in \mathcal{R}$,
- $x_{t'}^{[m]}(x_t - f_t)$ 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. $\square$

The code snippet corresponding to this tag is a part of the file `crystalline.tex` and is located in lines 183–199 (see updates for more information).

```
\begin{lemma}
\label{lemma-describe-divided-power-envelope}
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 $r_t \in B$, $r_0 \in I$.
\end{lemma}
\begin{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 \ref{dpa-remark-divided-power-polynomial-algebra}.
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 $K = \Ker(\Psi)$ 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 \ref{dpa-lemma-kernel}.
Let $K' \subset B\langle x_t \rangle$ be the ideal
generated by the elements
\begin{enumerate}
\item $\delta_m(\sum r_t x_t - r_0)$ where $m > 0$ and
$(r_0, r_t) \in \mathcal{R}$,
\item $x_{t'}^{[m]}(x_t - f_t)$ where $m > 0$ and $t', t \in I$.
\end{enumerate}
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 \ref{dpa-lemma-kernel} (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.
\end{proof}
```

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