## Tag `07ME`

Chapter 51: Crystalline Cohomology > Section 51.7: Divided power schemes

Lemma 51.7.4. Let $(U', T', \delta') \to (S'_0, S', \gamma')$ and $(S_0, S, \gamma) \to (S'_0, S', \gamma')$ be morphisms of divided power schemes. If $(U', T', \delta')$ is a divided power thickening, then there exists a divided power scheme $(T_0, T, \delta)$ and $$ \xymatrix{ T \ar[r] \ar[d] & T' \ar[d] \\ S \ar[r] & S' } $$ which is a cartesian diagram in the category of divided power schemes.

Proof.Omitted. Hints: If $T$ exists, then $T_0 = S_0 \times_{S'_0} U'$ (argue as in Divided Power Algebra, Remark 23.3.5). Since $T'$ is a divided power thickening, we see that $T$ (if it exists) will be a divided power thickening too. Hence we can define $T$ as the scheme with underlying topological space the underlying topological space of $T_0 = S_0 \times_{S'_0} U'$ and as structure sheaf on affine pieces the ring given by Lemma 51.5.3. $\square$

The code snippet corresponding to this tag is a part of the file `crystalline.tex` and is located in lines 1543–1557 (see updates for more information).

```
\begin{lemma}
\label{lemma-fibre-product}
Let $(U', T', \delta') \to (S'_0, S', \gamma')$ and
$(S_0, S, \gamma) \to (S'_0, S', \gamma')$ be morphisms of
divided power schemes. If $(U', T', \delta')$ is a divided power
thickening, then there exists a divided power scheme $(T_0, T, \delta)$
and
$$
\xymatrix{
T \ar[r] \ar[d] & T' \ar[d] \\
S \ar[r] & S'
}
$$
which is a cartesian diagram in the category of divided power schemes.
\end{lemma}
\begin{proof}
Omitted. Hints: If $T$ exists, then $T_0 = S_0 \times_{S'_0} U'$
(argue as in Divided Power Algebra, Remark \ref{dpa-remark-forgetful}).
Since $T'$ is a divided power thickening, we see that $T$
(if it exists) will be a divided power thickening too.
Hence we can define $T$ as the scheme with underlying topological
space the underlying topological space of $T_0 = S_0 \times_{S'_0} U'$
and as structure sheaf on affine pieces the ring given
by Lemma \ref{lemma-affine-thickenings-colimits}.
\end{proof}
```

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