# The Stacks Project

## Tag 07MU

Remark 54.24.9 (Base change isomorphism). The map (54.24.8.1) is an isomorphism provided all of the following conditions are satisfied:

1. $p$ is nilpotent in $A'$,
2. $\mathcal{F}'$ is a crystal in quasi-coherent $\mathcal{O}_{X'/S'}$-modules,
3. $X' \to S'_0$ is a quasi-compact, quasi-separated morphism,
4. $X = X' \times_{S'_0} S_0$,
5. $\mathcal{F}'$ is a flat $\mathcal{O}_{X'/S'}$-module,
6. $X' \to S'_0$ is a local complete intersection morphism (see More on Morphisms, Definition 36.51.2; this holds for example if $X' \to S'_0$ is syntomic or smooth),
7. $X'$ and $S_0$ are Tor independent over $S'_0$ (see More on Algebra, Definition 15.57.1; this holds for example if either $S_0 \to S'_0$ or $X' \to S'_0$ is flat).

Hints: Condition (1) means that in the arguments below $p$-adic completion does nothing and can be ignored. Using condition (3) and Mayer Vietoris (see Remark 54.24.2) this reduces to the case where $X'$ is affine. In fact by condition (6), after shrinking further, we can assume that $X' = \mathop{\rm Spec}(C')$ and we are given a presentation $C' = A'/I'[x_1, \ldots, x_n]/(\bar f'_1, \ldots, \bar f'_c)$ where $\bar f'_1, \ldots, \bar f'_c$ is a Koszul-regular sequence in $A'/I'$. (This means that smooth locally $\bar f'_1, \ldots, \bar f'_c$ forms a regular sequence, see More on Algebra, Lemma 15.27.17.) We choose a lift of $\bar f'_i$ to an element $f'_i \in A'[x_1, \ldots, x_n]$. By (4) we see that $X = \mathop{\rm Spec}(C)$ with $C = A/I[x_1, \ldots, x_n]/(\bar f_1, \ldots, \bar f_c)$ where $f_i \in A[x_1, \ldots, x_n]$ is the image of $f'_i$. By property (7) we see that $\bar f_1, \ldots, \bar f_c$ is a Koszul-regular sequence in $A/I[x_1, \ldots, x_n]$. The divided power envelope of $I'A'[x_1, \ldots, x_n] + (f'_1, \ldots, f'_c)$ in $A'[x_1, \ldots, x_n]$ relative to $\gamma'$ is $$D' = A'[x_1, \ldots, x_n]\langle \xi_1, \ldots, \xi_c \rangle/(\xi_i - f'_i)$$ see Lemma 54.2.4. Then you check that $\xi_1 - f'_1, \ldots, \xi_n - f'_n$ is a Koszul-regular sequence in the ring $A'[x_1, \ldots, x_n]\langle \xi_1, \ldots, \xi_c\rangle$. Similarly the divided power envelope of $IA[x_1, \ldots, x_n] + (f_1, \ldots, f_c)$ in $A[x_1, \ldots, x_n]$ relative to $\gamma$ is $$D = A[x_1, \ldots, x_n]\langle \xi_1, \ldots, \xi_c\rangle/(\xi_i - f_i)$$ and $\xi_1 - f_1, \ldots, \xi_n - f_n$ is a Koszul-regular sequence in the ring $A[x_1, \ldots, x_n]\langle \xi_1, \ldots, \xi_c\rangle$. It follows that $D' \otimes_{A'}^\mathbf{L} A = D$. Condition (2) implies $\mathcal{F}'$ corresponds to a pair $(M', \nabla)$ consisting of a $D'$-module with connection, see Proposition 54.17.4. Then $M = M' \otimes_{D'} D$ corresponds to the pullback $\mathcal{F}$. By assumption (5) we see that $M'$ is a flat $D'$-module, hence $$M = M' \otimes_{D'} D = M' \otimes_{D'} D' \otimes_{A'}^\mathbf{L} A = M' \otimes_{A'}^\mathbf{L} A$$ Since the modules of differentials $\Omega_{D'}$ and $\Omega_D$ (as defined in Section 54.17) are free $D'$-modules on the same generators we see that $$M \otimes_D \Omega^\bullet_D = M' \otimes_{D'} \Omega^\bullet_{D'} \otimes_{D'} D = M' \otimes_{D'} \Omega^\bullet_{D'} \otimes_{A'}^\mathbf{L} A$$ which proves what we want by Proposition 54.21.3.

The code snippet corresponding to this tag is a part of the file crystalline.tex and is located in lines 4740–4811 (see updates for more information).

\begin{remark}[Base change isomorphism]
\label{remark-base-change-isomorphism}
The map (\ref{equation-base-change-map}) is an isomorphism provided
all of the following conditions are satisfied:
\begin{enumerate}
\item $p$ is nilpotent in $A'$,
\item $\mathcal{F}'$ is a crystal in quasi-coherent
$\mathcal{O}_{X'/S'}$-modules,
\item $X' \to S'_0$ is a quasi-compact, quasi-separated morphism,
\item $X = X' \times_{S'_0} S_0$,
\item $\mathcal{F}'$ is a flat $\mathcal{O}_{X'/S'}$-module,
\item $X' \to S'_0$ is a local complete intersection morphism (see
More on Morphisms, Definition \ref{more-morphisms-definition-lci}; this
holds for example if $X' \to S'_0$ is syntomic or smooth),
\item $X'$ and $S_0$ are Tor independent over $S'_0$ (see
More on Algebra, Definition \ref{more-algebra-definition-tor-independent};
this holds for example if either $S_0 \to S'_0$ or $X' \to S'_0$ is flat).
\end{enumerate}
Hints: Condition (1) means that in the arguments below $p$-adic completion
does nothing and can be ignored.
Using condition (3) and Mayer Vietoris (see
Remark \ref{remark-mayer-vietoris}) this reduces to the case
where $X'$ is affine. In fact by condition (6), after shrinking
further, we can assume that $X' = \Spec(C')$ and we are given a presentation
$C' = A'/I'[x_1, \ldots, x_n]/(\bar f'_1, \ldots, \bar f'_c)$
where $\bar f'_1, \ldots, \bar f'_c$ is a Koszul-regular sequence in $A'/I'$.
(This means that smooth locally $\bar f'_1, \ldots, \bar f'_c$ forms
a regular sequence, see More on Algebra,
Lemma \ref{more-algebra-lemma-Koszul-regular-flat-locally-regular}.)
We choose a lift of
$\bar f'_i$ to an element $f'_i \in A'[x_1, \ldots, x_n]$. By (4) we see that
$X = \Spec(C)$ with $C = A/I[x_1, \ldots, x_n]/(\bar f_1, \ldots, \bar f_c)$
where $f_i \in A[x_1, \ldots, x_n]$ is the image of $f'_i$.
By property (7) we see that $\bar f_1, \ldots, \bar f_c$ is a Koszul-regular
sequence in $A/I[x_1, \ldots, x_n]$. The divided power envelope of
$I'A'[x_1, \ldots, x_n] + (f'_1, \ldots, f'_c)$ in $A'[x_1, \ldots, x_n]$
relative to $\gamma'$ is
$$D' = A'[x_1, \ldots, x_n]\langle \xi_1, \ldots, \xi_c \rangle/(\xi_i - f'_i)$$
see Lemma \ref{lemma-describe-divided-power-envelope}. Then you check that
$\xi_1 - f'_1, \ldots, \xi_n - f'_n$ is a Koszul-regular sequence in the
ring $A'[x_1, \ldots, x_n]\langle \xi_1, \ldots, \xi_c\rangle$.
Similarly the divided power envelope of
$IA[x_1, \ldots, x_n] + (f_1, \ldots, f_c)$ in $A[x_1, \ldots, x_n]$
relative to $\gamma$ is
$$D = A[x_1, \ldots, x_n]\langle \xi_1, \ldots, \xi_c\rangle/(\xi_i - f_i)$$
and $\xi_1 - f_1, \ldots, \xi_n - f_n$ is a Koszul-regular sequence in the
ring $A[x_1, \ldots, x_n]\langle \xi_1, \ldots, \xi_c\rangle$.
It follows that $D' \otimes_{A'}^\mathbf{L} A = D$. Condition (2)
implies $\mathcal{F}'$ corresponds to a pair $(M', \nabla)$
consisting of a $D'$-module with connection, see
Proposition \ref{proposition-crystals-on-affine}.
Then $M = M' \otimes_{D'} D$ corresponds to the pullback $\mathcal{F}$.
By assumption (5) we see that $M'$ is a flat $D'$-module, hence
$$M = M' \otimes_{D'} D = M' \otimes_{D'} D' \otimes_{A'}^\mathbf{L} A = M' \otimes_{A'}^\mathbf{L} A$$
Since the modules of differentials $\Omega_{D'}$ and $\Omega_D$
(as defined in Section \ref{section-quasi-coherent-crystals})
are free $D'$-modules on the same generators we see that
$$M \otimes_D \Omega^\bullet_D = M' \otimes_{D'} \Omega^\bullet_{D'} \otimes_{D'} D = M' \otimes_{D'} \Omega^\bullet_{D'} \otimes_{A'}^\mathbf{L} A$$
which proves what we want by
Proposition \ref{proposition-compute-cohomology-crystal}.
\end{remark}

## Comments (0)

There are no comments yet for this tag.

There are also 2 comments on Section 54.24: Crystalline Cohomology.

## Add a comment on tag 07MU

Your email address will not be published. Required fields are marked.

In your comment you can use Markdown and LaTeX style mathematics (enclose it like $\pi$). A preview option is available if you wish to see how it works out (just click on the eye in the lower-right corner).

All contributions are licensed under the GNU Free Documentation License.

In order to prevent bots from posting comments, we would like you to prove that you are human. You can do this by filling in the name of the current tag in the following box. So in case this where tag 0321 you just have to write 0321. Beware of the difference between the letter 'O' and the digit 0.

This captcha seems more appropriate than the usual illegible gibberish, right?