The Stacks project

Remark 60.24.9 (Base change isomorphism). The map (60.24.8.1) 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 37.59.2; 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 15.61.1; 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 60.24.2) this reduces to the case where $X'$ is affine. In fact by condition (6), after shrinking further, we can assume that $X' = \mathop{\mathrm{Spec}}(C')$ 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 15.30.17.) 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 $X = \mathop{\mathrm{Spec}}(C)$ 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 60.2.4. 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 60.17.4. 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 60.17) 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 60.21.3.


Comments (0)

There are also:

  • 2 comment(s) on Section 60.24: Some further results

Post a comment

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 toolbar).

Unfortunately JavaScript is disabled in your browser, so the comment preview function will not work.

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 input field. As a reminder, this is tag 07MU. Beware of the difference between the letter 'O' and the digit '0'.