Lemma 60.5.3. In Situation 60.5.1.

1. $\text{CRIS}(C/A)$ has finite products (but not infinite ones),

2. $\text{CRIS}(C/A)$ has all finite nonempty colimits and (60.5.2.1) commutes with these, and

3. $\text{Cris}(C/A)$ has all finite nonempty colimits and $\text{Cris}(C/A) \to \text{CRIS}(C/A)$ commutes with them.

Proof. The empty product, i.e., the final object in the category of divided power thickenings of $C$ over $(A, I, \gamma )$, is the zero ring viewed as an $A$-algebra endowed with the zero ideal and the unique divided powers on the zero ideal and finally endowed with the unique homomorphism of $C$ to the zero ring. If $(B_ t, J_ t, \delta _ t)_{t \in T}$ is a family of objects of $\text{CRIS}(C/A)$ then we can form the product $(\prod _ t B_ t, \prod _ t J_ t, \prod _ t \delta _ t)$ as in Divided Power Algebra, Lemma 23.3.2. The map $C \to \prod B_ t/\prod J_ t = \prod B_ t/J_ t$ is clear. However, we are only guaranteed that $p$ is nilpotent in $\prod _ t B_ t$ if $T$ is finite.

Given two objects $(B, J, \gamma )$ and $(B', J', \gamma ')$ of $\text{CRIS}(C/A)$ we can form a cocartesian diagram

$\xymatrix{ (B, J, \delta ) \ar[r] & (B'', J'', \delta '') \\ (A, I, \gamma ) \ar[r] \ar[u] & (B', J', \delta ') \ar[u] }$

in the category of divided power rings. Then we see that we have

$B''/J'' = B/J \otimes _{A/I} B'/J' \longleftarrow C \otimes _{A/I} C$

see Divided Power Algebra, Remark 23.3.5. Denote $J'' \subset K \subset B''$ the ideal such that

$\xymatrix{ B''/J'' \ar[r] & B''/K \\ C \otimes _{A/I} C \ar[r] \ar[u] & C \ar[u] }$

is a pushout, i.e., $B''/K \cong B/J \otimes _ C B'/J'$. Let $D_{B''}(K) = (D, \bar K, \bar\delta )$ be the divided power envelope of $K$ in $B''$ relative to $(B'', J'', \delta '')$. Then it is easily verified that $(D, \bar K, \bar\delta )$ is a coproduct of $(B, J, \delta )$ and $(B', J', \delta ')$ in $\text{CRIS}(C/A)$.

Next, we come to coequalizers. Let $\alpha , \beta : (B, J, \delta ) \to (B', J', \delta ')$ be morphisms of $\text{CRIS}(C/A)$. Consider $B'' = B'/ (\alpha (b) - \beta (b))$. Let $J'' \subset B''$ be the image of $J'$. Let $D_{B''}(J'') = (D, \bar J, \bar\delta )$ be the divided power envelope of $J''$ in $B''$ relative to $(B', J', \delta ')$. Then it is easily verified that $(D, \bar J, \bar\delta )$ is the coequalizer of $(B, J, \delta )$ and $(B', J', \delta ')$ in $\text{CRIS}(C/A)$.

By Categories, Lemma 4.18.6 we have all finite nonempty colimits in $\text{CRIS}(C/A)$. The constructions above shows that (60.5.2.1) commutes with them. This formally implies part (3) as $\text{Cris}(C/A)$ is the fibre category of (60.5.2.1) over $C$. $\square$

## Comments (6)

Comment #6002 by Simon Paege on

The empty product of thickenings is not $C$ but the zero ring, right? In general there shouldn't be a canonical map $B \to C$ for a thickening $B$.

Comment #7048 by nkym on

I guess that in the proof of (1), 07GV should be cited instead of 07GX.

Comment #7052 by nkym on

In fact, I am now wondering if the usual product construction in the proof of (1) is legitimate since it does not seem to assure that $p$ is nilpotent in $\prod B_t$. For example, how about setting $(A, I, \gamma) = (\mathbb{Z}_{(p)}, 0, \emptyset)$, $C = \mathbb{Z}/p\mathbb{Z}$, $T$ to be the set of positive integers, and $(B_t, J_t, \delta_t) = (\mathbb{Z}/p^t\mathbb{Z}, p\mathbb{Z}/p^t\mathbb{Z}, (px\mapsto \frac{p^n}{n!}x^n)_{n>0})$ with the obvious isomorphism $C\to B_t/J_t$, $\emptyset$ meaning $(0\to 0)_{n>0}$?

Comment #7053 by on

Yes, good catch. Your example shows that the category $\text{CRIS}(C/A)$ does not have infinite products. Finite products do exist. In your example you could also have used $(A, I, \gamma)$ to be $A = \mathbf{Z}_{(p)}$ and $I = (p)$ with the usual divided powers. I will fix this the next time I go through all the comments. Thanks.

There are also:

• 5 comment(s) on Section 60.5: Affine crystalline site

## 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 07HN. Beware of the difference between the letter 'O' and the digit '0'.