Example 60.22.1. Let $A = \mathbf{Z}_ p$ with divided power ideal $(p)$ endowed with its unique divided powers $\gamma $. Let $C = \mathbf{F}_ p[x, y]/(x^2, xy, y^2)$. We choose the presentation

Let $D = D_{P, \gamma }(J)^\wedge $ with divided power ideal $(\bar J, \bar\gamma )$ as in Section 60.17. We will denote $x, y$ also the images of $x$ and $y$ in $D$. Consider the element

We note that $p\tau = 0$ as

in $D$. We also note that $\text{d}\tau = 0$ in $\Omega _ D$ as

Finally, we claim that $\tau \not= 0$ in $D$. To see this it suffices to produce an object $(B \to \mathbf{F}_ p[x, y]/(x^2, xy, y^2), \delta )$ of $\text{Cris}(C/S)$ such that $\tau $ does not map to zero in $B$. To do this take

with the obvious surjection to $C$. Let $K = \mathop{\mathrm{Ker}}(B \to C)$ and consider the map

One checks this satisfies the assumptions (1), (2), (3) of Divided Power Algebra, Lemma 23.5.3 and hence defines a divided power structure. Moreover, we see that $\tau $ maps to $uv$ which is not zero in $B$. Set $X = \mathop{\mathrm{Spec}}(C)$ and $S = \mathop{\mathrm{Spec}}(A)$. We draw the following conclusions

$H^0(\text{Cris}(X/S), \mathcal{O}_{X/S})$ has $p$-torsion, and

pulling back by Frobenius $F^* : H^0(\text{Cris}(X/S), \mathcal{O}_{X/S}) \to H^0(\text{Cris}(X/S), \mathcal{O}_{X/S})$ is not injective.

Namely, $\tau $ defines a nonzero torsion element of $H^0(\text{Cris}(X/S), \mathcal{O}_{X/S})$ by Proposition 60.21.3. Similarly, $F^*(\tau ) = \sigma (\tau )$ where $\sigma : D \to D$ is the map induced by any lift of Frobenius on $P$. If we choose $\sigma (x) = x^ p$ and $\sigma (y) = y^ p$, then an easy computation shows that $F^*(\tau ) = 0$.

## Comments (2)

Comment #4913 by Riccardo Pengo on

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