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Example 58.22.1. 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

\[ C = P/J = \mathbf{Z}_ p[x, y]/(x^2, xy, y^2, p) \]

Let $D = D_{P, \gamma }(J)^\wedge $ with divided power ideal $(\bar J, \bar\gamma )$ as in Section 58.17. We will denote $x, y$ also the images of $x$ and $y$ in $D$. Consider the element

\[ \tau = \bar\gamma _ p(x^2)\bar\gamma _ p(y^2) - \bar\gamma _ p(xy)^2 \in D \]

We note that $p\tau = 0$ as

\[ p! \bar\gamma _ p(x^2) \bar\gamma _ p(y^2) = x^{2p} \bar\gamma _ p(y^2) = \bar\gamma _ p(x^2y^2) = x^ py^ p \bar\gamma _ p(xy) = p! \bar\gamma _ p(xy)^2 \]

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

\begin{align*} \text{d}(\bar\gamma _ p(x^2) \bar\gamma _ p(y^2)) & = \bar\gamma _{p - 1}(x^2)\bar\gamma _ p(y^2)\text{d}x^2 + \bar\gamma _ p(x^2)\bar\gamma _{p - 1}(y^2)\text{d}y^2 \\ & = 2 x \bar\gamma _{p - 1}(x^2)\bar\gamma _ p(y^2)\text{d}x + 2 y \bar\gamma _ p(x^2)\bar\gamma _{p - 1}(y^2)\text{d}y \\ & = 2/(p - 1)!( x^{2p - 1} \bar\gamma _ p(y^2)\text{d}x + y^{2p - 1} \bar\gamma _ p(x^2)\text{d}y ) \\ & = 2/(p - 1)! (x^{p - 1} \bar\gamma _ p(xy^2)\text{d}x + y^{p - 1} \bar\gamma _ p(x^2y)\text{d}y) \\ & = 2/(p - 1)! (x^{p - 1}y^ p \bar\gamma _ p(xy)\text{d}x + x^ py^{p - 1} \bar\gamma _ p(xy)\text{d}y) \\ & = 2 \bar\gamma _{p - 1}(xy) \bar\gamma _ p(xy)(y\text{d}x + x \text{d}y) \\ & = \text{d}(\bar\gamma _ p(xy)^2) \end{align*}

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

\[ B = \mathbf{F}_ p[x, y, u, v]/(x^3, x^2y, xy^2, y^3, xu, yu, xv, yv, u^2, v^2) \]

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

\[ \delta _ p : K \longrightarrow K,\quad ax^2 + bxy + cy^2 + du + ev + fuv \longmapsto a^ pu + c^ pv \]

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

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

  2. 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 58.21.3. 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 (1)

Comment #4913 by Riccardo Pengo on

I think that should be mapped to , and not to . Am I mistaken?

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  • 1 comment(s) on Section 58.22: Two counter examples

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