60.22 Two counter examples
Before we turn to some of the successes of crystalline cohomology, let us give two examples which explain why crystalline cohomology does not work very well if the schemes in question are either not proper over the base, or singular. The first example can be found in [BO].
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
\[ C = P/J = \mathbf{Z}_ p[x, y]/(x^2, xy, y^2, p) \]
Let $D = D_{P, \gamma }(J)^\wedge $ with divided power ideal $(\bar J, \bar\gamma )$ as in Section 60.17. 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
$H^0(\text{Cris}(X/S), \mathcal{O}_{X/S})$ 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 $F^*(\tau ) = 0$.
The next example shows that even for affine $n$-space crystalline cohomology does not give the correct thing.
Example 60.22.2. Let $A = \mathbf{Z}_ p$ with divided power ideal $(p)$ endowed with its unique divided powers $\gamma $. Let $C = \mathbf{F}_ p[x_1, \ldots , x_ r]$. We choose the presentation
\[ C = P/J = P/pP\quad \text{with}\quad P = \mathbf{Z}_ p[x_1, \ldots , x_ r] \]
Note that $pP$ has divided powers by Divided Power Algebra, Lemma 23.4.2. Hence setting $D = P^\wedge $ with divided power ideal $(p)$ we obtain a situation as in Section 60.17. We conclude that $R\Gamma (\text{Cris}(X/S), \mathcal{O}_{X/S})$ is represented by the complex
\[ D \to \Omega ^1_ D \to \Omega ^2_ D \to \ldots \to \Omega ^ r_ D \]
see Proposition 60.21.3. Assuming $r > 0$ we conclude the following
The cristalline cohomology of the cristalline structure sheaf of $X = \mathbf{A}^ r_{\mathbf{F}_ p}$ over $S = \mathop{\mathrm{Spec}}(\mathbf{Z}_ p)$ is zero except in degrees $0, \ldots , r$.
We have $H^0(\text{Cris}(X/S), \mathcal{O}_{X/S}) = \mathbf{Z}_ p$.
The cohomology group $H^ r(\text{Cris}(X/S), \mathcal{O}_{X/S})$ is infinite and is not a torsion abelian group.
The cohomology group $H^ r(\text{Cris}(X/S), \mathcal{O}_{X/S})$ is not separated for the $p$-adic topology.
While the first two statements are reasonable, parts (3) and (4) are disconcerting! The truth of these statements follows immediately from working out what the complex displayed above looks like. Let's just do this in case $r = 1$. Then we are just looking at the two term complex of $p$-adically complete modules
\[ \text{d} : D = \left( \bigoplus \nolimits _{n \geq 0} \mathbf{Z}_ p x^ n \right)^\wedge \longrightarrow \Omega ^1_ D = \left( \bigoplus \nolimits _{n \geq 1} \mathbf{Z}_ p x^{n - 1}\text{d}x \right)^\wedge \]
The map is given by $\text{diag}(0, 1, 2, 3, 4, \ldots )$ except that the first summand is missing on the right hand side. Now it is clear that $\bigoplus _{n > 0} \mathbf{Z}_ p/n\mathbf{Z}_ p$ is a subgroup of the cokernel, hence the cokernel is infinite. In fact, the element
\[ \omega = \sum \nolimits _{e > 0} p^ e x^{p^{2e} - 1}\text{d}x \]
is clearly not a torsion element of the cokernel. But it gets worse. Namely, consider the element
\[ \eta = \sum \nolimits _{e > 0} p^ e x^{p^ e - 1}\text{d}x \]
For every $t > 0$ the element $\eta $ is congruent to $\sum _{e > t} p^ e x^{p^ e - 1}\text{d}x$ modulo the image of $\text{d}$ which is divisible by $p^ t$. But $\eta $ is not in the image of $\text{d}$ because it would have to be the image of $a + \sum _{e > 0} x^{p^ e}$ for some $a \in \mathbf{Z}_ p$ which is not an element of the left hand side. In fact, $p^ N\eta $ is similarly not in the image of $\text{d}$ for any integer $N$. This implies that $\eta $ “generates” a copy of $\mathbf{Q}_ p$ inside of $H^1_{\text{cris}}(\mathbf{A}_{\mathbf{F}_ p}^1/\mathop{\mathrm{Spec}}(\mathbf{Z}_ p))$.
Comments (1)
Comment #1611 by Rakesh Pawar on