Is there a version of algebraic de Rham cohomology that can be used to calculate torsion classes?

Question

Much work has gone into the construction of cohomology theories which are defined on algebraic varieties (étale, crystalline, etc.) and comparison isomorphisms between them.

Say $X$ is an algebraic variety over $\mathbb Z$. I am interested in computing $H^*_{\rm sing}(X_{\mathbb C}, \mathbb{Z}_p)$ using a variant of the de Rham complex. However, the obvious integral version of de Rham cohomology, gives the completely wrong answer even for $X = \mathbb A^1$.

This is surprising to me, because (if I am reading the literature correctly, as a non-expert) the de Rham complex of $X_{\mathbb Z_p}$ computes the crystalline cohomology of $X_{\mathbb F_p}$. And crystalline cohomology is often claimed to be the "correct" $p$-adic replacement for the etale cohomology groups $H^*_{et}(X_{\overline{\mathbb F_p}}, \mathbb Z_l)$. However, it seems that crystalline cohomology/de Rham cohomology groups of affine space are "wrong", in the moral sense of not lining up with what I would naively expect.

Is there a correction/explanation for this discrepancy? Is there a variant of de Rham/crystalline cohomology which computes the "right" cohomology for $\mathbb A^1$? Morally, is there a reason why we would want integral crystalline cohomology to carry so much extraneous torsion? Is there something I'm missing?

Answer 1

You should read the introduction to Bhargav Bhatt's lecture notes on prismatic cohomology: available here. This is a new cohomology theory introduced by Bhatt-Scholze (closely related to prior work by Bhatt-Morrow-Scholze) for explaining these torsion phenomena.

The presence of extra torsion in de Rham (or crystalline etc.) cohomology is unavoidable, as SashaP's answer demonstrates. These cohomology theories are only supposed to be reasonable analogues of singular cohomology (i.e. Weil cohomology theories) when you use torsion free coefficients.

However, we can explain where the extra torsion "comes from" (relative to the "true" etale/singular cohomology with e.g. $\mathbf F_p$ coefficients) with a certain long exact sequence.

Let $X$ be a reasonable variety over (e.g.) $\mathbf Z$. Then we can consider its etale cohomology $H^i(X_{{\overline{\mathbf{Q}}}}, \mathbf Z_p)$. This agrees as a module with the singular cohomology of $X_{\mathbf{C}}$, so it's the "right answer" (in particular it's $\mathbf A_1$-invariant).

On the other hand, we can consider the algebraic de Rham cohomology of $X_{\mathbf Z_p}$ (this is essentially the algebraic de Rham cohomology of $X$ tensored with $\mathbf Z_p$), $H^i_{\mathrm{dR}}(X/\mathbf Z_p)$. This agrees with the crystalline cohomology of the special fiber - in particular it only depends on the special fiber, so we shouldn't expect it to capture the same information as the etale/singular cohomology of the generic fiber.

These two $\mathbf Z_p$-modules have the same rank by Fontaine's comparison theorem (at least in the proper case, but if you use some sort of log theory you can extend to the open case), but in general the de Rham cohomology has extra $p$-torsion.

To explain this, Bhatt-Scholze construct a third cohomology theory for schemes over $\mathbf Z_p$ (it's essentially a variant of the construction of crystalline cohomology), valued in modules over the ring $\mathbf Z_p[[u]]$. Let's call this $\mathcal{H}(X)$. Then:

• $\mathcal{H}(X)/u$ is isomorphic to de Rham cohomology of $X_{\mathbf{Z}_p}$ (equivalently, it's isomorphic to the crystalline cohomology of $X_{\mathbf{F}_p}$).

• $\mathcal{H}(X)[1/u]$ is non-canonically isomorphic to the etale cohomology with $\mathbf Z_p$ coefficients of $X_{{\overline{\mathbf Q}}}$ (equivalently to singular cohomology of $X_{\mathbf C}$), tensored with $\mathbf Z_p((u))$.

• $\mathcal{H}(X)$ (at least after some scalar extension) can in principle be computed using a "q-de Rham complex"

(I'm sweeping some homological algebra consideration, about e.g. exactness of modding out by u, under the rug - see the Bhatt-Scholze paper for precise statements. You really need to use derived categories everywhere.)

Answer 2

Let $k$ be a field of characteristics $p$ and $R$ be any ring where $p$ is not invertible.

Asuume that $F:Var_{k}\to D(R-mod)$ is a cohomology theory of smooth algebraic varieties over the field $k$ with values in $R$-modules that satisfies etale descent and has $F^0(\mathrm{Spec}\,k)=R[0]$ such that for any geometrically connected variety $X$ the structure morphism induces an isomorphism $R=F^0(\mathrm{Spec}\,k)\simeq F^0(X)$(these assumptions are satisfied for crystalline cohomology with torsion-free coefficients, e.g. over $W(\bar{k})$). Then at least one of the modules $F^1(\mathbb{A}^1_k)$ and $F^2(\mathbb{A}^1_k)$ must be non-zero.

Indeed, consider the etale Artin-Schreier $\mathbb{Z}/p$-cover given by $\mathbb{A}^1\to\mathbb{A}^1,t\mapsto t^p-t$ (a generator of the cyclic group acts by $t\mapsto t+1$). The Hoschschild-Serre spectral sequence looks like(group cohomology is taken in the category of $R$-modules)$$E_2^{i,j}=H^i(\mathbb{Z}/p,F^j(\mathbb{A}^1))\Rightarrow F^{i+j}(\mathbb{A}^1)$$ If $F^1(\mathbb{A}^1)=0$ then there is an injection(there are no differentials that could touch this term) $H^2(\mathbb{Z}/p,F^0(\mathbb{A}^1))=E^{2,0}_2\hookrightarrow F^2(\mathbb{A}^1)$. But we've assumed that $F^0(\mathbb{A}^1)=R$ with any automorphism acting trivially, so $H^2(\mathbb{Z}/p,F^0(\mathbb{A}^1))=R/p\neq 0$ hence $F^2(\mathbb{A}^1)$ is non-zero.