# flat/crystalline cohomology of abelian variety

Let $A/k$ be an abelian variety over an algebraically closed field and $\ell \neq \mathrm{char}\,k$.

In http://jmilne.org/math/articles/1986b.pdf, Theorem 15.1(b) it is proved that $$H^r_{et}(A, R) = \bigwedge^rH^1_{et}(A,R)\quad\text{for R = \mathbf{Z}_\ell,\mathbf{Q}_\ell,\mathbf{F}_\ell.}$$ The proof uses $$H^1_{et}(A,\mathbf{Z}_\ell) = \mathrm{Hom}_{cont}(\pi_1^{et}(A,0),\mathbf{Z}_\ell).$$

For $R = \mathbf{Z}/p\mathbf{Z}$ one has $$H^r_{et}(A,\mathbf{Z}/p\mathbf{Z}) = H^r_{fppf}(A,\mathbf{Z}/p),$$ since $\mathbf{Z}/p\mathbf{Z}$ is a smooth quasi-projective commutative group scheme. Using induction, the five lemma and short exact sequences $$0 \to \mathbf{Z}/p \to \mathbf{Z}/p^{n+1} \to \mathbf{Z}/p^n \to 0,$$ one gets $$H^r_{et}(A,\mathbf{Z}/p^n) = H^r_{fppf}(A,\mathbf{Z}/p^n)\quad\text{ for all n \geq 0,}$$ and hence $$H^r_{et}(A,\mathbf{Z}_p) = H^r_{fppf}(A,\mathbf{Z}_p).$$

One has $$\pi_1^{et}(A,0) = \prod_{\ell} T_\ell(A)$$ with $T_p(A) = \varprojlim_nA[p^n]^r$ with $A[p^n] = A[p^n]^0 \times A[p^n]^r$, with $A[p^n]^0$ local and $A[p^n]^r$ reduced (Mumford, Abelian Varieties, Chapter IV.18) for $p = \mathrm{char}\,k$.

By Lei Fu, Étale cohomology theory, Proposition 5.7.20, one has $$H^1_{et}(X,G) = \mathrm{Hom}_{cont}(\pi_1^{et}(X),G)$$ for a finite group $G$ and $X$ connected Noetherian.

Is there a similar isomorphism for crystalline cohomology or flat cohomology with $R = \mathbb{Z}_p,\mathbb{Q}_p,\mathbb{F}_p$, $p = \mathrm{char}\,k$?

Edit (26.04.2018): The answer below settles the question for $W(k)$ and $\mathrm{Quot}(W(k))$ coefficients and crystalline cohomology.

What is the analogue of $H^q_\mathrm{et}(\bar{A},\mu_{\ell^n}) = \mathrm{Hom}(\bigwedge^qA[\ell^n],\mu_{\ell^n})$ in crystalline cohomology?

• I think this formula holds if you use the crystalline fundamental group. But it is certainly false for the etale fundamental group. – Will Sawin Jul 12 '17 at 18:01
• alternately you could look at the top related question mathoverflow.net/questions/20381/… which contains an arguably similar isomorphism. – Will Sawin Jul 12 '17 at 18:03
• @WillSawin: Can you give me some references to start reading with? – TKe Jul 13 '17 at 10:02
• I'm not very good with references, but it depends on what your goal is. If you want to understand crystalline cohomology in the concrete possible way, you probably want to read about Dieudonne modules. Perhaps the Demazure reference in the linked question is a good place to start. – Will Sawin Jul 13 '17 at 11:14

## 1 Answer

In Illusie, Complexe de de Rham-Witt et cohomologie cristalline, p. 651, (7.1.1) it is proved that $H^*_{cris}(A/W) = \bigwedge^*H^1_{cris}(A/W)$.

Is $H^1_{cris}(A/W) = T_pA$?

I have found in http://citeseerx.ist.psu.edu/viewdoc/download?doi=10.1.1.295.7431&rep=rep1&type=pdf, p. 202, Theorem 5.7.1: For $A/W(k)$, $k$ algebraically closed of characteristic $p > 0$, there is a short exact sequence of $W = W(k)$-modules: $$0 \to H^1_{et}(A_k,\mathbf{Z}_p)\otimes W \to H^1_{cris}(A_k/W) \to \mathbf{D}(\hat{A}/W) \to 0$$

Edit (26.04.2018): This settles the question for $W(k)$ and $\mathrm{Quot}(W(k))$ coefficients.

What is the analogue of $H^q_\mathrm{et}(\bar{A},\mu_{\ell^n}) = \mathrm{Hom}(\bigwedge^qA[\ell^n],\mu_{\ell^n})$ in crystalline cohomology?

• To respond to your question: it is not $T_p A$, because that may have the wrong $W(k)$ rank. Instead it is the Dieudonne module associated to the $p$-divisible group of the abelian variety. This is a result of Mazur-Messing See Remark 3.11.2 of the same Illusie paper for another proof and for a reference to Mazur-Messing – user45150 Jul 13 '17 at 12:59
• B. MAZUR et W. MESSING , Universal Extensions and One Dimensional Crystalline Cohomology [Lecture Notes in Math., n° 370, Springer-Verlag, 1974). – user45150 Jul 13 '17 at 13:01
• Remark 3.11.2 is on p. 618 in Illusie. – TKe Jul 13 '17 at 13:32
• @user45150: Can you give me some references on the relation between $T_pA$ and $D(A(p))$? – TKe Jul 13 '17 at 14:04
• @TimoKeller Are you asking for a proof of the following fact: The Dieudonn\'{e} module associated to $A[p^\infty]$ is isomorphic to $H^1_{\text{crys}}(A/W)$? This is in the Illusie paper you cite in your answer, Section II.7.1. – Ben Lim Jul 16 '17 at 11:36