Let $A/\mathbf{F}_{p^n}$ be an Abelian variety with $\bar{A} = A \times_{\mathbf{F}_{p^n}} \bar{\mathbf{F}}_{p^n}$.

If $\ell \neq p$ is prime, there is a natural isomorphism $$\mathrm{H}^2_{\mathrm{et}}(\bar{A},\mu_{\ell^m}) = \mathrm{Hom}(\Lambda^2A[\ell^m],\mu_{\ell^m}).$$

What is the correct analogue (with fppf cohomology?) of this and of $T_\ell A = \varprojlim_mA[\ell^m]$ and of the Weil pairing $T_\ell A \times T_\ell A^t \to \mathbf{Z}_\ell(1)$ for $\ell = p$?

Edit: On the Weil pairing: See Oda, Tadao: *The first De Rham cohomology group and Dieudonne modules*. In: Ann. Sci. Éc. Norm. Supér. (4), 2 (1969), 63–135, p. 66 f., Theorem 1.1:

Let $f: \mathscr{A} \to \mathscr{A}'$ be an $X$-isogeny of Abelian schemes. The Weil pairing $$ \langle\cdot,\cdot\rangle_f: \ker(f) \times_X \ker(f^t) \to \mathbf{G}_m $$ is a non-degenerate and biadditive pairing of finite flat $X$-group schemes, i. e. it defines a canonical $X$-isomorphism $$ \ker(f^t) = (\ker(f))^t. $$ Moreover, it is functorial in $f$.

It remains the question on $$\mathrm{H}^2_{\mathrm{et}}(\bar{A},\mu_{\ell^m}) = \mathrm{Hom}(\Lambda^2A[\ell^m],\mu_{\ell^m})$$ for fppf cohomology.

Edit 2: For $\ell \neq \mathrm{char}\,k$, this can be proved as follows: $$\mathrm{H}_\mathrm{et}^q(\bar{A},\mu_{\ell^n}) = \Lambda^q\mathrm{H}_\mathrm{et}^1(\bar{A},\mu_{\ell^n}) \otimes \mu_{\ell^n}^{\otimes(-q+1)} = \Lambda^q\mathrm{Hom}(A[\ell^n],\mu_{\ell^n}) \otimes \mu_{\ell^n}^{\otimes(-q+1)} = \Lambda^q\mathrm{Hom}(A[\ell^n],\mathbf{Z}/\ell^n) \otimes \mu_{\ell^n} = \mathrm{Hom}(\Lambda^qA[\ell^n],\mu_{\ell^n}).$$ But the calculation of the $q$-th étale cohomology of $\bar{A}$ as an exterior algebra of $\mathrm{H}^1$ in http://jmilne.org/math/articles/1986b.pdf, Theorem 15.1(b) requires $\ell \neq \mathrm{char}\,k$.

The first De Rham cohomology group and Dieudonne modules. In: Ann. Sci. Éc. Norm. Supér. (4), 2 (1969), 63–135, p. 66 f., Theorem 1.1. What about the other questions? $\endgroup$ – TKe Mar 16 '17 at 16:52