See page 67 of the free online text for Proposition 2.27 below. All cohomology is Galois cohomology. Page 62 defines goodness, ordinaryness, and semistability. Page 65-66 defines $H^1_f$. The notation $\operatorname{ad} \rho$ refers to $\operatorname{End}(M_\rho)$, where $M_\rho$ is the module underlying the representation $\rho$, and $\operatorname{ad}^0 \rho$ refers to trace-zero endomorphisms. The notation $G_\ell$ refers to the local Galois group at $\ell$.
Proposition 2.27
Suppose that $\overline{\rho}: G_{\ell} \to \operatorname{GL}_2(k)$ is semistable (meaning flat or ordinary). Then
$$\left| H^1_f(G_{\ell}, \operatorname{ad}^0 \overline{\rho}) \right| \leq \left| H^0(G_{\ell}, \operatorname{ad}^0 \overline{\rho})\right| \left|k\right|$$
The given proof apparently uses the following result of Fontaine:
Proposition 2.28: Suppose that $\rho: G_\ell \to \operatorname{GL}_2(\mathcal O/\lambda^n)$ is good (i.e. flat, more or less). Then
$$\left| H^1_f(G_\ell, \operatorname{ad} \rho)\right| = \left| H^0(G_\ell, \operatorname{ad}^0 \rho)\right| (\left| \mathcal O/\lambda^n \right|)^2$$
Here $\mathcal O$ is a complete Noetherian local ring with maximal ideal $\lambda$ and residue field $k$. Here is the how proof goes assuming 2.28:
Note that if $\rho: G_{\ell} \to \operatorname{GL}_2(\mathcal O/\lambda^n)$ is good, then
$$H^1(G_\ell/I_\ell) \subset H^1_f(G_\ell, \operatorname{ad} \rho) \cap H^1(G_\ell, \mathcal O/\lambda^n)$$
by Lemma 2.22 d), which gives the result.
I do not understand this proof. Specifically:
- Why does goodness of $\rho$ give us the desired containment?
- Where did we use 2.28?
- Most importantly, why does the result follow from the given containment? I read Lemma 2.22 d), and it essentially just says if two modules are $\mathbb Z_\ell[I_\ell]$-isomorphic, then one is good iff the other one is.
Assume that $\overline{\rho}$ is good. Note that $\operatorname{ad}\overline{\rho}=\operatorname{ad}^0\overline{\rho} \oplus k$.
By the definition of $H^1_f$, it follows that $H^1_f(G_{\ell},\operatorname{ad}^0\overline{\rho})$ is the kernel of the map $H^1_f(G_{\ell},\operatorname{ad}\overline{\rho}) \rightarrow H^1(G_{\ell},k)$ (given by the trace map $\operatorname{ad}\overline{\rho} \rightarrow k$).
The domain has cardinality $|H^1_f(G_{\ell},\operatorname{ad}\overline{\rho})|= |H^0(G_{\ell},\operatorname{ad}^0\overline{\rho})||k|^2$, so it’s enough to show that the map is not zero (since it’s $k$-linear).
The trick is to see that this image contains $H^1(G_{\ell}/I_{\ell},k)$, which is nonzero. Indeed, $k$ is a direct factor of $\operatorname{ad}\overline{\rho}$ and unramified cocycles are flat (that may follow from Lemma 2.22(d)).
If I’m not mistaken, they would correspond to extensions of the form $M_{\rho} \otimes k^{\oplus 2}$, where $G_{\ell}$ acts on $k^{\oplus 2}$ in an unramified way and the Frobenius is a unipotent matrix.
With a good enough formalism for mod $\ell$ representations (which may or may not currently exist), I would expect that $H^1_f(G_{\ell},\operatorname{ad}\overline{\rho}) = H^1_f(G_{\ell},\operatorname{ad}^0\overline{\rho}) \oplus H^1_f(G_{\ell},k)$, and that the flat $k$-valued cocycles be exactly the unramified ones.