galois representations - One dimensional (phi,Gamma)-modules in char p

OK...I think I see how to do this now. In the end, I am seeing $(p-1)^2$ distinct $(phi,Gamma)$-modules which matches well with the Galois side.

To do this, let $D$ be any 1-dimensional etale $(phi,Gamma)$-module. Let $e$ be a basis, and set $phi(e)=h(T)e$ with $h(T) in F_p((T))^times$. Write $h(T) = h_0 T^a f(T)$ with $h_0 in F_p^times$ and $f(T) in F_p[[T]]$ with $f(0)=1$.

Changing basis from $e$ to $u(T)e$ with $u(T) in F_p((T))^times$ gives

$$phi(u(T)e) = u(T^p)h(T)e = frac{u(T^p)}{u(T)} h(T) (u(T)e).$$
I claim one can find $u(T)$ such that $u(T)/u(T^p)$ equals any element of $1+TF_p[[T]]$. Indeed, for such an element $g(T)$, the infinite product $prod_{j=1}^infty phi^j(g(T))$ (which hopefully converges since $g(0)=1$) works.

Thus, we can change basis so that $phi$ has the form $phi(e) = h_0 T^a e$ -- i.e. we can kill off the $f(T)$ term. Further, by making a change of basis of the form $e$ goes to $T^b e$, we may assume that $0 leq a < p-1$.

Now, we use the fact that the $phi$ and $Gamma$ actions commute (which is a strong condition even in dimension 1). Namely, let $gamma$ be a generator of $Gamma$, and set $gamma e = g(T) e$. Then $gamma phi e = phi gamma e$ implies

$$((1+T)^{chi(gamma)}-1)^a g(T) = g(T^p) T^a.$$
Comparing leading coefficients, we see this is only possible if $a=0$ and $g(T)$ is a constant.

Thus, $D$ has a basis $e$ so that $phi(e) = h_0 e$ and $gamma(e) = g_0 e$ with $h_0,g_0 in F_p^times$ as desired.

Does this look okay? Any takers for the 2-dimensional case?

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