TITLE: Semi-linear operators
QUESTION [13 upvotes]: If $V_1$ and $V_2$ are finite-dimensional vector spaces over a field $E$, each equipped with an $E$-linear operator $\phi$, we can tell if $V_1$ and $V_2$ are isomorphic as $\phi$-modules by comparing the Jordan canonical form of the $\phi$-operator of each space.
If $E$ is itself equipped with a $\phi$-operator, and now each $V_i$ has a semi-linear operator $\phi$, is there some way (algorithm?) to determine whether or not $V_1$ is isomorphic to $V_2$ as a $\phi$-module?
(I would be content for an answer in the case when $E=F_p((T))$ and $\phi(f(T)) = f(T^p)$.)

REPLY [3 votes]: This question is really interesting. As Brian said, there is no doubt that things are very sensitive to the specific $E$ and $\phi$, and I'd be interested in knowing how things work for just about any example. Here are a few things that you might find helpful, if you don't know them already.
Suppose $\phi$ is an automorphism of $E$ of finite order $n$, and suppose that we look at $E$-vector spaces with semi-linear operators of order $n$. If we let $F$ denote the fixed field of $\phi$, then $E/F$ is a cyclic Galois extension of order $n$, and an $E$-vector space with a semi-linear operator $\phi$ of order $n$ is, by Galois descent, the same as an $F$-vector space. More precisely, given an $F$-vector space $N$, then $E\otimes_F N$ is an $E$-vector space with a semi-linear operator $\phi\otimes 1$ of order $n$; and given an $E$-vector space $M$ with semi-linear operator $\phi$ of order $n$, the subset of $\phi$-invariants is an $F$-vector space. These two functors induce an equivalence of categories.
But things seem to be much richer if we don't require the semi-linear endos of $M$ to be of finite order, even if the automorphism $\phi$ of $E$ does have finite order. For instance, Dieudonne-Manin theory mentioned by Brian.
A few months ago, I wondered a bit about the case where $E=\mathbf{C}$ and $\phi$ is complex conjugation. Then if we write $\phi$ as a matrix $A$ with respect to a basis, a change of basis matrix $B$ changes $A$ to $\bar{B}A B^{-1}$. (Even the case of 1-dimensional vector spaces is not completely trivial, though it's not hard.) I asked quite a few people, and eventually someone was able to tell me, after doing a bit of poking around himself, that this equivalence relation is called consimilarity by some people, and there's a paper (from 1988!) by Hong and Horn giving a Jordan theorem for it. It's called "A canonical form for matrices under consimilarity". (NB I haven't read it. The style didn't really appeal to me, and at the time, I didn't care enough to get beyond that.)