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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 $\phi(f(T)) = f(T^p)$.)

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  • $\begingroup$ I don't know if this will help, but you can linearize $\phi_1:V_1\rightarrow V_1$ (as in definition 3.1.1 of Brinon-Conrad's notes) so that a $\phi$-module is then $V_1$ together with a linear map $\phi_1^\prime:\phi_1^*V_1\rightarrow V_1$. $\endgroup$
    – Rob Harron
    Commented Feb 24, 2010 at 4:27
  • $\begingroup$ Rob H., the problem is that it is hard to encode on some $V$ that it has the form $\phi^{\ast}(V_1)$ for some specified $V_1$. So I don't think it helps much to linearize like this. If I am missing some trick, that would be swell. $\endgroup$
    – BCnrd
    Commented Feb 24, 2010 at 6:25

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Rob, I am doubtful that in such generality (with $\phi$ presumably meant to act on $E$ by some unspecified endomorphism) there is a reasonable answer. The reason why Jordan canonical form "works" even if $E$ isn't algebraically closed is that one can appeal to the crutch of rational canonical form over any field and relate it to the Jordan form over an algebraic closure. That is, one gets "lucky" that passing to an extension field has no effect on the answer.

In contrast, this is not true for semilinear algebra in interesting cases. For example, already in the classical case $E = W(k)[1/p]$ with $k$ perfect of characteristic $p > 0$ and $\phi_E$ the usual Frobenius, if $k$ is algebraically closed then the category is semisimple and the Dieudonn\'e-Manin classification characterizations isomorphism classes in terms of Newton polygons. But for more general $k$ one has isoclinic decomposition (by suitable descent from $W(\overline{k})[1/p]$) yet semisimplicity fails. In other words, there are objects over $W(k)[1/p]$ which are non-isomorphic that become isomorphic over $W(\overline{k})[1/p]$. So any answer of "linear algebra" flavor would have to be sensitive to change of $k$.

To the extent anything useful could be said, one probably has to specify a particular class of pairs $(E, \phi_E)$. For example, in the classical case as above one has the linguistic answer of working with modules over the Dieudonn\'e ring for $k$. It is not such a nice ring. For finite $k$, so some iterates of $V$ and $F$ are actually $W(k)[1/p]$-linear, one can get a handle on the structure of this ring via central simple algebras. This already comes up in Tate's work on the $p$-part of his isogeny theorem for abelian varieties over finite fields (see the CM seminar notes on Honda-Tate theory on my webpage, or the original article by Milne-Waterhouse referenced there).

It sounds like the case you wish to consider is when $E$ is a field of characteristic $p > 0$ and $\phi_E$ is the $p$-power map (though I suspect you ultimately would not be happy with only $\mathbf{F} _p$-coefficients in your Laurent series field). So here the relevant ring is a 1-variable non-commutative algebra, the structure of which seems quite hard beyond the algebraically closed case (especially imperfect $E$); I doubt anything useful can be said in general. In the algebraically closed case there is a kind of "Jordan decomposition" which is explained early in an expose of Katz in SGA7 (modeled on the connected-etale sequence, so I think also ok for perfect $E$), and the non-commutative polynomial algebra $E[\phi]$ is like a non-commutative PID; my recollection is that one again has a Dieudonn\'e-Manin kind of result, but more in the spirit of the structure of finite abelian groups or rational canonical form. However, this is quite far-removed from the case you care about with imperfect $E$.

Moral of the story: it is generally quite difficult to identify isomorphism classes beyond very low-dimensional situations or "algebraically closed" hypotheses.

To put matters in perspective, think about the structure of 1-dimensional $p$-divisible groups over a field $E$ of characteristic $p$: if $E$ is algebraically closed (I think even just separably closed is enough) then we have Lazard's result that height is the unique invariant. Beyond that, one has no nice classification (aside from being able to shift computations to the Dieudonne module side when $E$ is perfect, where it quickly becomes apparent via basis transformation formulas that matters are quite tricky when $E \ne \overline{E}$). And since a $p$-divisible group involves an "infinite" amount of data, the gap between working over $E$ and working over $E_s$ (or $\overline{E}$) is gigantic in a way that isn't the case when working with schemes of finite type over the ground field.

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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.)

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