# Frobenius Semi-linear Algebra: 1

Today I want to explain some “well-known” facts in semilinear algebra. Here’s the setup. For safety we’ll assume ${k}$ is algebraically closed of characteristic ${p>0}$ (but merely being perfect should suffice for the main point later). Let ${V}$ be a finite dimensional vector space over ${k}$. Consider some ${p}$-semilinear operator on ${V}$ say ${\phi: V\rightarrow V}$. The fact that we are working with ${p}$ instead of ${p^{-1}}$ is mostly to not scare people. I think ${p^{-1}}$ actually appears more often in the literature and the theory is equivalent by “dualizing.”

All this means is that it is a linear operator satisfying the usual properties ${\phi(v+w)=\phi(v)+\phi(w)}$, etc, except for the scalar rule in which we scale by a factor of ${p}$, so ${\phi(av)=a^p\phi(v)}$. This situation comes up surprisingly often in positive characteristic geometry, because often you want to analyze some long exact sequence in cohomology associated to a short exact sequence which involves the Frobenius map or the Cartier operator. The former will induce a ${p}$-linear map of vector spaces and the latter induces a ${p^{-1}}$-linear map.

The facts we’re going to look at I’ve found in three or so papers just saying “from a well-known fact about ${p^{-1}}$-linear operators…” I wish there was a book out there that developed this theory like a standard linear algebra text so that people could actually give references. The proof today is a modification of that given in Dieudonne’s Lie Groups and Lie Hyperalgebras over a Field of Characteristic ${p>0}$ II (section 10).

Let’s start with an example. In the one-dimensional case we have the following ${\phi: k\rightarrow k}$. If the map is non-trivial, then it is bijective. More importantly we can just write down every one of these because if ${\phi(1)=a}$, then

$\displaystyle \begin{array}{rcl} \phi(x) & = & \phi(x\cdot 1) \\ & = & x^p\phi(1) \\ & = & ax^p \end{array}$

In fact, we can always find some non-zero fixed element, because this amounts to solving ${ax^p-x=x(ax^{p-1}-1)=0}$, i.e. finding a solution to ${ax^{p-1}-1=0}$ which we can do by being algebraically closed. This element ${b}$ obviously serves as a basis for ${k}$, but to set up an analogy we also see that ${Span_{\mathbb{F}_p}(b)}$ are all of the fixed points of ${\phi}$. In general ${V}$ will breakup into parts. The part that ${\phi}$ acts bijectively on will always have a basis of fixed elements whose ${\mathbb{F}_p}$-span consists of exactly the fixed points of ${\phi}$. Of course, this could never happen in linear algebra because finding a fixed basis implies the operator is the identity.

Let’s start by proving this statement. Suppose ${\phi: V\rightarrow V}$ is a ${p}$-semilinear automorphism. We want to find a basis of fixed elements. We essentially mimic what we did before in a more complicated way. We induct on the dimension of ${V}$. If we can find a single ${v_1}$ fixed by ${\phi}$, then we would be done for the following reason. We kill off the span of ${v_1}$, then by the inductive hypothesis we can find ${v_2, \ldots, v_n}$ a fixed basis for the quotient. Together these make a fixed basis for all of ${V}$.

Now we need to find a single fixed ${v_1}$ by brute force. Consider any non-zero ${w\in V}$. We start taking iterates of ${w}$ under ${\phi}$. Eventually they will become linearly dependent, so we consider ${w, \phi(w), \ldots, \phi^k(w)}$ for the minimal ${k}$ such that this is a linearly dependent set. This means we can find some coefficients that are not all ${0}$ for which ${\sum a_j \phi^j(w)=0}$.

Let’s just see what must be true of some fictional ${v_1}$ in the span of these elements such that ${\phi(v_1)=v_1}$. Well, ${v_1=\sum b_j \phi^j(w)}$ must satisfy ${v_1=\phi(v_1)=\sum b_j^p \phi^{j+1}(w)}$.

To make this easier to parse, let’s specialize to the case that ${k=3}$. This means that ${a_0 w+a_1\phi(w)+a_2\phi^2(w)=0}$ and by assumption the coefficient on this top power can’t be zero, so we rewrite the top power ${\phi^2(w)=-(a_0/a_2)w - (a_1/a_2)\phi(w)}$.

The other equation is

$\displaystyle \begin{array}{rcl} b_0w+b_1\phi(w) & = & b_0^p\phi(w)+b_1^p\phi^2(w)\\ & = & -(a_0/a_2)b_1^pw +(b_0^p-(a_1/a_2)b_1^p)\phi(w) \end{array}$

Comparing coefficients ${b_0=-(a_0/a_2)b_1^p}$ and then forward substituting ${b_1=-(a_0/a_2)^pb_1^{p^2}-(a_1/a_2)b_1^p}$. Ah, but we know the ${a_j}$ and this only involves the unknown ${b_1}$. So since ${k}$ is algebraically closed we can solve to find such a ${b_1}$. Then since we wrote all our other coefficients in terms of ${b_1}$ we actually can produce a fixed ${v_1}$ by brute force determining the coefficients of the vector in terms of our linear dependence coefficients.

There was nothing special about ${k=3}$ here. In general, this trick will work because it only involves the fact that applying ${\phi}$ cycled the vectors forward by one which allows us to keep forward substituting all the equations from the comparison of coefficients to get everything in terms of the highest one including the highest one which transformed the problem into solving a single polynomial equation over our algebraically closed field.

This completes the proof that if ${\phi}$ is bijective, then there is a basis of fixed vectors. The fact that ${V^\phi=Span_{\mathbb{F}_p}(v_1, \ldots, v_n)}$ is pretty easy after that. Of course, the ${\mathbb{F}_p}$-span is contained in the fixed points because by definition the prime subfield of ${k}$ is exactly the fixed elements of ${x\mapsto x^p}$. On the other hand, if ${c=\sum a_jv_j}$ is fixed, then ${c=\phi(c)=\sum a_j^p \phi(v_j)=\sum a_j^p v_j}$ shows that all the coefficients must be fixed by Frobenius and hence in ${\mathbb{F}_p}$.

Here’s how this is useful. Recall the post on the fppf site. We said that if we wanted to understand the ${p}$-torsion of certain cohomology with coefficients in ${\mathbb{G}_m}$ (Picard group, Brauer group, etc), then we should look at the flat cohomology with coefficients in ${\mu_p}$. If we specialize to the case of curves we get an isomorphism ${H^1_{fl}(X, \mu_p)\simeq Pic(X)[p]}$.

Recall the exact sequence at the end of that post. It told us that via the ${d\log}$ map ${H^1_{fl}(X, \mu_p)=ker(C-I)=H^0(X, \Omega^1)^C}$. Now we have a ridiculously complicated way to prove the following well-known fact. If ${E}$ is an ordinary elliptic curve over an algebraically closed field of characteristic ${p>0}$, then ${E[p]\simeq \mathbb{Z}/p}$. In fact, we can prove something slightly more general.

By definition, a curve is of genus ${g}$ if ${H^0(X, \Omega^1)}$ is ${g}$-dimensional. We’ll say ${X}$ is ordinary if the Cartier operator ${C}$ is a ${p^{-1}}$-linear automorphism (I’m already sweeping something under the rug, because to even think of the Cartier operator acting on this cohomology group we need a hypothesis like ordinary to naturally identify some cohomology groups).

By the results in this post we know that the structure of ${H^0(X, \Omega^1)^C}$ as an abelian group is ${\mathbb{Z}/p\oplus \cdots \oplus \mathbb{Z}/p}$ where there are ${g}$ copies. Thus in more generality this tells us that ${Jac(X)[p]\simeq Pic(X)[p]\simeq H^0(X, \Omega^1)^C\simeq \mathbb{Z}/p\oplus \cdots \oplus \mathbb{Z}/p}$. In particular, since for an elliptic curve (genus 1) we have ${Jac(E)=E}$, this statement is exactly ${E[p]\simeq \mathbb{Z}/p}$.

This point is a little silly, because Silverman seems to just use this as the definition of an ordinary elliptic curve. Hartshorne uses the Hasse invariant in which case it is quite easy to derive that the Cartier operator is an automorphism (proof: it is Serre dual to the Frobenius which by the Hasse invariant definition is an automorphism). Using this definition, I’m actually not sure I’ve ever seen a derivation that ${E[p]\simeq \mathbb{Z}/p}$. I’d be interested if there is a lower level way of seeing it than going through this flat cohomology argument (Silverman cites a paper of Duering, but it’s in German).