# linear independence of orbits via a set of transformations in char p

Let $T_1, \ldots, T_n \in GL(n,\mathbb{F}_p)$. Suppose for all $\vec{v} \in \mathbb{F}_p^n$ we have $\det (T_1 \vec{v}, T_2 \vec{v}, \ldots, T_n \vec{v}) = 0$. Now, let $k$ be a finite extension of $\mathbb{F}_p$. Is it true that $\det(T_1 \vec{v}, \ldots, T_n \vec{v})=0$ for all $\vec{v} \in k^n$?

I know the above is true in some special cases. For example, if the $T_i$ are powers of a single linear transformation $T$, if the $T_i$ are the elements of a finite abelian subgroup of $GL(n,\mathbb{F}_p)$ of order equal to $n$ coprime to $p$, or if $\mathbb{F}_p$ is replaced with $\mathbb{Q}$ and $k$ is real.

UPDATE: So the above, unfortunately, isn't true in general. But ultimately, as the comments below indicate, I want want to know about something rather specific (to which the above problem is related): Let $p$ be an odd prime. If $E$ is the group of units of a real abelian number field with Galois group $G$, does the cyclicity of $E\otimes k$ as a $k[G]$-module imply the cyclicity of $E \otimes \mathbb{F}_p$ as an $\mathbb{F}_p[G]$-module. I know that this is true if $p$ does not divide $|G|$.

• I was hoping something very general like the above would be true. Truth be told, I'm interested in something a bit more specific: the $T_i$ are the non-identity elements of an abelian group of order $n+1$. Also, if it helps, we may assume $\sum T_i = -1$. This is coming from the following number theoretic place: Let $E$ be the group of units of a real abelian number field with Galois group G. If $E \otimes k$ is a cyclic $k[G]$-module, is it true that $E \otimes \mathbb{F}_p$ is a cyclic $\mathbb{F}_p[G]$-module?
– Tim
Feb 6 '13 at 5:18
• I don't know if there's a principled way to make counterexamples. Keeping in mind what Jason said, you probably need to keep $p$ small. You could try $p = 2$ and $n = 3$, with the non-identity elements of a group isomorphic to $\mathbb{Z}/2 \oplus \mathbb{Z}/2$ (so try to find two idempotent transformations which commute). I don't know if this will work, but it should be easy enough to write a computer program to search. In the other direction, I assume you're already tried rephrasing the problem in a form where you could hit it with Hilbert 90? Feb 6 '13 at 15:05
• @Abhinav: Your counterexample for $p=2$ extends to every $p$. I think you can repeat factors in other diagonal entries to get counterexamples for every $n\geq p$. Feb 6 '13 at 16:26
• @Jason: Nice, I think you're right. Thanks.
– Tim
Feb 6 '13 at 16:57
• @Abhinav: The cyclicity of $E \otimes \mathbb{F}_p$ can indeed be attacked with Hilbert's theorem 90, and I have thought about this a good deal. But I'm not sure if all that applies to the question at hand, namely, does the cyclicity of $E \otimes k$ imply the cyclicity of $E \otimes \mathbb{F}_p$. The only hope of attacking this problem seems to a be through linear algebra and/or modular representation theory. It couldn't hurt to get my hands dirty, as you suggest though.
– Tim
Feb 6 '13 at 17:04

Not in general: here is a counterexample. Take $p = 2$, and consider the matrices $T_1 = [1,0,0; 0,0,0; 0,0,0], T_2 = [0,0,0; 0,1,0; 0,1,0], T_3 = [0,0,0; 1,0,0; 0,1,0]$. (I'm using semi-colons to separate rows; a bit of LaTeX trouble formatting the matrix ...)

Then if $v$ is the column vector $(x,y,z)$, we get that the matrix $(T_1 v, T_2 v, T_3 v)$ is $[x,0,0; 0,y,x; 0,y,y]$. It has determinant $xy^2 - yx^2$. Note that if $x,y$ are in $\mathbb{F}_2$, then $x^2 = x$ and $y^2 = y$, so this determinant is $0$. However, the polynomial $xy^2 - yx^2$ is not identically $0$, so there are values in $\overline{\mathbb{F}_2}$for which the determinant will not vanish.

Update:

These transformations are not invertible, but you can embed this example in 4 dimensions. Namely, define $T_1(x,y,z,w) = (x,y,z,w)$, $T_2(x,y,z,w) = (x,x+y,z,w)$, $T_3(x,y,z,w) = (x,y,y+z,y+w)$ and $T_4(x,y,z,w) = (x,y,x+z, y + w)$. Then these are all clearly invertible, and the determinant of the matrix of linear forms is $x^2 y(y - x)$, which vanishes as before if $x$ and $y$ are in $\mathbb{F}_2$, but not over the algebraic closure.

• An interesting example, thanks. But your $T_i$ aren't in $GL(3, \mathbb{F}_2)$. Perhaps this idea can be generalized though?
– Tim
Feb 6 '13 at 5:16
• Indeed, I forgot about invertibility! But I think one can embed this example in $GL_4$; see the update to my answer above. Feb 6 '13 at 13:20
• Awesome, thanks. Any thoughts on the special case mentioned in the comment to the original post?
– Tim
Feb 6 '13 at 14:12

If $n \leq p$, then obviously this is true. For variables $x_1,\dots,x_n$, form the vector $\vec{v} = (x_1,\dots,x_n)^\dagger$, and consider the polynomial $F(x_1,\dots,x_n) = \text{det}(T_1\vec{v},\dots,T_n\vec{v})$. This is a homogeneous degree $n$ polynomial in the variables $x_1,\dots,x_n$ with coefficients in $\mathbb{F}_p$.

Now prove the following lemma: for every integer $m\geq 0$, every homogeneous polynomial $F(x_0,\dots,x_m)$ of degree $\leq p$ that vanishes identically on $\mathbb{F}_p^{m+1}$ is the zero polynomial. This is proved by induction on $m$. For $m=0$, this is trivial. For $m=1$, this is a slight variant of the inhomogeneous analogue. Every inhomogeneous polynomial $F(1,x)$ that vanishes identically on $\mathbb{F}_p$ is a multiple of $x^p-x$. Since the degree is $\leq p$, this means that $F(x_0,x_1) = a(x_1^p-x_1x_0^{p-1})$ for some $a\in \mathbb{F}_p$. Since $F(0,1)$ is zero, also $a$ equals $0$, so that $F$ is the zero polynomial. That establishes the base case of the induction.

By way of induction, assume that $m\geq 2$, and assume the result for $m-1$. Write $F(x_0,\dots,x_m)$ as $$F_{0}(x_0,\dots,x_{m-1})x_m^n + \dots + F_{n-d}(x_0,\dots,x_{m-1})x_m^d + \dots + F_n(x_0,\dots,x_{m-1})x_m^0,$$
with each $F_{n-d}$ a homogeneous polynomial in $x_0,\dots,x_{m-1}$ of degree $n-d \leq p$. Plug in $(0,\dots,0,1)$ to conclude that the constant $F_0$ is zero. For fixed values $(a_0,\dots,a_{m-1})\in \mathbb{F}_p^{m}$ the resulting polynomial $F(a_0,\dots,a_{m-1},x) \in \mathbb{F}_p[x]$ vanishes identically on $\mathbb{F}_p$ by hypothesis. By the same argument as for $m=1$, $F(a_0,\dots,a_{m-1},x) = a(x^p-x)$ for some $a\in \mathbb{F}_p$. Since $F_0$ is zero, in fact $a$ equals $0$. Thus every $F_{m-d}(a_0,\dots,a_{m-1})$ equals $0$ for every $(a_0,\dots,a_{m-1})$ in $\mathbb{F}_p^m$. By the induction hypothesis, this forces every polynomial $F_{m-d}(x_0,\dots,x_{m-1})$ to be the zero polynomial. Therefore also $F(x_0,\dots,x_m)$ is the zero polynomial.

Finally, for $n=p+1$, there is a generalization of Abhinav's counterexample. Namely, the $p+1$ matrices $T_0,\dots,T_p$ defined by, $$T_0(\sum_{i=0}^px_i \mathbf{e}_i) = \sum_{i=0}^p x_i \mathbf{e}_i, \ \ T_0 = \text{Id},$$ and for $i=1,\dots,p$, $$T_i(\sum_{i=0}^p x_i\mathbf{e}_i) = \sum_{i=0}^p x_i \mathbf{e}_i + (x_p-ix_0)\mathbf{e}_i,$$ (these are all invertible except if $p=2$, where a shuffle as in Abhinav's answer works). The determinant is $F(x_0,x_1,\dots,x_p) = x_0x_p(x_p-1x_0)\cdots (x_p-(p-1)x_0)$, which vanishes identically on $\mathbb{F}_p^{p+1}$.