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 $|G|$. 



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


