Question

Let k be the algebraic closure of Fp and K = k(t), where t is an indeterminate. Let G be an elementary p-group of order p^2 generated by a, b. Show that
a → A =
1 1
0 1

b→ B =
1 t
0 1
defines a representation of G over K which is not equivalent to any representation of G over k.

Answer 1

Since $A^p = B^p = I$ and $AB = BA$, given two maps define a representation $D: G \to \mathrm{GL}_2(K)$. We shall prove a slightly stronger statement: For any extension field $L \supseteq K$, $D$ cannot be equivalent over $L$ to any representation of $G$ over $k$. Indeed, assume otherwise. Then there exists $U = \begin{pmatrix} a & b \\ c & d \end{pmatrix} \in \mathrm{GL}_2(L)$ which conjugates both $A$ and $B$ into $GL_2(k)$. By explicit matrix multiplication,

$$U^{-1}BU = e^{-1}\left( \begin{array}{cc}* & td^{2} \\ -tc^{2} & * \end{array} \right), U^{-1}AU = e^{-1}\left( \begin{array}{cc}* & d^{2} \\ -c^{2} & * \end{array} \right)$$

where $e = \det U = ad - bc \in L^*$. Therefore, $td^{2}e^{-1}$, $d^{2}e^{-1}$, $tc^{2}e^{-1}$, $c^{2}e^{-1}$ all belong to $k$. Now $c$, $d$ cannot both be zero, since $U \in \mathrm{GL}_2(L)$. If $c \neq 0$, we have $t = tc^{2}e^{-1}/c^{2}e^{-1} \in k$. If $d \neq 0$, we have similarly $t = td^{2}e^{-1}/d^{2}e^{-1} \in k$. This gives the desired contradiction.