Simple $KC_3$-modules where $\text{char}(K)=2$ I am trying to understand part of this book I’m reading and it’s trying to find the simple $KC_3$-modules where $K$ is a field of characteristic 2.
It says ‘$KC_3$ is semisimple and if $K$ contains a primitive cube root of unity $\omega $, there are 3 1-dimensional simple representations, on which the generator of $C_3$ acts as $1, \omega,$ or $ \omega^2 $.
I am trying to understand how they come to this. Firstly, how is it the case that the simple modules are necessarily 1-dimensional, and secondly, what does $KC_3$ being semisimple have to do with its simple modules?
 A: The fact that simple modules are dimensional boils down to linear algebra: a (finite dimensional) $K$-linear representation of $C_3$ is exactly the same as a (finite dimensional) $K$-linear vector space $V$ and a linear transformation $T: V \longrightarrow V$ such that $T^3 = 1$.
If $K$ contains a primitive $3$rd root of $1$, call it $\omega$, then $X^3 - 1$ splits completely into $(X-1)(X-\omega)(X-\omega^2)$ and is thus diagonalizable. As such, it admits a $T$-invariant one dimensional subspace, and hence if $V$ is simple it must coincide with this one dimensional $C_3$-submodule.
This also proves the claim you are quoting, since on such an eigenspace, $T$ acts by multiplication by one of $1$, $\omega$ or $\omega^2$. Note that $KC_3$ being semi-simple is independent of $K$ being algebraically closed: this is true because the characteristic of $K$ is coprime with the order of $C_3$, so Maschke's Theorem applies.
If $K= \mathbb F_2$, on the other hand, notice that the $KC_3$-module breaks down into the direct product $K \times K[\alpha]$ where $\alpha^2+\alpha+1=0$, and $K[\alpha]$ is a simple two dimensional $KC_3$-module since $t^2+t+1$ has no roots in the base field $K$.
