Let $k$ be an algebraically closed field of characteristic $p$, and $G$ be a finite group whose order is not divisible by $p$. I would like to prove the following: if $V$ is an irreducible representation of $kG$, then $\dim V \neq 0$ in $k$, i.e. $p$ does not divide the dimension of $V$.
Here is one explanation of why it is true: for representations over $\mathbb{C}$, the dimension of a representation must divide the order of the group. Then, there is a claim that the dimensions of the irreducible representations of $G$ over $k$ are equal to the dimensions of the irreducible representations of $G$ over $\mathbb{C}$. (I do not know why this claim is true). This then implies that over $kG$, the dimension of any irreducible representation $V$ divides $|G|$, and hence is coprime to $p$.
Is there a shorter argument, which avoids the character theory over $\mathbb{C}$?