A search brought up [this](https://mathoverflow.net/questions/252855/can-an-irreducible-representation-have-a-zero-character), with reference to a book by I. M. Isaacs. However, the proof in the book leverages on a lot of field theory knowledge. I am wondering, is there a simpler proof (or a proof requiring only materials covered in a standard representation theory course) of characters being non-zero, if I only care about about irreducible characters of a group $G$ over a field $F$ where $char(F)$ divides the order of $G$?



"Simple" proof of irreducible characters of finite groups being non-zero
Here is a solution, which may or may not be simple, based on a suggestion in Glasby’s comment to my question <https://math.stackexchange.com/questions/819466/the-division-algebras-arising-in-the-wedderburn-decomposition-of-a-finite-group>


First of all, since $G$ is finite, assuming $F$ has characteristic $p>0$, your irreducible representation is defined over a finite extension of the prime field so you can assume without loss of generality that $F$ is a finite field. (**Edit.** See Lemma 6 of <https://archives.maths.anu.edu.au/people/Kovacs/K095.pdf> for a proof). By Weddderburn theory and Wedderburn’s little theorem that any finite division ring is a field, we have that the image of $FG$ under you irreducible representation is isomorphic to $M\_k(L)$ where $L/F$ is a finite extension and your simple module is isomorphic to $L^k$ with the natural action of $M\_k(L)$ on it. If $A(g)$ is the $k\times k$ matrix over $L$ corresponding to $g\in G$, then the value of your character $\chi$ on $g$ is $tr\_{L/F}(tr(A))$ where $tr\_{L/F}(a)$ is the trace of left multiplication by $a$ on $L$ as a linear operator over $F$.


Thus the value of $\chi$ on any element of $FG$ is obtained by taking $tr\_{L/F}$ of the trace of the corresponding $k\times k$ matrix over $L$. By considering the rank $1$ matrix $aE\_{11}$ with $a\in L$, we see that if $tr\_{L/F}$ is not identically zero, then $\chi$ does not vanish on $FG$ and hence $G$. But this is easy since $tr\_{L/F}(a)$ is the sum of its Galois orbit. In fact, since $L/F$ is separable and Galois, as finite fields are perfect and all their finite extensions are cyclic, standard field theory tells you the trace form is non-degenerate. Alternatively, the normal basis theorem tells you that there is an element $a\in L$ whose Galois conjugates from an $F$-basis for $L$ and hence cannot sum to $0$. 


This is of course simpler in characteristic $0$ because $M\_k(L)$ is replaced by $M\_k(D)$ for a division algebra $D$ but $tr\_{D/F}(1)=\dim D$ doesn’t vanish.


