# Can an irreducible representation have a zero character?

Is there an example of the following situation : $F$ is a field, $G$ is a finite group, $\rho$ is an irreducible $F$-representation of $G$ and the character of this representation takes the zero value at every element of $G$ ? If I'm not wrong, $F$ cannot be algebraically closed (Robinson, A Course in the Theory of Groups, 8.1.9, p. 220) and must have a nonzero characteristic $p$ such that the degree of the representation is divisible by $p$ and such that $G$ is not a $p$-group (Robinson, exerc. 8.1.5, p. 222). But that doesn't solve the problem. Thanks in advance for the answers.

(Edit. Since I didn'get answers here, I asked the question on Mathoverflow :

• Won't $\rho(1) = I$ the identity of $GL_n(F)$ ? Oct 23, 2016 at 7:01
• Yes, so the value of the character at 1 is the trace of the identity and this trace is zero if the degree of $\rho$ is divisible by $p$ (the characteristic of $F$). Oct 23, 2016 at 7:14
• SImple remark: the $F$-algebra generated by $\rho(G)$ will also have trace identically zero, and will be a simple $F$-algebra.
– YCor
Oct 23, 2016 at 7:42
• @YCor Why is it clear that the trace is zero? If that is true, we are done, because I can reduce to the case that $F$ is a finite field. All simple algebras over a finite field are separable field extensions (Wedderburn's little theorem + finite fields are perfect) and the trace of a separable field extension is nonzero. Oct 23, 2016 at 15:02
• @DavidSpeyer because the $F$-subspace generated by a subgroup of $GL_n(F)$ is stable under multiplication, and hence equals the $F$-subalgebra it generates.
– YCor
Oct 23, 2016 at 15:40