This may be something very simple, but I really got stuck.
Let $G$ be a group and $F$ be a field. We define the set of all maps $F^G$ from $G$ to $F$.
We say that a character of $G$ in $F$ is a function $\sigma:G\longrightarrow F$ (i.e. $\sigma \in F^G$) such that
1) $\forall x,y\in G: \ \sigma (xy)=\sigma(x)\sigma(y).$
2) $\forall x\in G: \ \sigma (x)\neq 0_F$.
In other words $\sigma \in F^G$ is a character of $G$ in $F$ if $\sigma:G\longrightarrow U(F)=F^*$ is a group homomorphism (where $U(F)$ is the multiplicative group of the field $F$).
My question is why these are equivalent?
Obviously if $\sigma:G\longrightarrow F$ is a character then is $\sigma:G\longrightarrow U(F)$ is a group homomorphism.
And what about the converse?
Let's suppose that $\sigma:G\longrightarrow U(F)$ is a group homomorphism. Then 1) is immediate. But why $\sigma(x)\neq 0_F,\ \forall x\in G$?
If $\sigma: G \rightarrow U(F)$, then $\sigma(x)$ can not be $0_F$, since $0_F$ is not an element of $U(F)$, yielding 2). This finishes the proof of the converse.