This is a question from Pinter's A book of Abstract Algebra:
If domain A of the homomorphism $f$ is a field, and if the range of $f$ has more than one element, then $f$ is injective.
Here's my attempt:
Claim. If $\text{im} f \ne \{ 0\}$ then $f(1)\ne 0$.
Proof (by contrapositive). Suppose that $f(1)=0$. Then if $x\in A$ then $f(x)= f(x\cdot 1) = f(x)f(1)= f(x)\cdot 0=0$. Hence, $\text{im} f = \{ 0\}$.
To show that $f$ is injective, it is enough to show that $\text{ker} f =\{ 0\}$. Let $x \in \text{ker} f$. Then $f(x)=0$. We claim that $x=0$. Suppose that $x\ne 0$. Then there is an element $x^{-1} \in A$ such that $xx^{-1} = 1$. Now, $f(x)=0 \\ \Rightarrow f(x)f(x^{-1}) = 0 \cdot f(x^{-1}) \\ \Rightarrow f(xx^{-1})=0 \\ \Rightarrow f(1)=0$
Which is a contradiction!
Thus, $\text{ker} f =\{ 0\}$ . This completes the proof.
Is this okay or am I missing something? I was wondering if there was some easier way to do this.