prove that homomorphism (rings) from a field to ring is bijective or the zero homomorphism. $F$ is a field and $R$ is a ring.$\:\phi :F\rightarrow R$ is a ring homomorphism. I need to prove that it is bijective or it is $\phi =0$.
I tried to use some how the fact that I have opposites in F, but it didn't give me any result. 
I stuck after trying to write few strokes of the proof: 
If $\phi$ bijective, done. Otherwise $\exists a,b\in F,\:a\ne b\:;\:\phi \left(a\right)=\phi \:\left(b\right)$
Now I can write the homomorphism characteristics: $$\:\phi \left(ab\right)=\:\phi \left(a\right)\cdot \:\phi \left(b\right)=\:\phi \left(a\right)\cdot \:\phi \left(a\right)=\:\phi \left(a^2\right)$$
$$\:\phi \left(a+b\right)=\:\phi \left(a\right)+\:\:\phi \left(b\right)=\:\phi \left(a\right)+\phi \left(a\right)=\:\phi \left(2a\right)$$
Can somebdy give a hint for the right direction to prove that? 
 A: Consider the ideal $\ker \phi\subset F$. Since $F$ is a field, either $\ker \phi=0$ or $\ker \phi=F$. In the former case, $\phi$ is injective. In the latter case, $\phi=0$.
(Note: It is not true that $\phi$ is bijective. For example, consider the ring homomorphism $\phi\colon \mathbb Q\to \mathbb Q\oplus \mathbb Q$ given by $q\mapsto (q,0)$.)
A: Direct proof of injectivity:
If $a\neq b$ and $\phi\left(a\right)=\phi\left(b\right)$ then $$\phi\left(x\right)=\phi\left(x\left(a-b\right)\left(a-b\right)^{-1}\right)=\phi\left(x\right)\left[\phi\left(a\right)-\phi\left(b\right)\right]\phi\left(\left(a-b\right)^{-1}\right)=0$$
As you allready noted surjectivity is not necessary.
A: given a homomorphism H:F->R where F is a field and R is a ring.
we know than kernel of homomorphism will be a subset of F say K.
we also know that kernels are ideals => K is an ideal of F.
But a field has only trivial ideals =>K={0} or K=F.
if K={0} then we know that H is an isomorphism and hence by the defination of 
isomorphism it is bijective.
if K=F.
then entire F is the kernel hence it becomes the zero homomorphism.
