Show that $\varphi$ is injective. $\varphi : F \rightarrow R$ is a ring homomorphism, $F$ is a field and $R$ is a ring that is not the zero ring. Let $F$ be a field and $R$ a commutative ring where $R$ is not the zero ring. Suppose $\varphi : F \rightarrow R$ is a ring
homomorphism. Show that $\varphi$ is injective.
Any help would be appreciated. I feel pretty lost on this.
 A: 
$\varphi$ is zero homomorphism, or it is injective:  

Assume that $\varphi$ is not injective. Let  $0\neq x\in F$, such that $\varphi (x)=0$. Since $F$ is a field, $x$ has an inverse. So 
$$1=\varphi (1)=\varphi (xx^{-1}) =\varphi (x)\varphi (x)^{-1}=0.$$ 
Hence $\varphi$ is zero homomorphism.
A: The kernel of a ring homomorphism is an ideal. A field has exactly two ideals, the unit ideal and the zero ideal. So either $\ker{\varphi}$ is the zero ideal or $\ker{\varphi}$ is the unit ideal, $F$ itself. If $\ker{\varphi}$ is the zero ideal, then $\varphi$ is injective (a general result you should know). If $\ker{\varphi}$ is the unit ideal $F$, then every element of $F$ is sent to $0_R$. However, since $R$ is not the zero ring, $0_R\neq 1_R$. But by the definition of a ring homomorphism, $\varphi(1_F)=\varphi(1_R)$. Therefore $\ker{\varphi}$ cannot be the unit ideal, so $\varphi$ is injective. 
A: Do you know or can you show that $\phi(0_F) = 0_R$? (EDIT: see the comment below, this isn't actually sufficient. You have to show $\phi(a) = 0_R \implies a = 0_F.$) 
Once you have that, say $a,b \in F$. To show that $\phi$ is an injection, go to the definition. You just need to show that $\phi(a) = \phi(b) \implies a=b$. So let's suppose $\phi(a) = \phi(b)$. Now use the definition of a ring homomorphism to show $\phi(a-b) = 0$. Now if you can logically piece this together with the first statement, you've got it.   
