Textbook Solution: Since $\ker \varphi$ is an ideal of $K$, it follows from previous excerise that $\ker \varphi = \{0\}$. Hence $\forall a,b \in K$, $$ \varphi(a)=\varphi(b) \Rightarrow a-b \in \ker\varphi = \{0\} \Rightarrow a = b$$
I am really having a problem understanding this proof. Can someone explain it, or provide me an alternative proof? thank you for your time.
For anyone that might find this question helpful later on, heres my solution
Recall that if $\varphi$ is one to one then $\varphi$ is a monomorphism. Recall that by defnition of one to one, every element in $K$ must be mapped to exactly one element in $L$. Recall that a field is a field iff the only ideals are $\{0\}$ and the field itself. This means that the kernal of $\varphi$ is $\{0\}$. Hence $\forall a,b \in K$, $$\varphi(a) = \varphi(b) \Rightarrow 0 = \varphi(a)-\varphi(b) \Rightarrow a-b = 0 \Rightarrow a-b\in \ker\varphi$$ This means that $a = b$. Hence $\varphi$ is one to one since no two elements of $K$ map to the same element in $L$.