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Let $A \neq0$ be a ring. Then the following are equivalent
i) A is a field
ii) the only ideals in A are $0$ and $(1)$
iii) every homomorphism of A into a non-zero ring B is injective.
I have shown $\ \ \ i)\rightarrow ii) \ \ $ and $\ \ \ iii)\rightarrow i) \ \ $
But couldnt do $\ \ \ ii)\rightarrow iii) \ \ $
My thoughts: I think it is enough to show that for a homomorphism $\phi: A\rightarrow B,\ \ \ $ $Ker \ \phi=(0)$ but I dont know how can I show $Ker \ \phi\neq(1)$ which automatically gives us $Ker \ \phi=(0)\ $ as $\ Ker \ \phi$ can only be $(1)$ or $(0)$