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This question already has an answer here:

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 $(0)$

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marked as duplicate by rschwieb abstract-algebra Oct 19 '18 at 16:14

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

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    $\begingroup$ This is a repost of a question you deleted which had useful comments. That is bad form. You should have just edited the original question. $\endgroup$ – rschwieb Oct 19 '18 at 16:11
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Let $f:A\rightarrow B$ be a morphism, such that $B$ is not the zero ring $ker(f)$ is an ideal of $A$ we deduce that $Ker(f)=0$ or $(1)$, if $ker f=0$ it is injective, if $Ker f=(1)$ then the image is the zero ideal, this implies that $f(1)=0$ since $B$ has a unit, we deduce that and $f(1)=1=0$ and $B=0$. Contradiction.

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