# Ideals and Field [duplicate]

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 $$(0)$$

## marked as duplicate by rschwieb abstract-algebra StackExchange.ready(function() { if (StackExchange.options.isMobile) return; $('.dupe-hammer-message-hover:not(.hover-bound)').each(function() { var$hover = $(this).addClass('hover-bound'),$msg = $hover.siblings('.dupe-hammer-message');$hover.hover( function() { $hover.showInfoMessage('', { messageElement:$msg.clone().show(), transient: false, position: { my: 'bottom left', at: 'top center', offsetTop: -7 }, dismissable: false, relativeToBody: true }); }, function() { StackExchange.helpers.removeMessages(); } ); }); }); Oct 19 '18 at 16:14

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.