Cfrgtkky

This is not a question on how to prove the result

In proposition 1.2. of Atiyah's Introduction to Conmutative Algebra it says:

Let $A$ be a non trivial ring. The following are equivalent:

i) $A$ is a field

ii) $(0)$ and $(1)$ are the only ideals of $A$

iii) Every homomorphism from $A$ to a non trivial ring $B$ is injective

The problem here is with the translation. The spanish edition says "homomorfismo de $A$ en un anillo no nulo $B$" which may be understood as "homomorphism from $A$ onto a non trivial ring $B$" or "homomorphism from $A$ to a non trivial ring $B$".

I could only prove the proposition supposing the "onto" part and the book also does so. Does the english edition say "onto"?

Surjectivity is immaterial, in this case. Since ring homomorphisms are assumed to send $1$ to $1,$ then $1$ cannot be in the kernel of the homomorphism. Thus, if $(0)$ and $(1)$ are the only ideals of $A,$ then $(0)$ must be the kernel of the homomorphism, and so it is injective.

Added: We may assume surjectivity without loss of generality, and it sounds like Atiyah does. Suppose all surjective homomorphisms from $A$ are injective. Now take an arbitrary homomorphism $f:Ato B,$ for some ring $B.$ The image of $A$ under $f$ is a subring of $B,$ say $C.$ Then $g:Ato C$ defined by $g(x):=f(x)$ for all $xin A$ is a surjective homomorphism $Ato C.$ Thus, $g$ is injective by hypothesis, and so $f$ is injective, as desired.

Unless you assume a nontrivial homomorphism, it must be mentioned onto. Because every homomorphism from a field to a ring is either one one or maps everything to 0.