When is a Quotient Ring equal to the Zero Ring?

Question

I am reading Atiyah and MacDonald's 'Introduction to Commutative Algebra' and wished to check a point in one of their proofs. They state that if $A$ is a non-zero ring the following are equivalent:

(i.) $A$ is a field

(ii.) The only ideals in $A$ are $0$ and $(1)$

(iii.) Every homomorphism of $A$ to a non-zero ring $B$ is injective.

To prove that (iii.) implies (i.) they say 'Let $x$ be an element of $A$ which is not a unit. Then $(x)$ does not equal $(1)$, hence $B=A/(x)$ is not the zero ring'. To fill in the gaps here, is that because the principal ideal generated by 1 would return the original ring $A$ because you are just multiplying $1$ through by every element of the ring one-by-one, and then that gives you $B=A/A$, which is the zero ring? If anyone wants to fill in the gaps a bit more explicitly I would appreciate it.

Apologies about the basic question, as I am new to elementary abstract algebra.

Answer 1

$iii)\implies i):$ Every ideal $(x)$ is the kernel of the quotient homomorphism to $A/(x)$. The homomorphism being injective implies the only proper ideal is $(x)=(0)$. So for each proper ideal $(x)$, $x=0$. This means each $x\neq0$ is invertible ($(x)=A$) which means the ring is a field.

Answer 2

The step (iii.) $\Rightarrow$ (i.) is done by contradiction. Assume for the sake of contradiction that (iii.) holds but $A$ is not a field. Then there is a non-zero element $x\in A$ that is not a unit. Since it is not a unit, the ideal $I=(x)$ does not contain $1$, so $I$ is a proper ideal of $A$. Now consider the canonical homomorphism $f\colon A\to A/I$, $r\mapsto r+I$. Note that $f$ is not injective since $f(x)=0$ and $x\neq 0$. However, since $I$ is a proper ideal, the quotient $A/I$ is not the zero ring. (Note that $A/I$ is the zero ring if and only if $I=A$.) Hence, we got a contradiction to (iii.). We conclude that our assumption was false and that (iii.) implies (i.).