# When is a Quotient Ring equal to the Zero Ring?

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.

• I think you pretty much have it. Perhaps you can argue as follows: if $x$ is not invertible, then consider the ideal $\langle x\rangle \neq R$ generated by $x$ and let $f$ be the corresponding homomorphism with kernel as $\langle x\rangle$. Then you have a non-injective homomorphism, which contradicts the assumption. – Anurag A Dec 17 '18 at 20:17

$$iii)\implies i):$$ 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.
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.).