Proposition 1   Let $\mathcal{R}\neq 0$ be a commutative ring with identity. The ring $\mathcal{R}$ (as above) is a field if and only if $\mathcal{R}$ has exactly two ideals: $(0),\mathcal{R}$ .

Proof. ($\Rightarrow$ ) Suppose $\mathcal{R}$ is a field and let $\mathcal{A}$ be a non-zero ideal of $\mathcal{R}$ . Then there exists $r\in \mathcal{A}\subseteq \mathcal{R}$ with $r\neq 0$ . Since $\mathcal{R}$ is a field and $r$ is a non-zero element, there exists $s\in \mathcal{R}$ such that $$s\cdot r =1 \in \mathcal{R}$$ Moreover, $\mathcal{A}$ is an ideal, $r\in \mathcal{A}, s\in \mathcal{S}$ , so $s\cdot r =1 \in \mathcal{A}$ . Hence $\mathcal{A}=\mathcal{R}$ . We have proved that the only ideals of $\mathcal{R}$ are $(0)$ and $\mathcal{R}$ as desired.

($\Leftarrow$ ) Suppose the ring $\mathcal{R}$ has only two ideals, namely $(0),\mathcal{R}$ . Let $a\in \mathcal{R}$ be a non-zero element; we would like to prove the existence of a multiplicative inverse for $a$ in $\mathcal{R}$ . Define the following set: $$\mathcal{A}=(a)=\{r\in\mathcal{R} \mid r=s\cdot a, \text{ for some } s\in\mathcal{R}\}$$ This is clearly an ideal, the ideal generated by the element $a$ . Moreover, this ideal is not the zero ideal because $a\in \mathcal{A}$ and $a$ was assumed to be non-zero. Thus, since there are only two ideals, we conclude $\mathcal{A}=\mathcal{R}$ . Therefore $1\in \mathcal{A}=\mathcal{R}$ so there exists an element $s\in \mathcal{R}$ such that $$s\cdot a=1 \in \mathcal{R}$$ Hence for all non-zero $a\in \mathcal{R}$ , $a$ has a multiplicative inverse in $\mathcal{R}$ , so $\mathcal{R}$ is, in fact, a field.