characterization of field

Proposition 1.

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

Proof.

($\mathrm{\Rightarrow }$) Suppose $ℛ$ is a field and let$𝒜$ be a non-zero ideal of $ℛ$. Then thereexists $r\in 𝒜\subseteq ℛ$ with $r\ne 0$.Since $ℛ$ is a field and $r$ is a non-zero element,there exists $s\in ℛ$ such that

$$s\cdot r=1\in ℛ$$

Moreover, $𝒜$ is an ideal, $r\in 𝒜,s\in 𝒮$, so $s\cdot r=1\in 𝒜$. Hence$𝒜=ℛ$. We have proved that the only ideals of$ℛ$ are $(0)$ and $ℛ$ as desired.

($\mathrm{\Leftarrow }$) Suppose the ring $ℛ$ has only twoideals, namely $(0),ℛ$. Let $a\in ℛ$ be anon-zero element; we would like to prove the existence of amultiplicative inverse for $a$ in $ℛ$. Define thefollowing set:

$$𝒜=(a)=\{r\in ℛ\mid r=s\cdot a,\text{ forsome }s\in ℛ\}$$

This is clearly an ideal, the idealgenerated by the element $a$. Moreover, this ideal is not the zeroideal because $a\in 𝒜$ and $a$ was assumed to benon-zero. Thus, since there are only two ideals, we conclude$𝒜=ℛ$. Therefore $1\in 𝒜=ℛ$ so there exists an element $s\in ℛ$ such that

$$s\cdot a=1\in ℛ$$

Hence for all non-zero $a\in ℛ$, $a$ has a multiplicative inverse in $ℛ$, so $ℛ$ is, in fact, a field.∎