Let R be a commutative ring with identity. Prove that R is a field if and only if (0R) is a maximal ideal.

Question

I'm confused on how to go in both directions and how to start this proof.

Answer 1

Well, if $R$ is a field, then the only proper ideal is $\{ 0\}$, by virtue of the fact that $0 \ne a \in I$ for an ideal $I$ implies $ba \in I$ for all $b \in R$; then for $x \in R$ we have $x = (xa^{-1})a \in I$, so $R \subseteq I$, so $I = R$, so $I$ is not proper. Going in the other direction, suppose $\{ 0 \} = 0R$ is the only maximal ideal and let $a \in R$ be a non-unit element. Then $a \in aR$ since $1 \in R$, and $1 \notin aR$ since $a$ is not a unit; if $1 \in aR$, then $1 = ba$ for some $b \in R$, contradicting the fact that $a$ is a non-unit element. But then $aR$ is a proper ideal and $0R \subsetneq aR \subsetneq R$, so $0R$ cannot be maximal. Thus every $a \in R$ is a unit and hence $R$ is a field. QED.

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