Assume $R$ is a commutative ring with $1 \neq 0$. Prove that $R$ is a field iff $0$ is a maximal ideal.
Recall that an ideal $M$ of $R$ is maximal if $M \neq R$ and the only ideals containing $M$ are $R$ and $M$.
Recall that one way of defining a field is that it is a commutative ring with identity in which every nonzero element has an inverse (with respect to both $+$ and $\times$ of course)
This problem is another way of establishing that fields have trivial ideal structures. Note the analogy here with groups: Simple groups are special types of groups that have trivial normal subgroup structures. Fields are special types of rings that have trivial ideal structures.
$\textit{Proof}$ : "$\Longrightarrow$": Let $I$ be an ideal of $R$ s.t. $(0) \subset I \subset R$. So $\exists r \neq 0$ s.t. $r\in{I}$. Since $R$ is a field, multiplicative inverses exist, so $rr^{-1}\in{I} \Longrightarrow 1 \in{I}$. With the identity being in $I$, this means that $s \cdot 1 \in{I}$ $\forall s \in{R}$ since $I$ is, in particular, a left ideal. So, $s\in{I}$. But $s$ was arbitrary, so all elements lie in $I$, i.e. $I=R$.
$\therefore \quad (0)$ is a maximal ideal of $R$.
"$\Longleftarrow$": Assume that $(0)\in{R}$ is a maximal ideal. So, the only ideals containing $(0)$ are $(0)$ and $R$. So, $\forall r \neq 0$, the ideal generated by $0$ and $r$, $(0,r)=R$. In particular, if $r \neq 0$ then $1 \in{(0,r)}$. So, $1=a \cdot 0 + b\cdot r$ for some $a,b\in{R}$.
$br=1$
$b=r^{-1}$. Since $b$ and $r$ were arbitrary, inverses exist in general in $R$. So, $R$ is a commutative ring with identity $1 \neq 0$ in which every element has an inverse.
$\therefore \quad R$ is a field. $\blacksquare$
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