Let $R$ be a commutative ring.

Then prove that $R$ is a field if and only if $\{0\}$ is a maximal ideal of $R$.
 

Proof.

$(\implies)$: If $R$ is a field, then $\{0\}$ is a maximal ideal

Suppose that $R$ is a field and let $I$ be a non zero ideal:
\[ \{0\} \subsetneq I \subset R.\]

Then the ideal $I$ contains a nonzero element $x \neq 0$. Since $R$ is a field, we have the inverse $x^{-1}\in R$.
Then it follows that $1=x^{-1}x \in I$ since $x$ is in the ideal $I$.

Since $1\in I$, any element $r \in R$ is in $I$ as $r=r\cdot 1 \in I$.
Thus we have $I=R$ and this proves that $\{0\}$ is a maximal ideal of $R$.

$(\impliedby)$: If $\{0\}$ is a maximal ideal, then $R$ is a field

Let us now suppose that $\{0\}$ is a maximal ideal of $R$.
Let $x$ be any nonzero element in $R$.

Then the ideal $(x)$ generated by the element $x$ properly contains the ideal $\{0\}$.
Since $\{0\}$ is a maximal ideal, we must have $(x)=R$.

Since $1\in R=(x)$, there exists $y\in R$ such that $1=xy$.
This implies that the element $x$ is invertible. Therefore any nonzero element of $R$ is invertible, and hence $R$ is a field.

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