Theorem   If $R$ is a commutative ring with identity and $I$ is an ideal of $R$ with $V(I)=\emptyset$ , then $I=R$ .

Proof. Let $R$ be a commutative ring with identity and $I$ be an ideal of $R$ with $I \neq R$ . Then, by this theorem, there exists a maximal ideal $M$ of $R$ containing $I$ . Since $M$ is maximal, then $M$ is a proper prime ideal of $R$ . Thus, $M \in V(I)$ . The theorem follows.