# $V(I)=\emptyset$ implies $I=R$

Note that most of the notation used here is defined in the entry prime spectrum.

###### 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 (http://planetmath.org/EveryRingHasAMaximalIdeal), there exists a maximal ideal $M$ of $R$ containing $I$. Since $M$ is , then $M$ is a proper prime ideal of $R$. Thus, $M\in V(I)$. The theorem follows. ∎

Title $V(I)=\emptyset$ implies $I=R$ VIemptysetImpliesIR 2013-03-22 16:07:43 2013-03-22 16:07:43 Wkbj79 (1863) Wkbj79 (1863) 10 Wkbj79 (1863) Theorem msc 14A15 ProofThatOperatornameSpecRIsQuasiCompact