$V(I)=\mathrm{\varnothing }$ 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 $V\mathit{}\mathrm{(}I\mathrm{)}\mathrm{=}\mathrm{\varnothing }$, then $I\mathrm{=}R$.

Proof.

Let $R$ be a commutative ring with identity and $I$ be an ideal of $R$ with $I\ne R$. 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	$V(I)=\mathrm{\varnothing }$ implies $I=R$
Canonical name	VIemptysetImpliesIR
Date of creation	2013-03-22 16:07:43
Last modified on	2013-03-22 16:07:43
Owner	Wkbj79 (1863)
Last modified by	Wkbj79 (1863)
Numerical id	10
Author	Wkbj79 (1863)
Entry type	Theorem
Classification	msc 14A15
Related topic	ProofThatOperatornameSpecRIsQuasiCompact