$V(I)=\\emptyset$ 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 $I\eq R$ . 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.∎

单词	VIemptysetImpliesIR
释义	$V(I)=\\emptyset$ 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 $I\eq R$ . 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.∎
随便看	DeterministicPushdownAutomaton deterministic pushdown automaton DeterministicTuringMachine deterministic Turing machine DevelopableSurface developable surface Development development Deviance deviance Diagonal diagonal DiagonalEmbedding diagonal embedding DiagonalIntersection diagonal intersection Diagonalizable diagonalizable DiagonalizableOperator diagonalizable operator Diagonalization diagonalization DiagonalizationOfQuadraticForm diagonalization of quadratic form DiagonallyDominantMatrix

V⁢(I)=∅ implies I=R

Theorem.

Proof.

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