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# $V(I)=\emptyset$ 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, 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. ∎

Related:

ProofThatOperatornameSpecRIsQuasiCompact

Major Section:

Reference

Type of Math Object:

Theorem

Parent:

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## Comments

## $V(I)& and $\operatorname{Spec}(R)$

I didn't can find out somewhere in PM what $V(I)$ and $\operatorname{Spec}(R)$ are. That is a well-known notation in math? Anyway, what do mean V(I) and Spec(R)? Thanks!

perucho

## Re: $V(I)& and $\operatorname{Spec}(R)$

this is standard notation in commutative algebra/algebraic geometry...V(I) is defined in

http://planetmath.org/encyclopedia/ZariskiTopology.html

and Spec R at

http://planetmath.org/encyclopedia/PrimeSpectrum.html

Perhaps the owner of the entry can force some kind of link.

## Re: $V(I)& and $\operatorname{Spec}(R)$

Yes, this is standard notation, and yes, I should provide links to explanations of the notation so that those who are not familiar with commutative algebra and algebraic geometry can figure out what is going on in this proof. Thanks for the idea silverfish, and thanks perucho for pointing out this issue.

## Re: $V(I)& and $\operatorname{Spec}(R)$

Thanks to Silverfish and Wkbj79 by clarify me the referred notations. Indeed I didn't can find the respective cross references.

perucho

## mention axiom of choice?

I just realized that, within this entry (as well as at least one other one), I am assuming the axiom of choice. Should I state such? The reason that I ask is that I am concerned that it may become monotonous to read in these types of entries, "The axiom of choice is being aassumed." Also, it seems that most people who are active on PM accept the axiom of choice. I would appreciate any opinions you have. Thanks.

Warren

## Re: mention axiom of choice?

I personally prefer to see it mentioned, independent of whether i accept it or not. Sometimes one can assume a weaker axiom though and get

the result.

## Re: mention axiom of choice?

> Sometimes one can assume a weaker axiom though and get

> the result.

Are you referring specifically to the result $V(I)=\emptyset$ implies $I=R$? If that is the case, I would love to see a proof that does not use the axiom of choice (unless, of course, it's extremely messy).

## Re: mention axiom of choice?

No,i am not referring to that result. There is a book that deals with

the subject of axiom of choice and its equivalent formulations

and weaker forms.

http://consequences.emich.edu/conseq.htm

Also see:

http://at.yorku.ca/z/a/a/b/18.htm

## Re: mention axiom of choice?

There's no need to mention it, especially as it's already mentioned in the EveryRingHasAMaximalIdeal entry that you are referencing. I don't mention it in my entries unless I have a specific reason to. (In fact, in many cases, I wouldn't really know for sure if AC is needed or not.) There are also a great many entries by other editors that assume AC without saying so.

I'd be more interested in knowing which results need the Axiom Schema of Replacement (but, of course, I don't expect authors to know this, just as I don't expect them to know if AC is needed).

## Re: mention axiom of choice?

I have decided to go with not mentioning the axiom of choice within the entry. I think the fact that there is a string of posts attached to this entry that mention the axiom of choice should appease those who would have preferred to see it mentioned within the entry.

Warren