VSRP 1: On V(a)

So what exactly are prime ideals? They are ideals such that ab\in I\implies a\in I or b\in I.

Let a\in R, where R is a commutative ring. How is the Zariski topology motivated?

Let V(a) be the set of all prime ideals which contain a. All ideals that contain a contain the whole of aR. Hence all members of V(a) also contain the whole of aR.

Now take V(a,b), where a,b is any arbitrary two-element set in R. Clearly V(a,b)= V(a)\cap V(b). Note that for both these properties, there is nothing special that is true only for prime ideals and not for any other ideal containing the elements.

Now take V(r(aR)), where aR is the ideal generated by the element a. Here r(aR) is the radical of aR. It can be proved that V(aR)=V(a)=V(r(aR)). The inclusion of V(r(aR)) in the equality relation V(a)=V(aR) is what sets apart prime ideals containing a from any random ideal containing a.

Now we will go one step further. Let \{P_k\}=V(a) be the prime ideals containing a. We will prove that \bigcap\{P_k\}=r(aR). It is known that the nilradical of R is the intersection of all prime ideals in it. Now consider prime ideals in R/(a), where (a)=aR. The intersection of all prime ideals in R/(a) will again be the nilradical of R/(a). The inverse of all such prime ideals will be prime ideals containing (a) in R. The rest of the argument is easy to see from here.

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Graduate student

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