### 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.