$V(I)$ and $V(J)$ disjoint iff $I$ and $J$ are relatively prime

Question

Let $R=k[x_1,\ldots,x_n]$, where $k$ is algebraically closed, and let $I,J\subset R$ be ideals.

Show that $V(I)$ and $V(J)$ are disjoint if and only if $I$ and $J$ are relatively prime (meaning $I+J=(1)$).

Suppose $I+J=(1)$. Then for any $r\in R$, we have $r=f+g$ for some $f\in I, g\in J$. If there is an $a\in V(I)\cap V(J)$, then $f(a)=0=g(a)$ for all $f\in I, g\in J$. Hence $r(a)=0$ for all $r\in R$. But if $k$ is non-trivial, there is no such $a$. Hence $V(I)$ and $V(J)$ are disjoint.

Converely, if $V(I)\cap V(J)=\varnothing$, we have $V(I+J)=\varnothing$, so $I+J=k[x]=(1)$. [I know how to prove $V(I+J)=V(I)\cap V(J)$.]

Is this correct?

Answer 1

Yes, absolutely. Note that, in the converse part you are using Nullstellensatz, so $k$ being algebraically closed is important.