Let $R = \mathbb{C}[x_1,\ldots, x_n]$, and $I,J$ ideals in $R$ with the property that for all $a\in \mathbb{C}^n$, $f(a) = 0$ for all $f\in I$ if and only if $g(a) =0$ for all $g\in J$. I'd like to prove that $(I+J)/I$ is nil.
For $\bar{h} \in (I+J)/I$, we write $\bar{h} = h + I$, $h\in I+J$. Then I have to show there exists $n$ such that $h^n \in I$, which would give that an arbitrary $\bar{h}$ is nilpotent, which is what we want to show. I'd like to make sure my logic is correct:
Write $h = h_1+h_2$, where $h_1,h_2$ in $I,J$ respectively. Let $a\in Var(I)$. By definition, $f(a)=0$ for all $f\in I$, so we must have $g(a)=0$ for all $g\in J$. In particular, this gives $h(a)=h_1(a)+h_2(a) = 0$. Then, by Hilbert-Nullsetellensatz, we have $h\in Id(Var(I))=\sqrt{I}$, since $a$ was arbitrary. This completes the proof by definition of $\sqrt{I}$.
It just seemed a bit odd since $Var(I+J)=Var(I)\cap Var(J) \subset Var(I)$, so $\sqrt{I}\subset \sqrt{I+J}$, and so now we've showed $I+J\subset \sqrt{I} \subset \sqrt{I+J}$.
In any case, once we have this, how might we go about showing the ring is nilpotent?