# Non-radical ideal giving the empty set.

Let $R$ be the polynomial ring of $n$ variables over $\mathbb C$. It is known that a radical ideal $I (\ne R)$ defines a non-empty set $\mathbf V(I) \subset \mathbb C^n$.

I am looking for a counterexample. Can we have a non-radical ideal $I (\ne R)$ such that $\mathbf V(I)=\varnothing$?

No. A proper ideal $I$ is contained in a maximal ideal $\mathfrak{m}$, and we have $$I \subset \sqrt{I} \subset \sqrt{\mathfrak{m}}=\mathfrak{m}$$ as every maximal ideal is a radical ideal. Hence the radical of a proper ideal is again a proper ideal. Note that this argument even works in an arbitrary commutative ring.
Now $V(\sqrt{I})$ is non-empty by what you wrote, but then $V(I)$ is non-empty as well, because $V(I) \supset V(\sqrt{I})$ (as $I \subset \sqrt{I}$).
If $\mathbf V(I) = \emptyset$, then $\sqrt I = \mathbf I (\mathbf V(I)) = R$. (The first equality is a version of the Nullstellentsatz.)
I let you prove that an ideal of radical $R$ is itself $R$.