Inclusion of open sets in closed sets of Zariski topology

Question

For the Zariski topology on the $n$-dimensional affine space, is there any closed set other than the entire space that contains a non empty open set? I think no because open sets are usually "bigger" than closed sets because closed sets are solutions to a system of polynomials and open set are non-solutions.

Answer 1

This is true if the grounded field is $\mathbb{C}$, a closed subset of $\mathbb{C}^n$ is $V(I)$ where $I$ is an ideal of $\mathbb{C}[X_1,...,X_n]$ supposed that $V(I)$ contains an open subset $U$, for every $P\in I$, the restriction of $P$ on $U$ is zero, this implies that all the partial derivatives of $P$ are zero and $P$ is constant and is zero, thus $I=0$ and $V(I)=\mathbb{C}^n$.

Answer 2

I think the answer is that there is non, as you can prove every open set intersects any other not empty open set and that is equivalent to no closed set contains any open set.

Answer 3

The following discussion is strictly classical.

A non-empty topological space is irreducible iff the only closed set with non-empty interior is the total set.

Let $k$ be any field (algebraically closed or not). An algebraic set $X\subset\mathbb{A}^n_k$ is irreducible if and only if $I(X)\subset k[x_1,\dots,x_n]$ is a prime ideal (true independently of the Nullstellensatz). We also have that $I(\mathbb{A}_k^n)=(0)$ if and only if $|k|=+\infty$. It holds that $\mathbb{A}_k^n$ is irreducible if and only if $n>0$ or $|k|=+\infty$: this set is a singleton if $n=0$, so suppose $n>0$. If $k$ is finite, then $\mathbb{A}_k^n$ is a finite $\mathrm{T}_1$ space and hence it is discrete with more than one point, hence disconnected, hence reducible. If $k$ is infinite, then $I(\mathbb{A}_k^n)=(0)$ is prime, so $\mathbb{A}_k^n$ is irreducible.