Can the zero set of an irreducible polynomial contain a non-empty Zariski open subset? 
Let $k$ be an algebraically closed field and $f \in k[x_1,...,x_n]$ be an irreducible polynomial.
Is it possible that $Z(f)$, the zero set of $f$, contains a non-empty Zariski open subset of $\mathbb A^n(k)$ ? (By $\mathbb A^n(k)$ I  don't mean $\mathrm{Spec}(k[x_1,...,x_n])$, rather only the closed points.)

I know that it is impossible for $n=1$. Also when $k=\mathbb C$, I know it is impossible for all $n$, in fact then the zero set can't even contain any non-empty Euclidean open set. But I'm not sure what happens in other cases.
 A: If $U \subseteq \mathbf{A}^{n}$ is nonempty and open, then $\text{dim }U=\text{dim }\overline{U}=\text{dim }\mathbf{A}^{n}=n$. 
Now if $U \subseteq Z(f)$, then $\text{dim }U \leq \text{dim }Z(f) = n-1$, which is impossible.
Here are the relevant facts used from Ch. 1, Sec. 1 in Hartshorne (valid over any algebraically closed field): 
$\bullet$ The dimension of a quasi-affine variety is the same as its closure (Proposition 1.10) 
$\bullet$ $\mathbf{A}^{n}$ is irreducible (its ideal is prime: $I(\mathbf{A^{n}})=0$), and a nonempty open subset of an irreducible space is dense (Exercise 1.6) 
$\bullet$ The dimension of $\mathbf{A}^{n}$ is $n$ (Proposition 1.9) 
$\bullet$ If $Y \subseteq X$ are topological spaces, then $\text{dim }Y \leq \text{dim }X$ (Exercise 1.10a)

$\bullet$ $Z(f)$ has dimension $n-1$ (Proposition 1.13) 

A: Suppose that $Z(f)$ contains the open subset $U$, the complementary $C$ of $U$ is closed and is $V(I)$ this implies that $Z(f)\cup V(I)=V(fI)$ is the whole space the theorem of zero implies that $If=0$ contradiction
