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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.

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    $\begingroup$ No for dimension reasons. $\endgroup$ – Alex Youcis Aug 29 '19 at 19:53
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    $\begingroup$ If you mean the scheme version, no, since non-empty open subsets of the integral scheme $\Bbb A^n_k$ are of dimension $n$ but $V(f)$ is of dimension $n-1$ for any nonzero $f\in k[x_1,\cdots,x_n]$. If you mean $Z(f)=\{x\in k^n\mid f(x)=0\}$, then yes, take $k=\Bbb F_2$, $n=1$, and $f=x^2-x$. $\endgroup$ – KReiser Aug 29 '19 at 19:53
  • $\begingroup$ @KReiser Your last example fails to have $k$ algebraically closed. $\endgroup$ – Servaes Aug 29 '19 at 20:01
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    $\begingroup$ @Servaes Ugh, this is what I get for going too fast. Thanks for pointing it out - with $k$ algebraically closed this is always true even if you take the non-scheme version. $\endgroup$ – KReiser Aug 29 '19 at 20:04
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    $\begingroup$ As originally written, your post read $\Bbb A^n_k$, which is usually meant to be $\operatorname{Spec} k[x_1,\cdots,x_n]$, instead of $k^n=\Bbb A^n_k(k)$. You later edited your post to make it read the latter, which is the "non-scheme" version. As $A^n_k(k)$ is not always dense in $\Bbb A^n_k$ (see the example from the first comment), it's worth pointing out that the formulation matters. $\endgroup$ – KReiser Aug 29 '19 at 20:24
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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)


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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

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  • $\begingroup$ I'm not sure why this was downvoted ... unless I'm missing anything, this is a perfectly valid answer ... $\endgroup$ – user102248 Aug 29 '19 at 20:22

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