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.

• No for dimension reasons. – Alex Youcis Aug 29 '19 at 19:53
• 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 $f=x^2-x$. – KReiser Aug 29 '19 at 19:53
• @KReiser Your last example fails to have $k$ algebraically closed. – Servaes Aug 29 '19 at 20:01
• @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. – KReiser Aug 29 '19 at 20:04
• 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. – KReiser Aug 29 '19 at 20:24

2 Answers

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)

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

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