Let $k$ be an infinite field. Assume that the index of the algebraic closure $\bar{k}$ over $k$ is strictly greater than $2$. Let $U$ be a non-empty open subset of some affine space over $k$. Is it true that $U$ always contain a $k$-rational point?



Rational points on open subsets of affine space
Here is a short proof that, for an infinite field $k$, and all non-zero polynomials $F \in k[x\_1,\ldots,x\_n]$ in $n$ variables, there exists an $n$-tuple $a\_1,\ldots,a\_n \in k$ such that
$$
F(a\_1,\ldots,a\_n) \neq 0.
$$
We do induction on $n$. For $n=1$, the assertion follows from the fact that $F$ has finitely many zeros in $k$. Now suppose we have proven the assertion for all $m<n$. Then we write $F$ as
$$
F=C\_d x\_n^d + C\_{d-1} x\_n^{d-1} + \ldots + C\_0,
$$
where the $C\_i$ are polynomials in $x\_1,\ldots,x\_{n-1}$. According to the induction hypothesis there exists an $(n-1)$-tuple $a\_1,\ldots,a\_{n-1} \in k$ such that not all $C\_i$ vanish at the value $(a\_1,\ldots,a\_{n-1})$.


Let $f$ be $F$ with the first $n-1$ variables specialized to $a\_1,\ldots,a\_{n-1}$, then $f$ is a non-zero polynomial by the choice of the $a\_i$ above. Then according to the base case of our induction, there exists $a\_n \in k$ such that
$$
f(a\_n) = F(a\_1,\ldots,a\_n)
$$
is non-zero, which is what we wanted to prove.


