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?
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1$\begingroup$ Over any infinite field $k$, the set of rational points in affine space is Zariski dense. $\endgroup$– Daniel LoughranCommented Mar 9, 2017 at 14:39
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$\begingroup$ @DanielLoughran Could you please suggest a reference for this statement. $\endgroup$– user43198Commented Mar 9, 2017 at 14:40
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5$\begingroup$ By making $U$ smaller if need be, one may assume that $U$ is the complement of a hypersurface $f=0$ in some affine $n$-space $\mathbb{A}^n$, with $f$ some non-zero polynomial in $n$ variables. So you're asking whether a non-zero polynomial in $n$ variables assumes non-zero values for certain rational input parameters. That the answer to this is 'yes' is easily proved using induction on $n$ (here you clearly need that $k$ is an infinite field). Not really research-level, though... $\endgroup$– R.P.Commented Mar 9, 2017 at 15:14
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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.
