Existence of Countably Infinite Varieties Over Uncountable Fields Let $k$ be an uncountable field. Can a countably infinite set in $k^n$ be obtained as the zero set of a system of polynomial equations? What if $k$ is algebraically closed or a real-closed field?
I imagine the answer is no in general, and according to this post there are no countably infinite subsets in $\mathbb{R}^n$ that can be obtained as the zero locus of a single polynomial because of the Tarski–Seidenberg theorem. 
 A: Yes, this is is possible.  For instance, let $k$ be a field of rational functions over $\overline{\mathbb{Q}}$ in uncountably many indeterminates and consider the polynomial $f(x,y)=y^2-x^3-x$.  I claim that if $(a,b)\in k^2$ and $f(a,b)=0$ then $a,b\in\overline{\mathbb{Q}}$.  Indeed, suppose $(a,b)\not\in\overline{\mathbb{Q}}$ and let $F=\overline{\mathbb{Q}}(a,b)\subset k$.  Then $F$ is the function field of an elliptic curve $E$ over $\overline{\mathbb{Q}}$, and is also contained in the subfield $\overline{\mathbb{Q}}(x_1,\dots,x_n)\subset k$ generated by finitely many of the indeterminates.  The inclusion $F\to \overline{\mathbb{Q}}(x_1,\dots,x_n)$ induces a nonconstant rational map $\mathbb{P}^n\to E$ over $\overline{\mathbb{Q}}$.  However, no such rational map exists (for instance, restricting to any line in $\mathbb{P}^n$ it would have to be constant), which is a contradiction.
Thus the zero set of $f(x,y)$ in $k^2$ is the same as its zero set over $\overline{\mathbb{Q}}$, and in particular is countably infinite.

On the other hand, it is not possible if $k$ is algebraically closed or real-closed.  If $k$ is algebraically closed and $V\subset k^n$ is a variety with infinitely many points, note that there is a countable subfield $k_0\subset k$ over which $V$ is defined.  Letting $A$ be the coordinate ring of $V$ over $k_0$, the ring $A$ has positive dimension since $V$ has infinitely many points.  Let $P$ be a non-maximal prime ideal and let $k_1$ be the field of fractions of $A/P$.  Since $P$ is not maximal, $k_1$ is not algebraic over $k_0$; let $x\in k_1$ be transcendental over $k_0$.  Then since $k$ is algebraically closed, for any $a\in k$ which is transcendental over $k_0$, there is an embedding of $k_1$ into $k$ which fixes $k_0$ and sends $x$ to $a$.  Every such embedding gives a point of $V$, and so $V$ is uncountable (and in fact $|V|=|k|$).
The case that $k$ is real-closed follows from the Tarski-Seidenberg theorem exactly as over $\mathbb{R}$, since the Tarski-Seidenberg theorem applies to all real-closed fields.
