Zariski topology analogue for non-algebraically closed fields

Solution 1:

You are right with $\bar{k}$ and you essentially gave a proof.

Let $K$ be an algebraic field extension of $k$ (e.g. $K=\bar{k}$). Then the Zariski topology on $k^n$ is induced by that of $K^n$.

Proof. The trivial direction: any closed subset of $k^n$ is the intersection of a closed subset of $K^n$ with $k^n$. This is true for any extension $K/k$.

Conversely, let $Z(F)$ be a closed subset of $K^n$ defined by a polynomial $F$ with coefficients in $K$. Then $F$ has coefficients in some finite extension $L/k$ (take the extension generated by the coefficients of $F$!). If $L/k$ is separable, we let $f$ be the product of all conjugates (in an obvious sense) of $F$. Then $f$ has coefficients in $k$ and $Z(F)\cap k^n=Z(f)$ is closed.

If $L/k$ is not separable, then $k$ has positive characteristic $p$, and some power $F^{p^r}$ has coefficients in the separable closure $L_s$ of $k$ in $L$ and we are reduced to the separable extension case. So $Z(F)\cap k^n$ is closed in $k^n$. As any closed subset of $K^n$ is an intersection of $Z(F)$, the claim is proved.

In the transcendental case, surprisingly (for me) this is still true. The extension $K/k$ is algebraic over some purely transcendental extension of $k$. By the algebraic extension case we can suppose $K$ itself is purely transcendental and even of finite type: $K=k(s_1, \dots, s_m)$. Let $F\in K[T_1, \dots, T_n]$. Chasing the denominators in $K$ (which doesn't change $Z(F)\subseteq K^n$), we can suppose $F\in k[s_1, \dots, s_m][T_1, \dots, T_n]$, so we can write $$ F(T_1, \dots, T_n)=\sum_{\nu\in \mathbb N^m} f_\nu(T_1,\dots, T_n)s^\nu, \quad f_v\in k[T_1, \cdots, T_n].$$ Let $(a_1, \cdots, a_n)\in k^n$. Then $F(a_1,\dots, a_n)=0$ if and only if $$f_\nu(a_1, \cdots, a_n)=0, \quad \forall \nu\in \mathbb N^m.$$ So $Z(F)\cap k^n=\cap_{\nu} Z(f_\nu)$ is closed in $k^n$.