Let $k\subseteq K$ be an extension of fields. Assume further that $k$ is algebraically closed.
Now, let $f_1,...,f_m$ be some polynomials in $k\lbrack x_1,...,x_n\rbrack$, and write $V_k(f_1,...,f_m)$ for the vanishing set of these polynomials in $k^n$, and $V_K(f_1,...,f_m)$ for the vanishing set of these polynomials in $K^n$.
As $k^n\subseteq K^n$, we have $V_k(f_1,...,f_m)\subseteq V_K(f_1,...,f_m)$. Is it always true that $V_k(f_1,...,f_m)$ is a dense subset of $V_K(f_1,...,f_m)$ in the Zariski topology?
Using Galois descent, I can prove that this is true when the extension $K/k$ is Galois. I think that if $K/k$ is only separable, a slight elaboration on my argument should give the result. However, I have no idea how to deal with non-separable extensions! Maybe some fancy argument using schemes does the trick?
Let $P \in K[x_1,\ldots,x_n]$ be nonzero on $V_K(f_1,\ldots,f_m)$. We want to show that there exists a point $z \in k^n$ such that $P(z) \neq 0$ and every $f_i(z)$ is zero.
Let $z_0 \in K^n$ be a common root of the $f_i$ with $P(z_0) \neq 0$. Let $A$ be the finitely generated sub-$k$-algebra of $K$ (hence an integral domain) generated by the coefficients of $P$, the coordinates of $z_0$ and $1/P(z_0)$.
In particular, $A$ is noetherian and nonzero so has a maximal ideal $\mathfrak{m}$ and $k \rightarrow A/\mathfrak{m}$ is an isomorphism. Then the inverse image $z_1$ in $k$ of $z_0$ is a common root of the $f_i$ and has invertible (hence nonzero) image under $P \pmod{\mathfrak{m}}$ so $P(z_1) \neq 0$.