k-schemes determined by k-rational points

Question

Are $k$-schemes determined by their $k$-rational points? More precisely, if $X$ and $Y$ are $k$-schemes, such that $X(k)=Y(k)$. Is then automatically $X$ isomorphic to $Y$?

Answer 1

No, not in general. For instance, let $K$ be any nontrivial field extension of $k$, $X=\emptyset$, and $Y=\operatorname{Spec}K$. Then $X$ and $Y$ both have no $k$-points, so $X(k)$ and $Y(k)$ are the same, and this can even be induced by a morphism $X\to Y$. But $X$ and $Y$ are not isomorphic.