Affine open sets of a scheme after a base change

Question

Consider a $k$-scheme $X$ where $k$ is a generic field. If we have a field extension $k\subseteq K$, then we can construct the fibered product $X^K:=X\times_{\operatorname{Spec}k}\operatorname{Spec} K$. We know that if $U\subseteq X$ is an affine open of $X$ then $U\times_{\operatorname{Spec}k}\operatorname{Spec} K$ is an affine open of $X^K$, but is it true that every affine open of $X^K$ has this form?

Answer 1

Expanding on my comment: since you are allowing arbitrary $k$-schemes, consider $X = \operatorname{Spec} K$. This is a perfectly good integral affine $k$-scheme, which is even of finite type if $K$ is a finite extension of $k$. As a topological space, $X$ is a point. However, the base change $X_K = \operatorname{Spec} K \otimes_k K$ is often more complicated: for instance, if $K$ is a separable extension of degree $d > 1$, then $X_K$ is the disjoint union of $d$ copies of $\operatorname{Spec} K$; so in this case $X_K$ has many affine open subschemes that do not come from open subschemes of $X$.

Answer 2

No. Take $V\subset \mathbb{P}^1_K$ be the open affine subset being the complement of a point $q$, not defined over $k$.

If $U$ is an open affine subset of $\mathbb{P}^1_k$ then the scalar extension gives the complement of points of $\mathbb{P}^1_K$, defined by a homogeneous polynomial in $k$. This is not the case for $V$.