Let $K/k$ be an extension of fields, let $X_0$ be a scheme over $k$, and let $X:=X_0\times_k\mathrm{Spec}\;K$, so that $X$ is defined over $k$. In this scenario, I often see the phrase "$k$-rational points of $X$," which confuses me because this would require a map $K\to k$, but no such map exists. Does this phrase implicitly mean the $k$-rational points of $X_0$? Or, does it mean the subset of $K$-rational points of $X$ defined by $k$-rational points of $X_0$ via $\mathrm{Spec}\;K\to\mathrm{Spec}\;k\to X_0$ and the universal property of the fiber product? Perhaps these two answers are saying the same thing, as I'm just canonically identifying the $k$-rational points of $X_0$ with a subset of the $K$-rational points of $X$.
One example I have in mind is that $X_0$ is the affine scheme associated to a finite dimensional vector space $V$ over $k$. Then $X$ is the affine scheme associated to $V\otimes_kK$, and either the "$k$-rational points of $X$" are the points of vector space $V$, or those of the subspace $V\otimes_k1\subset V\otimes_kK$.
Is my interpretation of the phrase "$k$-rational points of $X$" correct? If so, do we view these points as a subset of the $K$-rational points of $X$ as I've described above?
Best Answer
This is an abuse of language. As your linked example shows, it occurs in the algebraic groups literature (and I would say probably only in that literature). This is because for various reasons classical language persisted much longer in this area than in other parts of algebraic geometry.
In any case, if $X$ is obtained by base-change to $K$ of $X_0$ over $k$, then "the $k$-points of $X$" will precisely mean "the $k$-points of $X_0$".
As you noted in the question, tensoring with $K$ gives a canonical embedding of $X_0(k)$ (the $k$-points of $X$ as a $k$-scheme) into $X(K)$ (the $K$ points of $X$ as a $K$-scheme), and in this way we may regard the $k$-points of $X$ as a subset of the $K$-points of $X$. This gives some (slight) justification for this abuse of language.
Note also that $X(K)$ (the $K$-points of $X$ as a $K$-scheme) is canonically identified with $X_0(K)$ (the $K$-points of $X_0$ as a $k$-scheme), just using the universal property. Thus the above canonical embedding of $X_0(k)$ in $X(K)$ may also be regarded as the (more obvious) embedding of $X_0(k)$ into $X_0(K)$.
Finally, as Keenan Kidwell notes in comments, this whole discussion very much relies on the choice of the $k$-scheme $X_0$ underlying the $K$-scheme $X$. (If you chose a different $X_0'$, say, giving rise to the same $K$-scheme $X$, then $X_0'(k)$ would likely be identified with a different subset of $X(K)$ to $X_0(k)$.)