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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?

Jared
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  • @MartinBrandenburg: In the first paragraph of this paper, the authors use the phrase "$\mathbb{F}_p$-rational points of $G$" where $G$ is an $\overline{\mathbb{F}_p}$-algebraic group defined over $\mathbb{F}_p$. Also, when you say "$T$ is defined over $S$", do you mean we have a map of rings $S\to T$ giving $\mathrm{Spec};T$ the structure of an $S$-scheme? – Jared Jan 09 '14 at 22:00
  • Dear @Jared, I didn't look at the linked paper, but I think the confusion is arising from trying to mix classical with scheme-theoretic terminology. A variety over a field $k$ is a separated $k$-scheme of finite type (possibly also (geometrically) reduced and (geometrically) integral depending on convention). The structural morphism to the base field is part of the data of a $k$-variety. It's not a $\bar{k}$-variety with a ``$k$-structure." A $\bar{k}$-scheme of finite type with $k\neq\bar{k}$ has no $k$-rational points. – Keenan Kidwell Jan 09 '14 at 22:05
  • Anyway, probably the most natural interpretation is the set of $k$-points of $X_0$, which does inject into the $K$-points of $X$: $X_0(k)\hookrightarrow X_0(K)=(X_0\times_kK)(K)=X(K)$. – Keenan Kidwell Jan 09 '14 at 22:19
  • @KeenanKidwell: Thanks for your response, although I'm not sure I understand. Are you saying that, in my notation, $X$ is already a $k$-scheme with structure morphism given by $X\to X_0\to\mathrm{Spec};k$, and we should use this view of $X$ when describing its $k$-rational points? – Jared Jan 09 '14 at 22:20
  • It's true that $X$ is a $k$-scheme via the composite $X\rightarrow X_0\rightarrow\mathrm{Spec}(k)$, but it's probably not going to be of finite type unless $\bar{k}$ is finite over $k$. I'm saying that the modern theory provides a completely satisfactory theory of varieties over $k$, without any reference to $\bar{k}$, whereas, to my knowledge, the classical theory works over $\bar{k}$, and then involves a theory of ``$k$-structures." The scheme $X_0$ is not necessarily the only $k$-scheme which becomes isomorphic to $X$ over $\bar{k}$, – Keenan Kidwell Jan 09 '14 at 22:27
  • so the set $X_0(k)\subseteq X(K)$ is not intrinsic to $X$. I'm not sure what else you're looking for. If $A$ is a $k$-algebra and $x:A\rightarrow k$ is a $k$-algebra map (i.e. a $k$-point of $\mathrm{Spec}(A)$), then by tensoring with $K$ we get a $K$-point of $\mathrm{Spec}(A\otimes_kK)$. Note I was using $\bar{k}$ above, but what I said about $k$-points injecting into $K$-points works for any extension $K$ of $k$. – Keenan Kidwell Jan 09 '14 at 22:29
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    @MartinBrandenburg: Dear Martin, For various historical reasons, the theory of algebraic groups uses classical language much more than other areas of algebraic geometry. (Maybe this is not really true any more, but classical language persisted for much longer.) So phrases like ``$k$-structure'' appear, and cause confusion to those raised on schemes (as is happening here)! Regards, – Matt E Jan 12 '14 at 02:38

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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 $X_0(k)$.)

Matt E
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