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Let $X$ be a scheme over a field $k$, and consider a point $x\in X$, i.e. a point in the underlying topological space of $X$. Does there exist a field extension $K$ of $k$ such that $x\in X(K)$?

I have a hard time relating the notions of "actual point" and "rational point" to one another. Every $K$-rational point $Spec(K) \to X$ has an image in $X$, which is an actual point. My question is about the converse, can every actual point be realized as (the image of) a $K$-rational point of some (big enough?) field extension $K$ of $k$?

mathfan24
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    Yes, but not uniquely. (Also I personally prefer to not think of the points of the so-called underlying topological space as “actual” – if anything, it is the rational points that are “actual”!) – Zhen Lin Jan 16 '23 at 15:18

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Sure: for a $k$-scheme $X$ and a field $K$, giving a map $\operatorname{Spec}(K)\to X$ is equivalent to choosing a point $x\in X$ and an injection $\mathcal{O}_x/\mathfrak{m}_x\to K$. This is Hartshorne Exercise II.2.7; the map on topological spaces is clear and the injection gives you the map between structure sheafs.

In particular, every point of $X$ is a $\mathcal{O}_x/\mathfrak{m}_x$-rational point.

imtrying46
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  • But still, turning $x \in |X|$ (the underlying top. space of $X$) into a $K$-rational point requires a choice of a morphism $O_x/\mathfrak m_x \to K$. If $k$ is not algebraically closed, there is no canonical way of doing this. If $k$ is algebraically closed and $x$ a maximal point, then there is really only one morphism $O_x/\mathfrak m_x \cong k \to k$. In this situation, the set of $k$-valued points is naturally isomorphic to the set of maximal points of the underlying topological space of $X$. – LurchiDerLurch Jan 16 '23 at 13:38
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    Did I say anything at odds with what you are saying? All I said is that by choosing $K=\mathcal{O}_x/\mathfrak{m}_x$, $x$ becomes $K$-rational, because then we automatically have a map $\mathcal{O}_x/\mathfrak{m}_x\to K=\mathcal{O}_x/\mathfrak{m}_x$, namely the identity. – imtrying46 Jan 16 '23 at 13:56
  • My point is that if $K$ is not algebraically closed, saying that $O_x/\mathfrak m_x = K$ is not a good idea because there is no canonical isomorphism. Saying that "every point of $X$ is a $O_x/\mathfrak m_x$-rational point" feels wrong to me because there is no natural way to turn $x \in |X|$ into a $O_x/\mathfrak m_x$-rational point. – LurchiDerLurch Jan 16 '23 at 14:07
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    But yeah, this is quite subtle. Take $A = \Bbb Q[i]$ for example. This is a field, hence it has only one maximal ideal. Consider $X = Spec(A)$ as schemes over $\mathbb Q$. Now there are two $\mathbb Q$-linear morphisms $A \to \bar {\Bbb Q}$, so there are two elements in $X(\bar {\mathbb Q})$, and one isn't better than the other. – LurchiDerLurch Jan 16 '23 at 14:18
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    I see what your saying and completely agree. But not that technically my answer avoids this subtlety, because I'm not saying that $K$ is just some field isomorphic to the residue field at $x$, but actually $is$ the residue field at $x$. So in this case, we have a canonical map :) – imtrying46 Jan 16 '23 at 17:40