Wikipedia defines a rational point as "A rational point on an algebraic variety is a point whose coordinates belong to a given field"
But Hartshorne defines a $k-$rational point on a scheme $X$ over a field $k$ as a point $x \in X$ such that $\mathscr O_{X,x}/ \mathfrak m_x \approx k$.
First question, is the isomorphism $k \rightarrow \mathscr O_{X,x}/ \mathfrak m_x$ required to be the composition $k \rightarrow \mathscr O_X (X) \rightarrow \mathscr O_{X,x} \rightarrow \mathscr O_{X,x}/ \mathfrak m_x$ or can the above morphism not be an isomorphism and we can just consider $x$ a rational point if there exists any isomorphism $k \xrightarrow{\approx}\mathscr O_{X,x}/ \mathfrak m_x$?
Secondly, the definition from wikipedia considers two fields $k' \subset k$ and defines what it means for a point to be $k'-$rational if your variety is over $k$, how do we extend this relativity to Hartshornes definition?
If you have a scheme over $k$ then you naturally have a scheme over $k'$ because of the map $\text{Spec}(k) \rightarrow \text{Spec}(k')$ and then we can talk about $k'-$rational points as per the local ring definitions. Is this the correct way of looking at things?
My final question is why does the Wikipedia and Hartshorne notions of rational points coincide?
If we think about the case $X = \text{Spec}(\mathbb R[x,y]/(x^2 -y))$ over $\mathbb Q$ is there a bijection
$$\{(x,y) \in \mathbb Q \times \mathbb Q : x^2 = y \} \rightarrow \{\mathfrak p \in X : \mathfrak p \text{ is a } \mathbb Q-\text{rational point and a closed point of X}\}?$$
Thank you in advance for answering my question!
Best Answer
As in the last time the first issue came up, you have misread Hartshorne's definition and Hartshorne failed to be as specific as he ought to have been. Hartshorne's text actually says "$k(x)=k$," which is intended to mean that the map is the canonical map $k\to \mathcal{O}_{X,x}/\mathfrak{m}_x$ you write. (A less confusing way to write this may be to specify that a $k$-rational point for a scheme over $k$ is a map $\operatorname{Spec} k\to X$ so that the composition with the structure map $X\to \operatorname{Spec} k$ is the identity.)
Next, Wikipedia actually gives two definitions: the first which depends on an embedding and a choice of algebraic closure, and then a second which is the same as the Hartshorne definition. The first is (at least to me) a little poorly explained: it leaves open the possibility that for a variety over $k$ you could ask about $l$ points for $l$ a subfield of $k$, which is not a thing. Such $l$ points aren't intrinsically defined (for instance, the point $(x-1)\in \Bbb A^1_\Bbb C$ can be moved to $(x-i\pi)\in \Bbb A^1_\Bbb C$ by several obvious automorphisms) in contrast to asking about points corresponding to extension fields.
Your question about looking at the composite map $X\to \operatorname{Spec} k\to \operatorname{Spec} k'$ is reasonable, but it fails to give you any interesting answer. No point of a $k$-variety will be a $k'$-point unless $k'=k$ (and even then you may not have any). The fact that $X$ is a $k$-variety says that the residue field of every point of $X$ is some (possibly trivial) extension of $k$, and unless $k=k'$, none of these can be $k'$. For instance, in your final example, there are no points on the right hand side of your equation in your second-to-last line.
To understand why these two definitions coincide, it's enough to work affine-locally. Then we have both an embedding $X\to \Bbb A^n_k$ and a map $\operatorname{Spec} F\to X$ picking out the $F$-point. Writing down the map on rings corresponding to the composition $\operatorname{Spec} F\to \Bbb A^n_k$, this is a map $k[x_1,\cdots,x_n]\to F$, and we can think of the coordinates of our $F$-point as the images of $x_1,\cdots,x_n$ in $F$. As the