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Suppose $X \to k$ is a scheme over a field. A point $x$ is defined to be rational(by Hartshorne) if $k(x) = \mathcal{O}_{X,x}/{\mathfrak{m}_x}$ is isomorphic to $k$.

My questions are: 1) Does he mean that the canonical map($k \to \mathcal{O}_{X,x} \to \mathcal{O}_{X,x}/{\mathfrak{m}_x}$) is an isomorphism?(or will any isomorphism do?)

2) Does the fact that they are isomorphic imply that the canonical map is an isomorphism?

3) If 2) does not hold in general does it hold if $X$ is locally of finite type over $k$?

Jehu314
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    What is "the canonical map"? The answer is almost certainly yes, but if you're really trying to dig in to this question you should include your definitions so that the answers can be more helpful. – KReiser Sep 19 '19 at 05:19
  • The composition $k \to \mathcal{O}{X,x} \to \mathcal{O}{X,x}/{\mathfrak{m}_x}$. That is, the map induced on the stalks. – Jehu314 Sep 19 '19 at 05:24
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    If you mean that the map $k\to A$ is the one given by the structure morphism, then the answer is yes - this is map makes $A$ a $k$-algebra (with a distinguished copy of $k$) and so any ideal must also be a $k$-submodule (via the action of the distinguished copy of $k$). Since $\mathfrak{m}_x$ can't contain the distinguished copy of $k$ (otherwise it would contain $1$), the quotient map doesn't touch the distinguished copy of $k$ and you can see that everything's exactly what you expect. – KReiser Sep 19 '19 at 05:34
  • Do you mean that the existence of SOME isomorphism implies that the canonical map is an isomorphism? – Jehu314 Sep 19 '19 at 08:05
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    I dug out my copy of Hartshorne to check exactly what he said, and the first and only time he defines a rational point is on page 80 (exercise II.2.8, final two lines), and he says that a rational point is one so that "$k(x)=k$". So the canonical map should be the isomorphism involved. This is also true if one requires that the isomorphism be as $k$-algebras, which one should do if one is working in the category of schemes over a field. Basically, these definitions are/should be constructed to prohibit the issues you're considering. – KReiser Sep 19 '19 at 08:58
  • Thanks, it makes sense when you consider everything in $k-Alg$. – Jehu314 Sep 19 '19 at 09:02
  • Great. Would you like me to post that comment as an answer? – KReiser Sep 19 '19 at 09:11
  • That would be good. – Jehu314 Sep 19 '19 at 10:10

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From the comments:

I dug out my copy of Hartshorne to check exactly what he said, and the first and only time he defines a rational point is on page 80 (exercise II.2.8, final two lines), and he says that a rational point is one so that "$k(x)=k$". So the canonical map should be the isomorphism involved. This is also true if one requires that the isomorphism be as $k$-algebras, which one should do if one is working in the category of schemes over a field. Basically, these definitions are/should be constructed to prohibit the issues you're considering.

KReiser
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