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Wanderer
Wanderer
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It is certainly true for schemes of finite type over $k$ (algebraically closed) that the closed points are exactly the $k$-points. To see this, notice that if $x \in X$ is any point, then the closure $\overline{\{x\}}$, equipped with its reduced subscheme structure, is integral and has dimension equal to the transcendence degree of its function field over $k$ (Hartshorne, exercise 3.20 in chapter 2). I hope that's clear enough?

For $k$-scheme which are not (locally) of finite type, this doesn't work.., as Martin shows below.

It is certainly true for schemes of finite type over $k$ (algebraically closed) that the closed points are exactly the $k$-points. To see this, notice that if $x \in X$ is any point, then the closure $\overline{\{x\}}$, equipped with its reduced subscheme structure, is integral and has dimension equal to the transcendence degree of its function field over $k$ (Hartshorne, exercise 3.20 in chapter 2). I hope that's clear enough?

For $k$-scheme which are not (locally) of finite type, this doesn't work...

It is certainly true for schemes of finite type over $k$ (algebraically closed) that the closed points are exactly the $k$-points. To see this, notice that if $x \in X$ is any point, then the closure $\overline{\{x\}}$, equipped with its reduced subscheme structure, is integral and has dimension equal to the transcendence degree of its function field over $k$ (Hartshorne, exercise 3.20 in chapter 2). I hope that's clear enough?

For $k$-scheme which are not (locally) of finite type, this doesn't work, as Martin shows below.

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Wanderer
Wanderer
  • 5.2k
  • 40
  • 49

It is certainly true for schemes of finite type over $k$ (algebraically closed) that the closed points are exactly the $k$-points. To see this, notice that if $x \in X$ is any point, then the closure $\overline{\{x\}}$, equipped with its reduced subscheme structure, is integral and has dimension equal to the transcendence degree of its function field over $k$ (Hartshorne, exercise 3.20 in chapter 2). I hope that's clear enough?

As forFor $k$-scheme which are not (locally) of finite type, I don't knowthis doesn't work...

It is certainly true for schemes of finite type over $k$ (algebraically closed) that the closed points are exactly the $k$-points. To see this, notice that if $x \in X$ is any point, then the closure $\overline{\{x\}}$, equipped with its reduced subscheme structure, is integral and has dimension equal to the transcendence degree of its function field over $k$ (Hartshorne, exercise 3.20 in chapter 2). I hope that's clear enough?

As for $k$-scheme which are not (locally) of finite type, I don't know.

It is certainly true for schemes of finite type over $k$ (algebraically closed) that the closed points are exactly the $k$-points. To see this, notice that if $x \in X$ is any point, then the closure $\overline{\{x\}}$, equipped with its reduced subscheme structure, is integral and has dimension equal to the transcendence degree of its function field over $k$ (Hartshorne, exercise 3.20 in chapter 2). I hope that's clear enough?

For $k$-scheme which are not (locally) of finite type, this doesn't work...

Source Link
Wanderer
Wanderer
  • 5.2k
  • 40
  • 49

It is certainly true for schemes of finite type over $k$ (algebraically closed) that the closed points are exactly the $k$-points. To see this, notice that if $x \in X$ is any point, then the closure $\overline{\{x\}}$, equipped with its reduced subscheme structure, is integral and has dimension equal to the transcendence degree of its function field over $k$ (Hartshorne, exercise 3.20 in chapter 2). I hope that's clear enough?

As for $k$-scheme which are not (locally) of finite type, I don't know.