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I know for a scheme $X$ locally of finite type over a field $k$, $k$-rational points are closed ponits.

If we remove the assumption that $X$ is locally of finite type over $k$, are there some $k$-rational points which are not closed point?

LOCOAS
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  • Closely related: https://math.stackexchange.com/questions/2969148. It is mentioned to consider $Spec(\Bbb C(T))$ over $Spec(\Bbb C)$. – Watson Apr 19 '19 at 10:16

2 Answers2

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Reducing to the affine case, your question comes down to the following: if $A$ is an algebra over a field $k$, not necessarily finitely generated, and $\mathfrak p$ is a prime of $A$ such that $k \rightarrow A_{\mathfrak p}/\mathfrak p A_{\mathfrak p}$ is an isomorphism, is it possible for $\mathfrak p$ to be a nonmaximal ideal? The answer is no.

By hypothesis, the composition $k \xrightarrow{i} A \xrightarrow{\pi} A/\mathfrak p \xrightarrow{h} \operatorname{Quot}(A/\mathfrak p) = A_{\mathfrak p}/\mathfrak p A_{\mathfrak p}$ is an isomorphism. Since

$$h \circ (\pi \circ i)$$

is a bijection, it is in particular surjective, which implies that $h$ must be surjective. The surjectivity of

$$A/\mathfrak p \rightarrow \operatorname{Quot}(A/\mathfrak p)$$ implies that $A/\mathfrak p$ is a field, i.e. $\mathfrak p$ is maximal.

D_S
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    I think we cannot reduced to the affine case. Let $U$ be an affine open subset of a scheme $X$. Then it is not always correct that closed points in $U$ are closed points in $X$ unless $X$ is Jacobson. – LOCOAS Apr 19 '19 at 02:58
  • The previous reasoning works, for example, for quasi-compact schemes see here. – Armando j18eos Apr 20 '19 at 08:37
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Let $x\in X$ be a $k$-rational point and $X=\bigcup U_i$ be an affine open cover (not necessarily finite). Then, $$ \overline{\{ x\}}^{(X)}=\overline{\{ x\}}^{(X)}\cap \left( \bigcup U_i\right) = \bigcup \left( \overline{\{ x\}}^{(X)}\cap U_i\right)=\bigcup_{U_i\ni x} \left( \overline{\{ x\}}^{(U_i)}\right). $$ If the affine case is verified, $\overline{\{ x\}}^{(U_i)}=\{x\}$ since $x $ is also $k$-rational in $U_i$. So, $$ \overline{\{ x\}}^{(X)}=\{x\}. $$ Therefore we can reduce to the affine case.

prime_1019
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