I know that if we have a scheme $X$ over an algebraically closed field of characteristic $0$, $k$, there exists at least $1$ rational point, i.e, a map $\text{Spec}(k)\rightarrow X$. Let us assume now that $\pi: X\rightarrow \text{Spec}(A)$ is a $\text{Spec}(A)$-scheme, where $A$ is a $k$-algebra.. My question is if there exists a map $\text{Spec}(A)\rightarrow X$ such that it is a section of the structural map. Do we need to impose an additional property on the morphism $\pi: X\rightarrow\text{Spec}(A)$?

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    $\begingroup$ For any $k$ algebra $A$, you have a natural map $\operatorname{Spec} A\to\operatorname{Spec} k$ and take the composite. $\endgroup$
    – Mohan
    Jun 27, 2021 at 21:14
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    $\begingroup$ Your question is not so clear. Do you have in mind the situation where $X$ is a scheme over $\operatorname{Spec} A$ and you're looking for an $A$-rational point on $X$, i.e. a map $\operatorname{Spec} A\to X$ so that the composite with the structure morphism $X\to\operatorname{Spec} A$ is the identity? Because these are hard to come by - e.g. $V(x^2+(1-a)y^2+(1-b)z^2)\subset \Bbb P^2_{\Bbb A^2_k}$ as a scheme over $\Bbb A^2_k$ where $x,y,z$ are coordinates on $\Bbb P^2$ and $a,b$ are coordinates on $\Bbb A^2_k$ has no section over the generic point of $\Bbb A^2$ and thus no $\Bbb A^2$-point. $\endgroup$
    – KReiser
    Jun 27, 2021 at 21:38
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    $\begingroup$ Please specifiy the question - where exactly do you replace $k$ by $A$? (Is $X$ defined over $A$, or is the point over $A$, or both?) Also, have you looked at a single example? $\endgroup$ Jun 27, 2021 at 21:40
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    $\begingroup$ Your statement in the algebraically closed case is wrong without additional hypotheses. Obviously you need to assume $X$ is nonempty, but you also need some finiteness condition (such as $X$ being finite type over $k$). $\endgroup$ Jun 27, 2021 at 22:19
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    $\begingroup$ @DanielHast consider $k\subset k(t)$ in the standard fashion: taking Spec, we find a quasi-compact $k$-scheme with no $k$-points. Some sort of algebraic finiteness condition is necessary, and once you add quasicompactness you're pretty much at finite type. $\endgroup$
    – KReiser
    Jun 30, 2021 at 10:05

1 Answer 1


Question: "My question is if there exists a map Spec(A)→X such that it is a section of the structural map. Do we need to impose an additional property on the morphism $X→Spec(A)$?"

Answer: Affine schemes: Assume $\pi: X:=Spec(B) \rightarrow S:=Spec(A)$ has a section $s:S \rightarrow X$ with $\pi \circ s =Id_S$. It follows we get maps of rings

$$A \rightarrow^f B \rightarrow^g A$$

with $g \circ f =Id_A$ hence the map $g:B \rightarrow A$ is surjective with $I:=ker(g)$ and $A \cong B/I$. Hence if $X$ is affine you must assume $S\subseteq X$ is a closed subscheme. The map $\phi:= f \circ g$ satisfies

$$ \phi^2 =\phi.$$

Hence you get an idempotent endomorphism of rings $\phi: B \rightarrow B$. It follows there is an isomorphism of rings

$$B \cong I \oplus B/I$$


$$(u,v)\times (x,y):=(ux+uy+vx,vy)$$

for all $(u,v),(x,y)\in I\oplus B/I$. Here $B/I$ acts on $I$ via $f$. This is a strong condition on $B$.

Example: If $g:X \rightarrow S$ is a separated and smooth morphism of relative dimension $n$ (see Hartshorne, "Smooth morphisms") and if $i:S \rightarrow X$ is a section of $g$, it follows $i$ is a "regular embedding".

A closed embedding $i: S \rightarrow X$ is a regular embedding of dimension $d$ iff the ideal sheaf defining $S$ is locally generated by by a regular sequence of length $d$.

Example: If $X:=Spec(B)$ is an affine scheme and $I \subseteq B$ is an ideal generated by a regular sequence, it follows the closed subscheme $S:=V(I) \subseteq X$ is a complete intersection. If $I:=(h)$ is generated by an element $h$ which is a non-zero divisor, it follows $S=V(I)$ is a complete intersection.

Projective space bundles: Let $E$ be a finitely generated and projective $A$-module with corresponding $\mathcal{O}_S$-module $\mathcal{E}$. If $\pi:X:=\mathbb{P}(\mathcal{E}^*)\rightarrow S$, there is a 1-1 correspondence between sections $s:S \rightarrow \mathbb{P}(\mathcal{E}^*)$ of $\pi$ and rank one quotients

$$\phi_s: \mathcal{E}^* \rightarrow L \rightarrow 0$$

with $L\in Pic(S)$, modulo an equivalence relation. Two quotients $L,L'$ are equivalent iff there is an ismorphism $\psi: L \cong L'$ such that the two obvious diagrams commute. This is HH.Prop.II.7.12.

Example: If $A:=k$ is any field, it follows a section $x: Spec(k) \rightarrow \mathbb{P}(E^*)$ corresponds 1-1 to a line $l_x \subseteq E$. Hence there is a 1-1 correspondence

$$ \mathbb{P}(E^*)(k) \cong \{l \subseteq E:\text{ such that $l$ is a line.} \}$$


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