# Rational points of schemes.

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 $$\pi: X\rightarrow\text{Spec}(A)$$?

• For any $k$ algebra $A$, you have a natural map $\operatorname{Spec} A\to\operatorname{Spec} k$ and take the composite. Jun 27, 2021 at 21:14
• 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. Jun 27, 2021 at 21:38
• 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? Jun 27, 2021 at 21:40
• 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 $k$). Jun 27, 2021 at 22:19
• @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. Jun 30, 2021 at 10:05

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

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

with

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

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

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