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 $\pi: X\rightarrow\text{Spec}(A)$?
 A: 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
$$ \mathbb{P}(E^*)(k) \cong \{l \subseteq E:\text{ such that $l$ is a line.} \}$$
