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There's a definition from synthetic geometry that says that a projective space is (\textit{$P,L,I$}),
where $\textit{P}$ is the set of points, $\textit{L}$, set of lines and $\textit{I}$ is the incidence relation, that is, tells if a point is in a line or not, and should satisfy the following axioms: $\\ $

  1. For every pair of distinct points $A$, $B$ there is a unique line that's incident to both of them, we call it $AB$
  2. If $A$, $B$, $C$ and $D$ are distinct points such that the lines $AB$ and $CD$ have a common point (that's a point which's incident to both of them), then $AC$ and $BD$ also have a common point.
  3. Each line is incident to at least 3 points.

But then we have the definition from projective geometry, that's, if $S$ is a graded ring, consider $Proj(S)$ to be set of homogeneous prime ideals not contaning the irrelevant ideal. If $A$ is a ring then we call the $Proj(A[x_0,...,x_n])$ to be the projective n-space over $A$, since rings of polynomials are graded by homogeneous elements.

I would like to know what are the points and lines in $Proj(A[x_0,...,x_n])$.

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  • $\begingroup$ Linear spaces are given by homogeneous linear forms. So a line is given by the ideal generated by $n-1$ general linear forms, and a point is given by the ideal generated by $n$ general linear forms; specifically $(a_0:\cdots a_n)$ is cut out by $\langle x_0 - a_0,\cdots, x_n - a_n \rangle$. $\endgroup$ Jul 7, 2020 at 20:21
  • $\begingroup$ Sorry, I didn't get it. Are you assuming that $A$ is a field? Other thing, what's a n general linear form? And what would be the relation of incidence? $\endgroup$
    – eipi10
    Jul 7, 2020 at 20:24

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$\operatorname{Proj}(A[x_0,...x_n])$ is not a "projective space" in the sense of synthetic geometry. If $k$ is an $A$-algebra that is a field, then the set of $k$-points of $\operatorname{Proj}(A[x_0,...x_n])$ (that is, the morphisms $\operatorname{Spec} k\to \operatorname{Proj}(A[x_0,...x_n])$ over $\operatorname{Spec} A$) is a projective space in the sense of synthetic geometry, since it can be naturally identified with the projective space of the vector space $k^{n+1}$. Explicitly, a point $(a_0,\dots,a_n)\in k^{n+1}$ determines a homomorphism of graded $A$-algebras $A[x_0,\dots,x_n]\to k[t]$ which maps each $x_i$ to $a_it$, and this determines a morphism $\operatorname{Spec} k\cong \operatorname{Proj}(k[t])\to\operatorname{Proj}(A[x_0,...x_n])$. It can be shown that this morphism is unchanged if you multiply $(a_0,\dots,a_n)$ by a scalar, and that every $k$-point of $\operatorname{Proj}(A[x_0,...x_n])$ arises in this way, so that $k$-points of $\operatorname{Proj}(A[x_0,...x_n])$ are naturally in bijection with lines through the origin in $k^{n+1}$.

To be explicit, the set of lines through the origin in $k^{n+1}$ forms a projective space in the sets of synthetic geometry in the following way. A "point" is a line through the origin in $k^{n+1}$, and a "line" is a plane through the origin in $k^{n+1}$. The incidence relation is just containment: a "point" is on a "line" if the corresponding line through the origin is a subset of the corresponding plane through the origin.

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  • $\begingroup$ Thank you, I was looking for it and didn't find for a long time. $\endgroup$
    – eipi10
    Jul 7, 2020 at 20:27

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