# Relation between projective space and graded rings

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 $$Proj(A[x_0,...,x_n])$$.

• 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 $\langle x_0 - a_0,\cdots, x_n - a_n \rangle$. Jul 7, 2020 at 20:21
• 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? Jul 7, 2020 at 20:24

$$\operatorname{Proj}(A[x_0,...x_n])$$ 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.