For $X = S^1$, finding $X({\mathbf{Q}})$
- Finding one point in $X({\mathbf{Q}})$ and using it to “generate” all of $X({\mathbf{Q}})$.
- ${\mathbf{A}}^1_{/ {{\mathbf{Q}}}} $ as a moduli space for $X({\mathbf{Q}})$.
Defining projective space
- In coordinates: e.g. ${\mathbf{P}}^2_{/ {{\mathbf{C}}}} $ is ${\left[ {a:b:c} \right]} = \left\{{{\left[ {a,b,c} \right]} \in {\mathbf{C}}^3}\right\}/ \sim$ where ${\left[ {a,b,c} \right]}\sim {\left[ {\lambda a, \lambda b, \lambda c} \right]}$ for $\lambda \in U({\mathbf{C}})$, so the space of lines through $\mathbf{0}$.
- Topological description: ${\mathbf{RP}}^n = {\mathbf{R}}^{n+1}/C_2$ and ${\mathbf{CP}}^n = {\mathbf{C}}^{n+1}/C_2$ for $C_2 \curvearrowright k^n$ the antipodal action.
  - E.g. ${\mathbf{RP}}^1 {\mathbf{P}}^1({\mathbf{R}}) = {\mathbf{P}}({\mathbf{R}}^1) = {\mathbf{R}}^2/{\mathbf{G}}_m \cong S^1$ is the space of lines in ${\mathbf{R}}^2$ and ${\mathbf{CP}}^1 = {\mathbf{P}}^1({\mathbf{C}}) = {\mathbf{P}}({\mathbf{C}}^1) = {\mathbf{C}}^2/{\mathbf{G}}_m \cong S^2$. Here ${\mathbf{G}}_m({\mathbf{R}}) = {\mathbf{R}}^{\times}= U({\mathbf{R}}) = {\mathbf{R}}\setminus\left\{{0}\right\}$ and similarly ${\mathbf{G}}_m({\mathbf{C}}) = {\mathbf{C}}^{\times}= {\mathbf{C}}\setminus\left\{{0}\right\}$.
  - ${\mathbf{A}}^1\hookrightarrow{\mathbf{P}}^1$ by ${\left[ {x} \right]} \mapsto {\left[ {x: 1} \right]}$ or ${\left[ {x} \right]}\mapsto {\left[ {1: x} \right]}$, so ${\mathbf{P}}^1 = {\mathbf{A}}^1{ \coprod_{x\mapsto 1/x} } {\mathbf{A}}^1$. Here we think of ${\left[ {1: 0} \right]}$ as the “point (hyperplane) at infinity” and ${\left[ {0: 1} \right]}$ as zero.
- For ${\mathbf{P}}^2$: to parameterize all lines, fix $\mathbf{0}\in {\mathbf{A}}^3$, cast a ray and intersect with the plane $\left\{{z=1}\right\} \cong {\mathbf{A}}^2$; this is the embedding ${\left[ {a, b} \right]}\mapsto {\left[ {a,b,1} \right]}$. You get every line this way except for an ${\mathbf{P}}^1$ worth in the $z=0$ plane, so ${\mathbf{P}}^2 \cong {\mathbf{A}}^2{\textstyle\coprod}{\mathbf{P}}^1$. Similarly ${\mathbf{P}}^1 \cong {\mathbf{A}}^1 {\textstyle\coprod}{\mathbf{P}}^0 = {\mathbf{A}}^1{\textstyle\coprod}{\operatorname{pt}}$ by projecting onto the $y=1$ line. Leads to a CW decomposition.
Examples:
- Count points in coordinates in ${\mathbf{A}}^1(K)$ for $K = { \mathbf{F} }_q$ to get ${\mathbf{A}}^1({ \mathbf{F} }_q) = q$.
- Count points in coordinates to get ${\mathbf{P}}^1(K) = q+1$. Need decomposision ${\mathbf{P}}^1 = {\mathbf{A}}^1 {\textstyle\coprod}{\left[ {1: 0} \right]}$.
Polynomial functions on projective space: the homogenization procedure
- Homogeneous of degree $d$ iff $F(\lambda x, \lambda y, \lambda z) = \lambda^2 F(x, y, z)$.