$\mathbb{P}^1 = \{[a_0 : a_1]\}$. The standard open cover:

• $U_0 = \{a_0 \neq 0\} = \{[1:t] \mid t \in k\} \cong \mathbb{A}^1$, where $t = a_1/a_0$.
• $U_1 = \{a_1 \neq 0\} = \{[s:1] \mid s \in k\} \cong \mathbb{A}^1$, where $s = a_0/a_1$.

$\mathbb{P}^1 = U_0 \cup U_1$ with overlap $U_0 \cap U_1 = \{t \neq 0\} = \{s \neq 0\}$ and transition $s = 1/t$. The "extra" point $[0:1] \in U_1 \setminus U_0$ is the point at infinity (where $t \to \infty$). Think of $\mathbb{P}^1$ as $\mathbb{A}^1 \cup \{\infty\}$, the one-point compactification.

Over $\mathbb{C}$, $\mathbb{P}^1(\mathbb{C})$ is the Riemann sphere $\mathbb{C} \cup \{\infty\} \cong S^2$.

$\mathbb{P}^2 = \{[x:y:z]\}$ with three standard charts:

• $U_x = \{x \neq 0\} \cong \mathbb{A}^2$ via $[1:s:t]$.
• $U_y = \{y \neq 0\} \cong \mathbb{A}^2$ via $[u:1:v]$.
• $U_z = \{z \neq 0\} \cong \mathbb{A}^2$ via $[a:b:1]$.

The "line at infinity" relative to $U_z$ is $\{z = 0\} \cong \mathbb{P}^1$.

$V(ax + by + cz) \subseteq \mathbb{P}^2$ is a projective line, isomorphic to $\mathbb{P}^1$. In the affine chart $z \neq 0$, this becomes the affine line $V(aX + bY + c) \subseteq \mathbb{A}^2$ (where $X = x/z$, $Y = y/z$). The projective line adds one "point at infinity" in the direction of the line.

Any two distinct lines in $\mathbb{P}^2$ meet in exactly one point — this is why projective geometry is more natural than affine geometry.

$V(x^2 + y^2 - z^2) \subseteq \mathbb{P}^2$. In the chart $z = 1$, this is the affine circle $x^2 + y^2 = 1$. The "points at infinity" are $[x:y:0]$ with $x^2 + y^2 = 0$, i.e., $[1:i:0]$ and $[1:-i:0]$ over $\mathbb{C}$.

Over $\mathbb{C}$, all smooth conics in $\mathbb{P}^2$ are isomorphic to $\mathbb{P}^1$ (via the parametrization $[s:t] \mapsto [s^2 - t^2 : 2st : s^2 + t^2]$).

The homogenization of $y^2 = x^3 - x$ is $Y^2 Z = X^3 - XZ^2$. The projective curve

$E = V(Y^2 Z - X^3 + XZ^2) \subseteq \mathbb{P}^2$

is an elliptic curve. It has a single point at infinity $[0:1:0]$ (set $Z = 0$: $0 = X^3$, so $X = 0$, giving $[0:1:0]$). This point serves as the identity element for the group law on $E$.

Over $\mathbb{C}$, $E(\mathbb{C}) \cong \mathbb{C}/\Lambda$ as a complex manifold (a torus), where $\Lambda$ is a lattice. This is not rational — it has genus 1.

The Fermat curve of degree $d$ is

$F_d = V(x^d + y^d - z^d) \subseteq \mathbb{P}^2.$

• $d = 1$: a line ($\cong \mathbb{P}^1$, genus 0).
• $d = 2$: a conic ($\cong \mathbb{P}^1$, genus 0).
• $d = 3$: an elliptic curve (genus 1). The integer points on $F_3$ are related to Fermat's Last Theorem for $n=3$.
• $d \geq 4$: a curve of genus $g = \frac{(d-1)(d-2)}{2} \geq 3$. By Faltings' theorem (Mordell conjecture), $F_d(\mathbb{Q})$ is finite for $d \geq 4$.

The Veronese embedding of degree $d$ is

$\nu_d : \mathbb{P}^n \hookrightarrow \mathbb{P}^N, \quad N = \binom{n+d}{d} - 1$

sending $[x_0:\cdots:x_n]$ to all monomials of degree $d$. For $n = 1$, $d = 2$:

$\nu_2 : \mathbb{P}^1 \to \mathbb{P}^2, \quad [s:t] \mapsto [s^2 : st : t^2].$

The image is the conic $V(xz - y^2)$. This shows that every smooth conic is isomorphic to $\mathbb{P}^1$.

For $n = 1$, $d = 3$: the image $\nu_3(\mathbb{P}^1) \subseteq \mathbb{P}^3$ is the rational normal curve of degree 3 (the projective closure of the twisted cubic).

The key property: the Veronese embedding converts degree-$d$ hypersurfaces in $\mathbb{P}^n$ into hyperplane sections in $\mathbb{P}^N$. This linearizes the problem of studying high-degree varieties.

The Segre embedding is

$\sigma : \mathbb{P}^m \times \mathbb{P}^n \hookrightarrow \mathbb{P}^{(m+1)(n+1)-1}$

sending $([x_0:\cdots:x_m], [y_0:\cdots:y_n]) \mapsto [x_i y_j]_{0 \leq i \leq m, 0 \leq j \leq n}$.

For $m = n = 1$:

$\sigma : \mathbb{P}^1 \times \mathbb{P}^1 \hookrightarrow \mathbb{P}^3, \quad ([s:t],[u:v]) \mapsto [su:sv:tu:tv].$

The image is $V(xw - yz) \subseteq \mathbb{P}^3$, a smooth quadric surface. This quadric is ruled by two families of lines:

• Fix $[s:t]$, vary $[u:v]$: a line on the quadric.
• Fix $[u:v]$, vary $[s:t]$: another line.

Every smooth quadric surface in $\mathbb{P}^3$ is isomorphic to $\mathbb{P}^1 \times \mathbb{P}^1$.

The Grassmannian $\mathbb{G}(r, n)$ parametrizes $r$-dimensional linear subspaces of $k^n$ (equivalently, $(r-1)$-planes in $\mathbb{P}^{n-1}$). It is a projective variety via the Plücker embedding:

$\mathbb{G}(r, n) \hookrightarrow \mathbb{P}^{\binom{n}{r}-1}, \quad W \mapsto [\text{Plücker coordinates}].$

For example, $\mathbb{G}(2, 4)$ parametrizes lines in $\mathbb{P}^3$. The Plücker embedding sends it to $\mathbb{P}^5$, where the image is the smooth quadric hypersurface $V(p_{01}p_{23} - p_{02}p_{13} + p_{03}p_{12})$.

$\dim \mathbb{G}(r,n) = r(n-r)$. So $\dim \mathbb{G}(2,4) = 4$ — the space of lines in $\mathbb{P}^3$ is 4-dimensional.

The affine hyperbola $V(xy - 1) \subseteq \mathbb{A}^2$ "looks" like two separate branches. Homogenize: $XY - Z^2 = 0$ in $\mathbb{P}^2$. The points at infinity ($Z = 0$) are $[1:0:0]$ and $[0:1:0]$ — the two branches of the hyperbola "meet at infinity." The projective curve is a smooth conic $\cong \mathbb{P}^1$.

Similarly, the parabola $V(y - x^2)$ becomes $YZ - X^2 = 0$, which has one point at infinity $[0:0:1]$ (actually $[0:1:0]$). The projective closure is again a smooth conic $\cong \mathbb{P}^1$.

All smooth conics are projectively equivalent — the distinction between ellipse, parabola, hyperbola is purely affine.