On $\mathbb{P}^1_k$ with $k = \bar{k}$, closed points correspond to $a \in k$ (the point $[a:1]$) plus $\infty = [1:0]$. Typical divisors:

• $D = 3 \cdot [0] - 2 \cdot [\infty]$: degree $1$.
• $D = [0] + [1] + [\infty]$: degree $3$, effective.
• $D = -[0]$: degree $-1$, not effective.

Over $\mathbb{R}$: the point $[i:1] \in \mathbb{P}^1_\mathbb{C}$ is not a point of $\mathbb{P}^1_\mathbb{R}$, but the closed point corresponding to $x^2 + 1 \in \mathbb{R}[x]$ has degree $[\mathbb{C}:\mathbb{R}] = 2$.

On $\mathbb{P}^1$ with coordinate $t = x/y$:

• $\operatorname{div}(t) = [0] - [\infty]$: a simple zero at $0$, simple pole at $\infty$.
• $\operatorname{div}(t^2) = 2[0] - 2[\infty]$: double zero at $0$, double pole at $\infty$.
• $\operatorname{div}(t - 1) = [1] - [\infty]$.
• $\operatorname{div}\left(\frac{t}{t-1}\right) = [0] - [1]$: zero at $0$, pole at $1$, regular at $\infty$.
• $\operatorname{div}(t^2 - 1) = [1] + [-1] - 2[\infty]$.

Notice: $\deg(\operatorname{div}(f)) = 0$ always. This is a theorem!

Let $E: y^2 = x^3 - x$ over $\mathbb{C}$ with identity $O = [0:1:0]$.

• $\operatorname{div}(x) = (0,0) + P_1 + P_{-1} - 3O$ where $P_{\pm 1} = (\pm 1, 0)$... but wait, we need $\deg = 0$. Actually $\operatorname{div}(x)$ is the divisor of $x$ viewed as a function $E \to \mathbb{P}^1$. The function $x$ has zeros at $(0,0)$ (where $x = 0$, and $y^2 = 0$ so $y = 0$ too — but we must account for the intersection multiplicity). More carefully:

$\operatorname{div}(x) = 2 \cdot (0,0) - 2O$

since $x$ has a double zero at $(0,0)$ (the tangent to $E$ at $(0,0)$ is vertical) and a double pole at $O$.

• $\operatorname{div}(y) = (0,0) + (1,0) + (-1,0) - 3O$: degree $0$. ✓

On $\mathbb{P}^1$: any two points are linearly equivalent ($[a] - [b] = \operatorname{div}\left(\frac{t - a}{t - b}\right)$). So every degree-$d$ divisor is equivalent to $d \cdot [\infty]$, and:

$\operatorname{Pic}(\mathbb{P}^1) \cong \mathbb{Z}, \quad \operatorname{Pic}^0(\mathbb{P}^1) = 0.$

For an elliptic curve $E$ with origin $O$, the Abel–Jacobi map gives:

$E(k) \xrightarrow{\sim} \operatorname{Pic}^0(E), \quad P \mapsto [P - O].$

This is an isomorphism of groups! The group law on $E$ (chord-tangent construction) corresponds to addition of divisor classes. So $\operatorname{Pic}^0(E) \cong E$ as a group variety.

Over $\mathbb{C}$: $\operatorname{Pic}^0(E) \cong \mathbb{C}/\Lambda$ (a complex torus), and $\operatorname{Pic}(E) \cong \mathbb{Z} \oplus E(\mathbb{C})$.

For a smooth curve $C$ of genus $g$:

$\operatorname{Pic}^0(C) \cong \operatorname{Jac}(C)$

the Jacobian variety, an abelian variety of dimension $g$. Over $\mathbb{C}$, $\operatorname{Jac}(C) \cong \mathbb{C}^g / \Lambda$ where $\Lambda$ is a lattice determined by the period matrix.

• $g = 0$: $\operatorname{Jac} = 0$ (a point).
• $g = 1$: $\operatorname{Jac}(E) \cong E$ (the curve is its own Jacobian).
• $g = 2$: $\operatorname{Jac}(C)$ is a $2$-dimensional abelian surface. The curve $C$ embeds in $\operatorname{Jac}(C)$ via $P \mapsto [P - P_0]$.
• $g = 3$: $\operatorname{Jac}(C)$ is a $3$-fold, and the theta divisor $\Theta \subseteq \operatorname{Jac}(C)$ is a surface isomorphic to $\operatorname{Sym}^2(C)$ (for non-hyperelliptic $C$).

On $\mathbb{P}^1$: $\mathcal{O}(d \cdot [\infty]) = \mathcal{O}_{\mathbb{P}^1}(d)$, the usual twisting sheaf. So $\operatorname{Pic}(\mathbb{P}^1) = \{\mathcal{O}(d) \mid d \in \mathbb{Z}\} \cong \mathbb{Z}$, matching the divisor picture.

On $\mathbb{P}^n$: a hypersurface $H = V(F)$ of degree $d$ gives a divisor with $\mathcal{O}(H) = \mathcal{O}(d)$. Different degree-$d$ hypersurfaces give linearly equivalent divisors.

• $\mathbb{P}^1$ ($g = 0$): $K = -2[\infty]$, $\deg K = -2$. The differential $dt$ has a double pole at $\infty$: $dt = d(1/s) = -s^{-2}ds$. So $\operatorname{div}(dt) = -2[\infty]$.

• Elliptic curve ($g = 1$): $K \sim 0$ (trivial). The regular differential $dx/y$ has no zeros or poles. So $\operatorname{div}(dx/y) = 0$.

• Genus-2 curve $y^2 = f(x)$, $\deg f = 5$ or $6$: the differential $dx/y$ has $\operatorname{div}(dx/y) = \sum_i W_i - \text{polar terms}$ where $W_i$ are the Weierstrass points. $\deg K = 2$.

• Smooth plane curve of degree $d$: $K_C \sim (d - 3)H|_C$ by adjunction. $\deg K = d(d-3)$, $g = \frac{(d-1)(d-2)}{2}$.