An axiomatic structure for a smooth projective curve $C/k$: it bundles the function field, divisor group, degree-zero Picard group $\operatorname{Pic}^0(C)$, set of rational points, principal divisor map, class map, genus, and the basic compatibilities between them.
- FunField : Type u_2
- DivGroup : Type u_3
- divGroupInst : AddCommGroup self.DivGroup
- Pic0 : Type u_4
- pic0Inst : AddCommGroup self.Pic0
- RatPoint : Type u_5
- principalDiv_in_kernel (f : self.FunField) : f ≠ 0 → self.classMap (self.principalDiv f) = 0
- pointToDivisor_injective : Function.Injective self.pointToDivisor
- isAlgebraic_implies_mem_range_algebraMap (f : self.FunField) : IsAlgebraic k f → f ∈ Set.range ⇑(algebraMap k self.FunField)
- genusVal : ℕ
- exists_two_ratPoints_of_algClosed : IsAlgClosed k → ∃ (P : self.RatPoint) (Q : self.RatPoint), P ≠ Q
- principalDiv_mul (f g : self.FunField) : f ≠ 0 → g ≠ 0 → self.principalDiv (f * g) = self.principalDiv f + self.principalDiv g
Instances For
Axiom (degree formula): if $f \in k(C)$ has principal divisor $(P) - (Q)$ with $P \neq Q$, then for any injective $k$-algebra homomorphism $\varphi: k(t) \hookrightarrow k(C)$, the extension $k(C)/\varphi(k(t))$ has degree $1$.
If $k(C) \cong k(t)$ as $k$-algebras, then every degree-zero divisor on $C$ is principal.
Over an algebraically closed field, a smooth projective curve has at least two distinct rational points.
If $f \in k(C)$ has a nonzero principal divisor, then $f$ is transcendental over $k$ (an algebraic element would lie in $k \subseteq k(C)$ and have trivial divisor).
If there is a function $f \in k(C)$ with divisor $(P) - (Q)$ for distinct rational points $P, Q$, then any injective $k$-algebra map $\varphi: k(t) \to k(C)$ is automatically surjective.
The field extension $k(C)/\varphi(k(t))$ has degree one whenever there is a function with divisor $(P) - (Q)$ and $\varphi: k(t) \hookrightarrow k(C)$ is injective.
A smooth projective curve $C/k$ is isomorphic to $\mathbb{P}^1_k$ iff its function field $k(C)$ is $k$-algebra isomorphic to the rational function field $k(t)$.
Instances For
The Picard group $\operatorname{Pic}^0(C)$ is trivial iff every degree-zero divisor class is the identity.
Instances For
If $k(C) \cong k(t)$, then $\operatorname{Pic}^0(C)$ embeds (injectively) into the class group of $k[t]$. Used as a stepping stone to show $\operatorname{Pic}^0(C)$ is trivial.
If $k(C) \cong k(t)$ as $k$-algebras, then $\operatorname{Pic}^0(C)$ is trivial.
If $k(C) \cong k(t)$, then every degree-zero divisor $D$ on $C$ is principal: $D = \operatorname{div}(f)$ for some nonzero $f \in k(C)$.
Re-export of Pic0_injective_to_classGroup_of_iso: an injective additive map $\operatorname{Pic}^0(C) \hookrightarrow \operatorname{Cl}(k[t])$ exists whenever $k(C) \cong k(t)$.
If $C \cong \mathbb{P}^1_k$, then every degree-zero divisor on $C$ is principal.
If $C \cong \mathbb{P}^1_k$, then the divisor-class image of any degree-zero divisor $D$ vanishes in $\operatorname{Pic}^0(C)$.
Same statement as principalDiv_of_iso_P1, packaged as a lemma: every degree-zero divisor on $C \cong \mathbb{P}^1$ is principal.
If $C \cong \mathbb{P}^1_k$ then $\operatorname{Pic}^0(C) = 0$.
Bundles the injective embedding $\operatorname{Pic}^0(C) \hookrightarrow \operatorname{Cl}(k[t])$ provided by an isomorphism $k(C) \cong k(t)$. Since the target is trivial, this concretely produces the zero map.
Instances For
Over an algebraically closed field, a smooth projective curve has two distinct rational points.
If $f \in k(C)$ has divisor $(P) - (Q)$ with $P \neq Q$, then $f$ is transcendental and induces an injective $k$-algebra homomorphism $k(t) \hookrightarrow k(C)$ by $t \mapsto f$.
Any injective $\varphi: k(t) \to k(C)$ is automatically surjective if there is a function on $C$ with divisor $(P) - (Q)$: the field extension $k(C)/\varphi(k(t))$ has degree one.
A bijective $k$-algebra homomorphism $k(t) \to k(C)$ gives the isomorphism $C \cong \mathbb{P}^1_k$.
If $C$ admits a function with divisor $(P) - (Q)$ for distinct rational points, then $C \cong \mathbb{P}^1_k$.
If $\operatorname{Pic}^0(C) = 0$, then for any two rational points $P, Q$, the degree-zero divisor $(P) - (Q)$ is principal: there is a function $f \in k(C)$ with $\operatorname{div}(f) = (P) - (Q)$.
Over an algebraically closed field, if $\operatorname{Pic}^0(C) = 0$ then $C \cong \mathbb{P}^1_k$.
Main result: over an algebraically closed field, a smooth projective curve $C$ is isomorphic to $\mathbb{P}^1_k$ iff $\operatorname{Pic}^0(C)$ is trivial.