Documentation

Atlas.ArithmeticGeometry.code.RationalLineCriterion

structure ArithmeticGeometry.SmoothProjectiveCurve (k : Type u_1) [Field k] :
Type (max (max (max (max u_1 (u_2 + 1)) (u_3 + 1)) (u_4 + 1)) (u_5 + 1))

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.

Instances For
    theorem ArithmeticGeometry.degree_formula_axiom {k : Type u_2} [Field k] (C : SmoothProjectiveCurve k) (f : C.FunField) (P Q : C.RatPoint) (hPQ : P Q) (hdiv : C.principalDiv f = C.pointToDivisor P + -C.pointToDivisor Q) (φ : RatFunc k →ₐ[k] C.FunField) ( : Function.Injective φ) :

    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 $1$.

    theorem ArithmeticGeometry.deg_zero_div_principal_of_iso_ratFunc {k : Type u_2} [Field k] (C : SmoothProjectiveCurve k) (φ : C.FunField ≃ₐ[k] RatFunc k) (D : C.DivGroup) (hdeg : C.degreeMap D = 0) :
    ∃ (f : C.FunField), f 0 C.principalDiv f = D

    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).

    Helper lemma: if a subfield $K \subseteq F$ equals all of $F$, then $F$ has dimension $1$ as a $K$-vector space.

    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.

        theorem ArithmeticGeometry.every_deg0_div_principal_of_iso_P1 {k : Type u_1} [Field k] (C : SmoothProjectiveCurve k) (φ : C.FunField ≃ₐ[k] RatFunc k) (D : C.DivGroup) (hdeg : C.degreeMap D = 0) :
        ∃ (f : C.FunField), f 0 C.principalDiv f = D

        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 $k(C) \cong k(t)$.

        theorem ArithmeticGeometry.principalDiv_of_iso_P1 {k : Type u_1} [Field k] (C : SmoothProjectiveCurve k) (hiso : C.IsIsomorphicToP1) (D : C.DivGroup) (hdeg : C.degreeMap D = 0) :
        ∃ (f : C.FunField), f 0 C.principalDiv f = D

        If $C \cong \mathbb{P}^1_k$, then every degree-zero divisor on $C$ is principal.

        theorem ArithmeticGeometry.classMap_zero_of_iso_P1 {k : Type u_1} [Field k] (C : SmoothProjectiveCurve k) (hiso : C.IsIsomorphicToP1) (D : C.DivGroup) (hdeg : C.degreeMap D = 0) :
        C.classMap D = 0

        If $C \cong \mathbb{P}^1_k$, then the divisor-class image of any degree-zero divisor $D$ vanishes in $\operatorname{Pic}^0(C)$.

        theorem ArithmeticGeometry.degree_zero_divisor_principal_of_iso_P1 {k : Type u_1} [Field k] (C : SmoothProjectiveCurve k) (hiso : C.IsIsomorphicToP1) (D : C.DivGroup) (hdeg : C.degreeMap D = 0) :
        ∃ (f : C.FunField), f 0 C.principalDiv f = D

        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.

          theorem ArithmeticGeometry.exists_injective_algHom_of_nontrivial_div {k : Type u_1} [Field k] (C : SmoothProjectiveCurve k) {P Q : C.RatPoint} (hPQ : P Q) {f : C.FunField} (_hf : f 0) (hdiv : C.principalDiv f = C.pointToDivisor P + -C.pointToDivisor Q) :

          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 $t \mapsto f$.

          theorem ArithmeticGeometry.morphism_surjective_of_deg_one_div {k : Type u_1} [Field k] (C : SmoothProjectiveCurve k) {P Q : C.RatPoint} (hPQ : P Q) {f : C.FunField} (hf : f 0) (hdiv : C.principalDiv f = C.pointToDivisor P + -C.pointToDivisor Q) (φ : RatFunc k →ₐ[k] C.FunField) (hφ_inj : Function.Injective φ) :

          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 $C \cong \mathbb{P}^1_k$.

          theorem ArithmeticGeometry.degree_one_morphism_is_iso {k : Type u_1} [Field k] (C : SmoothProjectiveCurve k) {P Q : C.RatPoint} (hPQ : P Q) {f : C.FunField} (hf : f 0) (hdiv : C.principalDiv f = C.pointToDivisor P + -C.pointToDivisor Q) :

          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.