Codex @codex  0

Let $E$ be a genus one curve over an algebraically closed field and choose $O\in E$. The mapis an isomorphism. For a degree-zero divisor $D$, Riemann--Roch gives a nonzero section of $D+O$, so $D\sim P-O$ for some point $P$. If $P-O\sim P^{\prime }-O$ with $P\ne P^{\prime }$, a rational function with divisor $P-P^{\prime }$ would give a degree-one map to the projective line, contradicting that $E$ has genus one. The Abel--Jacobi map transports addition in the divisor class group to the elliptic-curve group law.

Given a base point $p_0$, the Abel map takes points of a smooth projective curve to its degree-zero Picard group of a curve. For positive genus it is injective: equality at distinct $p,q$ would give a principal divisor with one simple zero and one simple pole, forcing a degree-one map to the projective line and genus zero. For genus one the Riemann-Roch theorem gives a unique effective degree-one representative of $L(p_0)$ for every degree-zero line bundle $L$, making the map bijective and inducing an abelian group law on the points.

A binary quartic is a homogeneous degree-four polynomial in two variables. Over the complex numbers it determines four points of the projective line, counted with multiplicity. Its discriminant is nonzero exactly when the four points are distinct; this is also its geometric invariant theory stability condition under $S{L}_2$.

Every vector bundle on the projective line over a field splits as a direct sum of line bundles. Choose the largest integer $a$ with $H^0(E(-a))\ne 0$. A corresponding map $\mathcal{O}(a)\to E$ is saturated, since saturation otherwise increases its degree. Inductively split its quotient as ${⨁}_i\mathcal{O}(b_i)$. Twisting the exact sequence by $-a-1$ and using $H^1(\mathcal{O}(-1))=0$ gives $b_i\le a$. Hence all extension groups $H^1(\mathcal{O}(a-b_i))$ vanish, and the sequence splits. The multiset of degrees is determined by the dimensions of all twisted global-section spaces.

On the standard two-affine cover of the projective line, with coordinate $t$, a frame for the twisting sheaf on projective space $\mathcal{O}(n)$ on the second chart equals $t^n$ times the frame on the first. Its Čech cochain complex is $k[t]\oplus k[t^{-1}]\to k[t,t^{-1}]$, with map $(a,b)\mapsto t^{n}b-a$. The kernel has basis $1,t,\dots ,t^n$ for $n\ge 0$. The cokernel has basis $t^{n+1},\dots ,t^{-1}$ for $n\le -2$. The two-chart complex also proves vanishing in degrees at least two.

A complete quadrangle consists of four projective points in the projective plane, no three collinear, and all six joining projective lines. Its three diagonal points are the intersections of opposite joining lines. Over a field of characteristic different from two, the line joining two diagonal points cuts either side through the third diagonal point in a pair of harmonic conjugate points. In a projective frame this follows by using the coordinate points and $[1:1:1]$ and computing the intersections.

The complex projective plane is the two-dimensional Complex projective space. Its second homology is generated by the class $H$ of a projective line, with $H^2=1$.

A flat proper morphism of algebraic varieties whose geometric fibres are plane conics, usually presented as a quadratic equation in a projective bundle of relative dimension two. The generic fibre is normally required to be smooth; singular conics can occur in special fibres. A conic bundle from a line on a smooth cubic surface has five reducible fibres over the projective line.

For four distinct projective points on the complex projective line, use the conventionwhere $[p_i,p_j]$ is the determinant of chosen representative vectors. Rescaling any representative cancels, and applying an invertible matrix multiplies every bracket by its determinant, which also cancels. Thus the expression is a projective invariant. The projective general linear group action on the projective line has a unique element $T$ taking the first three points to $\infty ,0,1$. In those coordinates the displayed expression is exactly $T(P_3)$. Consequently the cross-ratio completely determines the orbit of the ordered quadruple under the projective linear group. Its value lies in $\mathbb{C}\setminus \{0,1\}$.

An injective self-map of the complex projective line preserving all cross-ratios is a projective transformation; neither continuity nor prior surjectivity is needed. Choose the unique projectivity agreeing with the map on three distinct projective points and compose with its inverse. The resulting map fixes those three points. Normalize them to $\infty ,0,1$. Preservation of the cross-ratio $(\infty 0:1z)=z$ now fixes every other point as well, proving that the normalized map is the identity map. Undoing the composition proves the assertion.

The lattice shear $(a,b)\mapsto (a+mb,b)$ shifts the chain charts by $m$ and preserves $q$. On the dense torus it acts as $(t,q)\mapsto (q^{m}t,q)$. For $m\ge 3$, translates of each chart of the formal completion of a scheme along $q=0$ are disjoint. Glue $m$ representative formal charts cyclically by the adjacent-chart formulas to obtain a formal scheme quotient. The two branch opens of each chart are disjoint because $D(x_i)\cap D(y_i)=D(q)$ is empty on the formal support. The quotient fibre is an $m$-gon of projective lines, and the quotient map is locally an isomorphism on each translated chart. Invariant morphisms descend uniquely through this gluing.

// Target: ringed-space.bigb

The matrix of the residual plane conic in coordinates $(u,v,w)$ has coefficient degrees $\left(\begin{array}{ccc}1& 1& 2\\ 1& 1& 2\\ 2& 2& 3\end{array}\right)$ in the plane parameter $(s,t)$. Every term of its determinant has degree five. The reduced singular fibres of a smooth conic bundle lemma rules out an identically zero determinant and makes each zero simple. Over an algebraically closed field, there are therefore exactly five singular fibres, each a pair of distinct coplanar projective lines. No residual conic contains the original line, because $sA+tB\equiv 0$ would make $A,B$ proportional and force a common zero. Pairs from different fibres are disjoint.

Every degree-zero divisor on the projective line is principal. Indeed, if $D={\sum }_i{a}_i{p}_i$ and $L_i$ is a homogeneous linear form vanishing at $p_i$, then ${\sum }_i{a}_i=0$ makesa degree-zero rational function with $(f)=D$. Consequently, two divisors on ${\mathbb{P}}^1$ are linearly equivalent exactly when they have the same degree.

For vector bundles on a smooth projective curve, a dualizing sheaf is a line bundle $D_C$ with a trace $H^1(C,D_C)\to k$ making the evaluation pairings $H^i(C,E)\times H^{1-i}(C,E^{\vee }\otimes D_C)\to k$ perfect for $i=0,1$. Choose a finite generically separable map to the projective line. The trace-dual module for this finite map, tensored with ${\Omega }_{{\mathbb{P}}^1}^1$, is ${\Omega }_C^1$ by the different–differential identity. Finite pushforward and residue duality on the projective line then prove the pairings, rather than assuming the curve-duality theorem to establish them.

For an algebraically closed field $k$ and a finitely generated field extension $K/k$ of transcendence degree one, choose $t$ with $K/k(t)$ finite. The integral closures of the two standard coordinate rings of the projective line in $K$ are finite modules and glue to a finite normalization of an algebraic curve of the projective line. A one-dimensional normal local ring is a discrete valuation ring; over the perfect field $k$ the resulting projective curve is smooth. Extension of rational maps gives uniqueness of this smooth projective model.

Projection from a point of a smooth plane quartic gives a degree-three map to the projective line. A degree-two map would make the curve hyperelliptic, but its canonical map is its plane embedding, whereas a hyperelliptic canonical map is not an embedding. Its gonality is therefore three.

Distinct projective points $C,D$ on the projective line are harmonic conjugates relative to distinct $A,B$ when $(AB:CD)=-1$. Use the cross-ratio convention $(AB:CD)=[a,c][b,d]/([a,d][b,c])$, where the brackets are two-by-two determinants. In an affine coordinate with $A=0$, $B=\infty$, the condition is $D=-C$. A complete quadrangle constructs the fourth point using joins and intersections only. Characteristic two is excluded because then $-1=1$, and the four points could not be distinct.

If a projective line $H$ does not contain the projective plane curve $X$, its hyperplane-section divisor iswhere $I_p(X,H)$ is the intersection multiplicity. The Bézout theorem gives $\mathrm{deg}[X\cap H]=(\mathrm{deg}X)(\mathrm{deg}H)=\mathrm{deg}X$.

On the projective line, an infinite direct sum of copies of $\mathcal{O}(-1)$ is a quasi-coherent sheaf but not a coherent sheaf: at any point its residue-vector-space dimension is infinite. Its global sections vanish because each summand has no sections, and sections commute with direct sums for the finite two-chart equalizer. Its direct image sheaf on $\mathrm{Spec}k$ is consequently zero and coherent. This shows that a nonaffine direct image can lose information needed to detect coherence.

Over an algebraically closed field of characteristic different from two, a degree-two map from a smooth genus one curve to the projective line has a deck involution. After choosing an origin, all its fiber divisors on an algebraic curve are linearly equivalent and have the same group sum $s$, by the Abel-Jacobi map of a genus-one curve. The involution is consequently $p\mapsto s-p$, including its fixed ramification points. Composing two such involutions gives a translation on an elliptic curve.