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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.

Normalize the target in the function field of the generically finite Gauss map of a theta divisor; its finite map has an intrinsic branch locus. For a nonhyperelliptic smooth projective curve, this locus is the projective dual variety of the canonical curve. For a hyperelliptic curve of genus at least three, it is the union of the dual hypersurface of its canonical rational normal curve and the $2g+2$ hyperplanes corresponding to branch points. Locally, ramification exchanges two roots of a hyperplane section, or exchanges the two lifts of a root passing through a branch point. Biduality then recovers the curve or its double-cover data.

For a nonhyperelliptic smooth projective curve of geometric genus four, the canonical map is an embedding of degree six. Dimension counts $10>h^0(2K)=9$ and $20>h^0(3K)=15$ supply a quadric and a cubic not divisible by it. The quadric is irreducible because the curve is nondegenerate. Degree six and unmixedness of a complete intersection then identify the intersection with the curve as a scheme.

If a line bundle on a smooth projective curve has both $h^0(\mathcal{L})>0$ and $h^0({\omega }_C\otimes {\mathcal{L}}^{-1})>0$, the displayed inequality holds. Write these dimensions as $r,s$. The product dimension bound for sections on a curve gives $r+s-1\le g$, and the Riemann-Roch theorem gives $r-s=\mathrm{deg}\mathcal{L}+1-g$. Adding gives $2r\le \mathrm{deg}\mathcal{L}+2$.

On a smooth projective curve, the exact sequence $0\to \mathcal{O}(D-P)\to \mathcal{O}(D)\to k(P)\to 0$ increases the Euler characteristic by one. Iteration for positive and negative coefficients gives $\chi (\mathcal{O}(D))=\mathrm{deg}D+1-g$. Serre duality identifies $h^1(\mathcal{O}(D))$ with $h^0(\mathcal{O}(K-D))$, proving $\ell (D)-\ell (K-D)=\mathrm{deg}D+1-g$.

For a smooth projective curve, the degree-zero Picard group consists of line bundles of degree zero up to isomorphism, with tensor product as addition. A nonzero rational section of a line bundle identifies each line bundle with $\mathcal{O}(D)$ for a divisor on an algebraic curve; degree is the sum of multiplicities weighted by residue-field degrees. Over an algebraically closed field, every degree-zero divisor is a sum of differences of points.

For an effective divisor on an algebraic curve $D$ on a smooth projective curve, the Zariski tangent space to its symmetric product of a curve is $H^0({\mathcal{O}}_D(D))$. The derivative of the Abelian sum map is the connecting homomorphism of $0\to {\mathcal{O}}_X\to {\mathcal{O}}_X(D)\to {\mathcal{O}}_D(D)\to 0$. Its dual under Serre duality is restriction of holomorphic differential forms to $D$, including all jets at repeated points. Hence its rank is $\mathrm{deg}D+1-h^0({\mathcal{O}}_X(D))$, and its image annihilates $H^0(K_X(-D))$.

For $d\in {\mathbb{Q}}^{\times }$, the smooth projective curve $X^3+Y^3+d{Z}^3=0$, with identity $O=(1:-1:0)$, is a genus-one elliptic curve. Its partial derivatives have no common projective zero. The invertible linear projective transformationsends it to ${\mathcal{Y}}^2\mathcal{Z}={\mathcal{X}}^3-432d^2{\mathcal{Z}}^3$, taking $O$ to $(0:1:0)$. Indeed $(X+Y{)}^3+3(X+Y)(X-Y{)}^2=4(X^3+Y^3)=-4d{Z}^3$. In affine coordinates $u=X/Z,v=Y/Z$, this gives $x=-12d/(u+v)$ and $y=-36d(u-v)/(u+v)$, with inverse $u=(y-36d)/(6x)$, $v=(-y-36d)/(6x)$. Thus diagonal sums of two cubes are naturally studied through elliptic-curve addition formula.

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.

Let $P=p_1+\cdots +p_k$ and $Q=q_1+\cdots +q_k$ be disjoint sums of distinct points on a smooth projective curve. A nonzero function with principal divisor $P-Q$ exists exactly when $L(Q-P)\ne 0$: a nonzero member makes $(f)+Q-P$ an effective divisor of degree zero, forcing equality. Evaluation $f\mapsto (f(p_1),\dots ,f(p_k))$ on $L(Q)$ has kernel of a linear map $L(Q-P)$.

If $\ell (Q)=k$, a choice of basis $f_1,\dots ,f_k$ turns the criterion into $\det (f_j(p_i))=0$. This determinant is not identically zero. Starting with $L(Q)$, repeatedly choose a point where some nonzero function in the current space does not vanish; evaluation then lowers the dimension of that space by one. After $k$ choices the joint kernel is zero. Thus a generic tuple of points gives an invertible evaluation matrix, and no function with divisor $P-Q$. On a genus one curve, Riemann-Roch theorem gives $\ell (Q)=k$, so this obstruction applies for every $k>0$.

Every rational map from a smooth projective curve to a projective variety extends uniquely to a morphism. At a missing point, the discrete valuation ring of the curve lets one divide homogeneous coordinates by their smallest valuation, leaving regular coordinates of which at least one is a unit.

The Fermat cubic $x^3+y^3+z^3=0$ is a smooth projective curve: its three first partial derivatives vanish simultaneously only at the forbidden zero vector. The genus of a smooth plane curve formula gives genus one.

This is the degree-three case of a Fermat equation; in general $x^n+y^n+z^n=0$ defines a Fermat curve.

The embedding line bundle has at least four independent global sections. If it is nonspecial, the Riemann-Roch theorem gives $h^0=6-g$, so $g\le 2$. If it is special, the Clifford inequality for curves gives $h^0\le 3$, a contradiction. The smooth projective curve need not span its full ambient projective space.

For a smooth projective curve with $H^0(C,{\mathcal{O}}_C)=k$, its genus is ${\dim }_k{H}^1(C,{\mathcal{O}}_C)$, equivalently ${\dim }_k{H}^0(C,{\Omega }_C^1)$ by Serre duality. It equals the arithmetic genus $1-\chi ({\mathcal{O}}_C)$. Without the condition on constants, the general Riemann-Roch theorem instead uses $\chi ({\mathcal{O}}_C)=h^0({\mathcal{O}}_C)-h^1({\mathcal{O}}_C)$.

Suppose a line bundle $L$ of degree $g-1$ on a smooth projective curve has $h^0(L)=h^1(L)=r$. Choose $r$ distinct points where evaluation is invertible for both $H^0(L)$ and $H^0(K_X\otimes L^{-1})$. Such choices exist simultaneously: each determinant defines a nonempty open subset of $X^r$, since independent sections cannot vanish at every point. Let $\xi \in H^1({\mathcal{O}}_X)=H^0(K_X{)}^{\ast }$ be a sum of nonzero weighted evaluation functionals at these points. The Serre duality matrix of $\xi ⌣-$ factors as the product of the two invertible evaluation matrices and an invertible diagonal matrix. Its determinant is nonzero. This gives the nonzero leading determinant needed for the Riemann singularity theorem.

A morphism of algebraic varieties whose source and target are algebraic curves. A nonconstant morphism between smooth projective curves is finite and surjective; its degree is the degree of the induced extension of function fields.

If a divisor on an algebraic curve $D$ on a smooth projective curve has negative degree of a divisor, then its Riemann-Roch space is zero. Otherwise a nonzero $f\in L(D)$ would make $(f)+D$ an effective divisor of degree $\mathrm{deg}D<0$, since a principal divisor has degree zero. This is impossible. In genus one the Riemann-Roch theorem consequently gives $\ell (D)=\mathrm{deg}D$ for every divisor of positive degree.

The genus-degree formula givesbecause the nonsingular plane cubic is already its own normalization of an algebraic curve. If it were birational to the projective line, the two smooth projective curves would have isomorphic function fields. The uniqueness of the finite normalization model of a one-variable function field would then make them isomorphic. But the projective line has genus of a smooth projective curve zero. Therefore the cubic is not rational. This obstruction works in the stated positive characteristics as well as in characteristic zero.

An isogeny of elliptic curves over $k$ is a nonconstant morphism $\varphi :E_1\to E_2$ defined over $k$ and preserving the identity. Such a morphism is a group homomorphism: apply the Mumford rigidity lemma to $\varphi (P+Q)-\varphi (P)-\varphi (Q)$, whose restriction to either coordinate axis is zero. A nonconstant morphism of smooth projective curves is finite and surjective, so its degree of an isogeny is the positive integerDegrees multiply under composition. A separable isogeny has degree equal to the number of geometric points in its kernel of an isogeny. In positive characteristic of a field, inseparable isogenies also occur, for example the Frobenius isogeny; their total degree includes inseparable multiplicity, so counting geometric kernel points alone is insufficient. Every isogeny of degree $d$ has a dual isogeny $\hat{\varphi }$ satisfying $\hat{\varphi }\varphi =[d]$ and $\varphi \hat{\varphi }=[d]$.

Pointwise addition and negation make the homomorphism group of elliptic curves $H={\mathrm{Hom}}_k(E_1,E_2)$ an abelian group. Its zero is the constant identity map; every other element is an isogeny, because its nonconstant image is the whole target curve. Define $q(0)=0$ and $q(\varphi )=\mathrm{deg}\varphi$ otherwise. We prove the positive definiteness of degree on elliptic-curve homomorphisms, including arbitrary characteristic.

Let $L={\mathcal{O}}_{E_2}(O)$, a line bundle associated to a divisor of degree one. On $E_2\times E_2$ write $m(P,Q)=P+Q$, $d(P,Q)=P-Q$ and let ${\pi }_i$ be the projections. The divisor proof of the degree parallelogram law givesHere is a direct justification. The degree-two Weierstrass coordinate $x$ has a double pole at $O$, and $x(P)=x(Q)$ precisely when $Q=P$ or $Q=-P$. Consequentlywhere $D_{+}$ is the diagonal and $D_{-}$ the graph of negation. Since $D_{+}=d^{-1}(O)$ and $D_{-}=m^{-1}(O)$, this principal divisor gives the displayed line bundle identity. This remains valid in characteristic two: the general Weierstrass equation of an elliptic curve of a smooth algebraic curve has a nontrivial generic involution $(x,y)\mapsto (x,-y-a_{1}x-a_3)$, so the two graphs still occur. This identity is also the symmetric-line-bundle form of the Theorem of the square.

Pull back the bundle identity along $(\varphi ,\psi ):E_1\to E_2\times E_2$ and take degrees of line bundles. For a nonzero map, $\mathrm{deg}({\varphi }^{\ast }L)=q(\varphi )$; for a zero map its pullback is trivial and has degree zero. Thus the degree parallelogram law isUsing bundles rather than pulling back the rational function itself is essential when $\varphi =\pm \psi$, where that function can vanish identically. Negation preserves degree, so $q(-\varphi )=q(\varphi )$. Applying the parallelogram identity to $n\varphi$ and $\varphi$ gives the quadratic degree recursion for elliptic multiplication, and induction from $q(0)=0$ yields $q(n\varphi )=n^{2}q(\varphi )$ for every integer $n$.

Define the bilinear degree pairing for elliptic-curve homomorphisms byIt is symmetric and odd in each argument. Two applications of the parallelogram law giveAdding proves $B(x+z,y)=B(x,y)+B(z,y)$, and symmetry proves additivity in the other argument. Also $B(\varphi ,\varphi )=q(\varphi )$. Therefore $q$ is an integer-valued quadratic form. The integral polarization is $2B(\varphi ,\psi )=q(\varphi +\psi )-q(\varphi )-q(\psi )$; $B$ itself may be half-integral.

Finally $q(\varphi )>0$ for every nonzero $\varphi$, since an isogeny has positive finite degree. If $n\varphi =0$ with $n\ne 0$, the identity $q(n\varphi )=n^{2}q(\varphi )$ forces $\varphi =0$, so $H$ is a torsion-free abelian group. On any finite-dimensional rational span in $H\otimes \mathbb{Q}$, the pairing has a rational matrix and is positive on every nonzero rational vector. By density it is nonnegative on real vectors; if that rational matrix were singular, its kernel would contain a nonzero rational vector, contradicting strict positivity. Thus its real extension is a positive-definite quadratic form. This establishes the required positivity as well as the quadratic identity. For example, on the subgroup of multiplication maps $[n]$, $q([n])=n^2$; on the endomorphism ring of an elliptic curve with complex multiplication it is the imaginary-quadratic field norm, explaining the lattice geometry behind isogeny degrees.

At a point $P$ of the smooth algebraic curve $X$, the local ring ${\mathcal{O}}_{X,P}$ is a discrete valuation ring. Embed the projective variety $Y$ in ${\mathbb{P}}^N$. A rational map of algebraic varieties to $Y$ has homogeneous coordinates $[f_0:\cdots :f_N]$ in the function field $k(X)$, not all zero. Let $v_P$ be the local valuation, choose a uniformizer $u$, and set $m={\min }_{f_i\ne 0}{v}_P(f_i)$. The coordinates $u^{-m}{f}_i$ are regular at $P$, and at least one is a unit. On a sufficiently small open subset containing $P$, all these functions are regular and that coordinate stays nonzero. Their ratios therefore define a morphism of algebraic varieties into the corresponding affine chart of ${\mathbb{P}}^N$. Every homogeneous equation defining $Y$ vanishes on the original dense open, hence vanishes identically on this neighborhood. The local extension consequently lands in $Y$.

Two such extensions agree wherever the original rational map of algebraic varieties is defined. Their equalizer is closed, because $Y$ is a separated variety, and contains a dense open. Thus they agree throughout their overlap and glue. The rational map has a unique extension to a morphism on all of $X$. Projectivity is needed for the target; the source need not be projective.

For the function field correspondence, let $K/k$ be a finitely generated field extension of transcendence degree one. Choose a transcendence basis consisting of one element $t$. The extension $K/k(t)$ is finite: finitely many algebraic generators each have finite degree. Take the integral closures of $k[t]$ and $k[t^{-1}]$ in $K$. The finiteness assumption allowed in the paper makes both finite modules over their original coordinate rings. Integral closure commutes with localization, so these two rings agree over $k[t,t^{-1}]$ and glue to the finite normalization model of a one-variable function field $C_K\to {\mathbb{P}}^1$. This is a finite morphism, hence projective, and its function field is $K$. Its one-dimensional normal local rings are discrete valuation rings, so it is regular. Since the algebraically closed field $k$ is perfect, regularity here implies smoothness. Thus $C_K$ is a smooth projective curve.

Conversely the function field of a smooth projective curve is finitely generated over $k$ and has transcendence degree one. Two such curves with isomorphic function fields have mutually inverse birational maps. The extension argument just proved makes both maps morphisms; their composites equal the identity on a dense open, and hence everywhere by separatedness. This proves uniqueness of the smooth projective model. More precisely, pullback gives a contravariant correspondence between nonconstant morphisms of smooth projective curves and $k$-embeddings of their function fields. A field embedding defines the map on affine coordinates near the generic point, and the extension argument completes it. Constant morphisms are excluded from this correspondence because they do not induce embeddings of function fields.

Finally let $f:C\to D$ be nonconstant. Its image is dense, and pullback gives a finite extension $k(D)\subset k(C)$: both fields have transcendence degree one and the latter is finitely generated. Normalize $D$ in $k(C)$, applying the allowed finite-integral closure result on affine charts. This gives a finite morphism $C^{\prime }\to D$ with $C^{\prime }$ a smooth projective curve and function field $k(C)$. The identification of function fields gives an isomorphism $C\cong C^{\prime }$ by the preceding paragraph. Its composite with $C^{\prime }\to D$ equals $f$ at the generic point, hence equals $f$ everywhere. Every nonconstant morphism of these curves is finite.