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For a smooth plane quartic $C\subset {\mathbb{P}}^2$, the Adjunction formula gives $K_C\cong {\mathcal{O}}_C(1)$. Hence every line section $C\cap \ell$ is a canonical divisor.

An elliptic curve works over every algebraically closed field, including one algebraic over a finite field. Give it explicitly in ${\mathbb{P}}^2$ byIn the first case the only point at infinity is $O=[0:1:0]$, where the derivative with respect to $z$ is nonzero. In the affine chart $z=1$, a singular point of an algebraic variety would have $y=0$, $x^3-x=0$, and $3x^2-1=0$. The possible roots $x=0,1,-1$ cannot satisfy the last equation in characteristic different from $2$: at $0$ it is $-1$, and at $\pm 1$ it is $2$. In the second case the three homogeneous partial derivatives are $x^2,z^2,y^2$, up to harmless signs, and cannot vanish simultaneously at a projective point. Thus both cubics are smooth varieties. A reducible plane cubic would have intersecting positive-degree components by Bézout theorem, and be singular at an intersection; therefore these smooth cubics are irreducible varieties. The genus-degree formula gives $g=(3-1)(3-2)/2=1$. With the marked point $O$ they are elliptic curves.

For a smooth algebraic curve, its divisor class group is its Picard group. Its degree-zero subgroup isThe map $P\mapsto [P-O]$ gives an isomorphism of abelian groupsTo recall why it is a bijection, Riemann-Roch theorem in geometric genus one says every degree-one divisor class has exactly one independent section: a canonical divisor has degree zero, so the dual degree-$-1$ term has no section. Thus the class contains one effective divisor on an algebraic curve of degree one, namely a point $P$, and that point is unique. Translating by $-O$ gives the claimed bijection; this is precisely the chord-and-tangent group law on the elliptic curve.

Choose a prime integer $\ell$ different from $\mathrm{char}k$. The multiplication-by-n morphism $[\ell ]:E\to E$ has derivative $\ell$ times the identity at $O$, so it is nonconstant. A nonconstant morphism of algebraic varieties between smooth projective curves is surjective: its image is closed by proper morphism and cannot have algebraic dimension zero. Since $k$ is an algebraically closed field, this makes $[\ell ]$ surjective on $k$-points. ThereforeYet $E(k)$ is infinite. For each $x\in k$, the affine equation has a solution for $y$ in the algebraically closed field; since $k$ is infinite, this gives infinitely many points.

By the Fundamental theorem of finitely generated abelian groups, a finitely generated abelian group $A$ satisfying $A=\ell A$ has no free part and is a finite torsion group of order prime to $\ell$. The infinite group $E(k)$ cannot have this property while being finitely generated. Thus ${\mathrm{Pic}}^0(E)$ is not finitely generated. A subgroup of a finitely generated abelian group is finitely generated, soThis proof uses divisibility and infinitude, rather than uncountability, so it remains valid over countable algebraically closed fields.

For a smooth projective curve $C$, let $g=h^1(C,{\mathcal{O}}_C)$ be its geometric genus, let $K_C$ be a canonical divisor, and write $\ell (D)=h^0(C,{\mathcal{O}}_C(D))$ for the dimension of a vector space of the Riemann-Roch space. The Riemann-Roch theorem saysfor every divisor on an algebraic curve $D$ on $C$.

Here is the cohomological proof of Riemann-Roch for curves. At a closed point $P$, the local ring is a discrete valuation ring. A local uniformizer shows that increasing the allowed pole order by one gives the short exact sequence of sheavesThe last term is a length-one skyscraper sheaf; its precise identification with $k$ is noncanonical, but its sheaf cohomology has $h^0=1$ and $h^1=0$. All the relevant groups of sheaf cohomology are finite-dimensional because $C$ is a projective variety, and coherent sheaf cohomology above degree one vanishes on a algebraic curve. The long exact sequence in sheaf cohomology therefore givesSince $k$ is an algebraically closed field, every closed point has degree one. Iterating this identity for both positive and negative coefficients of $D$ provesEvery global regular function on a projective variety which is an irreducible variety is constant; hence $h^0(C,{\mathcal{O}}_C)=1$ and $\chi ({\mathcal{O}}_C)=1-g$.

Finally apply Serre duality for a smooth projective curve:Taking dimensions of a vector space and inserting this in the Euler characteristic of a coherent sheaf identity proves the displayed Riemann-Roch theorem. The substantial inputs from sheaf cohomology are finiteness and vanishing for coherent sheaves on a projective curve, and Serre duality; the change of Euler characteristic of a coherent sheaf is proved directly by the point exact sequence. In particular the argument handles negative as well as effective divisors on an algebraic curve.

The Appell–Humbert theorem identifies the positive Hermitian form $H_1$ with a holomorphic line bundle whose First Chern class is represented, in our linear-first convention, byIts integral over the square fundamental domain is $1$, so $\mathrm{deg}L=1$. In particular the oriented Chern pairing is $-E_1(1,i)=1$; using $E_1(1,i)$ without adjusting the Hermitian convention would give the wrong sign.

The complex torus has a nowhere-vanishing differential $dz$, so its canonical divisor is trivial and its genus is one. A nonzero section of a line bundle of negative degree would have an effective divisor of negative degree, which is impossible. Thus $H^0(X,L^{-1})=0$. The Riemann-Roch theorem and Serre duality giveTherefore $\boxed{h^0(X,L)=1}$. The choice of semicharacter of a complex lattice changes the degree-zero component of the line bundle, but not this calculation.

For a smooth projective curve $C$ of geometric genus $g$ and a divisor on an algebraic curve $D$, write $\ell (D)=h^0(C,{\mathcal{O}}_C(D))$ and let $K_C$ be a canonical divisor. The curve form of the Riemann-Roch theorem isHere $\mathrm{deg}{K}_C=2g-2$ and $\ell (K_C)=g$. In line bundle form,The usual curve convention is a smooth geometrically integral projective curve; geometric embedding tests may be checked after extending the ground field to an algebraic closure.

First use the canonical map to obtain the embedding. Its canonical divisor has degree six and $h^0(K_C)=4$. Over an algebraically closed ground field, every effective degree-two divisor on an algebraic curve $E$ satisfies $h^0(E)=1$: a second section would give a nonconstant rational function with poles bounded by $E$, hence a map $C\to {\mathbb{P}}^1$ of degree at most two. Degree one would make $C$ rational, and degree two would make it hyperelliptic, both excluded here.

The Riemann-Roch theorem now givesThus the complete linear system of divisors $|K_C|$ separates every pair of points and every tangent direction, including the tests $E=2p$. By the length-two criterion for a very ample linear system, it is a very ample linear system. Consequently the canonical map embeds $C$ in ${\mathbb{P}}^3$, with degree of a projective curve six. Its image is a nondegenerate projective variety, since the four canonical sections are linearly independent.

The space of homogeneous quadrics in four variables has dimension ten, while the Riemann-Roch theorem givesRestriction therefore has a nonzero kernel, giving a quadric $Q$ containing $C$. This quadric is irreducible: a reducible quadric is a union of two planes, or a double plane, and an integral curve lying in it would lie in a plane, contradicting nondegeneracy.

Similarly, homogeneous cubics form a twenty-dimensional space, whereasThere are at least five independent cubics vanishing on $C$. The multiples of $Q$ by linear forms give only four dimensions, so choose a cubic $F$ vanishing on $C$ which is not divisible by $Q$.

The polynomials $Q,F$ are a regular sequence, hence their intersection $D$ is a projective complete intersection of pure dimension one and degree $2\cdot 3=6$. It contains $C$, which already has degree six. The unmixedness of a complete intersection excludes embedded components. Equal degrees force $C$ to be the only irreducible component and force multiplicity one at its generic point. By generic reducedness with no embedded components, this makes $D$ reduced, so $D=C$ as schemes. We have provedThis is a canonical genus-four curve as a quadric-cubic intersection. The quadric may be smooth or a cone; the argument does not require it to be smooth. The same construction descends over the ground field when the curve is geometrically nonhyperelliptic, since the restriction maps and canonical embedding are defined there.

The Riemann-Roch theorem states, for a divisor $D$ on a smooth projective curve of genus $g$,where $\ell (D)=\dim H^0(C,\mathcal{O}(D))$ and $K_C$ is a canonical divisor, of degree $2g-2$. Negative-degree divisors have no nonzero global sections. The theorem's usual formulation here is over an algebraically closed field, as in the curve course.

For a divisor on an algebraic curve $D$ on a smooth projective curve $C$ of geometric genus $g$, and a canonical divisor $K_C$, the Riemann-Roch theorem states

LetAt a finite ramification point $P_{\alpha }=(\alpha ,0)$, the uniformizer $t=y$ from part b satisfies $x-\alpha =c_{\alpha }{t}^3+O(t^6)$. ConsequentlyAt every other affine point, $x-x(p)$ is a uniformizer and $y$ is a unit, so again ${\nu }_p(\omega )=0$.

At $P_{\infty }$, take the uniformizer $t=u=X/Y$ used above. Since $v=Z/Y=t^4$ times a local unit,It follows that ${\nu }_{P_{\infty }}(dx)=-4$ and ${\nu }_{P_{\infty }}(y^2)=-8$, so the valuation of a rational differential isThuswhose degree four agrees with $2g_X-2$. For the resulting canonical divisor $K_X=4P_{\infty }$, the functions $1,x,y$ have pole orders $0,3,4$ at $P_{\infty }$ and no other poles. They therefore belong to the canonical Riemann-Roch space. Since $\ell (K_X)=g_X=3$, they form a basis:Equivalently, multiplying by $\omega$ gives the basisof holomorphic differential forms, also matching the general description of holomorphic differentials on a smooth plane curve.

For any smooth projective curve of genus $g$, a canonical divisor has degree $2g-2$. If this plane curve had genus two, part (ii) would implyso $d^2-3d-2=0$. The quadratic discriminant is $17$, which is not a perfect square, so the equation has no integral solution. Thus a smooth projective plane curve cannot have genus two, as recorded by no smooth plane curve has genus two.

The intersection with a line $\ell$ is a hyperplane section:The canonical bundle of a smooth projective hypersurface formula, obtained from the Adjunction formula, gives for a smooth degree-four curve in ${\mathbb{P}}^2$Thus ${\mathcal{O}}_C(D)\cong K_C$, so the divisor class of $D$ is the canonical divisor class. Therefore $D$ is a canonical divisor. Its degree is also $4=2g-2$, consistently with $g=3$.

Choose a nonconstant rational function $t$ on $C$. Functions in $L(D)$ are integral over a finite-dimensional bounded-pole space over $k(t)$; equivalently, evaluation of sufficiently many principal parts embeds $L(D)$ into a finite-dimensional vector space. Hence $L(D)$ is finite dimensional.

If $E=D+\mathrm{div}(h)$, multiplication by $h$ gives the isomorphism $L(E)\to L(D)$. A canonical divisor is the divisor of a nonzero rational differential. Riemann-Roch theorem says $\ell (D)-\ell (K-D)=\mathrm{deg}D+1-g$. Taking $D=0$ gives $1-g=1+g-\mathrm{deg}K$, hence $\mathrm{deg}K=2g-2$.

For a smooth plane curve of degree $d$, the adjunction formula gives $K_C\sim (d-3)H{|}_C$. A cubic has $d=3$, so $K_C\sim 0$.