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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.

Work first where all three coordinates are finite and the joining line is not vertical. By the chord-and-tangent group law, the third intersection of $y=\lambda x+\mu$ with the elliptic curve is $-(P_1+P_2)$, whose first coordinate is also $x_3$. Substitution gives the monic polynomialIts roots are $x_1,x_2,x_3$, counted with intersection multiplicity. Comparing its coefficients with the elementary symmetric polynomials provesThis symmetric-coordinate identity for elliptic-curve addition is an identity of rational functions. It includes tangent cases by specialization. Literally assigning finite $s_i$ when a point is $O$ would be meaningless; such cases must be interpreted on the projective curve. A nonsingular equation in the printed short form has characteristic different from two.

Here is how the identity yields the quadratic form for the degree of an isogeny. With $u=x(P)$, $v=x(Q)$, write $s=u+v$ and $q=uv$. As an equation for $t=x(P+Q)$, it becomesBoth $x(P+Q)$ and $x(P-Q)$ satisfy it, because replacing $Q$ by $-Q$ does not change $v$. Thus its two branches are the sum and difference maps. Their common denominator is the square of $u-v$. The following divisor on an algebraic curve calculation keeps track of cancellations and exceptional points rigorously, instead of assuming degrees of displayed numerators always add.

On $E_2\times E_2$, equality of the two first coordinates means $Q=P$ or $Q=-P$. Each coordinate has a double pole at $O$. Consequently the principal divisor identity iswhere $D_{+}$ is the diagonal and $D_{-}$ is the graph of negation. Equivalently, the two zero divisors are the inverse images of $O$ under difference and sum, exactly the two branches identified above. Their generic multiplicities are one; the divisor identity also accounts for their intersections at 2-torsion points.

For nonzero isogenies of elliptic curves $\varphi ,\psi :E_1\to E_2$ with $\varphi \ne \pm \psi$, pull this identity back by $(\varphi ,\psi )$. Taking degrees of the resulting principal divisor on $E_1$ gives the divisor proof of the degree parallelogram law:If one map is zero, this is immediate. If $\varphi =\pm \psi$, it follows from $\mathrm{deg}([2]\circ \varphi )=4\mathrm{deg}\varphi$. This also proves the parallelogram law for a general Weierstrass equation of an elliptic curve in characteristic two: the same divisor identity uses the quotient by negation, even though the particular symmetric-coordinate formula above is unavailable.

Put $q(\varphi )=\mathrm{deg}\varphi$ and $q(0)=0$. Applying the parallelogram law to $n\varphi ,\varphi$ gives the recurrence $q((n+1)\varphi )+q((n-1)\varphi )=2q(n\varphi )+2q(\varphi )$, and hence $q(n\varphi )=n^{2}q(\varphi )$ for every integer $n$. Its polarization $B(\varphi ,\psi )=q(\varphi +\psi )-q(\varphi )-q(\psi )$ is symmetric and integer-valued. For fixed $\psi$, the parallelogram law makes $B(\cdot ,\psi )$ odd and gives $B(x+y,\psi )+B(x-y,\psi )=2B(x,\psi )$. Interchanging $x,y$ and adding proves additivity, since integers have no two-torsion. Thus $B$ has bilinearity, and $q$ is a positive quadratic form on $\mathrm{Hom}(E_1,E_2)$, with $q(\varphi )>0$ whenever $\varphi \ne 0$.

For a nonsingular projective integral curve, apply the Riemann-Roch theorem to the divisor on an algebraic curve $mP$. It gives $\ell (mP)\ge m\mathrm{deg}P+1-g$, where $g$ is the geometric genus. For sufficiently large $m$, this exceeds the dimension of the space of constant functions; over the usual geometrically integral convention that dimension is one. Thus $L(mP)$ contains a nonconstant rational function.

By the definition of $L(mP)$, the principal divisor on an algebraic curve of that function plus $mP$ is effective, so it has no pole away from $P$. Therefore there is a nonconstant rational function regular on $X\setminus \{P\}$. If $P$ is not a rational point, using its closed-point degree in the same inequality handles that case. A positive pole order at $P$ is unavoidable, since regular functions on a projective integral curve are constant in its constant field.

First consider the preliminary cohomological facts. The exam's term flabby means flasque sheaf: every restriction map is surjective. For open sets $V\subseteq U$, lift a section of ${\mathcal{F}}_3(V)$ to ${\mathcal{F}}_2(V)$ using the given surjectivity when ${\mathcal{F}}_1$ is flasque. Extend that lift to $U$ using the flasque sheaf ${\mathcal{F}}_2$, then project to ${\mathcal{F}}_3(U)$. This proves that ${\mathcal{F}}_3$ is a flasque sheaf.

One construction of a flasque resolution starts with the sheaf ${\mathcal{C}}^0(\mathcal{F})(U)={\prod }_{P\in U}{\mathcal{F}}_P$, with pointwise restriction. It is a sheaf of abelian groups with surjective restrictions, and sending a section to all its germs embeds $\mathcal{F}$ in it. Repeating on the cokernels constructs an exact sequencewith each ${\mathcal{I}}^j$ flasque. Define sheaf cohomology by the cohomology of the cochain complex of global sections of such a resolution; the groups are independent of the chosen flasque resolution. If $\mathcal{F}$ is itself flasque, the quotient result just proved makes every successive cokernel flasque. Applying the assumed surjectivity of sections to each successive short exact sequence of sheaves proves exactness of the global-section complex in positive degrees. Thus $H^i(X,\mathcal{F})=0$ for $i>0$.

Now let $K=k(X)$ be the function field of the smooth algebraic curve. Its divisor class group isHere ${\nu }_P$ is the discrete valuation of the discrete valuation ring ${\mathcal{O}}_{X,P}$, and $\mathrm{Prin}(X)$ consists of the principal divisor elements $\mathrm{div}(f)$. These sums have finite support: on a finite affine cover, represent $f$ as a fraction of regular functions; each nonzero numerator or denominator has only finitely many zeros on a curve, since a proper closed subset of a Noetherian one-dimensional irreducible space is finite.

For a divisor on an algebraic curve $D={\sum }_P{n}_{P}P$, define the line bundle associated to a divisor byIn this expression the valuation condition applies to nonzero $f$. If $t_P$ is a uniformizer, the stalk is $t_P^{-n_P}{\mathcal{O}}_{X,P}$. Shrinking around $P$ removes all other zeros and poles of $t_P$ and all other points in the support of $D$, so this also gives an actual local generator. Thus ${\mathcal{O}}_X(D)$ is a locally free sheaf of rank one. Multiplication gives an isomorphismon every stalk. If $D=\mathrm{div}(g)$, then ${\mathcal{O}}_X(D)=g^{-1}{\mathcal{O}}_X$, so the construction factors through a homomorphism $\mathrm{Cl}(X)\to \mathrm{Pic}(X)$.

It is injective: if ${\mathcal{O}}_X(D)$ is trivial, an isomorphism from ${\mathcal{O}}_X$ supplies a nonzero global rational generator $h$. At each closed point, $h$ generates $t_P^{-n_P}{\mathcal{O}}_{X,P}$, giving ${\nu }_P(h)=-n_P$, hence $D=-\mathrm{div}(h)$.

It is surjective: take a line bundle $\mathcal{L}$ and a nonzero rational section $s$, obtained by choosing a frame on any nonempty trivializing open set. If $e_i$ is a local frame, write $s=f_i{e}_i$ with $f_i\in K^{\ast }$. Since changes of frame are units, ${\nu }_P(f_i)$ is independent of the frame near $P$. A finite trivializing cover shows that these numbers have finite support. Set $D={\sum }_P{\nu }_P(f_i)P$. The map $f\mapsto fs$ gives ${\mathcal{O}}_X(D)\cong \mathcal{L}$, because at $P$ the condition ${\nu }_P(f)+{\nu }_P(f_i)\ge 0$ is exactly regularity of $fs$. Replacing $s$ by $gs$ changes $D$ by $\mathrm{div}(g)$, so this construction is well defined on classes. This proves the divisor class group and Picard group of a smooth curve identification

Put $\mathcal{Q}={\mathcal{K}}_X^{\ast }/{\mathcal{O}}_X^{\ast }$, the quotient in sheaves of abelian groups, where ${\mathcal{K}}_X^{\ast }$ is the sheaf of nonzero rational functions on an irreducible variety and ${\mathcal{O}}_X^{\ast }$ is the sheaf of units of the structure sheaf. At a closed point, the discrete valuation induces an isomorphismThe kernel consists precisely of the units of the discrete valuation ring, and surjectivity follows by taking powers of a uniformizer.

A global section of $\mathcal{Q}$ is locally represented by nonzero rational functions. Since $X$ has a finite affine cover and is Noetherian, it is quasi-compact, so finitely many such representatives suffice. Each has only finitely many zeros and poles; consequently the valuations of the section define a finite divisor on an algebraic curve. A section with zero valuation at every closed point is zero in every stalk and therefore zero. This gives an injective mapIt is surjective as well. For $D={\sum }_P{n}_{P}P$, take a rational uniformizer $t_P$ at each point of its finite support. On a sufficiently small neighborhood of $P$, the function $t_P^{n_P}$ has the prescribed divisor there: remove the other points in its zero and pole set and the other support points of $D$. On the complement of the support use the rational function $1$. On overlaps these representatives have the same valuations at every point, so their ratios are units everywhere on the overlap, and their classes in $\mathcal{Q}$ agree. The sheaf gluing axiom now supplies a global section. ThusSince $X$ is irreducible, $\Gamma (X,{\mathcal{K}}_X^{\ast })=K^{\ast }$. Under this identification the map to $\Gamma (X,\mathcal{Q})$ is $f\mapsto \mathrm{div}(f)$. Its image is exactly the group of principal divisor elements, and hence

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

For a divisor on an algebraic curve $D$, define its Riemann-Roch space byThen

For a divisor on an algebraic curve $D={\sum }_p{a}_{p}p$ over an algebraically closed field, its degree isFor a nonzero rational function $f$, its principal divisor iswhere zeros have positive order and poles have negative order.

Now let $D={\sum }_i{a}_i{p}_i$ have degree zero on the projective line. Choose a homogeneous linear form $L_i(X_0,X_1)$ whose zero is $p_i$. Since ${\sum }_i{a}_i=0$, the expressionis homogeneous of degree zero and therefore defines a rational function on ${\mathbb{P}}^1$. Each $L_i$ has one simple zero, and the cancellation of total degree removes any common scaling ambiguity, soThus every degree-zero divisor on ${\mathbb{P}}^1$ is principal. If $E$ and $E^{\prime }$ have the same degree, then $E-E^{\prime }$ is principal, whenceThis is the divisor class on the projective line.

Write $p_{\infty }=[1:0]$. The coordinate $t=X_0/X_1$ has no critical zero or pole in the finite chart. Near infinity use $s=X_1/X_0=t^{-1}$; thenHence the rational differential has one double pole at infinity and no other zero or pole:which represents the canonical divisor of the projective line.

If $D\sim m{K}_{{\mathbb{P}}^1}$, then $D\sim -2m{p}_{\infty }$, and linear equivalence of divisors gives $L(D)\cong L(-2m{p}_{\infty })$. A function in the latter space has no finite poles, so it is a polynomial, and its degree is at most $-2m$. ThereforeThis calculation uses only the rational functions on ${\mathbb{P}}^1$, rather than the Riemann-Roch theorem.

Finally, suppose distinct $p,q\in C$ satisfy $p-q=(f)$. The function $f$ has one simple pole and therefore defines a degree-one finite morphism $f:C\to {\mathbb{P}}^1$. A degree-one finite morphism between smooth projective curves is an isomorphism. It would follow that $C\cong {\mathbb{P}}^1$ and hence that $C$ has genus zero, contrary to the hypothesis. Thus the principal divisor with one simple zero and one simple pole cannot occur on $C$:

For a divisor on an algebraic curve $D$ on a smooth projective curve $X$ of genus $g$, the Riemann-Roch theorem stateswhere $K_X$ is a canonical divisor. Taking $D=0$ gives $1-\ell (K_X)=1-g$, so $\ell (K_X)=g$. Taking $D=K_X$ then gives

To obtain a uniform projective embedding, choose a divisor $D$ of degree $2g+1$. Since every divisor appearing below has degree greater than $2g-2$, Riemann--Roch givesThe first two equalities show that the complete linear system of a divisor $|D|$ has no base point. The strict drops in the last two comparisons show respectively that its sections separate distinct points $P,Q$ and tangent directions at $P$. Thus $D$ is a very ample divisor, in accordance with the general fact that a high-degree divisor is very ample on a smooth projective curve, and its sections define a closed embeddingThe ambient dimension therefore depends only on $g$.

The Riemann-Hurwitz formula for a nonconstant morphism $h:C_1\to C_2$ of degree $n$ isChoose a smooth plane quartic $C$, so $g(C)=3$, and form the product of projective varieties $S=C\times C$. This is a smooth projective variety of dimension two. If $B\subset S$ is an irreducible curve, pass to its normalization $\tilde{B}$. At least one coordinate projection $\tilde{B}\to C$ is nonconstant, since otherwise $B$ would be a point. For that projection, Riemann--Hurwitz givesso the geometric genus of $B$ is at least three. Hence $S$ is the required surface; this is the product surface without low-genus curves construction.

Finally let $X\subset {\mathbb{P}}^2$ be a smooth plane curve of degree $d$ and let $p\notin X$. After a projective change of coordinates, take $p=[0:0:1]$. Projection away from $p$ isIts two homogeneous coordinate functions cannot vanish simultaneously on $X$, because their common zero in ${\mathbb{P}}^2$ is $p$. The criterion for a morphism of algebraic varieties therefore shows that the restrictionis a morphism. A fibre is the intersection with a line through $p$, and a general such line meets $X$ in $d$ points counted with multiplicity. Thus the projection of a plane curve from an exterior point has degree $d$.

By the genus of a smooth plane curve, $g(X)=(d-1)(d-2)/2$. Applying Riemann--Hurwitz to $\pi$ and its ramification divisor $R_{\pi }$ givesEvery ramification point contributes at least one to this degree, so

The units in the structure sheaf of a scheme form a multiplicative sheaf of abelian groups. On a smooth algebraic curve, a nonzero rational function is a unit at a point precisely when its discrete valuation there is zero. Its inclusion in the sheaf of nonzero rational functions on an irreducible variety relates divisor on an algebraic curve to the Picard group through the long exact sequence in sheaf cohomology.