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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 $K_C\sim 0$.

Lift $f$ to a doubly periodic meromorphic function on $\mathbb{C}$. Integrate it around a fundamental parallelogram whose boundary avoids all poles. Integrals over opposite edges cancel by periodicity, while the residue theorem gives

Thus the principal-part map from part (b) takes values in the codimension-one hyperplane on which the sum of residue coefficients is zero. The Mittag--Leffler existence criterion on a compact Riemann surface says that prescribed principal parts occur precisely when their residues pair trivially with every holomorphic one-form. On a complex torus the holomorphic one-forms are the scalar multiples of $dz$, so the single condition is exactly the displayed residue sum. Hence the image has dimension
${\sum }_i{m}_i-1$, and the kernel of constants has dimension one. For $n\ge 1$,Equivalently, this is the genus-one case of the Riemann-Roch theorem for a positive divisor.

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 smooth projective curve $C$ of geometric genus one, the Riemann-Roch theorem saysThe canonical divisor has degree zero and is principal because a nonzero regular differential has no zeros. Hence $K_C\sim 0$, so equivalentlyIn particular, $\ell (D)=\mathrm{deg}D$ when $\mathrm{deg}D>0$.

The group ${\mathrm{Pic}}^0(E)$ is the group of degree-zero divisor classes on $E$, with addition induced by addition of divisors. ConsiderFor surjectivity, let $D$ have degree zero. Since $\mathrm{deg}(D+(O_E))=1$, Riemann--Roch gives $\ell (D+(O_E))=1$. A nonzero element of this space makes $D+(O_E)$ linearly equivalent to an effective divisor of degree one, necessarily $(P)$ for some point $P$. Thus $[D]=[(P)-(O_E)]$.

For injectivity, suppose $(P)-(O_E)$ is a principal divisor. If $P\ne O_E$, its defining function would be nonconstant and would have at most one simple pole, whereas Riemann--Roch gives $\ell ((O_E))=1$, so every such function is constant. Therefore $P=O_E$, and $\iota$ is a bijection.

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 Riemann-Roch theorem statesOn a genus-one curve $K\sim 0$. If $D$ has degree zero, then $D+P_0$ has degree one, and Riemann--Roch gives $\ell (D+P_0)=1$. Its unique nonzero section has an effective divisor of degree one, say $P$, soIf also $D\sim Q-P_0$, then $P\sim Q$; a nonconstant function with divisor $P-Q$ would define a degree-one map to ${\mathbb{P}}^1$, forcing the genus to be zero. Hence $P=Q$.

This identifies $X$ with ${\mathrm{Pic}}^0(X)$ by $P\mapsto [P-P_0]$ and defines $P+Q=R$ throughFor the stated cubic, the line through $A=(2,1)$ and $B=(-2,-1)$ is $y=x/2$. Its third intersection is $(1/4,1/8)$, so reflection in the $x$-axis gives

Because $P_0$ is an inflection point, a line section is linearly equivalent to $3P_0$. A point $P$ is an inflection point exactly when some line has intersection divisor $3P$, equivalentlyUnder the group-law identification this is precisely $3(P-P_0)=0$, so the inflection points are exactly the three-torsion points.

Apply the Riemann-Roch theorem to the divisors $nO$. Since the genus is one and the canonical divisor is trivial, $\ell (nO)=n$ for $n\ge 1$. ChooseThen $x$ and $y$ have exact pole orders two and three at $O$. The seven functionslie in the six-dimensional space $L(6O)$, so they satisfy one relation. Comparing pole orders and completing squares and cubes gives a nonsingular Weierstrass equation of an elliptic curveThe functions $1,x,y$ define the morphism $P\mapsto (x(P):y(P):1)$ away from $O$. Their pole orders show that it extends with $O\mapsto (0:1:0)$. It has degree one and is therefore an isomorphism of smooth projective curves.

A smooth plane curve of degree $m$ has genusThe line bundle ${\mathcal{O}}_X(D)(t)$ has degree $d+mt$. The Riemann-Roch theorem therefore gives

For a divisor $D$ on a smooth projective curve $X$, defineThe Riemann-Roch theorem stateswhere $K_X$ is the divisor of a nonzero rational differential, $g$ is the genus of $X$, and $\mathrm{deg}D$ is the sum of the divisor coefficients because the ground field is algebraically closed.

Under the coordinate change $u=1/x$ and $v=y/x^n$, we haveHenceThis is a polynomial, and its value at zero is the nonzero leading coefficient of $f$. Thus the even-degree hyperelliptic model has two points $P_{+},P_{-}$ above $u=0$, distinguished by the two values of $v$.

Consider the rational differentialAt a finite branch point $x=a$, square-freeness gives a local parameter $y$, with $x-a$ a nonzero constant times $y^2$ to first order. Hence $dx/y$ is regular and nonzero there. It is also regular and nonzero at finite unramified points.

At infinity,soBecause $v(P_{\pm })\ne 0$ and $u$ is a local parameter at each point, $\omega$ has a zero of order $n-2$ at both. Therefore the canonical divisor of an even-degree hyperelliptic curve isand

Since $\mathrm{deg}{K}_X=2g-2$, it follows that

The function $x$ has a simple pole at each of $P_{+}$ and $P_{-}$. ConsequentlyThese $n-1$ functions are linearly independent. Riemann-Roch with $D=K_X$ gives $\ell (K_X)=g=n-1$, so

With respect to this basis, the canonical map of a hyperelliptic curve isIt identifies $(x,y)$ with $(x,-y)$ for general $x$, and hence is not an embedding.

If $X$ embedded in ${\mathbb{P}}^2$, its smooth plane image would have degree $d\ge 4$: degrees one and two have genus zero, and degree three has genus one, whereas $g=n-1\ge 2$. But the canonical map of a smooth plane curve of degree $d\ge 4$ is an embedding by adjunction, contradicting the preceding calculation. Therefore

Riemann-Roch theorem is $\ell (D)-\ell (K-D)=\mathrm{deg}D+1-g$. Putting $D=0$ and using $\ell (0)=1$, $\ell (K)=g$ gives $\mathrm{deg}K=2g-2$. If an effective degree-two divisor has $\ell (D)=2$, its pencil defines a degree-two map to ${\mathbb{P}}^1$, so the curve is hyperelliptic. For $g=2$, Riemann-Roch applied to $K$ supplies such a degree-two pencil. For nonhyperelliptic genus 3, the canonical system has dimension 2, separates points and tangents, and hence embeds $C$ as a plane quartic. On a genus-3 hyperelliptic curve, $\ell (2Q)=2$ exactly at the $2g+2=8$ ramification points.

Let $E$ be a genus one curve and choose $O\in E$. Since a canonical divisor is linearly equivalent to zero, Riemann--Roch gives $\ell (nO)=n$ for $n\ge 1$. Choose $x\in L(2O)\setminus L(O)$ and $y\in L(3O)\setminus L(2O)$. Their pole orders at $O$ are two and three. The seven functionsare linearly dependent because $\ell (6O)=6$, and comparison of pole orders puts the relation in Weierstrass form. The complete linear system $|3O|$ is very ample and realizes this relation as a smooth plane cubic. In characteristic other than two and three, completing the square and translating $x$ produce a Short Weierstrass form.

At every point of a smooth projective curve of genus $g$, exactly $g$ positive integers are Weierstrass gap. Indeed, each quotient $L(np)/L((n-1)p)$ has dimension zero or one, while Riemann--Roch gives $\ell (np)=n+1-g$ for all $n>2g-2$.