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An algebraic curve is a one-dimensional algebraic variety. On a smooth algebraic curve, the local ring at every closed point is a discrete valuation ring, which allows orders of zeros and poles to be recorded as a divisor on an algebraic curve.

At a point of a smooth algebraic curve, its local ring is a discrete valuation ring. A local parameter is a generator $t$ of its maximal ideal, equivalently an element of valuation one. It measures the order of a zero of a function or pole by expressing functions as powers of $t$ times units. Over the complex numbers its analytic counterpart is a local coordinate vanishing simply at the point. For an elliptic curve in a Weierstrass equation of an elliptic curve, $t=-x/y$ is the standard parameter at the identity.

At every point of a smooth algebraic curve, the local ring is a discrete valuation ring. Its valuation measures the order of vanishing of a rational function at that point.

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.

Write $g_2=60G_4$ and $g_3=140G_6$, with the lattice sums taken over nonzero lattice elements. Reindexing the normally convergent series shows that the Weierstrass elliptic function $\wp$ is an even function, and consequently ${\wp }^{\prime }$ is an odd function. Expanding $(z-w{)}^{-2}$ near zero and cancelling the odd lattice sums gives the Laurent coefficients of the Weierstrass elliptic functionThus the elliptic function ${\wp }^{\prime 2}-4{\wp }^3+60G_4\wp$ has no pole at zero and its constant term is $-140G_6$. By periodicity it has no poles anywhere. An entire elliptic function is bounded on a fundamental parallelogram and hence everywhere, so Liouville theorem makes it constant. This proves the Weierstrass elliptic differential equation

We first verify that the cubic is smooth, rather than presupposing it when using its group law. There are three nonzero two-torsion points of a complex torus $h_j\in (\frac{1}{2}\Lambda )/\Lambda$. Oddness and periodicity imply ${\wp }^{\prime }(h_j)=0$. Since ${\wp }^{\prime }$ has a triple pole and no other poles on the torus, value multiplicity of an elliptic function says that these three distinct zeros exhaust its zeros and are simple. Put $e_j=\wp (h_j)$. Each $\wp -e_j$ has a double zero at $h_j$, since ${\wp }^{\prime }(h_j)=0$ and the zero of ${\wp }^{\prime }$ is simple. If $e_i=e_j$ with $i\ne j$, this same function would have at least four zeros counted with multiplicity, although its only pole is double. Therefore the half-period values of the Weierstrass elliptic function are distinct. They are roots of $4x^3-g_{2}x-g_3$, so all its roots are distinct and the affine curve is a smooth algebraic curve. Its projective closure isIt has the single point $O=(0:1:0)$ at infinity. The derivative with respect to $Z$ of the defining homogeneous polynomial is nonzero at $O$, proving smoothness there too.

For any finite value $a$, value multiplicity of an elliptic function gives exactly two solutions of $\wp (z)=a$, counting multiplicities. Evenness supplies the pair $z,-z$. At a nonzero half-period these coincide and the zero is double; elsewhere they are distinct. An even elliptic function $f$ therefore descends to a meromorphic function of $x=\wp (z)$ on the Riemann sphere. To justify descent at a half-period, put $z=h+u$: both $f(h+u)$ and $\wp (h+u)$ are even in $u$, and $\wp (h+u)-\wp (h)$ is $u^2$ times a nonvanishing even analytic function. The local Laurent series of $f$ is consequently meromorphic in the coordinate $x-e$. At zero the same argument uses $1/x=z^2+O(z^6)$. Every meromorphic function on the Riemann sphere is a rational function, proving that even elliptic functions are rational in the Weierstrass function.

For an arbitrary elliptic function $f$, its even part $(f(z)+f(-z))/2$ is $A(\wp )$. Its odd part divided by ${\wp }^{\prime }$ is an even meromorphic function, hence is $B(\wp )$ with $B$ rational. We have therefore proved the elliptic function-field decompositionUniqueness follows by taking even and odd parts. In particular the field of elliptic functions is $\mathbb{C}(\wp ,{\wp }^{\prime })$.

Define $\Phi (0)=O$ and $\Phi (z)=(\wp (z),{\wp }^{\prime }(z))$ elsewhere on $\mathbb{C}/\Lambda$. Periodicity makes the map well defined. Near zero, the projective chart $Y\ne 0$ has coordinates $X/Y=\wp /{\wp }^{\prime }=-z/2+O(z^5)$ and $Z/Y=1/{\wp }^{\prime }=-z^3/2+O(z^7)$, so the extension is holomorphic. The fiber description above proves injectivity: the two candidates $z,-z$ are distinguished by ${\wp }^{\prime }$ unless they coincide at a half-period. It also proves surjectivity, since every $x$ occurs and the two derivatives are the two allowed $y$ values, with $y=0$ at a branch value.

Finally, a nonvertical line $y=ax+b$ pulls back to the elliptic function ${\wp }^{\prime }-a\wp -b$, whose only pole is triple at zero. Its three zeros $z_1,z_2,z_3$, counted with intersection multiplicity, satisfy $z_1+z_2+z_3=0$ modulo $\Lambda$ by the zero-pole sum of an elliptic function. On the cubic, reflection $y\mapsto -y$ corresponds to $z\mapsto -z$. The chord-and-tangent group law therefore gives $\Phi (z_1)+\Phi (z_2)=\Phi (-z_3)=\Phi (z_1+z_2)$. Vertical lines give inverse pairs; tangencies are included through repeated zeros, and the identity cases follow from the extension at zero. This completes the Weierstrass uniformization by half-period values: $\Phi$ is a group isomorphism.

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

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.