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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.

A line subbundle is an injective map from a line bundle into a vector bundle whose quotient is locally free. Equivalently the injection splits locally. On a smooth algebraic curve, saturation of a rank-one coherent subsheaf produces a saturated line subbundle on a smooth 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.

For an affine plane algebraic curve $F(x,y)=0$, a point over an algebraic closure is nonsingular when the displayed condition holds. Its Zariski tangent space has dimension one, and a local coordinate can be chosen in the direction of a nonzero partial derivative. A smooth algebraic curve has all points nonsingular. Over a complete local field, the Hensel lemma lifts a nonsingular residue point by solving for a coordinate with unit partial derivative.

For each $x\in {\mathbb{F}}_p$, count the roots of $y^2=x^3+x^2+x+1$, and add the identity at infinity. A nonzero quadratic residue gives two roots, zero gives one, and a nonsquare gives none. Equivalently, with the Legendre symbol extended by ${\chi }_p(0)=0$,The numbers of affine $y$ values, in increasing $x$ order starting at zero, areso the answer isAll four reductions are smooth algebraic curves: the original PDF gives the elliptic-curve discriminant $-2^8$, so these odd primes are primes of good reduction.

An elliptic curve over $\mathbb{Q}$ has good reduction at $p$ if it admits an integral Weierstrass equation of an elliptic curve over ${\mathbb{Z}}_p$ whose reduced projective cubic is a smooth algebraic curve over ${\mathbb{F}}_p$. Equivalently, its Minimal Weierstrass equation has a unit elliptic-curve discriminant, or its smooth proper model is an elliptic curve over the valuation ring. The existence of an appropriate integral model matters: a nonminimal equation can have discriminant divisible by $p$ even when the curve has good reduction. Equivalently,

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.

Let $C$ be the smooth algebraic curve. A nonzero regular section of a line bundle has an effective zero divisor, so its degree of a line bundle is nonnegative. Degrees add under tensor products and change sign on taking the dual. We use genuine line subbundles, so their quotients in $E$ are locally free. Both bounds can be proved constructively.

For the upper bound, take a rational frame of $E^{\ast }$ over the function field of $C$. Its two members are rational sections with only finitely many poles. Choose an effective divisor $D$ large enough to bound the pole orders of both rational sections. They become global sections of $E^{\ast }\otimes {\mathcal{O}}_C(D)$ and define a sheaf mapwhich is an isomorphism at the generic point. It is therefore injective as a sheaf map: a section in its kernel would vanish generically in a locally free sheaf, which is a torsion-free sheaf. For any line subbundle $L\subset E$, at least one of its two component maps into ${\mathcal{O}}_C(D)$ is nonzero. Such a map is a nonzero global section of ${\mathcal{O}}_C(D)\otimes L^{-1}$. Its zero divisor on an algebraic curve is effective, soThis upper degree bound for line subbundles on a smooth curve depends only on a choice made for $E$, and applies to every one of its line subbundles.

For the lower-bound question we will actually construct subbundles ${\mathcal{O}}_C(-m{p}_0)$ for every sufficiently large integer $m$, where $p_0$ is a fixed point. First prove the globally generated vector bundle property rather than assuming that negative-degree inclusions automatically are subbundles.

Apply the upper-bound argument to the rank-two vector bundle $F=E^{\ast }\otimes K_C$, where $K_C$ is the canonical bundle. Let $B$ be its upper degree bound. Choose $m$ with $m-1>B$. For any point $p\in C$, a nonzero global section of $F(-m{p}_0+p)$ would give a nonzero mapIts image can be saturated to a saturated line subbundle on a smooth curve $M\subset F$. Locally the local ring of a smooth algebraic curve is a discrete valuation ring: dividing out the common power of a uniformizer makes the image a primitive rank-one summand, and its quotient is free. Thus the saturation is indeed a subbundle. The original map gives a nonzero global section of $M\otimes {\mathcal{O}}_C(-m{p}_0+p)$ and consequently $\mathrm{deg}M\ge m-1$, contradicting the upper bound. Hence $H^0(F(-m{p}_0+p))=0$ for every $p$. By Serre duality,The exact sequenceand its long exact sequence in sheaf cohomology show that evaluation $H^0(E(m{p}_0))\to E(m{p}_0){|}_p$ is onto for every $p$. This proves the global generation of positive twists on a smooth curve with one bound on $m$ that works at every point.

Write $W=H^0(E(m{p}_0))$ and $N=\dim W$. The evaluation map $W\otimes {\mathcal{O}}_C\to E(m{p}_0)$ is surjective with kernel bundle of rank $N-2$. Consider the incidence subsetIt is the total space of this kernel bundle, so $\dim Z=1+(N-2)=N-1$. Its projection into $W$ is closed because $C$ is proper, and its image has dimension at most $N-1<N$. Choose $s$ outside this proper closed subset. It vanishes nowhere. In a local trivialization at any point, one component of $s$ is a unit, so it can be completed to a basis. Therefore ${\mathcal{O}}_C\stackrel{s}{\to }E(m{p}_0)$ is a subbundle with a locally free rank-one quotient. Twisting back givesThis proves the unbounded negative degrees of line subbundles. The nowhere-vanishing construction guarantees a subbundle at every point, including the zeros that might otherwise have occurred in a generic meromorphic inclusion. If the curve has several connected components, apply these constructions on each component and add the degrees; the same upper-bound and no-lower-bound conclusions follow.

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.

Let $V\subseteq U$ be open subsets, and let $s\in {\mathcal{F}}_3(V)$. The assumed flasque-kernel section-lifting lemma lifts $s$ to $\tilde{s}\in {\mathcal{F}}_2(V)$. Since ${\mathcal{F}}_2$ is a flabby sheaf, extend $\tilde{s}$ to a section on $U$, then project to ${\mathcal{F}}_3(U)$. Its restriction is $s$. Thus every restriction map of ${\mathcal{F}}_3$ is surjective, proving that the quotient of two flabby sheaves in a short exact sequence is flabby.

For the construction of sheaf cohomology, embed a sheaf of abelian groups $\mathcal{F}$ into the sheafby sending a section to all its germs. This is injective because a section with every germ zero is zero. Arbitrary families of stalk elements glue point by point, and restriction maps are projections; hence $C(\mathcal{F})$ is a flabby sheaf. Set ${\mathcal{Q}}^0=\mathcal{F}$, ${\mathcal{I}}^j=C({\mathcal{Q}}^j)$ and ${\mathcal{Q}}^{j+1}={\mathcal{I}}^j/{\mathcal{Q}}^j$. Compose the quotient map with the next embedding to obtain the flasque resolutionThe sheaf cohomology groups are the cohomology groups of the resulting cochain complex of global sections:The usual comparison of resolutions identifies this construction with the right derived functors of global sections; the construction above is the Godement resolution.

If $\mathcal{F}$ is itself a flabby sheaf, the quotient argument shows inductively that every ${\mathcal{Q}}^j$ is flabby. The assumed lifting result therefore makesan exact sequence for every $j$. The global-section complex is consequently exact in every positive degree. In particular,This establishes the vanishing from the section-lifting property rather than assuming it in advance.

For the smooth algebraic curve $X$, a divisor on an algebraic curve is a finite formal sum $D={\sum }_P{n}_{P}P$ of closed points with integer coefficients. At each closed point, the local ring of a smooth algebraic curve is a discrete valuation ring; write its normalized discrete valuation as $v_P$. For $f\in k(X{)}^{\ast }$ the principal divisor is $\mathrm{div}(f)={\sum }_P{v}_P(f)P$. Only finitely many coefficients are nonzero: zeros and poles are proper closed subsets in the Noetherian Zariski topology of a curve. The divisor class group is the abelian groupEquivalently, its elements are divisors on an algebraic curve modulo linear equivalence of divisors.

Write ${\Omega }_C^1={\Omega }_{C/k}^1$. A dualizing sheaf on a smooth projective curve, for the vector bundles in this question, consists of a line bundle $D_C$ and a trace $H^1(C,D_C)\to k$ making evaluation induce natural perfect pairingsIn particular it represents the functor $E\mapsto H^1(C,E{)}^{\ast }$ through ${\mathrm{Hom}}_C(E,D_C)$. We construct it by reducing to the proved residue duality on the projective line.

Choose a separating nonconstant rational function $t$ on $C$. Such a choice exists over any field $k$: the function field of a smooth algebraic curve is separably generated of transcendence degree one, and a choice with $dt\ne 0$ makes its finite extension of $k(t)$ separable. The map $(t:1)$ extends across poles using $(1:t^{-1})$ in the discrete valuation ring at each point. It gives a finite generically separable morphismIndeed it is nonconstant, hence has finite fibres, and is proper; a proper quasi-finite morphism is finite. This finite map is flat: locally its direct-image module is finite and torsion-free over a discrete valuation ring, and hence free. The same argument works componentwise if needed.

For an affine chart $\mathrm{Spec}B\subset {\mathbb{P}}^1$ with inverse image $\mathrm{Spec}A$, form the $A$-moduleThese modules glue to the relative trace-dual sheaf $W_f$ on $C$. There are two algebraic identities used here. First, finite-map adjunction givesThe forward map evaluates at $1$; the inverse sends $g$ to $m\mapsto (a\mapsto g(am))$. Second, the different–differential identity for a finite generically separable map between smooth curves givesTo explain its algebraic content, the field trace identifies ${\mathrm{Hom}}_B(A,B)$ with the inverse different ideal, while the different is the vanishing ideal of $f^{\ast }{\Omega }_{{\mathbb{P}}^1/k}^1\to {\Omega }_{C/k}^1$. These identities include wild ramification; one must not replace the different by just ramification index minus one. This is the identity allowed to be quoted in the question; the finite-map description is recorded in Stacks Project, finite morphisms, and its differential form in Stacks Project, Section 53.12.

Set $D_C=W_f\otimes f^{\ast }{\Omega }_{{\mathbb{P}}^1/k}^1$. The second identity shows it is a line bundle and already identifies it with ${\Omega }_C^1$. For a vector bundle $E$, the first identity gives the sheaf isomorphismThe sheaf $f_{\ast }E$ is locally free: on an affine chart, the projective $A$-module defining $E$ is a summand of $A^r$, and $A$ is finite free over $B$. Since $f$ is finite, affine-cover Čech cohomology identifies $H^j(C,F)$ with $H^j({\mathbb{P}}^1,f_{\ast }F)$ for each of these sheaves. Apply the already established duality on ${\mathbb{P}}^1$ to $f_{\ast }E$. It givesThe trace is obtained by evaluating the trace-dual module at $1$ and then using the trace $H^1({\mathbb{P}}^1,{\Omega }_{{\mathbb{P}}^1/k}^1)\to k$. The adjunction formulas show the displayed isomorphisms are induced by evaluation against this trace and are natural in $E$. They thus establish the requested dualizing property, not merely the existence of an invertible differential sheaf. Uniqueness follows from the Yoneda lemma applied to the representing functor on vector bundles.

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

Let the base be a smooth algebraic curve, the total surface be smooth, and the field be algebraically closed of characteristic different from two. At a rank-two singular fibre with kernel vector $v$, the fibre derivatives vanish at $[v]$. Smoothness requires $v^{\mathsf{T}}{M}^{\prime }v\ne 0$. The adjugate matrix is a nonzero multiple of $v{v}^{\mathsf{T}}$, so the determinant has a simple zero. At rank at most one, the projective kernel contains a line and the quadratic $v^{\mathsf{T}}{M}^{\prime }v$ has a zero on it; that point would make the total surface singular. Thus every singular fibre is a pair of distinct projective lines and its discriminant zero is simple.

A rank-one subsheaf of a vector bundle on a smooth algebraic curve is saturated when the quotient is torsion-free. Intersecting a one-dimensional subspace of the generic fibre with each stalk constructs a saturated subsheaf. The local ring of a smooth algebraic curve is a discrete valuation ring, so the subsheaf and its torsion-free quotient are locally free. Saturating a nonzero map from a line bundle enlarges that line bundle by an effective divisor, recording the map's zeros.

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.