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For $U_{i_0\dots i_q}=U_{i_0}\cap \cdots \cap U_{i_q}$, the Čech cochain complex has groupsTerms in ${\delta }^2$ cancel in pairs, and Čech cohomology is $\ker \delta /\mathrm{im}\delta$. A degree-zero cocycle is exactly a family of compatible local sections. The sheaf gluing axiom gives their unique global section, proving ${\check{H}}^0(\mathcal{U},\mathcal{F})\cong \mathcal{F}(X)$. If the cover has affine finite intersections, a quasi-coherent sheaf has no higher cohomology on those intersections by vanishing of quasi-coherent cohomology on an affine scheme. The acyclic cover theorem then identifies all Čech groups with sheaf cohomology. In particular, a finite affine open cover of a separated variety has this property.

For the sheaf of units of the structure sheaf, a multiplicative degree-one cocycle is a family $g_{ij}\in {\mathcal{O}}_X^{\ast }(U_i\cap U_j)$ satisfying $g_{ij}{g}_{jk}=g_{ik}$, with $g_{ji}=g_{ij}^{-1}$. It glues trivial rank-one free modules into an invertible sheaf. Changing the local frames multiplies $g_{ij}$ by a coboundary, and two sets of transition data give isomorphic line bundles precisely when their cocycles differ this way. Tensoring line bundles multiplies their cocycles. Thus line bundles trivialized by an open cover give the group isomorphismFor the remaining arguments, work over the algebraically closed ground field. On the irreducible variety $V$, put $\mathcal{Q}={\mathcal{K}}^{\ast }/{\mathcal{O}}_V^{\ast }$, a quotient of sheaves of abelian groups. A global section of $\mathcal{Q}$ is locally represented by rational functions $f_i$ whose ratios are regular units. The corresponding unit cocycle $f_j/f_i$ defines an invertible sheaf. A single global rational function has trivial cocycle. Conversely, every line bundle has a nonzero rational section: choose a nonzero vector in its one-dimensional fibre at the generic point and express it in local frames. This supplies such local $f_i$. If the associated line bundle is trivial, changing frames makes all $f_i$ restrictions of one rational function. Thereforeis exact. This is the Cartier-divisor description of the Picard group. The sheaf of nonzero rational functions on an irreducible variety is flasque, since all restrictions between nonempty open sets are the identity on $k(V{)}^{\ast }$. Apply the long exact sequence in sheaf cohomology to $1\to {\mathcal{O}}_V^{\ast }\to {\mathcal{K}}^{\ast }\to \mathcal{Q}\to 1$. Since $H^1(V,{\mathcal{K}}^{\ast })=0$, its connecting map has exactly the cokernel just computed, provingFinally the Segre description of a smooth quadric surface identifies $V$ with ${\mathbb{P}}^1\times {\mathbb{P}}^1$. The two rulings of a smooth quadric surface have classes $F_1,F_2$ generating $\mathrm{Pic}(V)\cong {\mathbb{Z}}^2$. For completeness, remove one line in each ruling: the remaining chart is the affine plane, with factorial coordinate ring $k[s,t]$ and trivial divisor class group. The localization sequence for the divisor class group makes $F_1,F_2$ generators, and their degrees on the two ruling lines prove independence. The hyperplane class, and hence the conic $C$, has bidegree $(1,1)$. By Picard-group localization on a smooth variety, the Picard group of a smooth affine quadric surface isExplicitly, the restriction of ${\mathcal{O}}_V(1,0)$ is nontrivial: if it were trivial on $U$, its rational trivialization would have divisor supported on $C$, forcing $(1,0)$ to be an integer multiple of $(1,1)$, which is impossible.

On the complex torus $\mathbb{C}/\Lambda$, the Weierstrass elliptic function $\wp$ is a degree-two map to the Riemann sphere, and its fibres are the pairs $\{z,-z\}$, with coalescence at the four branch points. An even elliptic function is constant on every fibre, so away from those branch points it descends to a meromorphic function $Q$ of $\wp$.

The descent also holds at a branch point. At a half-period $h$, evenness and periodicity give $f(h+u)=f(h-u)$, so its local Laurent series uses only even powers of $u$. Also $\wp (h+u)-\wp (h)$ is a local coordinate proportional to $u^2$. At $0$, use $1/\wp (z)\sim z^2$ instead. Thus $Q$ extends meromorphically across every branch point and infinity. Every meromorphic function on the Riemann sphere is a rational function, proving $f=Q\circ \wp$.

For a nonconstant rational function $Q=p/q$ in coprime form, $\mathrm{deg}Q=\max (\mathrm{deg}p,\mathrm{deg}q)$: a generic equation $p-wq=0$ has that many roots on the sphere. Degrees of holomorphic maps multiply under composition. Hence the degree isConstant functions have degree zero. Degree two therefore occurs exactly when $Q$ is a Möbius transformation:Conversely every such function is even, elliptic and of degree two.

Write the period lattice as $\Lambda =\mathbb{Z}{\omega }_1+\mathbb{Z}{\omega }_2$, with ${\omega }_1,{\omega }_2$ real-linearly independent and $\mathrm{Im}({\omega }_2/{\omega }_1)>0$. A meromorphic function $f$ is an elliptic function for this lattice if $f(z+\lambda )=f(z)$ for every $\lambda \in \Lambda$. If it has no poles, it is an entire function bounded on the closure of a fundamental parallelogram. Periodicity makes it bounded on all of $\mathbb{C}$, so the Liouville theorem proves that an elliptic function without poles is constant.

Here are the two contour identities needed for counting its zeros and their sum. Choose a positively oriented fundamental parallelogram $P$ whose boundary avoids all zeros and poles, and put $h=f^{\prime }/f$. The argument principle givesbecause $h$ is periodic and opposite edges cancel. Here $N,P_f$ count zeros and poles with multiplicity. For the weighted integral, let $J_i$ be the integral of $h$ along the edge from a vertex $c$ to $c+{\omega }_i$. Translating the opposite edges givesThe endpoint values of $f$ agree along either edge. A continuous logarithm along the edge therefore changes by $2\pi i{n}_i$, with $n_i\in \mathbb{Z}$, and $J_i=2\pi i{n}_i$. On the other hand, the residue theorem evaluates the left side as the sum of the zero positions minus the sum of the pole positions, both with multiplicity. Thus the elliptic divisor-sum identity isUnder the stated pole hypothesis there is just one pole class, represented by zero, of order $m$. ConsequentlyChanging representatives of the zero classes changes their weighted sum by an element of $\Lambda$, so the congruence is well defined.

The Weierstrass elliptic function isOn a bounded set of $z$, the summand is $O(|\omega {|}^{-3})$ for large $|\omega |$. The lattice sum of $|\omega {|}^{-3}$ converges in real dimension two, proving Normal convergence of the Weierstrass elliptic-function series on compact sets away from $\Lambda$. Hence $\wp$ is a holomorphic function there, and at zero its principal part is $z^{-2}$, so it has a double pole. Replacing $\omega$ by $-\omega$ shows that $\wp$ is even. Differentiating the normally convergent series giveswhose absolute convergence permits reindexing by any $\lambda \in \Lambda$. Thus ${\wp }^{\prime }$ is periodic and $\wp (z+\lambda )-\wp (z)$ is constant. Substituting $-z-\lambda$ for $z$ and using evenness changes that constant to its negative, so it is zero. This proves that $\wp$ is an elliptic function, with double poles precisely at the lattice points.

The preceding zero count applied to $\wp -a$ says that it has exactly two zeros modulo $\Lambda$, counted with multiplicity. Evenness pairs any zero $z$ with $-z$. If these are distinct classes they must both be simple. If they coincide, $2z\in \Lambda$, and locally $\wp (z+t)=\wp (z-t)$: the zero has even order, which the total count forces to be two. In particular such a zero cannot be a lattice point, where $\wp$ has a pole. Every finite fibre of the Weierstrass function consists of two opposite simple points or one double nonzero half-period point.

To obtain the rational representation, first consider an even elliptic function $g$. The preceding description proves that $\wp :\mathbb{C}/\Lambda \to {\mathbb{P}}^1$ has exactly the fibres $\{z,-z\}$, so $g$ defines a function $A$ of $w=\wp (z)$. Away from the branch values a local inverse of $\wp$ makes $A$ meromorphic. At a half-period $z_0$, the local expansion is $w-w_0=c{t}^2+O(t^4)$, with $c\ne 0$ because the zero has order exactly two. The Laurent series of $g(z_0+t)$ contains only even powers, and $t^2$ is a holomorphic local coordinate as a function of $w-w_0$. Thus $A$ is meromorphic at that value too. At zero the same argument uses $1/w=t^2+O(t^6)$, proving meromorphy at infinity. A meromorphic function on the Riemann sphere is a rational function: subtract its finitely many principal parts and use compactness to make the remainder constant. This proves that even elliptic functions are rational in the Weierstrass function.

For an arbitrary elliptic function, split $f=f_{+}+f_{-}$, where $f_{\pm }(z)=(f(z)\pm f(-z))/2$. The even part is $A(\wp )$. Since ${\wp }^{\prime }$ is odd and not identically zero, $f_{-}/{\wp }^{\prime }$ is an even meromorphic elliptic function, including at the zeros of ${\wp }^{\prime }$ where the quotient may have poles. It is therefore $B(\wp )$ for a rational function $B$. The elliptic function-field decomposition isThe representation is unique: its even and odd parts determine $A$ and $B$, and $\wp$ takes every value on the Riemann sphere.

A function $f:U\to {\mathbb{R}}^n$ has a Frechet derivative at $a\in U$ if there is a linear map $A:{\mathbb{R}}^m\to {\mathbb{R}}^n$ such thatThe remainder condition is a limit in all directions at once, not merely the existence of partial derivatives.

For the stated scalar two-variable result, fix $a=(a_1,a_2)$ and choose a small rectangle contained in the open set $U$. Split the increment into two coordinate segments. Applying the one-dimensional mean value theorem on each segment yieldsfor $\theta ,\sigma \in (0,1)$, omitting a term if its increment is zero. Subtract $h{f}_x(a)+k{f}_y(a)$. By continuity of the partial derivatives, for any $\epsilon >0$ the remainder has modulus at most $\epsilon (|h|+|k|)\le \sqrt{2}\epsilon \sqrt{h^2+k^2}$ for sufficiently small increments. This proves differentiability, with Jacobian matrix $(f_x(a),f_y(a))$.

Away from the origin the specified rational function has a nonzero denominator and is smooth. At the origin, $f(t,0)=t$ and $f(0,t)=2t^2$, so the partial derivatives are $f_x(0,0)=1$, $f_y(0,0)=0$. Any Frechet derivative there would therefore be $A(h,k)=h$. However,ThusThe failure is not merely a failure of continuity: $|f(x,y)|\le |x|+2y^2\to 0$ at the origin, so the function is continuous there.

A character of an algebra is a nonzero complex-linear multiplicative map $\phi :A\to \mathbb{C}$. The character space of an algebra ${\Phi }_A$ carries the Gelfand topology, namely pointwise convergence on $A$, equivalently the subspace weak-star topology in $A^{\ast }$.

For a unital complex Banach algebra, $\phi (1)=1$: multiplicativity and nonzeroness force this. If $a-\phi (a)1$ were invertible, applying $\phi$ to its product with its inverse would give $0=1$. Thus $\phi (a)\in {\sigma }_A(a)$. The Neumann series shows ${\sigma }_A(a)\subset \{|z|\le \Vert a\Vert \}$, so $|\phi (a)|\le \Vert a\Vert$ and the character of an algebra is automatically bounded. The same conclusion for a nonunital algebra follows by extending $\phi$ to its unitization of an algebra via $\tilde{\phi }(a,t)=\phi (a)+t$ and using the standard norm $\Vert (a,t)\Vert =\Vert a\Vert +|t|$.

In the algebra $R(K)$ let $u(z)=z$ and fix $\phi \in {\Phi }_{R(K)}$. Write $\lambda =\phi (u)$. If $\lambda \notin K$, the rational function $(u-\lambda {)}^{-1}$ belongs to $R(K)$, which is impossible because a character of an algebra cannot vanish on an invertible element. Thus $\lambda \in K$. For a rational function $r=p/q$ without poles in $K$, choose a reduced denominator having no zeros in $K$. Multiplicativity gives $\phi (r)=p(\lambda )/q(\lambda )=r(\lambda )$. Density of these rational functions and boundedness of $\phi$ then yield $\phi (h)=h(\lambda )$ for every $h\in R(K)$.

Conversely, evaluation at each $\lambda \in K$ is a character of an algebra. The map $\lambda \mapsto {\delta }_{\lambda }$ is continuous in the Gelfand topology because every $h\in R(K)$ is continuous; its inverse is $\phi \mapsto \phi (u)$, also continuous. Hence the character space of R(K) isThe identification includes the topology, not merely a bijection of sets.

Work over the complex scalar field. If $\lambda \in {\sigma }_A(x)$, the ideal generated by $x-\lambda 1$ is proper: otherwise commutativity would make $x-\lambda 1$ invertible. By Zorn lemma extend it to a maximal ideal $M$. This ideal is closed. Indeed, if its closure were all of $A$, it would contain an element of $M$ sufficiently close to $1$ to be invertible, since the invertible group is open; a proper ideal cannot contain such an element. Its closure is therefore proper, and maximality gives $\overline{M}=M$.

The quotient $A/M$ is a complex unital Banach algebra and a division algebra. The Gelfand-Mazur theorem identifies it with $\mathbb{C}$, giving a character of an algebra $\phi$ with $\phi (x)=\lambda$. Conversely, the argument in (a) shows that every $\phi (x)$ is in the spectrum. ThusHere, as usual, a unital Banach algebra has a nonzero identity; the zero algebra is excluded.

The holomorphic functional calculus assigns to every function holomorphic on an open neighbourhood $U$ of ${\sigma }_A(x)$ an element $f(x)\in A$. It is a unital algebra homomorphism, sends the coordinate function to $x$, agrees with evaluation of rational functions having no poles on the spectrum, depends only on the germ near the spectrum, and is continuous for uniform convergence on a fixed surrounding contour. Its formula isHere $\Gamma$ is a finite oriented contour cycle in $U\setminus {\sigma }_A(x)$, with winding number one on the spectrum and zero off $U$. A finite union of suitable polygonal boundary cycles suffices; no connectedness of $U$ is required. Such cycles exist by compactness of the spectrum, and the calculus is independent of the choice. The inverse is the resolvent of an element.

For the exponential, use a circle $|z|=R>\Vert x\Vert$. The uniformly convergent Neumann expansion $(z1-x{)}^{-1}={\sum }_{n\ge 0}{x}^n{z}^{-n-1}$ may be integrated term by term. The scalar Cauchy integral formula gives $\frac{1}{2\pi i}\int e^z{z}^{-n-1}dz=1/n!$, provingThis series converges absolutely in the Banach algebra norm.

A character of an algebra is bounded and takes the resolvent to $(z-\phi (x){)}^{-1}$. Applying it inside the contour integral and using the scalar Cauchy formula gives $\phi (f(x))=f(\phi (x))$. The character of an algebra description of the spectrum, now applied to $f(x)$, proves the spectral mapping theorem:In particular this spectrum is contained in $V$ when $V\supset f(U)$.

For the composition rule for holomorphic functional calculus, choose a relatively compact open neighbourhood $D$ of $f({\sigma }_A(x))$ whose closure is contained in $V$, and a surrounding contour cycle $\Delta$ in $V$ with winding number one on $D$. Choose a sufficiently small neighbourhood $\Omega$ of ${\sigma }_A(x)$ with $\overline{\Omega }\subset U$ and $f(\overline{\Omega })\subset D$, and a contour $\Gamma$ in $\Omega$ surrounding the spectrum. The neighbourhoods and cycles may be finite unions, obtained from finite covers of the relevant compact sets. For $\zeta \in \Delta$ the function $h_{\zeta }(z)=(\zeta -f(z){)}^{-1}$ is holomorphic near $\overline{\Omega }$. Locality and multiplicativity of the calculus giveInsert this in the formula for $g(f(x))$ and interchange the two vector-valued integrals. The scalar inner integral is $g(f(z))$, soThis proves equality as algebra elements. Equality of their character of an algebra values alone would be insufficient in a Banach algebra with nonzero radical.

Finally, $\Vert x\Vert <1$ places the spectrum in the open unit disk. On that disk $f(z)=\log (1-z)=-{\sum }_{n\ge 1}{z}^n/n$ is a holomorphic branch. Its calculus value is the norm-convergent series $y=-{\sum }_{n\ge 1}{x}^n/n$: one may integrate the uniformly convergent scalar series on a circle with $\Vert x\Vert <R<1$. Since $\exp (f(z))=1-z$, the composition rule gives the logarithm of an element near the identity formula

The prescribed-pole form of Runge theorem is this: let $K\subset \mathbb{C}$ be compact, and let $E\subset \hat{\mathbb{C}}\setminus K$ meet every connected component of the complement in the Riemann sphere. If $f$ is holomorphic on a neighbourhood of $K$, then for every $ϵ>0$ there is a rational function with all its poles in $E$ whose uniform distance from $f$ on $K$ is less than $ϵ$. A polynomial is regarded as having its only possible pole at infinity. The empty compact set is trivial, so assume $K\ne \varnothing$.

First suppose $\infty \in E$. Let $A\subset C(K)$ be the uniform rational approximation algebra with prescribed poles, the uniform closure of rational functions whose finite poles lie in $E$. It is a commutative unital Banach algebra, and contains $u(z)=z$. Setwhere the inverse in this definition is the pointwise continuous function on $K$. The set $S$ is relatively open: if the inverse at $\lambda$ belongs to $A$, the Neumann series gives inverses at nearby points. It is relatively closed: when ${\lambda }_n\to \lambda \notin K$, the corresponding scalar functions converge uniformly on $K$, and $A$ is closed.

Each bounded complementary component meets $E$ in a finite point $a$, and $(u-a1{)}^{-1}$ is an allowed rational function. The unbounded component meets $S$ because for $|\lambda |>\Vert u\Vert$ the geometric series for $(u-\lambda 1{)}^{-1}$ belongs to $A$. Being both open and closed in the complement, $S$ therefore contains every component, so it is the entire complement. Conversely, evaluation at any $z\in K$ prevents $u-z1$ from being invertible. HenceApply the holomorphic functional calculus in $A$ to $f$ near $K$. For every $z\in K$, the evaluation character of an algebra and the contour formula from (b) give $f(u)(z)=f(z)$. Thus $f{|}_K\in A$. By the definition of this uniform closure, the required rational approximations exist.

If $\infty \notin E$, choose a finite point $a\in E$ in the component containing infinity. The Möbius transformation $w=(z-a{)}^{-1}$ sends $K$ to a compact subset $K^{\prime }$ of the plane and sends $E$ to a set $E^{\prime }$ containing infinity. It preserves complementary components, so $E^{\prime }$ meets every component of the complement of $K^{\prime }$. The transformed function $F(w)=f(a+1/w)$ is holomorphic near $K^{\prime }$, since $0\notin K^{\prime }$. The case already proved approximates $F$ by rational functions with poles in $E^{\prime }$. Pulling them back gives rational approximations to $f$ on $K$ with poles only in $E$: a pole at infinity in the $w$-plane becomes a pole at $a$, and every other pole has its prescribed preimage. In particular the pullbacks do not acquire a pole at the original infinity, since $0\notin E^{\prime }$.

This proves the full prescribed-pole theorem, including the case where a pole at infinity is forbidden. If $\mathbb{C}\setminus K$ is connected, take $E=\{\infty \}$ to obtain the polynomial Runge theorem as a corollary.

For a nonzero isogeny of elliptic curves $\varphi :E_1\to E_2$, its degree of an isogeny is the degree of the induced extension of function fields. Since $\varphi$ commutes with negation, its first coordinate is a rational function of $x_1$ alone. Write it in lowest terms asThe coordinate maps $x_i:E_i\to {\mathbb{P}}^1$ are finite morphisms of degree two. Multiplying function field degrees in the commutative diagram $x_2\circ \varphi =R\circ x_1$ gives the degree of an isogeny from its x-coordinate map:This is the total degree, so the argument also covers an inseparable isogeny of elliptic curves. Counting distinct points in the kernel of an isogeny would give only its separable degree. The zero group homomorphism is assigned degree zero separately.

On a Short Weierstrass form $y^2=x^3+ax+b$ in characteristic of a field other than two, the elliptic-curve addition formula givesThere is no cancellation: at a root $r$ of $f(x)=x^3+ax+b$, the numerator equals $f^{\prime }(r{)}^2$, which is nonzero because the elliptic-curve discriminant is nonzero. The numerator has degree four and the denominator degree three. HenceFor completeness, the same answer holds in characteristic two, where the displayed short equation is not a nonsingular model. On a general Weierstrass equation of an elliptic curve usewhere $b_2=a_1^2+4a_2$, $b_4=a_1{a}_3+2a_4$, $b_6=a_3^2+4a_6$, and $b_8=a_1^2{a}_6+4a_2{a}_6-a_1{a}_3{a}_4+a_2{a}_3^2-a_4^2$. The resultant of this numerator and denominator is ${\Delta }^2$. Their resultant is therefore nonzero in every nonsingular case; even when the denominator drops degree, the numerator retains degree four.

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 $\varphi \ne 0$.

By the argument principle, the number of zeros minus poles, counted with multiplicity, in the fundamental parallelogram is $(2\pi i{)}^{-1}\oint f^{\prime }/fdz$. The logarithmic derivative is periodic with the same lattice, so the integrals over opposite sides cancel. Thus the numbers of zeros and poles are equal.

The Weierstrass elliptic function gives a degree-two map $\wp :\mathbb{C}/\Lambda \to {\mathbb{P}}^1$ whose fibers identify $z$ and $-z$. An even elliptic function is constant on these fibers, so it descends to a single-valued function $Q$ on the sphere. This is meromorphic even at a branch value: near a half-period the involution is $t\mapsto -t$, the Laurent series of $f$ has only even powers, and $\wp -{\wp }_0$ has a double zero and provides a local coordinate in $t^2$. At the image of $0$, $\wp \sim t^{-2}$ gives the same conclusion at infinity. A meromorphic function on the Riemann sphere is a rational function, so $\boxed{f=Q(\wp )}$.

A meromorphic antiderivative need not be elliptic. Take $f=\wp$. Its double poles have zero residue, so integrating on the punctured plane gives a single-valued primitive, since every closed-contour integral is a sum of zero residues. The primitive extends meromorphically over each pole, with local principal part $-1/z$. If $F$ were elliptic, it would have precisely one pole modulo the lattice, with residue $-1$, contradicting the zero sum of residues in a fundamental parallelogram. This is a nonconstant counterexample satisfying the stated premise; the preceding descent proves that even elliptic functions are rational in the Weierstrass function.

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.

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

The relevant invariant of a divisor is its Iitaka dimension, also called its Kodaira dimension. If two linearly independent global sections $s,t$ occur in $H^0(X,nH)$, their ratio is a nonconstant rational function. The sections $s^q,s^{q-1}t,\dots ,t^q$ are linearly independent, because a nonconstant rational function over an algebraically closed field is transcendental over that field. Thus some complete linear system of a divisor has positive-dimensional image, contradicting $\kappa (H)=0$.

Consequently $h^0(X,nH)=1$ for every $n>0$; the nonemptiness of $|H|$ supplies a nonzero section in every multiple. Its unique divisor is entirely fixed. Hence every movable part $M_n$ is zero.