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Choose a point $Q\in E$. Since $g(E)=1$, a nonzero regular differential has an effective divisor of degree $2g-2=0$, so $K_E\sim 0$. The Riemann-Roch theorem therefore givesChooseThe rational functions $x$ and $y$ have their only poles at $Q$, of orders two and three respectively.

The seven functionsbelong to the six-dimensional Riemann-Roch space $L(6Q)$, so they satisfy a nontrivial linear relation. Comparing pole orders shows that the coefficients of both $x^3$ and $y^2$ are nonzero. After rescaling, the relation has the Weierstrass formThe divisor $3Q$ is very ample, so its complete linear system embeds $E$ in ${\mathbb{P}}^2$ and the displayed relation cuts out its image; this is the Weierstrass construction from Riemann-Roch in genus one.

Because $\mathrm{char}k\ne 2$, completing the square removes the terms linear in $y$, and because $\mathrm{char}k\ne 3$, translating $x$ removes the quadratic term. Thus the equation becomesSince $k$ is algebraically closed, the cubic factors asSmoothness of $E$ says that the elliptic-curve discriminant is nonzero, equivalently the three roots are distinct. Hence

A standard sufficient hypothesis is that $X$ is both a Noetherian scheme and an integral scheme, and is regular in codimension one; in particular, a Noetherian normal scheme qualifies. A Weil divisor is then a finite sumover integral codimension-one closed subschemes $Z$. The local ring at the generic point of each $Z$ is a discrete valuation ring, so every nonzero rational function $g\in k(X{)}^{\times }$ has a principal divisorThe divisor class group is

The inclusion $L((n-1)p)\subseteq L(np)$ allows one additional coefficient: the coefficient of $t^{-n}$ in the local Laurent expansion at $p$, where $t$ is a uniformizer. The resulting linear mapis injective, so the dimension can increase by at most one.

It increases precisely when some rational function has a pole of exact order $n$ at $p$ and no other poles. By the definition of the Weierstrass gap set $G(p)$, this occurs exactly when $n\notin G(p)$. Hence

First, $1\in G(p)$. Otherwise there would be a nonconstant rational function with one simple pole and no others, defining a degree-one finite morphism $X\to {\mathbb{P}}^1$. Such a map is an isomorphism, which would make $X$ have genus zero.

Part (iii) says that there are exactly two gaps. If $2$ is a gap, they must therefore be $1$ and $2$. If $2$ is not a gap, choose a function with its only pole at $p$, of order two. Pole orders form an additive semigroup under multiplication of functions. If $3$ were also a nongap, the semigroup generated by $2$ and $3$ would contain every integer at least two, leaving only the single gap $1$, a contradiction. Hence $3$ is the second gap. This proves the Genus-two Weierstrass gap sequence

A Weil divisor on $X$ is a finite formal sumover integral codimension-one closed subschemes $Z$. Since $X$ is regular in codimension one, the local ring at the generic point of each $Z$ is a discrete valuation ring. A nonzero rational function $f\in k(X{)}^{\times }$ therefore defines the principal divisor $\mathrm{div}(f)={\sum }_Z{v}_Z(f)[Z]$. The divisor class group is

For an irreducible algebraic variety $X$ and $p\in X$, the local ring ${\mathcal{O}}_{X,p}$ consists of germs of rational functions regular on a neighbourhood of $p$. If $X\subseteq {\mathbb{A}}^n$ has ideal $I(X)$, its Zariski tangent space is

The given variety is the blowup of the affine plane at the origin. On the chart $W\ne 0$, put $W=1$; then $Y=XZ$, so $(X,Z)$ are free affine coordinates. On the chart $Z\ne 0$, put $Z=1$; then $X=WY$, so $(W,Y)$ are free affine coordinates. These two smooth affine-plane charts cover $V$, hence every point of $V$ is smooth.

If $(X,Y)\ne (0,0)$, the equation forcesConsequently $\pi$ restricts to an isomorphism away from the origin, and is therefore birational. Above the origin, however,so $\pi$ is not injective and cannot be an isomorphism of algebraic varieties. Birationality gives

For any morphism $\phi :V\to V^{\prime }$ with $V^{\prime }$ affine, its restriction to the exceptional curve $E\cong {\mathbb{P}}^1$ is constant: every regular function on a projective line is constant, and the affine coordinate functions of $V^{\prime }$ therefore have constant pullbacks. Since $E$ has more than one point, $\phi$ is not injective. If $V$ itself were affine, its identity morphism would contradict this conclusion, so $V$ is not affine.

Over $\mathbb{C}$, $V$ is not compact in the Euclidean topology: the sequence $(n,0,[1:0])$ has no convergent subsequence. Every complex projective variety is Euclidean compact. Since compactness is preserved by homeomorphisms, $V$ cannot be homeomorphic to a projective variety.

Let $\zeta =e^{2\pi i/n}$ and define Möbius transformationsThey satisfyand the $2n$ maps $r^k$ and $s{r}^k$ are distinct. They therefore give a faithful action of the dihedral group $D_{2n}$ on the Riemann sphere.

The rational functionis invariant under both $r$ and $s$, so it is constant on every orbit. Conversely, for nonzero finite $z_1,z_2$, put $u=z_1^n$ and $v=z_2^n$. Equality of the function values givesIf $u=v$, then $z_2={\zeta }^k{z}_1=r^k{z}_1$ for some $k$. If $uv=1$, then $z_2={\zeta }^k/z_1=r^{k}s{z}_1$ for some $k$. Finally, zero and infinity both map to infinity and are interchanged by $s$. HenceThis is the orbit-separating invariant for the standard dihedral action on the Riemann sphere.

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 $C$:

If $F:X\to Y$ is a nonconstant holomorphic map between compact connected Riemann surfaces, its local degree at $p$ is the integer $m_F(p)\ge 1$ for which suitable local coordinates giveThe valency theorem states thatis independent of $y\in Y$. This common value is the degree of a holomorphic map, denoted $\mathrm{deg}F$.

Now let $f$ be a nonconstant rational function of degree $d$. If its distinct finite poles have orders $m_1,\dots ,m_r$, and its pole order at infinity is $m_{\infty }\ge 0$, thenThe derivative $f^{\prime }$ has a pole of order $m_j+1$ at each finite pole. When $m_{\infty }>0$, the expansion $f(z)\sim c{z}^{m_{\infty }}$ shows that $f^{\prime }(z)\sim c{m}_{\infty }{z}^{m_{\infty }-1}$ at infinity. Hence the degree of the derivative of a rational function isIn the first case $r\ge 0$, while in the second $1\le r\le d$; thereforeFor every $d\ge 1$, the lower bound is attained by $f(z)=z^d$, whose derivative has degree $d-1$ (with a constant assigned degree zero). For distinct $a_1,\dots ,a_d$, the functionhas $d$ simple poles, degree $d$, and a derivative with $d$ double poles, so $\mathrm{deg}{f}^{\prime }=2d$. Thus both rational bounds are sharp for every $d$.

Let next $g$ be a nonconstant elliptic function for the period lattice $\Lambda$. Its degree is the total order of its poles in a fundamental parallelogram; by the valency theorem, this is also the degree of the induced map $\mathbb{C}/\Lambda \to \hat{\mathbb{C}}$. If these poles have orders $n_1,\dots ,n_s$, then $d=\mathrm{deg}g={\sum }_j{n}_j$, and $g^{\prime }$ has poles of orders $n_j+1$. The degree of the derivative of an elliptic function is consequentlySince every nonconstant elliptic function has at least one pole and $1\le s\le d$,

Let $d\ge 3$ be odd. The Weierstrass elliptic function supplies the lower-bound exampleIt has one pole modulo $\Lambda$, of order $d$, so its derivative has one pole of order $d+1$. For the upper bound, choose distinct points $a_1,\dots ,a_d$ modulo $\Lambda$ and nonzero constants $c_j$ with ${\sum }_j{c}_j=0$. The quasi-periodicity of the Weierstrass zeta function makeselliptic. It has exactly $d$ simple poles, while $g^{\prime }$ has $d$ double poles. Therefore the two bounds are attained for every required odd degree.

A polynomial factor $g$ of a polynomial $f$ satisfies $f=gh$ for another polynomial $h$. Common polynomial factors are cancelled when expressing a rational function in reduced form; its degree of a rational map of the Riemann sphere is determined after this cancellation.

A principal Cartier divisor is represented on every chart by the same nonzero rational function. It is linearly equivalent to zero and has zero intersection number with every complete curve.

For a nonzero rational function $f$ on a smooth projective curve $C$, its principal divisor isIt records the zeros of $f$ with positive multiplicity and its poles with negative multiplicity. Every principal divisor has degree zero.

A rational differential on a smooth projective curve is a rational section of its canonical bundle. In a local coordinate $t$ it has the form $f(t)dt$ for a rational function $f$.

A rational parametrization expresses the coordinates of an algebraic curve as rational functions of a parameter, with a rational inverse on a dense open subset. It identifies the function field with $K(t)$ and the smooth projective normalization of an algebraic curve with the projective line. A nonconstant parametrizing map without an inverse is a weaker notion; even this cannot exist for an elliptic curve in characteristic zero.

The Riemann-Roch space of a divisor $D$ is the finite-dimensional vector spaceIts elements are the rational functions whose poles are bounded by the negative coefficients of $D$.

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.

On $y^2=x^3+ax+b$ in characteristic of a field other than two, take the elementary symmetric polynomials $s_i$ of $x(P),x(Q),x(P+Q)$. If the line through $P,Q$ is $y=\lambda x+\mu$, its third intersection is $-(P+Q)$, so coefficient comparison gives $s_1={\lambda }^2$, $s_2=a-2\lambda \mu$, $s_3={\mu }^2-b$. Hence $(s_2-a{)}^2=4s_1(s_3+b)$. This is an identity of rational functions; tangent cases follow by specialization, and coordinates at infinity are handled through the projective addition map.

At a point $p$ of a smooth projective curve, a positive integer $n$ is a Weierstrass gap when no rational function has a pole of order exactly $n$ at $p$ and no other poles. Equivalently,