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For an algebraically closed field $k$ and a finitely generated field extension $K/k$ of transcendence degree one, choose $t$ with $K/k(t)$ finite. The integral closures of the two standard coordinate rings of the projective line in $K$ are finite modules and glue to a finite normalization of an algebraic curve of the projective line. A one-dimensional normal local ring is a discrete valuation ring; over the perfect field $k$ the resulting projective curve is smooth. Extension of rational maps gives uniqueness of this smooth projective model.

This applies normalization of an algebraic curve to the two local branches of a node.

Normalization separates a node's two branches; a rational nodal cubic has normalization ${\mathbb{P}}^1$ and two points above its node.

The genus-degree formula gives arithmetic genus three for a plane quartic. Each of its three double points has multiplicity of a plane curve at a point two, so the arithmetic genus drop under a point blowup is one at each point. The resulting integral projective curve $C^{\prime }$ hasFor its finite normalization of an algebraic curve $\nu :\tilde{C}\to C^{\prime }$, the arithmetic genus identity isBoth terms are nonnegative, so $g(\tilde{C})=0$. Over the algebraically closed field, the rationality of a smooth projective genus-zero curve identifies $\tilde{C}$ with the projective line: a point $P$ exists, and Riemann-Roch theorem gives a degree-one map from the two-dimensional space $L(P)$. Consequently $C$ is birational to ${\mathbb{P}}^1$. This proves the rationality of a plane quartic with three double points even when a double point is not an ordinary node.

The genus-degree formula givesbecause the nonsingular plane cubic is already its own normalization of an algebraic curve. If it were birational to the projective line, the two smooth projective curves would have isomorphic function fields. The uniqueness of the finite normalization model of a one-variable function field would then make them isomorphic. But the projective line has genus of a smooth projective curve zero. Therefore the cubic is not rational. This obstruction works in the stated positive characteristics as well as in characteristic zero.

Let $C$ be the projective cubic of a Weierstrass equation of an elliptic curve, allowing its elliptic-curve discriminant to vanish, and let $O=[0:1:0]$. Use only the smooth locus of a variety $C_{\mathrm{sm}}(K)$: the singular point, if present, is excluded. For $P,Q$, intersect their chord with $C$, using the tangent if $P=Q$ and counting intersection multiplicity. If the third intersection is $R$, define $P+Q=-R$, wherein general Weierstrass equation of an elliptic curve coordinates. The line through $R$ and $O$ gives this reflection. A vertical chord gives $P+(-P)=O$, and the tangent at $O$ meets $C$ three times at $O$, so $O$ is the identity. The construction is symmetric in $P,Q$. It stays in the nonsingular locus: a line through a singular point has intersection multiplicity at least two there, and therefore cannot also contain two smooth intersections counted with multiplicity.

For the nonsingular case, prove associativity by transporting a known abelian group law. The Abel-Jacobi map of a genus-one curveis bijective over an algebraic closure. Indeed, the Riemann-Roch theorem in genus one says that every divisor class of degree one has a unique effective representative consisting of one point: existence follows from $\ell (D)=1$, and uniqueness follows because two distinct representatives would produce a degree-one map to the projective line, impossible for a genus one curve. Subtracting $O$ gives the claimed bijection.

Any line section represents the same divisor class as $3O$, including tangencies. Thus $P+Q+R\sim 3O$ as divisors on an algebraic curve, while the vertical line gives $R+(-R)+O\sim 3O$. HenceAddition in the Picard group is associative, soThe chord-and-tangent group law is defined over $K$, so the group operation restricts to the $K$-rational points.

For completeness, a singular Weierstrass equation of an elliptic curve produces the smooth-locus group of a singular Weierstrass cubic, not an elliptic curve. Over an algebraic closure, its normalization of an algebraic curve is the projective line; deleting the two preimages of a nodal crossing gives the multiplicative algebraic group, while deleting the single preimage of a cusp gives the additive group. For example, on $y^2=x^3$, the coordinate $u=x/y$, with $u(O)=0$, makes the smooth-locus law addition. On $y^2=x^2(x+1)$ in characteristic different from two, put $t=y/x$ and $z=(t+1)/(t-1)$, with $z(O)=1$; the chord relation gives $z(P)z(Q)z(R)=1$, so the group law is multiplication of $z$. A nonsplit node gives the corresponding form of the multiplicative algebraic group over $K$. These descriptions also establish the singular-case group laws.

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

A degree-four projective plane curve. Its arithmetic genus is three by the genus-degree formula; singularities can lower the genus of its normalization of an algebraic curve.

Three distinct points of multiplicity two use the entire arithmetic genus three of an integral projective curve of degree four. One point blowup at each gives arithmetic genus zero. The normalization of an algebraic curve then has genus zero and is the projective line over an algebraically closed field, by rationality of a smooth projective genus-zero curve. This allows double points with one branch as well as ordinary nodes.

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.