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Let $E$ be a genus one curve over an algebraically closed field and choose $O\in E$. The mapis an isomorphism. For a degree-zero divisor $D$, Riemann--Roch gives a nonzero section of $D+O$, so $D\sim P-O$ for some point $P$. If $P-O\sim P^{\prime }-O$ with $P\ne P^{\prime }$, a rational function with divisor $P-P^{\prime }$ would give a degree-one map to the projective line, contradicting that $E$ has genus one. The Abel--Jacobi map transports addition in the divisor class group to the elliptic-curve group law.

For a smooth genus one curve with origin $O$, the map $P\mapsto [(P)-(O)]$ identifies geometric points with the degree-zero Picard group of a curve. Riemann-Roch theorem gives a unique effective representative for each divisor class of degree one: its Riemann-Roch space has dimension one. A line cuts a plane cubic in a divisor linearly equivalent to $3(O)$ when $O$ is the Weierstrass flex. The two successive line intersections in the chord-and-tangent group law therefore give addition of the corresponding divisor classes. Associativity follows by injectivity of the identification and associativity of divisor-class addition, including tangent intersection multiplicities.

Let $P=p_1+\cdots +p_k$ and $Q=q_1+\cdots +q_k$ be disjoint sums of distinct points on a smooth projective curve. A nonzero function with principal divisor $P-Q$ exists exactly when $L(Q-P)\ne 0$: a nonzero member makes $(f)+Q-P$ an effective divisor of degree zero, forcing equality. Evaluation $f\mapsto (f(p_1),\dots ,f(p_k))$ on $L(Q)$ has kernel of a linear map $L(Q-P)$.

If $\ell (Q)=k$, a choice of basis $f_1,\dots ,f_k$ turns the criterion into $\det (f_j(p_i))=0$. This determinant is not identically zero. Starting with $L(Q)$, repeatedly choose a point where some nonzero function in the current space does not vanish; evaluation then lowers the dimension of that space by one. After $k$ choices the joint kernel is zero. Thus a generic tuple of points gives an invertible evaluation matrix, and no function with divisor $P-Q$. On a genus one curve, Riemann-Roch theorem gives $\ell (Q)=k$, so this obstruction applies for every $k>0$.

If two smooth plane conics over $\mathbb{C}$ meet at four distinct points, their tangent incidence curve is a smooth projective genus one curve. Projection to $C_2$ is a double cover branched at those four intersections. The tangent equation becomes a quadratic whose discriminant cuts out $C_1\cap C_2$; its four simple zeros make the cover smooth and connected. The Riemann-Hurwitz formula then gives genus one.

Over an algebraically closed field of characteristic different from two, a degree-two map from a smooth genus one curve to the projective line has a deck involution. After choosing an origin, all its fiber divisors on an algebraic curve are linearly equivalent and have the same group sum $s$, by the Abel-Jacobi map of a genus-one curve. The involution is consequently $p\mapsto s-p$, including its fixed ramification points. Composing two such involutions gives a translation on an elliptic curve.

Over an algebraically closed field of characteristic different from two, a cubic Weierstrass model with distinct roots becomes this form by sending two roots to zero and one, then scaling $Y$ by a square root of the resulting nonzero coefficient. The third root becomes $\lambda \notin \{0,1\}$. The point $[0:1:0]$ is a flex and the curve is a genus one curve.

To analyze the genus one tangent incidence curve of two conics, choose projective coordinates in which $C_1$ has equation $XZ-Y^2=0$. Its points and tangent lines are parametrized by $[a:b]\in {\mathbb{P}}^1$ asFor $x=[X:Y:Z]\in C_2$, the incidence equation is a nonzero homogeneous quadratic in $a,b$, so the projection $E\to C_2$ is finite of degree two. Its discriminant vanishes exactly when $Y^2-XZ=0$, that is, at $C_1\cap C_2$. There are four such points; Bézout theorem says their intersection multiplicities sum to four, so all four intersections are transverse.

Locally on $C_2$, after choosing an affine tangent-parameter chart and completing the square, the incidence equation is $w^2=u(s)$, where $s$ is a local parameter and $u$ has a simple zero at each intersection. This is smooth, with ramification index two there; away from those points the roots are distinct and the cover is étale. The analogous chart covers a root at infinity. Thus $E$ is smooth everywhere, and the cover has exactly four ramification points. It is connected: the discriminant has odd valuation at each of its four zeros, so it cannot be a square in the function field of $C_2$. The associated quadratic extension is therefore a field, not two separate sheets.

A smooth plane conic over $\mathbb{C}$ is isomorphic to the projective line. Apply the Riemann-Hurwitz formula to this connected degree-two cover:Consequently $\boxed{g(E)=1}$, and $E$ is a smooth projective genus one curve.

The printed construction is not well defined: the next tangent must pass through $x_1$, rather than $x_0$. If $x_0\ne x_1$ and ${\ell }_1\ne {\ell }_0$, a second line through $x_0$ cannot also pass through $x_1$, because the unique line through both points is ${\ell }_0$. Here is a concrete counterexample satisfying all the conic hypotheses. In the affine chart takewith ${\ell }_0:x=1$, $x_0=(1,3\sqrt{3}/2)$ and $x_1=(1,-3\sqrt{3}/2)$. The second tangent through $x_0$ is $-23x+12\sqrt{3}y=31$. It does not contain $x_1$: its left side there is $-77$. The conics meet transversely at the four complex points with $x^2=32/5$, $y^2=-27/5$. Thus this is a defect in the original PDF, not just in its conversion.

For the corrected construction, let ${\iota }_{\ell }$ exchange the two points of $C_2$ on a fixed tangent line, and let ${\iota }_x$ exchange the two tangents to $C_1$ through a fixed point of $C_2$. Both projections from $E$ are degree-two morphisms to a smooth plane conic: for the line projection this follows from intersecting a line with $C_2$, which has no line component. Since $E$ is smooth and the ground field has characteristic zero, their quadratic function-field extensions define regular involutions on the whole curve. At a ramification point “the second” point is the same point, counted with multiplicity. Each switch is an involution of a degree-two map from a genus one curve. Hence the corrected step is the everywhere-defined automorphism $F={\iota }_x\circ {\iota }_{\ell }$.

Choose an origin $O$ on the genus one curve. The Abel-Jacobi map of a genus-one curve identifies $E$ with ${\mathrm{Pic}}^0(E)$ by $p\mapsto {\mathcal{O}}_E(p-O)$. Fibers of each degree-two projection are linearly equivalent Weil divisors, since they are pullbacks of points of ${\mathbb{P}}^1$. Thus their group sums are constant: for suitable $a,b\in E$,This also holds at the ramification points, where $2p=a$ or $2p=b$. Therefore the corrected step is a translation on an elliptic curve. If $F^n(p_0)=p_0$ for one point, then $n(b-a)=0$. It follows that $F^n(p)=p$ for every $p\in E$. The corrected construction has the Poncelet porism: one periodic orbit implies all orbits are periodic, with the same least period. The literal printed construction fails before this conclusion; the proof establishes the intended, explicitly corrected assertion.

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.

Let $E$ be a genus one curve and choose $O\in E$. Since a canonical divisor is linearly equivalent to zero, Riemann--Roch gives $\ell (nO)=n$ for $n\ge 1$. Choose $x\in L(2O)\setminus L(O)$ and $y\in L(3O)\setminus L(2O)$. Their pole orders at $O$ are two and three. The seven functionsare linearly dependent because $\ell (6O)=6$, and comparison of pole orders puts the relation in Weierstrass form. The complete linear system $|3O|$ is very ample and realizes this relation as a smooth plane cubic. In characteristic other than two and three, completing the square and translating $x$ produce a Short Weierstrass form.