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If a smooth degree-$d$ projective plane curve had genus two, the genus-degree formula would give $(d-1)(d-2)=4$, equivalently $d^2-3d-2=0$. Its discriminant is $17$, so it has no integral solution for the degree $d$.

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.

For a smooth projective curve of genus $g$, the Riemann-Roch theorem stateswhere $D$ is a divisor, $K$ a canonical divisor, and $\ell (D)=\dim H^0(V,{\mathcal{O}}_V(D))$. In particular, applying it to $D=0$ and $D=K$ gives $\ell (K)=g$ and $\mathrm{deg}K=2g-2$.

For a nonconstant morphism $f:C\to D$ of degree $d$ in characteristic zero, choose local parameters with $t\circ f=u^e$ up to a unit at a point of ramification index $e$. Pulling back a differential multiplies its local order by $e$ and adds $e-1$, because $d(u^e)=e{u}^{e-1}du$ and $e\ne 0$ in the field. ThusThe degree of a divisor of a pulled-back divisor is $d$ times its degree. Taking degrees and using the Riemann-Roch theorem gives the Riemann-Hurwitz formula

A smooth plane cubic has genus one by the genus-degree formula. In the displayed normal form, $F(0,X_1)=c{X}_1^3$ with $c\ne 0$: otherwise $X_0$ divides $F$ and the cubic is reducible. Therefore the line $X_0=0$ meets the cubic in the divisor $3P_0$, and ${\mathcal{O}}_V(1)\cong {\mathcal{O}}_V(3P_0)$. The elliptic curve group law identifies $P$ with the degree-zero divisor class $[P-P_0]$. Order three means $3P\sim 3P_0$. Consequently ${\mathcal{O}}_V(3P)\cong {\mathcal{O}}_V(1)$ and has a section with zero divisor $3P$.

Every section of ${\mathcal{O}}_V(1)$ is the restriction of a linear form: ambient linear forms form a three-dimensional space and restrict injectively, while Riemann-Roch theorem gives $h^0({\mathcal{O}}_V(1))=3$, since $\mathrm{deg}(K-3P_0)=-3$ forces the dual term to vanish. Thus this section is some nonzero $H$, proving $V\cap V(H)=\{P\}$, with intersection multiplicity three. This is the three-torsion points are flexes of a plane cubic property.

Projection forgetting $X_3$ defines a morphism $W\to V$: its only possible indeterminacy point $(0:0:0:1)$ does not satisfy the second equation. It is a double cover. Away from the zeros of $H_1{H}_2$, its two square roots are distinct. Each line intersects $V$ in three simple points, since it is nowhere tangent; the two triples are disjoint because the lines do not meet on $V$. In a local trivialization near any of these six points the covering equation is $s^2=ua(u)$ with $a(0)\ne 0$, so the ramification index is two. There is no other ramification. The Riemann-Hurwitz formula therefore gives

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.

Put $O=0_E$. Choose rational projective coordinates in which $O=(0:1:0)$ and its tangent is $Z=0$. Because $O$ is an inflection point of a plane cubic, that tangent intersects the cubic in the divisor $3O$. Consequently its cubic equation has the formHere $a\ne 0$, since the restriction to $Z=0$ is a nonzero multiple of $X^3$, and $b\ne 0$, since smoothness at $O$ requires a nonzero $Z$-derivative there. Completing the square in $Y$, by the rational linear change $Y^{\prime }=Y+(cX+eZ)/(2b)$, gives on $Z=1$Set $x^{\prime }=c_{0}x$ and $y^{\prime \prime }=c_0{y}^{\prime }$. ThenThese are invertible rational projective coordinate changes, and $O$ remains the unique point at infinity. Smoothness ensures that the resulting monic cubic has no repeated root. Thus this is a Weierstrass equation of an elliptic curve, without requiring any extraction of a square or cube root in $\mathbb{Q}$.

On this model write $-(x,y)=(x,-y)$ and $-O=O$. To add $P,Q$, take their joining line, or the tangent if $P=Q$, and let $R$ be the third intersection counted with multiplicity. DefineFor a vertical line its third intersection is $O$, so $P+(-P)=O$. The line through $O$ and an affine point is vertical, giving $P+O=P$. The tangent at $O$ has triple contact, giving $O+O=O$. This is the chord-and-tangent group law.

If $P,Q$ are rational, their joining line or tangent is rational. Its residual intersection point is rational: substituting the line in the cubic leaves a cubic whose other two roots, counted with multiplicity, are rational. Thus addition preserves $E(\mathbb{Q})$. The construction is symmetric in $P,Q$, so it is commutative.

For associativity, use divisor classes. A smooth plane cubic has geometric genus $1$, by the genus-degree formula. The Riemann-Roch theorem says that a divisor of degree $1$ on a genus-one curve has a one-dimensional space of sections. Therefore every degree-zero divisor class is uniquely represented by $P-O$: add $O$, take its unique effective degree-one representative, and then subtract $O$. Uniqueness also follows because a nonconstant function with its only pole a simple pole at $Q$ would contradict $\ell (Q)=1$.

Every line section is linearly equivalent to the tangent section $3O$. Thus, whenever $P,Q,R$ are collinear, including tangencies,A vertical section gives $[(-R)-O]=-[R-O]$. Hence the chord-and-tangent group law becomes ordinary addition in the degree-zero Picard group under the injective correspondence $P\mapsto [P-O]$. Associativity follows from associativity in that abelian group. Together with closure, identity and inverses already checked, this proves that $E(\mathbb{Q})$ is an abelian group.

Let the degree-$d$ projective hypersurface $X_d$ be defined by the homogeneous polynomial $F$. Multiplication by $F$ identifies its ideal sheaf with ${\mathcal{O}}_{{\mathbb{P}}^n}(-d)$, giving the ideal-sheaf sequence of a projective hypersurfaceIn particular, the kernel in the question is ${\mathcal{O}}_{{\mathbb{P}}^n}(-d)$.

For $n=2$, additivity of the Euler characteristic in this short exact sequence and the line-bundle formulagiveHere $H^2(X_d,{\mathcal{O}}_{X_d})=0$ by the preceding cohomological-dimension argument. Moreover $H^0({\mathbb{P}}^2,\mathcal{O}(-d))=H^1({\mathbb{P}}^2,\mathcal{O}(-d))=0$ for $d>0$, so the long exact sequence gives $H^0(X_d,{\mathcal{O}}_{X_d})\cong k$. Consequentlywhich is the genus-degree formula for a projective plane curve.

A plane cubic is a projective plane curve defined by a nonzero homogeneous polynomial of degree three. Its arithmetic genus is one by the genus-degree formula, and its Hilbert polynomial is $3m$. A singular cubic can have normalization of smaller genus; the Hilbert polynomial measures the arithmetic genus.

Blow up each distinct singular point of an integral projective curve in the projective plane once. The arithmetic genus drop under a point blowup and the genus-degree formula give arithmetic genus $(d-1)(d-2)/2-{\sum }_P{m}_P(m_P-1)/2$ for the strict transform. Its arithmetic genus is nonnegative, proving the bound in every characteristic. Infinitely near singularities need not have been resolved.

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.