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If a projective line $H$ does not contain the projective plane curve $X$, its hyperplane-section divisor iswhere $I_p(X,H)$ is the intersection multiplicity. The Bézout theorem gives $\mathrm{deg}[X\cap H]=(\mathrm{deg}X)(\mathrm{deg}H)=\mathrm{deg}X$.

Suppose there were distinct singular points of an algebraic variety $P,Q$. Let $L$ be their joining projective line. The homogeneous restriction $F{|}_L$ has degree $3$. At a singular point of an algebraic variety of a plane algebraic curve, the local defining equation has no constant or linear term, so its restriction to any line through that point vanishes to order at least $2$, unless the restriction is identically zero. Therefore a nonzero $F{|}_L$ would have at least four zeros counted with multiplicity, two at $P$ and two at $Q$, which a degree-$3$ homogeneous polynomial on ${\mathbb{P}}^1$ cannot have.

Hence $F{|}_L=0$. After choosing coordinates with $L=V(z)$, this says that every monomial of $F$ contains $z$, so $z$ divides $F$. That contradicts the irreducible polynomial hypothesis on the cubic. ThusThis elementary version of the Bézout theorem argument works in every characteristic, without assuming that the partial derivatives of the cubic have their characteristic-zero form.

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.

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.

Let $H=V(L)$ for a nonzero homogeneous linear polynomial $L$. Because $H$ does not contain $X$, the restriction $L{|}_X$ is a nonzero section of ${\mathcal{O}}_X(1)$. Its zero divisor is the hyperplane-section divisor of a projective plane curvewhere $I_p(X,H)$ is the local intersection multiplicity; equivalently, it is the order of vanishing of $L{|}_X$ at $p$.

The curves $X$ and $H$ have no common component and have degrees $d$ and $1$. The Bézout theorem therefore gives

Choose a point $p\in C$ and project from $p$ to a line not containing $p$. Every line through $p$ meets the quartic in four points counted with intersection multiplicity by Bézout theorem; removing the fixed intersection at $p$ leaves a generic fibre of three points. This projection from a point on a plane curve therefore defines a nonconstant morphismof degree three, so the gonality is at most three.

It cannot be one, because a degree-one morphism between smooth projective curves is an isomorphism of algebraic varieties, whereas $C$ has genus three and the projective line has genus zero. It cannot be two either: that would make $C$ a hyperelliptic curve. The canonical map of a hyperelliptic curve is not a closed embedding, but for a smooth plane quartic the canonical map of a smooth plane curve is exactly its given embedding $C\hookrightarrow {\mathbb{P}}^2$, since $K_C\cong {\mathcal{O}}_C(1)$. This contradiction proves

For an algebraically closed field $k$, the strong Hilbert Nullstellensatz says that every ideal $J\subseteq k[x_1,\dots ,x_n]$ satisfiesIts weak form says that $V(J)=\varnothing$ exactly when $J=(1)$. Now let $S=\mathbb{C}[x_0,\dots ,x_n]$ and let $I\subseteq S$ be a homogeneous ideal. The affine zero set of $I$ is the affine cone over $V_{+}(I)$ together with the origin. Hence, when the projective zero set is empty, eitheror the affine zero set is exactly the origin. In the latter case the Hilbert Nullstellensatz givesthe irrelevant ideal of projective space. Equivalently, some power $(x_0,\dots ,x_n{)}^N$ is contained in $I$. These are precisely the alternatives in the Projective Nullstellensatz.

Let $V\subseteq {\mathbb{P}}^3$ be a smooth quadric surface. After a projective linear transformation, the Segre embedding identifiesChoose distinct points $p,q\in {\mathbb{P}}^1$. The two fibresare the promised disjoint curves on a smooth quadric surface: each is a smooth projective line, and $C_p\cap C_q=\varnothing$. By the Bézout theorem, any two nonempty curves in the projective plane intersect. An isomorphism of algebraic varieties preserves intersections and takes projective curves to projective curves, so these disjoint curves prove

Next let $W$ be a smooth projective curve and let $f:W⇢Y\subseteq {\mathbb{P}}^N$ be a rational map to a projective variety. At a point $P$ outside its initial domain, writeusing elements of the function field $\mathbb{C}(W)$. The local ring of a smooth algebraic curve at $P$ is a discrete valuation ring. If $m={\min }_i{v}_P(f_i)$, multiplying every coordinate by a function of valuation $-m$ makes all coordinates regular at $P$ and at least one a unit there. They therefore define a morphism near $P$. Repeating this at the finitely many missing points proves the extension of a rational map from a smooth projective curve.

Smoothness is essential. Let $\nu :{\mathbb{P}}^1\to C$ be the normalization of a nodal curve for a rational nodal cubic $C$. The inverse is a rational map $C⇢{\mathbb{P}}^1$ on the smooth locus. If it extended to a morphism $g:C\to {\mathbb{P}}^1$, then $g\circ \nu$ would agree with the identity on a dense Zariski-open subset of ${\mathbb{P}}^1$ and hence everywhere. But the two distinct points $a,b\in {\mathbb{P}}^1$ above the node would satisfya contradiction. This is the failure of rational-map extension on a singular curve.

Finally take the Fermat cubic curveand define the projective hypersurfaceBecause $F$ and $xyz$ have no common irreducible factor, the bihomogeneous polynomial $sF+txyz$ is irreducible, so $Z$ is an algebraic variety. Letbe the second coordinate projection. Every homogeneous cubic in three variables has a projective zero over $\mathbb{C}$, so every fibre is nonempty and $\pi$ is a surjective morphism. Over $p=[1:0]$ the fibre is the Fermat cubic curve, which is smooth and has genusOver $q=[0:1]$ the fibre isthe union of exactly three distinct projective lines, hence exactly three irreducible components. This is the cubic pencil with a triangular member required by the question.

For a degree-$d$ plane curve $C$ and a smooth point $p\in C$, the pencil of lines through $p$ induces a morphism $C\to {\mathbb{P}}^1$. A general line has one fixed intersection at $p$, so Bézout theorem makes the morphism have degree $d-1$.