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Let $H$ denote a hyperplane-section divisor of a projective plane curve. The Adjunction formula for a smooth degree-$d$ projective plane curve statesSince $\mathrm{deg}(H{|}_C)=d$, this gives the canonical degree of a smooth plane curveCombining this with $\mathrm{deg}{K}_C=2g-2$ gives the genus of a smooth plane curve

Homogenization of the affine equation gives the projective completionThis is the plane model y plus x cubed plus xy cubed equals zero of the Klein quartic: after the coordinate relabelling $[A:B:C]=[Y:X:Z]$, its equation is $A^{3}B+B^{3}C+C^{3}A=0$.

Its first partial derivatives areIf one of $X,Y,Z$ vanishes at a common zero of these three derivatives, the displayed equations successively force all three coordinates to vanish, which is impossible in projective space. If $XYZ\ne 0$, multiplying the three derivative equations givesagain a contradiction. The Jacobian criterion therefore proves that $C$ is smooth.

The rational function $y=Y/Z$ defines a rational map of projective varieties $C⇢{\mathbb{P}}^1$. Because $C$ is a smooth projective curve, it extends uniquely to a morphismFor a generic finite value of $y$, its fibre is given by the cubicso the degree of a holomorphic map is $3$. The discriminant of a depressed cubic is

At $y=0$ the fibre consists of $(0,0)$. Since $F_y(0,0)=1$, the holomorphic implicit function theorem givesso $x$ is a local coordinate and $\pi$ has ramification index of a holomorphic map $3$ there. Each of the seven distinct roots of $4y^7+27$ gives one double, but not triple, root of the cubic in $x$, hence seven further ramification points of index $2$.

It remains to inspect the points at infinity. They areNear $Q$, set $u=X/Y$ and $v=Z/Y$. The equation becomes $u^{3}v+u+v^3=0$, so $u=-v^3+O(v^{10})$, while $y=1/v$ has a simple pole. Thus $e_Q=1$. Near $P$, set $s=Y/X$ and $t=Z/X$. Now $t+s^3+s{t}^3=0$, so $t=-s^3+O(s^{10})$ andThus $y$ has a double pole at $P$ and $e_P=2$. These calculations are the ramification of the x-coordinate on the Klein quartic after the coordinate relabelling above.

The total ramification contribution isThe Riemann-Hurwitz formula for the degree-three map to the projective line now givesand therefore

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

The two pairs of regular functionsdefine maps to the projective line on the loci where their respective coordinates do not vanish simultaneously. Those loci cover $Y$, since simultaneous failure would force $x=y=z=w=0$. On their overlap the equation $xy=zw$ says that the two projective points are equal. They therefore glue to a morphism

Let $[A:B]$ be the homogeneous coordinates on ${\mathbb{P}}_k^1$. The pullback of the hyperplane divisor $V(A)$ has local equation $x$ on the first chart and $z$ on the second. It is therefore exactly $D_Y=V(x,z)$. Compatibility of the pullback of a sheaf of modules with the line bundle associated to a divisor givesThis is the ruling morphism of the punctured three-dimensional affine quadric cone associated with $D_Y$.

A projective variety is smooth when every local ring is regular, equivalently when its Jacobian has the expected rank at every point. The affine curve $y=x^3$ is smooth because the derivative of $y-x^3$ with respect to $y$ is $1$, while its projective closure $Y{Z}^2=X^3$ is singular at $[0:1:0]$.

The genus is ${\dim }_k{H}^1(C,{\mathcal{O}}_C)$. Intersect the Fermat cubic threefold with the plane $X_3=X_4=0$. This gives the smooth plane cubic $X_0^3+X_1^3+X_2^3=0$, whose genus of a smooth plane curve is $(3-1)(3-2)/2=1$.

An irreducible plane conic is smooth: a singular quadratic form in three variables has rank at most two and factors over the algebraically closed field, contradicting irreducibility. Projection from any point of the conic, or its degree-two complete linear system, then identifies it with the projective line.

Finally, a nonzero ternary quadratic form is determined up to nonzero scalar by its six coefficients. Thus generalized conics are parametrized bijectively by ${\mathbb{P}}^5$.

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 $C$:

Fordefine its action on the projective line $F\cup \{\infty \}$ by the Möbius transformationwith the usual conventions when the denominator vanishes or $x=\infty$. Matrix multiplication agrees with composition, so this is a group action. If $A$ fixes every projective point, then it fixes $0$ and $\infty$, forcing $b=c=0$, and fixing $1$ then gives $a=d$. Thus the kernel consists exactly of the nonzero scalar matrices $Z$. The induced group homomorphismis therefore injective. This is the projective general linear group action on the projective line.

For $F={\mathbb{F}}_4$,The scalar subgroup has order $|Z|=3$, soThe four classes represented byform a subgroup $P$ of order four, hence a Sylow subgroup for the prime two. Its action is translation $x\mapsto x+b$ on ${\mathbb{F}}_4$ and fixes $\infty$. For $b\ne 0$, characteristic two makes this permutation a product of two disjoint transpositions on the four finite points. It is therefore even, andThis realizes the Sylow 2-subgroup of PGL2 over F4.

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.

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.

This smooth projective plane curve has no real rational points: a sum of three real squares is zero only when all three coordinates vanish, which is forbidden in projective space. It is therefore not isomorphic over $\mathbb{R}$ to the projective line, which has rational points. Over $\mathbb{C}$ it becomes a nonsingular conic isomorphic to a projective line. Thus the ground field changes whether a geometric genus-zero curve admits a projective-line parametrization.

Under $Q\cong {\mathbb{P}}^1\times {\mathbb{P}}^1$, the fibres of the two coordinate projections are the two rulings of the smooth quadric surface. Two distinct fibres in one ruling are disjoint projective lines, while one fibre from each ruling meets in one point.

For any field $F$, a projective linear group element is uniquely determined by its images of three distinct points of the projective line, and any ordered distinct target triple is attainable. Given representatives $v_z,v_x$ forming a basis and $v_y=A{v}_z+B{v}_x$, the matrix with columns $A{v}_z,B{v}_x$ sends $(0,1,\infty )$ to $(x,y,z)$. Distinctness forces $A,B\ne 0$. The matrices realizing one projective map differ by a nonzero scalar; over ${\mathbb{F}}_p$ there are $p-1$ such matrices.

A matrix $\left(\begin{array}{cc}a& b\\ c& d\end{array}\right)$ acts on the projective line by the Möbius transformation $x\mapsto (ax+b)/(cx+d)$, with the pole sent to infinity and infinity sent to $a/c$ when $c\ne 0$. The action is transitive because $\left(\begin{array}{cc}t& -1\\ 1& 0\end{array}\right)$ sends infinity to any finite $t$. The stabilizer subgroup of infinity consists of $\left(\begin{array}{cc}a& b\\ 0& a^{-1}\end{array}\right)$ with $a\ne 0$. The orbit-stabilizer theorem gives $|{\mathrm{SL}}_2({\mathbb{F}}_q)|=q(q^2-1)$. Its kernel is the scalar matrices of determinant one: a transformation fixing infinity has $c=0$, fixing zero then forces $b=0$, and fixing one gives $a=d$. Thus the kernel is $\{I,-I\}$, with these matrices coinciding in characteristic two.