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First, $1\in G(p)$. Otherwise there would be a nonconstant rational function with one simple pole and no others, defining a degree-one finite morphism $X\to {\mathbb{P}}^1$. Such a map is an isomorphism, which would make $X$ have genus zero.

Part (iii) says that there are exactly two gaps. If $2$ is a gap, they must therefore be $1$ and $2$. If $2$ is not a gap, choose a function with its only pole at $p$, of order two. Pole orders form an additive semigroup under multiplication of functions. If $3$ were also a nongap, the semigroup generated by $2$ and $3$ would contain every integer at least two, leaving only the single gap $1$, a contradiction. Hence $3$ is the second gap. This proves the Genus-two Weierstrass gap sequence

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 three coordinate axes are the three irreducible components ofTheir common intersection is the origin. Likewise, if ${\ell }_1,{\ell }_2,{\ell }_3$ are distinct linear forms defining the three lines in ${\mathbb{A}}_k^2$, thenand its three irreducible components have only the origin in common. Any isomorphism of algebraic varieties would permute irreducible components and therefore map the origin of $V$ to the origin of $W$.

Every defining polynomial $xy,xz,yz$ has zero differential at the origin, soThe defining polynomial of $W$ is homogeneous of degree three, so its differential also vanishes at the origin, givingAn isomorphism induces an isomorphism of Zariski tangent spaces, which is impossible here. Hence $V$ and $W$ are not isomorphic. This is the tangent-space obstruction between two unions of three lines.

For an irreducible algebraic variety $X$ and $p\in X$, the local ring ${\mathcal{O}}_{X,p}$ consists of germs of rational functions regular on a neighbourhood of $p$. If $X\subseteq {\mathbb{A}}^n$ has ideal $I(X)$, its Zariski tangent space is

The given variety is the blowup of the affine plane at the origin. On the chart $W\ne 0$, put $W=1$; then $Y=XZ$, so $(X,Z)$ are free affine coordinates. On the chart $Z\ne 0$, put $Z=1$; then $X=WY$, so $(W,Y)$ are free affine coordinates. These two smooth affine-plane charts cover $V$, hence every point of $V$ is smooth.

If $(X,Y)\ne (0,0)$, the equation forcesConsequently $\pi$ restricts to an isomorphism away from the origin, and is therefore birational. Above the origin, however,so $\pi$ is not injective and cannot be an isomorphism of algebraic varieties. Birationality gives

For any morphism $\phi :V\to V^{\prime }$ with $V^{\prime }$ affine, its restriction to the exceptional curve $E\cong {\mathbb{P}}^1$ is constant: every regular function on a projective line is constant, and the affine coordinate functions of $V^{\prime }$ therefore have constant pullbacks. Since $E$ has more than one point, $\phi$ is not injective. If $V$ itself were affine, its identity morphism would contradict this conclusion, so $V$ is not affine.

Over $\mathbb{C}$, $V$ is not compact in the Euclidean topology: the sequence $(n,0,[1:0])$ has no convergent subsequence. Every complex projective variety is Euclidean compact. Since compactness is preserved by homeomorphisms, $V$ cannot be homeomorphic to a projective variety.

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 $C$:

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.

Part (iii) givestwo complex projective lines meeting at one point. If $x\ne 0$, the fiber ${\mathbb{P}}^1\times \{x\}$ misses the blowup center. The blowup map is an isomorphism of algebraic varieties away from its center, sofor every nonzero closed point $x\in {\mathbb{A}}^1$.

If distinct points $p,q$ on a smooth projective curve satisfy $(f)=p-q$, then $f:C\to {\mathbb{P}}^1$ is a finite morphism of degree one. Such a morphism between smooth projective curves is an isomorphism, so $C$ has genus zero. Hence no divisor $p-q$ on a curve of positive genus is principal.

The union of the three coordinate axes in ${\mathbb{A}}^3$ and a union of three distinct lines through the origin in ${\mathbb{A}}^2$ are not isomorphic. Their origins are intrinsically the unique points common to all three irreducible components, but their Zariski tangent spaces there have dimensions three and two respectively.