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A complete quadrangle consists of four projective points in the projective plane, no three collinear, and all six joining projective lines. Its three diagonal points are the intersections of opposite joining lines. Over a field of characteristic different from two, the line joining two diagonal points cuts either side through the third diagonal point in a pair of harmonic conjugate points. In a projective frame this follows by using the coordinate points and $[1:1:1]$ and computing the intersections.

Two neighboring smooth cones can be normalized so that the three rays are $(1,0),(0,1),(-1,-r)$, with $r\ge 1$. Completeness forces the negative sign and their primitive relation has weights $(1,r,1)$. Thus these surfaces are precisely the weighted projective planes $\mathbb{P}(1,1,r)$. For $r=1$ the surface is the projective plane; for $r>1$ its unique singularity is the cyclic quotient surface singularity $\frac{1}{r}(1,1)$.

For two projective triangles in a projective plane over a field, perspectivity from a point implies perspectivity from a line. The usual nondegenerate form assumes that the corresponding-side intersections are well defined and distinct. Its dual is the converse, by point-line duality. The result holds in the real projective plane; it is not an axiom valid in every abstract incidence plane.

One linear algebra proof explains the collinearity. Choose a representative $o$ of the center. Rescale the paired representatives to $p^{\prime }=p+\alpha o$, $q^{\prime }=q+\beta o$, and $r^{\prime }=r+\gamma o$, with nonzero coefficients in a nondegenerate configuration whose vertices differ from the center. The first corresponding-side intersection is represented by $x=\beta p-\alpha q=\beta p^{\prime }-\alpha q^{\prime }$. Similarly choose $y=\gamma q-\beta r$ and $z=\alpha r-\gamma p$. Then $\gamma x+\alpha y+\beta z=0$, so the three representatives are linearly dependent. Their projective points are consequently collinear.

Hyperplanes in $\mathbb{P}(V)$ are kernels of nonzero linear functionals modulo scalar multiplication. Their parameter space is therefore $\mathbb{P}(V^{\ast })$, of the same dimension. In particular lines in a projective plane are parametrized by another projective plane.

The torus and Klein bottle are not homeomorphic because only the former is orientable. Adding a projective plane to either by connected sum of surfaces makes a nonorientable surface with Euler characteristic minus one. The classification of compact surfaces identifies both with three projective planes. Thus connected-sum cancellation can fail even for closed connected surfaces.

The Fano plane is the projective plane over the finite field ${\mathbb{F}}_2$. It has seven points and seven lines, each line contains three points, and two distinct lines meet in one point. A line-point incidence matrix $N$ therefore satisfies $N{N}^{\mathsf{T}}=2I+J$, making it invertible over the real numbers and useful for the transplantation theorem.

Distinct lines in the projective plane are in general position when no three are concurrent. Their union is a reduced plane curve with exactly $\left(\genfrac{}{}{0ex}{}{d}{2}\right)$ singular points, all ordinary pairwise crossings, if there are $d$ lines. An explicit family is $y=t_{i}x+t_i^{2}z$ for distinct complex $t_i$: the intersection of the $i$th and $j$th lines is $[-t_i-t_j:-t_i{t}_j:1]$, and substituting it into a third equation gives $-(t_k-t_i)(t_k-t_j)\ne 0$.

For six distinct points on an irreducible conic in the complex projective plane, the three intersections of opposite sides of the resulting hexagon are collinear. Form the two cubics given by the products of alternate side equations. They meet at the six conic points and the three opposite-side intersections. Applying residual intersection through a curve pencil with degrees three and two places the last three points on a line. This is the conic counterpart of Pappus's hexagon theorem for two lines.

First make the invertible linear change of homogeneous coordinatesThe original equation becomesThis takes the flex $[0:0:1]$ to the point at infinity $[0:1:0]$ of a Weierstrass model.

Put $h=\omega -1$ and choose $s\in \mathbb{C}$ with $s^2=h^3$. Both are nonzero. DefineSince $U=hX+Z$, the three factors on the right are $hX$, $h(X-Z)$ and $h(X-\lambda Z)$, whereThe resulting equation is preciselythe Legendre form of an elliptic curve. In particular, $\lambda \ne 0,1$.

Combining the two changes gives an explicit projective transformation from the original coordinates:Its determinant is $-\sqrt{3}/(hs)\ne 0$, so this is a valid transformation of the whole projective plane, including the points where an affine coordinate would be undefined.

Bézout's theorem says that two projective plane curves over $\mathbb{C}$ of degrees $d,e$ with no common irreducible component have total intersection multiplicity $de$, so at most $de$ distinct intersection points. Let the irreducible curve have reduced homogeneous equation $F$ of positive degree $d$. In characteristic zero some partial derivative is nonzero. For $d>1$ this derivative has degree $d-1$ and cannot share a component with the irreducible degree-$d$ polynomial $F$. Every singular point belongs to their intersection. Bézout bounds their number by $d(d-1)$, proving there are finitely many singular points. For $d=1$ a partial is a nonzero constant and there are none.

For a smooth complex projective plane curve of degree $d$,Project generically from a point outside the curve to a projective line. Its degree is $d$, by intersection with a generic line through the center. Ramification occurs when such a line is tangent. In coordinates for projection $[x:y:z]\mapsto [x:z]$, these points satisfy $F=F_y=0$. The polar has degree $d-1$, and Bézout gives total ramification $d(d-1)$; generic projection makes the ramification simple, or the intersection multiplicities can be read as the corresponding ramification orders. The Riemann-Hurwitz formula yields $2g-2=-2d+d(d-1)$ and hence the formula. The smoothness qualification is essential for geometric genus: singular plane curves have the same arithmetic degree-genus expression, while the genus of their normalization can be smaller.

The classification of compact surfaces states that every compact connected topological surface without boundary is homeomorphic to exactly one of the sphere, a connected sum of surfaces consisting of $g\ge 1$ tori, or a connected sum of $h\ge 1$ real projective planes. The torus family is orientable and the projective-plane family is nonorientable; their respective genera are unique. Including $g=0$ makes the sphere the genus-zero orientable case. If boundaries are allowed, the number of removed open discs is an additional invariant.

A square model of the torus has one vertex, two edges and one face, so its Euler characteristic is zero. The connected-sum formula givesThe real projective plane has a bigon model with one vertex, one edge and one face, hence characteristic one. Therefore

A polygonal model of the real projective plane is a bigon with boundary word $aa$, with the equally labelled arrows identified. Its two vertices become one, its sides form one edge, and it has one face. ThusEquivalently, glue antipodal boundary points of a disc.

The classification of compact surfaces says that every compact connected surface without boundary is homeomorphic to exactly one of the following: the sphere, an orientable surface obtained as the connected sum of $g\ge 1$ tori, or a nonorientable surface obtained as the connected sum of $k\ge 1$ real projective planes. The integers $g$ and $k$ are unique, and the orientable and nonorientable families do not overlap. Including $g=0$ makes the sphere the genus-zero member of the orientable family.

Their Euler characteristics are $2-2g$ and $2-k$, respectively. Thus orientability and the Euler characteristic determine the closed connected surface. If boundaries are allowed, one additionally specifies the number of boundary circles and removes that many open discs from a closed model.

Let $u=(1,1,1)$ and $v=(1,\alpha ,\beta )$. The vector representing $r_{23}$ satisfiesConsequently the three representative vectors are linearly dependent, so $r_{12},r_{13},r_{23}$ are collinear in the projective plane. Equivalently their three-by-three determinant vanishes. This is the two-line configuration of Pappus's hexagon theorem.

The smoothness criterion for a toric variety says that every cone must be regular: its primitive ray generators must be part of a lattice basis of $N$. Such a $d$-dimensional cone gives the chart ${\mathbb{A}}^d\times ({\mathbb{C}}^{\ast }{)}^{n-d}$; all charts are smooth precisely under this criterion.

Here is a terminating construction of a smooth fan subdivision. First subdivide compatibly into simplicial polyhedral cones. One method is the barycentric subdivision: choose a rational interior ray in each nonzero cone and perform the star subdivisions in decreasing dimension. Its cones correspond to chains of faces, hence are simplicial. For a simplicial cone with primitive generators $v_i$, let its multiplicity be the index of their integer span in the saturated lattice $N\cap {\mathrm{span}}_{\mathbb{R}}\sigma$.

If a cone is not regular, choose a nonregular face $\tau$ minimal under inclusion. Its half-open fundamental parallelepiped contains a nonzero point of the toric lattice $w={\sum }_i{\alpha }_i{v}_i$, $0\le {\alpha }_i<1$. Every coefficient is positive: otherwise $w$ lies in the span of a regular proper face, whose generators extend to a lattice basis, forcing all the remaining coefficients to be integers and hence zero. Replace $w$ by the primitive vector $w^{\prime }$ on its ray; its coefficients ${\beta }_i$ still satisfy $0<{\beta }_i<1$. Star-subdivide the entire fan at this ray. In each cone containing $\tau$, the new maximal cones replace one of these $v_i$ by $w^{\prime }$, so multilinearity of the determinant makes the new multiplicity ${\beta }_i$ times the old one. It is a positive integer strictly smaller than the old multiplicity. Other cones are unchanged. Select $\tau$ inside a maximal cone of largest multiplicity. The number of maximal cones having that largest multiplicity decreases, and once it reaches zero the largest multiplicity decreases. This lexicographic integer descent terminates at multiplicity one. It proves every finite rational fan has a smooth subdivision, not merely a simplicial one; this is multiplicity descent in toric desingularization.

For the specified threefold, abbreviate the primitive rays byIts boundary two-dimensional faces are $ab,bc,cd,da$. For example their supporting normals, chosen inward, are $(0,1,-1),(1,1,-2),(1,-1,2),(0,0,1)$; each is zero on the named pair and positive on the other two rays. These normals are primitive. Equivalently the greatest common divisor of the two-by-two minors for each ray pair is one, so these faces are regular; the rays themselves are primitive as well. The full cone is three-dimensional with four extremal rays, hence not simplicial and not regular. The orbit-cone correspondence now shows the closed torus-fixed point is its only singular point: every other orbit belongs to the smooth chart of a proper face.

Define the first subdivision by its maximal conesHere a string of letters denotes their nonnegative span; all faces are included. This is the triangulation along diagonal $ac$ of a positive cross-section. Both determinants have absolute value one, so it is smooth. The subdivision has the same support as the original cone, hence its toric morphism is proper and birational and is a resolution. No new rays were added, so it has no exceptional surfaces. Its internal wall $ac$ gives a compact exceptional curve isomorphic to ${\mathbb{P}}^1$, by the two-ray quotient fan at that wall. The only proper coarsening is the original nonregular cone, so this resolution is minimal in the stated fan sense.

For a second resolution, setStart with the other diagonal $bd$. The cone $abd$ is regular, but $bcd$ has determinant of absolute value two. Its primitive interior point $e$ divides it into the three regular cones shown, each with determinant of absolute value one. Therefore ${\Sigma }_2$ is again a smooth resolution and is different from ${\Sigma }_1$.

This fan has exactly one new ray, so its only exceptional surface is $E_e\cong {\mathbb{P}}^2$. To identify it, quotient $N$ by $\mathbb{Z}e$. The three neighboring rays are $\bar{b},\bar{c},\bar{d}$, their sum is zero, and consecutive pairs are bases because $bce,cde,dbe$ are regular. Their quotient fan is the simplex fan of the projective plane. In fact its normal bundle is ${\mathcal{O}}_{{\mathbb{P}}^2}(-2)$: the chart $U_{bcd}$ before subdivision is ${\mathbb{A}}^3/\{\pm 1\}$ since its index-two lattice extension is generated by $(b+c+d)/2$, and the new radial parameter is the square of the ordinary radial parameter of $\mathcal{O}(-1)$. It has the transition factors of $\mathcal{O}(-2)$.

For minimality, the ray $e$ cannot be deleted in a smooth coarsening. Deleting an interior ray in this triangulation merges all three incident cones into $bcd$, which is nonregular, or merges further into the original nonsimplicial cone. If $e$ is retained, deleting a wall between its three incident cones gives a nonconvex union unless all three are merged, again giving $bcd$. The remaining internal wall $bd$ separates $abd$ and $bde$; their union has four extremal rays $a,b,d,e$, since $a+e=b+d$, and is not regular. Thus no proper coarsening is smooth. These are the two minimal toric resolutions of a four-ray threefold cone.

The factorization can be made completely explicit. Starting with ${\Sigma }_1$, add $e=a+c$ in the wall $ac$, producing a smooth fan ${\Sigma }_A$ with maximal cones $abe,bce,ade,cde$. Next addin the wall $ae$. The resulting common fan has maximal conesStarting instead with ${\Sigma }_2$, the star subdivision at $w=b+d$ in the wall $bd$ gives exactly the same six cones. Each insertion is the sum of the generators of a smooth two-dimensional cone. Such a star subdivision is the toric blowup along an orbit closure of its smooth invariant curve: in local coordinates the center has ideal $(x_1,x_2)$, and the two blowup charts with ratios $x_2/x_1$ and $x_1/x_2$ give the two new cones. We have the zigzag of blowup morphismsThe required birational map is the composite of the inverses of the first two morphisms with the last. This supplies individual smooth toric blowups, rather than appealing to an abstract factorization theorem.

Write the projective plane as $\mathbb{P}(V)$ for a three-dimensional complex vector space $V$. A projective line is the zero set of a nonzero linear functional $\ell \in V^{\ast }$. Two such functionals give the same line exactly when they differ by a nonzero scalar. Hence the parameter space of lines is the dual projective space $\mathbb{P}(V^{\ast })$. A choice of basis identifies $V^{\ast }$ with ${\mathbb{C}}^3$, giving $\boxed{({\mathbb{P}}^2{)}^{\vee }\cong {\mathbb{P}}^2}$. The description as $\mathbb{P}(V^{\ast })$ is intrinsic; the identification with the original $\mathbb{P}(V)$ requires a choice.

The even-degree subring is the second Veronese subringwhere $u,v,w$ have degree one in the regraded ring. The canonical invariance of the Proj construction under passage to a Veronese subring givesOn homogeneous points this is the degree-two Veronese embeddingIt is an isomorphism onto the closed subschemeThus the quotient homomorphism $k[u,v,w]\twoheadrightarrow R^{(2)}$ supplies the requested closed immersion of $\mathrm{Proj}{R}^{(2)}$ into the projective plane.

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.