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Two irreducible varieties $X$ and $Y$ are birational if they contain nonempty Zariski-open subsets $U\subseteq X$ and $V\subseteq Y$ that are isomorphic. Equivalently, their function fields are isomorphic:For an irreducible variety,An isomorphism of function fields preserves transcendence degree, so $\dim X=\dim Y$. This is the dimension from the function field.

Now let $K/\mathbb{C}$ be a finitely generated field extension. Choose field generators $a_1,\dots ,a_m$ and setThen $A$ is a finitely generated integral $\mathbb{C}$-algebra and $\mathrm{Frac}(A)=K$. The affine irreducible varietyhas function field $K$. Embed $X_0$ as a closed affine subvariety of some affine space ${\mathbb{A}}^N$, identify that affine space with the standard open chart of projective space ${\mathbb{P}}^N$, and take the projective closure $X$ of $X_0$. The closure of an irreducible topological space is irreducible, and $X_0$ is dense and open in $X$. Thus $X$ is a projective variety andThis constructs the projective model of a finitely generated field.

For the curveputIn its function field, the defining equation givesHence $\mathbb{C}(X)=\mathbb{C}(t)$, so $X$ is rational. Equivalently, this is the normalization of y squared equals x times x minus one squared, whose smooth projective normalization is ${\mathbb{P}}^1$ and has geometric genus zero.

A smooth projective curve in the projective plane of degree $d$ has, by the genus of a smooth plane curve formula,Its genus can be zero only for $d=1$ or $d=2$. Conversely, a projective line and every smooth conic over $\mathbb{C}$ are rational, so each is birational to $X$. Therefore the required degrees are exactly

Finally, in ${\mathbb{A}}_{x,y,z}^3$ takeThe polynomial $F$ is irreducible: as a quadratic in $y$, it could factor only if $x(x-1{)}^2$ were a square polynomial, which it is not. Thus $S$ is an irreducible affine hypersurface of dimension two. Its function field isso $S$ is birational to ${\mathbb{A}}^2$.

The partial derivatives areSolving $F=F_x=F_y=0$ gives $x=1$, $y=0$, with $z$ arbitrary. By the Jacobian criterion,This is an irreducible subvariety of dimension one, giving the requested singular cylinder over a nodal curve.

The two singly marked sides of the first triangle are glued with their endpoints at $A$ matched. A cross-section parallel to the remaining side has its two endpoints identified, becoming a circle; the apex becomes its cone point. Thus this quotient triangle is a disc, whose boundary is the doubly marked edge.

In the second triangle, the two triply marked sides run consecutively along the boundary from $E$ through $D$ to $F$. Pairing these sides in the indicated directions introduces one reversal; its boundary word is $aa{b}^{-1}$, where $b$ is the remaining doubly marked edge. This is a Möbius band with boundary $b$. One can obtain the usual rectangle model by cutting from the identified triple-edge seam to the midpoint of $b$ and opening the triangle.

The double markings glue the disc boundary to the boundary of that Möbius band. Capping a Möbius band by a disc givesThis description also verifies that the quotient is a closed surface, including the vertices. As a check, the vertex classes are $\{A\}$ and $\{B,C,D,E,F\}$; there are three edge classes and two faces. The Euler characteristic is $2-3+2=1$, consistent with the projective plane. The Möbius band exhibits nonorientability directly.

Use the same $4\times 4$ square grid with four centre-based triangles per small square, but identify boundary points byThis gives the real projective plane. To see the model explicitly, a radial homeomorphism from the square, centred at $(1/2,1/2)$, to the closed disc carries the boundary involution to the antipodal map. The map from that disc to the projective planeis continuous and surjective: every one-dimensional vector subspace of ${\mathbb{R}}^3$ has a unit vector representative in the closed upper hemisphere. Its only identifications are antipodal boundary points. The compact-to-Hausdorff continuous bijection theorem therefore identifies this disc quotient space with ${\mathbb{RP}}^2$.

The square triangulation descends to an abstract simplicial complex $K_2$. Small-square centres remain distinct interior vertices; each triangle has one centre and two adjacent grid vertices. Boundary pairings identify whole grid edges, while preserving distinct vertices in each triangle, so this is a finite triangulation of the real projective plane. Its counts are $33$ vertices, $96$ edges and $64$ triangles, giving Euler characteristic one.

Choose $\sigma$ to be the interior triangle with vertices $(1/4,1/4)$, $(1/2,1/4)$ and $(3/8,3/8)$. Removing its relative interior and cutting from $(0,0)$ to $(1/4,1/4)$ gives the same nine-sided cut disc as before. Parametrize it by a convex polygon, transporting the boundary identifications. Its boundary word is nowThe boundary involution pairs the two $a$ sides and the two $b$ sides in matching traversal directions; the $c$ pair restores the slit. Leave $d,e,f$ unpaired. The quotient space isthe real projective plane with an open disc removed. In particular it is a compact connected surface with boundary consisting of one three-edge circle.

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 point $[a:b:c]$ of the projective plane corresponds to the dual line $aX+bY+cZ=0$. A line given by those coefficients corresponds back to the point $[a:b:c]$. Incidence is reversed: $p\in \ell$ if and only if ${\ell }^{\ast }\in p^{\ast }$. Thus collinear point triples become concurrent line triples, preserving the essential combinatorics of incidence geometry.

A projective conic is the zero locus in a projective plane of a nonzero homogeneous degree-two polynomial. The equation is considered up to multiplication by a nonzero scalar. Over the complex numbers, a symmetric matrix for the quadratic form has rank three, two, or one: these give a nonsingular conic, two distinct projective lines, or one line counted twice. Diagonalization of a quadratic form and rescaling its nonzero diagonal entries prove this classification. The smooth model $XZ-Y^2=0$ is parametrized by the degree-two Veronese map $[s:t]\mapsto [s^2:st:t^2]$.

An ordered projective frame consists of $n+2$ projective points such that any $n+1$ representative vectors are linearly independent. After mapping the first $n+1$ points to the coordinate axes, the last point has every coordinate nonzero; rescaling those axes sends it to $[1:\cdots :1]$. This proves existence and uniqueness of the projective transformation carrying one ordered frame to another. In the projective plane a frame consists of four points, no three collinear.

After choosing the affine chart $z\ne 0$ in the projective plane, its complement $z=0$ is the projective line at infinity. Affine parallel lines meet at the same projective point on this line, corresponding to their common direction. The choice of this line is part of the affine chart, not an intrinsic distinction preserved by every projective transformation.

A projective plane curve is a projective curve embedded in the projective plane, usually given by one homogeneous equation.

A projective triangle is an ordered triple of noncollinear projective points in a projective plane, together with their three joining projective lines. In the real projective plane this means that three chosen representative vectors are linearly independent. Pairing the vertices of two such triangles determines corresponding sides and the three corresponding-vertex joins.

On a genus-one curve, for distinct $p,q,r$, the Riemann-Roch theorem gives $\ell (p+q+r)=3$. Choose $f$ with simple poles exactly at $p,q$ and $g$ with simple poles exactly at $q,r$. Pole comparison proves $1,f,g$ independent. Multiplying these coordinates by a vanishing local parameter extends $F$ across every pole, with images $[0:1:0]$, $[0:b:c]$, $[0:0:1]$, where $bc\ne 0$. The ten homogeneous cubic monomials evaluate in $L(3p+3q+3r)$, of dimension nine, so a nonzero homogeneous cubic vanishes on the image. The three chosen points map to the line with first coordinate zero. This connects linear systems to projective plane equations by a dimension count.

For this toric cone, one smooth fan subdivision has maximal cones $abc,acd$ and no exceptional divisors. Another has $abd,bce,cde,dbe$, with $e=a+c=(b+c+d)/2$. Its only exceptional divisor is a projective plane, since the quotient fan at $e$ has the three rays $\bar{b},\bar{c},\bar{d}$ with sum zero and unimodular neighboring cones. Both subdivisions admit no proper smooth coarsening. Their common smooth refinement is obtained from the first by adding $e$ in $ac$ and then $w=a+e$ in $ae$, and from the second by adding $w=b+d$ in $bd$. Each step is a toric blowup along an orbit closure.