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For an affine variety, every coherent ideal sheaf is quasi-coherent; vanishing of quasi-coherent cohomology on an affine scheme therefore gives $H^1(X,\mathcal{I})=0$.

For the projective-space complement, assume $n\ge 1$ and choose two distinct closed points $P,Q\in X$. Take the ideal sheaf of two closed points $\mathcal{I}={\mathfrak{m}}_P\cap {\mathfrak{m}}_Q$ on $X$. The codimension-two extension of regular functions on a normal variety givesOne can see this directly: on every standard affine chart of ${\mathbb{P}}^n$, a rational function written in lowest terms cannot have a nonconstant denominator, because an irreducible polynomial factor of the denominator would define a pole along a codimension-one hypersurface, and such a hypersurface is not removed by $Z$. The extended function is constant because every global regular function on projective space is constant. Now the short exact sequencesends $k$ diagonally into $k^2$ on global sections. Its cokernel is $k$, and the long exact sequence in sheaf cohomology injects that cokernel into $H^1(X,\mathcal{I})$. ThusThe assumption $n\ge 1$ is necessary: ${\mathbb{P}}^0$ is already affine and has no such example.

Let $\mathfrak{m}$ be the maximal ideal of the local ring $A$. Every open subset of $\mathrm{Spec}A$ containing the closed point $\mathfrak{m}$ is the whole spectrum of a commutative ring: it contains a principal open subscheme $D(a)$ with $a\notin \mathfrak{m}$, and that $a$ is a unit, so $D(a)=\mathrm{Spec}A$.

Given $f:\mathrm{Spec}A\to {\mathbb{P}}_{\mathbb{Z}}^n$, choose a standard affine open subscheme $D_{+}(x_i)$ containing $f(\mathfrak{m})$. Its preimage is consequently all of $\mathrm{Spec}A$. The affine-target adjunction for schemes expresses $f$ in this chart by elements $b_j\in A$ for $j\ne i$, the images of $x_j/x_i$. It is represented by homogeneous coordinates with $a_i=1$ and $a_j=b_j$.

Conversely, a tuple $(a_0,\dots ,a_n)$ with some $a_i\in A^{\times }$ defines a morphism of schemes into $D_{+}(x_i)$ by $x_j/x_i\mapsto a_j/a_i$. Choosing another unit entry gives the same morphism of schemes, since the usual projective space transition functions identify the ratios. Multiplying all entries by one unit does not change any ratio. If two such tuples define the same morphism of schemes, choose a unit entry $a_i$ in the first and a unit entry $a_j^{\prime }$ in the second. In the second chart, the function $x_i/x_j$ pulls back to $a_i^{\prime }/a_j^{\prime }$. Because the whole map lies in $D_{+}(x_i)$, this ratio is a unit, so $a_i^{\prime }$ is a unit too. Equality in this chart gives $a_j/a_i=a_j^{\prime }/a_i^{\prime }$ for every $j$, hence $a_j^{\prime }=(a_i^{\prime }/a_i)a_j$.

Thus the correspondence is exactlyThis is the projective coordinates over a local ring description.

For a general ring, the key open-neighbourhood argument fails. Even a tuple generating the unit ideal need not have any unit entry. For example, over $A=k\times k$ the pair $((1,0),(0,1))$ defines a map to ${\mathbb{P}}^1$ whose two points have images $[1:0]$ and $[0:1]$. Neither coordinate is a unit, and no common unit multiple changes that fact. The map lies in no single standard chart. More generally, maps into projective space correspond to invertible sheaf quotients of $A^{n+1}$; the quotient need not be a free rank-one module outside the local ring case.

A line bundle $\mathcal{L}$ is a globally generated line bundle when the evaluation mapis surjective. Equivalently, its global sections span each fibre of the line bundle; locally at every point, some section is a generator.

Choose a finite generating family $s_0,\dots ,s_n$. On the open set where $s_i$ generates $\mathcal{L}$, the ratios $s_j/s_i$ are regular functions, giving a morphism of schemes into the standard chart $D_{+}(x_i)$ of projective space. The ratios obey the usual transition rules on overlaps, so these chart maps glue toThe tuple is computed using any local trivialization of $\mathcal{L}$; changing that trivialization multiplies all entries by the same invertible function. The construction has ${\phi }^{\ast }\mathcal{O}(1)\simeq \mathcal{L}$ and pulls back the coordinate sections to the chosen $s_i$.

There is a finiteness qualification for an arbitrary $X$: global generation alone need not provide such a finite family. It does if $X$ is quasi-compact, since the open sets on which individual sections generate have a finite subcover. Without that hypothesis, take $X={\coprod }_{m\ge 1}{\mathbb{P}}_k^m$ and let $\mathcal{L}$ restrict to $\mathcal{O}(1)$ on each component. This line bundle is globally generated, but any $N$ global sections have a common zero on a component ${\mathbb{P}}^m$ with $m\ge N$. Thus this is a globally generated line bundle without finite generators. This is finite global generation on a quasi-compact scheme. The finite-family construction is automatic in the projective case asked next.

For projective nonsingular $X$, put $V=⟨s_0,\dots ,s_n⟩$. The length-two criterion for a very ample linear system says that $\phi$ is a closed immersion precisely when, after extending to an algebraic closure, $V$ separates distinct points and tangent directions. Equivalently, the evaluationis surjective for every length-two geometric closed subscheme $E\subseteq X$. Two distinct points give point separation; a nonreduced length-two subscheme supported at one point gives separation of a direction in the Zariski tangent space. For the complete space $V=H^0(X,\mathcal{L})$, this says exactly that $\mathcal{L}$ is a very ample line bundle. For a chosen smaller $V$, it is the chosen linear system of divisors which must be a very ample linear system; mere global generation is insufficient.

The resolution principle for sheaf cohomology states that an exact sequencewith $H^q(X,{\mathcal{A}}^j)=0$ for every $q>0$ computes $H^i(X,\mathcal{F})$ as the cohomology of $\Gamma (X,{\mathcal{A}}^{\bullet })$. In particular one may use a flasque resolution.

An affine morphism $\varphi :X\to Y$ is one for which the inverse image of every affine open subset of $Y$ is affine. We use vanishing of quasi-coherent cohomology on an affine scheme: a quasi-coherent sheaf has zero higher sheaf cohomology on an affine variety. Take a flasque resolution $\mathcal{F}\to {\mathcal{I}}^{\bullet }$. Each direct image sheaf ${\varphi }_{\ast }{\mathcal{I}}^j$ is flasque, because restriction on $V\subseteq U\subseteq Y$ is restriction on ${\varphi }^{-1}V\subseteq {\varphi }^{-1}U$. On an affine open $U\subseteq Y$, the augmented complex of sections isIts positive-degree cohomology vanishes by the affine theorem, and its degree-zero kernel is $\Gamma ({\varphi }^{-1}U,\mathcal{F})$. Checking exactness on this affine basis shows that ${\varphi }_{\ast }{\mathcal{I}}^{\bullet }$ is a flasque resolution of ${\varphi }_{\ast }\mathcal{F}$. The global-section complexes are identical, proving cohomology under an affine morphism:

For projective space, put $S=k[x_0,\dots ,x_n]$ with its usual grading and $U_i=\{x_i\ne 0\}$. Construct the twisting sheaf on projective space by setting its sections on $U_i$ to be the degree-$m$ part $(S_{x_i}{)}_m$, viewed as a module over $(S_{x_i}{)}_0$. It is free of rank one with generator $x_i^m$, even when $m$ is negative. On overlaps the generators differ by the unit $(x_i/x_j{)}^m$, so these modules glue to a line bundle $\mathcal{O}(m)$.

Assume first $n\ge 1$. These local modules all embed in the graded fraction field, and agreement on overlaps identifies global sections with ${\bigcap }_i(S_{x_i}{)}_m$. Since $S$ is a unique factorization domain and at least two distinct variables occur, a reduced fraction belonging to every $S_{x_i}$ has no nonconstant denominator. Thus the intersection is $S_m$. Counting monomials by stars and bars givesHere $h_n^i(m)={\dim }_k{H}^i({\mathbf{P}}^n,\mathcal{O}(m))$. The printed formula needs the convention that a counting binomial coefficient is zero when its upper integer is smaller than its nonnegative lower integer; the generalized polynomial convention for negative upper integers would give incorrect dimensions. For $n=0$, every twist on the single point ${\mathbf{P}}^0$ is trivial, so $h_0^0(m)=1$ for every $m$ and its higher cohomology is zero. This case supplies the base of dimension induction; the displayed top-cohomology formula below is for $n\ge 1$.

The standard $n+1$ affine opens have affine finite intersections. The acyclic cover theorem and affine vanishing identify sheaf cohomology with a Čech cochain complex ending in degree $n$. In particular $H^i({\mathbf{P}}^n,\mathcal{O}(m))=0$ for $i>n$.

Let $H=\{x_n=0\}\cong {\mathbf{P}}^{n-1}$, with inclusion $i$. Multiplication by $x_n$ gives the hyperplane exact sequence for twisting sheavesExactness is checked in the local trivializations, where a hyperplane equation is a non-zero-divisor and the quotient is its restriction to $H$. Since a closed immersion is an affine morphism, the first part identifies the cohomology of $i_{\ast }{\mathcal{O}}_H(m)$ with that on $H$.

For $n=1$, the long exact sequence in sheaf cohomology and $H\cong {\mathbf{P}}^0$ giveStarting from the given vanishing for sufficiently large $m$ and descending gives $h_1^1(m)=0$ for $m\ge -1$ and $h_1^1(m)=-m-1$ for $m\le -2$.

Now let $n\ge 2$, and assume the formulas in dimension $n-1$. Restrictionis surjective for every $m$: for $m\ge 0$ it is polynomial restriction, and for $m<0$ both groups are zero since $\dim H\ge 1$. The long exact sequence in sheaf cohomology therefore makes $H^1({\mathbf{P}}^n,\mathcal{O}(m-1))\to H^1({\mathbf{P}}^n,\mathcal{O}(m))$ injective. For $2\le j<n$, dimension induction gives $H^{j-1}(H,{\mathcal{O}}_H(m))=0$, so the same sequence makesinjective. Composing these injections up to a sufficiently large twist, where the target is zero by the given hypothesis, proves $H^j({\mathbf{P}}^n,\mathcal{O}(m))=0$ for every $m$ and $0<j<n$.

The remaining part of the long exact sequence in sheaf cohomology is consequentlywhere the last zero uses the affine-cover dimension bound on $H$. Hence $h_n^n(m-1)=h_n^n(m)+h_{n-1}^{n-1}(m)$. Using dimension induction and eventual vanishing, downward summation yieldsby telescoping Pascal's identity; for $m\ge -n$ it is zero. Thus, without invoking Serre duality in the induction,Finally, the canonical bundle of projective space is ${\omega }_{{\mathbf{P}}^n}\cong \mathcal{O}(-n-1)$, obtained by taking determinants in the dual Euler sequence. Serre duality therefore predictsIt pairs the two computed extreme-degree dimensions and pairs zero intermediate groups with zero intermediate groups, exactly as required.

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

PutBy the Jacobian criterion, a projective plane curve is singular only at a common projective zero of $F$ and all its partial derivatives. HereTheir simultaneous vanishing forces $X_0=X_1=X_2=0$, which is not a point of projective space. Hence the Fermat quartic $C$ is nonsingular. In particular it is a smooth plane quartic and has geometric genus $(4-1)(4-2)/2=3$.

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.

Let $R$ be a discrete valuation ring with uniformizer $\pi$, fraction field $K$, and normalized discrete valuation $\nu :K^{\times }\to \mathbb{Z}$. A map $\mathrm{Spec}K\to {\mathbb{P}}_{\mathbb{Z}}^n$ is a projective pointwith at least one nonzero coordinate. Put $m={\min }_{a_i\ne 0}\nu (a_i)$ and set $b_i={\pi }^{-m}{a}_i$. Then every $b_i$ lies in $R$, and at least one $b_j$ is a unit.

On the standard affine chart $U_j$ of projective space, the ratios $b_i/b_j$ all lie in $R$. They therefore define a map $\mathrm{Spec}R\to U_j\subseteq {\mathbb{P}}_{\mathbb{Z}}^n$ whose generic restriction is the original point. This proves existence in the DVR case. The assumed separatedness of ${\mathbb{P}}_{\mathbb{Z}}^n\to \mathrm{Spec}\mathbb{Z}$, through the valuative criterion for separatedness, gives uniqueness. Thus the morphism satisfies the requested DVR form of the valuative criterion for properness.

The projective bundle of a real or complex vector bundle $E\to X$ has fiber the space of one-dimensional subspaces of $E_x$. It carries a tautological bundle $\lambda \subset p^{\ast }E$; in the real case this is the projectivization of a real vector bundle.

A projective complete intersection is a closed subscheme of a projective space defined by a homogeneous regular sequence of positive degrees. A length-$r$ sequence in $k[t_0,\dots ,t_N]$ gives dimension $N-r$ when $r\le N$. Sheafifying its successive quotient sequences gives exact restriction sequences of twisting sheaves on projective space, allowing computation of its sheaf cohomology. It need not be smooth or reduced.

For a local ring $A$, every map from $\mathrm{Spec}A$ into projective space lands in one standard chart: the inverse image of a chart containing the closed point is the whole spectrum. This gives homogeneous coordinates with at least one unit entry, unique up to a common unit. For a nonlocal ring, coordinates may instead merely generate the unit ideal, and general maps can involve a nontrivial invertible sheaf quotient.

The projective plane is the two-dimensional projective space ${\mathbb{P}}^2$.

A locally closed affine algebraic set in a projective space, equivalently an open subset of a closed subset of projective space cut out by homogeneous polynomials. Reducibility is allowed for an algebraic set; an irreducible variety with such an embedding is a quasi-projective variety.

An irreducible variety which admits a locally closed embedding into projective space. Open subsets of affine varieties and blowups of algebraic varieties along closed centres give important examples.