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For regular homogeneous equations of degrees $d_1,\dots ,d_r$ in ${\mathbb{P}}^n$, the normal bundle of a regular zero locus is ${⨁}_i\mathcal{O}(d_i){|}_V$. The Euler sequence gives $K_{{\mathbb{P}}^n}=\mathcal{O}(-n-1)$, hence $K_V=\mathcal{O}({\sum }_i{d}_i-n-1){|}_V$. Equations with unreduced multiplicities do not have this conclusion for the underlying reduced submanifold.

Over any field, the canonical bundle of projective space is $\mathcal{O}(-n-1)$. Indeed, dualizing the Euler sequence gives $0\to {\Omega }_{{\mathbf{P}}^n}^1\to \mathcal{O}(-1{)}^{\oplus (n+1)}\to \mathcal{O}\to 0$; taking determinants gives the result. Hence Serre duality identifies $H^i({\mathbf{P}}^n,\mathcal{O}(m))$ with the dual of $H^{n-i}({\mathbf{P}}^n,\mathcal{O}(-m-n-1))$.

For $n\ge 1$, the cotangent form of the Euler sequence is $0\to {\Omega }_{{\mathbb{P}}^n/k}^1\to \mathcal{O}(-1{)}^{\oplus (n+1)}\to \mathcal{O}\to 0$. The cohomology of twisting sheaves on projective space and the long exact sequence in sheaf cohomology give $H^1({\mathbb{P}}_k^n,{\Omega }^1)\cong k$, with every other cohomology group zero. Therefore the Euler characteristic of a coherent sheaf is $-1$. The nonzero class is the connecting image of the constant section $1$.

Use homogeneous coordinates $X_0,\dots ,X_n$. Multiplication by these degree-one sections defines the sheaf morphismOn $U_i=D_{+}(X_i)$, write $t_j^{(i)}=X_j/X_i$, with $t_i^{(i)}=1$, and let $E_j^{(i)}$ be the standard vector in component $j$ using the local line bundle frame $X_i^{-1}$ of $\mathcal{O}(-1)$. Then $ϵ(E_j^{(i)})=t_j^{(i)}$, so $ϵ$ is surjective since $ϵ(E_i^{(i)})=1$. Its kernel is a locally free sheaf with basisIndeed any vector in the kernel is uniquely a sum of these vectors, by solving for its $i$-th component.

The Kähler differential sheaf ${\Omega }_{{\mathbf{P}}^n/k}^1$ has basis $d{t}_j^{(i)}$ on $U_i$, since this chart is a polynomial affine scheme. Define the local map $d{t}_j^{(i)}\mapsto F_j^{(i)}$. To check that these isomorphisms glue, on $U_i\cap U_{\ell }$ put $u=t_i^{(\ell )}$ and $v=t_j^{(\ell )}$. ThenThe twisting sheaf on projective space frame changes by $E_j^{(i)}=u^{-1}{E}_j^{(\ell )}$, henceFor $j=\ell$, use $t_{\ell }^{(\ell )}=1$, $d{t}_{\ell }^{(\ell )}=0$ and $F_{\ell }^{(\ell )}=0$; the same formula applies. Thus the differential and kernel frames have identical transition matrices. We obtain the cotangent form of the Euler sequence:This construction works over any field, including positive characteristic.

The Kähler differential sheaf has rank $n$. Taking determinant line bundles in this locally split short exact sequence of sheaves givesThe left side is the tensor product of $n+1$ copies of $\mathcal{O}(-1)$, and $\det (\mathcal{O})=\mathcal{O}$. ConsequentlyIt is the canonical bundle of projective space. The determinant identity can also be seen directly by taking the wedge of the kernel basis followed by a lift of the quotient basis; changing that lift by a kernel vector leaves the wedge unchanged.

On the standard affine charts $U_i=\{X_i\ne 0\}$, use the formal frame $e_i=X_i^m$ and glue rank-one free sheaves byThe transition functions are regular units on overlaps and satisfy the cocycle identity. This constructs the twisting sheaf on projective space $\mathcal{O}(m)$ as an invertible sheaf for every integer $m$, including negative $m$.

On a nonempty $U$, a section has local expressions $h_i{e}_i$, where $h_i\in \mathcal{O}(U\cap U_i)$. Pulling to the punctured affine cone gives $h_i(X/X_i)X_i^m$. Their agreement on overlaps makes them one homogeneous rational function $r$ of degree $m$, regular on ${\pi }^{-1}U$. To make its polynomial representation explicit, write one nonzero local coefficient as $p(y)/q(y)$ in the affine coordinates. Homogenize $p,q$ to degrees $a,b$; the resulting rational expression isIf the exponent of $X_i$ is negative, move its power to the denominator. Cancel common factors; the greatest common divisor of homogeneous polynomials can be chosen homogeneous. Thus $r=F/G$ with $F,G$ coprime homogeneous polynomials, $G\ne 0$, and $\mathrm{deg}F-\mathrm{deg}G=m$.

Conversely, for such a homogeneous rational function regular on ${\pi }^{-1}U$, its coefficient $r/X_i^m$ is regular on $U\cap U_i$: restrict to the slice $X_i=1$. These coefficients obey the displayed transition rule and hence define a section. This proves the homogeneous rational sections of a twisting sheaf description in both directions, for arbitrary open $U$ rather than only a standard chart.

Now lift a regular degree-zero rational function $f=P/Q$ to the punctured cone. The quotient rule givesThis is a rational function homogeneous of degree $-1$. It is independent of the chosen representation because differentiation is a derivation of the rational function field. It is regular wherever $f$ is regular: a derivation of a polynomial ring extends to each localization by the quotient rule, and regular functions are locally such fractions. The preceding homogeneous-section description therefore provesZeros of a particular displayed denominator do not invalidate this argument; regularity is a property of the rational function, which can have another local representation.

The map $f\mapsto ({\partial }_{0}f,\dots ,{\partial }_{n}f)$ is a $k$-linear sheaf derivation with values in $\mathcal{O}(-1{)}^{\oplus (n+1)}$. The universal property of Kähler differentials consequently definesMultiplication by the coordinate section $X_i\in \Gamma ({\mathbb{P}}^n,\mathcal{O}(1))$ defines the other mapFor a degree-$d$ homogeneous polynomial, differentiating each monomial gives ${\sum }_i{X}_i{\partial }_{i}P=dP$. Applying this to the equal-degree numerator and denominator gives ${\sum }_i{X}_i{\partial }_{i}f=0$ for degree-zero $f$, so $\sigma D=0$. This identity is valid in every characteristic.

To prove all the exactness assertions, fix $U_i$ and put $y_j=X_j/X_i$ for $j\ne i$. The Kähler differential sheaf is free there on $d{y}_j$, because a derivation of the polynomial ring is determined freely by its values on the coordinates. Trivialize each $\mathcal{O}(-1)$ with frame $X_i^{-1}$. In this frame,where $e_j$ now denotes the $j$th coordinate vector of the direct sum. The map $\sigma$ is surjective, since its $i$th coefficient is one. Its kernel consists exactly of tuples with $b_i=-{\sum }_{j\ne i}{y}_j{b}_j$, so the $n$ displayed vectors $e_j-y_j{e}_i$ form a free basis of that kernel. The map $D$ sends the differential basis bijectively to this kernel basis and is therefore injective. Exactness on these charts proves the cotangent Euler sequence in homogeneous coordinatesThe proof uses no division by an integer and hence no characteristic-zero assumption.

For $n\ge 1$, the cohomology of twisting sheaves on projective space givesIn particular, $H^0(\mathcal{O})=k$, $H^{q>0}(\mathcal{O})=0$, and $H^q(\mathcal{O}(-1))=0$ for every $q$. The long exact sequence in sheaf cohomology of the displayed Euler sequence beginsand gives zero in the remaining degrees. ThusThe nonzero group is generated by the connecting image of the section $1$ of $\mathcal{O}$. If $n=0$ is admitted, ${\mathbb{P}}^0$ is a point and its cotangent sheaf is zero, so the sole requested group $H^0$ is zero.

Use the quotient convention for projective space. To a scheme $T$, associate the set of pairswhere $\mathcal{L}$ is a line bundle, modulo isomorphisms $\mathcal{L}\to {\mathcal{L}}^{\prime }$ carrying $q$ to $q^{\prime }$. Pullback defines the contravariant functor. On $T=\mathrm{Spec}A$, this means rank-one projective module quotients of $A^{n+1}$; the quotient need not be free. Thus merely listing generating tuples modulo multiplication by a unit would omit points over rings with nontrivial Picard group.

This functor is represented by ${\mathbb{P}}_{\mathbb{Z}}^n$. For each $i$, the locus where $q(e_i)$ generates $\mathcal{L}$ is open. These loci cover $T$, since a surjection onto a rank-one free stalk must have some coefficient a unit in its local ring. On this locus, trivialize by $q(e_i)$ and writeThese functions specify a map to the standard chart $U_i=\mathrm{Spec}\mathbb{Z}[t_j:j\ne i]$. Their transition rules are exactly those of projective space, so the chart maps glue. Conversely the universal quotient on projective space pulls back to the given pair, and the construction is natural and unique. This proves that projective space represents invertible quotients.

Write its universal exact sequence asA first-order change of the quotient, with target trivialized locally, is a map $\delta q:V\otimes \mathcal{O}\to \mathcal{O}(1)$. Changing the target trivialization by $1+\epsilon h$ changes $\delta q$ by $hq$. Restriction to $\mathcal{K}$ removes this ambiguity, and every local map $\mathcal{K}\to \mathcal{O}(1)$ extends because the displayed sequence splits locally. Consequently the tangent sheaf of a scheme isThis identification can also be seen on $U_i$: the first-order change of $t_j$ is $\delta q(e_j)-t_j\delta q(e_i)$ after setting $q(e_i)=1$. Thus these maps are precisely the coordinate tangent directions.

Apply $\mathcal{H}om(-,\mathcal{O}(1))$ to the universal sequence. Since its terms are locally free and it splits locally, this gives the Euler sequenceIt remains exact after any base change. Taking its dual of a sheaf over a field $k$ givesFor $n=1$, the first term has rank one; taking determinants yieldsExplicitly, for $u=t^{-1}$ on the overlap of the two affine charts, $du=-t^{-2}dt$, the same transition as $\mathcal{O}(-2)$ after changing one local frame's sign.

The holomorphic normal bundle is the quotient $N_V=T^{1,0}M{|}_V/T^{1,0}V$, of rank $r$. The tangent sequence is a short exact sequence of holomorphic vector bundles:Taking determinant line bundles gives $\det (T^{1,0}M){|}_V\cong \det (T^{1,0}V)\otimes \det N_V$. Dualizing and rearranging proves adjunction for a smooth submanifold:The determinant identity follows locally by adjoining lifts of a quotient frame to a subbundle frame; changing the lifts adds only off-diagonal blocks, so does not change the determinant.

For a smooth divisor, choose reduced local defining functions $f_i$, with $f_i=u_{ij}{f}_j$ and $u_{ij}$ holomorphic and nowhere zero. The holomorphic line bundle associated to a divisor $[V]={\mathcal{O}}_M(V)$ has local frames $e_i=f_i^{-1}$, with $e_j=u_{ij}{e}_i$. The products $f_i{e}_i$ define its canonical section vanishing on $V$. Along $V$,Thus $v\mapsto d{f}_i(v)e_i$ patches to a map $T^{1,0}M{|}_V\to [V]{|}_V$, whose kernel is $T^{1,0}V$ and which is surjective because each reduced defining function has nonzero differential there. Hence $N_V\cong [V]{|}_V$. This is the normal bundle of a smooth analytic hypersurface. The divisor construction uses a closed hypersurface: an arbitrary nonclosed embedded hypersurface need not be a divisor. For example, a punctured line in ${\mathbb{C}}^2$ cannot be the support of a divisor, since its missing limit point would also lie in every local holomorphic zero set containing that line.

Let $s$ be the section of $E$ defining $V$. The intended hypothesis is that it is a regular zero locus: $ds$ has rank $r$ on $V$. In a local trivialization, differentiation of the component functions definesThis is intrinsic: differentiating a change-of-frame matrix introduces terms multiplied by $s$, which vanish on $V$. Its kernel is $T^{1,0}V$, so it induces an isomorphism of holomorphic normal bundles. ConsequentlyThis is the normal bundle of a regular zero locus. Here $\det E=L_1\otimes \cdots \otimes L_r$.

Regularity is essential if “vanishing” is read only as equality of underlying sets. On ${\mathbb{P}}^2$, the section $Z_0^2$ of $\mathcal{O}(2)$ vanishes set-theoretically on a smooth line $V$. Its derivative vanishes on $V$, andThus smoothness of the underlying zero set alone does not establish the claimed normal-bundle isomorphism. The preceding proof applies when the equations define the reduced smooth submanifold, equivalently are transverse to zero.

For Complex projective space, put $\mathcal{O}(a)=[aH]$. At a line $\ell \subset {\mathbb{C}}^{n+1}$, the tangent space is $\mathrm{Hom}(\ell ,{\mathbb{C}}^{n+1}/\ell )$. Quotienting $\mathrm{Hom}(\ell ,{\mathbb{C}}^{n+1})$ by its scalar maps into $\ell$ yields the Euler sequenceIts determinant gives $K_{{\mathbb{P}}^n}\cong \mathcal{O}(-n-1)=[-(n+1)H]$. For a smooth complete intersection defined regularly by the homogeneous equations, apply the normal-bundle result to $E={⨁}_i\mathcal{O}(d_i)$:This is the canonical bundle of a regular projective complete intersection. The same regularity qualification is needed here: the squared-line example would otherwise predict ${\mathcal{O}}_{{\mathbb{P}}^1}(-1)$ for its canonical bundle, whereas the line has $K_V={\mathcal{O}}_{{\mathbb{P}}^1}(-2)$.

For the Sheaf of relative Kähler differentials on $X={\mathbb{P}}_{\mathbb{C}}^3$, the cotangent form of the Euler sequence isThe last arrow is onto because at least one homogeneous coordinate is invertible on each standard affine chart. Locally its kernel is the rank-three module generated by the differentials of the three affine coordinates, yielding the displayed Sheaf of relative Kähler differentials. The cohomology of twisting sheaves on projective space gives $H^q(X,{\mathcal{O}}_X(-1))=0$ for every $q$, while $H^0(X,{\mathcal{O}}_X)=\mathbb{C}$ and its positive-degree groups vanish. The long exact sequence in sheaf cohomology therefore gives the cotangent-sheaf cohomology of projective space in this dimension:The isomorphism for $q=1$ is the connecting map from the constant global sections of ${\mathcal{O}}_X$. The Euler characteristic of a coherent sheaf is the alternating sum of dimensions, explaining the minus sign.

For the curve, put $S=\mathbb{C}[t_0,t_1,t_2,t_3]$. It is a unique factorization domain, so irreducibility of $F$ makes $(F)$ a prime ideal. Also $F$ does not divide $G$: otherwise irreducibility of $G$ would make them associates, contrary to their distinct degrees. Thus the image of $G$ is a nonzero element of the integral domain $S/(F)$. It follows that $(F,G)$ is a regular sequence. Its Koszul complex is an exact sequence, and graded sheafification giveswhere $i:Z\hookrightarrow X$ is the closed immersion. The signs make the composite $F(-G)+GF=0$. The shifts come respectively from $5+7$, $5$, and $7$.

Let ${\mathcal{I}}_Z$ be the ideal sheaf of a closed subscheme. Split this Koszul resolution intoFor all integers $d$, $H^1(X,{\mathcal{O}}_X(d))=H^2(X,{\mathcal{O}}_X(d))=0$. Also $H^1(X,{\mathcal{O}}_X)=H^2(X,{\mathcal{O}}_X)=0$. The two long exact sequences in sheaf cohomology, together with sheaf cohomology under a closed inclusion, consequently identifyBy Serre duality, $H^3(X,{\mathcal{O}}_X(-m))\cong S_{m-4}^{\ast }$ for $m\ge 4$. Counting degree-$d$ monomials in four variables gives $\dim S_d=\left(\genfrac{}{}{0ex}{}{d+3}{3}\right)$. Thus the source has dimension $\left(\genfrac{}{}{0ex}{}{11}{3}\right)=165$, and the two target spaces have dimensions $\left(\genfrac{}{}{0ex}{}{4}{3}\right)=4$ and $\left(\genfrac{}{}{0ex}{}{6}{3}\right)=20$. The rank-nullity theorem proves the requested bound:

In fact equality holds. Under the Serre duality pairings, the dual of this map isIf $Ga=Fb$, primeness of $F$ and $F\nmid G$ imply $F\mid a$. But $a$ has degree one and $F$ has degree five, forcing $a=0$, and then $b=0$. The dual map is injective, so the original map is surjective and its kernel has dimension $141$. Thus the lower bound is attained for every pair allowed in the question. No smoothness assumption is needed. The regular sequence cuts out a projective complete intersection of dimension one. Also $H^0(X,{\mathcal{I}}_Z)=H^1(X,{\mathcal{I}}_Z)=0$, by the negative twists and the intermediate cohomology vanishing in the first short exact sequence, so the second gives $H^0(Z,{\mathcal{O}}_Z)=\mathbb{C}$. This is the genus of a complete-intersection space curve: the first cohomology has dimension $141$, equal to its arithmetic genus.

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.

The cohomology of twisting sheaves on projective space isIn particular, $H^i(X,{\mathcal{O}}_X)=H^i(X,{\mathcal{O}}_X(1))=0$ for $i>0$, while $H^0(X,{\mathcal{O}}_X)=k$ and $H^0(X,{\mathcal{O}}_X(1))\cong k^{r+1}$.

Apply the long exact sequence in cohomology to the displayed Euler sequence. Its degree-zero part isand all later terms vanish. The first map sends $1$ to the tuple of homogeneous coordinates and is injective. Therefore

The Euler sequence on ${\mathbb{P}}^m$ isThe Whitney product formula for the Total Chern class therefore givesin $A^{\ast }({\mathbb{P}}^m)=\mathbb{Z}[H]/(H^{m+1})$.

With the quotient convention, projective space represents the functor of surjections from a rank-$n+1$ free sheaf to a line bundle, modulo isomorphisms of the target compatible with the map. On an affine base the target may be a nontrivial rank-one projective module, so a tuple of elements generating the unit ideal is only the description when that target is trivial. Where the image of the $i$th basis vector generates the target, divide the other images by it to get the usual affine coordinates. Infinitesimal variations of the quotient form $\mathcal{H}om(\ker q,\mathcal{L})$ and give the Euler sequence.