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For a projective complete intersection curve cut out by a homogeneous regular sequence of positive degrees $a,b$ in ${\mathbb{P}}_k^3$, the Koszul resolution gives Hilbert polynomial $abm+ab(4-a-b)/2$. Hence its arithmetic genus is $1+ab(a+b-4)/2$. The same resolution shows $H^0(C,{\mathcal{O}}_C)=k$, and the dimension of a scheme is one, so this genus equals ${\dim }_k{H}^1(C,{\mathcal{O}}_C)$. Smoothness and irreducibility of the curve are not required. For degrees five and seven, the value is $141$.

The Koszul complex on $f_1,\dots ,f_r$ has term $K_i={\bigwedge }^i{R}^r$ and differential $\partial (e_{j_1}\wedge \cdots \wedge e_{j_i})={\sum }_a(-1{)}^{a-1}{f}_{j_a}{e}_{j_1}\wedge \cdots \wedge \widehat{e_{j_a}}\wedge \cdots \wedge e_{j_i}$. Its degree-zero homology is $R/(f_1,\dots ,f_r)$. For a regular sequence it is exact in positive degrees, hence a Koszul resolution of that quotient. The variables of a polynomial ring form a regular sequence, so this resolves the coefficient field in length equal to the number of variables.

For a regular sequence $f_1,\dots ,f_r$ in a commutative ring $R$, its Koszul complex is the finite free projective resolution of $R/(f_1,\dots ,f_r)$ with terms ${\bigwedge }^j{R}^r$ and boundary map contracting by $(f_1,\dots ,f_r)$. Exactness follows by induction: append $f_r$ using the mapping cone for multiplication by $f_r$, which is injective on the preceding quotient. For a polynomial ring viewed as a bimodule, use the regular sequence $X_{i,\ell }-X_{i,r}$ in the enveloping algebra.

For an ideal generated by a finite regular sequence in a commutative ring, the self-Ext functor algebra is exterior on the degree-one dual generators. Contraction chain maps on the Koszul resolution represent those generators, square to zero, and anticommute. The exterior square-zero relations also apply in characteristic two.

Embed a smooth plane conic in ${\mathbf{P}}^3$ through a plane. The plane equation and conic equation form a regular sequence of degrees one and two, so its normal sheaf is ${\mathcal{O}}_C(1)\oplus {\mathcal{O}}_C(2)$. Under $C\cong {\mathbf{P}}^1$, hyperplane degree is two, hence this normal bundle is ${\mathcal{O}}_{{\mathbf{P}}^1}(2)\oplus {\mathcal{O}}_{{\mathbf{P}}^1}(4)$. The normal exact sequence splits as well because $H^1({\mathbf{P}}^1,\mathcal{O}(2))=0$.

If a projective complete intersection $Z\subseteq {\mathbf{P}}^n$ is cut out by a homogeneous regular sequence of degrees $d_1,\dots ,d_c$, then $\mathcal{I}/{\mathcal{I}}^2\cong {⨁}_j{\mathcal{O}}_Z(-d_j)$ and ${\mathcal{N}}_{Z/{\mathbf{P}}^n}\cong {⨁}_j{\mathcal{O}}_Z(d_j)$. The classes of the defining equations form the conormal basis: the first syzygies of a regular sequence are the Koszul complex relations, whose coefficients vanish modulo the defining ideal. Thus the generator classes have no residual linear relations. Dualizing gives the stated normal sheaf.

The identification with ${\mathbf{P}}^1$ means that $C$ is a smooth plane conic. An arbitrary singular or nonreduced conic over an algebraically closed field cannot be identified with the projective line, so this is a necessary qualification implicit in the requested description.

Choose an equation $\ell =0$ for $L\subset {\mathbf{P}}^3$ and a degree-two equation $q=0$ whose restriction to $L$ defines $C$. Its homogeneous ideal in ${\mathbf{P}}^3$ is $(\ell ,q)$. These equations form a regular sequence: $\ell$ is a non-zero-divisor in the polynomial ring, and modulo $\ell$ the ring is a polynomial ring in three variables, in which the nonzero quadratic $q$ is again a non-zero-divisor. The generator classes therefore form a basis of the conormal sheaf, with their grading shifts:One can verify the absence of relations from the two-generator Koszul complex: a relation between $\ell ,q$ is a multiple of $(q,-\ell )$, whose coefficients become zero modulo ${\mathcal{I}}_C$. This is the normal sheaf of a projective complete intersection calculation. Dualizing the conormal sheaf gives

These twists are restrictions from ${\mathbf{P}}^3$, not intrinsic degree-one twists on ${\mathbf{P}}^1$. A line in the plane cuts the degree-two curve in two points counted with multiplicity, so the degree of a line bundle ${\mathcal{O}}_C(1)$ is two. The Picard group of the projective line then givesEquivalently, the degree-two Veronese embedding $[s:t]\mapsto [s^2:st:t^2]$ pulls each linear coordinate back to a quadratic. Thus the normal bundle of a smooth plane conic isThe same splitting follows from the normal sequence for the two embeddings:Its outer terms are ${\mathcal{O}}_{{\mathbf{P}}^1}(4)$ and ${\mathcal{O}}_{{\mathbf{P}}^1}(2)$. The extension group class lies in ${\mathrm{Ext}}^1(\mathcal{O}(2),\mathcal{O}(4))=H^1({\mathbf{P}}^1,\mathcal{O}(2))=0$, by the preceding Čech cohomology calculation, so the sequence splits. Both methods distinguish the ambient twist from the intrinsic degree on the curve.

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.

Let $R$ be a commutative ring and $f=(f_1,\dots ,f_r)$. The Koszul complex packages these elements and their relations in a finite chain complex of free modules. With $E=R^r$ and basis $e_1,\dots ,e_r$, set $K_i(f;R)={\bigwedge }^{i}E$ for $0\le i\le r$, and defineDeleting two distinct basis elements in the two possible orders gives opposite signs and the same coefficient, so ${\partial }^2=0$. Equivalently $K(f;R)$ is the tensor product of chain complexes of $r$ two-term complexes $[R\stackrel{f_i}{\to }R]$ in degrees one and zero. The tensor differential is $\partial (u\otimes v)=\partial u\otimes v+(-1{)}^{\mathrm{deg}u}u\otimes \partial v$. This fixes the sign convention. On the exterior algebra, the differential is a graded derivation determined by $\partial e_i=f_i$.

For an $R$-module $M$, define the Koszul complex with module coefficients by $K(f;M)=K(f;R){\otimes }_{R}M$. Its degree-zero Koszul homology isIts higher Koszul homology measures the failure of these equations to form a regular sequence on a module. Each $f_j$ annihilates every homology group. Indeed the Koszul homotopy for multiplication by a generator is $h_j(u)=e_j\wedge u$, and the graded product rule givesFor coefficients in $M$ the same identity holds after tensoring. Thus multiplication by $f_j$ is zero on homology, and the homology groups are naturally modules over $R/(f)$. If $(f)=R$, choose ${\sum }_j{a}_j{f}_j=1$; then ${\sum }_j{a}_j{h}_j$ is a contracting homotopy, so the entire complex is contractible. These identities are also useful after localization of a ring: wherever one generator is a unit, the complex has zero homology.

The construction is functorial under a ring homomorphism, and base change gives $K(f;R){\otimes }_{R}S=K(\bar{f};S)$, since its terms are free with the displayed basis and differential. An invertible change of generators $f_i={\sum }_j{u}_{ij}{g}_j$ gives a chain isomorphism by sending $e_i$ to ${\sum }_j{u}_{ij}{e}_j$ and extending to exterior powers. In particular, permuting the generators changes only the exterior signs, not the isomorphism class of the complex.

The essential exactness theorem is that a regular sequence on a module gives zero positive Koszul homology. Recall that $f_i$ must act injectively on $M/(f_1,\dots ,f_{i-1})M$, with final quotient nonzero. Put $f^{\prime }=(f_1,\dots ,f_{r-1})$. Adding the final two-term complex identifies $K(f;M)$ with the mapping cone of multiplication by $f_r$ on $K(f^{\prime };M)$, with the chosen tensor signs. The long exact sequence in homology consequently has segmentsFor $r=1$, positive homology is exactly $\ker (f_1:M\to M)$. By induction the preceding homology vanishes above zero. The final injectivity assumption kills $H_1(f;M)$, while the exact sequence kills every $H_i$ for $i>1$. Thereforefor a regular sequence on a module. In particular $K(f;R)$ is a finite free resolution of $R/(f)$ when $f$ is a regular sequence, with ranks $\left(\genfrac{}{}{0ex}{}{r}{i}\right)$ and length $r$.

There is a precise converse under local finiteness assumptions. Suppose $(R,\mathfrak{m})$ is a Noetherian local ring, $0\ne M$ is finitely generated, and all $f_i\in \mathfrak{m}$. If positive Koszul homology vanishes, the same exact sequence makes multiplication by $f_r$ surjective on every $H_i(f^{\prime };M)$ for $i>0$. These modules are finitely generated because the ring is Noetherian and the terms of the complex are finite modules. The Nakayama lemma gives $H_i(f^{\prime };M)=0$. Induction makes $f^{\prime }$ regular on $M$, and the degree-one part of the exact sequence makes $f_r$ injective on $H_0(f^{\prime };M)$. The final quotient is nonzero, again by the Nakayama lemma. This proves the Koszul acyclicity criterion in a Noetherian local ring:The hypotheses matter: a unit ideal makes the complex contractible but cannot be a proper regular sequence.

Examples make both sides visible. For $R=k[x,y]$ and $f=(x,y)$, the Koszul resolution isThe signs agree with $\partial (e_1\wedge e_2)=x{e}_2-y{e}_1$. For $R=k[x]$ and $f=(x,x)$, the degree-one cycles are $R(1,-1)$ and the boundaries are $xR(1,-1)$; hence $H_1\cong R/(x)$, while $H_2=0$. The repeated element has exposed a relation that the first element already kills.

The Koszul complex also computes derived functors. If $f$ is a regular sequence on $R$, its free resolution givesHere no regularity of $f$ on $M$ is assumed. For $S=k[x_1,\dots ,x_r]$, tensor the Koszul resolution on the variables with $k$. Every differential becomes zero, soThe top group is nonzero, showing that a free resolution of $k$ cannot be shorter than $r$.

Finally, wedging complementary exterior degrees gives a perfect pairing $K_i{\otimes }_R{K}_{r-i}\to {\bigwedge }^r{R}^r\cong R$. It identifies the dual cochain complex with the degree-reversed Koszul complex, after the appropriate signs. This self-duality of the Koszul complex shows, for a regular sequence, thatThus the same explicit complex simultaneously records quotient equations, regularity, relations, Tor functor computations and Ext functor computations. Its finiteness and exterior structure are what make it especially effective in commutative algebra.

Let $u_n$ be the rationalized universal class of degree $n$. The rational cohomology of an integral Eilenberg–MacLane space isHere $\Lambda$ is an exterior algebra. For $n=1$, $K(\mathbb{Z},1)\simeq S^1$. The standard path-loop spectral-sequence calculation supplies the induction: inthe total space is contractible and the fundamental fiber class transgresses to $u_n$. An odd exterior fiber generator gives an even polynomial base generator. An even polynomial fiber generator gives an odd exterior base generator; the differential on its $k$th power has coefficient $k$, which is invertible over $\mathbb{Q}$. The multiplicative spectral sequence then has no remaining positive-degree classes in the total space. This is the rational transgression calculation for Eilenberg–MacLane spaces; it includes the absence of additional base generators.

For $m=2n+1\ge 3$, choose $S^m\to K(\mathbb{Z},m)$ representing the integral fundamental class. The ring calculation shows that this map is a rational homology equivalence: both spaces have rational cohomology only in degrees zero and $m$. The rational Whitehead theorem for simply connected spaces identifies their rational homotopy groups. Since the target has only ${\pi }_m=\mathbb{Z}$, the rational homotopy groups of a sphere in odd dimension areFor $n=0$, this follows directly from ${\pi }_1(S^1)=\mathbb{Z}$ and the contractible universal cover of the circle, which makes every higher homotopy group zero.

Let $M={\mathbb{CP}}^2\#{\mathbb{CP}}^2$, with the standard complex orientations. It is simply connected by the Seifert-van Kampen theorem applied to the punctured summands. Classes $a,b\in H^2(M;\mathbb{Q})$ can be chosen from the two summands. Their cross product vanishes, while their squares equal the oriented top class:Thus the cohomology ring of the connected sum of two complex projective planes isThe two relations also kill all cubic monomials, so its dimensions are $1,2,1$ in degrees $0,2,4$ and zero otherwise.

We use the Sullivan minimal model dictionary: for a simply connected finite-type space, the dual of its degree-$i$ generator space is ${\pi }_i(M)\otimes \mathbb{Q}$. The following free graded-commutative differential algebra is the Sullivan model of the connected sum of two complex projective planes:It is minimal because all differentials of generators are decomposable.

To verify that no further generators are required, observe that $ab,a^2-b^2$ is a regular sequence in $\mathbb{Q}[a,b]$. The first polynomial is a nonzerodivisor. If $(a^2-b^2)h$ is divisible by $ab$, restricting to each coordinate axis forces $h$ to vanish on both axes, hence to be divisible by $ab$. The second polynomial is therefore a nonzerodivisor modulo the first. The Koszul complex of this regular sequence is exactly $\mathcal{M}$, so its cohomology is the quotient ring above, with no additional odd cohomology.

For completeness, choose rational polynomial forms representing $a,b$ on $M$. Their product and the difference of their squares are exact; choose degree-three primitives for them. Sending $r,s$ to those primitives defines a differential-algebra map from $\mathcal{M}$ to the rational polynomial forms on $M$. It induces the specified cohomology-ring isomorphism and hence is a quasi-isomorphism. This verifies the model directly, rather than assuming that a cohomology presentation alone automatically determines all rational homotopy.

The model has exactly two degree-two and two degree-three generators. Consequently the rational homotopy groups of the connected sum of two complex projective planes are

First use the canonical map to obtain the embedding. Its canonical divisor has degree six and $h^0(K_C)=4$. Over an algebraically closed ground field, every effective degree-two divisor on an algebraic curve $E$ satisfies $h^0(E)=1$: a second section would give a nonconstant rational function with poles bounded by $E$, hence a map $C\to {\mathbb{P}}^1$ of degree at most two. Degree one would make $C$ rational, and degree two would make it hyperelliptic, both excluded here.

The Riemann-Roch theorem now givesThus the complete linear system of divisors $|K_C|$ separates every pair of points and every tangent direction, including the tests $E=2p$. By the length-two criterion for a very ample linear system, it is a very ample linear system. Consequently the canonical map embeds $C$ in ${\mathbb{P}}^3$, with degree of a projective curve six. Its image is a nondegenerate projective variety, since the four canonical sections are linearly independent.

The space of homogeneous quadrics in four variables has dimension ten, while the Riemann-Roch theorem givesRestriction therefore has a nonzero kernel, giving a quadric $Q$ containing $C$. This quadric is irreducible: a reducible quadric is a union of two planes, or a double plane, and an integral curve lying in it would lie in a plane, contradicting nondegeneracy.

Similarly, homogeneous cubics form a twenty-dimensional space, whereasThere are at least five independent cubics vanishing on $C$. The multiples of $Q$ by linear forms give only four dimensions, so choose a cubic $F$ vanishing on $C$ which is not divisible by $Q$.

The polynomials $Q,F$ are a regular sequence, hence their intersection $D$ is a projective complete intersection of pure dimension one and degree $2\cdot 3=6$. It contains $C$, which already has degree six. The unmixedness of a complete intersection excludes embedded components. Equal degrees force $C$ to be the only irreducible component and force multiplicity one at its generic point. By generic reducedness with no embedded components, this makes $D$ reduced, so $D=C$ as schemes. We have provedThis is a canonical genus-four curve as a quadric-cubic intersection. The quadric may be smooth or a cone; the argument does not require it to be smooth. The same construction descends over the ground field when the curve is geometrically nonhyperelliptic, since the restriction maps and canonical embedding are defined there.

For central $t_1,\dots ,t_n$, define the Koszul complex on central ring elements using formal exterior symbols:with zero terms outside $0\le k\le n$ and differentialThe terms are free bimodules with the formal symbols commuting with coefficients. Centrality makes $d$ a bimodule map, and the terms in $d^2$ cancel in pairs because the $t_i$ commute. This defines the Koszul complex over a possibly noncommutative ring without using an undefined exterior algebra of arbitrary one-sided modules. In particular $K(\mathbf{t}){\otimes }_{R}M$ is a well-defined chain complex of left modules.

A regular sequence on a module means that multiplication by $t_i$ is injective on $M/(t_1,\dots ,t_{i-1})M$ for each $i$, usually with the additional convention that the final quotient is nonzero. The homology vanishing below uses only the injectivity conditions. For one element, the chain complex is $M\stackrel{t_1}{\to }M$, with zero first homology and zeroth homology $M/t_{1}M$.

For the induction let $C=K(t_1,\dots ,t_{n-1}){\otimes }_{R}M$. Adjoining the last generator identifies the new Koszul complex with the mapping cone of multiplication by $t_n$ on $C$. In the convention $\mathrm{Cone}(t_n{)}_k=C_k\oplus C_{k-1}$ its differential is $(a,b)\mapsto (da+t_{n}b,-db)$; the identification is $a+e_n\wedge b$. The long exact sequence in homology of this mapping cone givesThe induction hypothesis is $H_k(C)=0$ for $k>0$ and $H_0(C)=N=M/(t_1,\dots ,t_{n-1})M$. Regularity makes $t_n:N\to N$ injective. Hence the new positive homology is zero and its zeroth homology is $N/t_{n}N$. The augmentation to this quotient induces these homology isomorphisms, giving the quasi-isomorphismwhere the right side is placed in degree zero. The results used are the long exact sequence in homology of a degreewise short exact sequence of complexes and its mapping cone form, together with the stated induction; no unmentioned acyclicity criterion is needed.

For $R=\mathbb{Z}[X]$, $(p,X)$ is a regular sequence: $p$ is a non-zero-divisor in $R$, and multiplication by $X$ is injective in $(\mathbb{Z}/p\mathbb{Z})[X]$, even when $p$ is composite. The Koszul resolution of $R/(p,X)$ isWith $N=\mathbb{Z}/q\mathbb{Z}$ and $X$ acting as zero, applying the Hom functor gives the cochain complexThus, more generally, the Ext groups between polynomial-ring residue modules areand vanish in all degrees above two. For the equal-modulus case,These are $R$-modules through $X=0$ and reduction modulo $p$. If the multiplicative self-Ext functor is desired, the Koszul self-Ext algebra is the exterior algebra on two degree-one generators over $\mathbb{Z}/p\mathbb{Z}$. The two contractions on the Koszul resolution lift these classes, square to zero and anticommute, so this description also holds for composite $p$ and characteristic two. For coprime $p,q$, multiplication by $p$ is invertible on $N$, so both its kernel and cokernel vanish:

For a finitely generated module $V={⨁}_d{V}_d$ over the graded ring $S=k[X_1,\dots ,X_n]$, with each $X_i$ of degree one, its Poincare series of a graded module isThis is also its Hilbert series. One usually takes $d\ge 0$; allowing an integer grading also permits finitely many negative degrees. Homogeneous generators of degrees $a_1,\dots ,a_h$ give a surjection ${⨁}_{j}S(-a_j)\to V$, where the graded shift is defined by $V(-a{)}_d=V_{d-a}$. Consequently every $V_d$ is a finite-dimensional vector space and $V_d=0$ for all sufficiently negative $d$.

The Hilbert-Serre theorem in the standard grading saysFor a nonnegatively graded $V$, the numerator is in $\mathbb{Z}[t]$. Equivalently, the only possible pole of this rational expression is at $t=1$, with order at most $n$.

We prove the Hilbert-Serre theorem by induction on the number of variables. For $n=0$, $S=k$ and $V$ is a finite-dimensional graded module; thus $P_V(t)$ is a Laurent polynomial. Suppose $n>0$ and putThe Hilbert basis theorem makes $S$ a Noetherian ring, so $K$ is a finitely generated module, as is $C$. Both are graded modules killed by $X_n$, hence finitely generated over $S^{\prime }=k[X_1,\dots ,X_{n-1}]$.

The graded exact sequenceidentifies the kernel and cokernel of multiplication by $X_n$. Taking the alternating sum of the finite-dimensional degree components yieldssoBy induction the right side has denominator $(1-t{)}^{n-1}$, which proves the required denominator $(1-t{)}^n$. Since $(1-t{)}^n{P}_V(t)$ has no negative powers when $V$ is nonnegatively graded, its Laurent polynomial numerator is an ordinary polynomial in that case.

For $n\ge 1$, this also implies the usual Hilbert polynomial consequence. Writing $Q(t)={\sum }_j{q}_j{t}^j$ and usinggives, for all sufficiently large $d$,a polynomial in $d$ of degree at most $n-1$. For $n=0$, the graded pieces are eventually zero.

For the final request, a free resolution of $V$ is an exact sequencewith each $F_i$ a free module. In the graded setting we take finite direct sums of shifts $S(-a)$ and maps preserving degree. The resolution has length at most $n$ if $F_i=0$ for every $i>n$. The syzygy modules are the successive kernels which record the relations among generators, then relations among those relations, and so on.

Put $\mathfrak{m}=(X_1,\dots ,X_n)$, the homogeneous maximal ideal with $S/\mathfrak{m}=k$. Choose a homogeneous $k$-basis of $V/\mathfrak{m}V$ and lift it to $V$. These lifts generate $V$ by the graded Nakayama lemma: a bounded-below graded module $W$ satisfying $W=\mathfrak{m}W$ must be zero, since a nonzero homogeneous element of least degree could not be a sum of variables times elements of lower degrees. Apply this to the quotient by the submodule generated by the chosen lifts. This gives a surjection $F_0\to V$ inducing an isomorphism modulo $\mathfrak{m}$.

Its kernel is finitely generated because $S$ is a Noetherian ring. Repeat the construction for that kernel, then for each subsequent kernel. We obtain a minimal graded free resolution, meaningIndeed, at each stage the free cover induces an isomorphism modulo $\mathfrak{m}$, so its kernel lies in $\mathfrak{m}$ times its source. The sequence is exact by construction, though it may at first appear infinite.

The Koszul complex on $X_1,\dots ,X_n$ iswith basis vectors $e_j$ given degree one, and differentialThe hat means omission of that factor. Every pair of terms in ${\partial }^2$ cancels with opposite signs, so this is a chain complex. The sequence $X_1,\dots ,X_n$ is a regular sequence: after quotienting by the first $j-1$ variables, the next variable is a non-zero-divisor in the remaining polynomial ring.

The Koszul complex is consequently exact in positive degrees and has $H_0=k$, so it is a Koszul resolution of $k$ of length $n$. For completeness, one proves this by induction: for one variable it is $0\to S\stackrel{X_1}{\to }S\to S/(X_1)\to 0$. Appending $X_j$ constructs the mapping cone of multiplication by $X_j$ on the previous chain complex. Its homology is zero in positive degrees because $X_j$ acts injectively on $S/(X_1,\dots ,X_{j-1})$; its degree-zero homology is the further quotient. Here a mapping cone combines a chain complex with a shifted copy and adds the given multiplication map to the differential, thereby measuring its kernel and cokernel on homology.

The Tor functor ${\mathrm{Tor}}_i^S(V,k)$ is defined as the degree-$i$ homology of $F_{\bullet }{\otimes }_{S}k$ for a free resolution $F_{\bullet }$ of $V$. It can equally be computed as the homology of $V{\otimes }_S{K}_{\bullet }$, using the Koszul resolution of $k$. To see the equality, form the double complex $F_{\bullet }{\otimes }_S{K}_{\bullet }$. Taking homology first along $F_{\bullet }$ leaves $V{\otimes }_S{K}_{\bullet }$, since each $K_i$ is a free module; taking homology first along $K_{\bullet }$ leaves $F_{\bullet }{\otimes }_{S}k$, since each $F_i$ is free. The finite sums along each total degree identify both with the homology of the total complex. This also explains the symmetry used in computing the Tor functor.

Since $K_i=0$ for $i>n$, we have ${\mathrm{Tor}}_i^S(V,k)=0$ for $i>n$. On the other hand, all differentials of the minimal graded free resolution become zero after tensoring with $k=S/\mathfrak{m}$, soThus $F_i/\mathfrak{m}{F}_i=0$ for $i>n$, and the graded Nakayama lemma implies $F_i=0$ for those $i$. We have proved the required Hilbert syzygy theorem:where some initial $F_i$ may be zero. The argument supplies finitely generated graded free modules, which is stronger than merely giving an ungraded free resolution.

Put $R_j=S/I_j$ and $R_0=S$. The hypothesis says multiplication by $F_j$ on the graded ring $R_{j-1}$ is injective for $j=2,3$. Checking homogeneous elements suffices: an arbitrary annihilated element splits into homogeneous components, each annihilated separately because $F_j$ and $I_{j-1}$ are homogeneous. Also $F_1\ne 0$ because it has degree $d_1>0$, and $S$ is an integral domain. Thus $F_1,F_2,F_3$ is a regular sequence, and the suggested short exact sequences becomeSheafification and twisting preserve exactness. With $X_0={\mathbb{P}}_{\mathbb{C}}^4$, the resulting sequences, interpreted on the ambient projective space via the closed inclusions, areCohomology under a closed immersion identifies the displayed sheaf cohomology with that on $X_j$. For every integer $m$, cohomology of twisting sheaves on projective space gives $H^q(X_0,\mathcal{O}(m))=0$ for $1\le q\le 3$. Inducting through the long exact sequence in sheaf cohomology givesIndeed the two adjacent groups have degrees $q$ and $q+1$ on $X_{j-1}$, both in its vanishing range. This is intermediate cohomology vanishing for a projective complete intersection.

We simultaneously prove $H^0(X_j,{\mathcal{O}}_{X_j}(m))=0$ for $m<0$ and $H^0(X_j,{\mathcal{O}}_{X_j})=\mathbb{C}$. Both hold on $X_0$ by the same projective-space formula. For $j\le 3$, the preceding stage has $H^1(X_{j-1},\mathcal{O}(m-d_j))=0$. If $m<0$, both degree-zero groups on that stage vanish, so the new one vanishes. If $m=0$, the left degree-zero group vanishes because $-d_j<0$, while the middle group is $\mathbb{C}$. Therefore the natural restriction of constant functions is an isomorphism at every stage:These are isomorphisms of $\mathbb{C}$-algebras, not just vector spaces. The global regular functions on a positive-dimensional projective complete intersection are constants even if the complete intersection is singular or nonreduced; smoothness and reducedness were not assumed.

A derivation of an algebra is a $k$-linear map $D:A\to A$ satisfying $D(ab)=D(a)b+aD(b)$. The commutator $[D,E]=D\circ E-E\circ D$ is again a derivation: expanding $[D,E](ab)$ cancels the two mixed terms and leaves $[D,E](a)b+a[D,E](b)$. The commutator on endomorphisms is bilinear, antisymmetric, and satisfies the Jacobi identity by cancellation of its twelve triple-composition terms. Therefore ${\mathrm{Der}}_k(A)$ is a Lie algebra.

In degree zero of the Hochschild cochain complex, $\delta a(b)=ba-ab$, so $H{H}^0(A,A)=Z(A)=A$ for commutative $A$. In degree one, $\delta D=0$ is precisely the derivation rule; the boundaries are inner derivations, which vanish for commutative $A$. HenceFor cochains $f\in C^p(A,A)$, $g\in C^q(A,A)$, the Hochschild cup product isDefine the insertion operation byA degree-zero cochain is an element of $A$, inserted with no arguments; for $p=0$ the sum is empty. For the Gerstenhaber bracket we use the left graded Leibniz rule convention, compatible with the unsigned Hochschild cup product just displayed:Both degree-zero inputs have bracket zero. Another common insertion convention writes $f\circ g-(-1{)}^{(p-1)(q-1)}g\circ f$; the two brackets differ by $(-1{)}^{(p-1)(q-1)}$. With an unsigned Hochschild cup product, that convention uses the corresponding right graded Leibniz rule. The distinction matters for a degree-two cochain bracketed with a function. Either consistent convention gives the same degree-one Lie bracket and the same derivation action on functions.