Past exam of the mathematics course of the University of Cambridge / 2016 / iii

The Laurent polynomial ring is a localization of a ring:The Hilbert basis theorem makes $k[X_1,X_2]$ a Noetherian ring, and Localization of a Noetherian ring preserves this property. Thus $\boxed{R\text{ is Noetherian}}$.

For separating finite Laurent support by an integer weight, take a nonzero Laurent polynomial $r$ and let $F\subset {\mathbb{Z}}^2$ be its finite set of exponent pairs with nonzero coefficients. Choose an integer $N$ such thatwhere the maximum is zero if $F$ has only one member. Set $(a_1,a_2)=(1,N)$. If $m_1+N{m}_2=m_1^{\prime }+N{m}_2^{\prime }$, thenso $m_2=m_2^{\prime }$ and then $m_1=m_1^{\prime }$. Consequently all powers of $T$ in ${\theta }_{(1,N)}(r)$ are distinct. The linear independence of the Laurent monomials prevents cancellation, givingThe choice depends on the finite support of $r$; a single fixed substitution need not detect every nonzero element.

For Noetherianity of subalgebras of a one-variable Laurent polynomial ring, let $A$ be a unital $k$-subalgebra of $k[T,T^{-1}]$. If $A=k$, it is already a Noetherian ring. Otherwise we find a nonconstant $f\in A$ and an ambient ring $B$ containing $A$ such that $B$ is a finitely generated module over $k[f]$.

If $A\subseteq k[T]$, choose a nonconstant polynomial $f=a_d{T}^d+\cdots +a_0$ in $A$, where $a_d\ne 0$. Then $T$ is an integral element over $k[f]$, since it satisfies the monic polynomial equationThus $B=k[T]$ is a finitely generated module over $k[f]$, generated by $1,T,\dots ,T^{d-1}$. The case $A\subseteq k[T^{-1}]$ is identical after replacing $T$ by $T^{-1}$.

In the remaining case $A$ is contained in neither of these two polynomial rings. There is an element with a positive exponent and an element with a negative exponent. If either already has both, use it as $f$. Otherwise their sum has both, since the positive powers of the first and the negative powers of the second cannot cancel. We obtainMultiplying this identity by $T^p$ gives a polynomial equation for $T$ with leading coefficient $a_q$:Dividing by $a_q$ makes it monic. Similarly, multiplication by $T^{-q}$ gives a monic equation for $T^{-1}$ after division by $a_{-p}$. Both generators are therefore integral elements over $k[f]$, so $B=k[T,T^{-1}]$ is a finitely generated module over $k[f]$.

In all three cases $k[f]$ is a Noetherian ring, as a quotient of a one-variable polynomial ring. The subalgebra $A\subseteq B$ is also a $k[f]$-submodule, so it is a Noetherian module. Every ideal of $A$ is a $k[f]$-submodule of $A$ and is therefore finitely generated over $k[f]$, hence over $A$. This proves

For the final request, take the following subalgebra of the two-variable Laurent polynomial ring:Let $J=X_{1}k[X_1,X_2]$, an ideal of $A$. We haveas $k$-vector spaces. In particular, the classes of $X_1{X}_2^j$, for $j\ge 0$, are linearly independent. If $J$ had finitely many generators as an $A$-module, their classes would span $J/J^2$ over $A/J=k$, contradicting its infinite dimension. Hence $J$ is not finitely generated and $A$ is not a Noetherian ring. This illustrates why the preceding result is special to one Laurent variable.

The Weak Hilbert Nullstellensatz states that if $k$ is an algebraically closed field, every maximal ideal of $k[Y_1,\dots ,Y_s]$ isEquivalently, every proper ideal has a common zero. We prove the more general field statement, the Zariski lemma: a field finitely generated as a $k$-algebra is a finite field extension of $k$.

Write such a field as $L=k[b_1,\dots ,b_s]$. Choose a maximal set of algebraically independent elements $t_1,\dots ,t_r$ among these generators and denote the remaining generators by $c_1,\dots ,c_h$. Each $c_j$ is an algebraic element over $K=k(t_1,\dots ,t_r)$, so $L/K$ is a finite field extension. Clearing the denominators in their monic equations, choose a nonzero $d\in k[t_1,\dots ,t_r]$ such that each $c_j$ is an integral element overBecause $L$ is a field and $d^{-1}\in L$, we have $L=D[c_1,\dots ,c_h]$. Thus $L$ is an integral extension of $D$.

A subring over which a field is integral must itself be a field. Indeed, for $0\ne a\in D$, an integral equation for $a^{-1}$ multiplied by $a^{m-1}$ expresses $a^{-1}$ as an element of $D$. But if $r>0$, the ring $D$ is not a field: choose an irreducible polynomial $q(t_1)\in k[t_1]$ which does not divide $d$. There are infinitely many monic irreducible polynomials in $k[t_1]$, by the usual product-plus-one argument, whereas only finitely many can divide the nonzero $d$. The prime ideal $(q)$ stays proper after inverting $d$, so $q$ has no inverse in $D$. Therefore $r=0$, proving that $L/k$ is a finite field extension, and in particular an algebraic field extension.

Apply the Zariski lemma to the residue field $k[Y_1,\dots ,Y_s]/\mathfrak{m}$ of a maximal ideal. When $k$ is an algebraically closed field, this field is $k$ itself. The images $a_i$ of $Y_i$ then give $\mathfrak{m}=(Y_1-a_1,\dots ,Y_s-a_s)$, proving the Weak Hilbert Nullstellensatz.

For the real Laurent polynomial ring, the same argument applies because it is the finitely generated algebraEvery residue field at a maximal ideal is consequently a finite field extension of $\mathbb{R}$, and so is $\mathbb{R}$ or $\mathbb{C}$. Here we use the fundamental theorem of algebra, which makes $\mathbb{C}$ the algebraic closure of $\mathbb{R}$ of degree two. The images of $X_1,X_2$ must be nonzero since these elements are units. It follows that the answer iswith $(\alpha ,\beta )$ and $(\overline{\alpha },\overline{\beta })$ defining the same maximal ideal. If both coordinates are real, evaluation has image $\mathbb{R}$; otherwise its image is $\mathbb{C}$. Thus every displayed kernel is maximal, and the Zariski lemma shows that this list is exhaustive.

For explicit generating sets of an ideal, there are three cases. For real nonzero $a,b$,If $\alpha =a\in {\mathbb{R}}^{\times }$ and $\beta =s+it$ with $t\ne 0$,If $\alpha =u+iv$ with $v\ne 0$ and $\beta =s+it$, thenThe last two quotients are $\mathbb{C}$; the nonzero coordinate condition ensures that the Laurent inverses exist in them. There are no further coincidences in the classification: equal kernels induce an $\mathbb{R}$-isomorphism between the evaluation image fields. The only $\mathbb{R}$-automorphisms of $\mathbb{C}$ are the identity and complex conjugation, so the two pairs must agree or be conjugate.

To prove the intersection assertion, fix a prime ideal $P$ and an element $f\notin P$. The integral domain $B=R/P$ has nonzero image $\bar{f}$. Its localization of a ring $B[{\bar{f}}^{-1}]$ is nonzero and still a finitely generated algebra over $\mathbb{R}$, so it has a maximal ideal $\mathfrak{n}$. By the Zariski lemma, the field $L=B[{\bar{f}}^{-1}]/\mathfrak{n}$ is finite over $\mathbb{R}$.

The image of $B$ in $L$ is a subring containing $\mathbb{R}$ and is a finite-dimensional $\mathbb{R}$-vector space. It is an integral domain, hence a field: multiplication by any nonzero element is an injective linear map of a finite-dimensional vector space and therefore surjective. The kernel of $B\to L$ is thus a maximal ideal which avoids $\bar{f}$, since $\bar{f}$ is invertible in $L$. Its inverse image in $R$ is a maximal ideal containing $P$ but avoiding $f$. As this works for every $f\notin P$,This is the Jacobson ring property for this Laurent polynomial ring.

For a nonzero finitely generated module $M$, its set of associated primes isHere ${\mathrm{Ann}}_R(m)=\{a\in R:am=0\}$ is the annihilator of the element. The Laurent polynomial ring $R$ is a Noetherian ring, so the nonempty collection of annihilators of nonzero elements of $M$ has a maximal member, say $P={\mathrm{Ann}}_R(m)$. It is proper because $m\ne 0$.

If $ab\in P$ and $b\notin P$, then $bm\ne 0$ andMaximality forces ${\mathrm{Ann}}_R(bm)=P$, hence $a\in P$. Thus $P$ is a prime ideal, proving $\boxed{{\mathrm{Ass}}_R(M)\ne \varnothing }$. This argument in fact only needs that $R$ is Noetherian, not the finite generation of $M$.

A proper ideal $Q$ is $P$-primary when its radical of an ideal is $P$ andEquivalently, $ab\in Q$ and $b\notin P$ imply $a\in Q$; equivalently, every zero divisor of $R/Q$ is nilpotent. These formulations distinguish the actual primary ideal property from the weaker assertion that there is only one minimal prime ideal.

For the final example, set $u=X_1-1$ and $v=X_2-1$. ThenSince $R/P\cong \mathbb{R}[X_2,X_2^{-1}]$ is an integral domain, $P$ is a prime ideal. Moreover $I\subseteq P$ and $u^2\in I$, so $\sqrt{I}=P$. Consequently $P$ is the unique minimal prime ideal over $I$.

Nevertheless $uv\in I$, while $u\notin I$ and no power of $v$ belongs to $I$. For the first assertion, reduce modulo $v$: the image of $I$ is $(u^2)$ in $\mathbb{R}[X_1,X_1^{-1}]$, and $u$ is not a multiple of $u^2$ because $u$ is a nonunit in this integral domain. For the second, $v\notin P$, so $v^j\notin P\supseteq I$ for every $j\ge 1$. Thus $I$ is not a $P$-primary ideal: a unique minimal prime does not imply primary. In $R/I$, the nonzero class of $u$ is killed by $v$, a zero divisor which is not nilpotent.

Put $u=X_1-1$ and $v=X_2-1$. Evaluation at $(1,1)$ is surjective, andso its kernel is the prime ideal $\boxed{P=(u,v)}$. In particular $P/P^2$ is naturally an $R/P=\mathbb{Z}$-module, generated by the classes of $u,v$.

For integers $a,b$, including negative integers, reduction modulo $P^2$ givesFor negative powers, this follows from $(1+u{)}^{-1}\equiv 1-u$ and $(1+v{)}^{-1}\equiv 1-v$; the mixed product $uv$ vanishes modulo $P^2$. Therefore a Laurent polynomial $f$ satisfiesThe formal derivatives obey the product rule, so $D_i(ab)=\theta (a)D_i(b)+\theta (b)D_i(a)$. If $a,b\in P$, both terms vanish. Hence both $D_i$ vanish on $P^2$ and define a mapIt sends $[u]$ to $(1,0)$ and $[v]$ to $(0,1)$, and the displayed expansion gives its inverse. Thus the conormal module has the explicit basis

Let $A=R_P$ be the localization at a prime ideal, with maximal ideal $\mathfrak{m}=P{R}_P$. Every nonzero integer lies outside $P$ and is therefore inverted. The residue field isIndeed there is an explicit identificationinverting $1+u$ and $1+v$ has no effect in this local ring, and a polynomial denominator lies outside $(u,v)$ precisely when its constant term is nonzero. Conversely, clearing rational denominators turns every such fraction into a fraction from $R_P$.

For the Krull dimension, the prime ideal correspondence for localization preserves the strict chainof prime ideals in $R$. They are prime because the successive quotients are the integral domains $R$, $\mathbb{Z}[X_2,X_2^{-1}]$, and $\mathbb{Z}$. They remain distinct after localization at a prime ideal because they are all contained in $P$. Hence $\dim A\ge 2$.

For the upper bound, $R$ is a Noetherian ring by the Hilbert basis theorem and Localization of a Noetherian ring. Since $P$ is generated by two elements, the Krull height theorem gives $\mathrm{ht}P\le 2$. The prime ideal correspondence for localization identifies $\dim R_P$ with $\mathrm{ht}P$, so

A Noetherian local ring $(B,\mathfrak{n})$ with residue field $\kappa$ is a regular local ring ifBy the Nakayama lemma, the dimension on the left is the minimal number of generators of $\mathfrak{n}$; this is also called the embedding dimension.

The localization of a module is exact and commutes with quotients and products of ideals. Applying it to the already computed conormal module gives the cotangent space of a local ringIts dimension is two, equal to $\dim A$. Therefore $\boxed{R_P\text{ is a regular local ring}}$. Notice that $P/P^2$ has a basis over $\mathbb{Z}$, whereas the localized cotangent space of a local ring has a basis over the residue field $\mathbb{Q}$.

An integral domain $A$ is an integrally closed domain if every element of its fraction field which is an integral element over $A$ belongs to $A$. Explicitly, if $z\in \mathrm{Frac}A$ satisfiesthen $z\in A$. Thus $\boxed{A\text{ is integrally closed exactly when its integral closure in its fraction field is }A}$.

Suppose $A$ is a unique factorization domain and $z=a/b$ is integral, with $a,b\in A$, $b\ne 0$, and with no common irreducible element as a factor. Multiplying its monic equation by $b^d$ givesHence $b$ divides $a^d$. Every irreducible element dividing $b$ is a prime element, so it would divide $a$, contrary to the choice of the fraction. Therefore $b$ is a unit, and $z\in A$. We have proved

The going-down theorem is the following. Let $A\subseteq B$ be an integral extension of integral domains with $A$ an integrally closed domain. IfthenWe first prove this for a module-finite ring extension and then remove that extra hypothesis. We use the Lying-over theorem, the going-up theorem, and the incomparability theorem for integral extensions: primes exist over every base prime, prime chains can be extended upwards, and distinct primes over the same base prime cannot contain one another. The going-up theorem follows by applying the Lying-over theorem to the quotient by the lower prime. These results concern arbitrary integral extensions; none assumes the going-down theorem.

For the finite case, let $K=\mathrm{Frac}A$ and $L=\mathrm{Frac}B$. Since $B$ is module-finite, $L/K$ is a finite field extension. Take a finite normal field extension $E/K$ containing $L$, obtained as a splitting field of the minimal polynomials of finitely many field generators. The extension $E/K$ need not be a separable field extension. Let $C$ be the integral closure of $A$ in $E$. Then $B\subseteq C$, and $C$ is integral over both $A$ and $B$.

We need one auxiliary fact: the finite group $G={\mathrm{Aut}}_K(E)$ acts transitively on the prime ideals of $C$ over any fixed prime ideal $\mathfrak{p}$ of $A$. First observe that an element $w\in E$ fixed by all of $G$ is a purely inseparable algebraic element over $K$. Indeed, all $K$-embeddings of $K(w)$ into an algebraic closure extend to $K$-automorphisms of the normal field extension $E$. Hence the minimal polynomial of an algebraic element $w$ has only one distinct root. In characteristic zero this says $w\in K$; in characteristic $p>0$ it says $w^{p^e}\in K$ for some $e\ge 0$.

Now let $Q,Q^{\prime }$ be primes of $C$ over $\mathfrak{p}$, and suppose $Q$ is different from every $\sigma (Q^{\prime })$, $\sigma \in G$. By the incomparability theorem for integral extensions, $Q$ is contained in none of these conjugate primes. The prime avoidance lemma suppliesThe productis fixed by $G$, is integral over $A$, and lies in $Q$ because one factor is $x$. Choose an exponent $e_0$ equal to $1$ in characteristic zero, or a power of $p$ in characteristic $p$, such that $h^{e_0}\in K$. Since $A$ is an integrally closed domain, $h^{e_0}\in A$. It also lies in $Q\cap A=\mathfrak{p}$, hence in $Q^{\prime }$. Because $Q^{\prime }$ is a prime ideal, some factor $\sigma (x)$ lies in $Q^{\prime }$. This means $x\in {\sigma }^{-1}(Q^{\prime })$, contradicting our choice. Thus $Q=\sigma (Q^{\prime })$ for some $\sigma$, proving transitivity. The power $e_0$ accounts for the possible purely inseparable field extension and makes the argument valid in every characteristic.

Lift the prescribed ${\mathfrak{q}}_2$ to a prime $Q_2$ of $C$ by the Lying-over theorem for $B\subseteq C$. Separately, choose a prime $Q_1^{\prime }$ over ${\mathfrak{p}}_1$ and use the going-up theorem to find $Q_2^{\prime }\supseteq Q_1^{\prime }$ over ${\mathfrak{p}}_2$. Transitivity gives $\sigma \in G$ with $\sigma (Q_2^{\prime })=Q_2$. ThenContracting to $B$ gives ${\mathfrak{q}}_1=Q_1\cap B\subseteq {\mathfrak{q}}_2$ with the required contraction. This proves the going-down theorem for the module-finite case.

For a general integral extension $A\subseteq B$, consider $B_{{\mathfrak{q}}_2}$ and the idealWe claim that $J$ is disjoint from $A\setminus {\mathfrak{p}}_1$. Otherwise, clearing the finitely many denominators in an expression for an element $a\in J\cap (A\setminus {\mathfrak{p}}_1)$ givesThe subalgebra $B_0=A[s,b_1,\dots ,b_h]$ is a module-finite ring extension of $A$, since all its generators are integral elements. Apply the finite case to ${\mathfrak{q}}_2\cap B_0$: there is a prime $Q_1\subseteq {\mathfrak{q}}_2\cap B_0$ contracting to ${\mathfrak{p}}_1$. The displayed identity puts $sa$ in $Q_1$, but $a\notin Q_1$. Thus $s\in Q_1\subseteq {\mathfrak{q}}_2$, a contradiction.

Localize $B_{{\mathfrak{q}}_2}$ further at the multiplicative subset $A\setminus {\mathfrak{p}}_1$. The extension of $J$ is a proper ideal, by the disjointness just proved. Choose a maximal ideal containing it and contract back to a prime ideal of $B_{{\mathfrak{q}}_2}$. This prime contains $J$ and avoids $A\setminus {\mathfrak{p}}_1$. Its contraction ${\mathfrak{q}}_1$ to $B$ lies inside ${\mathfrak{q}}_2$ by the prime ideal correspondence for localization, and its contraction to $A$ is exactly ${\mathfrak{p}}_1$. This completes the proof of the going-down theorem without any finite generation assumption on $B$.

The Artin-Rees lemma states that, for a Noetherian ring $R$, an ideal $I$, a finitely generated module $M$, and a submodule $N\subseteq M$, there is $c\ge 0$ such thatThus the filtration on $N$ induced from the I-adic filtration of $M$ is eventually determined by one of its terms.

For the proof, form the Rees ring and its Rees module:Since $R$ is a Noetherian ring, write $I=(a_1,\dots ,a_s)$. Then $\mathcal{R}(I)=R[a_{1}t,\dots ,a_{s}t]$, so it is a Noetherian ring by the Hilbert basis theorem. Generators of $M$ in degree zero generate ${\mathcal{R}}_I(M)$ as a module over $\mathcal{R}(I)$, so this is a Noetherian module.

The graded submoduleis therefore finitely generated. Choose homogeneous generators $z_{\alpha }{t}^{d_{\alpha }}$ with $d_{\alpha }\le c$ for some $c$. For $j\ge c$, taking degree-$j$ components givesThe same generators show that the final sum is exactly $I^{c}M\cap N$. This proves the Artin-Rees lemma. If $L=0$, one may simply take $c=0$.

Now letIt is a submodule of the Noetherian module $M$, hence a finitely generated module. Apply the Artin-Rees lemma with $N=K$ and $j=c+1$. Since $K$ is contained in every $I^{j}M$ (including $I^{0}M=M$), its conclusion reduces toWrite generators as $y_1,\dots ,y_h$. There is a matrix $A=(a_{ij})$ with entries in $I$ such that $y_i={\sum }_j{a}_{ij}{y}_j$. Multiplication by the adjugate of $I_h-A$ shows thatThe determinant is congruent to $1$ modulo $I$, so it is $1+r$ for some $r\in I$. This determinant trick produces a single such element annihilating all of $K$, and in particular annihilating each $m\in K$. If $K=0$, use $r=0$.

Conversely, if $(1+r)m=0$ with $r\in I$, then $m=-rm$. Iterating gives $m=(-r{)}^{j}m\in I^{j}M$ for every $j\ge 1$. Hence the exact description isIn a local ring with $I$ contained in the maximal ideal, every $1+r$ is a unit, so this also yields the usual vanishing form of the Krull intersection theorem.

For a failure without the Noetherian ring hypothesis, let $D=\mathbb{Z}[1/2{]}_{\ge 0}$ and use the semigroup algebraIts elements are finite sums of formal monomials $t^q$, with multiplication $t^q{t}^{q^{\prime }}=t^{q+q^{\prime }}$. Each ring in the union is a polynomial ring and the inclusions are injective, so $R$ is an integral domain. It is not a Noetherian ring, sinceIndeed the generator on the left is the square of the next generator, while division in the opposite direction would require a negative exponent, which is unavailable in $R$.

Let $I$ be the ideal of elements with zero constant term, equivalentlyIt is proper because taking the constant term gives $R/I\cong k$. Every generator $t^q$ is $(t^{q/2}{)}^2$, with $t^{q/2}\in I$, so $I^2=I$. Thus this is a nonzero idempotent ideal, with $I^j=I$ for every $j\ge 1$. TakeFor every $r\in I$, the constant term of $1+r$ is $1$, so $1+r\ne 0$. Since $R$ is an integral domain and $t\ne 0$, we have $(1+r)t\ne 0$. This gives all three requested objects and proves that no element of the required form annihilates $m$.

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.