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Use the unique curvature $-1$ hyperbolic metric on a closed Riemann surface of genus $g\ge 2$, supplied by the uniformization theorem. This is the implicit setting for the partition and spectral theorems; a genus-zero or genus-one surface does not have this hyperbolic metric.

A partition is a pants decomposition by disjoint essential simple closed geodesics, whose complement is a union of pairs of pants. A pair of pants is a sphere with three disks removed, furnished here with geodesic boundary. Its Euler characteristic is $-1$, so there must be $2g-2$ pants. Every pant has three boundaries and every cutting curve occurs twice, giving $3g-3$ cutting curves. The Bers pants decomposition theorem asserts that a constant $B_g$ depending only on $g$ bounds the lengths of all cuffs in some such pants decomposition of every closed genus-$g$ hyperbolic surface.

To see what geometry the cuffs determine, cut each pair of pants along its three perpendicular seams into two congruent right-angled hyperbolic hexagons. The three alternate sides are ${\ell }_1/2,{\ell }_2/2,{\ell }_3/2$, where the ${\ell }_i>0$ are its boundary lengths. The right-angled hyperbolic hexagon identity determines the seam $s_i$ opposite the half-cuff ${\ell }_i/2$ byWe use the standard existence and uniqueness theorem for a right-angled hyperbolic hexagon with prescribed positive alternating side lengths. Thus the three boundary lengths determine the pair of pants up to isometry.

Fix a topological pants decomposition, label its cuffs, and choose reference seam endpoints and orientations. Gluing two boundaries of the same length requires a translation along that boundary; its signed distance is a twist ${\tau }_i$. The gluing graph, the $3g-3$ positive cuff lengths, and the $3g-3$ twists therefore determine the unmarked hyperbolic surface up to isometry. A full boundary translation has period ${\ell }_i$, so for an unmarked gluing one may take $0\le {\tau }_i<{\ell }_i$, with endpoints identified. For a marked surface the whole real twist records the number of Dehn twists, and the Fenchel–Nielsen coordinates areThe topological gluing data are necessary when the decomposition is not fixed. Different markings or decomposition graphs can describe the same unmarked surface.

Teichmüller space ${\mathcal{T}}_g$ consists of pairs $(X,f)$, where $X$ is a closed genus-$g$ Riemann surface and $f:{\Sigma }_g\to X$ is an orientation-preserving marking, modulo the equivalence $(X,f)\sim (Y,h)$ when a conformal orientation-preserving map $a:X\to Y$ has $a\circ f$ homotopic to $h$. The Fenchel–Nielsen coordinates identify it with ${\mathbb{R}}_{>0}^{3g-3}\times {\mathbb{R}}^{3g-3}$, of real dimension $6g-6$. The moduli space of Riemann surfaces forgets the marking by quotienting by the mapping class group.

The Wolpert generic spectral rigidity theorem says that the locus of closed genus-$g$ hyperbolic surfaces possessing an isospectral but nonisometric partner lies in a locally real-analytic exceptional locus of lower dimension in Teichmüller space. Thus a generic surface is determined, up to isometry, by its unmarked length spectrum, equivalently by its Laplacian spectrum. Here an isometry may reverse orientation; the length spectrum cannot distinguish the two orientations of the same metric. The equivalence of the two spectra is the compact hyperbolic Selberg trace formula.

The simplifying result is the Buser finite length spectrum theorem: for every $g\ge 2$ and $\epsilon >0$ there is $L=L(g,\epsilon )$ such that two closed genus-$g$ hyperbolic surfaces with hyperbolic systoles at least $\epsilon$ have the same entire unmarked length spectrum if their length multisets up to $L$ agree. Multiplicities are included. We will count primitive unoriented closed geodesics; counting all iterates is equivalent by successively removing shorter iterates. The theorem is uniform on the thick part of moduli space, not merely a cutoff chosen separately for a particular pair.

Here is a proof using compactness and polynomial trace equations. We state the subsidiary facts and explain their role. First, Mumford's compactness theorem makes the genus-$g$ thick moduli space compact. It can also be seen from the Bers pants decomposition theorem: cuffs in a Bers decomposition lie in $[\epsilon ,B_g]$, there are finitely many pants decomposition graphs, and unmarked twists may be reduced modulo their cuff lengths. These data lie in finitely many compact boxes of Fenchel–Nielsen coordinates and cover the thick moduli space.

Second, these finite compact boxes give a compact family $K$ of marked hyperbolic structures on a fixed smooth surface ${\Sigma }_g$ representing every member of the thick moduli space. We use the standard smooth dependence of the glued metrics and their hyperbolic holonomy representations on Fenchel–Nielsen coordinates. Choose smooth representatives over finitely many parameter neighborhoods and conjugate the hyperbolic holonomy representation using a fixed lifted Riemannian orthonormal frame. Passing to finite compact subboxes gives a compact set of representations, together with a compact family of representative smooth metrics. Changing the markings between boxes causes no problem: all are markings from the same fixed ${\Sigma }_g$, and the union is finite.

Third, uniformization theorem identifies each marked hyperbolic surface with ${\mathbb{H}}^2/\rho ({\pi }_1{\Sigma }_g)$ for a faithful discrete hyperbolic holonomy representation into ${\mathrm{PSL}}_2(\mathbb{R})$. We use the standard lifting fact that a closed orientable hyperbolic holonomy representation admits a lift to ${\mathrm{SL}}_2(\mathbb{R})$, and lifts may be chosen continuously on small parameter neighborhoods. Choose lifts on finitely many such neighborhoods, allowing the finitely many possible sign choices. A representation is described by the matrices of $2g$ standard fundamental group generators. The polynomial encoding of hyperbolic geodesic lengths uses, for each fixed group word $w$, its squared matrix traceis a polynomial in their entries: products are polynomial, and the inverse of a determinant-one matrix is its polynomial adjugate. For a nontrivial hyperbolic word, the hyperbolic translation length obeysThus equality of the positive lengths is exactly equality of these trace polynomials, independent of lift signs.

Fourth, the length spectrum of a closed hyperbolic surface is locally finite with finite multiplicities. We need the uniform finiteness of short geodesic classes across the compact marked family. More uniformly, for each finite $b$, only finitely many unoriented primitive conjugacy classes can have length at most $b$ in any metric of $K$. To justify uniformity, compactness of the smooth metrics gives a common bilipschitz equivalence comparison with one fixed metric $h_0$ on ${\Sigma }_g$. If ${\ell }_h(\alpha )\le b$, the minimizing geodesic has $h_0$ length at most $Cb$, so ${\ell }_{h_0}(\alpha )\le Cb$. Local finiteness for $h_0$ then gives a finite list. Local finiteness itself follows from proper discontinuity of the cocompact group of the hyperbolic holonomy representation: conjugate a geodesic axis to meet a fixed compact fundamental set, and a bounded-length axis gives a group element moving that compact set a bounded distance, of which there are only finitely many.

Finally, the Hilbert basis theorem says that a polynomial ring in finitely many real variables is Noetherian. Consequently a decreasing sequence of sets cut out by polynomial equations eventually stabilizes. Indeed, their vanishing ideals form an increasing chain and stabilize. Finite unions of such sets are still algebraic: one can take all products of one defining polynomial from each component. This finiteness fact is the engine that turns arbitrarily many spectral comparisons into finitely many.

Enumerate the primitive unoriented conjugacy classes as ${\alpha }_1,{\alpha }_2,\dots$. For each $n$ letand let $F_n$ be the finite set of primitive classes that can have length at most $b_n$ anywhere in $K$. Define $Z_n$ in the finite-dimensional space of pairs of generator matrices as follows: it consists of pairs $(\rho ,\sigma )$ for which there exists an injectionand also an injection satisfying the analogous equations with $\rho$ and $\sigma$ interchanged. There are only finitely many injections. Each choice imposes finitely many polynomial equations; taking their finite unions and then intersecting the two directions proves that $Z_n$ is a real affine algebraic set. On determinant-one representations these equations are exactly the desired matching of lengths. Defining the polynomials on the ambient matrix-entry space is harmless; the argument will only apply them to representations in $K$.

Put $W_n=Z_1\cap \cdots \cap Z_n$. The sets $W_n$ form a decreasing chain of affine algebraic sets, hence for some $N$,For representations in $K$, membership in this intersection is equivalent to equality of the full length spectra. One direction follows by matching equal lengths, including their multiplicities. For the other, fix a length $\ell$. All primitive classes of length $\ell$ on the first surface occur in some finite initial segment. Its injective matching shows that the second surface has at least that multiplicity at $\ell$; the reverse injection gives the reverse inequality. Local finiteness makes both multiplicities finite. Doing this for every $\ell$ proves multiset equality.

TakeIf two surfaces in $K$ have equal length multisets up to $L$, each initial segment through $n\le N$ can be matched injectively on the other surface. Its matching classes lie in $F_n$, since the matched lengths are at most $b_n$. The reverse matching is available as well. Thus the pair belongs to $W_N$, hence to every $W_n$, and the preceding paragraph yieldsAll choices of the compact family were made using only $g$ and $\epsilon$, so the resulting cutoff has the asserted dependence. If the thick family is empty the assertion is vacuous. This proves Buser's finite length spectrum theorem without claiming that equal full length spectra always force isometry; exceptional isospectral pairs are compatible with the theorem and are precisely why Wolpert's theorem is a generic statement.

Use the following finite-length version of the Hilbert-Serre theorem. Let $S={⨁}_{n\ge 0}{S}_n$ be a graded algebra generated over an Artinian ring $A=S_0$ by finitely many homogeneous elements $x_1,\dots ,x_r$ of positive degrees $e_1,\dots ,e_r$. For a finitely generated nonnegatively graded $S$-module $M$, define its Poincare series of a graded module byEvery component has finite length, and the theorem saysFor a module whose grading is merely bounded below, the same statement holds with a Laurent-polynomial numerator. When the generators all have degree one, the denominator is $(1-t{)}^r$.

Here is an induction proof. An Artinian ring is Noetherian by question 1, so the Hilbert basis theorem makes $S$ Noetherian. If there are no positive-degree generators, $S=A$ and a finite homogeneous generating set for $M$ occupies only finitely many degrees. Each component is a finite module over the Artinian ring $A$, hence has finite length of a module, and $P_M$ is a polynomial.

For $r>0$, put $x=x_r$, $e=e_r$, $N=\{m\in M:xm=0\}$ and $C=M/xM$. Both are finitely generated graded modules annihilated by $x$, hence modules over $S/(x)$, which is generated by the first $r-1$ homogeneous elements. Their multiplication exact sequence isHere $M(-e{)}_n=M_{n-e}$. Additivity of length degree by degree givesThis is the Hilbert series multiplication exact sequence. By induction, the right-hand side has denominator ${\prod }_{i<r}(1-t^{e_i})$. Division by $1-t^e$ completes the proof of the Hilbert-Serre theorem. The finite-component and finite-generation claims also follow from the finite set of positive-degree algebra and module generators; no analytic convergence of a series is involved.

For the local invariant, use the usual Noetherian local ring hypothesis of Hilbert–Samuel growth dimension. Locality alone does not guarantee finite lengths or polynomial growth; the printed question leaves this finiteness assumption implicit. For instance, in the localization at a prime ideal of the polynomial ring $k[x_1,x_2,\dots ]$ at $(x_1,x_2,\dots )$, the vector space $\mathfrak{m}/{\mathfrak{m}}^2$ has the infinitely many independent classes of the variables, so its length is not finite. Write $\mathfrak{m}=P$ and $\kappa =R/\mathfrak{m}$. Its associated graded ringis generated over $\kappa$ in degree one, because $\mathfrak{m}$ is finitely generated. Thus its Hilbert series is rational with a denominator that is a power of $1-t$. After cancelling factors, writeDefine $d(R)=d$, the pole order at $t=1$, with $d=0$ when $P_G$ is a polynomial.

Equivalently, the Hilbert–Samuel functionhas generating series $P_G(t)/(1-t)$ and eventually agrees with a polynomial of degree $d$. Indeed, coefficients of $(1-t{)}^{-d-1}$ are $\left(\genfrac{}{}{0ex}{}{n+d}{d}\right)$, and multiplication by $q(t)$ leaves leading term $q(1)n^d/d!$. Consequently $d(R)$ is the degree of the cumulative Hilbert-Samuel polynomial, rather than the degree of the individual graded-component function; the latter has degree $d-1$ when $d>0$. This pole/growth invariant also equals Krull dimension by the local dimension theorem, although that theorem is not required to define it here.

For an example, take $R=k[x,y{]}_{(x,y)}$. Its associated graded ring is $k[x,y]$ with the ordinary degree grading: degree $n$ has the $n+1$ monomials $x^i{y}^{n-i}$ as a basis. ThereforeThe length formula also follows directly by counting monomials of total degree at most $n$ in $R/(x,y{)}^{n+1}$.

The height of a prime ideal is the supremum of the lengths of strict chains of prime ideals ending at $P$:The last equality follows from the prime ideal correspondence for localization. Here is a length-growth proof of the Krull height theorem that makes the bound explicit.

We first establish a prime-chain lower bound for local length. For a Noetherian local ring $(B,\mathfrak{m})$ admitting a prime chain of length $d$, put $H_B(t)=\mathrm{length}(B/{\mathfrak{m}}^t)$. Thenfor some $c>0$. For $d=0$, use $H_B(t)\ge 1$. For $d>0$, quotient by the first prime ideal of the chain; this only decreases $H_B$ and reduces us to a local integral domain $B$ withTake $0\ne a\in P_1$ and put $C=B/(a)$. It has a prime chain of length $d-1$, so induction bounds $H_C(t)$ below by a positive multiple of $t^{d-1}$.

The Artin-Rees lemma applied to $(a)\subseteq B$ gives, for a fixed integer $c_0\ge 1$,Because $a$ is a non-zero-divisor, the first term in the exact sequencehas length at least $H_B(t-c_0)$. HenceIterate this inequality about $t/(2c_0)$ times, while its arguments remain at least $t/2$. This gives the desired positive multiple of $t^d$. The particular Artin-Rees lemma inclusion used here follows directly from finite generation: the Rees algebra ${⨁}_{j\ge 0}{\mathfrak{m}}^j{t}^j$ is Noetherian, by the Hilbert basis theorem, and its graded submodule ${⨁}_j((a)\cap {\mathfrak{m}}^j)t^j$ has finitely many homogeneous generators. A bound on their degrees gives $c_0$.

Now localize at the given minimal prime ideal $P$. Write $B=R_P$, $\mathfrak{m}=P{R}_P$, and $J=I{R}_P$. The only prime ideal of $B$ containing $J$ is $\mathfrak{m}$, so $\sqrt{J}=\mathfrak{m}$. Since $\mathfrak{m}$ is finitely generated, some ${\mathfrak{m}}^q\subseteq J$. Thus $B/J$ has finite module length, say $L$. The ideal $J$ is generated by at most $n$ elements.

For $n>0$, products of these generators give a surjection from $\left(\genfrac{}{}{0ex}{}{j+n-1}{n-1}\right)$ copies of $B/J$ onto $J^j/J^{j+1}$. Summing the lengths gives the generator bound for primary-ideal lengthSince $J\subseteq \mathfrak{m}$, we have $J^t\subseteq {\mathfrak{m}}^t$, and thereforeThe prime-chain lower bound for local length now excludes any chain of length greater than $n$. ConsequentlyFor $n=0$, minimality over the zero ideal says that $P$ is a minimal prime ideal, so its height is zero. This also handles that boundary case.

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.

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 $\mathbb{Q}$.

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.

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.

A right Noetherian ring satisfies the ascending chain condition on right ideals; equivalently, every right ideal is a finitely generated module.

Here is a noncommutative Hilbert basis theorem proof adapted to the stated hypothesis. Set $F_{-1}=0$ andThe equality follows inductively from $A+xA=A+Ax$, by moving one coefficient past one $x$ at a time. These spaces form an exhaustive filtered algebra structure on $B$, with $F_n{F}_m\subseteq F_{n+m}$. This does not assert uniqueness of the displayed expressions or the existence of a coefficient-moving automorphism.

For a right ideal $I\subseteq B$, defineEach $L_n$ is a right ideal of $A$. In fact, if $a\in L_n$ and $c\in A$, write $c{x}^n=x^{n}b+u$ with $b\in A$, $u\in F_{n-1}$; then $ac{x}^n=(a{x}^n)b+au\in I+F_{n-1}$. Also $L_n\subseteq L_{n+1}$ by right multiplication by $x$. Since $A$ is a right Noetherian ring, this ascending chain stabilizes at some $L_N$.

Choose finite generators $a_{nj}$ for each $L_n$, $0\le n\le N$, and choose $f_{nj}\in I$ with $f_{nj}-a_{nj}{x}^n\in F_{n-1}$. These finitely many elements generate $I$ as a right ideal. To see this, induct on $m$ for $f\in I\cap F_m$. Write $f=a{x}^m+u$ with $u\in F_{m-1}$; then $a\in L_m$. Put $d=\min (m,N)$ and express $a={\sum }_j{a}_{dj}{c}_j$. Write $c_j{x}^d=x^d{b}_j+u_j$ with $u_j\in F_{d-1}$. The differencelies in $I\cap F_{m-1}$, so the induction applies. At $m=0$ the remainder is zero. Hence $B$ is right Noetherian.

For the quantum torus, take $q\in k^{\times }$ and the convention $YX=qXY$. Begin with the polynomial ring $k[X]$, which is Noetherian by the Hilbert basis theorem. Adjoining $X^{-1}$ preserves the hypothesis because it commutes with $k[X]$, and gives $k[X^{\pm 1}]$. Adjoin $Y$ next. The relation $Yf(X)=f(qX)Y$ and its inverse coefficient-moving relation give $A+YA=A+AY$ for $A=k[X^{\pm 1}]$. Finally adjoin $Y^{-1}$ to $k[X^{\pm 1}][Y;\sigma ]$, where $\sigma (X)=qX$. On a monomial, $Y^{-1}{X}^i{Y}^j=q^{-i}{X}^i{Y}^{j-1}$; when $j=0$ this is in $A{Y}^{-1}$, and when $j>0$ it is in $A$. The reverse inclusion follows by the same relation. The preceding argument applies at each step, proving the quantum torus is right Noetherian. Nonzero $q$ is required for this notation.

For a noncommutative ring, a prime ideal of a noncommutative ring means a proper two-sided ideal $P$ such that $UV\subseteq P$ for two-sided ideals $U,V$ implies $U\subseteq P$ or $V\subseteq P$. Equivalently, $aAb\subseteq P$ implies $a\in P$ or $b\in P$. This definition does not require $A/P$ to be a noncommutative domain.

Retain the right Noetherian ring hypothesis for the last assertion. More generally, the ascending chain condition on two-sided ideals suffices. We claim that every proper two-sided ideal $I$ contains a product of finitely many prime ideals of a noncommutative ring, each containing $I$. If not, choose a maximal counterexample $I$. It cannot be a prime ideal of a noncommutative ring. Thus there are two-sided ideals $U,V$ strictly containing $I$ with $UV\subseteq I$: add $I$ to the two witnesses for failure of the defining condition for a prime ideal of a noncommutative ring. By maximality, both $U$ and $V$ contain products of finitely many prime ideals of a noncommutative ring containing them. Concatenating these products gives a product inside $UV\subseteq I$, a contradiction.

Apply the claim to $I=0$ in a nonzero $A$, obtaining $P_1\cdots P_r=0$. Every prime ideal of a noncommutative ring $Q$ contains one of the $P_i$, by repeated application of the definition of a prime ideal of a noncommutative ring. For the prime radical of a noncommutative ring $N$, it follows thatIndeed, $N\subseteq P_i$ gives $N^r\subseteq P_1\cdots P_r=0$, and ${\bigcap }_i{P}_i\subseteq Q$ for every $Q$ gives equality of the intersections. If $A=0$, the empty intersection is the whole zero ring and the conclusion is immediate.

The final assertion is false for arbitrary algebras without the preceding chain condition. For example, in the commutative ring $k[z_1,z_2,\dots ]/(z_i^2:i\ge 1)$ the nilradical is $(z_1,z_2,\dots )$, its only prime ideal, but the product of any number of distinct $z_i$ is nonzero. Thus this nilradical is not a nilpotent ideal.