Lech’s multiplicity conjecture

Proof. For each j, the complex F[1/z_j] is exact. A bounded exact complex of projective modules is contractible, so clearing the finitely many denominators in a contracting homotopy shows that some power of z_j acts nullhomotopically on F. Choose a common exponent a_0. It follows that (z_1^{a_0},\ldots,z_h^{a_0}) annihilates every H_i(F), and therefore so does I^{h(a_0-1)+1}.

Put K_a=K(z_1^a,\ldots,z_h^a;C). If a\geq a_0, each multiplication by z_j^a is nullhomotopic on F and also on its tensor product with any of the other Koszul factors. Splitting the resulting mapping cones gives

The splittings can be chosen to preserve the inclusion of the copy of F in Koszul degree zero.

The homology of F\otimes K_a is annihilated by (z_1^a,\ldots,z_h^a), since these multiplications are nullhomotopic on K_a. In the bounded double-complex spectral sequence obtained by taking Koszul homology first, all positive Koszul rows have length zero by (2.1). The remaining row has terms

of finite length. Hence every H_n(F\otimes K_a) has finite length, and its length is zero when n>h. The summand F[h] in (2.3) proves the assertions: H_0(F) is a summand of H_h(F\otimes K_a), and H_i(F) for i>0 is a summand of H_{h+i}(F\otimes K_a).□

Proof. Set J_a=(z_1^a,\ldots,z_h^a). Since I^{h(a-1)+1}\subseteq J_a, the I-adic filtration of C/J_a is finite. There is a graded surjection

The sum of the lengths of the homogeneous pieces on each side is a^h\mu. Additivity and nonnegativity imply that every homogeneous piece of its kernel has length zero. For a given degree n, choose a>n. The relations Z_i^a have no effect in degree n, and J_a\subseteq I^{n+1}, so the degree-n map is precisely (2.4). This proves the first assertion. It also gives

Write P=(C/I)[Z_1,\ldots,Z_h] and let N be the kernel in (2.4). The degree-zero component of a homogeneous localization of N, as used on any standard projective chart or its intersections, is a filtered colimit of homogeneous pieces of N. All these pieces are annihilated by I and have length zero. The assumed colimit property therefore gives length zero for every such localized component. Tensoring with the restriction of V(t) preserves this property, since that restriction is finite projective over the chart ring. The finite standard affine Cech complexes for projective space and E consequently have termwise length-zero kernels, and restriction is an isomorphism modulo length zero on their cohomology.

To check finiteness without a hypothesis on C/I, resolve V(t) on \PP_\ZZ^{h-1} by finitely many finite sums of twists of the structure sheaf. Such a resolution exists by taking a finite graded free resolution of a finitely generated graded module representing V(t) over \ZZ[Z_1,\ldots,Z_h]. This polynomial ring is regular of finite dimension and therefore has finite global dimension [28, Tags 07NF and 00OE]; its finitely generated graded projectives are graded free, as follows by lifting a homogeneous basis modulo (Z_1,\ldots,Z_h) and applying graded Nakayama. The sheaf resolution is locally split because its final cokernel is a vector bundle, so it remains exact after arbitrary base change. The cohomology of every twist on \PP_{C/I}^{h-1} is finite free over C/I. A finite spectral sequence for the resolution proves the required finiteness. Finally, for t\geq0 the polynomial degree-t piece is the space of global sections of \cO(t) and its higher cohomology vanishes. Combining these facts with (2.4) proves the last assertions.□

Proof. On a standard affine chart U_i, L is generated by the nonzerodivisor z_i. The chart complex of D consists of finite projective modules and becomes contractible after inverting z_i: away from E, both F_Y and G identify with the pullback of F. Clearing denominators in a contracting homotopy gives an exponent m_i for which multiplication by z_i^{m_i} is nullhomotopic on this chart. Trivializing L shows that the same exponent works for every twist. It also works on all intersections contained in this chart. Choose m at least as large as every m_i.

The map D(j+m)\to D(j) is zero on the internal-homology page of the Cech double complex, since on each intersection it is identified with a nullhomotopic multiplication by a local equation to the power m. The Cech filtration has at most h steps. The filtered-complex observation above therefore shows that D(j+hm)\to D(j) is zero on hyperhomology. Every larger twist map factors through one of these, so M_D=hm works.

For a\in I^{M_D}, multiplication by a on D(j) factors as

Thus I^{M_D} annihilates its hyperhomology. If M\geq M_D, the long exact sequence for

injects \mathbb H_k(Y,D(j)) into the hyperhomology of the quotient. That quotient has a finite filtration by E-layers, each of whose terms is a finite sum of bundles V(t)|_E. Their sheaf cohomology has finite length by Lemma 2.3. The two finite spectral sequences, first for each layer and then for the filtration, give finite length for the quotient’s hyperhomology and hence for that of D(j).□

Proof. Comparison on finite thickenings. For each M\geq1 there is a natural comparison

Filter its two sides by powers of I and L, respectively. At the lth layer, 0\leq l<M, comparison for each free summand is

Lemma 2.3 says that this is an isomorphism on homology modulo length zero. The finite filtration, together with the finite complex F, proves the same assertion for (2.9). These comparisons commute with all transition maps as M increases.

The geometric tower. We identify the eventual images of these two towers. On the geometric side put

The triangles

are compatible with the transition N\to M for N\geq M. Lemma 2.4 implies that H_k(A)\to H_k(B_M) is injective once M is sufficiently large. It also says that the transition on the third terms of (2.10) is zero on homology once N-M is sufficiently large. Exactness then gives

The reverse inclusion in this equality uses only the compatible maps from A to all B_N.

The Koszul tower. For the algebraic side first use J_a=(z_1^a,\ldots,z_h^a) and A_a=F\otimes K_a. The Koszul augmentation gives a compatible map

which is an isomorphism on homology modulo length zero. Indeed its cone is obtained by tensoring the bounded free complex F with the cone of K_a\to C/J_a; all homology of the latter cone has length zero by (2.1).

The inclusion F\subseteq A_a is the subcomplex in Koszul degree zero. For a\geq a_0 it is split on homology by (2.3). Let Q_a=A_a/F. The standard transition K_b\to K_a, for b\geq a, is the identity in degree zero and multiplies the basis vector indexed by T by \prod_{j\in T}z_j^{b-a}. Filter Q_a increasingly by Koszul degrees 1,\ldots,p, for p=1,\ldots,h. Its associated graded complexes are copies of shifts of F, and the transition on each such piece has a nonempty index set T. Choose t with I^tH(F)=0, as in Lemma 2.2. For b-a\geq t every transition is zero on the homology of these graded pieces. A composition of h such transitions is therefore zero on H(Q_a). For example, H(Q_b)\to H(Q_a) is zero when b-a\geq ht. The long exact sequences for 0\to F\to A_a\to Q_a\to0 now give, for a sufficiently large and then b sufficiently large,

Passing through the Koszul augmentations proves this assertion modulo length zero for the tower F/J_aF.

From Koszul powers to ordinary powers. Here is explicitly why it passes to the ordinary-power tower P_M=F/I^M F. The containments

give all the required quotient maps. Choose one sufficiently large a. For M\geq h(a-1)+1, the composite H_k(F)\to H_k(P_M)\to H_k(F/J_aF) proves injectivity of the first map modulo length zero. Next fix such an M, choose a\geq M large, and choose b\geq a so that the image from level b to level a in the J-tower is the image of H_k(F). For N\geq h(b-1)+1, the transition factors through

Its homology image is consequently contained in the image of H_k(F) modulo length zero; compatibility gives the opposite containment. Thus the eventual image in the P-tower identifies with H_k(F) modulo length zero.

Choose M and then N large enough for both this description and (2.11). The compatible comparisons P_N\to B_N and P_M\to B_M in (2.9) identify the homology images of P_N\to P_M and B_N\to B_M modulo length zero. These images identify with H_k(F) and H_k(A), respectively, modulo length zero. All their lengths are finite by Lemma 2.2, (2.5), and Lemma 2.4. Therefore

for every k, which proves (2.7).

Tensoring by a vector bundle. The homology comparison is established. To prove (2.8), first compare consecutive twists of F_Y. The quotient of F_Y(j) by F_Y(j+1) has term \cO_E(j)^{b_i} in degree i, so its Euler characteristic is

Euler additivity shows that \chi_\lambda(F_Y(j)) is independent of j\in\ZZ. Finally resolve V on \PP_\ZZ^{h-1} by a finite complex of finite sums of twists, as in Lemma 2.3. The resolution remains exact on Y. Tensoring with F_Y and taking Euler characteristics expresses \chi_\lambda(F_Y\otimes V) as the alternating sum of the corresponding twist characteristics. The alternating sum of the ranks of the resolution is \rank V, giving (2.8).□

Proof. Put n=h-1. If n=0, use the rank-one bundle on \PP_\ZZ^0. For n>0, multiplication of binary linear forms defines

The map is proper and has finite geometric fibers, since a binary form has only finitely many ordered factorizations into linear factors, up to the scalars already removed by the projective coordinates. Hence f is finite. On every field fiber, source and target are smooth of dimension n. Corresponding local rings have equal dimensions, since the residue field extension is finite and the dimensions are n minus the respective residue-field transcendence degrees. Thus the source is Cohen–Macaulay and satisfies the dimension criterion for flatness over the regular target [28, Tag 00R4]. This gives flatness on each field fiber. Since both schemes are flat over \ZZ, the fiberwise flatness criterion [28, Tag 039D] gives flatness of f. Over \QQ the finite extension of function fields is separable. A geometric generic form has n distinct factors and exactly n! ordered factorizations. Its reduced geometric generic fiber therefore has n! points, so the generic degree, and hence the constant rank of f, is n!.

Define

Finite flatness makes this a vector bundle of rank n!, and its formation commutes with arbitrary base change. The projection formula reduces the cohomology of V(a) to that of \cO(a-1-r_1,\ldots,a-1-r_n) on (\PP_A^1)^n. On \PP_A^1, the bundle \cO(m) has free cohomology only in degree zero for m\geq0, only in degree one for m\leq-2, and none for m=-1; its Euler characteristic is m+1. These computations commute with any base change and their complexes split over \ZZ. Tensoring the complexes for the n factors proves the asserted cohomological degree and product formula over every A. For A=C/I, a free module of rank q has length q\mu. The statements on E follow from Lemma 2.3.□

Proof of Theorem 2.1. The homology assertions are Lemma 2.2. It remains to prove the estimate. We choose the bundle roots to serve two purposes: their positions force the hyperhomology of G\otimes V into degree zero, modulo length zero, while their limiting values make every layer sum with 1\leq i\leq h have the same leading term (h-1)!s^h. The alternating-rank identity will then isolate b_0. The derivative of the following polynomial provides exactly these roots:

and write \rho_1>\cdots>\rho_{h-1} for the roots of H'(x). Rolle’s Theorem gives

For all sufficiently large s, depending only on h, round s\rho_j to integers r_j so that

Let V be the bundle of Lemma 2.6, and put D=G\otimes V with G as in (2.6).

We claim that

In the lth E-layer of a twist of D, the term in homological degree i is V(l-is)|_E^{b_i}. If l\geq0, then (2.13) implies

Indeed a root with index j\geq i is strictly less than -is, and for i=0 all roots are negative. Thus the layer’s sheaf cohomology in degree v>i has length zero. In the finite hyperhomology spectral sequence no term of negative total degree i-v survives modulo length zero. Filtering by the layers l=0,\ldots,M-1 shows that

For M large, Lemma 2.4 makes the map on hyperhomology from D(M) to D zero. The quotient long exact sequence therefore injects \mathbb H_k(Y,D) into \mathbb H_k(Y,D/D(M)), proving length zero for k<0.

If l<0, then

For i\geq1 this follows because every j\leq i-1 has r_j>-(j+1)s\geq-is>l-is; for i=0 the assertion is vacuous. Consequently the corresponding layer has hyperhomology of length zero in total degrees greater than 1. Filtering the quotient with layers l=-M,\ldots,-1 gives

For large M the map from D to D(-M) is zero on hyperhomology. The long exact sequence gives a surjection

and hence length zero for k>0. Lemma 2.4 supplies finite length in all degrees. We have proved \chi_\lambda(D)=\lambda\mathbb H_0(Y,D)\geq0, as claimed.

The inclusion F_Y\otimes V\subseteq G\otimes V has, in degree i, quotient (L^{-is}/\cO_Y)\otimes V^{b_i}, with successive layers V(l)|_E^{b_i} for l=-is,\ldots,-1. All Euler characteristics here are finite. Additivity, Lemma 2.5, and Lemma 2.6 give

The sum for i=0 is empty.

We make the error in these sums explicit. Set

On the fixed interval [-h,0], (2.13) implies \|P_s-P\|_\infty=O_h(s^{-1}), while P and its derivative are bounded in terms of h alone. The elementary error estimate for left-endpoint Riemann sums consequently gives, uniformly for i=1,\ldots,h,

Since \sum_{i=1}^h(-1)^i b_i=-b_0, substitution into (2.15) yields

The error bound and the threshold for s depend only on h. The homotopy exponents and truncation levels used earlier may depend on C, F, and V; they establish exact comparisons and vanishing modulo length zero and do not enter this numerical error. This proves (2.2) for all sufficiently large s. Increase A_h, if necessary, to be at least the threshold for s. For the remaining positive integers s, its right side is nonpositive because b_0\leq\sum_i b_i, so nonnegativity of length proves the same inequality. For h=1 the empty-product conventions give the argument directly, with exact sums and no error. The proof is complete.□

Proof. A finitely generated submodule of a direct sum is contained in a finite subsum. If the summands have length zero, additivity gives length zero for each finite subsum, and (3.2) gives length zero for the direct sum. Both a sum of submodules and a directed colimit are quotients of direct sums, proving the first assertion. The common annihilating power ensures that those direct sums belong to the length category.

For the second assertion, each finitely generated submodule of M is contained in some M_n. Apply (3.2) and monotonicity.

For the third assertion, let N\subseteq M be finitely generated over \Lambda. Representatives of a finite generating set lie in a common stage M_{n_0}. Consequently N is contained in the image of M_n in M for every n\geq n_0. Monotonicity for submodules and quotients gives

The right side tends to zero. Now take the supremum over N.□

Proof. The field E=LK' is a finite separable extension of the perfect field K', and is therefore perfect. Since it contains L and is contained in L_{\mathrm{perf}}, it equals L_{\mathrm{perf}}. The ring A' is normal: any integral equation and the finitely many fractions involved in an element of \operatorname{Frac}(A') descend to some normal ring A^{1/q}.

Choose a K-basis b_1,\ldots,b_r of L with b_j\in D. Its trace-dual basis \beta_1,\ldots,\beta_r exists by separability. Since (A\setminus\{0\})^{-1}D=L, there is 0\neq g\in A with g\beta_j\in D for every j. The same two bases are trace-dual after extending scalars to K'.

Every c\in C is integral over A', as is g\beta_j. The trace \operatorname{Tr}_{E/K'}(g\beta_jc) is therefore integral over A' and lies in K', so normality puts it in A'. The trace-dual expansion now reads

Taking q-th roots inside the perfect ring C gives the second inclusion, because (A'D)^{1/q}=A'D^{1/q}=B_q.□

Proof. The assertion is immediate when M=0. Otherwise put J=\Ann_D M, a \mathfrak d-primary ideal, and let b be a number of generators of M. Define

The module M_q is obtained by flat scalar extension from A^{1/q} to A' of D^{1/q}\otimes_D M. Under the qth power isomorphism D^{1/q}\cong D, the latter module corresponds to \Phi_D^n(M). It is a finite-length module over the Noetherian ring A^{1/q}, so finite-stage normalization gives

Here the lengths over A^{1/q} and D^{1/q} agree because their residue fields agree.

By Lemma 3.3, multiplication by g^{1/q} defines a D-linear map C\to B_q whose composite B_q\hookrightarrow C\to B_q is multiplication by g^{1/q}. After tensoring with M, this shows that K_q=\ker(M_q\to N_q) is killed by g^{1/q}. The finite normalized length in (3.8) permits the same kernel–cokernel calculation as in (3.4). Since M is a quotient of (D/J)^b, it gives

The last estimate is the Hilbert–Samuel growth argument from (3.4), with J in place of H: J^{[q]} contains a power of \mathfrak d with exponent O(q), and \dim D/(g)\leq h-1 if g is a nonunit. If g is a unit, the error is zero.

The submodules N_q increase to C\otimes_D M and are all killed by J, hence by one fixed power of \mathfrak m_A. Additivity, (3.8), and Lemma 3.2 now give the assertion by taking the limit.□

Proof. In equal characteristic, choose a coefficient field and a presentation B=k[[X_1,\ldots,X_t]]/I, and set

The map between the power series rings is flat. For example, this follows from the local flatness criterion modulo (X_1,\ldots,X_t): the variables form regular sequences in both rings, and the resulting map of coefficient fields is flat. The quotient map is therefore flat as well and has the asserted maximal ideal and residue field.

In the remaining case the residue characteristic is p>0, and the Cohen Structure Theorem [28, Tag 032A] gives a presentation B=V[[X_1,\ldots,X_t]]/I, where V is a complete discrete valuation ring with uniformizer p and residue field k. We explain why V admits a complete discrete valuation extension V' with the same uniformizer and residue field \overline{k}, even when k is imperfect. Given a simple algebraic residue extension with irreducible minimal polynomial \overline f, lift \overline f to a monic polynomial f\in V[X]. The algebra V[X]/(f) is finite free over V, and its quotient by p is the desired residue field. Every maximal ideal lies over (p), so this algebra is local with maximal ideal generated by p. The element p is a nonzerodivisor, and the algebra has dimension one. It is consequently a discrete valuation ring. This argument does not require \overline f to be separable.

Repeat this construction along a tower of simple algebraic extensions whose union is \overline{k}. At limit stages take unions. Every nonzero element of such a union is p^a times a unit for an integer a\geq0; hence every nonzero ideal is generated by an element of least valuation. The union is again a discrete valuation ring with uniformizer p. Completing the final union gives the required V'. The extension V\to V' is flat because V' is torsion-free over V. Now use

Flatness of the power series extension follows from the same local flatness criterion, this time with coefficient map V\to V'.

Finally, for any flat local map with extended maximal ideal, a composition series of a finite-length module remains a composition series after tensoring: each residue-field factor becomes the target residue field. In particular

The two Hilbert–Samuel functions, and therefore their degrees and multiplicities, agree.□

Proof. Completion preserves dimension and Hilbert–Samuel multiplicity, and a flat local map induces a flat map on completions. We can thus first assume that R and S are complete. Put d=\dim R and f=\dim(S/\mathfrak mS). The dimension formula for a flat local map gives \dim S=d+f. Choose Q minimal over \mathfrak mS with \dim(S/Q)=f. Going-down gives \dim S_Q\geq d, whereas

All these inequalities are equalities. Since S is excellent, Nagata’s theorem gives e(S_Q)\leq e(S). The map R\to S_Q is flat local, its closed fiber has dimension zero, and \dim S_Q=d. Replacing S_Q by its completion preserves these properties. Thus a proof for this new target proves the original inequality.

Choose a finite prime filtration of the R-module R, with factors R/\mathfrak p_j, allowing repetitions. Flatness gives a filtration of S with factors S/\mathfrak p_jS. The flat local map

has the same zero-dimensional closed fiber, so its source and target have equal dimension. The additivity formula for multiplicity yields

It therefore suffices to prove the inequality for the indicated domain sources. Their quotients and the corresponding targets remain complete. If d=0, each such source is a field and has multiplicity one, while a nonzero zero-dimensional local target has multiplicity equal to its positive length. This case is settled.

For d>0, apply Lemma 4.1 to the target. The resulting composite remains flat local and has the same target dimension and multiplicity. This proves the reduction. The argument also appears in [20, Lemma 2.2].□

Proof. Constructing T. Let k be the residue field of R. Choose a complete regular normalization A\subset R with residue field k, such that R is finite over A. In equal characteristic, take a coefficient field k\subset R and a system of parameters a_1,\ldots,a_d, giving A=k[[a_1,\ldots,a_d]]. In mixed characteristic, choose a coefficient Cohen ring V\subset R and a system of parameters p,a_2,\ldots,a_d, giving A=V[[a_2,\ldots,a_d]]. The Cohen Structure Theorem and completeness give finiteness in both cases; the maps from these power series rings are injective by dimension. No generic separability is required here.

Apply the Cohen Factorization Theorem of Avramov–Foxby–Herzog [2] to A\to S:

where U is complete, A\to U is flat local, and its closed fiber is regular; see [20, Theorem 3.1]. Regularity ascends along this map because both A and the closed fiber are regular [28, Tag 031E]. Thus U is a regular local ring, with residue field L.

Set T=R\otimes_A U. This is finite over U, and multiplication of the two given maps to S induces a surjection T\to S. Its closed fiber over U is

The ring R/\mathfrak m_A R is local Artinian with residue field k. Its radical is nilpotent, so the displayed fiber is local Artinian with residue field L. A finite algebra over a complete local ring is a product of complete local rings; this fiber shows that there is only one factor. Hence T is complete local with residue field L. Base change shows that R\to T is flat and

Its closed fiber is therefore regular of dimension b, and the flat dimension formula gives \dim T=d+b=h=\dim U.

Figure 1 records the two operations in the construction: the finite normalization A\subset R and the Cohen factorization of A\to S.

Preserving multiplicity. To check the multiplicity assertion, lift a regular system of parameters of T/\mathfrak mT to w_1,\ldots,w_b\in\mathfrak t. Lifting regular sequences under a flat local map [28, Tag 00MG] shows that w is a T-regular sequence and T/(w) is flat over R. Completeness defines the evaluation map

The local flatness criterion modulo the variables [28, Tag 00ML] makes this map flat. Its extended maximal ideal is \mathfrak t, because the images of w generate the maximal ideal of the closed fiber. As in the last paragraph of Lemma 4.1, the Hilbert–Samuel functions are unchanged under a flat local map with extended maximal ideal. Adjoining formal variables preserves multiplicity, as follows also from the associated graded polynomial extension. Therefore

The resolution and the perfect kernel. Start with a minimal free T-resolution of S, possibly infinite. Each syzygy is flat over R: inductively, the kernel of a surjection between R-flat modules with flat quotient is flat. Tensoring with k over R consequently preserves exactness and yields a minimal free resolution of S/\mathfrak mS over the regular local ring T/\mathfrak mT. That ring has global dimension b. Minimality, or Nakayama’s Lemma, therefore forces the original resolution to vanish in degrees greater than b. Thus \operatorname{pd}_T S\leq b.

The closed fiber S/\mathfrak mS is zero-dimensional, so the image of J in T/\mathfrak mT is primary for its maximal ideal. Choose in this image a system of b parameters and lift them to elements of J. Since the fiber is regular, this is a regular sequence on the fiber; flat regular-sequence lifting makes its lifts a T-regular sequence. Consequently

This proves perfection and identifies the length of the minimal resolution. The assertions include b=0, when the regular sequence is empty and the resolution has only its degree-zero term.

Components and the supported complex. Because R is a finite torsion-free A-module, it embeds in a finite free A-module. Flatness gives an embedding of T in a finite free U-module. The contractions to U of the associated primes of T are thus zero. In particular every minimal prime P of T contracts to zero, and T/P is finite integral over U. It follows that \dim(T/P)=\dim U=h.

The field L is infinite, so a minimal reduction of \mathfrak n generated by d elements exists. Lift its generators to x in T. The ideal J+(x) is \mathfrak t-primary. At a prime outside its vanishing set, either the localized resolution resolves zero and is split exact, or one of the x_i is a unit and its Koszul complex is contractible. Hence F_\bullet is contractible off \mathfrak t. Minimality of the resolution and x_i\in\mathfrak t put all its differential entries in \mathfrak t. Its degree-zero term is T, and

where d>0 is used.

Local rings over the generic point of R. Finally put K=\operatorname{Frac}(A) and K_R=\operatorname{Frac}(R). Localizing at A\setminus\{0\} gives

Here U\otimes_A K is regular, and K_R/K is a finite field extension. Express this field extension as a finite tower of simple algebraic extensions. After base change each step is defined by a monic polynomial, which is a nonzerodivisor even if the intermediate ring is not reduced. The resulting local rings are quotients of regular local rings by regular sequences, hence complete intersections. Every Q contracting to zero in R belongs to this localization, proving the first assertion of (iv). In mixed characteristic the fields have characteristic zero, so K_R/K is separable. The displayed extension is then finite etale, and its local rings are regular.

Every minimal prime of S contracts to a minimal prime of R by flat going-down; that prime is zero. This applies to the primes of T minimal over J. Moreover every minimal prime of T avoids A\setminus\{0\} by the preceding torsion-freeness argument, so in mixed characteristic its localization is a zero-dimensional regular local ring, namely a field. Thus T is generically reduced. At a prime Q minimal over J, the regular local ring T_Q is a domain, which also shows that Q contains exactly one minimal prime of T.□

Proof. The first assertion is [20, Lemma 4.2]. For the second, we use the following complete-intersection consequence from [20, Section 2.3 and Lemma 4.2]: a primary ideal of finite projective dimension in a local complete intersection of characteristic p has Hilbert–Kunz multiplicity equal to its colength. Since J is perfect, JT_Q is such an ideal in T_Q. Consequently

The Hilbert–Samuel associativity formula gives the conclusion. No assumption that x extend to a parameter system of T is involved.□

Proof. Lemma 4.2 reduces the problem to complete rings of the same positive dimension, with R a domain and the residue field of S algebraically closed. Use Proposition 4.3 to obtain T,J,P_\bullet,F_\bullet and the integers d,b,h=d+b. Set

This is a \mathfrak t-primary ideal.

Fix a minimal prime P of T and put D=T/P, with maximal ideal \mathfrak d. This is a complete local domain of dimension h and has the same algebraically closed residue field. Choose a minimal reduction (z_1,\ldots,z_h) of \mathfrak d. There is an integer c\geq0 such that

Take the D-algebra C and its normalized length \lambda from Theorem 3.1. For I=(z)C the length hypotheses of Theorem 2.1 hold with

For a power q=p^n, let F_\bullet(q) be the complex obtained by raising every matrix entry of the differentials of F_\bullet to its qth power. This is scalar extension by the nth Frobenius endomorphism of T and has the same ranks as F_\bullet. Its differential entries lie in \mathfrak t^{[q]}. Their images in C therefore lie in

Frobenius preserves the identities defining a contracting homotopy on each localization where F_\bullet is contractible. Subsequent scalar extension preserves those identities as well. Since (z) is \mathfrak d-primary, the ideals I and \mathfrak tC have the same radical. Consequently F_\bullet(q)\otimes_T C is contractible locally off V(I).

The first differential of the resolution P_\bullet has entries generating J, and P_0=T. Thus

Apply Theorem 2.1 with s=q-c, for q>c. Its degree-zero rank is one, and the ranks of all its terms are fixed as q varies. The Hilbert–Kunz quotient formula from Theorem 3.1 and Frobenius scaling give

Dividing by q^h and letting q tend to infinity proves

Every component of T has dimension h by Proposition 4.3. The associativity formulas for Hilbert–Kunz [13, Theorem 3.14] and Hilbert–Samuel multiplicities have the same generic length coefficients, so

In particular this summation does not assume that T is reduced.

We now verify the hypotheses of Lemma 5.1. The ring T is complete of characteristic p, its residue field is algebraically closed and therefore perfect, and J is perfect. For every minimal prime Q of J with \dim(T/Q)=d, Proposition 4.3 shows that T_Q is a complete intersection. The chosen x is a system of parameters of S=T/J, and its ideal is a reduction of \mathfrak n. Hence

as required.□

Proof. Put d=\dim R. If d=0, then R is a field and F_0 is a nonzero free module, so \chi_\infty^R(F)=\rank_R F_0\geq1=e(R). Assume d>0. Splitting off contractible free summands makes F minimal without changing its homology or Dutta multiplicity. Write b_i=\rank_R F_i; in particular b_0\geq1.

First suppose that the residue field is algebraically closed. Choose a minimal reduction I=(z_1,\ldots,z_d) of \mathfrak m. As in the proof of Theorem 5.2, there is a fixed integer c\geq0 with

Use the characteristic-p algebra C=R_{\mathrm{perf}} and normalized length \lambda from Theorem 3.1. For any short complex G over R, right exactness gives

Lemma 3.4 and (5.1) therefore give

Let q=p^n. The entries of the differentials of \Phi_R^nF lie in \mathfrak m^{[q]}\subseteq\mathfrak m^q\subseteq I^{q-c}. Frobenius preserves a contracting homotopy at every localization where F is contractible. Thus \Phi_R^nF again has finite-length homology; it is minimal and has nonzero zeroth homology, so it is short. Scalar extension to C preserves the same contracting homotopies; since I is \mathfrak m-primary, the extended complex is contractible locally off V(IC). The defining Dutta limit, with its index shifted by n, now gives

Moreover \sum_i(-1)^ib_i=0, since F becomes exact over the fraction field of R. Apply Theorem 2.1 to C\otimes_R\Phi_R^nF with parameter ideal IC, depth s=q-c, and \mu=e(I,R)=e(R). For q>c it gives

After division by q^d, the right side tends to b_0e(R). Since b_0\geq1, this proves the corollary in this case.

For an arbitrary residue field, Lemma 4.1 gives a complete flat local extension (R,\mathfrak m)\to(S,\mathfrak n) with \mathfrak n=\mathfrak mS, algebraically closed residue field, and \dim S=d, e(S)=e(R). Put G=S\otimes_R F. This is again a minimal short complex. Frobenius scalar extension commutes with this base change, and finite-length module lengths are preserved, so

this invariance is also recorded in [14, Section 4.17]. The ring S need not be a domain, so we pass to its components by a prime filtration rather than apply the preceding case directly.

Choose a finite prime filtration of the S-module S with factors D_j=S/P_j, and put h_j=\dim D_j. Additivity of Euler characteristic for the bounded free complex \Phi_S^nG gives

The complexes in the sum have finite-length homology because the contracting homotopies off \mathfrak n persist after tensoring. The Hochster–Huneke Frobenius homology estimate, in the form [3, Theorem 5.2], gives constants C_j such that

Indeed, its complex has length d and finite-length homology, and its finite coefficient module D_j has dimension h_j. When h_j<d, every homology length in this summand is O(p^{nh_j}), so its contribution divided by p^{nd} tends to zero.

When h_j=d, the ring D_j is a complete local domain with the same algebraically closed residue field. The complex D_j\otimes_SG is short over D_j: its homology is supported at the maximal ideal because the localized contracting homotopies persist, and its zeroth homology is nonzero because its matrices remain minimal and b_0\geq1. Inspecting differential matrices gives

Lengths of its homology over S and D_j agree. Taking limits, applying the already proved domain case to these factors, and using additivity of top-dimensional Hilbert–Samuel multiplicity along the prime filtration, we obtain

This completes the proof.□

Proof. Multiplication by f induces a map A/H^{j-1}\longrightarrow A/H^j whose cokernel is A/(H^j,f). Consequently

For large j these are Hilbert–Samuel polynomials. The polynomial on the right has leading coefficient e(H,A)/(t-1)!, and the one on the left has leading coefficient e(H,A/(f))/(t-1)!. Comparing them proves the assertion, including the case t=1.□

Proof. Choose a\geq1 with \mathfrak b^a\subseteq L+(x_1,\ldots,x_d), and set r_N=d(N-1)+1. The pigeonhole principle for monomials gives

In particular, on any fixed nonzero quotient A of B of dimension t\leq d-1, monotonicity of multiplicity under inclusion of primary ideals gives

This also applies when t=0.

To use this estimate, we will pass from B to fixed quotients of smaller dimension. Quotienting by an element of L that lowers the dimension by one can only increase the multiplicity of H_N, by Lemma 6.1. In the first case we will reach dimension d-1. In the second case we will first reach dimension d, choosing the quotient so that its localization at \mathfrak q is the residue field of B_{\mathfrak q}. The component B/\mathfrak q will therefore have coefficient one in the associativity formula, while every other component of that quotient will admit one more dimension-lowering quotient.

The ring B is catenary and equidimensional. Thus, for every prime \mathfrak p\subset B, one has \dim B/\mathfrak p=h-\operatorname{ht}\mathfrak p. We will also use the following consequence. A list of j partial parameters in B has all its minimal primes of height j: its quotient has dimension h-j, which gives the lower bound on these heights, and the generalized principal ideal theorem gives the upper bound. In particular its quotient is equidimensional.

For the first assertion, select h-d+1 partial parameters v_1,\ldots,v_{h-d+1} in L. Here is the required prime avoidance argument. After j<h-d+1 choices, a minimal prime of the preceding ideal has height j and quotient dimension h-j\geq d. It cannot contain L, because \dim B/L<d. Hence the next element can be chosen in L outside all these finitely many primes. Each choice lowers the dimension by one. Applying Lemma 6.1 successively, with H_N fixed, gives

The quotient on the right is fixed and has dimension d-1, so (6.2) proves the first assertion.

For the second assertion put c=h-d=\operatorname{ht}\mathfrak q. We choose v_1,\ldots,v_c\in\mathfrak q that are partial parameters in B and generate the maximal ideal of the regular local ring B_{\mathfrak q}. To justify both conditions simultaneously, suppose that v_1,\ldots,v_j have been chosen, with j<c, and that their images in

are linearly independent over \kappa(\mathfrak q). Let E_j be the contraction to B of

Since this cotangent space has dimension c, the ideal \mathfrak q is not contained in E_j. Nor is it contained in any minimal prime of (v_1,\ldots,v_j): those primes have height j<c. Prime avoidance, with the single possibly nonprime ideal E_j, therefore supplies an element v_{j+1} in \mathfrak q outside their union. This preserves both conditions. At the end, Nakayama’s Lemma gives

For c=0 this is the empty list.

Set A=B/(v_1,\ldots,v_c), which is equidimensional of dimension d. Repeated application of Lemma 6.1, followed by associativity of multiplicity, gives

where primes of A are identified with primes of B. The prime \mathfrak q is one of these minimal primes, and its coefficient is

On this component H_N becomes (x_1^N,\ldots,x_d^N). The parameter-ideal power formula therefore makes its contribution exactly N^d e((x_1,\ldots,x_d),B/\mathfrak q).

For any other prime \mathfrak p in (6.3), the two distinct primes \mathfrak p and \mathfrak q have the same height c and are incomparable. Choose w_{\mathfrak p}\in\mathfrak q\setminus \mathfrak p. Its image is nonzero in the complete local domain B/\mathfrak p of dimension d, and its quotient has dimension d-1. It belongs to the image of every H_N. Thus Lemma 6.1 and (6.2) give

The number of these components and their coefficients in (6.3) are fixed and finite. Summing proves the second assertion.□

Proof. Write \mathfrak t for the maximal ideal of T, and recall that \dim T=h, \dim S=d>0, S=T/J, and

where P is a finite free resolution of S, of length at most h-d. The complex F has length at most h, has F_0=T, and is contractible away from the closed point. Its differentials have entries in \mathfrak t, and its ranks have alternating sum zero. The images of x_1,\ldots,x_d form a minimal reduction of the maximal ideal of S; in particular \sqrt{J+(x_1,\ldots,x_d)}=\mathfrak t. The residue field, denoted by k, is algebraically closed.

Fix a minimal prime P_0 of T and put B=T/P_0, with maximal ideal \mathfrak b. By Proposition 4.3, B is a complete local domain of mixed characteristic, has dimension h, and has residue field k. Choose a parameter ideal (u_1,\ldots,u_h) that reduces \mathfrak b.

Finite extensions that make the differentials deep. Fix an integer q>h and choose roots z_i=u_i^{1/q} in an algebraic closure of \operatorname{Frac}(B). The ring B'=B[z_1,\ldots,z_h] is a finite domain extension of B. It is complete local, has residue field k, and has dimension h. The ideal (z_1,\ldots,z_h) is primary for its maximal ideal. Since \mathfrak b is integral over (u_1,\ldots,u_h), every element of \mathfrak b is integral over (z_1,\ldots,z_h)^qB'.

Choose a\geq1 with p^a\in(z_1,\ldots,z_h)B', where p is the residue characteristic. Apply Heitmann’s integral-extension form of the Briançon–Skoda Theorem [11], as stated in [12, Proposition 2.8] to the generating list

of the ideal I=(z_1,\ldots,z_h)B'. This list has n=h+1 members and satisfies the theorem’s hypothesis p\in\sqrt{(p^a,z_1)B'}. With k'=q-h-1\geq0, its conclusion sends each element integral over I^{n+k'}=I^q into I^{k'+1}=I^{q-h} after an integral extension. Here redundant generators are allowed by the theorem’s statement. Apply it to a finite generating set of \mathfrak b. Taking a prime above zero in each extension if necessary gives integral domain extensions, which can be embedded in a common algebraic closure of \operatorname{Frac}(B). Adjoin the finitely many integral coefficients witnessing the resulting memberships. We obtain a domain D finite over B such that

We may replace D by its normalization: it remains finite by excellence, and the inclusion persists. A finite algebra over a complete local ring is a product of complete local rings; since D is a domain, there is just one factor. Thus D is complete local of dimension h, and its residue field is k because it is finite over the algebraically closed field k.

Put r=\rank_B D. For every \mathfrak b-primary ideal H, the finite-module multiplicity formula and equality of residue fields give

Indeed, a \operatorname{Frac}(B)-basis of D\otimes_B\operatorname{Frac}(B) can be cleared of denominators to give an injection B^r\longrightarrow D with torsion cokernel. This cokernel has dimension less than h, so it does not contribute to multiplicity. Moreover, lengths of finite-length D-modules are the same over B and over D because the residue fields agree. Applying (6.5) to (u_1,\ldots,u_h) and using the parameter-ideal power formula gives

A lower bound for the normalized length. Use the algebra C and additive length \lambda supplied for D by Theorem 3.1, and put \mu=e((z_1,\ldots,z_h),D). Extend F along T\longrightarrow B\longrightarrow D\longrightarrow C. Its entries belong to (z_1,\ldots,z_h)^{q-h}C by (6.4). Local contractions survive this base change. The ideals \mathfrak tD and (z_1,\ldots,z_h)D have the same radical, so the required support is also preserved. Theorem 2.1, with s=q-h and zeroth rank one, therefore gives a constant K depending only on h and the fixed ranks of F such that

for all sufficiently large q. Together with (6.6), this reads

In particular, K is independent of the choices of D, C, and \lambda as q varies.

The power rule for the quotient lengths. For now keep q, D, C, and \lambda fixed. For every list of positive integers a_1,\ldots,a_d, consider

Before base change this complex is contractible away from the closed point: outside V(J) the resolution P is split exact, and outside V(x_1,\ldots,x_d) the Koszul complex is contractible. Its length is at most h. Lemma 2.2 therefore shows that its positive homology has \lambda-length zero and that its zeroth homology has finite \lambda-length. Thus

To check the power rule without any finiteness assumption on partial Koszul complexes, fix all exponents but a_i, and let E denote P_C tensored with the remaining Koszul factors. For positive integers v,w, the composable scalar maps x_i^v,x_i^w on E give the mapping-cone triangle

Every cone includes a positive power of each x_j, so all three have finite homology lengths by the preceding argument. Additivity of \lambda in their long exact homology sequence gives additivity of Euler characteristic. Hence that Euler characteristic is linear in each positive exponent. Applying this to all d exponents proves

No assertion that the homology of E itself has finite length was needed.

An upper bound from a prime filtration. Fix a finite prime filtration

All Q_j contain J; the list records repetitions. Set

Tensoring each filtration sequence with M_N gives a right exact sequence

Every module here is killed by J+(x_1^N,\ldots,x_d^N), whose extension to D is primary for its maximal ideal. Their lengths are finite: for example, each S_j\otimes_T M_N is a quotient of a finite direct sum of C/(J,x_1^N,\ldots,x_d^N)C. The image of the first arrow is a quotient of its source, so additivity and nonnegativity of \lambda yield

Summing these inequalities, applying the primary-ideal bound from Theorem 3.1, and then applying (6.5), we obtain

Thus this use of the prime filtration requires only right exactness of tensor product, not flatness of C over T.

We apply Lemma 6.2 to each term of the last sum. Since Q_j\supseteq J, one has \dim T/Q_j\leq d. If \dim T/Q_j<d, then also \dim B/Q_jB<d. If P_0 is not contained in Q_j, its image in the domain T/Q_j is nonzero, and hence

In either case the first assertion of the Lemma gives e(H_{j,N},B)=O(N^{d-1}).

The remaining case is \dim T/Q_j=d and P_0\subseteq Q_j. Such a Q_j is minimal over J: a strictly smaller prime over J would give a quotient of dimension greater than d. The generic regularity in Proposition 4.3 gives that T_{Q_j} is regular. It follows that Q_j contains exactly one minimal prime of T, namely P_0, and

Indeed the localization of P_0 is the unique minimal prime of the regular local domain T_{Q_j} and is therefore zero. The second assertion of Lemma 6.2 now gives

The two limits and the component sum. Divide (6.9) by rN^d and use (6.8). Let N tend to infinity while keeping the chosen q,D,C,\lambda fixed. The preceding asymptotic estimates imply

The right side does not involve any auxiliary extension or length algebra. Combine this bound with (6.7) and then let q tend to infinity. Uniformity of K gives

There is no exchange of limits: the N-limit was taken separately for each fixed choice of the data belonging to q.

Finally sum (6.12) over the minimal primes P_0 of T. The ring T is equidimensional and generically reduced by Proposition 4.3; its generic local lengths are therefore all one, and associativity gives

Every Q_j of dimension d belongs to exactly one of these components, because T_{Q_j} is regular. Thus summation counts each occurrence of Q_j in the filtration exactly once. The number of occurrences of a fixed such prime Q is \length_{T_Q}(S_Q), as seen by localizing the filtration at Q. The multiplicity formula for S consequently yields

The equality e(T)=e(R) from Proposition 4.3 finishes the proof.□

Proof. Choose a coefficient field k\subset R, and identify it with its image in S. Put L=S/\mathfrak n. A coefficient field of S containing this copy of k can be chosen as follows. Lift a transcendence basis of L/k to S. These lifts generate a copy of the corresponding rational function field, because a polynomial with nonzero residue is a unit whenever it occurs as a denominator. The remaining algebraic extension of residue fields is separable. Hensel’s lemma lifts each simple algebraic root, and a maximal compatible choice, obtained by Zorn’s lemma, lifts all of L. We henceforth identify this coefficient field with L.

Choose a presentation R=k[[X_1,\ldots,X_a]]/I and set

The compatible maps from R and L to S give a local map R'\to S. Extension of coefficient fields on power series rings is flat; for example, the local criterion of flatness applies to the regular sequence of variables and the field extension on their quotients. Thus R\to R' is flat, its maximal ideal is \mathfrak m'=\mathfrak mR', and its residue field is L. Filtering by powers of the maximal ideal gives

In particular, \dim R'=\dim R and e(R')=e(R).

The flat dimension formula for the original map shows that S/\mathfrak mS is Artinian. Since \mathfrak m'S=\mathfrak mS, this quotient is a finite-dimensional L-vector space. Lift an L-basis to s_1,\ldots,s_q\in S. Successively expressing a remainder modulo \mathfrak m'S, then modulo (\mathfrak m'S)^2, and so on, expresses every element of S as \sum_i a_i s_i with a_i\in R'. Indeed, the coefficient corrections at step j lie in (\mathfrak m')^j, so they converge in R', and the remainders tend to zero in S: the \mathfrak m'S-adic and \mathfrak n-adic topologies agree. Hence S is finite over R'.

Let P_\bullet be a free resolution of k over R. Flatness of R' over R makes P_\bullet\otimes_R R' a free resolution of L over R'. Therefore

A finite module over a Noetherian local ring with this vanishing is free. Thus R'\to S is finite free, necessarily of positive rank.□

Proof. Choose a minimal reduction t_1,\ldots,t_d of \mathfrak m_0. The induced map

is finite, by completeness and the parameter property. It is injective: a nonzero kernel in the regular local domain A_0 would force the dimension of its finite algebra R_0 to be less than d. Since the residue fields agree, the finite-module multiplicity formula and the reduction property give

Here no domain or Cohen–Macaulay hypothesis on R_0 is needed. For completeness, a finite module of rank r over A_0 contains a copy of A_0^r with torsion cokernel; the cokernel has dimension less than d, so additivity of multiplicity gives the last equality.

We describe the finite system. Throughout the construction, an equality in a presented module means a coordinate equality with an additional vector witnessing membership in its relation submodule. Thus, for a matrix U, the assertion w\in\operatorname{im}U is encoded by w=Uc with a new vector of unknowns c. The number of coordinates and assertions below is always fixed and finite.

The system records the algebra structures, freeness, and locality. The numerical conditions use the source’s generic rank for a lower multiplicity bound and the dimensions of the target’s associated graded pieces for an upper bound.

The source algebra and its generic rank. Fix a presentation

The entries of a matrix U of this same size are unknowns. On A^u, specify a bilinear multiplication by vectors c_{ij}\in A^u giving the products of standard basis vectors, and specify a unit vector \epsilon\in A^u. Impose the following identities modulo \operatorname{im}U: commutativity and associativity on basis vectors, the two unit identities on basis vectors, and the assertion that a relation column multiplied by any basis vector is again in \operatorname{im}U. These identities give a unital commutative A-algebra structure on

They are polynomial equations in the entries of U, the multiplication table, \epsilon, and the membership witnesses.

Impose also the vanishing of all (u-r+1)-minors of U. If there are no minors of that size, this condition is empty. Over the fraction field of A, it gives

These equations are satisfied by the original presentation in view of (7.1). A solution may give U smaller rank, but this only increases the rank of its cokernel. To turn that rank bound into a multiplicity bound, we will retain the source reduction (t) as well.

The target as a free source-module. Let n>0 be the original free rank of S_0 over R_0. Define the underlying module of the new target to be S=R^n. Give it an R-bilinear multiplication by specifying

and a unit \delta\in R^n. Represent these coefficients and this unit by vectors over A. Impose commutativity, associativity, and the unit identities on the R-basis \eta_1,\ldots,\eta_n. The underlying A-module is presented by the block diagonal matrix with n copies of U, so these are again finitely many polynomial equations, using the source multiplication table and relation witnesses. The structural map is a\mapsto a\delta. Consequently, in every solution, S is a unital commutative R-algebra which is free of rank n as an R-module.

The local ideals and the source reduction. Choose finite generating lists for \mathfrak m_0 and \mathfrak n_0 that include the respective images of t_1,\ldots,t_d. Represent corresponding lists in R and S by vectors of unknowns, with their first d entries defined to be t_i\epsilon and t_i\delta. Denote the generated ideals by \mathfrak a and \mathfrak b. Fix exponents N_R,N_S and j for which the original rings satisfy the following conditions, and impose them on the new data:

Such exponents exist because (t) is primary for both original maximal ideals and reduces \mathfrak m_0. For each standard A-module generator e_i of R, impose an expression

where a_\nu is the source ideal list. Impose the analogous expressions for the un standard A-module generators of S, using its unit and its ideal list. Thus R/\mathfrak a and S/\mathfrak b are spanned by their units as A-modules. All these conditions reduce to identities on finite lists of generators and monomials. Expanding products with the multiplication tables, and representing each coefficient by a vector, encodes them as polynomial equations.

We first check their consequences, before imposing the target Hilbert-function conditions. By (7.2), R is a nonzero finite A-algebra, and S=R^n is also nonzero and finite over A. In a finite algebra over the local ring A, every maximal ideal contains the extended ideal (t). The first two conditions in (7.3) therefore show that every maximal ideal of R contains \mathfrak a, and every maximal ideal of S contains \mathfrak b. In particular, both ideals are proper. Since R/\mathfrak a is spanned by the unit and annihilated by (t), it is a nonzero quotient of K, hence is K. Thus \mathfrak a is the unique maximal ideal of R. The same argument gives S/\mathfrak b=K and uniqueness of \mathfrak b.

The map R\to S is local. Indeed, an element of \mathfrak a is nilpotent modulo (t)R, so its image belongs to \sqrt{(t)S}=\mathfrak b; contraction of \mathfrak b is then \mathfrak a. Positive generic rank implies that R is faithful as an A-module, so A\to R is injective. The map R\to S is also injective: if a\delta=0, multiplication by a on S is zero, whereas S=R^n is faithful. Finiteness now gives \dim R=\dim S=d. Both rings are complete, since they are finite over complete A and their maximal-ideal topologies agree with their (t)-adic topologies by (7.3).

Because (t)\subseteq\mathfrak a, the last inclusion in (7.3) is equality. It makes (t) a reduction of \mathfrak a, whence

The rank equality has the same proof as (7.1). Thus the reduction relation converts the lower rank bound into the required source multiplicity bound. Neither a nonvanishing minor nor an additional condition on the length of the residue quotient is required.

An upper bound for the target Hilbert function. Write y_1,\ldots,y_q for the chosen original target ideal list. There is a graded surjection

Let I be its homogeneous kernel. Fix a degree-compatible monomial order. Choose finitely many monic homogeneous polynomials f_1,\ldots,f_b\in I whose leading monomials generate the initial ideal J=\operatorname{in}(I). Write

Here B_\ell is a fixed finite set. Replace the coefficients c_{\ell\beta} by unknowns C_{\ell\beta}\in A and require, on the new target ideal list b_1,\ldots,b_q, that

Membership is expressed using the finitely many degree a_\ell+1 monomials in the ideal list and coefficient vectors in S. These are again polynomial equations, and the original data provide a solution.

For any transferred solution, reduce the coefficients C_{\ell\beta} modulo (t). Equation (7.5) gives a homogeneous relation in the kernel of K[Z]\to\gr_{\mathfrak b}S with the same leading monomial Z^{\alpha_\ell}. The leading coefficient is still 1. The higher (t)-adic terms of the other coefficients do not contribute: their products with a degree-a_\ell monomial lie in \mathfrak b^{a_\ell+1}. It follows that the new initial ideal contains the monomial ideal J, interpreted over K. The transferred target may have additional initial relations; this containment is the direction needed for an upper bound. Standard monomials outside J thus span each new graded piece, whereas they form a basis of the corresponding original graded piece. Consequently

Summation and the equality of dimensions give

Finally, every coefficient in the foregoing presentations, tables, lists, and membership witnesses is an unknown; only the sizes, exponents, monomials, and the distinguished coefficients 1 have been fixed. Accordingly the entire system has coefficients in \ZZ[t_1,\ldots,t_d]. It is finite, the original data solve it in A_0, and (7.4) and (7.6) prove the assertion.□

Proof. Put B=(L[t_1,\ldots,t_d]_{(t)})^h, whose completion is L[[t]]. Artin approximation [1, Theorem 1.10] gives a solution in B. The henselization is the filtered direct limit of pointed etale neighborhoods of the origin. Thus the finitely many solution entries and identities descend to one such neighborhood with an L-rational point over t=0; this is also the explicit conclusion of [1, Corollary 2.1].

Shrink about that point. An etale presentation then has the form

where the distinguished point is (t,z)=(0,c), with c\in L^m, G_i(0,c)=0, and g(0,c)\Delta(0,c)\ne0. The solution entries are elements of E. Choose polynomial representatives with denominators powers of g\Delta. For each of the finitely many solution identities, clear denominators and choose a polynomial expression in the ideal (G_1,\ldots,G_m) which witnesses that identity. One may multiply by a further power of g\Delta before doing so.

Let W\subset L be the \ZZ-algebra generated by the following finite set: the coefficients of G_i and g, the coordinates of c, the coefficients of the representatives and ideal-membership witnesses just chosen, and the inverse of g(0,c)\Delta(0,c). This is a nonzero finitely generated \ZZ-algebra of characteristic zero. The same presentation defines an etale W[t]-algebra

Evaluation at (0,c) gives a W-valued point. All solution identities hold in E_W, by the chosen witnesses.

Choose a maximal ideal of W and let k_0 be its residue field. This field is finite. Indeed, it is a field finitely generated as a \ZZ-algebra. If it had characteristic zero, Zariski’s lemma over \mathbb Q would make it a number field. Finitely many ring generators of a number field are contained in its ring of integers with one nonzero integer inverted, which cannot contain the inverse of every rational prime. This is impossible for a field of characteristic zero. In positive characteristic, Zariski’s lemma makes k_0 a finite extension of its finite prime field.

Let K be an algebraic closure of k_0. The distinguished point specializes to a K-rational point of E_W\otimes_W K over t=0. The inverted Jacobian and denominator remain nonzero there. The etale lifting property, or multivariable Hensel’s lemma, lifts this point to a K[t]-algebra map

Explicitly, the point lifts uniquely over each successive quotient K[[t]]/(t)^a, since the Jacobian is invertible, and the compatible lifts give the displayed map by completeness. Applying it to the solution entries preserves all the polynomial identities and proves the assertion.□

Proof. By Lemma 4.2, it suffices to consider complete rings of the same dimension with algebraically closed target residue field. Lemma 7.1 then permits us to assume that the map is finite free and that both rings have the same algebraically closed coefficient field L.

If the common dimension is zero, equality of residue fields and freeness of rank n\ge1 give e(S)=\length_S S=n\length_R R\ge e(R). Suppose instead that d>0 and that e(R)>e(S). Lemma 7.2 produces a finite polynomial system with a solution over L[[t_1,\ldots,t_d]]. Lemma 7.3 gives a solution over K[[t_1,\ldots,t_d]] for an algebraically closed field K of positive characteristic. The resulting finite free local map \widetilde R\to\widetilde S satisfies

contradicting Theorem 5.2.□

Proof of Theorem 1.1. The reductions of Section 4 reduce the assertion to a complete, equal-dimensional flat local map with domain source and algebraically closed target residue field; the dimension-zero case is immediate. A source of positive characteristic is covered by Theorem 5.2, a source of mixed characteristic by Theorem 6.3, and a source of residue characteristic zero by Theorem 7.4. These cases exhaust the possibilities. Each reduction preserves the source multiplicity and preserves or decreases the target multiplicity, or recombines the inequalities by additivity. Thus the inequality holds for the original map.□