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 \[ F\otimes_C K_a\simeq \bigoplus_{T\subseteq\{1,\ldots,h\}}F[|T|]. \tag{3}\] 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 (1). The remaining row has terms \[F_i\otimes_C C/(z_1^a,\ldots,z_h^a),\qquad 0\leq i\leq h,\] 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 (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 \[(C/I)[Z_1,\ldots,Z_h]/(Z_1^a,\ldots,Z_h^a) \longrightarrow\mathop{\mathrm{gr}}_I(C/J_a).\] 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 (4). This proves the first assertion. It also gives \[ \lambda(C/I^M)=\mu\binom{M+h-1}{h}\qquad(M\geq1). \tag{5}\]

Write \(P=(C/I)[Z_1,\ldots,Z_h]\) and let \(N\) be the kernel in (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 \(\mathbb P_\mathbb Z^{h-1}\) 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 \(\mathbb Z[Z_1,\ldots,Z_h]\). This polynomial ring is regular of finite dimension and therefore has finite global dimension [28]; 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 \(\mathbb P_{C/I}^{h-1}\) 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 \(\mathcal O(t)\) and its higher cohomology vanishes. Combining these facts with (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 \[D(j)\xrightarrow{a}D(j+M_D)\longrightarrow D(j).\] Thus \(I^{M_D}\) annihilates its hyperhomology. If \(M\geq M_D\), the long exact sequence for \[0\longrightarrow D(j+M)\longrightarrow D(j) \longrightarrow D(j)/D(j+M)\longrightarrow0\] 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 4. 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 \[ F/I^M F\longrightarrow \mathbf R\Gamma(Y,F_Y/L^M F_Y). \tag{9}\] Filter its two sides by powers of \(I\) and \(L\), respectively. At the \(l\)th layer, \(0\leq l<M\), comparison for each free summand is \[\mathop{\mathrm{gr}}_I^l C\longrightarrow\mathbf R\Gamma(E,\mathcal O_E(l)).\] Lemma 4 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 (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 \[A=\mathbf R\Gamma(Y,F_Y),\qquad B_M=\mathbf R\Gamma(Y,F_Y/L^M F_Y).\] The triangles \[ A\longrightarrow B_M\longrightarrow \mathbf R\Gamma(Y,F_Y\otimes L^M)[1]\longrightarrow A[1] \tag{10}\] are compatible with the transition \(N\to M\) for \(N\geq M\). Lemma 5 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 (10) is zero on homology once \(N-M\) is sufficiently large. Exactness then gives \[ \operatorname{im}\bigl(H_k(B_N)\to H_k(B_M)\bigr) =\operatorname{im}\bigl(H_k(A)\to H_k(B_M)\bigr) \cong H_k(A). \tag{11}\] 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 \[A_a\longrightarrow F/J_aF\] 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 (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 (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 3. 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, \[\operatorname{im}\bigl(H_k(A_b)\to H_k(A_a)\bigr) =\operatorname{im}\bigl(H_k(F)\to H_k(A_a)\bigr) \cong H_k(F).\] 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 \[ I^{h(a-1)+1}\subseteq J_a\subseteq I^a \tag{12}\] 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 \[P_N\longrightarrow F/J_bF\longrightarrow F/J_aF\longrightarrow P_M.\] 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 (11). The compatible comparisons \(P_N\to B_N\) and \(P_M\to B_M\) in (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 3, (5), and Lemma 5. Therefore \[\lambda\mathbb H_k(Y,F_Y)=\lambda H_k(F)\] for every \(k\), which proves (7).

Tensoring by a vector bundle. The homology comparison is established. To prove (8), first compare consecutive twists of \(F_Y\). The quotient of \(F_Y(j)\) by \(F_Y(j+1)\) has term \(\mathcal O_E(j)^{b_i}\) in degree \(i\), so its Euler characteristic is \[\left(\sum_i(-1)^i b_i\right) \sum_v(-1)^v\lambda H^v(E,\mathcal O_E(j))=0.\] Euler additivity shows that \(\chi_\lambda(F_Y(j))\) is independent of \(j\in\mathbb Z\). Finally resolve \(V\) on \(\mathbb P_\mathbb Z^{h-1}\) by a finite complex of finite sums of twists, as in Lemma 4. 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 \(\mathop{\mathrm{rank}}V\), giving (8). ◻

Proof. Put \(n=h-1\). If \(n=0\), use the rank-one bundle on \(\mathbb P_\mathbb Z^0\). For \(n>0\), multiplication of binary linear forms defines \[f:(\mathbb P_\mathbb Z^1)^n\longrightarrow\mathbb P_\mathbb Z^n, \qquad f^*\mathcal O(1)=\mathcal O(1,\ldots,1).\] 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]. This gives flatness on each field fiber. Since both schemes are flat over \(\mathbb Z\), the fiberwise flatness criterion [28] gives flatness of \(f\). Over \(\mathbb Q\) 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 \[V=f_*\mathcal O(-1-r_1,\ldots,-1-r_n).\] 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 \(\mathcal O(a-1-r_1,\ldots,a-1-r_n)\) on \((\mathbb P_A^1)^n\). On \(\mathbb P_A^1\), the bundle \(\mathcal O(m)\) 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 \(\mathbb Z\). 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 4. ◻

Proof of Theorem 2. The homology assertions are Lemma 3. 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: \[H(x)=\prod_{i=1}^h(x+i),\] and write \(\rho_1>\cdots>\rho_{h-1}\) for the roots of \(H'(x)\). Rolle’s Theorem gives \[-(j+1)<\rho_j<-j.\] For all sufficiently large \(s\), depending only on \(h\), round \(s\rho_j\) to integers \(r_j\) so that \[ -(j+1)s<r_j<-js,\qquad |r_j-s\rho_j|\leq1. \tag{13}\] Let \(V\) be the bundle of Lemma 7, and put \(D=G\otimes V\) with \(G\) as in (6).

We claim that \[ \chi_\lambda(D)\geq0. \tag{14}\] In the \(l\)th \(E\)-layer of a twist of \(D\), the term in homological degree \(i\) is \(V(l-is)|_E^{b_i}\). If \(l\geq0\), then (13) implies \[\#\{j:r_j>l-is\}\leq i.\] 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 \[\lambda\mathbb H_k(Y,D/D(M))=0\qquad(k<0).\] For \(M\) large, Lemma 5 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 \[\#\{j:r_j>l-is\}\geq i-1.\] 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 \[\lambda\mathbb H_k(Y,D(-M)/D)=0\qquad(k>1).\] For large \(M\) the map from \(D\) to \(D(-M)\) is zero on hyperhomology. The long exact sequence gives a surjection \[\mathbb H_{k+1}(Y,D(-M)/D)\longrightarrow\mathbb H_k(Y,D),\] and hence length zero for \(k>0\). Lemma 5 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 \((L^{-is}/\mathcal O_Y)\otimes V^{b_i}\), with successive layers \(V(l)|_E^{b_i}\) for \(l=-is,\ldots,-1\). All Euler characteristics here are finite. Additivity, Lemma 6, and Lemma 7 give \[ 0\leq (h-1)!\lambda H_0(F) +\mu\sum_{i=0}^h(-1)^i b_i \sum_{l=-is}^{-1}\prod_{j=1}^{h-1}(l-r_j). \tag{15}\] The sum for \(i=0\) is empty.

We make the error in these sums explicit. Set \[P_s(x)=\prod_{j=1}^{h-1}(x-r_j/s),\qquad P(x)=\prod_{j=1}^{h-1}(x-\rho_j)=\frac{H'(x)}h.\] On the fixed interval \([-h,0]\), (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 \(i=1,\ldots,h\), \[\begin{align*} s^{-h}\sum_{l=-is}^{-1}\prod_j(l-r_j) &=\frac1s\sum_{l=-is}^{-1}P_s(l/s)\\ &=\int_{-i}^0P(x)\,dx+O_h(s^{-1})\\ &=\frac{H(0)-H(-i)}h+O_h(s^{-1}) =(h-1)!+O_h(s^{-1}). \end{align*}\] Since \(\sum_{i=1}^h(-1)^i b_i=-b_0\), substitution into (15) yields \[\lambda H_0(F)\geq \mu s^h\left(b_0-O_h\left(\frac{\sum_i b_i}{s}\right)\right).\] 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) 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 (17) 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 (17) 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 \[0\leq\lambda_\infty(N)\leq\lambda_\infty(M_n) \qquad(n\geq n_0).\] 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 \[gc=\sum_{j=1}^r \operatorname{Tr}_{E/K'}(g\beta_jc)b_j\in A'D=B_1.\] 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. In equal characteristic, choose a coefficient field and a presentation \(B=k[[X_1,\ldots,X_t]]/I\), and set \[B'=\overline{k}[[X_1,\ldots,X_t]]/ I\overline{k}[[X_1,\ldots,X_t]].\] 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] 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 \[B'=V'[[X_1,\ldots,X_t]]/IV'[[X_1,\ldots,X_t]].\] 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 \[\mathop{\mathrm{\ell}}_{B'}(B'/\mathfrak b^N B') =\mathop{\mathrm{\ell}}_B(B/\mathfrak b^N)\quad(N\geq1).\] 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 \[\dim S\geq\dim(S/Q)+\dim S_Q\geq f+d=\dim S.\] 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 \[R/\mathfrak p_j\longrightarrow S/\mathfrak p_jS\] has the same zero-dimensional closed fiber, so its source and target have equal dimension. The additivity formula for multiplicity yields \[e(R)=\sum_{\dim(R/\mathfrak p_j)=d}e(R/\mathfrak p_j),\qquad e(S)=\sum_{\dim(R/\mathfrak p_j)=d}e(S/\mathfrak p_jS).\] 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 12 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]. ◻

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 \(A\to S\): \[A\longrightarrow (U,\mathfrak u)\twoheadrightarrow S,\] where \(U\) is complete, \(A\to U\) is flat local, and its closed fiber is regular; see [20]. Regularity ascends along this map because both \(A\) and the closed fiber are regular [28]. 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 \[T/\mathfrak uT=(R/\mathfrak m_A R)\otimes_k L.\] 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 \[T/\mathfrak mT\cong U/\mathfrak m_A U.\] 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] shows that \(w\) is a \(T\)-regular sequence and \(T/(w)\) is flat over \(R\). Completeness defines the evaluation map \[R[[W_1,\ldots,W_b]]\longrightarrow T,\qquad W_i\longmapsto w_i.\] The local flatness criterion modulo the variables [28] 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 12, 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 \[e(T)=e(R[[W_1,\ldots,W_b]])=e(R).\]

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 \[b\leq\operatorname{grade}_T J \leq\operatorname{pd}_T S\leq b.\] 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 \[\sum_i(-1)^i\mathop{\mathrm{rank}}F_i =\left(\sum_i(-1)^i\mathop{\mathrm{rank}}P_i\right)(1-1)^d=0,\] 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 \[T\otimes_A K \cong (U\otimes_A K)\otimes_K K_R.\] 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. Put \(d=\dim R\). If \(d=0\), then \(R\) is a field and \(F_0\) is a nonzero free module, so \(\chi_\infty^R(F)=\mathop{\mathrm{rank}}_R F_0\geq1=e(R)\). Assume \(d>0\). Splitting off contractible free summands makes \(F\) minimal without changing its homology or Dutta multiplicity. Write \(b_i=\mathop{\mathrm{rank}}_R F_i\); 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 17, there is a fixed integer \(c\geq0\) with \[\mathfrak m^q\subseteq I^{q-c}\qquad(q\geq c).\] Use the characteristic-\(p\) algebra \(C=R_{\mathrm{perf}}\) and normalized length \(\lambda\) from Theorem 8. For any short complex \(G\) over \(R\), right exactness gives \[C\otimes_R H_0(G)=H_0(C\otimes_R G),\qquad \Phi_R^n(H_0(G))=H_0(\Phi_R^nG).\] Lemma 11 and (24) therefore give \[\lambda H_0(C\otimes_R G)=\chi_\infty^R(G).\]

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 \[\lambda H_0(C\otimes_R\Phi_R^nF) =\chi_\infty^R(\Phi_R^nF)=q^d\chi_\infty^R(F).\] Moreover \(\sum_i(-1)^ib_i=0\), since \(F\) becomes exact over the fraction field of \(R\). Apply Theorem 2 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 \[q^d\chi_\infty^R(F) \geq e(R)(q-c)^d \left(b_0-\frac{A_d}{q-c}\sum_{i=0}^d b_i\right).\] 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 12 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 \[\chi_\infty^S(G)=\chi_\infty^R(F);\] this invariance is also recorded in [14]. 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 \[\chi_S(\Phi_S^nG) =\sum_j\chi_S(D_j\otimes_S\Phi_S^nG).\] 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], gives constants \(C_j\) such that \[\mathop{\mathrm{\ell}}_S H_i(D_j\otimes_S\Phi_S^nG) \leq C_jp^{n\min\{h_j,d-i\}} \qquad(0\leq i\leq d).\] 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 \[D_j\otimes_S\Phi_S^nG \cong\Phi_{D_j}^n(D_j\otimes_SG).\] 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 \[\chi_\infty^R(F)=\chi_\infty^S(G) =\sum_{h_j=d}\chi_\infty^{D_j}(D_j\otimes_SG) \geq\sum_{h_j=d}e(D_j)=e(S)=e(R).\] This completes 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 \[R'=L[[X_1,\ldots,X_a]]/I L[[X_1,\ldots,X_a]].\] 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 \[\mathop{\mathrm{gr}}_{\mathfrak m'}R' \cong (\mathop{\mathrm{gr}}_{\mathfrak m}R)\otimes_k L.\] 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 \[\mathop{\mathrm{Tor}}^{R'}_1(L,S) = H_1(P_\bullet\otimes_R S) = \mathop{\mathrm{Tor}}^R_1(k,S)=0.\] 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 \[A_0=L[[t_1,\ldots,t_d]]\longrightarrow R_0\] 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 \[ r=e(R_0)=e((t),R_0)=\mathop{\mathrm{rank}}_{A_0}R_0. \tag{37}\] 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 \[A_0^v\xrightarrow{U_0}A_0^u\longrightarrow R_0\longrightarrow0.\] 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 \[R=\operatorname{coker}U.\] 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 \[ \mathop{\mathrm{rank}}_A R=u-\mathop{\mathrm{rank}}U\ge r>0. \tag{38}\] These equations are satisfied by the original presentation in view of (37). 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 \[\eta_i\eta_j=\sum_{h=1}^n d_{ijh}\eta_h, \qquad d_{ijh}\in R,\] 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: \[ \mathfrak a^{N_R}\subseteq(t)R,\qquad \mathfrak b^{N_S}\subseteq(t)S,\qquad \mathfrak a^{j+1}\subseteq(t)\mathfrak a^j. \tag{39}\] 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 \[e_i=c_i\epsilon+\sum_\nu a_\nu w_{i\nu}, \qquad c_i\in A,\quad w_{i\nu}\in R,\] 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 (38), \(R\) 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 (39) 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 (39).

Because \((t)\subseteq\mathfrak a\), the last inclusion in (39) is equality. It makes \((t)\) a reduction of \(\mathfrak a\), whence \[ e(R)=e((t),R)=\mathop{\mathrm{rank}}_A R\ge r. \tag{40}\] The rank equality has the same proof as (37). 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 \[L[Z_1,\ldots,Z_q]\longrightarrow\mathop{\mathrm{gr}}_{\mathfrak n_0}S_0, \qquad Z_i\longmapsto y_i+\mathfrak n_0^2.\] 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 \[f_\ell=Z^{\alpha_\ell} +\sum_{\beta\in B_\ell}c_{\ell\beta}Z^\beta, \qquad |\beta|=|\alpha_\ell|=a_\ell, \quad Z^\beta<Z^{\alpha_\ell}.\] 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 \[ b^{\alpha_\ell} +\sum_{\beta\in B_\ell}C_{\ell\beta}b^\beta \in\mathfrak b^{a_\ell+1}. \tag{41}\] 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 (41) gives a homogeneous relation in the kernel of \(K[Z]\to\mathop{\mathrm{gr}}_{\mathfrak b}S\) 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 \[\dim_K\mathfrak b^a/\mathfrak b^{a+1} \le \dim_L\mathfrak n_0^a/\mathfrak n_0^{a+1} \quad(a\ge0).\] Summation and the equality of dimensions give \[ e(S)\le e(S_0). \tag{42}\]

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 \(\mathbb Z[t_1,\ldots,t_d]\). It is finite, the original data solve it in \(A_0\), and (40) and (42) prove the assertion. ◻

Proof. Put \(B=(L[t_1,\ldots,t_d]_{(t)})^h\), whose completion is \(L[[t]]\). Artin approximation [1] 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].

Shrink about that point. An etale presentation then has the form \[E=\bigl(L[t,z_1,\ldots,z_m]/(G_1,\ldots,G_m)\bigr)_{g\Delta}, \qquad \Delta=\det\left(\frac{\partial G_i}{\partial z_j}\right),\] 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 \(\mathbb Z\)-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 \(\mathbb Z\)-algebra of characteristic zero. The same presentation defines an etale \(W[t]\)-algebra \[E_W=\bigl(W[t,z]/(G_1,\ldots,G_m)\bigr)_{g\Delta}.\] 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 \(\mathbb Z\)-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 \[E_W\otimes_W K\longrightarrow K[[t]].\] 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 13, it suffices to consider complete rings of the same dimension with algebraically closed target residue field. Lemma 23 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 \(e(S)=\mathop{\mathrm{\ell}}_S S=n\mathop{\mathrm{\ell}}_R R\ge e(R)\). Suppose instead that \(d>0\) and that \(e(R)>e(S)\). Lemma 24 produces a finite polynomial system with a solution over \(L[[t_1,\ldots,t_d]]\). Lemma 25 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 \[e(\widetilde R)\ge e(R)>e(S)\ge e(\widetilde S),\] contradicting Theorem 17. ◻