Proof. Lift a homogeneous \(F\)-basis of \(B/A_{>0}B\) to \(B\) and map the free module on these lifts to \(B\). This map is surjective: in each degree, an element not yet expressed in the lifts differs from such an expression by a sum of positive-degree multiples of elements of smaller degree, to which induction applies. Its kernel is finitely generated because \(A\) is Noetherian. Repeating the construction for successive kernels gives a free resolution. At each stage, reduction modulo \(A_{>0}\) identifies the chosen free generators with a basis of the corresponding module modulo \(A_{>0}\), so the next kernel lies in \(A_{>0}\) times that free module. Thus every differential between free modules has positive-degree entries. Tensoring with \(F\) makes these differentials zero and proves the assertion about \(\mathop{\mathrm{Tor}}\).

A minimal resolution of a proper homogeneous ideal \(H\), followed by the inclusion \(H\hookrightarrow A\) and the quotient \(A\twoheadrightarrow A/H\), is a minimal resolution of \(A/H\): the additional entries also have positive degree since \(H\subseteq A_{>0}\). This proves (4), including \(H=0\).

For a monomial quotient, carry out the same construction with multihomogeneous lifts. The initial free module is \(A\) in multidegree zero. A free module whose generator multidegrees are nonnegative has no component in a multidegree outside \(\mathbb Z_{\geq0}^r\); neither does its kernel or a quotient of that kernel. Induction therefore gives the last assertion. ◻

Proof. For \(\sigma=\{i_1<\cdots<i_p\}\) the differential is \[d(be_\sigma)=\sum_{q=1}^p(-1)^{q-1}y_{i_q}b\, e_{\sigma\setminus\{i_q\}}.\] The complex \(K_\bullet(A)\) is the tensor product over \(F\) of the one-variable complexes \[0\longrightarrow F[y_i](-1) \xrightarrow{\ y_i\ }F[y_i]\longrightarrow0.\] Each has homology \(F\) in position zero and no other homology. A complex of vector spaces splits as its homology with zero differential plus contractible two-term summands. Tensoring these splittings shows that \(K_\bullet(A)\) is a free resolution of \(F\).

To compare this computation with a resolution of \(B\), take any graded free resolution \(F_\bullet\to B\) and form the first-quadrant double complex \(F_\bullet\otimes_A K_\bullet(A)\). Augmenting either factor gives maps from its total complex to \(F_\bullet\otimes_A F\) and to \(B\otimes_A K_\bullet(A)\), respectively. The kernels have exact columns or exact rows, since the terms of the other factor are free. Their total complexes are exact. For exact columns, eliminate a cycle’s component of greatest horizontal index by a vertical boundary; the correction introduces terms only in the next column to the left. There are finitely many components on each total-degree diagonal, so iteration eliminates the cycle. The argument for exact rows interchanges the two indices. Both augmentation maps consequently induce homology isomorphisms. This identifies the homology of \(B\otimes_A K_\bullet(A)=K_\bullet(B)\) with \(\mathop{\mathrm{Tor}}^A(B,F)\), preserving the internal grading. ◻

Proof. Write \(Q=E/W\). In degree \(d\geq0\), let \(E_{\succeq\alpha}\) be the sum of the positions \(Vy^\gamma\) with \(|\gamma|=d\) and \(\alpha\preceq\gamma\), and define \(E_{\succ\alpha}\) similarly with strict inequality. Their images in \(Q_d\) give a finite descending filtration. Its quotient at \(\alpha\) is \[ \frac{\operatorname{im}(E_{\succeq\alpha}\to Q_d)} {\operatorname{im}(E_{\succ\alpha}\to Q_d)} \simeq \frac{E_{\succeq\alpha}} {E_{\succ\alpha}+(W_d\cap E_{\succeq\alpha})} \simeq (V/V_\alpha)y^\alpha. \tag{7}\] The last isomorphism takes the coefficient at \(\alpha\). Summing its dimensions proves (5); both sides vanish for \(d<0\).

Fix an internal degree \(j\geq0\). We give all terms of \(K_\bullet(Q)_j\) a common filtration indexed by the degree-\(j\) exponents. For a subset \(\sigma\), put \(\mathbf{1}_\sigma=\sum_{i\in\sigma}e_i\). In the summand with exterior label \(e_\sigma\), place the coefficient position \(\alpha\) at \[\alpha+\mathbf{1}_\sigma.\] More explicitly, the filtration subspace at \(\beta\) in that summand is the image of the span of all \(Vy^\alpha\) satisfying \(\beta\preceq\alpha+\mathbf{1}_\sigma\). The full filtration subspace is the direct sum over the labels \(\sigma\). Addition compatibility means that the relative order of the positions within one summand agrees with their order in \(Q_{j-|\sigma|}\).

A differential deleting \(i\in\sigma\) multiplies its coefficient by \(y_i\). It sends the position with exponent \(\alpha\) to that with exponent \(\alpha+e_i\), and \[(\alpha+e_i)+\mathbf{1}_{\sigma\setminus\{i\}} =\alpha+\mathbf{1}_\sigma.\] Thus the differential preserves the filtration. By (7), its map on the corresponding associated graded positions is, up to the Koszul sign, \[V/V_\alpha\longrightarrow V/V_{\alpha+e_i}, \qquad [v]\longmapsto[v].\] This is well defined by \(V_\alpha\subseteq V_{\alpha+e_i}\). Consequently the associated graded complex is exactly \(K_\bullet(E/\operatorname{in}_{\prec}W)_j\), decomposed by the total exponent \(\beta\).

For completeness, if \(d:C\to C'\) is a filtration-preserving linear map between finite-dimensional filtered spaces, the filtration induced on \(\ker d\) gives an injection \[\operatorname{gr}(\ker d) \hookrightarrow\ker(\operatorname{gr}d).\] Indeed a leading class represented by an element of \(\ker d\) maps to zero, and the induced filtration on the kernel makes this inclusion injective. Since taking associated graded preserves dimension, it follows that \(\operatorname{rank}d\geq\operatorname{rank}(\operatorname{gr}d)\). At a fixed term of a complex, the homology dimension is the term dimension minus the ranks of its incoming and outgoing maps. Applying the rank inequality to these two maps proves that passing to the associated graded complex cannot decrease this homology dimension. Lemma 6 now gives (6). For \(j<0\) all the relevant Koszul terms vanish, so the same assertion holds. ◻

Proof. Commutativity of the \(t_i\) and centrality of \(k\) make the stated right \(T\)-action well defined. The map \[ \Delta\otimes_kT\longrightarrow N, \qquad b\otimes y^\alpha\longmapsto bt^\alpha \tag{11}\] is a graded vector-space isomorphism by (8). It is right \(T\)-linear because \((bt^\alpha)t_i=bt^{\alpha+e_i}\). Thus \(N\) is right \(T\)-free of rank \(m\), with a basis in degree zero, and \[ N_{>0}=NT_{>0}. \tag{12}\]

The left \(S\)-action commutes with the right \(T\)-action by associativity of multiplication in \(N\). Moreover, \(N=\Delta\otimes_\mathbb CS\) as a left \(S\)-module, so \(N\) is left \(S\)-free and therefore flat. Each element of \(I\) is central in \(N\), whence \(IN\) is a two-sided ideal. In particular it is a right \(T\)-submodule, and \[Q\simeq\Delta\otimes_\mathbb C(S/I).\] Taking a complex basis of \((S/I)_d\) proves (9).

Let \(F_\bullet\to S/I\) be a minimal graded \(S\)-free resolution. Since \(S\) is commutative, regard its terms as right \(S\)-modules and tensor with the \((S,T)\)-bimodule \(N\). Flatness gives a resolution \[F_\bullet\otimes_SN\longrightarrow Q\] by graded free right \(T\)-modules. An entry \(f\) of an original differential acts on \(N\) by left multiplication. This map is right \(T\)-linear, again by associativity.

Choose a \(k\)-basis \(b_1,\ldots,b_m\) of \(\Delta\), which is a right \(T\)-basis of \(N\) under (11). If \(f\) is nonzero and has degree \(e>0\), then each \(fb_u\) belongs to \(N_e\). By the graded decomposition (8), its coefficients in the basis \(b_1,\ldots,b_m\) are homogeneous polynomials in \(T\) of degree \(e\). They therefore lie in \(T_{>0}\). Every nonzero original differential entry has positive degree, so the transferred resolution is minimal. Each summand \(S(-s)\) of \(F_p\) becomes \(m\) summands \(T(-s)\). Lemma 5 proves (10) for every \(p,j\). If \(I=S\), both quotients and their resolutions may be taken to be zero, and the same formulas hold. ◻

Proof. If \(I=S\), take \(M=S\). Assume henceforth that \(I\) is proper. Apply Theorem 3 to the given regular sequence. It supplies \(k\), \(\Delta\), and commuting degree-one forms \(t_h\) satisfying the basis property (8) and homogeneous identities \[ t_h^{a_h}=c_hf_h+\sum_{\ell<h}B_{h\ell}f_\ell +\sum_{\ell>h}C_{h\ell}t_\ell, \qquad c_h\in k^\times, \tag{15}\] where the multipliers belong to \(N=\Delta[x_1,\ldots,x_n]\). In particular \(C_{h\ell}\) is homogeneous of degree \(a_h-1\) when nonzero. Define \(T=k[y_1,\ldots,y_n]\), \(m=\dim_k\Delta\), \(W=IN\), and \(Q=N/W\) as in Lemma 8.

Under (11), view \(N\) as the free \(T\)-module with coefficient space \(V=\Delta\). In each total degree, order exponents by placing the lexicographically smallest tuple \[(\alpha_n,\ldots,\alpha_1)\] first. This order is compatible with addition. Apply Proposition 7 to the homogeneous \(T\)-submodule \(W\) with this order.

For an exponent \(\alpha\), let \(V_\alpha\subseteq\Delta\) be its leading coefficient space. It is closed under left multiplication by \(\Delta\): if \(w\in W\) witnesses a coefficient \(b\) at \(\alpha\) with no earlier coefficients, then \(\lambda w\in W\) witnesses \(\lambda b\). Here \(W\) is left \(\Delta\)-linear because it is an ideal of \(N\). Hence \(V_\alpha\) is a left ideal of \(\Delta\). Since \(\Delta\) is a division algebra, \[V_\alpha=0\quad\hbox{or}\quad\Delta.\] Explicitly, if \(0\ne b\in V_\alpha\), every \(c\in\Delta\) is \((cb^{-1})b\) and therefore also belongs to \(V_\alpha\).

The set \[\mathcal U=\{\alpha\in\mathbb Z_{\geq0}^n:V_\alpha=\Delta\}\] is upward closed under addition of exponent vectors, since the leading object is a \(T\)-submodule. Define \(M\subseteq S\) to be the monomial ideal with exponent set \(\mathcal U\), and define \(M_k\subseteq T\) by the same exponent set. Then \[ N/\operatorname{in}_{\prec}W \simeq\Delta\otimes_k(T/M_k) \simeq(T/M_k)^{\oplus m} \tag{16}\] as graded right \(T\)-modules.

We next verify the power containment. Since every \(f_\ell\) lies in \(I\), Equation (15) gives \[t_h^{a_h}-\sum_{\ell>h}C_{h\ell}t_\ell\in W.\] Expand each \(C_{h\ell}\) in the left \(\Delta\)-basis \(t^\alpha\). Right multiplication by \(t_\ell\) turns a term \(bt^\alpha\) into \(bt^{\alpha+e_\ell}\), using only commutativity of the forms. Every resulting exponent has a positive coordinate of index greater than \(h\). It consequently lies strictly after \(a_he_h\) in the chosen order, because \(a_he_h\) has zero coordinates at all indices greater than \(h\). The displayed element therefore has leading position \(a_he_h\) with coefficient \(1\). Thus \(a_he_h\in\mathcal U\) for every \(h\), proving \(P\subseteq M\).

It remains to compare the numerical invariants. The scalar extension of \(S/M\) from \(\mathbb C\) to \(k\), with \(x_i\) renamed \(y_i\), is \(T/M_k\). Its Hilbert function is unchanged. Tensoring the Koszul complex over \(\mathbb C\) with the field \(k\) also preserves exactness and commutes with homology, so \[\begin{align*} \dim_k(T/M_k)_d&=\dim_\mathbb C(S/M)_d,\\ \beta^T_{p,j}(T/M_k)&=\beta^S_{p,j}(S/M). \end{align*}\] Equations (9) and (16), together with (5), now give \[m\dim_\mathbb C(S/I)_d =\dim_k Q_d =\dim_k(N/\operatorname{in}_{\prec}W)_d =m\dim_\mathbb C(S/M)_d.\] Likewise, (10), (6), and (16) yield, for every \(p\geq0\) and every \(j\in\mathbb Z\), \[m\beta^S_{p,j}(S/I) =\beta^T_{p,j}(Q) \leq\beta^T_{p,j}(N/\operatorname{in}_{\prec}W) =m\beta^S_{p,j}(S/M).\] Cancelling the positive integer \(m\) proves both assertions. ◻

Proof. The expansion of \(D(x^u)\) has coefficient \(1\) at \(x^u\), and every other monomial is lexicographically larger than \(x^u\). Thus \(D\) is unitriangular on each finite-dimensional graded piece and is invertible. Multiplication by a variable other than \(q\) commutes with \(D\), while \[ qD(x^u)=D(qx^u)+\zeta^{u_\ell}\xi D(x^u). \tag{18}\] Since \(K\) is spanned by monomials and is an ideal, these identities show that \(D(K)\) is an ideal. The invertibility and degree preservation of \(D\) give the Hilbert-function equality.

To check preservation of \(P\), first observe that \[\prod_{s=0}^{a_\ell-1}(q-\zeta^s\xi) =q^{a_\ell}-\xi^{a_\ell}\in P.\] Here the second term belongs to \(P\) because \(a_h\le a_\ell\). If a monomial is divisible by \(q^{a_\ell}\), its image under \(D\) is divisible by this displayed polynomial. If it is divisible by another generator of \(P\), that generator remains a factor of its image. Consequently \(D(P)\subseteq P\). In each degree this inclusion is an inclusion between vector spaces of the same finite dimension, so it is an equality.

It remains to compare Betti numbers. If \(K=S\), both quotients are zero. Otherwise, choose a minimal exponent-multigraded free resolution \(F_\bullet\to S/K\) with \(F_0=S\), as in Lemma 5. A nonzero differential entry from a generator of multidegree \(u\) to one of multidegree \(v\) has the form \(c x^{u-v}\), where \(u\ge v\) coordinatewise. Keeping the ordinary graded free modules, replace this entry by \[ c\frac{D(x^u)}{D(x^v)}. \tag{19}\] The ratio is a polynomial: all unchanged monomial factors are nested, as are the products in (17).

For each free summand whose generator has multidegree \(v\), define a linear change on its coefficient polynomials by \[ C_v(x^{w-v})=\frac{D(x^w)}{D(x^v)},\qquad w\ge v. \tag{20}\] More explicitly, the right-hand side is \[\left(\prod_{i\ne\ell}x_i^{w_i-v_i}\right) \prod_{s=v_\ell}^{w_\ell-1}(q-\zeta^s\xi).\] It has the original monomial \(x^{w-v}\) with coefficient \(1\) and only lexicographically larger additional terms. Hence \(C_v\) is an invertible graded vector-space map. These maps need not be \(S\)-linear; the new differential defined by (19) is \(S\)-linear.

The changes on the free summands intertwine the old and new differentials. Indeed, for a source coefficient \(x^{w-u}\) and a nonzero entry \(c x^{u-v}\), the required equality is precisely \[\frac{D(x^w)}{D(x^u)} \left(c\frac{D(x^u)}{D(x^v)}\right) =c\frac{D(x^w)}{D(x^v)}.\] Thus the new maps form a complex that is exact in every positive homological degree. At \(F_0=S\) the change is \(C_0=D\), so its image from \(F_1\) is \(D(K)\). We have obtained a free resolution of \(S/D(K)\) with the same ordinary graded free modules as \(F_\bullet\). Every replaced nonzero entry still has its original positive degree, so this resolution is minimal. Its free ranks in each degree give the asserted equality of all graded Betti numbers. ◻

Proof. If \(K\) is stable in the box, take \(H=K\). Otherwise choose a failed transfer from \(q=x_\ell\) to \(\xi=x_h\), with \(h<\ell\), and form the map \(D\) above. Replace \(K\) by \[K'=\operatorname{in}_{\mathrm{lex}}D(K),\] where the leading monomial is the lexicographically largest one. By Lemma 11, \(P\subseteq D(K)\), and hence \(P\subseteq K'\). That lemma and Proposition 7 show that the Hilbert function is preserved and \[\beta^S_{p,j}(S/K)\le\beta^S_{p,j}(S/K') \qquad(p\ge0,\ j\in\mathbb Z).\] We prove that repeated replacements terminate by measuring progress in every lexicographic prefix.

Fix a degree \(d\) and a prefix \(E\) of the degree-\(d\) monomials, listed from greatest to least. Let \(\pi_E\) be the projection onto their span. Gaussian elimination with this ordered list of columns gives \[\#\{m\in E:m\in K'\} =\dim_\mathbb C\pi_E(D(K)_d).\] For the monomials \(m\in K\cap E\), the vectors \(\pi_E(D(m))\) are linearly independent: their coordinates in their original monomial positions form a triangular matrix with diagonal entries \(1\). Consequently \[ \#(K'\cap E)\ge\#(K\cap E) \tag{21}\] for every such prefix in every degree.

There is a strict inequality for at least one prefix. Choose a monomial \(x^u\in K\) inside the box witnessing the failed transfer, so \[m'=\frac{\xi}{q}x^u \quad\text{lies inside the box but does not belong to }K.\] Its existence implies \(1\le u_\ell<a_\ell\). Let \(E\) be the degree-\(|u|\) prefix ending at \(m'\). Every monomial in \(K\cap E\) is strictly larger than \(m'\), so its \(D\)-image has zero coefficient at \(m'\). On the other hand, the coefficient at \(m'\) of \(D(x^u)\) is \[-\sum_{s=0}^{u_\ell-1}\zeta^s =-\frac{1-\zeta^{u_\ell}}{1-\zeta}\ne0.\] Thus \(\pi_E(D(x^u))\) is independent of the projected vectors already used for \(K\cap E\), proving strictness in (21).

Every monomial ideal containing \(P\) is determined by the monomials it includes in the finite box \(0\le u_i<a_i\). There are therefore only finitely many such ideals. The nondecreasing prefix counts, with at least one strict increase at every replacement, prevent a repetition. Continuing whenever stability fails must consequently stop at an ideal \(H\) stable in the box. The Hilbert-function equalities and all Betti inequalities persist along this finite sequence. ◻

Proof. We first prove the assertion about lex shadows, without assuming stability of \(X\). Empty shadows need no argument. For \(v\in\mathcal B_r(d+1)\), with \(d\ge0\), its lex largest degree-\(d\) predecessor is \[p(v)=v-e_{\max\{h:v_h>0\}}.\] Every predecessor belongs to the bounded box. Moreover, the map \(p\) preserves weak lex order. Indeed, suppose \(v>w\), with first difference at coordinate \(k\). Subtraction at the last nonzero coordinate cannot change an earlier coordinate of \(v\). If the operation changes a coordinate of \(w\) before \(k\), then \(p(v)>p(w)\) already at that coordinate. Otherwise the strict inequality at \(k\) can disappear only if \(v_k=w_k+1\) and the tail of \(v\) after \(k\) is zero. Equality of total degrees then says that the tail of \(w\) contains exactly one unit. Subtracting its last nonzero coordinate gives \(p(v)=p(w)\). In all other cases the first strict inequality remains. Membership of \(v\) in the shadow of a lex segment is equivalent to membership of \(p(v)\) in that segment, proving that its shadow is lex.

We prove the three inequalities simultaneously by induction on \(r\). If \(\mathcal B_r(d)\) is empty, they are immediate. For \(r=1\), each degree box has at most one vector, so \(X=Y\). Assume henceforth that \(r>1\). A prime will indicate lex replacement or shadow in the first \(r-1\) coordinates.

Each slice \(X(s)\) is stable, and stability between slices gives \[ \partial'X(s)\subseteq X(s-1) \qquad(1\le s\le b_r). \tag{23}\] Replace each slice by \(Z(s)=\operatorname{Lex}'(X(s))\), and let \(Z\subseteq\mathcal B_r(d)\) be the set with these slices. By the induction hypothesis and (23), \[|\partial'Z(s)|\le|\partial'X(s)|\le|X(s-1)|=|Z(s-1)|.\] The shadow on the left and the slice on the right are lex segments in the same degree box. Consequently \[ \partial'Z(s)\subseteq Z(s-1) \qquad(1\le s\le b_r). \tag{24}\] Thus a unit in the last coordinate of a member of \(Z\) may be moved to any earlier coordinate with available capacity, while remaining in \(Z\).

The zero slice. If \(Z(0)=\mathcal B_{r-1}(d)\), then \(|Y(0)|\le|Z(0)|=|X(0)|\). Otherwise let \(v\) be the largest vector of \(\mathcal B_{r-1}(d)\) omitted from \(Z(0)\); all members of \(Z(0)\) are strictly larger than \(v\).

We claim that every \((u,s)\in Z\) is strictly larger than \((v,0)\). Equality is excluded by the choice of \(v\). If instead \((u,s)<(v,0)\), let \(k<r\) be their first differing coordinate. Then \(u_k<v_k\) and earlier coordinates agree. Also \(s>0\), since every member of the lex slice \(Z(0)\) is larger than \(v\). We will move the \(s\) units of the last coordinate into the first \(r-1\) coordinates to obtain a vector \(w\le v\).

Let \[c=\sum_{k<j<r}(b_j-u_j)\] be the available capacity strictly after coordinate \(k\). If \(s\le c\), distribute all \(s\) units among those positions; the result still has \(w_k=u_k<v_k\). If \(s>c\), fill all those positions and put the remaining \(s-c\) units at coordinate \(k\). Since \(|u|+s=|v|=d\) and the prefixes before \(k\) agree, \[ s-c =v_k-u_k-\sum_{k<j<r}(b_j-v_j) \le v_k-u_k. \tag{25}\] The remaining units therefore fit at \(k\). If the inequality in (25) is strict, then \(w_k<v_k\). If equality holds, every coordinate of the tail of \(v\) is full, and the constructed vector is exactly \(v\). In either case \(w\le v\). All the moves stay within the bounds, so repeated use of (24) gives \(w\in Z(0)\), a contradiction. This proves the claim.

There are thus at least \(|Z|=|X|\) vectors preceding \((v,0)\). The lex segment \(Y\) of that cardinality also consists entirely of vectors preceding \((v,0)\). Its zero slice is therefore contained in \(Z(0)\), proving \[ |Y(0)|\le|Z(0)|=|X(0)|. \tag{26}\]

The top slice. Put \(B=\sum_{h=1}^r b_h\) and let \(\rho(u)=(b_1-u_1,\ldots,b_r-u_r)\). The reflected complement \[X^*=\rho\bigl(\mathcal B_r(d)\setminus X\bigr) \subseteq\mathcal B_r(B-d)\] is stable. In fact, if an allowed upward transfer took a member of \(X^*\) outside \(X^*\), reflection would give an upward transfer from a member of \(X\) to a vector outside \(X\), contradicting stability. Reflection reverses lex order, so \(Y^*=\rho(\mathcal B_r(d)\setminus Y)\) is the lex segment of cardinality \(|X^*|\).

The zero-slice inequality, already proved in \(r\) coordinates, applies to \(X^*,Y^*\). Their zero-slice cardinalities are \[\begin{split} |X^*(0)|&=|\mathcal B_{r-1}(d-b_r)|-|X(b_r)|,\\ |Y^*(0)|&=|\mathcal B_{r-1}(d-b_r)|-|Y(b_r)|. \end{split}\] It follows that \(|X(b_r)|\le|Y(b_r)|\).

The shadow. We have established both end-slice comparisons. To recover the shadow bound, we separate its zero slice from the remaining vectors. For any stable set \(X\), \[ |\partial X|=|\partial'X(0)|+|X|-|X(b_r)|. \tag{27}\] Indeed, its shadow vectors with last coordinate zero are precisely \(\partial'X(0)\). Every shadow vector \(v\) with positive last coordinate has \(v-e_r\in X\): if it was obtained as \(v=u+e_h\) with \(h<r\), transfer one unit of the positive last coordinate of \(u\) to coordinate \(h\). This transfer is within bounds because \(v\) is within bounds. Stability gives \(v-e_r\in X\). Thus adding \(e_r\) bijects the members of \(X\) outside its top slice with shadow vectors having positive last coordinate, proving (27).

The same identity holds for the stable set \(Y\). Its zero slice is a lex segment, and (26) gives \(Y(0)\subseteq\operatorname{Lex}'(X(0))\). The induction hypothesis and monotonicity of shadows under inclusion imply \[|\partial'X(0)| \ge\bigl|\partial'\operatorname{Lex}'(X(0))\bigr| \ge|\partial'Y(0)|.\] Combining this with the top-slice inequality in (27) proves \(|\partial X|\ge|\partial Y|\) and completes the induction. ◻

Proof. Write \(Y_d=\operatorname{Lex}(X_d)\). Since \(H\) is an ideal, \(\partial X_d\subseteq X_{d+1}\). Lemma 13 therefore gives \[|\partial Y_d|\le|\partial X_d| \le|X_{d+1}|=|Y_{d+1}|.\] Both \(\partial Y_d\) and \(Y_{d+1}\) are lex segments in \(\mathcal B_r(d+1)\), so \(\partial Y_d\subseteq Y_{d+1}\). Thus multiplying a monomial of \(\Lambda H\) by a variable either gives a monomial of the next lex segment or leaves the box and enters \(P_r\). Products of monomials already in \(P_r\) stay in \(P_r\). This proves the ideal assertion. Each \(Y_d\) is stable and has cardinality \(|X_d|\), which proves stability and equality of Hilbert functions. ◻

Proof. Write \(d_p(B)_j:K^R_p(B)_j\to K^R_{p-1}(B)_j\) for the Koszul differential. Since its homology computes \(\mathop{\mathrm{Tor}}^R(B,\mathbb C)\), \[\beta^R_{p,j}(B) =\dim_\mathbb CK^R_p(B)_j -\operatorname{rank}d_p(B)_j -\operatorname{rank}d_{p+1}(B)_j.\] This proves the rank formula. The Koszul complex of \(U\) embeds termwise into that of \(B\), and the image of each of its differentials embeds into the corresponding differential image for \(B\). For the quotient \(V\), the termwise surjections give \[\operatorname{im}d_p(V)_j =\text{the image of }\operatorname{im}d_p(B)_j \text{ in }K^R_{p-1}(V)_j.\] Thus each of the two ranks can only decrease under either operation. Direct sums and grading shifts commute with the Koszul complex. Finally, if all the variables annihilate \(B\), every Koszul differential is zero. ◻

Proof. By Proposition 14, \(G\) is an ideal stable in the box and has the same Hilbert function as \(H\). If \(H=A\), then \(G=A\) and both comparisons are equalities. For proper \(H\), the ideal comparison implies the quotient comparison in positive homological degrees by Lemma 5: \[ \beta^A_{p+1,j}(A/H)=\beta^A_{p,j}(H), \qquad \beta^A_{p+1,j}(A/G)=\beta^A_{p,j}(G). \tag{30}\] The two quotient Betti numbers in homological degree zero are both \(1\) when \(j=0\) and \(0\) otherwise. We prove the two comparisons simultaneously by induction on \(r\), establishing the ideal comparison and then using (30). When \(r=1\), a monomial ideal containing \(x_1^{a_1}\) is a power ideal or the full ring, and \(H=G\).

Assume \(r>1\) and \(H\) proper, and put \[R=\mathbb C[x_1,\ldots,x_{r-1}],\qquad z=x_r,\qquad a=a_r.\] For \(s\ge0\), define coefficient ideals \[H^{(s)}=\{u\in R:uz^s\in H\},\qquad G^{(s)}=\{u\in R:uz^s\in G\}.\] These are ascending monomial ideals containing the earlier prescribed powers, and they equal \(R\) for \(s\ge a\). For \(s<a\), their stability in the earlier-variable box follows by applying stability with the last exponent fixed at \(s\). For \(1\le s<a\) we also have \[ R_{>0}H^{(s)}\subseteq H^{(s-1)},\qquad R_{>0}G^{(s)}\subseteq G^{(s-1)}. \tag{31}\] To see the first inclusion, it is enough to multiply a monomial of \(H^{(s)}\) by an earlier variable \(x_i\). If its earlier-variable exponent, or that exponent after multiplication by \(x_i\), is outside the box, the earlier powers already imply the required membership. Otherwise stability moves one factor of \(z\) to \(x_i\). The proof for \(G\) is identical.

We next pass from ideals over \(A\) to modules over \(R\). Multiplication by \(z\) is injective on \(H\), since \(H\) is an ideal in the domain \(A\). Set \[B_H=H/zH,\qquad B_G=G/zG.\] Reducing a minimal \(A\)-free resolution of \(H\) modulo \(z\) gives a minimal \(R\)-free resolution of \(B_H\). Here is a direct exactness check. Lift a cycle in positive homological degree modulo \(z\) to an element \(y\) of the free resolution, and write \(dy=zv\). Applying the next differential shows that \(v\) is a cycle, using injectivity of \(z\) on the free modules; in degree zero, apply the augmentation and use injectivity of \(z\) on \(H\) instead. Exactness gives \(v=du\), so \(y-zu\) is a cycle and hence a boundary. Its reduction modulo \(z\) is the original cycle. Right exactness identifies the cokernel with \(B_H\), and all differential entries remain of positive degree or become zero, so minimality is preserved. The same argument applies to \(G\). Therefore \[ \beta^A_{p,j}(H)=\beta^R_{p,j}(B_H),\qquad \beta^A_{p,j}(G)=\beta^R_{p,j}(B_G). \tag{32}\] The injective multiplication maps also give \[\dim_\mathbb C(B_H)_d=\dim_\mathbb CH_d-\dim_\mathbb CH_{d-1},\qquad \dim_\mathbb C(B_G)_d=\dim_\mathbb CG_d-\dim_\mathbb CG_{d-1}.\] Consequently \(B_H\) and \(B_G\) have equal Hilbert functions.

The monomial decomposition by powers of \(z\) describes these modules more precisely; this is the decomposition appearing in [7]. The coefficient of \(z^0\) in \(H/zH\) is \(H^{(0)}\), and its coefficient of \(z^s\) for \(s\ge1\) is \(H^{(s)}/H^{(s-1)}\). Since \(H^{(s)}=R\) for \(s\ge a\), this gives an isomorphism of graded \(R\)-modules \[ B_H\cong H^{(0)} \oplus\bigoplus_{s=1}^{a-1} (H^{(s)}/H^{(s-1)})(-s) \oplus(R/H^{(a-1)})(-a). \tag{33}\] There are no terms for \(s>a\). The shifts record that multiplication by \(z^s\) raises ordinary degree by \(s\). By (31), all middle summands are annihilated by the variables of \(R\), as illustrated in Figure 1. The decomposition is \(R\)-linear because multiplying by an earlier variable preserves the exponent of \(z\). The middle summands still contribute to \(\mathop{\mathrm{Tor}}\), but their Koszul differentials have rank zero. Thus they contribute nothing to the deficits. Lemma 15 yields \[ D^R_{p,j}(B_H) =D^R_{p,j}(H^{(0)}) +D^R_{p,j-a}(R/H^{(a-1)}), \tag{34}\] and the identical formula holds for \(G\).

It remains to compare the two endpoint contributions. Write \(\Lambda'\) for lex replacement with the earlier prescribed powers in \(R\). The induction hypothesis applies to both \(H^{(0)}\) and \(H^{(a-1)}\). It gives an ideal comparison for the first and a quotient comparison for the second. Their respective lex replacements have the same Hilbert functions, and hence the same Koszul term dimensions. Subtracting the Betti inequalities from these dimensions gives \[\begin{align*} D^R_{p,j}(H^{(0)}) &\ge D^R_{p,j}(\Lambda'H^{(0)}), \tag{35}\\ D^R_{p,j-a}(R/H^{(a-1)}) &\ge D^R_{p,j-a}(R/\Lambda'H^{(a-1)}). \tag{36}\end{align*}\] This also covers any coefficient ideal equal to \(R\): its lex replacement is \(R\), the corresponding ideal comparison is equality, and its quotient is zero.

The slice inequalities of Lemma 13 imply \[ G^{(0)}\subseteq\Lambda'H^{(0)},\qquad \Lambda'H^{(a-1)}\subseteq G^{(a-1)}. \tag{37}\] Indeed, fix an earlier-variable degree \(e\ge0\). For the first inclusion, apply the zero-slice inequality to the included exponent sets of \(H\) and \(G\) in total degree \(e\). For the second, apply the top-slice inequality in total degree \(e+a-1\). Each slice of \(G\) is already a lex segment in the earlier-variable box, so its cardinality comparison with the corresponding slice of \(H\) becomes the stated containment in the appropriate lex replacement. All monomials outside the earlier-variable box belong to both ideals. This proves the inclusions in every degree.

Apply Lemma 15 to the first inclusion in (37) and to the quotient map furnished by the second: \[R/\Lambda'H^{(a-1)}\longrightarrow R/G^{(a-1)}.\] Together with (35) and (36), this gives \[\begin{align*} D^R_{p,j}(H^{(0)})&\ge D^R_{p,j}(G^{(0)}),\\ D^R_{p,j-a}(R/H^{(a-1)}) &\ge D^R_{p,j-a}(R/G^{(a-1)}). \end{align*}\] Adding and using (34), we obtain \[D^R_{p,j}(B_H)\ge D^R_{p,j}(B_G).\] The equal Hilbert functions of \(B_H\) and \(B_G\) imply \[\dim_\mathbb CK^R_p(B_H)_j =\binom{r-1}{p}\dim_\mathbb C(B_H)_{j-p} =\dim_\mathbb CK^R_p(B_G)_j,\] with the binomial coefficient interpreted as zero outside its range. Subtracting the deficit inequality from these equal dimensions gives \(\beta^R_{p,j}(B_H)\le\beta^R_{p,j}(B_G)\). Equation (32) proves the ideal comparison over \(A\), and (30) completes the quotient comparison. Every step applies to every \(p\ge0\) and \(j\in\mathbb Z\); in particular, negative internal degrees and Koszul terms outside their homological range introduce no additional cases. ◻

Proof of Theorem 1. If \(I=S\), then \(L_d=S_d\) is allowed in every degree and \(J=S\); all quotient Betti numbers are zero. Assume \(I\ne S\). Proposition 9 gives a monomial ideal \(M\supseteq P\) with the Hilbert function of \(I\) and \[\beta^S_{p,j}(S/I)\le\beta^S_{p,j}(S/M) \qquad(p\ge0,\ j\in\mathbb Z).\] Proposition 12 gives a box-stable monomial ideal \(H\supseteq P\), with the same Hilbert function, and \[\beta^S_{p,j}(S/M)\le\beta^S_{p,j}(S/H).\] By Proposition 14, the degreewise box-lex replacement \(G=\Lambda H\) is an ideal, with the same Hilbert function. Theorem 16 gives \[\beta^S_{p,j}(S/H)\le\beta^S_{p,j}(S/G).\]

It remains to check that \(G\) is the ideal prescribed in (1). In degree \(d\), list all monomials in decreasing lex order and remove those belonging to \(P_d\). The remaining list is exactly \(\mathcal B_n(d)\) in decreasing lex order. The monomials of \(G_d\) outside \(P_d\) are the first \[q_d=\dim_\mathbb CI_d-\dim_\mathbb CP_d\] positions of this remaining list. This integer lies between zero and \(|\mathcal B_n(d)|\), because \(H\supseteq P\) has the same Hilbert function as \(I\).

Every box prefix is obtained by intersecting some full-list prefix with the box. Thus a lex-segment subspace \(L_d\) with \(\dim_\mathbb C(L_d+P_d)=\dim_\mathbb CI_d\) exists. Enlarging that full-list prefix to the largest one of the same resulting dimension adds only monomials already in \(P_d\), and does not change the sum. Consequently \(L_d+P_d=G_d\) for every \(d\), including the empty and full box cases. Hence \(J=G\). The displayed comparisons prove all the asserted Betti bounds for this single \(J\). ◻

Proof of Corollary 2. If \(I=S_F\), the prescription gives \(J=S_F\), and all quotient Betti numbers vanish. We may therefore assume that \(I\) is proper.

First let \(F=\mathbb C\). After a linear change of coordinates, one may arrange that \[f_1,\ldots,f_c,x_{c+1},\ldots,x_n\] is a regular sequence; the list of variables is empty when \(c=n\). Write \(\overline f_i\) for the images of the \(f_i\) in \(\overline S=\mathbb C[x_1,\ldots,x_c]\). They form a full regular sequence of degrees \(a_1,\ldots,a_c\). Theorem 1, applied in \(\overline S\) to every homogeneous ideal containing this sequence, gives both the Eisenbud–Green–Harris and lex-plus-powers assertions for \(\overline f_1,\ldots,\overline f_c\).

The Artinian reduction of Caviglia–Maclagan [6] transfers the Hilbert-function assertion to \(f_1,\ldots,f_c\). Theorem 4.1 of Caviglia–Kummini [5] transfers the Betti assertion as well. Their formulation uses Betti numbers of ideals. For any proper homogeneous ideal \(K\) in a standard graded polynomial ring \(A\), the exact sequence \(0\to K\to A\to A/K\to0\) gives \[\beta^A_{p+1,j}(A/K)=\beta^A_{p,j}(K) \qquad(p\ge0,\ j\in\mathbb Z).\] The remaining quotient Betti number is \(\beta^A_{0,0}(A/K)=1\). Thus their convention is equivalent to the quotient convention used here. The lexicographic embedding modulo \(P_F\) in that reduction selects exactly the prescribed \(q_d\) greatest monomials outside \(P_F\) in each degree. A linear change of coordinates preserves Hilbert functions and graded Betti numbers, so this proves the corollary over \(\mathbb C\).

Now let \(F\) be any characteristic-zero field. Choose homogeneous generators \(g_1,\ldots,g_r\) of \(I\), and let \(F_0\subseteq F\) be the field generated over \(\mathbb Q\) by all coefficients of the \(g_i\) and the \(f_i\). Set \[S_0=F_0[x_1,\ldots,x_n],\qquad I_0=(g_1,\ldots,g_r,f_1,\ldots,f_c)\subseteq S_0.\] Then \(I_0S_F=I\). The sequence \(f_1,\ldots,f_c\) is regular in \(S_0\): after tensoring each multiplication map with the faithfully flat extension \(F/F_0\), it is the corresponding injective map over \(S_F\). The finitely generated characteristic-zero field \(F_0\) admits an embedding into \(\mathbb C\). Extending by this embedding preserves regularity, so the case just proved applies to \[S_{\mathbb C}=\mathbb C\otimes_{F_0}S_0,\qquad I_{\mathbb C}=I_0S_{\mathbb C}.\] No embedding of \(F\) into \(\mathbb C\) is needed.

Let \(J_{\mathbb C}\) be the resulting lex-plus-powers ideal. It is a monomial ideal, so let \(J_0\subseteq S_0\) be generated by the same monomials and put \(J_F=J_0S_F\). For \(E=F\) or \(E=\mathbb C\), let \(S_E=E\otimes_{F_0}S_0\). Graded field base change preserves the dimensions of homogeneous pieces. It also gives, for \(M_0=S_0/I_0\) and for \(M_0=S_0/J_0\), \[E\otimes_{F_0}\mathop{\mathrm{Tor}}^{S_0}_p(M_0,F_0)_j \;\cong\; \mathop{\mathrm{Tor}}^{S_E}_p(E\otimes_{F_0}M_0,E)_j.\] For example, this follows by tensoring the Koszul complex of the variables with \(E\); its graded homology commutes with this flat extension. Consequently the Hilbert equalities and Betti inequalities over \(\mathbb C\) give the same equalities and inequalities over \(F\). Finally, the monomials of \(P_F\) and the lexicographic order are determined by their exponents, while the dimensions defining \(q_d\) are unchanged by these field extensions. Hence \(J_F\) is precisely the degreewise ideal \(J\) prescribed in the corollary. ◻