The Artinian Lex-Plus-Powers Betti Theorem

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 \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 (3.1), 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 \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

The complex K_\bullet(A) is the tensor product over F of the one-variable complexes

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 \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

The last isomorphism takes the coefficient at \alpha. Summing its dimensions proves (3.2); 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

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

Thus the differential preserves the filtration. By (3.4), its map on the corresponding associated graded positions is, up to the Koszul sign,

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

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 3.2 now gives (3.3). 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

is a graded vector-space isomorphism by (4.1). 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

The left S-action commutes with the right T-action by associativity of multiplication in N. Moreover, N=\Delta\otimes_\C S 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

Taking a complex basis of (S/I)_d proves (4.2).

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

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 (4.4). If f is nonzero and has degree e>0, then each fb_u belongs to N_e. By the graded decomposition (4.1), 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 3.1 proves (4.3) 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 2.1 to the given regular sequence. It supplies k, \Delta, and commuting degree-one forms t_h satisfying the basis property (4.1) and homogeneous identities

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 4.1.

Under (4.4), view N as the free T-module with coefficient space V=\Delta. In each total degree, order exponents by placing the lexicographically smallest tuple

first. This order is compatible with addition. Apply Proposition 3.3 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,

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

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

as graded right T-modules.

We next verify the power containment. Since every f_\ell lies in I, Equation (4.8) gives

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 \C to k, with x_i renamed y_i, is T/M_k. Its Hilbert function is unchanged. Tensoring the Koszul complex over \C with the field k also preserves exactness and commutes with homology, so

Equations (4.2) and (4.9), together with (3.2), now give

Likewise, (4.3), (3.3), and (4.9) yield, for every p\geq0 and every j\in\Z,

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

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

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 3.1. 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

The ratio is a polynomial: all unchanged monomial factors are nested, as are the products in (5.1).

For each free summand whose generator has multidegree v, define a linear change on its coefficient polynomials by

More explicitly, the right-hand side is

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 (5.3) 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

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

where the leading monomial is the lexicographically largest one. By Lemma 5.2, P\subseteq D(K), and hence P\subseteq K'. That lemma and Proposition 3.3 show that the Hilbert function is preserved and

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

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

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

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

Thus \pi_E(D(x^u)) is independent of the projected vectors already used for K\cap E, proving strictness in (5.5).

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

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

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 (6.2),

The shadow on the left and the slice on the right are lex segments in the same degree box. Consequently

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

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,

The remaining units therefore fit at k. If the inequality in (6.4) 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 (6.3) 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

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

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

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,

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 (6.6).

The same identity holds for the stable set Y. Its zero slice is a lex segment, and (6.5) gives Y(0)\subseteq\operatorname{Lex}'(X(0)). The induction hypothesis and monotonicity of shadows under inclusion imply

Combining this with the top-slice inequality in (6.6) 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 6.1 therefore gives

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 \Tor^R(B,\C),

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

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 6.2, 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 3.1:

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 (7.2). 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

For s\ge0, define coefficient ideals

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

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

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

The injective multiplication maps also give

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 [?, Lemma 3.7]. 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

There are no terms for s>a. The shifts record that multiplication by z^s raises ordinary degree by s. By (7.3), 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 \Tor, but their Koszul differentials have rank zero. Thus they contribute nothing to the deficits. Lemma 7.1 yields

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

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 6.1 imply

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 7.1 to the first inclusion in (7.9) and to the quotient map furnished by the second:

Together with (7.7) and (7.8), this gives

Adding and using (7.6), we obtain

The equal Hilbert functions of B_H and B_G imply

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 (7.4) proves the ideal comparison over A, and (7.2) completes the quotient comparison. Every step applies to every p\ge0 and j\in\Z; in particular, negative internal degrees and Koszul terms outside their homological range introduce no additional cases.□

Proof of Theorem 1.1. 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 4.2 gives a monomial ideal M\supseteq P with the Hilbert function of I and

Proposition 5.3 gives a box-stable monomial ideal H\supseteq P, with the same Hilbert function, and

By Proposition 6.2, the degreewise box-lex replacement G=\Lambda H is an ideal, with the same Hilbert function. Theorem 7.2 gives

It remains to check that G is the ideal prescribed in (1.1). 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

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_\C(L_d+P_d)=\dim_\C I_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 1.2. 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=\C. After a linear change of coordinates, one may arrange that

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=\C[x_1,\ldots,x_c]. They form a full regular sequence of degrees a_1,\ldots,a_c. Theorem 1.1, 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, Proposition 10] 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

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 \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 \Q by all coefficients of the g_i and the f_i. Set

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 \C. Extending by this embedding preserves regularity, so the case just proved applies to

No embedding of F into \C is needed.

Let J_{\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=\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,

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 \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.□