Rank formulas for homological shifts
of polymatroidal ideals
Caius Folkerts
Abstract. We give a partition formula for the canonical rank of every homological
shift ideal of a polymatroidal ideal, over any field. Applied to powers, it determines
the exact interval of Rees degrees needed to generate each homological shift module
and yields for , with a sharp bound in every positive homological degree. Mixed
powers and fixed polymatroidal twists obey the same bound because their ranks
enter additively. The formula arises from the matroid independence complexes in
multigraded Koszul homology; its inequalities also identify the rank of the total
Betti support and show that the shift modules are integrally closed ideals of the
corresponding Rees algebras.
1. Introduction
The homological shift ideals of a monomial ideal record the multidegrees occurring in its
minimal resolution. For polymatroidal ideals they are again polymatroidal, as proved by
Cid-Ruiz, Matherne, and Shapiro [3], so a rank function encodes each set of syzygy
degrees. We determine that function explicitly from the rank of the original ideal. For
powers, the formula describes exactly when new generators appear in the associated
Rees module and when multiplication by the ideal supplies every subsequent
piece.
In particular, it proves the stabilization bound conjectured by Ficarra and Lu [2]; the
th homological shift module of the powers is generated in Rees degrees at most . The
bound is sharp for every , holds over every field, and remains valid with a fixed
polymatroidal factor throughout.
Let , where is finite and is any field. For a monomial ideal , set
Write , , and . A polymatroidal ideal is generated in one degree, and its generator
exponents satisfy the discrete polymatroid exchange axiom. The zero and unit ideals have
no positive homological shifts.
Theorem 1.1.If is polymatroidal, then for every and every ,
Consequently is generated as a module over in Rees degrees at most . This bound
is sharp for every . More generally, if is any fixed polymatroidal ideal, then
The exact rank formula explains both the uniform bound and the earlier stabilization
that can occur for an individual ideal. Recall the rank description of a discrete
polymatroid [1, Propositions 1.2–1.4 and Theorem 3.4]. For a polymatroidal ideal of
degree , its canonical upper rank is
This function is nonnegative, integral, normalized, monotone, and submodular, with .
The generator exponents are exactly
Conversely, each such rank function defines a polymatroidal ideal, denoted
.
A coordinate with occurs in no generator. Removing it preserves the minimal
resolution up to polynomial extension. For the rank formula we therefore use the active
ground set, where
For a nonempty , let be its set partitions into nonempty blocks, and define
Theorem 1.2 (Exact homological rank formula).For every , is an integral polymatroid
rank function. Moreover,
In the nonzero case, is the canonical upper rank of the shift ideal.
The formula comes from viewing a Koszul basis vector as a permitted set
of coordinate deletions. In each multidegree these sets form a matroid, whose
top homology is nonzero precisely when it has no coloops. This condition gives
inequalities on the exponents, and uncrossing them produces the partition minimum
.
For , the rank is . Each positive partition defect is therefore a positive integer
multiplied by , and truncation at removes its effect once . Tracking the defects determines
the exact generation interval in Theorem 4.4. Thus the same formula gives the sharp
uniform threshold and the complete stabilization behavior of each homological shift
module. It also organizes all homological degrees into a single rank function by adding one
coordinate, and its inequalities, viewed as linear weights on the Rees algebra, show that
the shift modules are integrally closed. These consequences use the same partition data
throughout.
2. The Koszul matroid
In a fixed multidegree, a Koszul basis vector records a set of variables whose exponents are
lowered by one. We therefore ask which simultaneous deletions leave a monomial in the
ideal. The lower inequalities below express the available room for deletion on every
subset of variables. Their submodularity will make the permitted deletions a
matroid.
We first record two basic facts in the form used below. Products of polymatroidal ideals
correspond to sums of their ranks
Indeed, the integral polymatroid sum theorem states that the Minkowski sum has rank
and that every integer point in the sum decomposes into integer points of the two
summands [1, Proposition 1.4]. For a point of total degree , both summands must be
bases. In particular, .
Every also has a linear resolution, since decreasing reverse lexicographic order gives
linear quotients. To verify this, take an earlier generator and a later generator . At their
last differing coordinate , . The exchange axiom applied to gives with . The
coordinate occurs before , so is earlier than , and divides . Thus the order
has linear quotients. The exact sequences for adjoining one generator, together
with the Koszul resolutions of the variable-generated colon ideals, give a -linear
resolution.
Lemma 2.1 (Lower-inequality membership).Put and . For ,
Proof.Necessity follows from a base and . For sufficiency, choose an integer vector
satisfying all and maximal under coordinatewise increase; such a vector exists
because is feasible and the box is finite. If , maximality gives a set with . The union
of upper-tight sets is upper-tight, by submodularity of . Let be the union of these ,
or if there are none. Then and . This is a base whose monomial divides , because
□
The constraint function in our application need not be monotone; adding a variable
increases its exponent sum, but can also increase the required lower bound. The following
form of the matroid construction keeps precisely the hypothesis that survives, namely
submodularity.
Lemma 2.2 (A submodular constraint matroid).Let be submodular with , and let
. The family
is the independence family of a matroid on . Its rank on is
Monotonicity of is not required.
Proof.The family is nonempty and hereditary. Choose an inclusion-maximal feasible
. For every , failure to add gives a set with . Tight sets for are closed under unions;
the left side is modular and the right side is submodular. Their union is tight and
contains . Therefore
If , use . For every feasible and every , . Thus every maximal feasible subset of
has the cardinality in (2.3). The hereditary-family characterization of a matroid now
applies; if independent sets with admitted no augmentation from , then would be
a maximal independent subset of of smaller than maximum size. □
The rank formula for the unit ideal is immediate, so assume . Fix and with
. If , the degree- part of the Koszul complex of is zero. Otherwise put and
In homological degree , the degree- Koszul basis consists of for the faces of size .
Removing an exterior variable is exactly the oriented simplicial boundary of that face.
The empty face corresponds to Koszul degree zero, so the degree shift is . Thus
This is a nonnegative integer-valued submodular function, because is supermodular.
Also and . Consequently is a matroid independence complex of rank at most . Here is
the number of deletions allowed inside ; the total-degree condition leaves room for at most
deletions in all. This is why only the top homology of the resulting matroid can contribute
to .
3. The homological rank
Lemma 3.1 (Homology of a matroid independence complex).Let be a matroid of
rank . Over every field, its independence complex has reduced homology only in degree
. Its homology in degree is nonzero if and only if has no coloops. Rank zero uses
the convention .
Proof.Induct on the ground-set size. Loops can be deleted without changing the
complex. A coloop makes the complex a cone, so all reduced homology vanishes. If
the rank is zero, the assertion is the stated convention. In the remaining case choose
a nonloop , necessarily a noncoloop. Deletion has rank , contraction has rank , and
The reduced Mayer–Vietoris sequence and induction give concentration in degree
and a short exact sequence
Contraction does not create coloops; for ,
Thus has no coloops when has none, and the last term is nonzero by induction.
The same argument includes using reduced degree . □
For the matroid , Lemma 3.1 says that (2.5) is nonzero exactly when its rank is and it
has no coloops. Since , this is equivalent to
Since the matroid rank is already at most , the minimum in (2.3) equals
precisely when every term is at least . For a set meeting , the strongest condition
comes from choosing , which contributes the final . Hence (3.1) is equivalent to
Put , so that records the exponents remaining after removing the first copy of every
variable in the support. Since for every nonempty , the sets disjoint from satisfy the same
inequality automatically. Thus (3.2) becomes
The bounds need not themselves form a rank function; the submodular inequality can
fail for disjoint sets. But is additive across the blocks of every partition, so all partition
sums are forced bounds as well. Taking their minimum gives the equivalent system
Indeed, (3.3) implies the bound for every partition, and the one-block partition gives for the converse.
Lemma 3.2 (The partition rank).is a nonnegative, integer-valued, normalized,
monotone, submodular function. Moreover, for ,
Proof.Let for nonempty . It is nonnegative and monotone and satisfies the
submodular inequality whenever the intersection is nonempty. Restrict an optimal
partition of to , omitting empty intersections. Its cost cannot increase, proving .
For submodularity, take optimal partitions of and , as a multiset of blocks. Replace
intersecting incomparable blocks by . This cannot increase total cost; it preserves the
multiplicity with which each element is covered; and it strictly increases the integer .
The latter is bounded, so uncrossing terminates in a laminar multiset. Every element
has multiplicity at most two. Consequently its outer blocks partition and its inner
blocks partition . If a block occurs twice, one copy is outer and one is inner. Hence
the final cost is at least , proving submodularity.
For (3.5), let be an optimal partition of and let be the number of its blocks disjoint
from . Restricting the other blocks to gives
The middle inequality holds because each of the disjoint blocks contains an element of .
The last inequality follows by adding the disjoint blocks of successively to ;
submodularity bounds each marginal increase by , and the increases telescope.
□
One direction follows from . For the other, apply (3.6) to to obtain
Finally, at total degree , Lemma 2.1 says that membership in is equivalent to
Combining these facts gives the following field-independent description, with the
inequalities required for every
Here linearity of the resolution has excluded all other total degrees.
3.1. A rank interpolation.
The inequality (3.5) is what makes the homological bounds into rank functions. The
first interpolation below gives each shift separately, while the second places all shifts in one
polymatroid by adding a coordinate for homological degree. Although the minimum of two
submodular functions is not usually submodular, (3.5) orders their marginal
increments and thus controls the mixed case in which the two sets select different
branches.
Lemma 3.3 (Rank interpolation).Let be integral polymatroid ranks on such that
For , the function
is a polymatroid rank. If , then so is the function on defined by
Proof.The hypothesis gives . Normalization, integrality, nonnegativity, and
monotonicity follow, including monotonicity of between sets of its two forms. For
submodularity of , if both values select the same branch, use submodularity of that
branch. The function taking value on nonempty sets and zero at the empty set is
submodular as well. In a mixed case the required inequality follows from
The same inequality, with and interchanged, proves the mixed case of
submodularity for . Its two unmixed cases use the submodularity of and . □
For completeness, an integral polymatroid rank is always attained on every subset by
one of its integer bases. Choose a coordinate order placing that subset first, let be its
successive initial sets, and take the greedy vector . Monotonicity makes it nonnegative and
integral, and submodularity shows for every by summing diminishing marginal
increments. It has and attains on each initial set. This proves the asserted canonicality
without any additional compatibility hypothesis.
Now . If the inequality is strict, (3.8) has no solutions. If equality holds, its solutions
are exactly all integer bases of ; these form the generator set of . This proves
Theorem 1.2.
Lemma 3.4 (Multiplication of shifts).For any two polymatroidal ideals and every
,
Proof.The zero and unit cases are immediate. Otherwise take a generator
multidegree of and a base , and work on the union of the active ground sets. Put
and . If a nonempty set meets the active ground set of , then (3.3), applied on that
ground set, gives
If misses it, then . When , ; otherwise , since is positive on every nonempty
subset of its active ground set. Thus satisfies (3.3) for . The vector has the required
total degree and its monomial belongs to , so it is a shift multidegree of the product.
□
4. Powers and partition defects
For a partition of a nonempty set , write
Both are nonnegative integers; the first measures failure of rank additivity across
the blocks, while the second depends only on the partition. In the homological
rank they become the slope and intercept of an affine function of the power,
so stabilization occurs when every positive-slope term reaches the homological
cutoff.
Let and , and use the union of their active coordinates. For , the rank is positive on
every nonempty subset. Theorem 1.2 gives
If , every term with is at least . The remaining minimum is independent of , so
The total-rank criterion in Theorem 1.2 makes the two consecutive shifts
simultaneously zero or nonzero. In the latter case, the polymatroid sum identity (2.1) gives
This proves both equalities in Theorem 1.1, the ordinary case being .
Lemma 3.4 makes the displayed direct sums Rees modules. Iterating the equality from
degree proves their generation bound.
For ordinary powers, delete the inactive coordinates of and put
Since the one-block partition has zero defect, this maximum is always defined.
Specializing (4.1) to gives
In particular,
Corollary 4.1 (Several colors).Let be polymatroidal ideals. If , then
Thus the corresponding multigraded shift module is generated in the box , with the
zero-degree pieces interpreted in the usual way.
Proof.Apply the mixed-power theorem with the fixed polymatroidal product , and
repeat while a coordinate exponent exceeds . □
4.1. Sharpness and projective dimension.
Proposition 4.2.For each , let and let be the squarefree Veronese ideal of degree
in . Then
Therefore no universal generation bound smaller than is possible.
Proof.The rank is . On every proper subset , all partitions have zero defect and the
singleton partition has intercept zero. Formula (4.3) therefore gives .
On , the one-block partition has defect zero and intercept . Every other partition
has defect one, with least intercept zero at the singleton partition. Hence
The total-rank criterion makes the shift zero exactly when . At , its singleton
ranks are and its total rank is , so its only base is . This proves both assertions,
including the case . □
Corollary 4.3 (Exact projective dimensions).For an active nonzero proper
polymatroidal ideal of rank ,
In particular the eventual value is , and is reached by .
Proof.Theorem 1.2 says that exactly when . Substituting the partition expression
gives (4.5). Every positive-defect term is at least , while the least zero-defect term is
. This proves the stabilization assertion. □
4.2. The exact interval of generator degrees.
The same partition formula determines the precise generation degree for each individual
ideal. Continue to assume that is nonzero and proper, and that is its active ground set.
Fix . For nonempty put
For a given partition with positive defect, the inequality gives the first power at which
that partition can no longer lower the final rank. The last partition to reach this value
determines the threshold. Accordingly, define
In particular . Although arose from powers with , its expression in (4.3) is also defined
at , where it is identically zero.
Theorem 4.4 (Exact generation interval).If , the entire th shift module is zero.
Otherwise set
Then . The first nonzero homological shift is . The Rees degrees occurring among the
minimal monomial generators of are exactly
Thus is the exact least generation bound for this ideal and homological degree.
Proof.The least zero-defect contribution to (4.3) is . Consequently
It equals its final value precisely when . Moreover, if , every positive-defect term
increases by at least one between and , while all terms are integral and the constant term
is strictly larger than . Hence
The total-rank criterion gives exactly when . If the final value is less than , all
pieces vanish. Otherwise the first nonzero degree is , which is positive because
.
For , both consecutive shifts are nonzero, so their displayed rank functions
are canonical. The product has canonical rank , whereas has canonical rank .
Equations (4.8) and (4.10) show that these ideals are equal exactly when . At
every , choose a set with . Its rank increases strictly, and a base attaining the
new rank on lies outside the preceding product. Such a base requires a new
module generator in degree . It cannot come from a still earlier piece, since the
module action puts every product from an earlier piece in . All generators in the
first nonzero piece are also required. This proves the exact interval assertion.
□
The proof also specifies the new minimal monomial generators themselves. All
bases of occur. For , a base of gives a new module generator exactly when
Indeed, the right-hand bounds are the exact rank description of . Since all candidate
minimal generators in this Rees degree have ordinary degree , no further divisibility among
them can remove any of these new generators.
For example, take and . Its rank is
The singleton partition gives , hence and . On , the partition formula gives , so . Since
every threshold is at most , the exact generation interval is . Concretely, , while lies
outside .
5. The total Betti support
The ranks of the individual shifts fit into one polymatroid when an extra coordinate
records homological degree. The homogenized Betti support was shown to be a
polymatroid in [3]; the interpolation in Lemma 3.3 identifies its rank explicitly and gives
a direct proof.
Theorem 5.1 (Homogenized total Betti support).Let be a nonzero proper
polymatroidal ideal on its active ground set , let , and put
In the polynomial ring , define
Then is a polymatroidal ideal generated in degree . Its canonical rank on is
Equivalently, its bases are exactly
This description is independent of the field.
Proof.The projective-dimension equality follows from Theorem 1.2. Lemma 3.3,
applied to , shows that is a rank function.
An integer base of satisfies and . Set . Its inequalities are exactly
By (3.8), these are equivalent to . Conversely every such Betti multidegree
satisfies every base inequality of . This proves (5.3), and hence (5.1) and its asserted
polymatroidality. The greedy attainment argument in Section 3 proves canonicality
of the displayed rank. □
When is principal, and is a loop of this polymatroid. Variables inactive in can
likewise be reattached as loops. The unit ideal gives the rank-zero singleton base, with the
same interpretation. Thus no extra positivity assumption on individual generator
coordinates is involved.
6. Integral closure inside the Rees algebra
For fixed , Lemma 3.4 realizes the shift module as an ideal of the Rees algebra, since
every shift multidegree belongs to the corresponding power of . The partition
inequalities describe it as an intersection of valuation ideals; each bounds a linear
weight nonnegative on the Rees semigroup. This description will prove integral
closure.
Theorem 6.1.For a polymatroidal ideal , the monomial ideal
is integrally closed in for every and every field.
Proof.Separate the zero and unit ideals, whose positive shift ideals vanish. Otherwise let
be the rank on the active ground set , and let . Write a Rees monomial as . For a
nonempty and a partition of , set
We claim that membership in is equivalent to membership in together with
and, for every nonempty and every ,
To see this, put . The lower-inequality membership criterion of Lemma 2.1, applied to
the shift rank, says
If , its inequality is impossible, exactly as required for a zero shift ideal. Otherwise it is
the ordinary membership criterion for . For , expanding the minimum defining yields the
Rees-membership inequalities and all of (6.3). The case gives the second inequality of
(6.2). This proves the claim.
Each weight in (6.1)–(6.3) is an integer linear function nonnegative on the Rees
exponent semigroup. For , nonnegativity on a polynomial variable is immediate. On a
generator of Rees degree one, meaning with , it follows from
The other two weights are the Rees degree and the excess ordinary degree.
For any one of these linear weights , assign to a nonzero polynomial in the
Rees algebra the minimum -weight of its monomials. This extends to a valuation
on its fraction field. In fact, initial forms of nonzero polynomials multiply to
a nonzero initial form in the Laurent polynomial domain, so the minimum is
additive under multiplication. The ideal of elements of valuation at least a fixed
integer is integrally closed; if an element of smaller valuation satisfied an integral
equation with coefficients in successive powers of that ideal, the leading power of
the element would have strictly smaller valuation than every other term, an
impossibility.
By the membership claim, is the intersection of precisely these valuation ideals, with
their displayed thresholds. An intersection of integrally closed ideals is integrally closed,
proving the theorem. Inactive variables have weight zero and do not affect this argument.
□
Corollary 6.2 (Twisted multi-Rees ideals).Let be nonzero polymatroidal ideals,
with the unit ideal allowed, and let
For every , the ideal
is integrally closed. This includes the multidegrees with some . A zero twist gives
the zero ideal, with the same conclusion.
Proof.Let be the ranks on their common ground set, and put and . Formulas
(1.3)–(1.4) remain valid if inactive coordinates are retained; an inactive coordinate
is optimally split off as a singleton of partition cost , so
where is the active set. Thus and are simply extended by rank-zero loops. The
same formulas include .
For , the membership inequalities for are
The second inequality ranges over , and the third over nonempty and . They follow by
expanding (6.4) for , exactly as in the preceding theorem. Every linear weight on the left is
nonnegative on the exponent semigroup of ; on a generator from the th factor this is the
base inequality or the nonnegative partition defect for , and it is immediate on the
polynomial variables. Hence the same intersection-of-valuations proof applies. The
multiplication inclusion of Lemma 3.4, and the containment , justify its displayed
realization as an ideal of . □
References
[1]
J. Herzog and T. Hibi, Discrete polymatroids, Journal of Algebraic Combinatorics 16 (2002),
239–268. arXiv math/0307236.
[2]
A. Ficarra and D. Lu, Polymatroidal ideals and their asymptotic syzygies, arXiv 2509.11977.
[3]
Y. Cid-Ruiz, J. P. Matherne, and A. Shapiro, Syzygies of polymatroidal ideals, arXiv
2507.13153v2, 25 September 2026.