# Polynomial division via initial terms – Part I (Homogenization)

$$\DeclareMathOperator{\In}{In} \DeclareMathOperator{\ld}{Ld} \DeclareMathOperator{\kk}{\mathbb{K}} \DeclareMathOperator{\ord}{ord} \DeclareMathOperator{\rnpos}{\mathbb{R}^n_{> 0}} \DeclareMathOperator{\rnzero}{\mathbb{R}^n_{\geq 0}} \DeclareMathOperator{\rr}{\mathbb{R}} \DeclareMathOperator{\scrB}{\mathcal{B}} \DeclareMathOperator{\scrI}{\mathcal{I}} \DeclareMathOperator{\scrJ}{\mathcal{J}} \DeclareMathOperator{\supp}{Supp} \DeclareMathOperator{\znplusonezero}{\mathbb{Z}^{n+1}_{\geq 0}} \DeclareMathOperator{\znplusonepos}{\mathbb{Z}^{n+1}_{> 0}} \DeclareMathOperator{\znpos}{\mathbb{Z}^n_{> 0}} \DeclareMathOperator{\znzero}{\mathbb{Z}^n_{\geq 0}} \DeclareMathOperator{\zz}{\mathbb{Z}}$$This is a continuation of an earlier post on polynomial division where we looked at division of polynomials via cancelling the leading term with respect to a monomial order. Here we look at “initial terms” of polynomials with respect to monomial orders, and show that, similar to the case of leading ideals, there can be only finitely many “initial ideals” for a given ideal of a polynomial ring. The problem with division via cancelling the initial term, as opposed to the cancellation of the leading term in (Step 1 of) the division algorithm, is that the initial exponent can indefinitely increase (whereas the termination of the original division algorithm is guaranteed by the fact that the leading exponent can not indefinitely decrease). This is the reason the initial-term-cancelling-division works for power series, as we have seen in the preceding post, but not for polynomials. There are (at least) two natural ways to proceed:

1. To treat the polynomials as power series, and apply the power series division algorithm even when the divisors and the dividend are polynomials. In this case the quotient and the remainder may not be polynomials, but it can be shown that the initial ideal of a polynomial ideal $$I$$ agrees with the initial ideal of the ideal of the ring of power series generated by $$I$$, and from this it follows that every ideal can have at most finitely many initial ideals, and every initial ideal comes from a weight.
2. To homogenize the original ideal with respect to a new variable, and consider initial-term-cancelling-division only by homogeneous polynomials. In this case for each homogeneous term, say of degree $$d$$, of the dividend, the corresponding term of the remainder can not escape the finite dimensional vector space of homogeneous polynomials of degree $$d$$, which ensures that the algorithm terminates in finitely many steps.

The first one will be pursued in a subsequent post. Here we take the second approach. Its shortcoming of the second approach is that, at least in the form described above, it applies only to homogeneous ideals. However, this division algorithm actually works whenever the divisors are homogeneous with respect to any grading on the polynomial ring which is “compatible” with monomials and has finite dimensional graded components. Moreover, the division can be adapted to any linear, but not necessarily well, order on $$\znpos$$ which is compatible with the addition on $$\znpos.$$

## Shallow monomial gradings on the polynomial Ring

A grading on a ring $$R$$ by a group $$\Sigma$$ (or in short, a $$\Sigma$$-grading) is a family $$(R_\sigma)_{\sigma \in \Sigma}$$ of additive subgroups of $$R$$ indexed by $$\Sigma$$ such that

• $$R = \oplus_{\sigma \in \Sigma} R_\sigma,$$ and
• for each $$\sigma, \tau \in \Sigma,$$ if $$f \in R_\sigma$$ and $$g \in R_\tau,$$ then $$fg \in R_{\sigma\tau}$$

We will work with rings containing a field $$\kk$$, and require our gradings to satisfy the additional property that

• each $$R_\sigma$$ is a vector space over $$\kk$$, or equivalently, $$\kk \subseteq R_\epsilon,$$ where $$\epsilon$$ is the unit of $$\Sigma.$$

The elements of $$R_\sigma,$$ $$\sigma \in \Sigma,$$ are called homogeneous of degree $$\sigma,$$ and given $$f \in R,$$ its homogeneous components are homogeneous elements $$f_\sigma$$ such that $f = \sum_{\sigma \in \Sigma} f_\sigma$We write $$\supp_\Sigma(f)$$ for the set of all elements in $$\Sigma$$ that “support” $$f,$$ i.e. $\supp_\Sigma(f) := \{\sigma \in \Sigma: f_\sigma \neq 0\}$A homogeneous ideal of $$R$$ is an ideal generated by homogeneous elements; equivalently, an ideal is homogeneous if and only if it contains all homogeneous components of each of its elements.

A grading on $$S := \kk[x_1, \ldots, x_n]$$ is monomial compatible, or simply a monomial grading, if every monomial in $$(x_1, \ldots, x_n)$$ is homogeneous, and it is shallow if for all $$\sigma$$ the dimension of $$S_\sigma$$ (as a vector space over $$\kk$$) is finite. Given any $$n$$-tuple of weights $$(w_1, \ldots, w_n)\in \rr^n,$$ the corresponding weighted degree defines a monomial compatible $$\rr$$-grading on $$S$$ which is shallow if every $$w_i$$ is positive. More generally, any linear map $$\pi: \rr^n \to \rr^m$$ induces a monomial compatible $$\rr^m$$-grading on $$S$$ which is shallow if and only if the restriction of $$\pi$$ to $$\zz^n$$ is injective. In particular, every monomial order on $$S$$ induces a shallow monomial compatible $$\zz^n$$-grading on $$S$$ such that $$\dim_\kk(S_\sigma) = 1$$ for each $$\sigma \in \zz^n.$$

Remark 1. If a grading on $$S := \kk[x_1, \ldots, x_n]$$ by a group $$\Sigma$$ is shallow, then the graded component of $$S$$ corresponding to the unit in $$\Sigma$$ must consist only of $$\kk.$$

## Initial-term-cancelling division by homogeneous elements with respect to shallow monomial gradings

Let $$\preceq$$ be a group order (i.e. an order which is compatible with addition) on $$\znpos$$ which is linear (but not necessarily well-) order. The notions of initial exponent/term/ideal with respect to $$\preceq$$ still makes sense for polynomials. The following observation, made in the previous post for monomial orders, also holds for linear orders with the same proof:

Lemma 1. If $$f_1, \ldots, f_N$$ are polynomials with mutually distinct values of $$\nu_\preceq$$, then the $$f_i$$ are linearly independent over $$\kk$$.

Fix an arbitrary shallow monomial compatible $$\Sigma$$-grading on $$S := \kk[x_1, \ldots, x_n].$$ The initial-term-cancelling-division algorithm in $$S$$ is the procedure for reducing a polynomial $$f$$ by a finite ordered collection of homogeneous (with respect to $$\Sigma$$) polynomials $$g_1, \ldots, g_N$$ as follows:

1. Set $$h := f$$
2. If there is $$i$$ such that $$\In_\preceq(g_i)$$ divides $$\In_\preceq(h)$$, then pick the smallest such $$i$$, and replace $$h$$ by $$h – qg_i$$, where $$q := \In_\preceq(h)/\In_\preceq(g_i)$$.
3. Otherwise $$\In_\preceq(h)$$ is not divisible by $$\In_\preceq(g_i)$$ for any $$i$$; in this case replace $$h$$ by $$h – \In_\preceq(h).$$
4. Repeat steps 2-3 until $$h$$ reduces to the zero polynomial.

Note that the grading and the linear order are not related – they are chosen completely independently of one another. Nevertheless, the division algorithm terminates:

Theorem 1. The above algorithm (of initial-term-cancelling-division by polynomials which are homogeneous with respect to shallow monomial gradings) stops in finitely many steps, and after the final step one has: $f = \sum_i q_ig_i + r$where

• either $$r = 0$$ or no monomial term in $$r$$ is divisible by $$\In_\preceq(g_i)$$ for any $$i$$, and
• $$\In_\preceq(\sum_i q_ig_i) \in \langle \In_\preceq(g_i): i = 1, \ldots, N \rangle$$; more precisely, there is a reordering $$i_1, \ldots, i_s$$ of $$\{i: q_i \neq 0\}$$ such that $\In_\preceq(\sum_j q_jg_j) = \In_\preceq(q_{i_1}g_{i_1}) \prec \In_\preceq(q_{i_2}g_{i_2}) \prec \cdots \prec \In_\preceq(q_{i_s}g_{i_s})$where all the $$\prec$$ orders on the right hand side are strict.

If $$g_1, \ldots, g_N$$ are elements of an ideal $$I$$ of $$\kk[x_1, \ldots, x_n]$$ such that $$\In_\preceq(g_1), \ldots, \In_\preceq(g_N)$$ generate $$\In_\preceq(I)$$, then the following are equivalent:

1. $$f \in I$$,
2. $$r$$ is the zero polynomial.

Proof. Let $$\sigma_1, \ldots, \sigma_s$$ be the elements of $$\supp_\Sigma(f).$$ Since the grading is monomial compatible, $$\In_\preceq(f) \in S_{\sigma_j}$$ for some $$j,$$ and if $$\In_\preceq(g_i)$$ divides $$\In_\preceq(f),$$ then since $$g_i$$ is homogeneous with respect to $$\Sigma,$$ both $$qg_i$$ and $$f – qg_i$$ are also in $$S_{\sigma_j},$$ where $$q := \In_\preceq(f)/\In_\preceq(g_i).$$ It follows that after every repetition of steps 2 and 3,

• each nonzero homogeneous component of $$h$$ has degree $$\sigma_k$$ for some $$k = 1, \ldots, s,$$ i.e. $\supp_\Sigma(h) \subseteq \supp_\Sigma(f)$
• there is at least one $$j$$ such that $$\nu_\preceq(h_{\sigma_j})$$ increases from the previous iteration.

Since the grading is shallow, each $$S_{\sigma_j}$$ is finite dimensional over $$\kk$$, and therefore Lemma 1 implies that the algorithm stops in finitely many steps. The other assertions of Theorem 1 are immediate.

A totally ordered grading on $$S := \kk[x_1, \ldots, x_n]$$ is a grading by a totally ordered group $$\Omega$$ (which in particular means that the ordering is compatible with the group law, i.e. if $$a \preceq b$$ then $$ac \preceq bc$$ and $$ca \preceq cb$$ for all $$c \in \Omega$$). If a grading is totally ordered, then every polynomial has a well defined initial/leading form which is simply the homogeneous component of respectively the lowest/highest degree. For an ideal of $$S,$$ its initial/leading ideal is the ideal generated by respectively the initial/leading forms of its elements. Every total order $$\preceq$$ on $$\znpos$$ induces a “refinement” of the grading by $$\Omega$$ in the following way: define $$\alpha \preceq’ \beta$$ if and only if

• either $$\ord_\Omega(x^\alpha)$$ is strictly smaller than $$\ord_\Omega(x^\beta)$$ with respect to the ordering on $$\Omega$$ (where $$\ord_\Omega(f)$$ is the smallest element (with respect to the ordering on $$\Omega$$) of $$\supp_\Omega(f)$$), or
• $$\ord_\Omega(x^\alpha) = \ord_\Omega(x^\beta)$$ and $$\alpha \preceq \beta$$

It is straightforward to check that $$\preceq’$$ is a total group order on $$\znpos.$$ It is a “weak refinement” of the grading by $$\Omega$$ in the sense that if $$\ord_\Omega(x^\alpha)$$ is less than $$\ord_\Omega(x^\beta)$$ with respect to the ordering on $$\Omega$$ then $$\alpha \prec’ \beta.$$ Note that if the grading by $$\Omega$$ is not monomial compatible, then there may exist polynomials $$f, g$$ such that $$\ord_\Omega(f)$$ is less than or equal to $$\ord_\Omega(g)$$ but $$\In_{\preceq’}(f)$$ is greater than $$\In_{\preceq’}(g)$$ with respect to $$\preceq’.$$

Theorem 1*. Let $$I$$ be an ideal of $$S$$ which is homogeneous with respect to a shallow monomial grading on $$S$$ by a group $$\Sigma.$$ Given a monomial grading of $$S$$ by a totally ordered group $$\Omega$$ and a linear order $$\preceq$$ on $$\znpos,$$ let $$\preceq’$$ be the above refinement of the $$\Omega$$-grading by $$\preceq.$$ If $$g_1, \ldots, g_N$$ are $$\Sigma$$-homogeneous elements of $$I$$ such that $$\In_{\preceq’}(g_1), \ldots, \In_{\preceq’}(g_N)$$ generate $$\In_{\preceq’}(I),$$ then $$\In_{\Omega}(g_1), \ldots, \In_{\Omega}(g_N)$$ generate $$\In_{\Omega}(I).$$

Proof. Pick $$f \in I.$$ Since $$\In_{\preceq’}(g_1), \ldots, \In_{\preceq’}(g_N)$$ generate $$\In_{\preceq’}(I),$$ there is $$i_1$$ such that $$\In_{\preceq’}(g_{i_1})$$ divides $$\In_{\preceq’}(f),$$ i.e. there is $$a_{i_1} \in \kk$$ and $$\alpha_{i_1} \in \znzero$$ such that $\In_{\preceq’}(f) = a_{i_1}x^{\alpha_{i_1}}\In_{\preceq’}(g_{i_1})$
Let
$h_1 := a_{i_1}x^{\alpha_{i_1}}g_{i_1},\quad f_1 := f – h_1$
Since the $$\Omega$$-grading is monomial compatible, it follows that $\ord_\Omega(h_1) = \ord_\Omega(x^{\alpha_{i_1}}) + \ord_\Omega(g_{i_1}) = \ord_\Omega(x^{\alpha_{i_1}}) + \ord_\Omega(\In_{\preceq’}(g_{i_1})) = \ord_\Omega(\In_{\preceq’}(f)) = \ord_\Omega(f)$
Note that

• The monomial compatibility of the $$\Omega$$-grading was used in the second and the last of the above equalities.
• we used “$$+$$” to denote the group operation on $$\Omega,$$ since $$S$$ is commutative and therefore every grading on $$S$$ can also be induced by a grading by an abelian group.

It follows from the above display that
$\ord_\Omega(f) \preceq_\Omega \ord_\Omega(f_1)$
where “$$\preceq_\Omega$$” is the ordering on $$\Omega.$$ Now, since the $$\Sigma$$-grading is monomial compatible, and since $$g_{i_1}$$ is $$\Sigma$$-homogeneous, it follows that
$\supp_\Sigma(f_1) \subseteq \supp_\Sigma(f)$
The shallowness of the $$\Sigma$$-grading then implies that if we keep repeating this process to construct $$f_2$$ from $$f_1$$ and $$f_3$$ from $$f_2$$ and so on, then from a step onward $$f_0 := f, f_1, f_2, \ldots,$$ would be linearly dependent over $$\kk.$$ On the other hand, since
$\nu_{\preceq’}(f_0) \prec’ \nu_{\preceq’}(f_1) \prec’ \cdots,$
Lemma 1 implies that as long as $$f_k$$ is nonzero, it would be linearly independent of $$f_0, \ldots, f_{k-1}.$$ Consequently, the process must stop, and there is $$k$$ such that $$f_k = 0.$$ In particular, there is $$m \geq 0$$ such that $\ord_\Omega(f_0) = \cdots = \ord_\Omega(f_m) \prec_\Omega \ord_\Omega(f_{m+1})$where $$\prec_\Omega$$ denotes “strictly smaller” with respect to $$\preceq_\Omega.$$ This means $\ord_\Omega(f) = \ord_\Omega(h_1) = \cdots = \ord_\Omega(h_{m+1})$and$\ord_\Omega(f – h_1 – \cdots – h_{m+1}) \succ_\Omega \ord_\Omega(f)$It follows that $\In_\Omega(f) = \In_\Omega(h_1 + \cdots + h_{m+1}) = \In_\Omega(h_1) + \cdots + \In_\Omega(h_{m+1})$ Since $$\In_\Omega(h_1) + \cdots + \In_\Omega(h_{m+1})$$ is in the ideal generated by $$\In_\Omega(g_{i_1}), \ldots, \In_\Omega(g_{i_{m+1}})$$, we are done!

We also have analogues of the corollaries to Theorem 1 from the preceding post. Corollary 1 below fails if we allow $$J$$ to be non-homogeneous, as evidenced by taking $$I := \langle x \rangle$$ and $$J := \langle x + x^2 \rangle$$. I suspect Corollary 2 remains true for non-homogeneous ideals, but need to think harder.

Corollary 1. Let $$I \supseteq J$$ be ideals of $$S := \kk[x_1, \ldots, x_n]$$ such that $$J$$ is homogeneous with respect to a shallow monomial grading on $$S.$$ If $$I \supsetneqq J$$, then $$\In_\preceq(I) \supsetneqq \In_\preceq(J)$$ for every linear group order $$\preceq$$ on $$\znpos.$$

Proof. Pick homogeneous $$g_1, \ldots, g_N \in J$$ such that $$\In_\preceq(g_i)$$, $$i = 1, \ldots, N$$, generate $$\In_\preceq(I)$$. Then Theorem 1 implies that $$g_1, \ldots, g_N$$ generate $$I$$, so that $$I = J$$.

Corollary 2. Let $$I$$ be an ideal of $$S := \kk[x_1, \ldots, x_n]$$ which is homogeneous with respect to a shallow monomial grading on $$S.$$ There can not exist linear group orders $$\preceq_1, \preceq_2$$ on $$\znpos$$ such that $$\In_{\preceq_1}(I) \subsetneqq \In_{\preceq_2}(I)$$.

Proof. Write $$J_k := \In_{\preceq_k}(I)$$, $$k = 1, 2$$. Assume to the contrary that $$J_1 \subsetneqq J_2$$. Then we can choose homogeneous $$g_1, \ldots, g_N \in I$$ such that $$\In_{\preceq_2}(g_i)$$, $$i = 1, \ldots, N$$, generate $$J_1$$. Choose $$f \in I$$ such that $$\In_{\preceq_2}(f) \not\in J_1.$$ Let $$r$$ be the remainder of the initial-term-cancelling-division of $$f$$ by $$g_1, \ldots, g_N$$ with respect to $$\preceq_2$$. Then $$r$$ is a nonzero element of $$I$$ such that $$\supp(r) \cap \supp(J_1) = \emptyset$$. On the other hand, if $$h_1, \ldots, h_M$$ are homogeneous elements in $$I$$ such that $$\In_{\preceq_1}(h_1), \ldots, \In_{\preceq_1}(h_M)$$ generate $$J_1,$$ then initial-term-cancelling-division of $$r$$ by $$h_1, \ldots, h_M$$ with respect to $$\preceq_1$$ implies that $$r \not\in I,$$ which is a contradiction. This completes the proof.

The notion of reduced initial basis described in the preceding post on division of power series also applies to an arbitrary linear group order on $$\znpos$$: a finite ordered collection of elements $$g_1, \ldots, g_N$$ from an ideal $$I$$ of $$S$$ is called a reduced initial basis with respect to a linear group order $$\preceq$$ if the following hold:

1. the initial terms of the $$g_i$$ generate $$\In_\preceq(I),$$
2. no monomial term of any $$g_i$$ is divisible by the initial term of any $$g_j$$ for $$j \neq i$$, and
3. the initial coefficient, i.e. the coefficient of the initial term, of each $$g_i$$ is $$1.$$

If $$I$$ is homogeneous with respect to a shallow monomial $$\Sigma$$-grading on $$S$$, then a reduced initial basis for $$I$$ can be produced by taking a minimal set of $$\Sigma$$-homogeneous elements of $$I$$ whose initial terms generate $$\In_\preceq(I)$$ and then replacing each element by the remainder of the initial-term-cancelling-division by other elements.

Corollary 3. Let $$I$$ be an ideal of $$S := \kk[x_1, \ldots, x_n]$$ which is homogeneous with respect to a shallow monomial $$\Sigma$$-grading on $$S$$, and $$\preceq, \preceq’$$ be linear group orders on $$\znpos$$ such that
$\In_\preceq(I) = \In_{\preceq’}(I)$
Then any reduced $$\Sigma$$-homogeneous initial basis of $$I$$ with respect to one of $$\preceq, \preceq’$$ is also a reduced initial basis with respect to the other one.

Proof. Indeed, pick a reduced initial basis $$g_1, \ldots, g_N$$ of $$I$$ with respect to $$\preceq$$. Let $$\alpha_i := \nu_{\preceq}(g_i),$$ $$i = 1, \ldots, n.$$ Fix $$i.$$ Since $$\In_{\preceq’}(I)$$ is also generated by $$x^{\alpha_1}, \ldots, x^{\alpha_N}$$, it follows that $$\In_{\preceq’}(g_i)$$ is divisible by some $$x^{\alpha_j}.$$ Since no monomial in the support of $$g_i$$ is divisible by $$x^{\alpha_j}$$ for $$j \neq i$$ it must be divisible by $$x^{\alpha_i}.$$ Since $$\alpha_i$$ is also in the support of $$g_i,$$ and $$g_i$$ is $$\Sigma$$-homogeneous and the grading by $$\Sigma$$ is shallow, Remark 1 above implies that the only monomial in $$g_i$$ divisible by $$x^{\alpha_i}$$ is $$x^{\alpha_i}$$ itself, so that $$\nu_{\preceq’}(g_i) = \alpha_i.$$ It follows that that $$\In_{\preceq’}(g_i),$$ $$i = 1, \ldots, N,$$ form a reduced initial basis of $$I$$ with respect to $$\preceq’$$ as well.

## Universal Initial Bases

Theorem 2. Let $$I$$ be an ideal of $$S := \kk[x_1, \ldots, x_n]$$ which is homogeneous with respect to a shallow monomial grading on $$S$$ by a group $$\Sigma.$$ There are finitely many distinct initial ideals of $$I$$ with respect to linear group orders. In particular, there is a finite subset of $$\Sigma$$-homogeneous elements of $$I$$ such that for every linear group order on $$\znpos,$$ the initial terms of polynomials in that subset generate the initial ideal of $$I.$$

Proof. Regarding the first assertion, Logar’s argument for the leading ideal case from the proof of Theorem 2 in the previous post on polynomial division applies almost immediately to the present situation – one only needs to choose homogeneous elements in every case and replace “$$\ld$$” with “$$\In$$”. The second assertion then follows from Corollary 3 to Theorem 1.

We now present two successively stronger versions of Theorem 2.

Theorem 2*. Theorem 2 continues to hold if “linear group orders on $$\znpos$$” are substituted by “totally ordered monomial gradings on $$S := \kk[x_1, \ldots, x_n].$$” In other words, if $$I$$ is a homogeneous ideal of $$S$$ with respect to a shallow monomial grading on $$S,$$ then there are finitely many distinct initial ideals of $$I$$ with respect to totally ordered monomial gradings on $$S.$$ In particular, there is a finite subset of $$\Sigma$$-homogeneous elements of $$I$$ such that for every totally ordered monomial gradings on $$S,$$ the initial forms of polynomials in that subset generate the initial ideal of $$I.$$

Proof. Theorem 2 implies that there is a finite subset of of $$\Sigma$$-homogeneous elements of $$I$$ such that for every linear group order on $$\znpos,$$ the initial terms of polynomials in that subset generate the initial ideal of $$I.$$ Now apply Theorem 1*.

Theorem 2**. Theorem 2* holds for all (i.e. not necessarily homogeneous) ideals of $$S := \kk[x_1, \ldots, x_n].$$ In other words, if $$I$$ is an ideal of $$S,$$ then there are finitely many distinct initial ideals of $$I$$ with respect to totally ordered monomial gradings on $$S.$$ In particular, there is a finite subset of elements of $$I$$ such that for every totally ordered monomial gradings on $$S,$$ the initial forms of polynomials in that subset generate the initial ideal of $$I.$$

Proof. Indeed, given an ideal $$I$$ of $$S,$$ take another variable $$x_0,$$ and consider the homogenization (in the usual sense) $$I^h$$ of $$I$$ with respect to $$x_0, \ldots, x_n),$$ i.e. $$I^h$$ is the ideal of $$\kk[x_0, \ldots, x_n]$$ generated by $f^h := x_0^{\eta(f)}f(x_1/x_0^{\eta_1}, \ldots, x_n/x_0^{\eta_n})$for all $$f \in I$$. Given a monomial grading on $$S$$ by a totally ordered group $$\Omega$$, consider the induced $$\Omega$$-grading on $$S’ := \kk[x_0, \ldots, x_n]$$ defined by: $S’_\omega := \{f \in S’: f|_{x_0 = 1} \in S_\omega\}$Since the $$\Omega$$-grading on $$S$$ is monomial compatible, so is the grading on $$S’.$$ For each $$f \in I,$$ it is straightforward to check that $(\In_\Omega(f^h))|_{x_0 = 1} = \In_\Omega(f)$ On the other hand, if $$f$$ is a homogeneous element of $$I^h,$$ then the monomials in the support of $$f$$ are in one-to-one correspondence with the monomials in the support of $$f|_{x_0 = 1}$$ and $(\In_\Omega(f))|_{x_0 = 1} = \In_\Omega(f|_{x_0 = 1})$By Theorem 2* there are homogeneous $$f_1, \ldots, f_N \in I^h$$ independent of $$\Omega$$ such that $$\In_\Omega(I^h)$$ is generated by $$\In_\Omega(f_i),$$ $$i = 1, \ldots, N.$$ It then follows from the above discussion that $$\In_\Omega(I)$$ is generated by $$\In_\Omega(f_i|_{x_0 = 1}),$$ $$i = 1, \ldots, N.$$ This proves Theorem 2**.

## Every initial ideal comes from a weight

In this section we consider weighted orders as opposed to weighted degrees from the previous post. Every $$\omega \in \rr^n$$ defines a weighted order, and for $$f \in \kk[x_1, \ldots, x_n]$$ we write $$\In_\omega(f)$$ for the corresponding initial term of $$f$$. Given an ideal $$I$$ of $$\kk[x_1, \ldots, x_n]$$ we similarly write $$\In_\omega(I)$$ for its leading ideal generated by $$\In_\omega(f)$$ for all $$f \in I$$.

Theorem 3. Let $$I$$ be an ideal of $$S := \kk[x_1, \ldots, x_n]$$ which is homogeneous with respect to a shallow monomial grading on $$S.$$ For every linear group order $$\preceq$$ on $$\znpos,$$ there is $$\omega = (\omega_1, \ldots, \omega_n) \in \zz^n$$ such that $$\In_\omega(I) = \In_\preceq(I)$$ (in particular, this means that $$\In_\omega(I)$$ is a monomial ideal). Moreover, one can ensure that

1. $$\omega_i \neq 0$$ for all $$i,$$
2. $$\omega_i > 0$$ if and only if $$e_i \succ 0$$,
3. $$\omega_i < 0$$ if and only if $$e_i \prec 0,$$

where $$e_i$$ is the standard $$i$$-th basis vector of $$\zz^n.$$

Proof. Fix $$g_1, \ldots, g_N \in I$$ such that $$\In_\preceq(g_i)$$, $$i = 1, \ldots, N$$, generate $$\In_\preceq(I)$$. Write $g_i = c_{i0} x^{\alpha_{i0}} + \cdots + c_{iN_i}x^{\alpha_{iN_i}}$with $$\alpha_{i0} = \nu_\preceq(g_i)$$. Let $$C$$ be the set of all $$\omega \in \rr^n$$ such that

1. $$\langle \omega, \alpha_{i0} – \alpha_{ij} \rangle \leq 0,$$ $$i = 1, \ldots, N,$$ $$j = 1, \ldots, N_i,$$
2. $$\omega_i \geq 0$$ for all $$i$$ such that $$e_i \succ 0$$,
3. $$\omega_i \leq 0$$ for all $$i$$ such that $$e_i \prec 0.$$

Note that $$C$$ is a convex polyhedral cone. We claim that $$\dim(C) = n$$. Indeed, $$C$$ is dual to the cone $$C’$$ generated by $$\alpha_{ij} – \alpha_{i0},\ i = 1, \ldots, N,\ j = 1, \ldots, N_i$$ and $$\epsilon_i e_i$$, $$i = 1, \ldots, n,$$ where $$\epsilon_i = \pm 1$$ depending on wheter $$e_i \succ 0$$ or $$e_i \prec 0.$$ (Notice the switch from $$\alpha_{i0} – \alpha_{ij}$$ from the first condition defining $$C$$ to $$\alpha_{ij} – \alpha_{i0}$$ in the definition of $$C’.$$) If $$\dim(C) < n$$, it would follow that $$C’$$ is not strongly convex. Since $$C’$$ is also rational, there would be nonnegative integers $$\lambda_{ij}$$, not all zero, such that $\sum_{i,j}\lambda_{ij}(\alpha_{ij} – \alpha_{i0}) = -\sum_k r_k\epsilon_ke_k$for some nonnegative integers $$r_1, \ldots, r_n$$. Since $$\preceq$$ is compatible with addition on $$\znpos,$$ it would follow that $\sum_{i,j}\lambda_{ij} \alpha_{ij} \preceq \sum_{i,j}\lambda_{ij} \alpha_{i0}$On the other hand, by construction $$\alpha_{ij} \succ \alpha_{i0}$$ for each $$i,j$$, so that $\sum_{i,j}\lambda_{ij} \alpha_{ij} \succ \sum_{i,j}\lambda_{ij} \alpha_{i0}$This contradiction shows that $$\dim(C) = n$$, as claimed. In particular this implies (see e.g. [Mon21, Proposition V.15]) that the topological interior $$C^0$$ of $$C$$ has a nonempty intersection with $$\zz^n$$, and if $$\omega \in C^0 \cap \zz^n$$, then $$\In_\omega(g_i) = c_{i0}x^{\alpha_{i0}}$$. Since $$\In_\preceq(I)$$ is generated by $$c_{i0}x^{\alpha_{i0}}$$, $$i = 1, \ldots, N,$$ it follows that $$\In_\omega(I) \supseteq \In_\preceq(I)$$. If this containment were proper, then since $$\In_\preceq(I)$$ is a monomial (and therefore, homogeneous in the usual sense) ideal, Corollary 1 to Theorem 1 would imply that $\In_\preceq(\In_\omega(I)) \supsetneqq \In_\preceq(\In_\preceq(I)) = \In_\preceq(I)$On the other hand, it is straightforward to check that $\In_\preceq(\In_\omega(I)) = \In_{\preceq_\omega}(I)$ where $$\preceq_\omega$$ is the binary relation on $$\znzero$$ defined as follows: $$\alpha \preceq_\omega \beta$$ if and only if either $$\langle \omega, \alpha \rangle < \langle \omega, \beta \rangle$$, or $$\langle \omega, \alpha \rangle = \langle \omega, \beta \rangle$$ and $$\alpha \preceq \beta$$. It is straightforward to check that $$\preceq_\omega$$ is a linear group order on $$\znpos,$$ and consequently, the two preceding displays would contradict Corollary 2 to Theorem 1. It follows that $$\In_\omega(I) = \In_\preceq(I),$$ as required.

Remark. I think there are stronger versions of Theorem 3 analogous to the stronger versions of Theorem 2 above. In particular, I think the following is true, although I am not able to prove it.

Theorem 3** (I don’t know whether it is true). Theorem 3 continues to hold even if $$I$$ is not homogeneous and “linear group orders on $$\znpos$$” are substituted by “totally ordered monomial gradings on $$S := \kk[x_1, \ldots, x_n].$$” In other words, if $$I$$ is any ideal of $$S,$$ then for every monomial grading on $$S$$ by a totally ordered group $$\Omega,$$ there is $$\omega = (\omega_1, \ldots, \omega_n) \in \zz^n$$ such that $$\In_\omega(I) = \In_\Omega(I).$$ Moreover, one can ensure that

1. $$\omega_i = 0$$ if and only if $$\deg_\Omega(x_i) = 0_\Omega,$$ where $$0_\Omega$$ is the identity element of $$\Omega,$$
2. $$\omega_i > 0$$ if and only if $$\deg_\Omega(x_i) \succ_\Omega 0_\Omega$$,
3. $$\omega_i < 0$$ if and only if $$\deg_\Omega(x_i) \prec_\Omega 0_\Omega,$$

where $$e_i$$ is the standard $$i$$-th basis vector of $$\zz^n.$$