# Division with power series

$$\DeclareMathOperator{\lcm}{lcm} \DeclareMathOperator{\In}{In} \DeclareMathOperator{\ld}{Ld} \DeclareMathOperator{\kk}{\mathbb{K}} \DeclareMathOperator{\nd}{ND} \DeclareMathOperator{\rnpos}{\mathbb{R}^n_{> 0}} \DeclareMathOperator{\rnzero}{\mathbb{R}^n_{\geq 0}} \DeclareMathOperator{\rr}{\mathbb{R}} \DeclareMathOperator{\scrB}{\mathcal{B}} \DeclareMathOperator{\scrI}{\mathcal{I}} \DeclareMathOperator{\scrJ}{\mathcal{J}} \DeclareMathOperator{\supp}{Supp} \DeclareMathOperator{\znplusonezero}{\mathbb{Z}^{n+1}_{\geq 0}} \DeclareMathOperator{\znplusonepos}{\mathbb{Z}^{n+1}_{> 0}} \DeclareMathOperator{\znpos}{\mathbb{Z}^n_{> 0}} \DeclareMathOperator{\znzero}{\mathbb{Z}^n_{\geq 0}} \DeclareMathOperator{\zz}{\mathbb{Z}}$$In this post we look at division in the ring $$\hat R := \kk[[x_1, \ldots, x_n]]$$ of formal power series over a field $$\kk$$. In contrast to division in polynomial rings discussed in the preceding post, while trying to divide by power series, one is forced to consider initial terms with respect to monomial orders.

The initial exponent $$\nu_\preceq(f)$$ of $$f \in \hat R$$ with respect to a monomial order $$\preceq$$ is the minimal element of $$\supp(f)$$ with respect to $$\preceq$$ (note: the minimal element is guaranteed to exist due to the well ordering property of a monomial order). The initial term $$\In_\preceq(f)$$ of $$f$$ is the monomial term $$cx^\alpha$$ where $$\alpha = \nu_\preceq(f)$$ and $$c$$ is the coefficient of $$x^\alpha$$ in $$f$$. Given an ideal $$I$$ of $$\hat R$$, its initial ideal $$\In_\preceq(I)$$ is the ideal generated by $$\{\In_\preceq(f): f \in I\}$$.

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

Proof. Indeed, pick the unique $$i$$ such that $$\nu_\preceq(f_i) = \min_\preceq\{\nu_\preceq(f_j): j = 1, \ldots, N\}$$. Then $$\nu_\preceq(f_i)$$ does not appear in the support of $$f_j$$ for any $$j \neq i$$, so it can not be cancelled by any $$\kk$$-linear combination of the $$(f_j)_{j \neq i}$$.

The division algorithm in $$\hat R$$ is completely analogous to that for polynomials, the only twist being that the process may not stop in finitely many steps, the quotient and remainders are defined as the (well-defined) limits of the quotients and remainders computed in the finite steps. The division of $$f \in \hat R$$ by an ordered collection $$g_1, \ldots, g_N$$ of elements in $$\hat R$$ goes as follows:

1. Initially set $$f_0 := f$$, and define $$r_0, q_{0,1}, \ldots, q_{0,N}$$ to be zero. ($$r_i$$ and the $$q_{i,j}$$ would be the finite step approximations of respectively the remainder and the quotients, and they will be updated in every step.)
2. Assuming $$f_i, r_i, q_{i,1}, \ldots, q_{i,N}$$ have been computed, proceed to the next step as follows:
1. If there is $$j$$ such that $$\In_\preceq(g_j)$$ divides $$\In_\preceq(f_i)$$, then pick the smallest such $$j$$, compute $$q’_{i,j} := \In_\preceq(f_i)/\In_\preceq(g_j)$$, and set$f_{i+1} := f_i – q’_{i,j}g_j,$$q_{i+1,j} := q_{i,j} + q’_{i,j},$$q_{i+1,k} := q_{i,k}\ \text{if}\ k \neq j,$ $r_{i+1} := r_i$
2. Otherwise $$\In_\preceq(f_i)$$ is not divisible by $$\In_\preceq(g_j)$$ for any $$j$$; in this case define$f_{i+1} := f_i – \In_\preceq(f_i),$$q_{i+1,j} := q_{i,j},\ j = 1, \ldots, N,$$r_{i+1} := r_i + \In_\preceq(f_i)$
3. Set $$r := \lim_{i \to \infty}r_i$$ and $$q_j := \lim_{i \to \infty}q_{i,j}$$, $$j = 1, \ldots, N.$$

Theorem 1. $$r$$ and the $$q_j$$ are well defined elements in $$\hat R$$ such that $f = \sum_j q_jg_j + r$where

• either $$r = 0$$ or no monomial term in $$r$$ is divisible by $$\In_\preceq(g_j)$$ for any $$j$$;
• $$\In_\preceq(\sum_j q_jg_j) \in \langle \In_\preceq(g_j): j = 1, \ldots, N \rangle$$; 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.

In addition, if $$g_1, \ldots, g_N$$ are elements of an ideal $$I$$ of $$\hat R$$ 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. $$r = 0$$.

In particular, $$g_1, \ldots, g_N$$ generate $$I$$ as well.

Proof. By construction at every step $$i$$, for each of $$r_i, q_{i,1}, \ldots, q_{i,N}$$, either it remains unchanged from the previous step or its initial exponent increases from the previous step. Lemma 1 then implies that for each degree $$d$$, if $$i$$ is sufficiently large, then all terms (if any) added to $$r_i$$ or $$q_{i,j}$$ have degree greater than $$d$$. This proves that $$r$$ and the $$q_j$$ are well defined power series, and similarly, it follows that $$\lim_{i \to \infty} f_i = 0.$$ All the remaining assertions of Theorem 1 are then immediate.

All the results of polynomial division discussed in the previous post have corresponding analogues for power series:

Corollary 1. Let $$I \supseteq J$$ be ideals of $$\hat R$$. If $$I \supsetneqq J$$, then $$\In_\preceq(I) \supsetneqq \In_\preceq(J)$$ for every monomial order $$\preceq$$.

Corollary 2. Let $$I$$ be an ideal of $$\hat R$$. There can not exist monomial orders $$\preceq_1, \preceq_2$$ such that $$\In_{\preceq_1}(I) \subsetneqq \In_{\preceq_2}(I)$$.

The proofs of Corollaries 1 and 2 are identical to that from the polynomial case, one only needs to replace “leading” (terms, ideals, etc) by the corresponding “initial” versions. We also have the analogue for a “reduced” Gröbner basis: a finite ordered collection of elements $$g_1, \ldots, g_N$$ from an ideal $$I$$ is called a reduced initial basis with respect to a monomial 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 $$1.$$

A reduced initial basis can be produced by taking a minimal set of elements of $$I$$ whose initial terms generate $$\In_\preceq(I)$$ and then replacing each element by the remainder of the division by other elements. Corollary 3 of the polynomial division is not completely true for power series, since even an ideal $$I$$ generated by a single power series $$\phi$$ has infinitely many distinct initial bases: for each polynomial $$f$$ without constant term, $$(1 + f(\phi))\phi$$ is a reduced initial basis of $$I.$$ However, a part of it remains true:

Corollary 3. The reducedness of an initial basis depends only on the initial ideal. In other words, if an ideal $$I$$ of $$\hat R$$ has the same initial ideal for two monomial orders, then any reduced initial basis of $$I$$ with respect to one of the monomial orders is a reduced initial basis with respect to the other monomial order as well.

Proof. Indeed, pick a reduced initial basis $$g_1, \ldots, g_N$$ of $$I$$ with respect to a monomial order $$\preceq$$ and a monomial order $$\preceq’$$ such that $$\In_{\preceq’}(I) = \In_\preceq(I).$$ Let $$\alpha_i := \nu_{\preceq}(g_i),$$ $$i = 1, \ldots, n.$$ Fix $$i.$$ Since $$\In_{\preceq’}(I)$$ is 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 $$g_i,$$ we must have 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.

The proof of the Universal Basis Theorem (i.e. Theorem 2 in the previous post) used the fact that the support of a polynomial is finite. To adapt this proof for power series we seek the help of Newton diagrams. Given a power series $$f$$, it follows from basic convex geometry that the convex hull of $$\supp(f) + \rnzero$$ is a convex rational polyhedron (see e.g. [Mon21, Corollary B.48 and Proposition V.29]. The Newton diagram $$\nd(f)$$ of $$f$$ is the union of compact faces of this polyhedron. We write $$f_{\nd}$$ for the “part” of $$f$$ supported at its Newton diagram; more precisely, if $$f = \sum_\alpha c_\alpha x^\alpha$$, then $f_{\nd} := \sum_{\alpha \in \nd(f)} c_\alpha x^\alpha$Note in particular that $$\supp(f_{\nd})$$ is finite.

Lemma 2. For any monomial order $$\preceq$$ and power series $$f$$, $$\In_\preceq(f) = \In_\preceq(f_{\nd}).$$

Proof. Every $$\alpha \in \supp(f) \setminus \nd(f)$$ is a convex rational combination $\alpha = \sum_{i=1}^N r_i(\alpha_i + \beta_i)$such that

• $$\alpha_i \in \nd(f)$$, $$\beta_i \in \rnzero$$,
• each $$\alpha_i$$ is a vertex of $$\nd(f)$$,
• the coordinates of all $$\beta_i$$ are rational
• not all $$\beta_i$$ are zero

Then clearing out the denominators of all $$r_i$$ and $$\beta_i$$ leads to an identity of the form $k\alpha = \alpha_{i_1} + \cdots + \alpha_{i_k} + \beta$where $$k$$ is a positive integer, $$\alpha_{i_j}$$ are vertices of $$\nd(f)$$, and $$\beta$$ is a nonzero element of $$\znzero$$. It is then clear from the defining properties of a monomial order that there is at least one $$j$$ such that $$\alpha \succ \alpha_{i_j}$$, concluding the proof.

Theorem 2 (Universal Basis Theorem). Every ideal $$I$$ of $$\hat R$$ has finitely many distinct initial ideals. In particular, there is a finite collection $$\scrB$$ of elements in $$I$$ such that for every monomial order $$\preceq$$ on $$\znzero$$, the initial ideal of $$I$$ with respect to $$\preceq$$ is generated by the initial terms of the elements in $$\scrB$$.

Proof. Assume to the contrary that there is an infinite collection $$\scrJ_0$$ of distinct initial ideals of $$I$$. Pick an arbitrary nonzero element $$f_1 \in I$$. Due to Lemma 2, out of the finitely many monomial terms of $$f_{1, \nd}$$, there must be one, say $$m_1$$, such that $$\scrJ_1 := \{J \in \scrJ_0: \langle m_1 \rangle \subsetneqq J\}$$ is infinite. Pick $$J_1 \in \scrJ_1$$ and a monomial order $$\preceq_1$$ such that to $$J_1 = \In_{\preceq_1}(I)$$. Since $$m_1 \in J_1$$, there is $$g_1 \in I$$ such that $$\In_{\preceq_1}(g_1) = m_1$$, and since $$\langle m_1 \rangle \neq J_1$$, there is $$h_1 \in I$$ such that $$\In_{\preceq_1}(h_1) \not\in \langle m_1 \rangle.$$ Let $$f_2$$ be the remainder of $$h_1$$ modulo division by $$g_1$$. Then $$f_2$$ is nonzero, and no monomial term in $$f_2$$ belongs to $$\langle m_1 \rangle$$. Since $$f_{2, \nd}$$ has finitely many monomial terms, due to Lemma 2 there must be a monomial term $$m_2$$ of $$f_{2, \nd}$$ such that $$\scrJ_2 := \{J \in \scrJ_1: \langle m_1, m_2 \rangle \subsetneqq J\}$$ is infinite. Now choose $$J_2 \in \scrJ_2$$ and a monomial order $$\preceq_2$$ such that to $$J_2 = \In_{\preceq_2}(I)$$. As in the preceding case, there are $$g_{21}, g_{22}, h_2 \in I$$ such that $$\In_{\preceq_2}(g_{21}) = m_1$$, $$\In_{\preceq_2}(g_{22}) = m_2$$, and $$\In_{\preceq_2}(h_2) \not\in \langle m_1, m_2 \rangle.$$ Define $$f_3$$ to be the remainder of $$h_2$$ modulo division by $$g_{21}, g_{22}$$. Then $$f_3$$ is nonzero, and no monomial term in $$f_3$$ belongs to $$\langle m_1, m_2 \rangle,$$ and we can proceed to define $$\scrJ_3$$ as above. In this way we get an infinite strictly increasing chain of monomial ideals: $\langle m_1 \rangle \subsetneqq \langle m_1, m_2 \rangle \subsetneqq \langle m_1, m_2, m_3 \rangle \subsetneqq \cdots$ which violates the Noetherianity of $$\kk[[x_1, \ldots, x_n]]$$ and completes the proof of the first assertion. The second assertion then follows from Corollary 3 to Theorem 1.

## Weighted orders in division

Every $$\omega \in \rr^n$$ defines a weighted order on $$\hat R$$, and for $$f \in \hat R$$ we write (by an abuse of notation) $$\omega(f)$$ and $$\In_\omega(f)$$ respectively for the corresponding order and initial term of $$f$$. Note that

• There can be $$\omega$$ and $$f$$ such that $$\omega(f) = -\infty$$, in which case $$\In_\omega(f)$$ is not defined!
• If $$\omega \in \rnzero,$$ then $$\omega(f) \geq 0$$ and $$\In_\omega(f)$$ is defined for all $$f \in \hat R.$$

Given an ideal $$I$$ of $$\hat R$$ we write $$\In_\omega(I)$$ for its initial ideal generated by $$\In_\omega(f)$$ for all $$f \in I$$ such that $$\In_\omega(f)$$ is defined.

Theorem 3. Let $$\preceq$$ be a monomial order and $$g_1, \ldots, g_N \in \hat R.$$ Then there is $$\omega \in \znpos$$ such that $\In_\omega(g_i) = \In_\preceq(g_i),\ i = 1, \ldots, N.$If in addition the $$g_i$$ are elements of an ideal $$I$$ such that $$\In_\preceq(g_i),$$ $$i = 1, \ldots, N,$$ generate $$\In_\preceq(I),$$ then$\In_\omega(I) = \langle \In_\omega(g_1), \ldots, \In_\omega(g_N)\rangle = \langle \In_\preceq(g_1), \ldots, \In_\preceq(g_N)\rangle = \In_\preceq(I)$In particular $$\In_\omega(I)$$ is a monomial ideal. Moreover, every $$f \in I$$ has an expression of the form $$f = \sum_i f_ig_i$$ such that

1. $$\nu_\preceq(f) = \min_\preceq\{\nu_\preceq(f_ig_i)\},$$ and
2. $$\omega(f) = \min\{\omega(f_ig_i)\}.$$

Proof. Write $g_{i,\nd} = c_{i0} x^{\alpha_{i0}} + \cdots + c_{iN_i}x^{\alpha_{iN_i}}$with $$\alpha_{i0} = \nu_\preceq(g_i)$$. Define $C := \{\omega \in \rnzero: \langle \omega, \alpha_{i0} – \alpha_{ij} \rangle \leq 0,\ i = 1, \ldots, N,\ j = 1, \ldots, N_i\}$Note that $$C$$ is a convex polyhedral cone. We claim that $$\dim(C) = n$$. Indeed, $$C$$ is dual to the cone $$C’$$ generated by $$\alpha_{ij} – \alpha_{i0},\ i = 1, \ldots, N,\ j = 1, \ldots, N_i$$ (notice the switch from $$\alpha_{i0} – \alpha_{ij}$$ in the formula for $$C$$ to $$\alpha_{ij} – \alpha_{i0}$$ in the formula for $$C’$$) and the standard basis vectors $$e_1, \ldots, e_n$$ of $$\rr^n$$. 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 $\sum_{i,j}\lambda_{ij}(\alpha_{ij} – \alpha_{i0}) = -\sum_k r_ke_k$for some nonnegative integers $$r_1, \ldots, r_n$$. Due to the second and third properties of a monomial order, 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 $$\znpos$$, and if $$\omega \in C^0 \cap \znpos$$, then $$\In_\omega(g_i) = c_{i0}x^{\alpha_{i0}}$$. This proves the first assertion. The remaining part of the theorem follows immediately from Lemma 3 below.

Remark. Given $$\omega \in \znpos$$ and elements $$g_1, \ldots, g_N$$ of an ideal $$I$$ such that $$\In_\omega(g_i) = \In_\preceq(g_i)$$ for each $$i$$ and $$\In_\preceq(g_i)$$, $$i = 1, \ldots, N$$, generate $$\In_\preceq(I)$$, if one only wanted to prove that $$\In_\omega(I) = \In_\preceq(I),$$ then one can give a much less involved argument than Lemma 3 following [Stu96, Chapter 1]. Indeed, since $$\In_\omega(g_i) = \In_\preceq(g_i)$$ for each $$i$$, it follows that $$\In_\omega(I) \supseteq \In_\preceq(I)$$. If this containment were proper, then 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 $$\alpha \preceq \beta$$. Since $$\omega \in \znzero$$, it follows that $$\preceq_\omega$$ is a monomial order, so that the last two displays would contradict Corollary 2 to Theorem 1. It follows that $$\In_\omega(I) = \In_\preceq(I),$$ as required.

Lemma 3. In the situation of Theorem 1 assume there is $$\omega$$ such that $$\omega(\In_\preceq(g_j)) = \omega(g_j)$$ (i.e. $$\omega(g_j)$$ is finite and $$\In_\preceq(g_j)$$ is a monomial with $$\omega$$-value equal to $$\omega(g_j)$$) for each $$j$$. Then

1. $$\omega(q_jg_j) \geq \omega(f)$$ for each $$j$$.
2. $$\omega(r) \geq \omega(f)$$.
3. Assume in addition $$\omega \in \rnpos$$, $$In_\preceq(g_j),$$ $$j = 1, \ldots, N,$$ generate $$\In_\preceq(I)$$, where $$I$$ is the ideal generated by $$g_1, \ldots, g_N,$$ and $$I$$ contains $$f.$$ Then $$\In_\omega(f)$$ is in the ideal generated by $$\In_\omega(g_j),$$ $$j = 1, \ldots, N.$$ (Note that the non-negativity of the coordinates of $$\omega$$ ensures that $$\In_\omega(f)$$ is well defined.) Moreover, $$f$$ has an expression of the form $$f = \sum_i f_ig_i$$ such that
1. $$\nu_\preceq(f) = \min_\preceq\{\nu_\preceq(f_ig_i)\},$$ and
2. $$\omega(f) = \min\{\omega(f_ig_i)\}.$$

Proof. It is straightforward to check that the first two assertions are true for the first step of the division algorithm, and $$\omega(f_1) \geq \omega(f)$$. Now substitute $$f$$ by $$f_1$$ and repeat to see that the first two assertions hold in general. It remains to prove the third assertion. Since $$f \in I,$$ and the initial forms the $$g_i$$ generate the inital ideal of $$I$$, by the division algorithm $$f$$ has an expression of the form $f = \sum_i f_ig_i$such that $$\nu_\preceq(f) = \min_\preceq\{\nu_\preceq(f_ig_i)\},$$ i.e. property 3.1 of Theorem 2 is satisfied. Assume property 3.2 is not satisfied. Then there must be a cancellation of the form $\sum_j \In_\omega(f_{i_j})\In_\omega(g_{i_j}) = 0$ where the sum is over all $$j$$ such that $$\omega(f_{i_j}) + \omega(g_{i_j})$$ is the smallest. We claim that $\nu_\preceq(f) \prec \nu_\preceq(f_{i_j}g_{i_j})$for each $$j$$ (where “$$\prec$$” denotes “strictly smaller with respect to $$\preceq$$”). Indeed, by Theorem 1, there is only one $$i$$ such that $$f_ig_i$$ has a monomial term with exponent $$\nu_\preceq(f)$$, so that this term can not cancel with any other monomial term in the support of $$f_kg_k$$ for $$k \neq i,$$ proving the claim. Now write $f = \sum_{i \neq i_j}f_ig_i + \sum_j (f_{i_j} – \In_\omega(f_{i,j}))g_{i_j} + f’$where $$f’ := \sum_j \In_\omega (f_{i_j})g_{i_j}.$$ Since the initial terms of the $$g_{i_j}$$ with respect to $$\omega$$ and $$\preceq$$ are “compatible”, and since $$\sum_j \In_\omega(f_{i_j})\In_\omega(g_{i_j}) = 0$$, it follows that $\nu_\preceq(f’) \succ \nu_\preceq (f_{i_j} g_{i_j})$for each $$j$$. Note in addition that $\nu_\preceq(f’) \succ \nu_\preceq(f)$ due to the above claim. Since $$f’$$ is in the ideal generated by the $$g_i$$, we can write $$f’ = \sum_i f’_ig_i,$$ where the $$f’_i$$ are the quotients produced by the division algorithm. It follows that $f = \sum_{i \neq i_j}f_ig_i + \sum_j (f_{i_j} – \In_\omega(f_{i,j}))g_{i_j} + \sum_i f’_ig_i$This suggests a double induction to track the change in this expression for $$f$$ in comparison to the original expression $$f = \sum_i f_ig_i$$ as follows: given $$h = (h_1, \ldots, h_N)$$ such that $$f = \sum_i h_ig_i$$, define $m(h) := \min_i\{\omega(h_ig_i)\}$and$\alpha(h) := \min_\preceq\{\nu_\preceq(\In_\omega(h_ig_i)): \omega(h_ig_i) = m(h)\}$Our discussion above implies that if property 3.2 is not satisfied, then starting from an expression of $$f$$ satisfying property 3.1, the value for the pair $$(m, \alpha)$$ can be increased (where the ordering on the pair is initially by $$m$$ and then by $$\preceq$$ on $$\alpha$$) while ensuring that property 3.1 remains true. Since each coordinate of $$\omega$$ is positive, for each value of $$m$$, the space of $$\omega$$-homogeneous elements of $$\omega$$-order (or degree) $$m$$ is finite dimensional, and therefore due to Lemma 1, it is not possible for only $$\alpha$$ to increase while keeping $$m$$ unchanged. Consequently at some point we must have an expression $$f = \sum_i h_ig_i$$ such that $$m(h) = \omega(f),$$ as required to complete the proof.

## (An adaptation of) Buchberger’s algorithm

Given two nonzero monomial terms $$ax^\alpha, bx^\beta$$, we define their lowest common multiple as $\lcm(ax^\alpha, bx^\beta) := x^{\max(\alpha, \beta)}$where the maximum is taken componentwise. Note that the coefficient of the lowest common divisor is by definition $$1$$, regardless of the coefficients of the original monomials. Buchberger’s algorithm for computing Gröbner bases of polynomial ideals adapts easily to the computation of initial ideals in the ring of power series (note that this “algorithm” is not really finite, since the division algorithm is defined only as the limit of an infinite process. For truly finite algorithms one needs something like Mora’s algorithm):

Input: a monomial order $$\preceq$$ and a finite ordered collection $$g_1, \ldots, g_s$$ of nonzero elements in $$\hat R$$. Denote the ideal generated by $$g_1, \ldots, g_s$$ in $$\hat R$$ by $$I$$.

Output: a finite ordered collection $$\scrB$$ of nonzero elements in $$I$$ whose initial terms generate $$\In_\preceq(I)$$.

Procedure:

1. Set $$\scrB_0 := (g_1, \ldots, g_s)$$.
2. Assuming $$\scrB_k = (h_1, \ldots, h_{s_k})$$ has been computed, compute $$\scrB_{k+1}$$ as follows: for each $$i, j$$, $$1 \leq i < j \leq k$$,
• Define $S(h_i, h_j) := \frac{\lcm(\In_\preceq(h_i), \In_\preceq(h_j))}{\In_\preceq(h_i)}h_i – \frac{\lcm(\In_\preceq(h_i), \In_\preceq(h_j))}{\In_\preceq(h_j)}h_j$
• Divide each $$S(h_i, h_j)$$ by $$\scrB_k$$. Define $$\scrB_{k+1}$$ to be the union of $$\scrB_k$$ and the set of all nonzero remainders of $$S(h_i, h_j)$$ by $$\scrB_k$$.
3. Repeat until you get $$\scrB_{k+1} = \scrB_k$$. Then $$\scrB := \scrB_k$$ is the desired output.

Theorem 4. The above algorithm stops in finitely many steps and the output is a finite collections of elements of $$I$$ whose initial terms generate $$\In_\preceq(I)$$.

Proof. Since we start with elements in $$I$$, it is clear that all new elements added to the original list are also from $$I$$. The process also stops in finitely many steps, since whenever a new element is added to the list, the ideal generated by their initial terms gets larger (due to Theorem 1), and the Noetherianity of $$\hat R$$ precludes infinite strictly increasing chains of ideals. It remains to show that $$(\In_\preceq(g): g \in \scrB)$$ generate $$\In_\preceq(I)$$. Enumerate the elements in $$\scrB$$ by $$g_1, \ldots, g_N.$$ We may assume without loss of generality that the coefficient of $$\In_\preceq(g_i)$$ is $$1$$, i.e. $$\In_\preceq(g_i)$$ is of the form $$x^{\alpha_i}$$. Pick $$f \in I.$$ Since the $$g_i$$ generate $$I$$, there is an expression $$f = \sum_i f_ig_i$$ with $$f_i \in \hat R.$$ Due to Theorem 3 there is $$\omega \in \znpos$$ such that $$\In_\omega(f) = \In_\preceq(f)$$ and $$\In_\omega(g_i) = \In_\preceq(g_i)$$ for each $$i$$. Define the pair $$(m, \alpha)$$ as in the proof of Lemma 3, i.e. $m(f_1, \ldots, f_N) := \min_i\{\omega(f_ig_i)\}$and$\alpha(f_1, \ldots, f_N) := \min_\preceq\{\nu_\preceq(\In_\omega(f_ig_i)): \omega(f_ig_i) = m(f_1, \ldots, f_N)\}$If $$m(f_1, \ldots, f_N) = \omega(f)$$, then it would follow that $$\In_\preceq(f) = \In_\omega(f)$$ is a sum of the form $$\sum_j \In_\omega(f_{i_j})\In_\omega(g_{i_j}) = \sum_j \In_\omega(f_{i_j})\In_\preceq(g_{i_j})$$ and therefore would be in the ideal generated by the $$\In_\preceq(g_i)$$. So assume $m(f_1, \ldots, f_N) < \omega(f)$Then there must be a cancellation of the form $\sum_j \In_\omega(f_{i_j})\In_\omega(g_{i_j}) = \sum_j \In_\omega(f_{i_j})\In_\preceq(g_{i_j}) = 0$ where the sum is over all $$j$$ such that $$\omega(f_{i_j}) + \omega(g_{i_j}) = m(f_1, \ldots, f_N).$$ It follows that $\sum_k \In_\preceq(\In_\omega(f_{i_{j,k}})\In_\preceq (g_{i_{j,k}})) = 0$ where the sum is over all $$k$$ such that $$\nu_\preceq(f_{i_{j,k}}g_{i_{j,k}}) = \alpha(f_1, \ldots, f_N) =: \alpha.$$ This implies that $$\In_\preceq(\In_\omega(f_{i_{j,k}})) = b_{i_{j,k}}x^{\alpha – \alpha_{i_{j,k}}}$$ with $$\sum_k b_{i_{j,k}} = 0.$$ Write $f = \sum_{i \neq i_j}f_ig_i + \sum_j (f_{i_j} – \In_\omega(f_{i,j}))g_{i_j} + f’_1 + f’_2$where $f’_1 := \sum_{i_j \neq i_{j,k}} \In_\omega(f_{i,j}) g_{i_j} + \sum_k (\In_\omega(f_{i_{j,k}}) – \In_\preceq(\In_\omega(f_{i_{j,k}})))g_{i_{j,k}}$and $f’_2 := \sum_k \In_\preceq(\In_\omega (f_{i_{j,k}})) g_{i_{j,k}}$Since $$b_{i_{j,1}} = -\sum_{k > 1} b_{i_{j,k}},$$ it follows that $f’_2 = \sum_k b_{i_{j,k}}x^{\alpha – \alpha_{i_{j,k}}}g_{i_{j,k}} = \sum_{k> 1} b_{i_{j,k}} (x^{\alpha – \alpha_{i_{j,k}}}g_{i_{j,k}} – x^{\alpha – \alpha_{i_{j,1}}}g_{i_{j,1}}) = \sum_{k > 1} b_{i_{j,k}}x^{\alpha – \beta_{i_{j,k}}}S(g_{i_{j,k}}, g_{i_{j,1}})$where $$x^{\beta_{i_{j,k}}}:= \lcm(x^{\alpha_{i_{j,k}}}, x^{\alpha_{i_{j,1}}}).$$ Due to Buchberger’s algorithm and Theorem 1, whenever $$S(g_{i_{j,k}}, g_{i_{j,1}})$$ is nonzero, it can be expressed as $$\sum_i h_ig_i$$ such that $\min_i(\nu_\preceq(h_ig_i)) = \nu_\preceq(S(g_{i_{j,k}}, g_{i_{j,1}}))$It is clear by construction of $$S(\cdot, \cdot)$$ that $\nu_\preceq (x^{\alpha- \beta_{i_{j,k}}}S(g_{i_{j,k}}, g_{i_{j,1}})) \succ \alpha$It follows that $$f$$ can be expressed as $$\sum_i \tilde f_i g_i$$ such that

• either $$m(\tilde f_1, \ldots, \tilde f_N) > m(f_1, \ldots, f_N)$$,
• or $$m(\tilde f_1, \ldots, \tilde f_N) = m(f_1, \ldots, f_N)$$ and $$\alpha(\tilde f_1, \ldots, \tilde f_N) > \alpha(f_1, \ldots, f_N).$$

To summarize: as long as $$m(f_1, \ldots, f_N) < \omega(f)$$, it is possible to find another expression $$f = \sum_i \tilde f_i g_i$$ such that the value for the pair $$(m, \alpha)$$ increases (where the ordering on the pair is initially by $$m$$ and then by $$\preceq$$ on $$\alpha$$). Since each coordinate of $$\omega$$ is positive, for each value of $$m$$, the space of $$\omega$$-homogeneous elements of $$\omega$$-order (or degree) $$m$$ is finite dimensional, and therefore due to Lemma 1, it is not possible for only $$\alpha$$ to increase while keeping $$m$$ unchanged. Consequently at some point we must arrive at an expression $$f = \sum_i h_ig_i$$ such that $$m(h_1, \ldots, h_N) = \omega(f),$$ which implies that $$\In_\preceq(f) = \In_\omega(f)$$ is in the ideal generated by the $$\In_\omega(g_i) = \In_\preceq(g_i),$$ $$i = 1, \ldots, N,$$ and completes the proof.