# Notes (part 2): S. Bosch, U Güntzer and R. Remmert, Non-Archimedean Analysis


• $$A^{(\infty)} := \bigoplus_{i=1}^\infty A$$
• the set of zero sequences, $$c(A) := \{(c_i): \lim_{i \to \infty} |c_i| = 0\}$$
• the set of bounded sequences, $$b(A) := \{(b_i)_i: \sup_i |b_i| < \infty\}$$

It is clear that $$A^{(\infty)} \subseteq c(A) \subseteq b(A).$$ If $$A$$ is a ring/algebra over a field/module over a ring, each of these sets inherits the structure from $$A.$$ We would always consider each of these sets equipped with the component-wise supremum norm:
$|(b_i)_i| := \sup_i |b_i|$
If $$A$$ is a normed ring (respectively, field), this makes these sets normed $$A$$-modules (respectively, vector spaces). If $$A$$ is complete, then so are $$b(A)$$ and $$c(A),$$ but not $$A^{(\infty)}.$$

## Examples/Facts:

1. Consider a valued field $$K$$ equipped with a nontrivial valuation.
1. $$c(K)$$ is a $$K$$-vector space of countable type, i.e. it has a dense linear subspace of at most countable dimension (e.g. $$K^{(\infty)}$$). It follows that any weakly $$K$$-cartesian family in $$c(K)$$ is at most countable. (For each element in such a family, replace it by an element of $$K^{(\infty)}$$ sufficiently close to it, then the resulting system would be a weakly cartesian subset of $$K^{(\infty)}.$$ Since weak cartesianity implies linear independence, and since $$K^{(\infty)}$$ has countable dimension, it follows that the original family must be countable.)
2. $$b(K)$$ has an uncountable orthonormal family. (The natural map
$(b(K))^\sim \to \prod_{i=1}^\infty K^\sim$
is surjective and the latter space has uncountable dimension over $$K^\sim.$$ Now pullback an uncountable orthonormal system from the image.)
3. It follows that there is no $$K$$-linear map $$b(K) \to c(K)$$ which is a homeomorphism.
4. On the other hand given any $$a \in K$$ with $$0 < |a| < 1,$$ the mapping
$(b_i)_i \mapsto (a^ib_i)_i$
defines a continuous injective $$K$$-linear map from $$b(K)$$ to $$c(K).$$
2. If a ring is complete, it does not necessarily mean that its field of fractions is complete.
1. E.g. take $$A := k[[x, y]]$$ with the usual (discrete) valuation corresponding to its maximal ideal. The valuation is also induced by the order, i.e.
$|f| := e^{-\ord(f)}$
Pick a power series $$f(t) \in k[[t]]$$ which is not a rational function in $$t,$$ i.e. which is not in the image of the natural map from $$k(t) \to k[[t]]$$ (e.g. one can take the power series for $$e^t,$$ at least if the characteristic of $$k$$ is zero). Then $$f(y^2/x)$$ is in the completion of the field of fractions of $$A,$$ but it can not be represented as $$g/h$$ for $$g, h \in A.$$ [To see this consider the edge of $$f(y^2/x)h$$ corresponding to the inner normal $$(2, 1).$$]
2. Let $$A := k \langle x \rangle$$ be the ring of strictly convergent power series over a field $$k$$ with a complete nontrivial valuation. Then $$A$$ is complete with respect to the Gauss norm. The field of fractions of $$A$$ is the field $$L := k \langle x \rangle [x^{-1}]$$ of strictly convergent meromorphic series. Take $$c \in k$$ such that $$0 < |c| < 1.$$ Then the sequence $$\sum_{j=1}^i c^i x^{-i},$$ $$i = 1, 2, \ldots,$$ is Cauchy but without limit in $$L.$$
3. The algebraic closure of a complete field does not have to be complete.
1. Take the field $$k((t))$$ of Laurent series, which is complete with respect to the valuation induced by $$\ord.$$ Its algebraic closure is the field of “meromorphic” Puiseux series, which is not complete, since the Cauchy sequence
$f_n := \sum_{j=1}^n t^{1/n + n}$
does not converge.
2. The algebraic closure of $$\qq_p$$ is not complete. Both this and the preceding example are special cases of the following:
Lemma [BGR 3.4.3/1]. Let $$K$$ be a field with a complete nontrivial valuation and $$K_a$$ be its algebraic closure. Assume $$[K_a:K] = \infty.$$ Then $$K_a$$ is not complete.
Proof. If the separable closure $$K_{sep}$$ of $$K$$ in $$K_a$$ had finite degree over $$K,$$ then it would be complete, and therefore, since $$K_{sep}$$ is dense in $$K$$ [BGR 3.4.1/6], it would follow that $$K_a = K_{sep}$$ is also of finite degree over $$K,$$ which is a contradiction. So $$[K_{sep}: K] = \infty.$$ Choose a sequence $$x_1 = 1, x_2, \ldots$$ of elements in $$K_{sep}$$ which are linearly independent over $$K$$. Choose $$c_2, c_3, \ldots \in K \setminus \{0\}$$ such that
• $$\lim_i |c_ix_i| = 0$$
• $$|c_{i+1}x_{i+1}| < \min\{|c_ix_i|, r(\sum_{j=2}^i c_jx_j)\}$$
(where $$r(\alpha)$$, for $$\alpha$$ separable over $$K$$ of degree > 1, is the minimum of $$|\alpha – \beta|$$ over all roots $$\beta \neq \alpha$$ of the minimal polynomial of $$\alpha$$ over $$K.$$) Note that $$r(\sum_{j=2}^i c_jx_j)$$ is well defined for $$j \geq 2,$$ since $$\sum_{j=2}^i c_jx_j \in K_{sep} \setminus K,$$ since all $$c_j$$ are nonzero and $$x_j$$ constitute a basis of $$K_{sep}$$ over $$K.$$ Then if $$\sum_{j=2}^\infty c_jx_j$$ converges to an element $$x \in K_a,$$ then by assumption,
$|x – \sum_{j=2}^i c_jx_j| = |\sum_{j=i+1}^\infty c_jx_j| \leq |c_{i+1}x_{i+1}| < r(\sum_{j=2}^{i}c_ix_i)$
so that Krasner’s lemma implies that $$K(x) \supseteq K(\sum_{j=2}^i c_jx_j).$$ It follows that $$K(x) \supseteq K(x_2, x_3, \ldots),$$ so that $$[K(x):K] = \infty,$$ which is a contradiction, since $$x$$ is algebraic over $$K.$$ This completes the proof. In fact, if $$x$$ is the element in the completion of $$K_a$$ which represents $$\sum_{j=2}^\infty c_jx_j,$$ then it follows (from the arguments in the proof of Krasner’s lemma [BGR 3.4.2/2]) that $$K(x)$$ indeed contains $$K(x_2, x_3, \ldots),$$ and in particular, $$x$$ is transcendental over $$K.$$
4. A (countable) basis of $$\tilde V$$ over $$\tilde K$$ may not lift to a Schauder basis of a vector space $$V$$ over a (valued field) $$K.$$ Let $$K$$ be a field with a nontrivial non-discrete valuation. Choose a Schauder basis $$\{x_1, x_2, \ldots,\}$$ of $$V := c(K)$$ such that $$|x_i| = 1$$ for all $$i.$$ Choose a sequence $$\lambda_2, \lambda_3, \ldots$$ in $$K$$ such that $$0 < |\lambda_i| < 1$$ and $$\lim_i \lambda_2 \lambda_3 \cdots \lambda_i \neq 0.$$ Set
$y_i := x_i – \lambda_{i+1}x_{i+1}$
Then $$\tilde y_i = \tilde x_i$$ for all $$i$$, i.e. $$\tilde y_1, \tilde y_2, \ldots,$$ constitute a $$\tilde K$$-basis of $$\tilde V.$$ However, $$x_1$$ can not be written as a convergent sum $$x_1 = \sum_{i=1}^\infty a_i y_i$$ because then
$x_1 = \sum_{i=1}^\infty a_i(x_i – \lambda_{i+1}x_{i+1}) = a_1x_1 + \sum_{i=2}^\infty (a_i – a_{i-1}\lambda_i)x_i$
which means $$a_1 = 1$$ and $$a_i = a_{i-1}\lambda_i = \lambda_2 \lambda_3 \cdots \lambda_i$$ for $$i \geq 2.$$ It follows that $$\lim_i |a_iy_i| = \lim_i|a_i| \neq 0.$$
5. A complete non perfect field. The field $$k((t))$$ of Laurent series is complete with respect to the valuation induced by order in $$t.$$ However, $$k((t))$$ does not contain a $$p$$-th root of $$t$$ if the characteristic $$p$$ of $$k$$ is positive.
6. There are valued fields $$K$$ admitting algebraic extensions $$L \neq K$$ such that $$K$$ is dense in $$L$$ with respect to the spectral norm. In particular, $$L$$ is not weakly cartesian over $$K.$$ In addition, there are such examples with $$L$$ purely inseparable over $$K$$, in which case the spectral norm on $$L$$ is a valuation.
1. E.g. one can take $$K$$ to be the separable closure of a complete non perfect field $$k$$ and $$L$$ to be the algebraic closure of $$k.$$ $$K$$ is dense in $$L$$ since there are separable elements arbitrarily close to every element in $$L.$$
2. For a slightly more explicit example, start with fields $$k$$ of characteristic $$p > 0$$ with a nontrivial valuation such that the field $$k^{p^{-1}}$$ of $$p$$-th roots over $$k$$ is of infinite degree over $$k.$$ E.g. one can take for $$k$$ the field of fractions (or, if completeness is required, the completion of the field of fractions) of the formal power series ring in infinitely many variables over the finite field of $$p$$ elements. Let $$L$$ be the field of fractions of the ring $$k^{p^{-1}}\langle t \rangle$$ of strictly convergent power series over $$k^{p^{-1}}$$ equipped with the Gauss norm. Let $$A$$ be the set of all series $$\sum_i a_i t^i \in k^{p^{-1}}\langle t \rangle$$ such that the coefficients $$a_i$$ generate a finite extension of $$k.$$ The field $$K$$ of fractions of $$A$$ is dense in $$L$$ (since e.g. it contains all polynomials over $$k^{p^{-1}}$$). Moreover, since $$L \subseteq K^{p^{-1}},$$ it follows that $$L$$ is algebraic over $$K$$ and the Gauss norm on $$L$$ is identical to the spectral norm of $$L$$ with respect to the Gauss norm on $$K.$$ To see that $$L \neq K,$$ choose a sequence $$h_i \in k^{p^{-1}}$$ such that $$\lim_i |h_i| = 0$$ and $$[k(h_1, h_2, \ldots, ): k] = \infty.$$ Then $$\sum_i h_it^i \in L \setminus K.$$
3. For an even more explicit example with discrete valuation and finite extension, start with a field $$k$$ of characteristic $$p > 0$$ such that there is $$f(t) \in k[[t]]$$ which is not algebraic over $$k(t)$$ (e.g. one can take $$k$$ and $$f$$ as in Schmidt’s example in the preceding post). Let $$K := k(x,y^p) \subset L := k(x,y).$$ As in that example, equip $$L$$ with the norm $$|\ |_f$$ induced by the valuation on $$k((t))$$ via (the pullback of) the homomorphism $$k[x,y] \to k[[t]]$$ given by $$x \mapsto t,$$ $$y \mapsto f(t).$$ Since $$g^p \in K$$ for each $$g \in L,$$ it follows that the norm on $$L$$ is identical to the spectral norm of $$L$$ with respect to the norm on $$K.$$ Writing $$f(t) := \sum_i a_it^i,$$ one gets that $$y – \sum_{j=1}^i a_ix^i \to 0$$ as $$i \to \infty,$$ so that $$K$$ is dense in $$L.$$
7. The above phenomenon is impossible if $$L/K$$ is separable or $$K$$ is perfect. Since the the algebraic closure of a perfect field $$K$$ is weakly $$K$$-cartesian [BGR 3.5.1/4].
8. Every perfect or complete field $$K$$ is weakly stable, i.e. each finite extension equipped with the spectral norm is weakly $$K$$-cartesian.
9. The preceding observation suggests that the weak stability of a field $$K$$ boils down to whether $$K_\infty := \bigcup_{n \geq 0} K_n$$ is weakly cartesian over $$K$$, where
$K_n := \{x \in \bar K: x^{p^n} \in K\}$
(where $$\bar K$$ is the algebraic closure of $$K$$), since $$K_\infty,$$ being perfect, is weakly stable (i.e. $$\bar K$$ is weakly cartesian over $$K_\infty$$). Whether $$K_\infty$$ is weakly $$K$$-cartesian depends on whether each $$K_n$$ is weakly $$K$$_cartesian, which in turn (via the Frobenius homomorphism $$x \mapsto x^{p^{n-1}}$$) depends on whether $$K_1 = K^{p^{-1}}$$ is weakly $$K$$-cartesian. This is the content of [BGR 3.5.3/1].
10. The field of rational functions over a field in finitely many variables is weakly stable with respect to the valuation induced by degree or order [BGR 3.5.3/4] (this essentially follows from the previous observation together with some computation).
11. A characteristic zero (in particular, separable) complete field $$K$$ and a finite algebraic extension $$L/K$$ which is weakly $$K$$-cartesian, but not $$K$$-cartesian; in other words, weak stability $$\not \Rightarrow$$ stability. Start with the field $$\qq_2$$ of $$2$$-adic numbers. Let $$\alpha_0 := 2, \alpha_1, \alpha_2, \ldots$$ be elements algebraic over $$\qq_2$$ such that $$\alpha_i^2 = \alpha_{i-1}$$ and $$k_i := \qq_2(\alpha_i) = \qq_2(\alpha_0, \ldots, \alpha_i).$$ Let $$K$$ be the completion of the field $$\bigcup_i k_i$$ and $$L := K(\sqrt 3).$$
Lemma [BGR 3.6.1]. $$[L:K] = 2,$$ $$|L| = |K|,$$ and $$L^\sim = K^\sim = \ff_2,$$ the finite field of two elements. In particular, $$L$$ is not $$K$$-cartesian.
The proof is mainly by computation. I tried to solve for $$x^2 = 3$$ over $$K.$$ It seems that one can write $$\sqrt 3$$ as an infinite sum of the form $$\sum_i c_i 2^{\alpha_i}$$ with $$|c_i| = 1$$ and $$0 = \alpha_0 < \alpha_1 < \alpha_2 < \cdots$$ such that $$\lim_i \alpha_i = 1.$$ This suggests that $$|\sqrt 3, K| = 2^{-1}$$ but there is no $$a \in K$$ such that $$|\sqrt 3 – a| = 2^{-1}.$$ In particular, $$K$$ is not strictly closed in $$L.$$
12. The above phenomenon is impossible in the following cases:
1. If the valuation on $$K$$ is discrete, since if $$K$$ is discretely valued and complete, then every finite dimensional normed vector space $$V$$ over $$K$$ is $$K$$-cartesian. (The completeness of $$K$$ implies that $$V$$ is weakly $$K$$-cartesian (so that every $$K$$-subspace of $$V$$ is closed) and then the discreteness of the valuation on $$K$$ implies that every closed $$K$$-subspace of $$V$$ is strictly closed, so that $$V$$ is $$K$$-cartesian.)
2. If $$K^\sim$$ has zero characteristic, since in that case $$K$$ is stable [BGR 3.6.2/13]. (The main reason – which follows from Zariski-Samuel Vol 2, Chapter VI, Sections 11 (remark following the proof of Theorem 19) and 12 (corollary to Theorem 24) – is that in that case $$\sum_i e_if_i = n$$ for all finite extensions of $$K$$.)

# Notes (part 1): S. Bosch, U Güntzer and R. Remmert, Non-Archimedean Analysis

## Semi-normed and normed groups


• $$\nu(0) = \infty$$
• $$\nu(x-y) \geq \min\{\nu(x), \nu(y)\}$$

Filtrations are in one-to-one correspondence with ultrametric functions or semi-norms, i.e. functions $$|\cdot|: G \to \rnneg$$ such that

• $$|0| = 0$$
• $$|x-y| \leq \max\{|x|, |y|\}$$

A semi-norm $$|\ |$$ is called a norm if $$\ker(|\ |) = 0.$$ The correspondence between filtrations and semi-norms is given by

• $$|\cdot| = \alpha^{-\nu(\cdot)}$$
• $$\nu(\cdot) = -\log_\alpha|\cdot|$$

where $$\alpha$$ is any fixed real number greater than $$1.$$ There are natural subgroups \begin{align*} G^0 & := \{g: \nu(g) \geq 0\} = \{g: |g| \leq 1\} \\ G^{\vee} &:= \{g: \nu(g) > 0\} = \{g: |g| < 1\} \end{align*}

Elementary properties of a semi-norm $$|\cdot|$$:

• $$|-y| = |y|$$
• $$|x+y| \leq \max\{|x|, |y|\}$$
• $$|x+y| = \max\{|x|, |y|\}$$ if $$|x| \neq |y|$$

A semi-norm $$|\ |$$ defines a pseudometric topology on $$G$$ via the distance function
$d(x,y) := |x-y|$
This makes $$G$$ into a topological group, i.e. the group operation $$(x,y) \mapsto x – y$$ is continuous.

$$G$$ is Hausdorff if and only if $$|\ |$$ is a norm.

Proposition 1. Every open subgroup of a topological group is also closed.

Proof. For any $$g \in G$$, the map $$x \mapsto x + g$$ is continuous and has a continuous inverse, so that it is a homeomorphism. Now if $$H$$ is a subgroup of $$G$$ and $$g \not\in H,$$ then $$g + H \subseteq G \setminus H.$$ If $$H$$ is open, then it follows that $$g + H$$ is an open neighborhood of $$g$$ completely contained in the complement of $$H,$$ so that $$H$$ is also closed.

Given a point $$a$$ on a semi-normed group $$G$$ and $$r > 0,$$ let
\begin{align*} B^0(a,r) & := \{x \in G: |x – a| \leq r\} \\ B^{\vee} (a,r) & := \{x \in G: |x – a| < r\} \end{align*}
respectively be the closed and the open ball of radius $$r$$ centered at $$a.$$ Note that

• $$B^\vee(a,r)$$ is open and $$B^0(a,r)$$ is closed in $$G$$ by definition of the topology on $$G,$$
• $$B^0(a,r) = a + G^0(r)$$ and $$B^{\vee}(a,r) = a + G^{\vee}(r)$$ where
\begin{align*} G^0(r) & := \{x \in G: |x| \leq r\} \\ G^{\vee}(r) & := \{x \in G: |x| < r\} \end{align*}
• The ultrametric inequality implies that
• $$x + G^\vee(r) = G^\vee(r)$$ for each $$x \in G^\vee(r)$$
• $$x + G^\vee(r) \subseteq x + G^0(r) = G^0(r)$$ for each $$x \in G^0(r)$$

These observations immediately implies the following results:

Proposition 2. Every ball $$B^0(a,r)$$ and $$B^\vee(a,r)$$ of positive radius $$r$$ is both open and closed in $$G.$$ Every sphere $$S(a,r) := \{x \in G: |x-a| = r\}$$ of positive radius $$r$$ is both open and closed in $$G.$$

Corollary 3. Every normed group $$G$$ is totally disconnected (i.e. the path components are precisely the singletons) in the ultrametric topology.

Given a subgroup $$H$$ of a semi-normed group $$G$$ and $$a \in G,$$ the distance from $$a$$ to $$H$$ is by definition:
$|a, H| := \inf_{y \in H} |a + y|$
The following observation seems to be important:

Proposition 3 (Section 1.1.4, Proposition 2). Let $$G$$ be a normed group and $$H$$ be a subgroup of $$G$$ which is “$$\epsilon$$-dense” for some nonnegative real number $$\epsilon < 1$$ in the sense that for all $$g \in G$$ there is $$h \in H$$ such that $$|g+h| \leq \epsilon|g|.$$ Then $$H$$ is dense in $$G.$$

Proof. Since $$G$$ is normed, it suffices to show that $$|x,H| = 0$$ for all $$x \in H.$$ If $$\epsilon = 0$$ it is obvious. So assume $$\epsilon > 0$$ and, in order to proceed by contradiction, that $$|x,H| > 0.$$ Since $$\epsilon < 1,$$ there is $$h \in H$$ such that $$|x+h| < \epsilon^{-1}|x,H|.$$ Then by the $$\epsilon$$-density, there is $$h’ \in H$$ such that
$|x+h+h’| \leq \epsilon|x+h| < |x,H|$
which gives the required contradiction and proves the proposition.

A subgroup $$H$$ of a normed group $$G$$ is called strictly closed if for each $$a \in G$$ there is $$y \in H$$ such that $$|a,H| = |a + y|.$$ Since in a normed space the a point is in the closure of a subset if and only if its distance from the subset is zero, it follows that a strictly closed subgroup is also closed. The ultrametric inequality implies that the ball groups $$G^0(r)$$ and $$G^\vee(r)$$ are strictly closed: we can take $$y = a$$ if $$a \in H$$ and $$y = 0$$ if $$a \not\in H$$ (where $$H$$ is either $$G^0(r)$$ or $$G^\vee(r)$$.

Proposition 4 (Section 1.1.5, Proposition 4). Let $$G$$ be a normed group and $$H$$ be a closed subgroup of $$G$$ such that $$|H \setminus \{0\}|$$ is discrete in $$\rpos.$$ Then $$H$$ is strictly closed.

Proof. Given $$x \in G,$$ we need to produce $$y \in H$$ such that $$|x+y| = |x,H|.$$ We may assume $$x \not \in H.$$ Since $$H$$ is closed, it follows that $$\epsilon := |x,H| > 0.$$ Pick $$\delta > 0$$. It suffices to show that $$V := \{|x+y’|: y’ \in H,\ \epsilon <|x+y’| < \epsilon + \delta\}$$ is finite. Indeed, if $$y’$$ is as above, then since $$|x+y’| > |x,H|,$$ there is $$y” \in H$$ such that $$|x+y’| > |x+y”|.$$ But then $$|y’-y”| = |x+y’|,$$ so that $$V \subseteq |H \setminus \{0\}|.$$ The discreteness assumption on $$|H \setminus \{0\}|$$ then implies that $$V$$ is finite, as required.

The functor $$G \to G^\sim$$: $$G^\sim := G^0/G^\vee.$$

Convergence: in a complete normed group a sum $$\sum_{k \geq 0} a_k$$ turns out to be convergent if and only if $$|a_k| \to 0$$ as $$k \to \infty.$$ It then turns out that if $$\sum_k a_k$$ converges, then $$\sum_k a_{\pi_k}$$ converges to the same limit for every bijection $$\pi: \znneg \to \znneg.$$

## Semi-normed, normed and valued rings

A pair $$(A, |\ |)$$, where $$A$$ is a commutative ring with identity, is called a semi-normed (respectively, normed) ring if

1. $$|\ |$$ is a semi-norm (respectively, norm) on the additive group $$A^+$$ of $$A$$. (Note: $$A^+$$ is simply $$A$$ considered an abelian group with respect to $$+$$.)
2. $$|xy| \leq |x||y|$$ for each $$x, y \in A$$
3. $$|1| \leq 1$$

$$|\ |$$ is called a valuation and $$A$$ is called a valued ring if

• $$|\ |$$ is a norm on $$A,$$ and
• property 2 above is always satisfied with an equality.

Note that for a valuation, property 3 above is redundant: it follows from the second property with equality.

## Strictly convergent power series

Let $$(A, |\ |)$$ be a semi-normed ring. A formal power series $$\sum_{k \geq 0} a_kX^k \in A[[X]]$$ is strictly convergent if $$\lim_{k \to \infty} |a_k| = 0.$$ The set of all strictly convergent power series over $$A$$ is denoted by $$A\langle X \rangle.$$ The Gauss norm on $$A\langle X \rangle,$$ which we also denote as $$|\ |,$$ is defined as:
$|\sum_{k \geq 0} a_k X^k| := \max_k |a_k|$

## Normed and faithfully normed modules

Let $$(A, |\ |)$$ be a normed ring. A pair $$(M, |\ |)$$ is normed $$A$$-module if

1. $$(M, |\ |)$$ is a normed group (with respect to addition in $$M$$), and
2. $$|ax| \leq |a||x|$$ for all $$a \in A,$$ $$x \in M.$$

If in addition $$A$$ is a valued ring and the second property above always holds with equality, then we say that $$(M, |\ |)$$ is a faithfully normed $$A$$-module.

Remark. If $$M \neq 0,$$ then the condition that “$$A$$ is a valued ring” in the definition of a “faithful norm” is unnecessary: indeed picking $$x \neq 0$$ in $$M,$$ so that $$|x| > 0.$$ Now for all $$a_1, a_2 \in A,$$ the chain of equalities
$|a_1a_2||x| = |a_1a_2x| = |a_1||a_2x| = |a_1||a_2||x|$
implies that $|a_1a_2 = |a_1||a_2|$

## Examples

1. Non-closed ideals.
1. A (non-complete) Noetherian example. Let $$A := k[x]$$ where $$k$$ is a field equipped with a nontrivial valuation. Consider the topology on $$A$$ induced by the Gauss norm.
Lemma. For each $$a \in K,$$ the ideal $$\mmm_a := \langle x – a \rangle$$ is closed in $$A$$ if and only if $$|a| \leq 1.$$
Proof. Since $$1 – x^n/a^n \in \mmm_a$$ for $$a \in K \setminus 0,$$ it follows that $$|1, \mmm_a| \leq |a|^{-n}.$$ Therefore if $$|a| > 1,$$ then $$|1, \mmm_a| = 0,$$ so that $$\mmm_a$$ is not closed. On the other hand, if $$|a| \leq 1,$$ then we claim that
$|1 + (x-a)f| \geq 1$
for all $$f.$$ Indeed, write $$f = \sum_{j=0}^n c_jx^j$$ so that
$|1 + (x-a)f| = \max\{|1 – ac_0|, |c_0 – ac_1|, \ldots, |c_{n-1} – ac_n|, |c_n|\}$
If $$|1-ac_0| < |1| = 1,$$ then one must have $$|ac_0| = |1| = 1,$$ so that $$|c_0| = 1/|a| \geq 1.$$ Continuing with $$|c_0 – ac_1|$$ and so on, it follows that if each of the terms $$|c_0 – ac_1|, \ldots, |c_{n-1} – ac_n|$$ is smaller than $$1$$, then $$|c_n| = |a|^{-n}$$ so that $$|1 + (x-a)f| = |c_n| = |a|^{-n} \geq 1.$$ It follows that $$1$$ is not in the closure $$\bar \mmm_a$$ of $$\mmm_a.$$ Since $$\bar \mmm_a$$ is an ideal of $$A$$ and $$\mmm_a$$ is a maximal ideal of $$A,$$ this implies that $$\bar \mmm_a = \mmm_a,$$ as required.
2. Contrast the above with the following:
1. If $$A$$ is a complete normed ring, then every maximal ideal is closed. This follows since if a maximal ideal $$\mmm$$ is not closed, then its closure, being an ideal, must contain $$1,$$ i.e. there are elements in $$\mmm$$ arbitrarily close to $$1.$$ However, since $$A$$ is complete, any element of the form $$1 + f$$ with $$|f| < 1$$ is invertible.
2. If $$A$$ is complete Noetherian normed ring, then every ideal is closed, at least if $$A$$ contains a field $$k$$ such that the restriction of the norm on $$k$$ is nontrivial. Indeed, then $$A$$ is a $$k$$-Banach algebra. Let $$\hat \qqq$$ be the completion of an ideal $$\qqq.$$ By Noetherianity there is a surjective $$A$$-module map $$\pi: A^n \to \hat \qqq$$ which is open due to the Open Mapping Theorem. Then $$\pi((A^\vee)^n)$$ is open in $$\hat \qqq.$$ Since $$\qqq$$ is dense in $$\hat \qqq,$$ it follows that $$\hat \qqq = \qqq + \pi((A^\vee)^n).$$ Then $$\hat \qqq = \qqq$$ by Nakayama’s lemma [BGR 1.2.4/6]. Thus $$\qqq$$ is complete, and therefore closed.
3. A quasi-Noetherian but complete example. Let $$A := k[[x_i: i \geq 1]]$$ be the ring of power series ring in infinitely many variables over a field $$k.$$ Then $$A$$ is a local ring with maximal ideal $$m$$ consisting of power series with zero constant term. Let $$m_0$$ be the ideal of $$A$$ generated by all $$x_i,$$ $$i \geq 1.$$ Then $$m_0 \subsetneq m,$$ since e.g. the element
$x_1 + x_2^2 + x_3^3 + \cdots$
can not be expressed as an $$A$$-linear combination of finitely many $$x_i.$$ However, the closure of $$m_0$$ is $$m,$$ e.g. with respect to the topology induced by valuation corresponding to the order of power series. This remains true if the topology on $$A$$ is induced by a weighted order such the weight of $$x_i$$ goes to $$\infty$$ as $$i \to \infty,$$ even though in that case $$A$$ is quasi-Noetherian, i.e. each ideal $$I$$ of $$A$$ has a (possibly infinite) sequence of elements $$a_i,$$ $$i \geq 1,$$ such that $$\lim_{i \to \infty}|a_i| = 0,$$ and in addition, every $$a \in I$$ can be expressed as a possibly infinite $$A$$-linear combination of the $$a_i.$$
2. “Unexpected” behaviour of residues: equivalent norm with transcendental residue. Take a normed ring $$A$$ such that there is real number $$\rho > 1$$ such that $$\rho^{-1} \not\in |A|.$$ (E.g. one can take $$A$$ to be a discrete valuation ring.) Take $$B :=A\langle X \rangle.$$ Let $$|\ |_1$$ be the Gauss norm on $$B$$ and define $$|\ |_2$$ on $$B$$ as:
$|\sum a_k X^k| := \max\{|a_0|, \rho|a_k|: k \geq 1\}$
Then $$|\ |_2$$ is also a norm (one needs $$\rho \geq 1$$ for the ultametric inequality to hold) and
$|\cdot|_1 \leq |\cdot|_2 \leq \rho|\cdot|_1$
so that $$|\ |_1$$ and $$|\ |_2$$ induce the same topology on $$B.$$ However,
$B^\sim_1 \cong A^\sim[X],\ \text{whereas}\ B^\sim_2 \cong A^\sim$
so that $$B^\sim_1$$ is transcendental over $$B^\sim_2.$$
3. “Unexpected” behaviour of residues: trivial residue. Take a cyclic/principal faithfully normed $$A$$-module $$M := Ax$$ such that $$|x|^{-1} \not\in |A|.$$ Then $$M^\vee = M^0,$$ so that $$M^\sim = 0.$$
4. A discrete valuation ring $$A$$ which is not Japanese, i.e. the field of fractions of $$A$$ has a finite algebraic extension $$K$$ such that the integral closure of $$A$$ in $$K$$ is not a finite $$A$$-module [F. K. Schmidt, Math. Z. 41, 443-450, 1936].
Take a prime $$p.$$ Set $$k := (\zz/p\zz)(t_0, t_1, t_2, \ldots)$$ where all $$t_i$$ are indeterminates. Take a new indeterminate $$T,$$ define
$f(T) := t_0 + t_1T + t_2T^2 + \cdots \in k[[T]]$
The lemma below implies that $$f$$ is not algebraic over $$k[T]$$, so that the homomorphism $$k[X,Y] \to k[[T]]$$ defined by $$x \mapsto T,$$ $$y \mapsto f(T)$$ is injective. Let $$K := k(X,Y)$$ and $$|\ |_f$$ be the valuation on $$K$$ induced by the standard valuation on $$k[[T]].$$ Let $$Q := k(X, Y^p) \subseteq K$$ and $$A$$ be the (discrete) valuation ring of the restriction of $$|\ |_f$$ to $$Q,$$ i.e.
$A := Q^0 := \{g \in Q: |g|_f \leq 1\}$
The following claim shows that the integral closure $$\bar A$$ of $$A$$ in $$K$$ is not a finite $$A$$-module, as required. The lemma used above is stated and proved after the claim.
Claim. $$\bar A = K^0 := \{g \in K: |g|_f \leq 1\}.$$ Moreover, $$K^0$$ is not a finite $$A$$-module.
Proof. Since $$char(K) = p > 0,$$ every $$g \in K^0$$ is a $$p$$-th root of some element in $$A.$$ On the other hand, if $$g \in K$$ satisfies an integral equation
$g^n + a_1g^{n-1} + \cdots + a_n = 0$
with $$a_i \in A$$ then one must have $$|g|_f \leq 1.$$ It follows that $$\bar A = K^0.$$ To prove the second assertion proceed by contradiction and assume $$e_1, \ldots, e_n \in K^0$$ generate $$K^0$$ as an $$A$$-module. Since $$1, Y, \ldots, Y^{p-1}$$ generate $$K$$ as a vector space over $$Q$$, there are expressions
$e_i = \sum_{j=0}^{p-1}c_{ij}Y^j,\ c_{ij} \in Q$
Since $$|X|_f < 1,$$ there is $$s \geq 1$$ such that $$|c_{ij}X^s|_f \leq 1$$ for each $$i,j.$$ Then each $$X^se_i$$ is in the $$A$$-module $$B$$ generated by $$1, Y, \ldots, Y^{p-1}.$$ It follows that
$X^s K^0 \subseteq B$
Let
$Z_s := X^{-(s+1)}(Y – t_0 – t_1X – \cdots – t_sX^s) \in K.$
Then $$Z_s \in K^0$$ and therefore
$X^sZ_s = X^{-1}(Y-t_0) – (t_1 + t_2X + \cdots + t_sX^{s-1}) \in B$
Since $$t_1 + t_2X + \cdots + t_sX^{s-1} \in A \subseteq B,$$ it follows that
$X^{-1}(Y – t_0) \in B = A + AY + \cdots + AY^{p-1}$
But this is impossible since $$X^{-1} \not\in A,$$ and $$1, Y, \ldots, Y^{p-1}$$ is a vector space basis of $$K$$ over $$Q.$$ This contradiction finishes the proof of the claim.
Lemma. Let $$k$$ be a field and $$k_0$$ be the prime field of $$k.$$ Let $$f := \sum_i c_i T^i \in k[[T]],$$ where $$T$$ is an indeterminate, be such that $$k_0(c_0, c_1, c_2, \ldots)$$ has infinite transcendence degree over $$k_0.$$ Then $$f$$ is not algebraic over $$k[T].$$
Proof. Assume to the contrary that there is a nonzero polynomial $$q(T,W) \in K[T,W]$$ such that $$q(T, f(T)) = 0.$$ Let $$k’$$ be the subfield of $$k$$ generated by all coefficients of $$q(T,W).$$ We claim that each coefficient $$c_n$$ of $$f$$ is algebraic over $$k’.$$ We proceed by induction on $$n$$. Write $$q = T^sq_0$$ where $$T$$ does not divide $$q_0(T,W).$$ Then $$q_0(T, f(T)) = 0,$$ and in addition, $$q_0(0,W)$$ is a nonzero polynomial over $$k’.$$ Since $$q(0,f(0)) = q(0,c_0) = 0,$$ it follows that $$c_0$$ is algebraic over $$k’,$$ so that the claim holds for $$n = 0.$$ For $$n \geq 1,$$ define
$q'(T,W) := q(T, c_0 + c_1T + \cdots + c_{n-1}T^{n-1} + WT^n) \in k'(c_0, \ldots, c_{n-1})[T,W]$
Since $$\deg_W(q’) \geq 1$$ it is straightforward to check by looking at the highest degree term in $$W$$ that
$\deg_W(q’) = \deg_W(q) > 0.$
In particular, $$q’$$ is a nonzero element in $$k'(c_0, \ldots, c_{n-1})[T,W]$$ such that
$q'(T, c_n + c_{n+1}T + \cdots) = q(T, f(T)) = 0$
The $$n = 0$$ case of the claim then implies that $$c_n$$ is algebraic over $$k'(c_0, \ldots, c_{n-1}),$$ so that $$c_n$$ is algebraic over $$k’$$ due to the induction hypothesis. This proves the claim and consequently, the lemma.
5. A bounded (in particular, continuous) bijective $$A$$-module map may not have a continuous inverse. Pick a field $$K$$ with a non-complete valuation. Pick the completion $$\hat K$$ of $$K$$ and take $$x \in \hat K \setminus K.$$ The map $$\phi: K^2 \to V := K + Kx$$ given by $$(a,b) \mapsto a + bx$$ is a continuous (in fact, bounded) and bijective $$K$$-vector space map. However, $$K$$ is not closed in $$V$$ so that $$\phi$$ is not an homeomorphism; in particular
1. $$\phi^{-1}:V \to K^2$$ is a $$K$$-vector space isomorphism which is not continuous (let alone, bounded)! Indeed, choose $$a_s \in K$$ such that $$a_s \to x$$ as $$s \to \infty.$$ Then $$a_s – x \to 0$$ but $$|\phi^{-1}(a_s -x)| = |(a_s, -1)| \geq 1.$$
2. Contrast the above with the following:
1. Every $$K$$-linear map from $$K^n$$ to a $$K$$-vector space is continuous.
3. Also note that $$\phi^{-1}$$ is not continuous, even though $$\ker(\phi^{-1}) = \{0\}$$ is closed in $$V.$$ In particular, the following fact does not always hold if $$K$$ is replaced by $$K^n$$ for $$n \geq 2$$: given a $$K$$-linear map $$\lambda: V \to K,$$ where $$V$$ is a normed $$K$$-vector space, if $$\ker(\lambda)$$ is closed, then $$\lambda$$ is bounded. (Indeed, if $$x \in V \setminus \ker(\lambda),$$ then $$\alpha := |x, \ker(\lambda)| > 0,$$ and the $$K$$-faithfulness of the norm implies that $$|ax, \ker(\lambda)| = |a|\alpha$$ for all $$a \in K.$$ Then for all $$y \in V \setminus \ker(\lambda),$$ writing $$y = ax + z$$ with $$z \in \ker(\lambda)$$ one has that $$|y| \geq |a|\alpha$$ so that $$|\lambda(y)| = |a||x| \leq (|x|/\alpha)|y|.$$)
4. In this example, define $$\lambda: V \to K$$ as
$\lambda(a + bx) := b$
Then $$\ker(\lambda) = K$$ is not closed in $$V.$$ Does it imply that $$V \to V/\ker(\lambda)$$ is not continuous? Well, the question is more basic. What is the topology to put on $$V/\ker(\lambda)$$? Since $$\ker(\lambda) = K$$ is dense in $$V,$$ it follows that in the residue topology every point has zero norm! So in this topology the open sets are only the whole space and the empty set. So the (surjective) map $$V/\ker(\lambda) \to K$$ is not continuous with respect to that semi-norm.
6. For every ring $$A$$ equipped with a degenerate valuation. there are $$A$$-module maps which are homeomorphisms (i.e. in particular, continuous), but not bounded. Let $$M_i, := Ae_i$$ $$i = 1, 2, \ldots,$$ be a sequence of free cyclic $$A$$-modules. Set $$M := \bigoplus M_i.$$ For each $$m \in \zz,$$ define an $$A$$-module norm $$|\ |_m$$ on $$M$$ by
$| \sum_{i=1}^n a_ie_i|_m := \max\{i^{-m}|a_i|\}$
Since $$|\ |_m \leq |\ |_{m-1},$$ the identity map $$(M, |\ |_{m-1}) \to (M, |\ |_m)$$ is bounded and continuous. The inverse map
$\iota_m: (M, |\ |_{m}) \to (M, |\ |_{m-1})$
is not bounded (look at the norms of $$e_i$$ for $$i \gg 1$$). However, if $$A$$ is a valued ring with a degenerate valuation, then there is $$m$$ such that $$\iota_m$$ is continuous. Indeed, in that case there are two possibilities:
1. If $$|\ | \geq 1$$ on $$A \setminus \{0\}$$ then whenever $$m \leq 0,$$ one has $$|\ |_m \geq 1$$ on $$M \setminus \{0\},$$ so that $$|\ |_m$$ induces the discrete topology on $$M$$ and every map from $$(M, |\ |_m)$$ is continuous.
2. If $$|\ | \leq 1$$ on $$A,$$ then $$|a_i| \leq |a_i|^{1/m}$$ for $$m \geq 1.$$ So that for $$m \geq 2$$
$i^{-m}|a_i| < \epsilon^m \Rightarrow |a_i|^{1/m} < i\epsilon \Rightarrow |a_i| < i^{m-1} \epsilon \Rightarrow |a_ie_i|_{m-1} < \epsilon$
which implies that $$\iota_m$$ is continuous, as claimed.
7. Contrast the preceding example with the following fact [BGR, Proposition 2.1.8/2]: if $$A$$ is equipped with a non-degenerate valuation, then for $$A$$-linear maps between faithfully normed $$A$$-modules, continuity is equivalent to boundedness. The key observation is [BGR, Proposition 2.1.8/1] which says that if the valuation of $$A$$ is non-degenerate, then for each faithfully normed $$A$$-module $$M$$ there is a fixed $$\rho > 1$$ such that for each $$x \in M\setminus \{0\},$$ there is $$c \in A$$ such that $$1 \leq |cx| < \rho.$$

# Polynomial division over valued fields – Part I (Chan-Maclagan’s Algorithm for Homogeneous Divisors)

$$\DeclareMathOperator{\In}{In} \DeclareMathOperator{\Incoeff}{In_coeff} \DeclareMathOperator{\Inexp}{In_exp} \DeclareMathOperator{\ld}{Ld} \DeclareMathOperator{\qq}{\mathbb{Q}} \DeclareMathOperator{\kk}{\mathbb{K}} \DeclareMathOperator{\ord}{ord} \newcommand{\preceqeq}{\preceq_{\equiv}} \DeclareMathOperator{\rnpos}{\mathbb{R}^n_{> 0}} \DeclareMathOperator{\rnzero}{\mathbb{R}^n_{\geq 0}} \DeclareMathOperator{\rr}{\mathbb{R}} \DeclareMathOperator{\scrB}{\mathcal{B}} \DeclareMathOperator{\scrM}{\mathcal{M}} \DeclareMathOperator{\supp}{Supp} \DeclareMathOperator{\znplusonezero}{\mathbb{Z}^{n+1}_{\geq 0}} \DeclareMathOperator{\znplusonepos}{\mathbb{Z}^{n+1}_{> 0}} \DeclareMathOperator{\znpos}{\mathbb{Z}^n_{> 0}} \DeclareMathOperator{\znzero}{\mathbb{Z}^n_{\geq 0}} \DeclareMathOperator{\zz}{\mathbb{Z}}$$
Today we discuss division of polynomials over valued fields, i.e. fields equipped with a (real) valuation, and consider orderings on monomials which incorporate that valuation. The goal is to arrive at the “universal basis theorem” which states that for every ideal there is a finite set of polynomials which determine the initial ideal with respect to all linear orders. As we have seen, the universal basis theorem for linear orders is harder to establish than that for monomial orders. The trend continues, and the case of linear orders over valued fields is even harder. In fact we do not even get to the universal basis theorem in this post – it will be covered in the next one. In this post we simply describe the division algorithm, and why/how it is different from the case of non-valued fields.

Example 1 [Chan-Maclagan2013]. Consider the $$p$$-adic valuation $$\nu_p$$ on $$\qq,$$ where $$p$$ is a prime number, and $$\nu_p$$ is defined on $$\qq$$ as: $\nu_p(p^qa/b) := q$ where $$a, b$$ are integers relatively prime to $$p$$, and $$q$$ is an arbitrary integer. This defines a $$\zz^3$$-grading on $$S := \qq[x,y]$$ with homogeneous components $S_{(q,\alpha, \beta)} := \{ax^\alpha y^\beta: a \in \qq,\ \nu_p(a) = q\}$Take any group order on $$\zz^2$$ and define an ordering $$\preceq’$$ on monomial terms of $$S$$ as follows: $$ax^\alpha y^\beta \preceq’ a’x^{\alpha’}y^{\beta’}$$ if and only if

• either $$\nu_p(a) < \nu_p(b),$$ or
• $$\nu_p(a) = \nu_p(b)$$ and $$(\alpha, \beta) \preceq (\alpha’, \beta’)$$

Note that $$\preceq’$$ is not a total order on the set of monomial terms on $$S,$$ since there are monomial terms $$T \neq T’$$ such that $$T \preceq’ T’$$ and $$T’ \preceq’ T,$$ e.g. take $$T := x$$ and $$T’ := qx$$ where $$q$$ is any integer relatively prime to $$p.$$ Nonetheless, we can try to follow the naive initial-term-cancelling division from the preceding post on $$S$$ with respect to $$\preceq’.$$ So take $$g_1 := x – py,$$ $$g_2 := y – px$$ and try dividing $$f := x$$ by $$g_1, g_2$$ following the naive initial-term-cancelling division strategy. The initial terms of $$g_1, g_2$$ with respect to $$\preceq’$$ are respectively $$x$$ and $$y.$$ It follows that

• after the first step one has: $$f = g_1 + py,$$
• after the second step one has: $$f = g_1 + pg_2 + p^2x,$$
• after the third step one has: $$f = (1+p^2)g_1 + pg_2 + p^3y,$$
• after the fourth step one has: $$f = (1+p^2)g_1 + (p+p^3)g_2 + p^4x,$$

and so on. It is clear that the process does not terminate. On the other hand, if we recognize the remainder $$p^2x$$ in the second step as $$p^2f,$$ then rearranging the terms in the second step one has: $f = \frac{1}{1-p^2}g_1 + \frac{p}{1-p^2}g_2$which seems to be a reasonable outcome for a division process. Chan and Maclagan [Chan-Maclagan2013] showed that this always works.

## Almost linear orders

An almost linear order on the set $$\scrM := \{ax^\alpha: a \in \kk \setminus \{0\},\ \alpha := (\alpha_1, \ldots, \alpha_n) \in \znzero\}$$ of monomial terms is a reflexive and transitive binary relation on $$\scrM$$ which is linear on all monomials with different exponents, i.e. every finite set of monomials with pairwise distinct exponents has a unique minimal element. Every linear order on $$\znzero$$ trivially lifts to a almost linear order on monomial terms which simply ignores the coefficients and applies the original linear order on the exponents. The relation $$\preceq’$$ from Example 1 is an example of a nontrivial almost linear order.

The notions of initial and leading monomial terms make sense for an almost linear order $$\preceq$$ on monomial terms, and therefore every step of the “initial term cancelling division” also makes sense. Recall that the initial term cancelling division of a polynomial $$f$$ by a finite ordered collection $$g_1, \ldots, g_N$$ of polynomials proceeds as follows:

1. Set $$h_0 := f$$
2. If $$h_j = 0$$ for some $$j \geq 0,$$ then stop.
3. Otherwise if there is $$i$$ such that $$\In_\preceq(g_i)$$ divides $$\In_\preceq(h_j)$$, then pick the smallest such $$i$$, and set $$h_{j+1} := h_j – qg_i$$, where $$q := \In_\preceq(h_j)/\In_\preceq(g_i)$$.
4. Otherwise $$\In_\preceq(h_j)$$ is not divisible by $$\In_\preceq(g_i)$$ for any $$i$$; in this case set $$h_{j+1} := h_j – \In_\preceq(h_j).$$
5. Repeat steps 2-4 with $$j + 1$$ in place of $$j.$$

Example 1 shows that for a general almost linear order $$\preceq$$ the above algorithm may not end even if the divisors are homogeneous with respect to a shallow monomial grading. However, it also suggests a way forward, namely to use remainders from earlier steps of the division. To fix notation, write $$r_j$$ for the remainder and $$q_{j,i}$$ to be the multiplier of $$g_i$$ after $$j$$ iterations of the algorithm. After the $$j$$-th iteration one has $f = \sum_i q_{j,i}g_i + r_j = \sum_i q_{j,i}g_i + r_{j,0} + r_{j,1}$where

• $$r_{j,1}$$ is the part of remainder that has already been computed; in particular, none of the monomials in the support of $$r_{j,1}$$ is divisible by $$\In(g_i)$$ for any $$i = 1, \ldots, s.$$
• $$r_{j,0}$$ is the part of the remainder “yet to be explored,” i.e. it is the $$h_j$$ from the above algorithm.

Let $$a_j \in \kk$$ and $$\alpha_j \in \znzero$$ be respectively the coefficient and the exponent of $$\In_\preceq(r_{j,0}),$$ i.e. $\In_\preceq(r_{j,0}) = a_j x^{\alpha_j}$(in the sequel we use $$\Incoeff(r_{j,0})$$ and $$\Inexp(r_{j,0})$$ to denote respectively $$a_j$$ and $$\alpha_j.$$) Let $$r’_{j,0} := r_{j,0} – \In_\preceq(r_{j,0}).$$ For the $$(j+1)$$-th iteration, Example 1 suggests that it is useful to look out for $$k < j$$ such that $$\alpha_k = \alpha_j.$$ Indeed, if such $$k$$ exists, then writing $$c_j := a_j/a_k,$$ the above equation for $$f$$ can be rearranged as
\begin{align*} f &= \sum_i q_{j,i}g_i + a_jx^{\alpha_j} + r’_{j,0} + r_{j,1} = \sum_i q_{j,i}g_i + c_j\In_\preceq(r_{k,0}) + r’_{j,0} + r_{j,1} \\ &= \sum_i q_{j,i}g_i + c_j(r_{k,0} – r’_{k,0}) + r’_{j,0} + r_{j,1} = \sum_i q_{j,i}g_i + c_j(f – \sum_i q_{k,i}g_i – r_{k,1} – r’_{k,0}) + r’_{j,0} + r_{j,1} \\ &= c_jf + \sum_i (q_{j,i} – c_j q_{k,i})g_i + (r’_{j,0} – c_jr’_{k,0}) + (r_{j,1} – c_jr_{k,1}) \end{align*}
If $$c_j \neq 1,$$ then it follows that
\begin{align} f &= \sum_i \frac{q_{j,i} – c_jq_{k,i}}{1-c_j}g_i + \frac{r’_{j,0} – c_jr’_{k,0}}{1-c_j} + \frac{r_{j,1} – c_jr_{k,1}}{1-c_j} \end{align}

## Almost linear monomial orders

We say that $$\preceq$$ is an almost linear group order or an almost linear monomial order if it is an almost linear order on the set of monomial terms which is also compatible with respect to product and sum of monomial terms in the following way:

1. if $$ax^\alpha \preceq bx^\beta,$$ then $$acx^{\alpha+\gamma} \preceq bcx^{\beta+\gamma}$$ for all $$c \in \kk\setminus\{0\}$$ and $$\gamma \in \znzero,$$
2. $$1 \preceq -1$$ and $$-1 \preceq 1,$$
3. for all $$a, b \in \kk,$$ either $$a \preceq (a+b)$$ or $$b \preceq (a+b).$$

At this point it is useful to introduce some notation: we will use $$\prec$$ to denote “strict ordering” with respect to $$\preceq,$$ and write $$\preceqeq$$ to denote “equivalence” with respect to $$\preceq,$$ i.e.

• $$a \prec b$$ if and only if $$a \preceq b$$ and $$b \not\preceq a$$.
• $$a \preceqeq b$$ if and only if $$a \preceq b$$ and $$b \preceq a$$,

Property 2 of almost linear group orders then can be rewritten as

• $$-1 \preceqeq 1.$$

Note that Property 1 implies the following seemingly stronger statement:

• if $$ax^\alpha \preceq bx^\beta,$$ and $$cx^\gamma \preceq dx^\delta,$$ then $$acx^{\alpha+\gamma} \preceq bdx^{\beta+\delta}$$ for all $$c,d \in \kk\setminus\{0\}$$ and $$\gamma, \delta \in \znpos.$$

Property 1 and 2 then imply that

• $$ax^\alpha \preceq bx^\beta$$ if and only if $$-ax^\alpha \preceq bx^\beta$$ if and only if $$ax^\alpha \preceq -bx^\beta$$ if and only if $$-ax^\alpha \preceq -bx^\beta.$$

Also due to Property 2 and 3, $$\preceq$$ can be consistently extended to $$\scrM \cup \{0\}$$ via declaring

• $$ax^\alpha \preceq 0$$ for all $$a \in \kk$$ and $$\alpha \in \znzero.$$

In presence of Properties 1 and 2, Property 3 implies that

• $$\preceq$$ restricts to a total binary relation on $$\kk,$$ i.e. for all $$a, b \in \kk,$$ either $$a \preceq b$$ or $$b \preceq a$$ (or both). Indeed, assume without loss of generality that $$a\preceq a+b.$$ Since $$b = (a+b) – a,$$ it follows that $$a+b \preceq b$$ or $$-a \preceq b.$$ Both of these possibilities imply that $$a \preceq b.$$
• If $$a \prec b,$$ then $$a \preceqeq a+b.$$ Indeed, Property 3 immediately implies that $$a \preceq a+b.$$ On the other hand, since $$a = (a+b) – b,$$ and $$b \not\preceq a,$$ we must have that $$a+b \preceq a.$$

If $$\preceq$$ is indeed an almost linear group order, then at each step of the initial-term-cancelling division algorithm, the initial term of $$h_j$$ strictly increases with respect to $$\preceq$$, i.e.
$\In_\preceq(h_j) \prec \In_\preceq(h_{j+1})$
It follows that in the last displayed equation for $$f$$ above $$c_j$$ can not be $$1,$$ so that the equation is well defined. In fact, it is straightforward to check that $$1 \prec c_j$$. The defining properties of almost linear group orders then imply that
$1 – c_j \preceqeq 1$
which in turn implies that
$\In_\preceq(r_{j,0}) \prec \In_\preceq(r’_{j,0} – c_jr’_{k,0}) \preceqeq \In_\preceq\left(\frac{r’_{j,0} – c_jr’_{k,0}}{1-c_j}\right)$
Consequently, if we set
$r_{j+1,0} := \frac{r’_{j,0} – c_jr’_{k,0}}{1-c_j}$
then
$\In_\preceq(r_{j,0}) \prec \In_\preceq(r_{j+1,0})$
We now rewrite the initial-term-cancelling-division algorithm for almost linear group orders to incorporate these observations. Recall that the aim of this algorithm is to find an expression$f = \sum_i q_ig_i + r$such that

• either $$r = 0$$ or no monomial term in $$r$$ is divisible by $$\In_\preceq(g_i)$$ for any $$i$$, and
• there is a reordering $$i_1, \ldots, i_s$$ of $$\{i: q_i \neq 0\}$$ such that $\In_\preceq(\sum_j q_jg_j) = \In_\preceq(q_{i_1}g_{i_1}) \prec \In_\preceq(q_{i_2}g_{i_2}) \prec \cdots \prec \In_\preceq(q_{i_s}g_{i_s})$

Each step of the algorithm below expresses $$f$$ as$f = \sum_j q_{j,i}g_i + r_{j,0} + r_{j,1}$where $$j$$ is the step, $$r_{j,0}$$ is the “part” of $$f$$ “still to be explored” and $$r_{j,1}$$ is the part of the remainder that has been computed. In particular, either $$r_{j,1} = 0$$ or no monomial term in $$r_{j,1}$$ is divisible by $$\In_\preceq(g_i)$$ for any $$i$$. The algorithm stops when $$r_{j,0} = 0.$$

Initial term cancelling division, version 2: Given a polynomial $$f$$ (the dividend), a finite ordered collection $$g_1, \ldots, g_N$$ of polynomials (the divisors) and an almost linear monomial order $$\preceq,$$ proceed as follows: start with $$j = 0.$$ Set $$q_{0,i} := 0$$ for all $$i,$$ $$r_{0, 0} := f$$ and $$r_{0, 1} := 0.$$ Now proceed as follows:

1. If $$r_{j,0} = 0,$$ then stop.
2. Otherwise check if there is $$k < j$$ such that
$\Inexp(r_{k,0}) = \Inexp(r_{j,0})$
If such $$k$$ exists, then necessarily one has
$c_j := \Incoeff(r_{j,0})/\Incoeff(r_{k,0}) \succ 1$
Then define
\begin{align*} q_{j+1,i} &:= \frac{q_{j,i} – c_jq_{k,i}}{1-c_j},\ i = 1, \ldots, N, \\ r_{j+1,0} &:= \frac{r’_{j, 0} – c_jr’_{k,0}}{1-c_j}, \\ r_{j+1,1} &:= \frac{r_{j, 1} – c_jr_{k,1}}{1-c_j} \end{align*}
where
$r’_{j,0} := r_{j,0} – \In_\preceq(r_{j,0})$
and $$r’_{k,0}$$ is defined similarly. Note that there is an ambiguity at this step in choosing $$k$$ if there are more than one choices. Below we will define a criterion for choosing $$k$$.
3. Otherwise if there is $$i$$ such that $$\In_\preceq(g_i)$$ divides $$\In_\preceq(r_{j,0})$$, then pick the smallest such $$i$$, and with $$q := \In_\preceq(r_{j,0})/\In_\preceq(g_i),$$ define
\begin{align*} r_{j+1,0} &:= r_{j,0} – qg_i, \\ r_{j+1,1} &:= r_{j,1}, \\ q_{j+1,k} &:= \begin{cases} q_{j,i} + q &\text{if}\ k = i,\\ q_{j,i} & \text{otherwise.} \end{cases} \end{align*}
4. Otherwise set
\begin{align*} r_{j+1,0} &:= r_{j,0}, \\ r_{j+1,1} &:= r_{j,1} + \In_\preceq(r_{j,0}), \\ q_{j+1,i} &:= q_{j,i},\ i = 1, \ldots, N. \end{align*}
5. Repeat steps 2-4 with $$j + 1$$ in place of $$j.$$

Earlier discussions on division via initial terms suggest that it might be prudent to consider the special case that the divisors $$g_1, \ldots, g_N$$ are homogeneous with respect to a shallow monomial grading $$\Sigma.$$ In that case, if the condition is step 2 is not satisfied at any iteration, then we claim that the algorithm must stop in finitely many steps. Indeed, if the condition in step 2 is not satisfied, then each time $$r_{j,0}$$ gets updated in step 3, a monomial with different exponent is added to the support of $$r_{j,0}.$$ Since $$g_i$$ are homogeneous with respect to $$\Sigma$$ the $$\Sigma$$-supports of all $$r_{j,k}$$ are contained in the $$\Sigma$$-support of $$f$$ and therefore due to the shallowness, the number of monomials in $$r_{j,0}$$ can grow indefinitely. Consequently $$r_{j,0}$$ must remain unchanged after a finite number of steps. But then step 4 can not repeat infinitely many times, since that would result in the number of monomials in $$r_{j,1}$$ getting indefinitely larger. Therefore, if the algorithm is to continue without stopping, the condition in step 2 must be satisfied at infinitely many iterations. Now comes a crucial observation of Chan and Maclagan [Chan-Maclagan2013]:

Criterion to pick $$k$$ in Step 2: In Step 2, if there are more than one choice for $$k < j$$ such that $$\Inexp(r_{k,0}) = \Inexp(r_{j,0}),$$ then choose any $$k$$ such that the number of monomials in $$r_{k,0}$$ which are not in $$r_{j,0}$$ is the smallest possible. Actually the only thing you really need to ensure stoppage in finitely many steps is: if it is possible, then $$k$$ should be chosen such that $$\supp(r_{k,0}) \subseteq \supp(r_{j,0}).$$

Theorem 1. If $$g_1, \ldots, g_N$$ are homogeneous with respect to a shallow monomial grading $$\Sigma$$ and the above criterion is used to pick $$k$$ in Step 2 of the initial term cancelling division by $$g_1, \ldots, g_N$$ with respect to an almost linear monomial order $$\preceq,$$ then the division algorithm stops in finitely many steps. The outcome of the algorithm satisfies the “aim” of the algorithm (described in the paragraphs preceding the version 2 of the division algorithm).

Proof. Since the $$g_i$$ are homogeneous, $$\supp_\Sigma(r_{j,0}) \subseteq \supp_\Sigma(f)$$ for each $$j.$$ Since the grading is shallow, there are finitely many possible choices for the usual support $$\supp(r_{j,0}) \subseteq \znzero$$ of $$r_{j,0}$$ consisting of the exponents of the monomials appearing in $$r_{j,0}.$$ Therefore there must be $$j_0$$ such that for each $$j \geq j_0,$$ if $$r_{j,0} \neq 0,$$ then there is already $$k < j$$ such that $$\supp(r_{k,0}) = \supp(r_{j,0}).$$ But then for each $$j \geq j_0,$$ if $$r_{j,0} \neq 0,$$ the Step 2 condition is satisfied at each step, and the above criterion for picking $$k$$ ensures that $$\supp(r_{k,0}) \subseteq \supp(r_{j,0})$$ so that the size of the support of $$r_{j+1,0}$$ decreases compared to that of $$r_{j,0}.$$ Since the support of a polynomial is finite, $$r_{j,0}$$ must become zero in finitely many steps beyond $$j_0.$$ To see that the “aim” is satisfied, note that the condition regarding the remainder is clear. For the condition regarding the quotients, it suffices to show that for each $$i, j,$$ either $$q_{j,i} = 0$$ or $$\In_\preceq(q_{j+1,i}) = \In_\preceq(q_{j,i}).$$ This is clear in Steps 3 and 4. For Step 2, this follows from the observation that $$c_j \succ 1$$ and, either $$q_{k,i} = 0$$ or $$\In_\preceq(q_{j,i}) = \In_\preceq(q_{k,i}).$$ This completes the proof of Theorem 1.

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 almost linear monomial order $$\preceq.$$

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 almost linear monomial orders $$\preceq_1, \preceq_2$$ 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 Chan-Maclagan 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 the Chan-Maclagan 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 earlier post on division of power series also applies to an almost linear monomial order: 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 an almost linear 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 $$1.$$

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 Chan-Maclagan 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 almost linear monomial orders 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 in the first post on polynomial divisions via initial terms 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.

Theorem 2 (Universal Bases – version 1). 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 almost linear monomial orders. In particular, there is a finite subset of $$\Sigma$$-homogeneous elements of $$I$$ such that for every almost linear monomial order, 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 earlier 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.

# 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 $$\zz^n.$$

# 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.

# Polynomial division and Universal bases

## Polynomial division

$$\DeclareMathOperator{\In}{In} \DeclareMathOperator{\ld}{Ld} \DeclareMathOperator{\kk}{\mathbb{K}} \DeclareMathOperator{\rnpos}{\mathbb{R}^n_{> 0}} \DeclareMathOperator{\rnzero}{\mathbb{R}^n_{\geq 0}} \DeclareMathOperator{\rr}{\mathbb{R}} \DeclareMathOperator{\scrB}{\mathcal{B}} \DeclareMathOperator{\scrI}{\mathcal{I}} \DeclareMathOperator{\scrJ}{\mathcal{J}} \DeclareMathOperator{\supp}{Supp} \DeclareMathOperator{\znpos}{\mathbb{Z}^n_{> 0}} \DeclareMathOperator{\znzero}{\mathbb{Z}^n_{\geq 0}} \DeclareMathOperator{\zz}{\mathbb{Z}}$$In this post we talk about division with respect to polynomials in more than one variables, which is a pretty cute algorithm that changed the face of a big part of mathematics, via e.g. Gröbner bases which we will also talk about. Dividing by polynomials in a single variable is straightforward – to divide $$f$$ by $$g$$, you pick a monomial $$m$$ such that the highest degree terms of $$f$$ and $$mg$$ are equal, then repeat the same process with $$f – mg$$ in place of $$f$$. Since the degree, as a nonnegative integer, can not drop infinitely many times, this algorithm stops after a finitely many steps, and you end up with polynomials $$q$$ (the quotient) and $$r$$ (the remainder) such that $$f = qg + r$$ and $$\deg(r) < \deg(f)$$. To divide by polynomials in more than one variables, you need an analogue of the “highest degree monomial term” of the single-variable case. A natural resolution comes via monomial orders.

A monomial order is a binary relation $$\preceq$$ on $$\znzero$$ such that

1. $$\preceq$$ is a total order,
2. $$\preceq$$ is compatible with the addition on $$\znzero$$, i.e. $$\alpha \preceq \beta$$ implies $$\alpha + \gamma \preceq \beta + \gamma$$ for all $$\gamma \in \znzero$$, and
3. $$0 \preceq \alpha$$ for all $$\alpha \in \znzero$$.

It turns out that in the presence of the first two conditions, the third one is equivalent to the following condition:

• $$\preceq$$ is a well order on $$\znzero$$

(For one direction: if $$\alpha \preceq 0$$ for some $$\alpha \in \znzero \setminus \{0\}$$, then $$\{\alpha, 2\alpha, 3\alpha, \ldots \}$$ is an infinite set with no minimal element with respect to $$\preceq$$. See [Mon21, Corollary B.47] for an elementary proof for the opposite direction.)

The support $$\supp(f)$$ of a polynomial $$f \in \kk[x_1, \ldots, x_n]$$, where $$\kk$$ is a field, is the set of all $$\alpha \in \znzero$$ such that $$x^\alpha$$ appears in $$f$$ with a nonzero coefficient. We write $$\delta_\preceq(f)$$ for the maximal element of $$\supp(f)$$ with respect to $$\preceq$$. The leading term $$\ld_\preceq(f)$$ of $$f$$ with respect to $$\preceq$$ is the monomial term $$cx^\alpha$$ where $$\alpha = \delta_\preceq(f)$$ and $$c$$ is the coefficient of $$x^\alpha$$ in $$f$$. The division algorithm in $$\kk[x_1, \ldots, x_n]$$ is the procedure for reducing a polynomial $$f$$ by a finite ordered collection of polynomials $$g_1, \ldots, g_N$$ as follows:

1. If there is $$i$$ such that $$\ld_\preceq(g_i)$$ divides $$\ld_\preceq(f)$$, then pick the smallest such $$i$$, and replace $$f$$ by $$f – hg_i$$, where $$h := \ld_\preceq(f)/\ld_\preceq(g_i)$$.
2. Otherwise $$\ld_\preceq(f)$$ is not divisible by $$\ld_\preceq(g_i)$$ for any $$i$$; in this case replace $$f$$ by $$f – \ld_\preceq(f)$$.
3. Repeat.

Since $$\delta_\preceq(f)$$ decreases at every step of the division algorithm, the well ordering property of $$\preceq$$ implies that the algorithm stops after finitely many steps. It is clear that after the algorithm stops, one has $f = \sum_i h_i g_i + r$ where either $$r = 0$$, or for every $$i$$, $$\ld_\preceq(g_i)$$ does not divide any term of $$r$$; we say that $$r$$ is the remainder of $$f$$ modulo division by the $$g_i$$. This immediately implies the following result:

Theorem 1. Let $$g_1, \ldots, g_N$$ be elements of an ideal $$I$$ of $$\kk[x_1, \ldots, x_n]$$ such that $$\ld_\preceq(g_1), \ldots, \ld_\preceq(g_N)$$ generate the leading ideal $$\ld_\preceq(I)$$ generated by $$\{\ld_\preceq(f): f \in I\}$$. Then for all $$f \in \kk[x_1, \ldots, x_n],$$ the following are equivalent:

• $$f \in I$$,
• the remainder of $$f$$ modulo $$g_1, \ldots, g_N$$ is zero.

A Gröbner basis of $$I$$ with respect to a monomial ordering $$\preceq$$ is simply a finite set of elements $$g_1, \ldots, g_N \in I$$ such that $$\ld_\preceq(g_i),$$ $$1 \leq i \leq N,$$ generate $$\ld_\preceq(I)$$.

For later use we record a few corollaries of the division algorithm.

Corollary 1. Let $$I \supseteq J$$ be ideals of $$\kk[x_1, \ldots, x_n]$$. If $$I \supsetneqq J$$, then $$\ld_\preceq(I) \supsetneqq \ld_\preceq(J)$$ for every monomial order $$\preceq$$.

Proof. If there are $$g_1, \ldots, g_N \in J$$ such that $$\ld_\preceq(g_i)$$, $$i = 1, \ldots, N$$, generate $$\ld_\preceq(I)$$, then Theorem 1 implies that $$g_1, \ldots, g_N$$ generate $$I$$, so that $$I = J$$.

Corollary 2. Let $$I$$ be an ideal of $$\kk[x_1, \ldots, x_n]$$. There can not exist monomial orders $$\preceq_1, \preceq_2$$ such that $$\ld_{\preceq_1}(I) \subsetneqq \ld_{\preceq_2}(I)$$.

Proof. Write $$J_k := \ld_{\preceq_k}(I)$$, $$k = 1, 2$$. Assume to the contrary that $$J_1 \subsetneqq J_2$$. Choose $$g_1, \ldots, g_N \in I$$ such that $$\ld_{\preceq_2}(g_i)$$, $$i = 1, \ldots, N$$, generate $$J_1$$. Choose $$f \in I$$ such that $$\ld_{\preceq_2}(f) \not\in J_1$$. Let $$r$$ be the remainder of the 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 $$h_1, \ldots, h_M \in I$$ are such that $$\ld_{\preceq_1}(h_1), \ldots, \ld_{\preceq_1}(h_M)$$ generate $$J_1$$, then division of $$r$$ by $$h_1, \ldots, h_M$$ with respect to $$\preceq_1$$ implies that $$r \not\in I$$, which is a contradiction.

Starting from a Gröbner basis of an ideal, one can discard “redundant” elements (whose leading terms are divisible by the leading term of some other basis element) and then replace each remaining element by the remainder of the division by the others to obtain a reduced Gröbner basis, i.e. a Gröbner basis such that

1. no monomial term of any basis element is divisible by the leading term of any other basis element, and
2. the leading coefficient, i.e. the coefficient of the leading term, of each basis element is $$1.$$

Corollary 3. The reduced Gröbner basis is uniquely determined up to a permutation by the leading ideal. In particular,

1. up to a permutation of the basis elements there is a unique reduced Gröbner basis for every monomial order, and
2. if an ideal has the same leading ideal for two monomial orders, then it also has the same reduced Gröbner basis (up to permutations) for those monomial orders.

Proof. Indeed, fix two reduced Gröbner bases $$g_1, \ldots, g_N$$ and $$h_1, \ldots, h_M$$ of an ideal $$I$$ with respect to a monomial order $$\preceq.$$ For each $$h_j$$, its leading term is divisible by the leading term of at least one $$g_i,$$ but there can not be more than one such $$i,$$ since then

• either the leading term of both such $$g_i$$ would be divisible by $$h_j,$$
• or there would be $$j’ \neq j$$ such that the leading term of $$h_{j’}$$ divides the leading term of one of the $$g_i$$, and therefore in turn divides the leading term of $$h_j$$

Each of these possibilities contradicts the reducedness of one of the above Gröbner bases. It follows that for each $$j \in \{1, \ldots, M\},$$ there is a unique $$i \in \{1, \ldots, N\}$$ such that $$\ld_\preceq(g_i)$$ divides $$\ld_\preceq(h_j).$$ Similarly for each \{i \in \{1, \ldots, N\},\} there is a unique \{j \in \{1, \ldots, M\}\) such that $$\ld_\preceq(h_j)$$ divides $$\ld_\preceq(g_i).$$ It follows that $$M = N$$ and after reordering the $$g_i$$ and the $$h_j$$ if necessary, one can ensure that $$\ld_\preceq(g_i) = \ld_\preceq(h_i),$$ $$i = 1, \ldots, N.$$ Now fix $$i$$ and divide $$h_i$$ by $$g_1, \ldots, g_N.$$ Since for $$j \neq i$$ the leading term of $$g_j$$ does not divide any monomial term of $$h_i,$$ after the first step the remainder is $$h_i – g_i,$$ and if $$h_i – g_i \neq 0,$$ then at the second step the leading term of $$h_i – g_i$$ must be divisible by the leading term of $$g_i.$$ But this is impossible, since the exponent of each monomial term of $$h_i – g_i$$ is smaller than $$\delta_\preceq(g_i) = \delta_\preceq(h_i).$$ It follows that $$h_i = g_i,$$ which proves the first assertion. For the second assertion, pick a monomial order $$\preceq’$$ such that $$\ld_{\preceq’}(I) = \ld_\preceq(I).$$ Let $$\alpha_i := \delta_{\preceq}(g_i),$$ $$i = 1, \ldots, n.$$ Fix $$i.$$ Since $$\ld_{\preceq’}(I)$$ is generated by $$x^{\alpha_1}, \ldots, x^{\alpha_N}$$, it follows that $$\ld_{\preceq’}(g_i)$$ 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 $$\delta_{\preceq’}(g_i) = \alpha_i.$$ It follows that $$g_1, \ldots, g_N$$ form a Gröbner basis of $$I$$ with respect to $$\preceq’$$ as well. It is clear that this is a reduced Gröbner basis, as required.

## Universal bases for Leading ideals

In this section we prove that as $$\preceq$$ varies over all possible monomial orders, there can be only finitely many leading ideals for a given ideal of a polynomial ring. The argument is elementary and very cute. I learned it from [Stu96, Chapter 1]. Mora and Robbiano [MoRobbiano88, Page 193] write that it is due to Alessandro Logar.

Theorem 2. Every ideal $$I$$ of $$\kk[x_1, \ldots, x_n]$$ has finitely many distinct leading ideals. In particular, there is a finite subset of $$I$$ such that for every monomial order, the leading terms of polynomials in that subset generate the leading ideal of $$I.$$

Proof. Assume to the contrary that there is an infinite collection $$\scrJ_0$$ of distinct leading ideals of $$I$$. Pick an arbitrary nonzero element $$f_1 \in I$$. Out of the finitely many monomial terms of $$f_1$$, 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 $$J_1 = \ld_{\preceq_1}(I)$$. Since $$m_1 \in J_1$$, there is $$g_1 \in I$$ such that $$\ld_{\preceq_1}(g_1) = m_1$$, and since $$\langle m_1 \rangle \neq J_1$$, there is $$h_1 \in I$$ such that $$\ld_{\preceq_1}(h_1) \not\in \langle m_1 \rangle.$$ 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 $$f_2$$ has finitely many monomial terms, there must be a monomial term $$m_2$$ of $$f_2$$ 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 $$J_2 = \ld_{\preceq_2}(I)$$. As in the preceding case, there are $$g_{21}, g_{22}, h_2 \in I$$ such that $$\ld_{\preceq_2}(g_{21}) = m_1$$, $$\ld_{\preceq_2}(g_{22}) = m_2$$, and $$\ld_{\preceq_2}(h_2) \not\in \langle m_1, m_2 \rangle.$$ 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.

A universal (leading) basis for an ideal $$I$$ of $$\kk[x_1, \ldots, x_n]$$ is an ordered collection $$\scrB$$ of elements in $$I$$ such that for every monomial order $$\preceq$$ on $$\znzero$$, the leading ideal of $$I$$ with respect to $$\preceq$$ is generated by the leading terms of the elements in $$\scrB$$. The above theorem implies that every ideal has a finite universal basis.

## Every leading ideal comes from a positive weighted degree

Weighted degrees are natural generalizations of degree. Every $$\omega \in \rr^n$$ defines a weighted degree, and for $$f \in \kk[x_1, \ldots, x_n]$$ we write $$\ld_\omega(f)$$ for the corresponding leading term of $$f$$. Given an ideal $$I$$ of $$\kk[x_1, \ldots, x_n]$$ we similarly write $$\ld_\omega(I)$$ for its leading ideal generated by $$\ld_\omega(f)$$ for all $$f \in I$$. In this section, also following [Stu96, Chapter 1], we prove the following result:

Theorem 3. For every ideal $$I$$ of $$\kk[x_1, \ldots, x_n]$$ and every monomial order $$\preceq$$, there is $$\omega \in \znpos$$ such that $$\ld_\omega(I) = \ld_\preceq(I)$$ (in particular, this means that $$\ld_\omega(I)$$ is a monomial ideal).

Remark. With a bit more work it can be more can be established (the proof is completely analogous to the proof of Theorem 3 from the next post): given a monomial order $$\preceq$$ and polynomials $$g_1, \ldots, g_N ,$$ there is $$\omega \in \znpos$$ such that $\ld_\omega(g_i) = \ld_\preceq(g_i),\ i = 1, \ldots, N.$If in addition the $$g_i$$ are elements of an ideal $$I$$ such that $$\ld_\preceq(g_i),$$ $$i = 1, \ldots, N,$$ generate $$\ld_\preceq(I),$$ then$\ld_\omega(I) = \langle \ld_\omega(g_1), \ldots, \ld_\omega(g_N)\rangle = \langle \ld_\preceq(g_1), \ldots, \ld_\preceq(g_N)\rangle = \ld_\preceq(I)$Moreover, every $$f \in I$$ has an expression of the form $$f = \sum_i f_ig_i$$ such that

1. $$\delta_\preceq(f) = \max_\preceq\{\delta_\preceq(f_ig_i)\},$$ and
2. $$\omega(f) = \max\{\omega(f_ig_i)\}.$$

Proof of Theorem 3. Fix a Gröbner basis $$g_1, \ldots, g_N$$ of $$I$$ with respect to $$\preceq$$. Write $g_i = c_{i0} x^{\alpha_{i0}} + \cdots + c_{iN_i}x^{\alpha_{iN_i}}$with $$\alpha_{i0} = \delta_\preceq(g_i)$$. Define $C := \{\omega \in \rnzero: \langle \omega, \alpha_{i0} – \alpha_{ij} \rangle \geq 0,\ i = 1, \ldots, N,\ j = 1, \ldots, N_i\}$Note that $$C$$ is a convex polyhedral cone. We claim that $$\dim(C) = n$$. Indeed, $$C$$ is dual to the cone $$C’$$ generated by $$\alpha_{i0} – \alpha_{ij},\ i = 1, \ldots, N,\ j = 1, \ldots, N_i$$, 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 $\sum_{i,j}\lambda_{ij}(\alpha_{i0} – \alpha_{ij}) = -\sum_k r_ke_k$for some nonnegative integers $$r_1, \ldots, r_n$$. Due to the second and third properties of a monomial order, it would follow that $\sum_{i,j}\lambda_{ij} \alpha_{i0} \preceq \sum_{i,j}\lambda_{ij} \alpha_{ij}$On the other hand, by construction $$\alpha_{i0} \succ \alpha_{ij}$$ for each $$i,j$$, so that $\sum_{i,j}\lambda_{ij} \alpha_{i0} \succ \sum_{i,j}\lambda_{ij} \alpha_{ij}$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 $$\ld_\omega(g_i) = c_{i0}x^{\alpha_{i0}}$$. Since $$\ld_\preceq(I)$$ is generated by $$c_{i0}x^{\alpha_{i0}}$$, $$i = 1, \ldots, N,$$ it follows that $$\ld_\omega(I) \supseteq \ld_\preceq(I)$$. If this containment were proper, then Corollary 1 to Theorem 1 would imply that $\ld_\preceq(\ld_\omega(I)) \supsetneqq \ld_\preceq(\ld_\preceq(I)) = \ld_\preceq(I)$On the other hand, it is straightforward to check that $\ld_\preceq(\ld_\omega(I)) = \ld_{\preceq_\omega}(I)$ 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, and taken toegther, the last two displays above contradict Corollary 2 to Theorem 1. Consequently, $$\ld_\omega(I) = \ld_\preceq(I)$$, as required.

# Lüroth’s theorem (a “constructive” proof)

Lüroth’s theorem (Lüroth 1876 for $$k = \mathbb{C}$$, Steinitz 1910 in general). If $$k \subseteq K$$ are fields such that $$k \subseteq K \subseteq k(x)$$, where $$x$$ is an indeterminate over $$k$$, then $$K = k(g)$$ for some rational function $$g$$ of $$x$$ over $$k$$.

I am going to present a “constructive” proof of Lüroth’s theorem due to Netto (1895) that I learned from Schinzel’s Selected Topics on Polynomials (and give some applications to criteria for proper polynomial parametrizations). The proof uses the following result which I am not going to prove here:

Proposition (with the set up of Lüroth’s theorem). $$K$$ is finitely generated over $$k$$, i.e. there are finitely many rational functions $$g_1, \ldots, g_s \in k(x)$$ such that $$K = k(g_1, \ldots, g_s)$$.

The proof is constructive in the following sense: given $$g_1, \ldots, g_s$$ as in the proposition, it gives an algorithm to determine $$g$$ such that $$K = k(g)$$. We use the following notation in the proof: given a rational function $$h \in k(x)$$, if $$h = h_1/h_2$$ with polynomials $$h_1, h_2 \in k[x]$$ with $$\gcd(h_1, h_2) = 1$$, then we define $$\deg_\max(h) := \max\{\deg(h_1), \deg(h_2)\}$$.

## Proof of Lüroth’s theorem

It suffices to consider the case that $$K \neq k$$. Pick $$g_1, \ldots, g_s$$ as in the proposition. Write $$g_i = F_i/G_i$$, where

• $$\gcd(F_i, G_i) = 1$$ (Property 1).

Without loss of generality (i.e. discarding $$g_i \in k$$ or replacing $$g_i$$ by $$1/(g_i + a_i)$$ for appropriate $$a_i \in k$$ if necessary) we can also ensure that

• $$\deg(F_i) > 0$$ and $$\deg(F_i) > \deg(G_i)$$ (Property 2).

Consider the polynomials $H_i := F_i(t) – g_iG_i(t) \in K[t] \subset k(x)[t], i = 1, \ldots, s,$ where $$t$$ is a new indeterminate. Let $$H$$ be the greatest common divisor of $$H_1, \ldots, H_s$$ in $$k(x)[t]$$ which is also monic in $$t$$. Since the Euclidean algorithm for computing $$\gcd$$ respects the field of definition, it follows that:

• $$H$$ is also the greatest common divisor of $$H_1, \ldots, H_s$$ in $$K[t]$$, which means, if $$H = \sum_j h_j t^j$$, then each $$h_j \in K$$ (Property 3).

Let $$H^* \in k[x,t]$$ be the polynomial obtained by “clearing the denominator” of $$H$$; in other words, $$H = H^*/h(x)$$ for some polynomial $$h \in k[x]$$ and $$H^*$$ is primitive as a polynomial in $$t$$ (i.e. the greatest common divisor in $$k[x]$$ of the coefficients in $$H^*$$ of powers of $$t$$ is 1). By Gauss’s lemma, $$H^*$$ divides $$H^*_i := F_i(t)G_i(x) – F_i(x)G_i(t)$$ in $$k[x,t]$$, i.e. there is $$Q_i \in k[x,t]$$ such that $$H^*_i = H^* Q_i$$.

Claim 1. If $$\deg_t(H^*) < \deg_t(H^*_i)$$, then $$\deg_x(Q_i) > 0$$.

Proof of Claim 1. Assume $$\deg_t(H^*) < \deg_t(H^*_i)$$. Then $$\deg_t(Q_i) > 1$$. If in addition $$\deg_x(Q_i) = 0$$, then we can write $$Q_i(t)$$ for $$Q_i$$. Let $$F_i(t) \equiv \tilde F_i(t) \mod Q_i(t)$$ and $$G_i(t) \equiv \tilde G_i(t) \mod Q_i(t)$$ with $$\deg(\tilde F_i) < \deg(Q_i)$$ and $$\deg(\tilde G_i) < \deg(Q_i)$$. Then $$\tilde F_i(t)G_i(x) – F_i(x) \tilde G_i(t) \equiv 0 \mod Q_i(t)$$. Comparing degrees in $$t$$, we have $$\tilde F_i(t)G_i(x) = F_i(x) \tilde G_i(t)$$. It is straightforward to check that this contradicts Propeties1 and 2 above, and completes the proof of Claim 1.

Let $$m := \min\{\deg_\max(g_i): i = 1, \ldots, s\}$$, and pick $$i$$ such that $$\deg_\max(g_i) = m$$. Property 2 above implies that $$\deg_t(H^*_i) = \deg_x(H^*_i) = m$$. If $$\deg_t(H^*) < m$$, then Claim 1 implies that $$\deg_x(H^*) < \deg_x(H^*_i) = m$$. If the $$h_j$$ are as in Property 3 above, it follows that $$\deg_\max(h_j) < m$$ for each $$j$$. Since $$H^* \not\in k[t]$$ (e.g. since $$t-x$$ divides each $$H_i$$), there must be at least one $$h_j \not \in k$$. Since adding that $$h_j$$ to the list of the $$g_i$$ decreases the value of $$m$$, it follows that the following algorithm must stop:

### Algorithm

• Step 1: Pick $$g_i := F_i/G_i$$, $$i = 1, \ldots, s$$, satisfying properties 1 and 2 above.
• Step 2: Compute the monic (with respect to $$t$$) $$\gcd$$ of $$F_i(t) – g_i G_i(t)$$, $$i = 1, \ldots, s$$, in $$k(x)[t]$$; call it $$H$$.
• Step 3: Write $$H = \sum_j h_j(x) t^j$$. Then each $$h_j \in k(g_1, \ldots, g_s)$$. If $$\deg_t(H) < \min\{\deg_\max(g_i): i = 1, \ldots, s\}$$, then adjoin all (or, at least one) of the $$h_j$$ such that $$h_j \not\in k$$ to the list of the $$g_i$$ (possibly after an appropriate transformation to ensure Property 2), and repeat.

After the last step of the algorithm, $$H$$ must be one of the $$H_i$$, in other words, there is $$\nu$$ such that $\gcd(F_i(t) – g_i G_i(t): i = 1, \ldots, s) = F_{\nu}(t) – g_{\nu}G_{\nu}(t).$

Claim 2. $$K = k(g_{\nu})$$.

Proof of Claim 2 (and last step of the proof of Lüroth’s theorem). For a given $$i$$, polynomial division in $$k(g_\nu)[t]$$ gives $$P, Q \in k(g_\nu)[t]$$ such that $F_i(t) = (F_{\nu}(t) – g_{\nu}G_{\nu}(t))P + Q,$ where $$\deg_t(Q) < \deg_t(F_{\nu}(t) – g_{\nu}G_{\nu}(t))$$. If $$Q = 0$$, then $$F_i(t) = (F_{\nu}(t) – g_{\nu}G_{\nu}(t))P$$, and clearing out the denominator (with respect to $$k[g_\nu]$$) of $$P$$ gives an identity of the form $$F_i(t)p(g_\nu) = (F_{\nu}(t) – g_{\nu}G_{\nu}(t))P^* \in k[g_\nu, t]$$ which is impossible, since $$F_{\nu}(t) – g_{\nu}G_{\nu}(t)$$ does not factor in $$k[g_\nu, t]$$. Therefore $$Q \neq 0$$. Similarly, $G_i(t) = (F_{\nu}(t) – g_{\nu}G_{\nu}(t))R + S,$ where $$R, S \in k(g_\nu)[t]$$, $$S \neq 0$$, and $$\deg_t(S) < \deg_t(F_{\nu}(t) – g_{\nu}G_{\nu}(t))$$. It follows that $F_i(t) – g_iG_i(t) = (F_{\nu}(t) – g_{\nu}G_{\nu}(t))(P – g_iR) + Q – g_iS.$ Since $$F_{\nu}(t) – g_{\nu}G_{\nu}(t)$$ divides $$F_{i}(t) – g_{i}G_{i}(t)$$ in $$k(x)[t]$$ and since $$\deg_t(Q – g_iS) < \deg_t(F_{\nu}(t) – g_{\nu}G_{\nu}(t))$$, it follows that $$Q = g_iS$$. Taking the leading coefficients (with respect to $$t$$) $$q_0, s_0 \in k(g_\nu)$$ of $$Q$$ and $$S$$ gives that $$g_i = q_0/s_0 \in k(g_\nu)$$, as required to complete the proof.

## Applications

The following question seems to be interesting (geometrically, it asks when a given polynomial parametrization of a rational affine plane curve is proper).

Question 1. Let $$k$$ be a field and $$x$$ be an indeterminate over $$k$$ and $$g_1, g_2 \in k[x]$$. When is $$k(g_1, g_2) = k(x)$$?

We now give a sufficient condition for the equality in Question 1. Note that the proof is elementary: it does not use Lüroth’s theorem, only follows the steps of the above proof in a special case.

Corollary 1. In the set up of Question 1, let $$d_i := \deg(g_i)$$, $$i = 1, 2$$. If the $$\gcd$$ of $$x^{d_1} – 1, x^{d_2} – 1$$ in $$k[x]$$ is $$x – 1$$, then $k(g_1, g_2) = k(t)$. In particular, if $$d_1, d_2$$ are relatively prime and the characteristic of $$k$$ is either zero or greater than both $$d_1, d_2$$, then $k(g_1, g_2) = k(x)$.

Remark. Corollary 1 is true without the restriction on characteristics, i.e. the following holds: “if $$d_1, d_2$$ are relatively prime, then $k(g_1, g_2) = k(x)$.” François Brunault (in a comment to one of my questions on MathOverflow) provided the following simple one line proof: $$[k(x): k(g_1, g_2)]$$ divides both $$[k(x): k(g_i)] = d_i$$, and therefore must be $$1$$.

My original proof of Corollary 1. Following the algorithm from the above proof of Lüroth’s theorem, let $$H_i := g_i(t) – g_i(x)$$, $$i = 1, 2$$, and $$H \in k(x)[t]$$ be the monic (with respect to $$t$$) greatest common divisor of $$H_1, H_2$$.

Claim 1.1. $$H = t – x$$.

Proof. It is clear that $$t-x$$ divides $$H$$ in $$k(x)[t]$$, so that $$H(x,t) = (t-x)h_1(x,t)/h_2(x)$$ for some $$h_1(x,t) \in k[x,t]$$ and $$h_2(x) \in k[x]$$. It follows that there is $$Q_i(x,t) \in k[x,t]$$ and $$P_i(x) \in k[x]$$ such that $$H_i(x,t)P_i(x)h_2(x) = (t-x)h_1(x,t)Q_i(x,t)$$. Since $$h_2(x)$$ and $$(t-x)h_1(x,t)$$ have no common factor, it follows that $$h_2(x)$$ divides $$Q_i(x,t)$$, and after cancelling $$h_2(x)$$ from both sides, one can write $H_i(x,t)P_i(x) = (t-x)h_1(x,t)Q’_i(x,t),\ i = 1, 2.$ Taking the leading form of both sides with respect to the usual degree on $$k[x,t]$$, we have that $(t^{d_i} – x^{d_i})x^{p_i} = a_i(t-x)\mathrm{ld}(h_1)\mathrm{ld}(Q’_i)$ where $$a_i \in k \setminus \{0\}$$ and $$\mathrm{ld}(\cdot)$$ is the leading form with respect to the usual degree on $$k[x,t]$$. Since $$\gcd(x^{d_1} – 1, x^{d_2} – 1) = x – 1$$, it follows that $$\mathrm{ld}(h_1)$$ does not have any factor common with $$t^{d_i} – x^{d_i}$$, and consequently, $$t^{d_i} – x^{d_i}$$ divides $$(t-x)\mathrm{ld}(Q’_i)$$. In particular, $$\deg_t(Q’_i) = d_i – 1$$. But then $$\deg_t(h_1) = 0$$. Since $$H = (t-x)h_1(x)/h_2(x)$$ is monic in $$t$$, it follows that $$H = t – x$$, which proves Claim 1.1.

Since both $$H_i$$ are elements of $$k(g_1, g_2)[t]$$, and since the Euclidean algorithm to compute $$\gcd$$ of polynomials (in a single variable over a field) preserves the field of definition, it follows that $$H \in k(g_1, g_2)[t]$$ as well (this is precisely the observation of Property 3 from the above proof of Lüroth’s theorem). Consequently $$x \in k(g_1, g_2)$$, as required to prove Corollary 1.

## References

• Andrzej Schinzel, Selected Topics on Polynomials, The University of Michigan Press, 1982.

# Math Research

## Overview

Starting from my PhD thesis Towards a Bezout-type Theory of Affine Varieties written at University of Toronto under the supervision of Pierre Milman, my research in math falls under two broad themes:

• Compactification of affine varieties
• Affine Bézout problem

I am in particular interested in one of the simplest cases of the first problem:

• Compactification of $$\mathbb{C}^2$$

## Papers

### Compactification of $$\mathbb{C}^2$$

#### Preprints

• Normal equivariant compactifications of $$\mathbb{G}^2_a$$ of Picard rank one, arXiv:1610.03563.
• Mori dream surfaces associated with curves with one place at infinity, arXiv:1312.2168.
• Analytic compactifications of $$\mathbb{C}^2$$ part II – one irreducible curve at infinity, arXiv:1307.5577.

### Affine Bézout problem

#### Preprints

• Intersection multiplicity, Milnor number and Bernstein’s theorem, arXiv:1607.04860

# Simplest singularity on non-algebraic normal Moishezon surfaces

A classical question in complex analytic geometry is to understand when a given analytic space is algebraic (i.e. analytiﬁcation of an algebraic scheme). A necessary condition for this to hold is that the transcendence degree of the ﬁeld of global meromorphic functions must be equal to the dimension of the space, i.e. the space has to be Moishezon. For dimension 2, it is a classical result that it is also suﬃcient, provided the space is nonsingular (Chow and Kodaira, 1952).

In general it is not clear how to determine algebraicity of normal (singular) Moishezon surfaces and our understanding of non-algebraic Moishezon surfaces, more precisely what prevents them from being algebraic, remains incomplete (Schröer [Sch00] gives a necessary and suﬃcient for algebraicity, but it is not very suitable for computation in a given case). We [Mon16] gave an example of a non-algebraic normal Moishezon surface $$X$$ which has the simplest possible singularity in the following sense: $$X$$ has only one singular point $$P$$, and the singularity at $$P$$

1. has multiplicity 2 and geometric genus 1,
2. is almost rational in the sense of [Ném07], and
3. is a Gorenstein hypersurface singularity which is minimally elliptic (in the sense of [Lau77]).

The claim that singularity of $$X$$ is the simplest possible is based on combining the preceding facts with the following observations:

• a Moishezon surface whose singularities are rational (i.e. with geometric genus zero) is algebraic [Art62], and
• minimally elliptic Gorenstein singularities form in a sense the simplest class of non-rational singularities.

The weighted dual graph of the resolution of singularity of $$X$$ at $$P$$ is of type $$D_{9,∗,0}$$ in the classiﬁcation of [Lau77] and the self-intersection number of its fundamental divisor is −2. It follows from [Lau77, Table 2] that the singularity at the origin of $$z^2 = x^5 + xy^5$$ is also of the same type.

## References

• [Art62] Michael Artin, Some numerical criteria for contractability of curves on algebraic surfaces, Amer. J. Math., 84:485–496,
• [Lau77] Henry B. Laufer, On minimally elliptic singularities, Amer. J. Math., 99(6):1257–1295, 1977.
• [Mon16] Pinaki Mondal, Algebraicity of normal analytic compactiﬁcations of $$\mathbb{C}^2$$ with one irreducible curve at inﬁnity, Algebra & Number Theory, 10(8), 2016.
• [Ném07] András Némethi, Graded roots and singularities, In Singularities in geometry and topology, pages 394–463. World Sci. Publ., Hackensack, NJ, 2007.
• [Sch00] Stefan Schröer, On contractible curves on normal surfaces, J. Reine Angew. Math., 524:1–15, 2000.