Loose edges and factorization theorems

TL;DR

This paper investigates the factorization properties of elements in regular local rings with specific geometric features in their Newton polyhedra, establishing conditions under which such elements factorize in the ring's completion.

Contribution

It introduces a new factorization criterion based on loose edges of Newton polyhedra and symbolic restrictions, extending understanding of factorization in regular local rings.

Findings

Factorization in the m-adic completion when the Newton polyhedron has a loose edge.

Coprimality of polynomials on the edge implies factorization.

Conditions linking geometric properties of Newton polyhedra to algebraic factorization.

Abstract

Let $ R $ be a regular local ring with maximal ideal $ \mathfrak{m} $. We consider elements $ f \in R $ such that their Newton polyhedron has a loose edge. We show that if the symbolic restriction of $f$ to such an edge is a product of two coprime polynomials, then $f$ factorizes in the $ \mathfrak{m} $-adic completion.

Equations

$$f=\sum _{\alpha \in {\mathbb{Z}}_{\ge 0}^n}{a}_{\alpha }{\underline{x}}^{\alpha },$$

$${\Delta }^{\xi }:=\{a\in \Delta :⟨\xi ,a⟩=\underset{b\in \Delta }{\min }⟨\xi ,b⟩\}$$

$$f{|}_A=\sum _{\alpha \in A}{\overline{a}}_{\alpha }{\underline{X}}^{\alpha }\in \mathbb{K}[X_1,\dots ,X_n],$$

$$\psi (\xi )=\underset{v\in V\setminus \{a,b\}}{\min }⟨\xi ,v⟩-⟨\xi ,a⟩,\xi \in {\mathbb{R}}_{\ge 0}^n.$$

$$\psi ({\xi }_0)>0.$$

$$\psi ({\xi }_1)<0.$$

$$\underset{i\in A}{\min }{c}_i/a_i+\underset{j\in B}{\min }{c}_j/b_j\ge 1.$$

$$\dim R_{w+z}=\dim R_w+\dim R_z-1.$$

$$l_v=\{\alpha \in {\mathbb{R}}_{\ge 0}^n:(⟨{\xi }_1,\alpha ⟩,\dots ,⟨{\xi }_{n-1},\alpha ⟩)=v\}.$$

$$G{R}_{z+i}+H{R}_{w+i}=R_{w+z+i}.$$

$$0⟶R_i\stackrel{\Phi }{⟶}{R}_{z+i}\times R_{w+i}\stackrel{\Psi }{⟶}{R}_{w+z+i},$$

$$g_w{h}_{z+i}+h_z{g}_{w+i}=f_{w+z+i}-F_i,$$

$$F_i=\sum _{\begin{array}{c}k+l=i,\\ k,l<i\end{array}}{g}_{w+k}{h}_{z+l}.$$

$$G{R}_{z_j+i}+H_j{R}_{w+i}=R_{w+z_j+i}$$

$$G{R}_{z_1+z_2+i}+H_1{H}_2{R}_{w+i}=R_{w+z_1+z_2+i}.$$

$$G{R}_{z_1+z_2+i}+H_1{H}_2{R}_{w+i}$$

$$G{R}_{z+i}+H{R}_{w+i}=R_{w+z+i}.$$

$$\omega (h):=\inf \{\omega ({\underline{x}}^{\alpha }):b_{\alpha }\ne 0\}.$$

$${\mathrm{gr}}_{\omega }(R):=\underset{w\in {\mathbb{Z}}_{\ge 0}^{n-1}}{⨁}{\mathcal{P}}_w/{\mathcal{P}}_w^{+},$$

$$g:=\sum _{\alpha \in {\mathbb{Z}}_{\ge 0}^n}{b}_{\alpha }{x}^{\alpha }\in R,$$

$${\varphi }_2:=(g_w+g_{w+i_1})(h_z+h_{z+i_1})=g_w{h}_z+(g_w{h}_{z+i_1}+h_z{g}_{w+i_1})+g_{w+i_1}{h}_{z+i_1}.$$

$$f=x^6{y}^2-z^4+xy{z}^4-x^7{y}^5{z}^2\in R.$$

$$f{|}_E=X^6{Y}^2-Z^4=(X^{3}Y-Z^2)(X^{3}Y+Z^2)\in \mathbb{K}[X,Y,Z]\cong {\mathrm{gr}}_{\omega }(R),$$

$$f=(x^{3}y\sqrt{1-x{y}^3{z}^2}-z^2\sqrt{1-xy})(x^{3}y\sqrt{1-x{y}^3{z}^2}+z^2\sqrt{1-xy})\in \hat{R}.$$

$$f=xyz+x^3{y}^3+x^3{z}^3+y^3{z}^3.$$

$$f=y^8+(x_1^3-x_2^2)y^3+x_1^5{x}_2^4{y}^2-x_1^{15}{x}_2^{18}\in R[y].$$

$$f{|}_E=X_1^5{X}_2^4{Y}^2-X_1^{15}{X}_2^{18}=X_1^5{X}_2^4(Y-X_1^5{X}_2^7)(Y+X_1^5{X}_2^7)\in \mathbb{K}[X_1,X_2,Y],$$

$$f=y^3+270y+540\in R[y].$$

$$F(x,y)=F_{a+b}(x,y)+F_{a+b+1}(x,y)+\dots ,$$

$$f,g\in \mathbb{K}[[x,y]],f=f_a+f_{a+1}+\dots ,g=g_b+g_{b+1}+\dots$$

Full text

The third author is supported by the DFG-project “Order zeta functions and resolutions of singularities” (DFG project number: 373111162)

1. Introduction

Notation. We denote by $\mathbb{R}_{\geq 0}$ (respectively $\mathbb{R}_{>0}$) the set of nonnegative (respectively positive) real numbers. The symbol $\langle\cdot,\cdot\rangle$ denotes the standard scalar product. We use a multi-index notation $\underline{x}^{\alpha}:=x_{1}^{\alpha_{1}}\cdots x_{n}^{\alpha_{n}}$ for $\alpha=(\alpha_{1},\dots,\alpha_{n})$. The ring of formal power series of variables $x_{1}$,…, $x_{n}$ with coefficients in a field $\mathbb{K}$ will be denoted by $\mathbb{K}[\![x_{1},\dots,x_{n}]\!]$. By $\mathbb{K}[\underline{x}^{\pm}]$ we mean the ring $\mathbb{K}[x_{1},x_{1}^{-1},\dots,x_{n},x_{n}^{-1}]$.

Let $R$ be a regular local ring with maximal ideal $\mathfrak{m}$ and residue field $\mathbb{K}=R/\mathfrak{m}$. Let $\widehat{R}$ be its $\mathfrak{m}$-adic completion. (Recall that $\widehat{R}=R$ if $R$ is complete). For example, this is the case $R=\mathbb{K}[\![x_{1},\dots,x_{n}]\!]$. The main results of the article are factorization theorems for elements in $R$, resp. for polynomials with coefficients in $R$, using convex geometry.

Let $(\underline{x})=(x_{1},\ldots,x_{n})$ be a regular system of parameters for $R$. As shown in [3, Proposition 2.1] (applied for $J=\{1,\ldots,n\}$), since $R$ is Noetherian and the map $R\subset\widehat{R}$ is faithfully flat, every element $f\in R$ can be expanded as a finite sum

[TABLE]

for $a_{\alpha}\in R^{\times}\cup\{0\}$. Here, the symbol $R^{\times}$ denotes the set of units in $R$.

Let $f=\sum_{\alpha\in\mathbb{Z}_{\geq 0}^{n}}a_{\alpha}\underline{x}^{\alpha}\in R$ be a nonzero element in $R$. We define the Newton polyhedron $\Delta(f)$ of $f$ as the convex hull of the set $\{\alpha:a_{\alpha}\neq 0\}+\mathbb{R}_{\geq 0}^{n}$. A set $\Delta\subset\mathbb{R}_{\geq 0}^{n}$ is called a Newton polyhedron if $\Delta=\Delta(f)$ for some nonzero $f\in~{}R$.

Given a Newton polyhedron $\Delta\subset\mathbb{R}^{n}_{\geq 0}$, for any $\xi\in\mathbb{R}_{\geq 0}^{n}$ we call the set

[TABLE]

a face of $\Delta$. A Newton polyhedron has a finite number of faces since it is defined by a finite set of exponents. A face $\Delta^{\xi}$ is compact if and only if $\xi\in\mathbb{R}_{>0}^{n}$. A face of dimension 0 (respectively 1) is called a vertex (respectively an edge). Following [9, p. 123], we call a compact edge of a Newton polyhedron a loose edge if it is not contained in any compact face of dimension $\geq 2$.

Several Newton polyhedra are drawn in the pictures that follow. The segments marked in blue are loose edges.

Fig. 1

Fig. 2

Fig. 3

The Newton polyhedron in Figure 1 does not have any loose edge. This is the typical situation. The Newton polyhedron in Figure 2 has a loose edge with the end point at $(0,0,d)$. Weierstrass polynomials $f\in\mathbb{K}[\![x_{1},\dots,x_{n}]\!][z]$ such that $\Delta(f)$ is of this type were studied in [1], [10], and [11].

In Figure 3 all compact edges are loose. A Newton polyhedron with this property is called in [5] a polygonal Newton polyhedron. Notice that the term polygonal Newton polyhedron can be a bit misleading since the union of compact edges in Figure 3 is not homeomomorphic to any polygon.

Every compact edge of a plane Newton polyhedron is loose as illustrated in Figure 4.

Fig. 4

Consider $f=\sum_{\alpha\in\mathbb{Z}_{\geq 0}^{n}}a_{\alpha}\underline{x}^{\alpha}\in R$. The symbolic restriction of $f$ to a compact face $A:=\Delta^{\xi}\subset\Delta(f)$, for some $\xi\in\mathbb{R}^{n}_{>0}$, is defined as the polynomial

[TABLE]

where $\overline{a}_{\alpha}\in\mathbb{K}$ is the image of $a_{\alpha}$ in the residue field of $R$. The element $f|_{A}$ is a polynomial since $A$ being a compact face implies that $A\cap\mathbb{Z}_{\geq 0}^{n}$ is finite. In section 3, we provide a connection to a certain graded ring which explains the use of capital letters $(X_{1},\ldots,X_{n})$.

Below are the main results of the paper.

Theorem 1.1**.**

Let $R$ be a regular local ring and let $f\in R$ be a nonzero element in $R$. Assume that the Newton polyhedron $\Delta(f)$ has a loose edge $E$. If $f|_{E}$ is a product of two relatively prime polynomials $G$ and $H$, where $G$ is not divided by any variable, then there exist $g$, $h\in\widehat{R}$ in the completion of $R$ such that $f=gh$ in $\widehat{R}$ and $g|_{E_{1}}=G$, $h|_{E_{2}}=H$ for some $E_{1}$, $E_{2}$ such that $E$ is the Minkowski sum of $E_{1}$ and $E_{2}$, $E=E_{1}+E_{2}$.

In the above theorem $E_{1}$ is a loose edge of $\Delta(g)$ parallel to $E$ and $E_{2}$ is a compact face of $\Delta(h)$ which is a loose edge parallel to $E$ or a vertex if $H$ is a monomial.

Remark 1.2**.**

The assumption of Theorem 1.1 that $G$ is not divisible by any variable cannot be omitted. Consider the power series $f=(x_{3}^{2}+x_{1}x_{2})(x_{3}+x_{1}x_{2})$. Its Newton polyhedron has a loose edge $E$ with endpoints $(1,1,1)$, $(2,2,0)$ and $f|_{E}=x_{1}x_{2}(x_{3}+x_{1}x_{2})$. The irreducible factors of $f$ are $f_{1}=x_{3}^{2}+x_{1}x_{2}$ and $f_{2}=x_{3}+x_{1}x_{2}$. Hence $f$ cannot be a product of power series $g$, $h$ such that $g|_{E_{1}}=x_{2}(x_{3}+x_{1}x_{2})$ and $h|_{E_{2}}=x_{1}$.

Corollary 1.3**.**

Assume that the Newton polyhedron of $f\in R$ has a loose edge and at least three vertices. Then $f$ is not irreducible in the completion $\widehat{R}$.

Corollary 1.4**.**

Assume that the Newton polyhedron of $f\in R$ has a loose edge $E$. If $f$ is irreducible in $\widehat{R}$, then $E$ is the only compact edge of $\Delta(f)$ and $f|_{E}=cF^{k}$, where $c\in\mathbb{K}^{\times}$ and $F\in\mathbb{K}[X_{1},\ldots,X_{n}]$ is an irreducible polynomial (with notations as before). Moreover, if the residue field $\mathbb{K}$ of $R$ is algebraically closed, then there exist $a,b\in\mathbb{K},k\in\mathbb{Z}_{\geq 0},\alpha,\beta\in\mathbb{Z}_{\geq 0}^{n}$ such that $f|_{E}=(a\underline{X}^{\alpha}+b\underline{X}^{\beta})^{k}$ with a primitive lattice vector $\alpha-\beta$.

We will say that a segment $E\subset\mathbb{R}^{n+1}$ is descendant if $E$ is parallel to some vector $c=(c_{1},\dots,c_{n},c_{n+1})$ such that $c_{i}\geq 0$ for $i=1,\dots,n$ and $c_{n+1}<0$.

Theorem 1.5**.**

Let $R$ be a regular local ring and let $f\in R[y]$ be a polynomial with coefficients in $R$, for some independent variable $y$. Assume that the Newton polyhedron $\Delta(f)\subset\mathbb{R}^{n+1}_{\geq 0}$ has a loose and descendant edge $E$. If $f|_{E}$ is a product of two relatively prime polynomials $G,H\in\mathbb{K}[X_{1},\ldots,X_{n}][y]$, where $G$ is monic with respect to $y$, then there exist $g,h\in\widehat{R}[y]$ such that $f=gh$, $g$ is a monic polynomial with respect to $y$ and $g|_{E_{1}}=G$, $h|_{E_{2}}=H$ for some $E_{1}$, $E_{2}$ such that $E=E_{1}+E_{2}$. Moreover, if $R=\mathbb{K}[\![x_{1},\dots,x_{n}]\!]$, then $g$ and $h$ are uniquely determined.

2. Proofs if $R=\mathbb{K}[\![x_{1},\dots,x_{n}]\!]$

At the beginning of this section we establish some results about loose edges.

Lemma 2.1**.**

Let $\Delta$ be a Newton polyhedron with a loose edge $E$ that has ends $a$, $b\in\mathbb{R}_{\geq 0}^{n}$. Then for every $c\in\Delta$ and every $\xi\in\mathbb{R}_{\geq 0}^{n}$ such that $\langle\xi,a\rangle=\langle\xi,b\rangle$ one has $\langle\xi,c\rangle\geq\langle\xi,a\rangle$.

Proof.

Let $V$ be the set of vertices of $\Delta$. If $V=\{a,b\}$ then Lemma 2.1 follows easily since every point of $\Delta$ can be written as $\lambda a+\mu b+z$ for some $z\in\mathbb{R}_{\geq 0}^{n}$ and $\lambda,\mu\geq 0$ such that $\lambda+\mu=1$. Hence in the rest of the proof assume that there exists a vertex of $\Delta$ different from $a$ or $b$ and consider the function

[TABLE]

Since the set of vertices is finite, this function is well defined and continuous.

Since $E$ is a compact face of $\Delta$, there exists $\xi_{0}\in\mathbb{R}_{>0}^{n}$ such that $\langle\xi_{0},a\rangle=\langle\xi_{0},b\rangle$ and $\langle\xi_{0},c\rangle>\langle\xi_{0},a\rangle$ for all $c\in\Delta\setminus E$. This yields

[TABLE]

Suppose that there exist $c\in\Delta$ and $\xi_{1}\in\mathbb{R}_{\geq 0}^{n}$ such that $\langle\xi_{1},c\rangle<\langle\xi_{1},a\rangle=\langle\xi_{1},b\rangle$. Observe that $c=\lambda_{1}v_{1}+\cdots+\lambda_{s}v_{s}+z$, for some $z\in\mathbb{R}_{\geq 0}^{n}$, $v_{1},\dots,v_{s}\in V$ and $\lambda_{i}\geq 0$ such that $\lambda_{1}+\cdots+\lambda_{s}=1$. If $\langle\xi_{1},v\rangle\geq\langle\xi_{1},a\rangle$ for every vertex $v$ of $\Delta$, then $\langle\xi_{1},c\rangle=\lambda_{1}\langle\xi_{1},v_{1}\rangle+\cdots+\lambda_{s}\langle\xi_{1},v_{s}\rangle+\langle\xi_{1},z\rangle\geq(\lambda_{1}+\cdots+\lambda_{s})\langle\xi_{1},a\rangle=\langle\xi_{1},a\rangle$, which contradicts the assumption $\langle\xi_{1},c\rangle<\langle\xi_{1},a\rangle$. Hence $\langle\xi_{1},v\rangle<\langle\xi_{1},a\rangle$ for at least one vertex $v\in V$. Thus

[TABLE]

It follows from the above that there exist $\xi$ in the segment, joining $\xi_{0}$ and $\xi_{1}$, such that $\psi(\xi)=0$. It means that there exists $v\in V\setminus\{a,b\}$ such that $\langle\xi,v\rangle=\langle\xi,a\rangle=\langle\xi,b\rangle\leq\langle\xi,d\rangle$ for all $d\in\Delta$. This implies that $\Delta^{\xi}$ is a compact (since $\xi\in\mathbb{R}_{>0}^{n}$) face of dimension $\geq 2$, which contains $E$. Therefore, we obtained a contradiction to the hypothesis that $E$ is a loose edge. ∎

Lemma 2.2**.**

Let $\Delta$ be a Newton polyhedron with a loose edge $E$ that has end points $a=(a_{1},\dots,a_{n})$, $b=(b_{1},\dots,b_{n})$. If $\min(a_{1},b_{1})=\dots=\min(a_{n},b_{n})=0$, then $a$ and $b$ are the only vertices of $\Delta$.

Proof.

By assumption there exist two nonempty and disjoint subsets $A$, $B$ of the set of indices $\{1,\dots,n\}$ such that $a_{i}>0$ for $i\in A$, $a_{i}=0$ for $i\in\{1,\dots,n\}\setminus A$, $b_{j}>0$ for $j\in B$ and $b_{j}=0$ for $j\in\{1,\dots,n\}\setminus B$.

Let $c=(c_{1},\dots,c_{n})$ be an arbitrary point of $\Delta$ different from $a$ and $b$. For any $i\in A$, $j\in B$ consider the vector $\xi_{ij}$, which has only two nonzero entries: $1/a_{i}$ at place $i$ and $1/b_{j}$ at place $j$ ($i\neq j$ since $A\cap B=\emptyset$). Then $\langle\xi_{ij},a\rangle=\langle\xi_{ij},b\rangle=1$ and $\langle\xi_{ij},c\rangle=c_{i}/a_{i}+c_{j}/b_{j}$. By Lemma 2.1 we get $c_{i}/a_{i}+c_{j}/b_{j}\geq 1$. It follows that

[TABLE]

Choose constants $\lambda\geq 0$, $\mu\geq 0$ such that $\lambda\leq\min_{i\in A}c_{i}/a_{i}$, $\mu\leq\min_{j\in B}c_{j}/b_{j}$ and $\lambda+\mu=1$. Then $c=\lambda a+\mu b+z$ for some $z\in\mathbb{R}_{\geq 0}^{n}$, hence $c$ cannot be a vertex of $\Delta$. ∎

Lemma 2.3**.**

Let $c\in\mathbb{Z}^{n}$ be a point with at least one positive and at least one negative coordinate. Then there exists a basis $\xi_{1}$,…, $\xi_{n}$ of the vector space $\mathbb{R}^{n}$ such that $\xi_{i}\in\mathbb{Z}_{\geq 0}^{n}$ for $i=1,\dots,n$ and $\langle\xi_{i},c\rangle=0$ for $i=1,\dots,n-1$.

Proof.

For any basis $\xi_{1}$, …, $\xi_{n}$ of $\mathbb{R}^{n}$ set $w=(w_{1},\dots,w_{n}):=(\langle\xi_{1},c\rangle,\dots,\langle\xi_{n},c\rangle)$. By the hypothesis of the lemma, for the standard basis $\xi_{1}=(1,0,\dots,0)$, …, $\xi_{n}=(0,\dots,0,1)$ of $\mathbb{R}^{n}$ there exist $i,j\in\{1,\dots,n\}$ such that $w_{i}$, $w_{j}$ have opposite signs. We will gradually modify the standard basis until we reach a basis such that only one coordinate of $w$ is nonzero. Below we outline the algorithm.

If there are only two nonzero entries $w_{j}$, $w_{k}$ of $w$ and $w_{j}+w_{k}=0$, then replace $\xi_{k}$ by $\xi_{k}+\xi_{j}$ and stop.

• 2.

If there are two nonzero entries $w_{j}$, $w_{k}$ of $w$ of opposite signs such that $|w_{j}|<|w_{k}|$, then replace $\xi_{k}$ by $\xi_{k}+\xi_{j}$ and go to step 1, otherwise go to step 3.

• 3.

Choose three nonzero entries $w_{j}$, $w_{k}$, $w_{l}$ of $w$ such that $w_{k}=w_{l}=-w_{j}$ (such a choice is possible since in this step all nonzero entries of $w$ have the same absolute value), replace $\xi_{k}$ by $\xi_{k}+\xi_{j}$ and go to step 1.

After every loop of the above algorithm $w_{k}$ is replaced by $w_{k}+w_{j}$. Hence the number $\sum_{i=1}^{n}|w_{i}|$ decreases and the algorithm must terminate. ∎

We encourage the reader to apply the algorithm from the proof in a simple case, for example for $c=(2,3,-4)$.

From now on up to the end of this section we fix a loose edge $E$. Let $c$ be a primitive lattice vector parallel to $E$. Applying Lemma 2.3 to $c$ we find ${n-1}$ linearly independent vectors $\xi_{1}$,…, $\xi_{n-1}\in\mathbb{Z}_{\geq 0}^{n}$ which are orthogonal to $E$. For every monomial $\underline{x}^{\alpha}$ we set $\omega(\underline{x}^{\alpha}):=(\langle\xi_{1},\alpha\rangle,\,\dots,\langle\xi_{n-1},\alpha\rangle)$. We call this vector a weight of $\underline{x}^{\alpha}$. Since $\omega(\underline{x}^{\alpha}\underline{x}^{\beta})=\omega(\underline{x}^{\alpha})+\omega(\underline{x}^{\beta})$ for any monomials $\underline{x}^{\alpha}$, $\underline{x}^{\beta}$, the ring $\mathbb{K}[x_{1},\dots,x_{n}]$ turns into a graded ring $\bigoplus_{w\in\mathbb{Z}_{\geq 0}^{n-1}}R_{w}$, where $R_{w}$ is spanned by monomials of weight $w$. We have that $R_{(\underline{0})}=\mathbb{K}$. Indeed, if $\underline{x}^{\alpha}\in R_{(\underline{0})}$ for some $\alpha\neq 0$ and $\beta$ is a point of $E$, then the points $\beta+t\alpha$, for $t>0$, belong to $E$, hence $E$ is unbounded.

All monomials of $R_{w}$ are of the form $\underline{x}^{a+ic}$ where $\underline{x}^{a}$ is a fixed monomial of $R_{w}$ and $i$ is an integer. This shows that $R_{w}$ is a finite-dimensional vector space over $\mathbb{K}$ since there is only a finite number of integers $i$ such that all coordinates of $a+ic$ are nonnegative. We denote by $\dim R_{w}$ the dimension of $R_{w}$ as a vector space over $\mathbb{K}$. It may happen that $\dim R_{w}=0$. Take for example the weight $\omega(x_{1}^{\alpha_{1}}x_{2}^{\alpha_{2}})=3\alpha_{1}+2\alpha_{2}$. Then there is no monomial of weight 1, hence $R_{1}\subset\mathbb{K}[x_{1},x_{2}]$ is a zero-dimensional vector space.

Let $M\subset\mathbb{Z}_{\geq 0}^{n-1}$ be the set of weights satisfying the condition: $z\in M$ if and only if there exists $\underline{x}^{\alpha}\in\mathbb{K}[\underline{x}^{\pm}]$ such that $\omega(\underline{x}^{\alpha})=z$ and $\langle\xi,\alpha\rangle\geq 0$ for all $\xi\in\mathbb{R}_{\geq 0}^{n}$ which are orthogonal to $E$. Observe that $M$ is closed under addition. Moreover for any $w\in\mathbb{Z}_{\geq 0}^{n-1}$ such that $\dim R_{w}>0$ we have $w\in M$.

Lemma 2.4**.**

Let $w\in\mathbb{Z}_{\geq 0}^{n-1}$ and $z\in M$. Assume that $R_{w}$ contains two coprime monomials. Then

[TABLE]

Proof.

For any $v\in\mathbb{Z}_{\geq 0}^{n-1}$ the dimension of the vector space $R_{v}$ is equal to the number of monomials ${\underline{x}}^{\alpha}$ such that $\omega({\underline{x}}^{\alpha})=v$, hence is equal to the number of lattice points, i.e. all points with integer coordinates, in the set

[TABLE]

Notice that $l_{v}$ is the intersection of the straight line $\{a+tc:t\in\mathbb{R}\}$ with $\mathbb{R}_{\geq 0}^{n}$, where $\omega({\underline{x}}^{a})=v$ and $c$ is a primitive lattice vector parallel to the edge $E$.

By the assumption, $R_{w}$ contains two coprime monomials ${\underline{x}}^{a}$ and ${\underline{x}}^{b}$. Hence $\min(a_{1},b_{1})=\dots=\min(a_{n},b_{n})=0$ which yields the partition of $\{1,\dots,n\}$ to three sets $A=\{i\in\{1,\dots,n\}:a_{i}=0,b_{i}>0\}$, $B=\{i\in\{1,\dots,n\}:a_{i}>0,b_{i}=0\}$ and $C=\{i\in\{1,\dots,n\}:a_{i}=0,b_{i}=0\}$. Since $A$ and $B$ are nonempty, $a$ and $b$ are the endpoints of the segment $l_{w}$. We may assume without loss of generality that the vector $c$ is pointed in the direction of $b-a$. If $\dim R_{w}=r+1$, then the lattice points of $l_{w}$ are $a$, $a+c$, …, $b=a+rc$.

If the lattice points of $l_{z}$ are $d$, $d+c$, …, $d+sc$ then $a+d$, $a+d+c$, …, $a+d+(r+s)c$ are the lattice points of $l_{w+z}$ (see Figure 5). It is easy to check that the points $a+d-c$ and $a+d+(r+s+1)c$ do not belong to $\mathbb{R}_{\geq 0}^{n}$. Hence $\dim R_{w+z}=r+s$. This ends the proof in the case $\dim R_{z}>0$.

a$$a+c$$a+rc$$d$$d+c$$d+sc

a+d$$a+d+c$$a+d+2c$$a+d+(r+s)c

Fig. 5

Now suppose that $\dim R_{z}=0$. Let ${\underline{x}}^{d}\in\mathbb{K}[\underline{x}^{\pm}]$ be any monomial with integer exponents such that $\omega({\underline{x}}^{d})=z$. Replacing this monomial by ${\underline{x}}^{d+kc}$, for suitably chosen integer $k$, we may assume that $d_{i_{0}}<0$ for some $i_{0}\in A$ and $d_{i}+c_{i}\geq 0$ for all $i\in A$.

Take the vector $v_{j}=\frac{1}{c_{i_{0}}}e_{i_{0}}-\frac{1}{c_{j}}e_{j}$, where $e_{1},\dots$, $e_{n}$ is the standard basis of $\mathbb{R}^{n}$ and $j\in B$. Then $v_{j}\in\mathbb{R}_{\geq 0}^{n}$ is orthogonal to $E$. By the assumption that $z\in M$ (use $\xi=v_{j}$ and $\alpha=d$ in the definition of $M$) we get the inequality $d_{i_{0}}/c_{i_{0}}-d_{j}/c_{j}=\langle v_{j},d\rangle\geq 0$. Hence $d_{j}>0$ for all $j\in B$, since, by the definitions of $A,B$ and the convention that $c$ points into the direction of $b-a$, we have $c_{j}<0$ for $j\in B$ and $c_{i_{0}}>0$ for $i_{0}\in A$, and since $d_{i_{0}}<0$, by assumption.

If $j\in C$ then $e_{j}$ is orthogonal to $E$. By the assumption that $z\in M$ (use $\xi=e_{j}$ and $\alpha=d$ in the definition of $M$) we get the inequality $d_{j}\geq 0$ for all $j\in C$.

Notice that $d_{j}+c_{j}<0$ for at least one $j\in B$, otherwise all exponents of ${\underline{x}}^{d+c}$ would be nonnegative, which would contradict $\dim R_{z}=0$.

The lattice points of $l_{w+z}$ are of the form $a+d+ic$, where $i\in\mathbb{Z}$. Since $d_{i_{0}}<0$, the point $a+d$ has a negative coordinate. Since $d_{j}+c_{j}<0$ for some $j\in B$, the point $a+d+(r+1)c=b+d+c$ has a negative coordinate. We claim that all coordinates of the points $a+d+ic$ are nonnegative, for $i=1,\dots,r$: If $j\in A$, then we have seen that $d_{j}+c_{j}\geq 0$ and $c_{j}>0$, and hence, $a_{j}+d_{j}+ic_{j}=a_{j}+(c_{j}+d_{j})+(i-1)c_{j}\geq 0$, for $i\geq 1$. On the other hand, if $j\in C$ then $a_{j}=b_{j}=0$, thus $c_{j}=0$, and we have seen that $d_{j}\geq 0$. This provides $a_{j}+d_{j}+ic_{j}=d_{j}\geq 0$, for $j\in C$. Finally, for $j\in B$, we have $d_{j}>0$ and $b_{j}=0$, and since $c=(b-a)/r$, we have that $a_{j}+d_{j}+ic_{j}=a_{j}+d_{j}-ia_{j}/r=(1-i/r)a_{j}+d_{j}>0$, for $i=1,\ldots,r$. This proves the claim. Hence $\dim R_{z+w}=r$ which finishes the proof. ∎

In the appendix, we outline an alternative proof for Lemma 2.4.

Lemma 2.5**.**

Let $G\in R_{w}$ and $H\in R_{z}$ be coprime polynomials. If $G$ is not divisible by any monomial then for every $i\in M$

[TABLE]

Proof.

Consider the following sequence

[TABLE]

where $\Phi\colon\eta\mapsto(\eta H,-\eta G)$ and $\Psi\colon(\psi,\varphi)\mapsto\psi G+\varphi H$. Take any $(\psi,\varphi)\in R_{z+i}\times R_{w+i}$ such that $\Psi(\psi,\varphi)=\psi G+\varphi H=0$. Since $G$ and $H$ are coprime, there exists $\eta\in R_{i}$ such that $\psi=\eta H$ and $\varphi=-\eta G$. Thus $\ker\Psi\subset{\rm im}\,\Phi$. The opposite inclusion is obvious. Moreover, it is easy to see that $\Phi$ is injective. It follows that the above sequence is exact.

By assumption $G\in R_{w}\subset\mathbb{K}[x_{1},\ldots,x_{n}]$ is neither a constant nor a monomial. Hence there appear at least two monomials in $G$ with non-zero coefficient and they can be chosen to be coprime as $G$ is not divisible by a monomial. Since both monomials are contained in $R_{w}$, we obtain that $R_{w}$ satisfies the hypothesis of Lemma 2.4. Hence we get $\dim{\rm Im}\,\Psi=\dim R_{z+i}+\dim R_{w+i}-\dim R_{i}=\dim R_{z+i}+(\dim R_{w}+\dim R_{i}-1)-\dim R_{i}=\dim R_{w}+\dim R_{z+i}-1=\dim R_{w+z+i}$, which implies that $\Psi$ is surjective. ∎

Proof of Theorem 1.1.

Suppose that $R=\mathbb{K}[\![x_{1},\dots,x_{n}]\!]$. Then, $\widehat{R}=R$ and $f\in\mathbb{K}[\![\underline{x}]\!]$ can be written as $f=\sum_{\alpha\in\mathbb{Z}_{\geq 0}^{n}}a_{\alpha}\underline{x}^{\alpha}$, with coefficients $a_{\alpha}\in\mathbb{K}$.

Since all monomials of $f|_{E}$ have the same weight, the polynomial $f|_{E}$ is homogeneous in a graded ring $\mathbb{K}[x_{1},\dots,x_{n}]$. Then by an elementary property of graded rings the equality $f|_{E}=GH$ implies that $G\in R_{w}$ and $H\in R_{z}$ for some $w$, $z\in\mathbb{Z}_{\geq 0}^{n-1}$. Let ${\underline{x}}^{\alpha}$ be any monomial which appears with nonzero coefficient in the power series $f$ and ${\underline{x}}^{\alpha_{0}}$ be a fixed monomial of $f|_{E}$. By Lemma 2.1 we have $\langle\xi,\alpha\rangle\geq\langle\xi,\alpha_{0}\rangle$ for every $\xi\in\mathbb{R}_{\geq 0}^{n}$ which is orthogonal to $E$. This yields $\omega({\underline{x}}^{\alpha-\alpha_{0}})\in M$. Since $\omega({\underline{x}}^{\alpha_{0}})=w+z$, we get ${\underline{x}}^{\alpha}\in R_{w+z+i}$ for some $i\in M$ and hence $f=\sum_{i\in M}f_{w+z+i}$, where $f_{w+z+i}\in R_{w+z+i}$.

Let $g_{w}:=G$ and $h_{z}:=H$. Then $f_{w+z}=g_{w}h_{z}$. Let us set $M$ in degree-lexicographic order. Using Lemma 2.5 we can find recursively $g_{w+i}\in R_{w+i}$ and $h_{z+i}\in R_{z+i}$ such that

[TABLE]

where

[TABLE]

If $g:=\sum_{i\in M}g_{w+i}$ and $h:=\sum_{i\in M}h_{z+i}$, then $f=gh$.

Let $\xi=\xi_{1}+\dots+\xi_{n-1}$, where $\xi_{1},\ldots,\xi_{n-1},\xi_{n}\in\mathbb{Z}_{\geq 0}^{n}$ is the basis of $\mathbb{R}^{n}$ obtained in Lemma 2.3. Then for $E_{1}:=\Delta(g)^{\xi}$ and $E_{2}:=\Delta(h)^{\xi}$ we have $g|_{E_{1}}=g_{w}$, $h|_{E_{2}}=h_{z}$ and $E=E_{1}+E_{2}$. ∎

Proof of Corollary 1.3.

Assume that $a=(a_{1},\dots,a_{n})$, $b=(b_{1},\dots,b_{n})$ are the ends of a loose edge $E$ of the Newton polyhedron $\Delta(f)$. Since $\Delta(f)$ has at least three vertices, Lemma 2.2 implies that $c=(\min(a_{1},b_{1}),\dots,\min(a_{n},b_{n}))$ has at least one nonzero coordinate. The monomials $\underline{x}^{a}$ and $\underline{x}^{b}$ appear in the polynomial $f|_{E}$ with nonzero coefficients and their greatest common divisor equals $\underline{x}^{c}$. Thus $G:=\underline{x}^{-c}\cdot f|_{E}$ is not divisible by any variable, so $G$ and $H:=\underline{x}^{c}$ are relatively prime. Using Theorem 1.1 we obtain that $f$ is not irreducible. ∎

Lemma 2.6**.**

Let $G\in R_{w}$ and $H_{j}\in R_{z_{j}}$ for $j=1,2$. Assume that for every $i\in M$

[TABLE]

for $j=1,2$. Then for every $i\in M$

[TABLE]

Proof.

By assumptions of the lemma we get

[TABLE]

Let us consider the case that $f\in R[y]$ is a polynomial with coefficients in $R$, for some independent variable $y$. Assume that $E\subset\mathbb{R}^{n+1}_{\geq 0}$ is a descendant loose edge of $\Delta(f)$. As before, we have $\mathbb{K}[x_{1},\dots,x_{n}][y]\cong\bigoplus_{w\in\mathbb{Z}_{\geq 0}^{n}}R_{w}$ (with the analogous notations). By definition, there exists a lattice vector $c=(c_{1},\dots,c_{n},c_{n+1})$ parallel to $E$ such that $c_{i}\geq 0$ for $i=1,\dots,n$ and $c_{n+1}<0$. Let $R_{w}^{\prime}:=R_{w}\cap\mathbb{K}[x_{1},\dots,x_{n}]$ for $w\in\mathbb{Z}_{\geq 0}^{n}$. Since every line parallel to $E$ intersects $\mathbb{R}^{n}\times\{0\}$ transversely, the dimension of $R_{w}^{\prime}$ is either 0 or 1.

We claim that for every $w\in M$, $z\in\mathbb{Z}_{\geq 0}^{n}$, $0\neq H\in R_{z}^{\prime}$ and $\psi\in R_{w+z}^{\prime}$ one has $\psi/H\in R_{w}^{\prime}$. To prove this claim it is enough to consider $H=\underline{x}^{\alpha}\in R_{z}^{\prime}$ and $\psi=\underline{x}^{\beta}\in R_{w+z}^{\prime}$. Then $\omega(\underline{x}^{\beta-\alpha})=w$. Denote $v_{i}:=e_{i}-\frac{c_{i}}{c_{n+1}}e_{n+1}\in\mathbb{R}_{\geq 0}^{n+1}$, where $e_{1},\dots,e_{n},e_{n+1}$ is the standard basis of $\mathbb{R}^{n+1}$. Since every vector $v_{i}$ is orthogonal to $E$ and $w\in M$, we have $0\leq\langle v_{i},\beta-\alpha\rangle=\beta_{i}-\alpha_{i}$ for $i=1,\dots,n$. Thus $\psi/H=\underline{x}^{\beta-\alpha}$ is a monomial with nonnegative exponents.

Lemma 2.7**.**

Let $G\in R_{w}$ and $H\in R_{z}$ be coprime polynomials. If $G$ is monic with respect to $y$, then for every $i\in M$

[TABLE]

Proof.

First, we prove (1) in the special case $G=y$ and $H\in R_{z}^{\prime}$.

Every $F\in R_{w+z+i}$ can be written in the form $F=y\phi+\psi$ where $\phi,\psi$ are polynomials and $\psi$ does not depend on $y$. Then $\phi\in R_{z+i}$ and $\psi=H\psi^{\prime}$ where $\psi^{\prime}=\psi/H\in R_{w+i}^{\prime}$. This gives (1) in the special case. All remaining cases follow from Lemma 2.5 and Lemma 2.6. ∎

Proof of Theorem 1.5.

Suppose that $R=\mathbb{K}[\![x_{1},\dots,x_{n}]\!]$. Then, $\widehat{R}=R$ and any element in $R[\![y]\!]=\mathbb{K}[\![\underline{x},y]\!]$ has an expansion with coefficients contained in the residue field $\mathbb{K}$.

Proceeding as in the proof of Theorem 1.1, but using Lemma 2.7 instead of Lemma 2.5 we obtain $\overline{g},\overline{h}\in\mathbb{K}[\![\underline{x},y]\!]$ such that $f=\overline{g}\overline{h}$, $\overline{g}|_{E_{1}}=G$ and $\overline{h}|_{E_{2}}=H$ for some segments $E_{1},E_{2}$, where $E=E_{1}+E_{2}$. The assumptions that the loose edge $E$ is descendant and the polynomial $G$ is monic with respect to $y$ imply that the Newton polyhedron of $\overline{g}$ has a vertex $(0,\dots,0,d)$ for some positive integer $d$. Therefore $\overline{g}(0,\dots,0,y)=vy^{d}$ for some $v\in\mathbb{K}[\![y]\!]$ such that $v(0)\neq 0$. It means that $\overline{g}$ fulfills the assumptions of the Weierstrass Preparation Theorem, which implies that $\overline{g}=u\hat{g}$, where $u$ is a power series such that $u(0)\neq 0$ and $\hat{g}\in\mathbb{K}[\![x_{1},\dots,x_{n}]\!][y]$ is a Weierstrass polynomial of degree $d$. Putting $g=\hat{g}$ and $h=u\overline{h}$ we get $f=gh$. Since we can also obtain $h$ using the polynomial division of $f$ by $g$ in the ring $\mathbb{K}[\![x_{1},\dots,x_{n}]\!][y]$, we conclude that $h$ is a polynomial with respect to $y$. ∎

3. Extending to regular local rings

After discussing the case of a formal power series ring over a field $\mathbb{K}$, we explain how the results can be extended to the completion of a regular local ring by using graded rings.

Let $(R,\mathfrak{m},\mathbb{K}=R/\mathfrak{m})$ be a regular local ring, let $(\underline{x})=(x_{1},\ldots,x_{n})$ be a regular system of parameters, and let $\widehat{R}$ be the $\mathfrak{m}$-adic completion of $R$. We consider $f\in R$ such that its Newton polyhedron has a loose edge $E$.

First, let us point out that Lemmas 2.1, 2.2, 2.3 are results on convex geometry. In fact, they are true for any $F$-subset $\Delta\subset\mathbb{R}^{n}_{\geq 0}$, which is closed convex subset such that $\Delta+\mathbb{R}^{n}_{\geq 0}=\Delta$, [7, p. 260]. In particular, they are independent of $R$. Furthermore, the proof of Corollary 1.3 only uses Theorem 1.1 and Lemma 2.2 and thus can be applied word by word.

Observation 3.1**.**

Recall that, after Lemma 2.3, we introduced a weight given by $\omega(\underline{x}^{\alpha}):=(\langle\xi_{1},\alpha\rangle,\,\dots,\langle\xi_{n-1},\alpha\rangle)$, where $\xi_{1}$,…, $\xi_{n-1}\in\mathbb{Z}_{\geq 0}^{n}$ are linearly independent vectors which are orthogonal to $E$. The map $\omega$ induces a monomial valuation (denoted also by $\omega$) on $R$ with values in $\mathbb{Z}_{\geq 0}^{n-1}$, i.e. for any nonzero $h=\sum_{\alpha\in\mathbb{Z}_{\geq 0}^{n}}b_{\alpha}\underline{x}^{\alpha}\in R$, where $b_{\alpha}\in R^{\times}\cup\{0\}$, we set

[TABLE]

We equip $\mathbb{Z}_{\geq 0}^{n-1}$ with the partial ordering defined by $\alpha\leq\beta\Leftrightarrow\beta=\alpha+\gamma$, for some $\gamma\in\mathbb{Z}_{\geq 0}^{n-1}$. The graded ring of $R$ associated to $\omega$ is defined as

[TABLE]

where $\mathcal{P}_{w}:=\{h\in R:\omega(h)\geq w\}$ and $\mathcal{P}_{w}^{+}:=\{h\in R:\omega(h)>w\}$. Note that $\mathcal{P}_{(\underline{0})}=R$ and $\mathcal{P}_{(\underline{0})}^{+}=\mathfrak{m}$, since $\omega(x_{i})>(\underline{0})$ for all $i=1,\ldots,n$. Hence, we have $\mathcal{P}_{(\underline{0})}/\mathcal{P}_{(\underline{0})}^{+}=\mathbb{K}$. Denote by $X_{i}\in\operatorname{gr}_{\omega}(R)$ the image of $x_{i}\in R$ in the graded ring. Then the initial form of $h$ with respect to $\omega$ is defined by $\operatorname{in}_{\omega}(h):=X_{i}$ if $h=x_{i}$ and otherwise, $\operatorname{in}_{\omega}(h):=\sum_{\alpha:\omega(\underline{x}^{\alpha})=\omega(h)}\overline{b}_{\alpha}\underline{X}^{\alpha}$, where $\overline{b}_{\alpha}\in\mathbb{K}$ is the image of $b_{\alpha}$ in the residue field. Since $\omega$ is a monomial valuation, we have $\mathcal{P}_{w}/\mathcal{P}_{w}^{+}=R_{w}$, for all $w\in\mathbb{Z}_{\geq 0}^{n-1}$, i.e., $\operatorname{gr}_{\omega}(R)=\bigoplus_{w\in\mathbb{Z}_{\geq 0}^{n-1}}R_{w}\cong\mathbb{K}[X_{1},\ldots,X_{n}]$.

In conclusion, we see that the second set of ingredients for the proofs of Theorems 1.1 and 1.5, Lemmas 2.4 and 2.5, are results on the graded ring $\operatorname{gr}_{\omega}(R)$ without the restriction $R=\mathbb{K}[\![\underline{x}]\!]$.

The essential part for extending the proof to any regular local ring, is the following construction to find a lift $g\in R$ for a given homogeneous element $G\in R_{w}\subset\operatorname{gr}_{\omega}(R)\cong\mathbb{K}[X_{1},\ldots,X_{n}]$, where $w\in\mathbb{Z}_{\geq 0}^{n-1}$: We can write $G$ as a finite sum $G=\sum_{\alpha\in\mathbb{Z}_{\geq 0}^{n}}\lambda_{\alpha}X^{\alpha}$, for $\lambda_{\alpha}\in\mathbb{K}$. For every $\alpha\in\mathbb{Z}_{\geq 0}^{n}$ with $\lambda_{\alpha}\neq 0$, we choose a unit $b_{\alpha}\in R^{\times}$ such that we have $b_{\alpha}\equiv\lambda_{\alpha}\mod\mathfrak{m}$. Otherwise, we put $b_{\alpha}:=0\in R$. We define

[TABLE]

which is an element in $R$ since all but finitely many coefficients $b_{\alpha}$ are zero. By construction, the image of $g$ in $\operatorname{gr}_{\omega}(R)$ is $G$, i.e., $\operatorname{in}_{\omega}(g)=G$.

In general, $g$ is not unique since it depends on a choice of a system of representatives in $R$ for the residue field $\mathbb{K}=R/\mathfrak{m}$. If $\mathbb{K}\subset R$, then we can uniquely choose $b_{\alpha}:=\lambda_{\alpha}$.

Using the above, we can adapt the proofs to obtain Theorems 1.1 and 1.5.

Proof of Theorem 1.1.

Let $w_{0}:=\omega(f)$. We have $f|_{E}=\operatorname{in}_{\omega}(f)$. By the hypothesis of the theorem, $\operatorname{in}_{\omega}(f)=GH\in R_{w_{0}}$. Since all monomials of $\operatorname{in}_{\omega}(f)$ have the same weight, it follows $G\in R_{w}$ and $H\in R_{z}$ for some $w$, $z\in\mathbb{Z}_{\geq 0}^{n-1}$ such that $w+z=w_{0}$. We set $G_{w}:=G$ and $H_{z}:=H$.

Let $f=\sum_{\alpha}a_{\alpha}\underline{x}^{\alpha}$ be a finite expansion of $f$ in $R$. Fix $\alpha_{0}\in\mathbb{Z}_{\geq 0}^{n}$ such that $a_{\alpha_{0}}\neq 0$ and $\omega(\underline{x}^{\alpha_{0}})=w_{0}$, i.e., $\underline{X}^{\alpha_{0}}$ is a monomial appearing in $\operatorname{in}_{\omega}(f)$. By Lemma 2.1, we have $\langle\xi,\alpha\rangle\geq\langle\xi,\alpha_{0}\rangle$ for every $\xi\in\mathbb{R}_{\geq 0}^{n}$ which is orthogonal to $E$. This yields $\omega({\underline{x}}^{\alpha-\alpha_{0}})\in M$. Since $\omega({\underline{x}}^{\alpha_{0}})=w+z$, we obtain $\omega({\underline{x}}^{\alpha})=w+z+i=w_{0}+i$, for some $i\in M$.

We choose a lift $g_{w}\in R$ (resp. $h_{z}\in R$) of $G_{w}\in R_{w}$ (resp. $H_{z}\in R_{z}$), as described before. We define our first approximation of $f$ as $\phi_{1}:=g_{w}\cdot h_{z}$. Note that $\operatorname{in}_{\omega}(f)=\operatorname{in}_{\omega}(\phi_{1})$. We introduce $f_{1}:=f-\phi_{1}$. By the previous, we have $w_{1}:=\omega(f_{1})=w_{0}+i_{1}$, for some $i_{1}\in M$ and $i_{1}\neq(\underline{0})$. In particular, $\operatorname{in}_{\omega}(f_{1})\in R_{w+z+i_{1}}\subset\operatorname{gr}_{\omega}(R)$. By assumption, $G_{w}=G$ is not divisible by a monomial, hence, by Lemma 2.5, there are $H_{z+i_{1}}\in R_{z+i_{1}}$ and $G_{w+i_{1}}\in R_{w+i_{1}}$ such that $\operatorname{in}_{\omega}(f_{1})=G_{w}H_{z+i_{1}}+H_{z}G_{w+i_{1}}.$ We choose $h_{z+i_{1}},g_{w+i_{1}}\in R$, as described before. The second approximation $\phi_{2}\in R$ of $f$ is then defined as

[TABLE]

By construction, $\omega(g_{w+i_{1}}h_{z+i_{1}})>w_{1}$. For $f_{2}:=f-\phi_{2}$, we obtain that $w_{2}:=\omega(f_{2})>w_{1}$. We proceed with the construction and finally get $g:=\sum_{i\in M}g_{w+i}$, $h:=\sum_{i\in M}h_{z+i}\in\widehat{R}$ with the property $f=g\cdot h$ in $\widehat{R}$.

The claim concerning $E=E_{1}+E_{2}$ follows with the same argument as in the case $R=\mathbb{K}[\![\underline{x}]\!]$. ∎

Finally, the proof of Theorem 1.5 for $R=\mathbb{K}[\![\underline{x}]\!]$ can easily be adapted to the general setting:

Proof of Theorem 1.5.

Proceeding as in the previous proof, but using Lemma 2.7 instead of Lemma 2.5 we obtain $\overline{g},\overline{h}\in\widehat{R}[\![y]\!]$ such that $f=\overline{g}\overline{h}$, $\overline{g}|_{E_{1}}=G$ and $\overline{h}|_{E_{2}}=H$ for some segments $E_{1},E_{2}$, where $E=E_{1}+E_{2}$. The assumptions that the loose edge $E$ is descendant and the polynomial $G$ is monic with respect to $y$ imply that the Newton polyhedron of $\overline{g}$ has a vertex $(0,\dots,0,d)$ for some positive integer $d$. Hence, the monomial $vy^{d}$ appears in an expansion of $\overline{g}$, for some unit $v\in\widehat{R}[\![y]\!]^{\times}$. The Weierstrass preparation theorem ([8, Theorem 9.2]) implies that there exist a unit $u\in\widehat{R}[\![y]\!]^{\times}$ and a Weierstrass polynomial $g\in\widehat{R}[y]$ of degree $d$ such that $\overline{g}=ug.$ We define $h:=u\overline{h}$ and obtain that $f=gh$. As before, $f,g\in\widehat{R}[y]$ imply $h\in\widehat{R}[y]$. ∎

4. Examples

Let us discuss some examples.

Example 4.1**.**

Let $R$ be a regular local ring (not necessarily complete) of dimension three with regular system of parameters $(x,y,z)$ and residue field $\mathbb{K}$. Suppose that $R$ contains a field of characteristic different from two. (E.g., $R=\mathbb{K}[x,y,z]_{\langle x,y,z\rangle}$ and $\mbox{char}(\mathbb{K})\neq 2$). Consider the element

[TABLE]

It is not hard to see that the Newton polyhedron $\Delta(f)$ has exactly two vertices, $v_{1}=(6,2,0)$ and $v_{2}=(0,0,4)$. Hence, $v_{1}$ and $v_{2}$ determine a loose edge $E$ of $\Delta(f)$. The symbolic restriction of $f$ to $E$ is

[TABLE]

where $\omega=\omega(\xi)$, for $\xi\in\mathbb{R}_{\geq 0}^{3}$ appropriately chosen such that $\Delta(f)^{\xi}=E$. (Here, and in the examples below, we leave the explicit computation of $\xi$ and $\omega$ as an exercise to the reader).

By Theorem 1.1, the factorization of $f|_{E}$ implies that there are $g,h\in\widehat{R}$ in the completion such that $f=gh$. Indeed, if we have a closer look at $f$, we observe that $f=x^{6}y^{2}(1-xy^{3}z^{2})-z^{4}(1-xy)$, which factors in the completion as

[TABLE]

Example 4.2**.**

Let $R$ be any regular local ring of dimension three with regular system of parameters $(x,y,z)$ and residue field $\mathbb{K}$. Let

[TABLE]

We observe that the Newton polyhedron is of the form as the one in Figure 3 at the beginning. In particular, it has three loose edges. If we consider the edge $E$ determined by the vertices $(1,1,1)$ and $(3,3,0)$, we see that the corresponding symbolic restriction is $f|_{E}=XYZ+X^{3}Y^{3}=XY(Z+X^{2}Y^{2})\in\mathbb{K}[X,Y,Z]$. Therefore, $f$ is not irreducible in the completion $\widehat{R}$, by Theorem 1.1.

Note that in the previous example, $R$ is allowed to have mixed characteristics, e.g., we might have $R=\mathbb{Z}[y,z]_{\langle p,y,z\rangle}$, where $x=p\in\mathbb{Z}$ is a prime number.

Example 4.3**.**

Let $R$ be a regular local ring of dimension two with regular system of parameters $(x_{1},x_{2})$ and residue field $\mathbb{K}$. Consider the Weierstrass polynomial

[TABLE]

In Figure 6 the corresponding Newton polyhedron is drawn.

Fig. 6

The edge $E$ determined by the vertices $(5,4,2)$ and $(15,18,0)$ is loose and descendant. Since the symbolic restriction of $f$ to $E$ is

[TABLE]

Theorem 1.5 implies that $f$ is not irreducible in $\widehat{R}[y]$.

Example 4.4**.**

Let $R=\mathbb{Z}_{p}$ be the $p$-adic integers, for some $p\in\mathbb{Z}$ prime. Let

[TABLE]

Note that $270=2\cdot 3^{3}\cdot 5$ and $540=2^{2}\cdot 3^{3}\cdot 5$ and the residue field of $R$ is $\mathbb{F}_{p}$.

1. (1)

If $p=2$, then $f=y^{3}+\epsilon_{1}py+\epsilon_{3}p^{2}$, for $\epsilon_{1},\epsilon_{3}\in\mathbb{Z}_{2}^{\times}$ units. The Newton polyhedron has three vertices $(0,3),(1,1),(2,0)$. If $E_{2}$ denotes the edge given by $(0,3)$ and $(1,1)$, then $f|_{E_{2}}=Y^{3}+PY=Y(Y^{2}+P)\in\mathbb{F}_{2}[P,Y].$ Hence, Theorem 1.5 implies that $f$ factors in $\widehat{\mathbb{Z}_{2}}[y]$. 2. (2)

If $p=3$, then $f=y^{3}+\mu_{1}p^{3}y+\mu_{3}p^{3}$, for $\mu_{1},\mu_{3}\in\mathbb{Z}_{3}^{\times}$ units. The Newton polyhedron has two vertices $(0,3)$ and $(3,0)$. Let $E_{3}$ be the corresponding compact edge. We have $f|_{E_{3}}=Y^{3}+P^{3}=(Y+P)^{3}\in\mathbb{F}_{3}[P,Y]$. Therefore, we cannot apply Theorem 1.5.

Note that if $p=5$, then $f=y^{3}+\rho_{1}py+\rho_{3}p$, for $\rho_{1},\rho_{3}\in\mathbb{Z}_{5}^{\times}$ units, which means that $f\in\mathbb{Z}_{5}[y]$ is irreducible in $\widehat{\mathbb{Z}_{5}}[y]$.

5. Relation with known results

Corollary 1.4 generalizes to $n>2$ variables the following well-known fact.

Theorem 5.1** ([6, Lemma 3.15]).**

Assume that a power series $f\in\mathbb{C}[\![x,y]\!]$ written as a sum of homogeneous polynomials $f=f_{d}+f_{d+1}+\dots$, where the subindex means the degree, is irreducible. Then $f_{d}$ is a power of a linear form.

Let $\mathbb{K}$ be a field and fix a weight $\omega(x^{i}y^{j}):=ni+mj$ for $n,m\in\mathbb{Z}_{>0}$. Given $0\neq F\in\mathbb{K}[\![x,y]\!]$, we will consider its decomposition in $\omega$-quasihomogeneous forms

[TABLE]

where the subindex means the $\omega$-weight.

Theorem 5.2** ([2, Lemma A1]).**

Assume that $F_{a+b}(x,y)=f_{a}(x,y)g_{b}(x,y)$, where $f_{a}$, $g_{b}\in\mathbb{K}[x,y]$ are quasihomogeneous and coprime. Then, there exist

[TABLE]

such that $F=fg$. Moreover if $f_{a}$ is an irreducible polynomial, then $f$ is an irreducible power series.

Theorem 1.1 was inspired by the method of the proof of Theorem 5.2.

Corollary 1.4 generalizes the following result of [4].

Theorem 5.3**.**

If $\phi\in\mathbb{C}\{x_{1},\dots,x_{n}\}$ is irreducible and has a polygonal Newton polyhedron $\Delta(\phi)$, then the polyhedron $\Delta(\phi)$ has only one compact edge $E$ and the polynomial $\phi|_{E}$ is a power of an irreducible polynomial.

We recall that a Newton polyhedron is called polygonal if all its compact edges are loose.

Theorem 1.5 generalizes the main result of [11] quoted below. For this, let us recall that a monic Weierstrass polynomial $P(Z)\in k[\![x_{1},\dots,x_{n}]\!][Z]$ of degree $d$ has an orthant associated polyhedron if the projection of its Newton polyhedron from the distinguished point $(0,\ldots,0,d)\in\mathbb{R}^{n+1}_{\geq 0}$ to the subspace corresponding to $(x_{1},\ldots,x_{n})$ is a polyhedron with exactly one vertex ([11, Definiton 2.2]).

Theorem 5.4**.**

*Let $P(Z)\in k[\![x_{1},\dots,x_{n}]\!][Z]$ be a monic Weierstrass polynomial. Assume that P(Z) has an orthant associated polyhedron and that $P|_{\Gamma}\in k[x_{1},\dots,x_{n},Z]$ is the product of two coprime monic polynomials $S_{1}$, $S_{2}\in k[x_{1},\dots,x_{n},Z]$, respectively, of degree $d_{1}$ and $d_{2}$. Then there exist two monic polynomials $P_{1}$, $P_{2}\in k[\![x_{1},\dots,x_{n}]\!][Z]$, respectively, of degrees $d_{1}$ and $d_{2}$ in $Z$ such that

i) $P=P_{1}P_{2}$,

ii) there is at least one $i\in\{1,2\}$ such that $P_{i}$ has an orthant associated polyhedron and if $\Gamma_{i}$ denotes the compact face of $\Delta(P_{i})$ containing the points $(0,\dots,0,d_{i})$, then $P_{i}|_{\Gamma_{i}}=S_{i}$ and $\Gamma_{i}$ is parallel to $\Gamma$.*

Remark. Let $f\in k[\![x_{1},\dots,x_{n}]\!][y]$ be a Weierstrass polynomial of degree $d$. Then $f$ has an orthant associated polyhedron if and only if $\Delta(f)$ has a loose edge with endpoint $(0,\dots,0,d)$. The reason for this is the following: $\Delta(f)$ not having a loose edge with endpoint $(0,\ldots,0,d)$ is equivalent to the existence of a compact face of dimension $\geq 2$, where one of the vertices is $(0,\ldots,0,d)$. The latter is equivalent to the condition that the projection of the Newton polyhedron from $(0,\ldots,0,d)$ to the subspace determined by $(x_{1},\ldots,x_{n})$ has at least two vertices which correspond to vertices of the considered compact face. We also refer to [10, Lemma 3.1].

Appendix: Alternative proof for Lemma 2.4

Recall the notations of section 2. We outline an alternative proof for

Lemma 2.4**.**

Let $w\in\mathbb{Z}_{\geq 0}^{n-1}$ and $z\in M$. Assume that $R_{w}$ contains two coprime monomials. Then

[TABLE]

Proof.

For any $v\in\mathbb{Z}_{\geq 0}^{n-1}$ the dimension of the vector space $R_{v}$ is equal to the number of monomials ${\underline{x}}^{\alpha}$ such that $\omega({\underline{x}}^{\alpha})=v$, hence is equal to the number of lattice points, i.e. all points with integer coordinates, in the set

[TABLE]

Notice that $l_{v}$ is the intersection of the straight line $\{a+tc:t\in\mathbb{R}\}$ with $\mathbb{R}_{\geq 0}^{n}$, where $\omega({\underline{x}}^{a})=v$ and $c$ is a primitive lattice vector parallel to the edge $E$.

By the assumption, $R_{w}$ contains two coprime monomials ${\underline{x}}^{a}$ and ${\underline{x}}^{b}$. Hence $\min(a_{1},b_{1})=\dots=\min(a_{n},b_{n})=0$. Thus there exist two nonempty and disjoint subsets $A$, $B$ of $\{1,\dots,n\}$ such that $a_{i}>0\Leftrightarrow i\in A$ and $b_{i}>0\Leftrightarrow i\in B$. Without loss of generality we may assume that the vector $c$ is pointed in the direction of $b-a$. Then

[TABLE]

for some positive integer $m$.

Now we will show that the set $l_{z}$ is nonempty and give its description.

Let $d\in\mathbb{Z}^{n}$ be such that $\omega({\underline{x}}^{d})=z$. Consider the system of inequalities

[TABLE]

Its solution space is an interval $[t_{1},\infty)$.

Let $a^{\prime}=d+t_{1}c$. Then $a^{\prime}_{j}\geq 0$ for all $j\in B$ and $a^{\prime}_{j_{0}}=0$ for some $j_{0}\in B$.

Consider the vectors $v_{j}=e_{j}-\frac{c_{j}}{c_{j_{0}}}e_{j_{0}}$ for $j\in\{1,\dots,n\}\setminus B$, where $e_{1}$, …$e_{n}$ is the standard basis of $\mathbb{R}^{n}$. These vectors belong to $\mathbb{R}_{\geq 0}^{n}$ and are orthogonal to $E$. By the assumption that $z\in M$ we get $a^{\prime}_{j}=\langle v_{j},a^{\prime}\rangle\geq 0$. It follows that $a^{\prime}\in\mathbb{R}_{\geq 0}^{n}$.

Using the same argument, but starting from the system of inequalities

[TABLE]

we show that there exist $t_{2}\in\mathbb{R}$ such that the point $b^{\prime}=d+t_{2}c$ belongs to $\mathbb{R}_{\geq 0}^{n}$ and $b^{\prime}_{k}=0$ for some $k\in A$.

It follows from the above that $a^{\prime},b^{\prime}\in l_{z}$ and

[TABLE]

for some rational numbers $t_{1}<t_{2}$.

Finally, $l_{w+z}$ is a segment with endpoints $a+a^{\prime}$, $b+b^{\prime}$ which can be written in the form

[TABLE]

Using (2)–(4) and counting integer points in the segments $[0,m]$, $[t_{1},t_{2}]$, $[t_{1},t_{2}+m]$ we get the lemma. ∎