Algebra/Chapter 10/Symmetric Polynomials/Monomial Ordering

Definition 1

Let $(A_{1},\ldots ,A_{n})$ be an N-tuple. Let us define:

{\vec {A}}{}^{n}=(A_{1},\ldots ,A_{n})

For functions ${\vec {F}}{}^{m},{\vec {G}}{}^{n}$ let us define:

{\vec {F}}{}^{m}({\vec {G}}{}^{n})={\bigl (}F_{1}(G_{1},\ldots ,G_{n}),\ldots ,F_{m}(G_{1},\ldots ,G_{n}){\bigr )}

This abbreviated notation will be of great use to us in the following pages and in the proof.

Definition 2

Let $\mathbb {F}$ be a field. Then $\mathbb {F} [{\vec {X}}{}^{n}]$ is the polynomial space in variables $X_{1},\ldots ,X_{n}$ with coefficients in $\mathbb {F}$ .

A monomial is a polynomial of the form $c\,X_{1}^{a_{1}}\!\cdots X_{n}^{a_{n}}$ , such that $c\in \mathbb {F}$ and $a_{1},\ldots ,a_{n}\in \mathbb {N}$ .

Definition 3

Let $X=X_{1}\cdots X_{n}$ , and let $a=(a_{1},\ldots ,a_{n})\in \mathbb {N} ^{n}$ be an exponent vector. Let us define:

$X^{a}=X_{1}^{a_{1}}\!\cdots X_{n}^{a_{n}}$
$\deg(X^{a})=a_{1}\!+\cdots +a_{n}$
The degree of a non-zero polynomial is equal to the maximum of the degrees of its compsing monomials.

Properties

Monomial multiplication maintains exponent vector addition:

{\begin{aligned}X^{a}\!\cdot X^{b}&=(X_{1}^{a_{1}}\!\cdots X_{n}^{a_{n}})\cdot (X_{1}^{b_{1}}\!\cdots X_{n}^{b_{n}})\\[5pt]&=(X_{1}^{a_{1}}\!\cdot X_{1}^{b_{1}})\cdots (X_{n}^{a_{n}}\!\cdot X_{n}^{b_{n}})\\[5pt]&=X_{1}^{a_{1}+b_{1}}\!\cdots X_{n}^{a_{n}+b_{n}}\\[5pt]&=X^{a+b}\end{aligned}}

Definition 4

Let $X^{a},X^{b}$ be monomials.

We say that $X^{a}$ is of lower order than $X^{b}$ (and denote it by $X^{a}\!\prec X^{b}$ ) if there exists an index $1\leq k\leq n$ such that

{\begin{cases}a_{i}=b_{i}&:\!1\leq i\leq k-1\\[3pt]a_{i}<b_{i}&:i=k\end{cases}}

In other words, the vectors $(a_{1},\ldots ,a_{n}),(b_{1},\ldots ,b_{n})$ have a lexicographic ordering.

In a polynomial $F$ , the monomial of maximal order is called the leading monomial, and is denoted by ${\text{L}}(F)$ .

Example

x_{1}^{7}x_{2}^{3}x_{3}^{10}\!\prec x_{1}^{7}x_{2}^{31}x_{3}\iff (7,3,10)\prec (7,31,1)

Lemma

Let $F,G\in \mathbb {F} [{\vec {X}}{}^{n}]$ be polynomials. Then ${\text{L}}(F\cdot G)={\text{L}}(F)\cdot {\text{L}}(G)$ .

Proof

Let $X^{a},X^{b},X^{f},X^{g}$ be monomials, with $X^{f}={\text{L}}(F),\,X^{g}={\text{L}}(G)$ .

1. Let us assume that $X^{a}\!\prec X^{f}$ . We will show that $X^{a+c}\!\prec X^{f+c}$ for all $X^{c}$ .
By definition, there exists an index $1\leq k\leq n$ such that

{\begin{cases}a_{i}(+\,c_{i})=f_{i}(+\,c_{i})&:\!1\leq i\leq k-1\\[3pt]a_{i}(+\,c_{i})<f_{i}(+\,c_{i})&:i=k\end{cases}}

2. Let us assume also that $X^{b}\!\prec X^{g}$ . We will show that $X^{a+b}\!\prec X^{f+g}$ .
By definition, there exist indexes $k_{1},k_{2}\in \{1,\ldots ,n\}$ such that respectively