Algebra/Chapter 10/Symmetric Polynomials/Elementary Symmetric Polynomials

Let there be a polynomial of degree ${\displaystyle n\geq 1}$

${\displaystyle P(z)=\sum _{k\,=\,0}^{n}a_{k}z^{k}\in \mathbb {C} [z]}$

By the Fundamental Theorem of Algebra, it has ${\displaystyle n}$ complex roots (with multiplicity). Then we can write:

${\displaystyle P(z)=a_{n}\prod _{k\,=\,1}^{n}(z-z_{k})}$

As we know, Vieta's formulae link between the coefficients and roots of a polynomial:

${\displaystyle {\begin{cases}z_{1}+\cdots +z_{n}=-{\dfrac {a_{n-1}}{a_{n}}}\\(z_{1}z_{2}+\cdots +z_{1}z_{n})+(z_{2}z_{3}+\cdots +z_{2}z_{n})+\cdots +z_{n-1}z_{n}={\dfrac {a_{n-2}}{a_{n}}}\\(z_{1}z_{2}z_{3}+\cdots +z_{1}z_{2}z_{n})+(z_{1}z_{3}z_{4}+\cdots +z_{1}z_{3}z_{n})+\cdots +(z_{2}z_{3}z_{4}+\cdots +z_{2}z_{n-1}z_{n})+\cdots +z_{n-2}z_{n-1}z_{n}=-{\dfrac {a_{n-3}}{a_{n}}}\\\vdots \\z_{1}\!\cdots z_{n}=(-1)^{n}{\dfrac {a_{0}}{a_{n}}}\end{cases}}}$

As we can see, these sums are symmetric polynomial, and are called elementary symmetric polynomials.

The elementary symmetric polynomials in variables ${\displaystyle X_{1},\ldots ,X_{n}}$, are defined as such:

${\displaystyle {\begin{aligned}E_{1}({\vec {X}}{}^{n})&=\sum _{1\leq i\leq n}X_{i}\\[5pt]E_{2}({\vec {X}}{}^{n})&=\sum _{1\leq i_{1}<i_{2}\leq n}\!\!\!\!X_{i_{1}}X_{i_{2}}\\&\,\,\,\vdots \\[5pt]E_{k}({\vec {X}}{}^{n})&=\sum _{1\leq i_{1}<\cdots <i_{k}\leq n}\!\!\!\!\!\!X_{i_{1}}\!\cdots X_{i_{k}}\\&\,\,\,\vdots \\[5pt]E_{n}({\vec {X}}{}^{n})&=X_{1}\!\cdots X_{n}\end{aligned}}}$