\documentclass[numbers=enddot,12pt,final,onecolumn,notitlepage]{scrartcl}%
\usepackage[headsepline,footsepline,manualmark]{scrlayer-scrpage}
\usepackage[all,cmtip]{xy}
\usepackage{amssymb}
\usepackage{amsmath}
\usepackage{amsthm}
\usepackage{framed}
\usepackage{comment}
\usepackage{color}
\usepackage[breaklinks=True]{hyperref}
\usepackage[sc]{mathpazo}
\usepackage[T1]{fontenc}
\usepackage{needspace}
\usepackage{tabls}
\usepackage{ytableau}
\usepackage{tikz}
\usepackage{pgfplots}
\usepackage[type={CC}, modifier={zero}, version={1.0},]{doclicense}
%TCIDATA{OutputFilter=latex2.dll}
%TCIDATA{Version=5.50.0.2960}
%TCIDATA{LastRevised=Friday, September 25, 2026 12:52:32}
%TCIDATA{SuppressPackageManagement}
%TCIDATA{<META NAME="GraphicsSave" CONTENT="32">}
%TCIDATA{<META NAME="SaveForMode" CONTENT="1">}
%TCIDATA{BibliographyScheme=Manual}
%TCIDATA{Language=American English}
%BeginMSIPreambleData
\providecommand{\U}[1]{\protect\rule{.1in}{.1in}}
%EndMSIPreambleData
\pgfplotsset{compat=1.13}
\theoremstyle{definition}
\newtheorem{theo}{Theorem}[subsection]
\newtheorem{exer}{Exercise}[subsubsection]
\newenvironment{theorem}[1][]
{\begin{theo}[#1]\begin{leftbar}}
{\end{leftbar}\end{theo}}
\newtheorem{lem}[theo]{Lemma}
\newenvironment{lemma}[1][]
{\begin{lem}[#1]\begin{leftbar}}
{\end{leftbar}\end{lem}}
\newtheorem{prop}[theo]{Proposition}
\newenvironment{proposition}[1][]
{\begin{prop}[#1]\begin{leftbar}}
{\end{leftbar}\end{prop}}
\newtheorem{defi}[theo]{Definition}
\newenvironment{definition}[1][]
{\begin{defi}[#1]\begin{leftbar}}
{\end{leftbar}\end{defi}}
\newtheorem{remk}[theo]{Remark}
\newenvironment{remark}[1][]
{\begin{remk}[#1]\begin{leftbar}}
{\end{leftbar}\end{remk}}
\newtheorem{coro}[theo]{Corollary}
\newenvironment{corollary}[1][]
{\begin{coro}[#1]\begin{leftbar}}
{\end{leftbar}\end{coro}}
\newtheorem{conv}[theo]{Convention}
\newenvironment{convention}[1][]
{\begin{conv}[#1]\begin{leftbar}}
{\end{leftbar}\end{conv}}
\newtheorem{quest}[theo]{Question}
\newenvironment{question}[1][]
{\begin{quest}[#1]\begin{leftbar}}
{\end{leftbar}\end{quest}}
\newtheorem{warn}[theo]{Warning}
\newenvironment{warning}[1][]
{\begin{warn}[#1]\begin{leftbar}}
{\end{leftbar}\end{warn}}
\newtheorem{conj}[theo]{Conjecture}
\newenvironment{conjecture}[1][]
{\begin{conj}[#1]\begin{leftbar}}
{\end{leftbar}\end{conj}}
\newtheorem{exam}[theo]{Example}
\newenvironment{example}[1][]
{\begin{exam}[#1]\begin{leftbar}}
{\end{leftbar}\end{exam}}
\newtheorem{exmp}[exer]{Exercise}
\newenvironment{exercise}[1][]
{\begin{exmp}[#1]\begin{leftbar}}
{\end{leftbar}\end{exmp}}
\newenvironment{statement}{\begin{quote}}{\end{quote}}
\newenvironment{fineprint}{\begin{small}}{\end{small}}
\iffalse
\newenvironment{proof}[1][Proof]{\noindent\textbf{#1.} }{\ \rule{0.5em}{0.5em}}
\newenvironment{convention}[1][Convention]{\noindent\textbf{#1.} }{\ \rule{0.5em}{0.5em}}
\newenvironment{question}[1][Question]{\noindent\textbf{#1.} }{\ \rule{0.5em}{0.5em}}
\fi
\let\sumnonlimits\sum
\let\prodnonlimits\prod
\let\cupnonlimits\bigcup
\let\capnonlimits\bigcap
\renewcommand{\sum}{\sumnonlimits\limits}
\renewcommand{\prod}{\prodnonlimits\limits}
\renewcommand{\bigcup}{\cupnonlimits\limits}
\renewcommand{\bigcap}{\capnonlimits\limits}
\usetikzlibrary{arrows,arrows.meta,decorations.markings}
\setlength\tablinesep{3pt}
\setlength\arraylinesep{3pt}
\setlength\extrarulesep{3pt}
\setlength\textheight{22.5cm}
\setlength\textwidth{14.8cm}
\newenvironment{verlong}{}{}
\newenvironment{vershort}{}{}
\newenvironment{noncompile}{}{}
\excludecomment{verlong}
\includecomment{vershort}
\excludecomment{noncompile}
\newcommand{\defn}[1]{{\color{darkred}\emph{#1}}}
\newcommand{\CC}{\mathbb{C}}
\newcommand{\RR}{\mathbb{R}}
\newcommand{\QQ}{\mathbb{Q}}
\newcommand{\NN}{\mathbb{N}}
\newcommand{\ZZ}{\mathbb{Z}}
\newcommand{\KK}{\mathbb{K}}
\newcommand{\set}[1]{\left\{ #1 \right\}}
\newcommand{\abs}[1]{\left| #1 \right|}
\newcommand{\tup}[1]{\left( #1 \right)}
\newcommand{\ive}[1]{\left[ #1 \right]}
\newcommand{\floor}[1]{\left\lfloor #1 \right\rfloor}
\newcommand{\mono}{\hookrightarrow}
\newcommand{\epi}{\twoheadrightarrow}
\newcommand{\iso}{\overset{\cong}{\to}}
\newcommand{\arinj}{\ar@{_{(}->}}
\newcommand{\arinjrev}{\ar@{^{(}->}}
\newcommand{\arsurj}{\ar@{->>}}
\newcommand{\arelem}{\ar@{|->}}
\newcommand{\arback}{\ar@{<-}}
\newcommand{\Ker}{\operatorname{Ker}}
\newcommand{\Coker}{\operatorname{Coker}}
\newcommand{\incpdftexpic}[1]{{\def\svgwidth{\columnwidth} \input{#1} }}
\definecolor{dbluecolor}{rgb}{0.01,0.02,0.7}
\definecolor{dgreencolor}{rgb}{0.2,0.4,0.0}
\definecolor{darkred}{rgb}{0.7,0,0}
\newtheoremstyle{plainsl}
{8pt plus 2pt minus 4pt}
{8pt plus 2pt minus 4pt}
{\slshape}
{0pt}
{\bfseries}
{.}
{5pt plus 1pt minus 1pt}
{}
\theoremstyle{plainsl}
\ihead{Math 701, Fall 2026, Lecture 2, version \today}
\ohead{page \thepage}
\cfoot{}
\begin{document}
\section*{Math 701 Fall 2026, Lecture 2: The classical families: $e,h,p,m$}

\textbf{website:}
\texttt{\href{https://www.cip.ifi.lmu.de/~grinberg/t/26fs}{\texttt{https://www.cip.ifi.lmu.de/\symbol{126}%
grinberg/t/26fs/}}}

\setcounter{section}{0}

\section{Symmetric polynomials}

So we have seen that symmetric polynomials can be useful. What can we learn
about them?

\subsection{Definitions and examples}

We begin by defining them.

\begin{convention}
We fix a commutative ring $K$ and a nonnegative integer $N\in\mathbb{N}$.

Throughout this chapter, $K$ and $N$ will be fixed.

We let $S_{N}$ denote the $N$-th symmetric group, i.e., the group of all
permutations of $\left[  N\right]  =\left\{  1,2,\ldots,N\right\}  $. (The
notation $\left[  n\right]  $ stands for the set $\left\{  1,2,\ldots
,n\right\}  $ whenever $n$ is an integer; in particular, $\left[  n\right]
=\varnothing$ whenever $n\leq0$.)
\end{convention}

\begin{definition}
\ \ 

\begin{enumerate}
\item[\textbf{(a)}] Let $\mathcal{P}$ be the polynomial ring $K\left[
x_{1},x_{2},\ldots,x_{N}\right]  $. This is a commutative $K$-algebra.

\item[\textbf{(b)}] The symmetric group $S_{N}$ acts on $\mathcal{P}$ by the
formula%
\[
\sigma\cdot f=f\left(  x_{\sigma\left(  1\right)  },x_{\sigma\left(  2\right)
},\ldots,x_{\sigma\left(  N\right)  }\right)  \ \ \ \ \ \ \ \ \ \ \text{for
any }\sigma\in S_{N}\text{ and }f\in\mathcal{P}.
\]
In other words, $\sigma\cdot f$ is obtained from the polynomial $f$ by
substituting $x_{\sigma\left(  i\right)  }$ for each $x_{i}$ (simultaneously).
For example, if $N=3$ and $f=x_{1}^{2}+x_{2}-x_{3}$, then%
\[
\underbrace{\operatorname*{cyc}\nolimits_{1,2,3}}_{\text{the cycle
}1\rightarrow2\rightarrow3\rightarrow1}\cdot\,f=f\left(  x_{2},x_{3}%
,x_{1}\right)  =x_{2}^{2}+x_{3}-x_{1}.
\]
So the group $S_{N}$ acts on $\mathcal{P}$ by permuting the variables. (See
\cite[Proposition 7.1.4 and Proposition 7.1.5]{21s} for a proof that this is
an actual group action by algebra automorphisms, i.e., we have $\sigma
\cdot\left(  fg\right)  =\left(  \sigma\cdot f\right)  \left(  \sigma\cdot
g\right)  $ for all $\sigma\in S_{N}$ and $f,g\in\mathcal{P}$.)

\item[\textbf{(c)}] A polynomial $f\in\mathcal{P}$ is said to be
\textbf{symmetric} if it satisfies%
\[
\sigma\cdot f=f\ \ \ \ \ \ \ \ \ \ \text{for all }\sigma\in S_{N}.
\]


\item[\textbf{(d)}] We let $\mathcal{S}$ be the set of all symmetric
polynomials $f\in\mathcal{P}$.
\end{enumerate}
\end{definition}

\begin{example}
Let $N=3$ and $K=\mathbb{Q}$, and let us rename the variables $x_{1}%
,x_{2},x_{3}$ as $x,y,z$ for brevity (I will often do this for $N=3$).

\begin{enumerate}
\item[\textbf{(a)}] We have $0,1\in\mathcal{S}$ and $x+y+z\in\mathcal{S}$ and
$\left(  x+1\right)  \left(  y+1\right)  \left(  z+1\right)  \in\mathcal{S}$
and%
\[
x^{2}y+y^{2}z+z^{2}x+y^{2}x+z^{2}y+x^{2}z\in\mathcal{S}.
\]


\item[\textbf{(b)}] We have $x+y\notin\mathcal{S}$ (since the transposition
$t_{1,3}\in S_{3}$ sends $x+y$ to $z+y\neq x+y$).

\item[\textbf{(c)}] We have $\left(  x-y\right)  \left(  y-z\right)  \left(
z-x\right)  \notin\mathcal{S}$. This polynomial $\left(  x-y\right)  \left(
y-z\right)  \left(  z-x\right)  $ is an example of an \textbf{antisymmetric}
polynomial, meaning a polynomial $f\in\mathcal{P}$ that satisfies $\sigma\cdot
f=\left(  -1\right)  ^{\sigma}f$ for each $\sigma\in S_{N}$, where $\left(
-1\right)  ^{\sigma}$ denotes the sign of a permutation $\sigma$.

\item[\textbf{(d)}] We have $\dfrac{1}{\left(  1-x\right)  \left(  1-y\right)
\left(  1-z\right)  }\notin\mathcal{S}$; this is symmetric but not a
polynomial. It is a symmetric power series.
\end{enumerate}
\end{example}

\begin{theorem}
The subset $\mathcal{S}$ is a $K$-subalgebra of $\mathcal{P}$.
\end{theorem}

\begin{proof}
This is just saying that $\mathcal{S}$ is closed under addition, scaling and
multiplication and contains $0$ and $1$. All of this is clear because $S_{N}$
acts on $\mathcal{P}$ by algebra automorphisms.
\end{proof}

\begin{definition}
The $K$-subalgebra $\mathcal{S}$ of $\mathcal{P}$ is called the \textbf{ring
of symmetric polynomials in }$N$ \textbf{variables over }$K$.
\end{definition}

We now shall define some specific symmetric polynomials. First, some notations:

\begin{definition}
\ \ 

\begin{enumerate}
\item[\textbf{(a)}] A \textbf{monomial} means an expression of the form
$x_{1}^{a_{1}}x_{2}^{a_{2}}\cdots x_{N}^{a_{N}}$ with $a_{1},a_{2}%
,\ldots,a_{N}\in\mathbb{N}$. (Recall that $0\in\mathbb{N}$.)

\item[\textbf{(b)}] The \textbf{degree} $\deg\mathfrak{m}$ of a monomial
$\mathfrak{m}=x_{1}^{a_{1}}x_{2}^{a_{2}}\cdots x_{N}^{a_{N}}$ is defined to be
$a_{1}+a_{2}+\cdots+a_{N}$.

\item[\textbf{(c)}] A monomial $\mathfrak{m}=x_{1}^{a_{1}}x_{2}^{a_{2}}\cdots
x_{N}^{a_{N}}$ is said to be \textbf{squarefree} if $a_{1},a_{2},\ldots
,a_{N}\in\left\{  0,1\right\}  $.

\item[\textbf{(d)}] A monomial $\mathfrak{m}=x_{1}^{a_{1}}x_{2}^{a_{2}}\cdots
x_{N}^{a_{N}}$ is said to be \textbf{primal} if there is at most one
$i\in\left[  N\right]  $ satisfying $a_{i}>0$ (that is, if $\mathfrak{m}$ is
$1$ or some power of some $x_{i}$).
\end{enumerate}
\end{definition}

\begin{definition}
\ \ 

\begin{enumerate}
\item[\textbf{(a)}] For each $n\in\mathbb{Z}$, define the $n$\textbf{-th
elementary symmetric polynomial} $e_{n}\in\mathcal{S}$ by%
\begin{align*}
e_{n}  &  =\sum_{\substack{\left(  i_{1},i_{2},\ldots,i_{n}\right)  \in\left[
N\right]  ^{n};\\i_{1}<i_{2}<\cdots<i_{n}}}x_{i_{1}}x_{i_{2}}\cdots x_{i_{n}%
}\\
&  =\sum_{\substack{L\subseteq\left[  N\right]  ;\\\left\vert L\right\vert
=n}}\ \ \prod_{i\in L}x_{i}\\
&  =\left(  \text{sum of all squarefree monomials of degree }n\right)  .
\end{align*}


\item[\textbf{(b)}] For each $n\in\mathbb{Z}$, define the $n$\textbf{-th
complete homogeneous symmetric polynomial} $h_{n}\in\mathcal{S}$ by%
\begin{align*}
h_{n}  &  =\sum_{\substack{\left(  i_{1},i_{2},\ldots,i_{n}\right)  \in\left[
N\right]  ^{n};\\i_{1}\leq i_{2}\leq\cdots\leq i_{n}}}x_{i_{1}}x_{i_{2}}\cdots
x_{i_{n}}\\
&  =\sum_{\substack{L\text{ is a multisubset of }\left[  N\right]
;\\\left\vert L\right\vert =n\text{ (counting multiplicities)}}}\ \ \prod
_{\substack{i\in L\\\text{(with multiplicities)}}}x_{i}\\
&  \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \left(
\begin{array}
[c]{c}%
\text{here we are using the language of multisets;}\\
\text{we shall mostly avoid this language}%
\end{array}
\right) \\
&  =\left(  \text{sum of all monomials of degree }n\right)  .
\end{align*}


\item[\textbf{(c)}] For each $n\in\mathbb{Z}$, define the $n$\textbf{-th
power-sum symmetric polynomial} (aka $n$\textbf{-th power sum})\textbf{
}$p_{n}\in\mathcal{S}$ by%
\begin{align*}
p_{n}  &  =%
\begin{cases}
x_{1}^{n}+x_{2}^{n}+\cdots+x_{N}^{n}, & \text{if }n>0;\\
1, & \text{if }n=0;\\
0, & \text{if }n<0
\end{cases}
\\
&  =\left(  \text{sum of all primal monomials of degree }n\right)  .
\end{align*}

\end{enumerate}
\end{definition}

\Needspace{12\baselineskip}
\begin{example}
\ \ 

\begin{enumerate}
\item[\textbf{(a)}] The $2$-nd elementary symmetric polynomial is%
\begin{align*}
e_{2}  &  =\sum_{\substack{\left(  i_{1},i_{2}\right)  \in\left[  N\right]
^{2};\\i_{1}<i_{2}}}x_{i_{1}}x_{i_{2}}=\sum_{\substack{\left(  i,j\right)
\in\left[  N\right]  ^{2};\\i<j}}x_{i}x_{j}\\
&  =\begin{array}[t]{@{}c*{3}{@{\;}c@{\;}c}@{}}
x_{1}x_{2} & + & x_{1}x_{3} & + & \cdots & + & x_{1}x_{N}\\
           & + & x_{2}x_{3} & + & \cdots & + & x_{2}x_{N}\\
           &   &           &   & \ddots &   & \vdots\\
           &   &           &   &        & + & x_{N-1}x_{N}.
\end{array}
\end{align*}


\item[\textbf{(b)}] The $2$-nd complete homogeneous symmetric polynomial is%
\begin{align*}
h_{2}  &  =\sum_{\substack{\left(  i_{1},i_{2}\right)  \in\left[  N\right]
^{2};\\i_{1}\leq i_{2}}}x_{i_{1}}x_{i_{2}}=\sum_{\substack{\left(  i,j\right)
\in\left[  N\right]  ^{2};\\i\leq j}}x_{i}x_{j}\\
&  =\begin{array}[t]{@{}c*{5}{@{\;}c@{\;}c}@{}}
x_{1}^{2} & + & x_{1}x_{2} & + & x_{1}x_{3} & + & \cdots & + & x_{1}x_{N-1} & + & x_{1}x_{N}\\
          & + & x_{2}^{2}  & + & x_{2}x_{3} & + & \cdots & + & x_{2}x_{N-1} & + & x_{2}x_{N}\\
          &   &           &   &           &   & \ddots &   & \vdots       &   & \vdots\\
          &   &           &   &           &   &        & + & x_{N-1}^{2}  & + & x_{N-1}x_{N}\\
          &   &           &   &           &   &        &   &              & + & x_{N}^{2}.
\end{array}
\end{align*}


\item[\textbf{(c)}] The $2$-nd power sum is%
\[
p_{2}=x_{1}^{2}+x_{2}^{2}+\cdots+x_{N}^{2}.
\]
Note that%
\[
h_{2}=e_{2}+p_{2}.
\]


\item[\textbf{(d)}] We have%
\[
e_{1}=h_{1}=p_{1}=x_{1}+x_{2}+\cdots+x_{N}.
\]


\item[\textbf{(e)}] We have%
\[
e_{0}=h_{0}=p_{0}=1,
\]
since the only monomial of degree $0$ is $1$.

\item[\textbf{(f)}] If $n<0$, then $e_{n}=h_{n}=p_{n}=0$, since there are no
monomials of negative degree.

\item[\textbf{(g)}] If $N=0$ (so $\mathcal{P}$ is a polynomial ring in $0$
variables), then%
\[
e_{n}=h_{n}=p_{n}=0\ \ \ \ \ \ \ \ \ \ \text{for all }n>0,
\]
but $e_{0}=h_{0}=p_{0}=1$. (Note that polynomials in $0$ variables are just
constants in a trenchcoat.)
\end{enumerate}
\end{example}

\begin{proposition}
For each integer $n>N$, we have $e_{n}=0$.
\end{proposition}

\begin{proof}
We defined $e_{n}$ as a sum over all $n$-tuples $\left(  i_{1},i_{2}%
,\ldots,i_{n}\right)  \in\left[  N\right]  ^{n}$ with $i_{1}<i_{2}%
<\cdots<i_{n}$. But there are no such $n$-tuples when $n>N$ (since any such
$n$-tuple must have $n$ distinct entries, but no $n$ distinct entries exist in
$\left[  N\right]  $). So the sum is empty and equals $0$.
\end{proof}

Thus, there are only $N$ \textquotedblleft interesting\textquotedblright%
\ elementary symmetric polynomials: $e_{1},e_{2},\ldots,e_{N}$. In contrast,
there are infinitely many \textquotedblleft interesting\textquotedblright%
\ complete homogeneous polynomials and power sums (when $N>0$): For example,
if $N=2$, then
\begin{align*}
h_{5}  &  =x_{1}^{5}+x_{1}^{4}x_{2}+x_{1}^{3}x_{2}^{2}+x_{1}^{2}x_{2}%
^{3}+x_{1}x_{2}^{4}+x_{2}^{5}\ \ \ \ \ \ \ \ \ \ \text{and}\\
p_{5}  &  =x_{1}^{5}+x_{2}^{5}.
\end{align*}


So we have three sequences of symmetric polynomials:%
\[
\left(  e_{0},e_{1},e_{2},\ldots\right)  ,\ \ \ \ \ \ \ \ \ \ \left(
h_{0},h_{1},h_{2},\ldots\right)  ,\ \ \ \ \ \ \ \ \ \ \left(  p_{0}%
,p_{1},p_{2},\ldots\right)  .
\]
Our first nontrivial result are three formulas that connect these sequences
recursively (the so-called \textbf{Newton--Girard identities}):

\begin{theorem}
[Newton--Girard identities]\label{thm.symp.newton}For any positive integer
$n$, we have%
\begin{align}
\sum_{j=0}^{n}\left(  -1\right)  ^{j}e_{j}h_{n-j} &
=0;\label{eq.thm.symp.newton.eh}\\
\sum_{j=1}^{n}\left(  -1\right)  ^{j-1}e_{n-j}p_{j} &  =ne_{n}%
;\label{eq.thm.symp.newton.ep}\\
\sum_{j=1}^{n}h_{n-j}p_{j} &  =nh_{n}.\label{eq.thm.symp.newton.hp}%
\end{align}


Expanded, these are saying that%
\begin{align*}
\underbrace{e_{0}}_{=1}h_{n}-e_{1}h_{n-1}+e_{2}h_{n-2}-e_{3}h_{n-3}\pm
\cdots+\left(  -1\right)  ^{n}e_{n}\underbrace{h_{0}}_{=1}  &  =0;\\
e_{n-1}p_{1}-e_{n-2}p_{2}+e_{n-3}p_{3}-e_{n-4}p_{4}\pm\cdots+\left(
-1\right)  ^{n-1}\underbrace{e_{0}}_{=1}p_{n}  &  =ne_{n};\\
h_{n-1}p_{1}+h_{n-2}p_{2}+h_{n-3}p_{3}+h_{n-4}p_{4}+\cdots+\underbrace{h_{0}%
}_{=1}p_{n}  &  =nh_{n}.
\end{align*}

\end{theorem}

\begin{example}
For $n=2$, the formula (\ref{eq.thm.symp.newton.ep}) says%
\[
e_{1}p_{1}-\underbrace{e_{0}}_{=1}p_{2}=2e_{2},
\]
so that%
\begin{align*}
p_{2} &  =e_{1}\underbrace{p_{1}}_{=e_{1}}-2e_{2}=e_{1}^{2}-2e_{2}\\
&  =\left(  x_{1}+x_{2}+\cdots+x_{N}\right)  ^{2}-2\sum_{i<j}x_{i}x_{j}.
\end{align*}
This is precisely the $e_{1}\left(  \lambda_{1}^{2},\lambda_{2}^{2}%
,\ldots,\lambda_{N}^{2}\right)  $ identity that we saw at the end of Lecture
1. Sadly, $e_{2}\left(  \lambda_{1}^{2},\lambda_{2}^{2},\ldots,\lambda_{N}%
^{2}\right)  $ cannot be computed this easily.
\end{example}

How do we prove Theorem \ref{thm.symp.newton}? I will only show the first
identity, i.e., (\ref{eq.thm.symp.newton.eh}), leaving the other two to
homework. To prove it, I will use power series in additional variables
$t,u,v$. The crucial fact here is the following:

\begin{proposition}
\label{prop.symp.eh-genfun}In addition to the indeterminates $x_{i}$ already
in $\mathcal{P}$, we introduce three further indeterminates $t,u,v$. Then:

\begin{enumerate}
\item[\textbf{(a)}] In the polynomial ring $\mathcal{P}\left[  t\right]  $, we
have%
\[
\prod_{i=1}^{N}\left(  1-tx_{i}\right)  =\sum_{n\in\mathbb{N}}\left(
-1\right)  ^{n}t^{n}e_{n}.
\]


\item[\textbf{(b)}] In the polynomial ring $\mathcal{P}\left[  u,v\right]  $,
we have%
\[
\prod_{i=1}^{N}\left(  u-vx_{i}\right)  =\sum_{n=0}^{N}\left(  -1\right)
^{n}u^{N-n}v^{n}e_{n}.
\]


\item[\textbf{(c)}] In the formal power series ring $\mathcal{P}\left[
\left[  t\right]  \right]  $, we have%
\[
\prod_{i=1}^{N}\dfrac{1}{1-tx_{i}}=\sum_{n\in\mathbb{N}}t^{n}h_{n}.
\]

\end{enumerate}
\end{proposition}

\begin{proof}
\textbf{(c)} For each $i\in\left[  N\right]  $, the geometric series formula
$\dfrac{1}{1-y}=1+y+y^{2}+y^{3}+\cdots$ yields%
\[
\dfrac{1}{1-tx_{i}}=1+tx_{i}+\left(  tx_{i}\right)  ^{2}+\left(
tx_{i}\right)  ^{3}+\cdots=\sum_{a\in\mathbb{N}}\left(  tx_{i}\right)  ^{a}.
\]
Multiplying these equalities over all the $i$'s yields%
\begin{align*}
\prod_{i=1}^{N}\dfrac{1}{1-tx_{i}}  &  =\prod_{i=1}^{N}\ \ \sum_{a\in
\mathbb{N}}\left(  tx_{i}\right)  ^{a}\\
&  =\sum_{\left(  a_{1},a_{2},\ldots,a_{N}\right)  \in\mathbb{N}^{N}%
}\underbrace{\left(  tx_{1}\right)  ^{a_{1}}\left(  tx_{2}\right)  ^{a_{2}%
}\cdots\left(  tx_{N}\right)  ^{a_{N}}}_{=t^{a_{1}+a_{2}+\cdots+a_{N}}%
x_{1}^{a_{1}}x_{2}^{a_{2}}\cdots x_{N}^{a_{N}}}\\
&  \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \left(  \text{by the product rule,
aka generalized distributivity}\right) \\
&  =\sum_{\left(  a_{1},a_{2},\ldots,a_{N}\right)  \in\mathbb{N}^{N}}%
t^{a_{1}+a_{2}+\cdots+a_{N}}x_{1}^{a_{1}}x_{2}^{a_{2}}\cdots x_{N}^{a_{N}}\\
&  =\sum_{n\in\mathbb{N}}\ \ \sum_{\substack{\left(  a_{1},a_{2},\ldots
,a_{N}\right)  \in\mathbb{N}^{N};\\a_{1}+a_{2}+\cdots+a_{N}=n}%
}\underbrace{t^{a_{1}+a_{2}+\cdots+a_{N}}}_{=t^{n}}x_{1}^{a_{1}}x_{2}^{a_{2}%
}\cdots x_{N}^{a_{N}}\\
&  =\sum_{n\in\mathbb{N}}t^{n}\underbrace{\sum_{\substack{\left(  a_{1}%
,a_{2},\ldots,a_{N}\right)  \in\mathbb{N}^{N};\\a_{1}+a_{2}+\cdots+a_{N}%
=n}}x_{1}^{a_{1}}x_{2}^{a_{2}}\cdots x_{N}^{a_{N}}}_{\substack{=\left(
\text{sum of all monomials of degree }n\right)  \\=h_{n}}}\\
&  =\sum_{n\in\mathbb{N}}t^{n}h_{n},
\end{align*}
which proves part \textbf{(c)}. \medskip

\textbf{(a)} Upon substituting $-t$ for $t$, the claim of part \textbf{(a)}
takes the form%
\[
\prod_{i=1}^{N}\left(  1+tx_{i}\right)  =\sum_{n\in\mathbb{N}}t^{n}e_{n}.
\]
This can be shown similarly to part \textbf{(c)}, where instead of the
geometric series formula we use the trivial formula $1+y=\sum\limits_{a\in
\left\{  0,1\right\}  }y^{a}$. Thus, $1+tx_{i}=\sum\limits_{a\in\left\{
0,1\right\}  }\left(  tx_{i}\right)  ^{a}$ for each $i\in\left[  N\right]
$.\ Multiplying this over all the $i$'s, we obtain%
\begin{align*}
\prod_{i=1}^{N}\left(  1+tx_{i}\right)   &  =\prod_{i=1}^{N}\ \ \sum
_{a\in\left\{  0,1\right\}  }\left(  tx_{i}\right)  ^{a}\\
&  =\sum_{\left(  a_{1},a_{2},\ldots,a_{N}\right)  \in\left\{  0,1\right\}
^{N}}\left(  tx_{1}\right)  ^{a_{1}}\left(  tx_{2}\right)  ^{a_{2}}%
\cdots\left(  tx_{N}\right)  ^{a_{N}}\\
&  =\sum_{\left(  a_{1},a_{2},\ldots,a_{N}\right)  \in\left\{  0,1\right\}
^{N}}t^{a_{1}+a_{2}+\cdots+a_{N}}x_{1}^{a_{1}}x_{2}^{a_{2}}\cdots x_{N}%
^{a_{N}}\\
&  =\sum_{n\in\mathbb{N}}\ \ \sum_{\substack{\left(  a_{1},a_{2},\ldots
,a_{N}\right)  \in\left\{  0,1\right\}  ^{N};\\a_{1}+a_{2}+\cdots+a_{N}%
=n}}\underbrace{t^{a_{1}+a_{2}+\cdots+a_{N}}}_{=t^{n}}x_{1}^{a_{1}}%
x_{2}^{a_{2}}\cdots x_{N}^{a_{N}}\\
&  =\sum_{n\in\mathbb{N}}t^{n}\underbrace{\sum_{\substack{\left(  a_{1}%
,a_{2},\ldots,a_{N}\right)  \in\left\{  0,1\right\}  ^{N};\\a_{1}+a_{2}%
+\cdots+a_{N}=n}}x_{1}^{a_{1}}x_{2}^{a_{2}}\cdots x_{N}^{a_{N}}}%
_{\substack{=\left(  \text{sum of all squarefree monomials of degree
}n\right)  \\=e_{n}}}\\
&  =\sum_{n\in\mathbb{N}}t^{n}e_{n},
\end{align*}
just as we wanted to prove. \medskip

\textbf{(b)} is similar to \textbf{(a)} (homework).
\end{proof}

Next time we will derive the first Newton--Girard formula from this.

\bigskip

\begin{thebibliography}{999}                                                                                              %


\bibitem[21s]{21s}Darij Grinberg, \textit{An Introduction to Algebraic
Combinatorics (Math 531, Winter 2024 lecture notes)}, 22 July 2026.\newline\url{http://www.cip.ifi.lmu.de/~grinberg/t/21s/lecs.pdf}
\end{thebibliography}


\end{document}