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\begin{document}
\section*{Math 701 Fall 2026, Lecture 4: Monomial symmetric polynomials}

\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}{1}\setcounter{subsection}{2}

\subsection{Partitions and monomial symmetric polynomials}

Partitions are one of the most classical combinatorial objects. They have
intrinsic interest (see, e.g., \cite[Chapter 4]{21s} for some of their
properties) and crucial connections to number theory and representation
theory, but we will be chiefly interested in them as indexing sets for
($K$-module) bases of $\mathcal{S}$. Let us define two variants of them:

\begin{definition}
\label{def.symp.partitions} \ 

\begin{enumerate}
\item[\textbf{(a)}] An \textbf{(integer) partition} means a weakly decreasing
finite tuple of positive integers, i.e., a tuple $\lambda=\left(  \lambda
_{1},\lambda_{2},\ldots,\lambda_{k}\right)  $ with $\lambda_{1},\lambda
_{2},\ldots,\lambda_{k}\in\left\{  1,2,3,\ldots\right\}  $ and $\lambda
_{1}\geq\lambda_{2}\geq\cdots\geq\lambda_{k}$. The \textbf{length} of this
partition is $k$, while its \textbf{size} is $\left\vert \lambda\right\vert
:=\lambda_{1}+\lambda_{2}+\cdots+\lambda_{k}$. More generally, the size of any
$k$-tuple $a=\left(  a_{1},a_{2},\ldots,a_{k}\right)  $ of integers is defined
to be $\left\vert a\right\vert :=a_{1}+a_{2}+\cdots+a_{k}$.

\item[\textbf{(b)}] An $N$\textbf{-partition} means a weakly decreasing
$N$-tuple of nonnegative integers, i.e., a tuple $\lambda=\left(  \lambda
_{1},\lambda_{2},\ldots,\lambda_{N}\right)  $ with $\lambda_{1},\lambda
_{2},\ldots,\lambda_{N}\in\mathbb{N}$ and $\lambda_{1}\geq\lambda_{2}%
\geq\cdots\geq\lambda_{N}$. The \textbf{size} of this $N$-partition is
$\lambda_{1}+\lambda_{2}+\cdots+\lambda_{N}$, and is denoted by $\left\vert
\lambda\right\vert $.
\end{enumerate}
\end{definition}

For example, $\left(  4,2,2,1\right)  $ is a partition, whereas $\left(
4,2,2,1,0\right)  $ is not a partition but is a $5$-partition (since
$N$-partitions are allowed to contain zeroes, but partitions aren't). Both of
these have size $9$. Note that the empty tuple $\left(  {}\right)  $ is also a
partition (of length $0$ and size $0$).

The two above versions of partitions are very close to each other:

\begin{proposition}
\label{prop.symp.partitions-padding} There is a bijection%
\begin{align*}
\left\{  \text{partitions of length }\leq N\right\}   &  \rightarrow\left\{
N\text{-partitions}\right\}  ,\\
\left(  \lambda_{1},\lambda_{2},\ldots,\lambda_{k}\right)   &  \mapsto\left(
\lambda_{1},\lambda_{2},\ldots,\lambda_{k},\underbrace{0,0,\ldots
,0}_{N-k\text{ zeroes}}\right)  ,
\end{align*}
which preserves the size of the respective partition. Namely, this bijection
pads the partition with zeroes (at its end) in order to increase its length to
$N$. The inverse map simply removes all zeroes at the end of the given $N$-partition.
\end{proposition}

\begin{definition}
\label{def.symp.partitions-padding}We shall use this bijection to identify the
$N$-partitions with the partitions of length $\leq N$. Thus, for example, the
partition $\left(  3,1,1\right)  $ becomes identified with the $3$-partition
$\left(  3,1,1\right)  $, with the $4$-partition $\left(  3,1,1,0\right)  $,
with the $5$-partition $\left(  3,1,1,0,0\right)  $, and so on. For another
example, the empty partition $\left(  {}\right)  $ is identified with the
$N$-partition $\left(  0,0,\ldots,0\right)  $ for each $N\in\mathbb{N}$.

We will also regard partitions as infinite sequences by padding them with
infinitely many zeroes at the end. So the partition $\left(  3,1,1\right)  $
becomes identified with the infinite sequence $\left(  3,1,1,0,0,0,\ldots
\right)  $. Thus, each partition can be viewed uniquely as an infinite weakly
decreasing sequence $\left(  \lambda_{1},\lambda_{2},\lambda_{3}%
,\ldots\right)  $ of nonnegative integers that has only finitely many nonzero
entries (which necessarily are all clustered at its start because it is weakly decreasing).
\end{definition}

We next define a few basic notations related to monomials:

\begin{definition}
\label{def.coeff-bracket}Let $\mathfrak{m}$ be a monomial, and let $f$ be a
polynomial or power series. Then, $\left[  \mathfrak{m}\right]  f$ denotes the
$\mathfrak{m}$-coefficient of $f$.

For instance, $\left[  x^{2}\right]  \left(  1+x\right)  ^{4}=6$.
\end{definition}

\begin{definition}
\label{def.symp.monomial-sort} Let $a=\left(  a_{1},a_{2},\ldots,a_{N}\right)
\in\mathbb{N}^{N}$ be an $N$-tuple of nonnegative integers.

\begin{enumerate}
\item[\textbf{(a)}] We let $x^{a}$ be the monomial $x_{1}^{a_{1}}x_{2}^{a_{2}%
}\cdots x_{N}^{a_{N}}$.

\item[\textbf{(b)}] We let $\operatorname*{sort}a$ be the $N$-partition
obtained from $a$ by sorting the entries in weakly decreasing order.
\end{enumerate}
\end{definition}

For instance, $\operatorname*{sort}\left(  1,0,2,4,1\right)  =\left(
4,2,1,1,0\right)  $ and $x^{\left(  1,0,2,4,1\right)  }=x_{1}^{1}x_{2}%
^{0}x_{3}^{2}x_{4}^{4}x_{5}^{1}=x_{1}x_{3}^{2}x_{4}^{4}x_{5}$.

The following simple lemma helps us feel at home with the notations we just introduced:

\begin{lemma}
\label{lem.symp.sort-a}Let $f\in\mathcal{P}$ be a polynomial. Then:

\begin{enumerate}
\item[\textbf{(a)}] For any $N$-tuple $a=\left(  a_{1},a_{2},\ldots
,a_{N}\right)  \in\mathbb{N}^{N}$ and any permutation $\sigma\in S_{N}$, we
have%
\[
\left[  x^{a}\right]  \left(  \sigma\cdot f\right)  =\left[  x^{a\circ\sigma
}\right]  f,
\]
where $a\circ\sigma$ denotes the $N$-tuple $\left(  a_{\sigma\left(  1\right)
},a_{\sigma\left(  2\right)  },\ldots,a_{\sigma\left(  N\right)  }\right)  $.

\item[\textbf{(b)}] The polynomial $f$ is symmetric (i.e., belongs to
$\mathcal{S}$) if and only if all $a\in\mathbb{N}^{N}$ satisfy%
\[
\left[  x^{a}\right]  f=\left[  x^{\operatorname*{sort}a}\right]  f.
\]

\end{enumerate}
\end{lemma}

\begin{proof}
\textbf{(a)} Let $a=\left(  a_{1},a_{2},\ldots,a_{N}\right)  \in\mathbb{N}%
^{N}$ be any $N$-tuple. The polynomial $\sigma\cdot f$ is obtained from $f$ by
substituting $x_{\sigma\left(  i\right)  }$ for each indeterminate $x_{i}$;
this substitution clearly transforms each monomial $x_{1}^{b_{1}}x_{2}^{b_{2}%
}\cdots x_{N}^{b_{N}}$ into $x_{\sigma\left(  1\right)  }^{b_{1}}%
x_{\sigma\left(  2\right)  }^{b_{2}}\cdots x_{\sigma\left(  N\right)  }%
^{b_{N}}$. In particular, this substitution transforms the monomial
\[
x^{a\circ\sigma}=x_{1}^{a_{\sigma\left(  1\right)  }}x_{2}^{a_{\sigma\left(
2\right)  }}\cdots x_{N}^{a_{\sigma\left(  N\right)  }}%
\ \ \ \ \ \ \ \ \ \ \left(  \text{since }a\circ\sigma=\left(  a_{\sigma\left(
1\right)  },a_{\sigma\left(  2\right)  },\ldots,a_{\sigma\left(  N\right)
}\right)  \right)
\]
into
\begin{align*}
x_{\sigma\left(  1\right)  }^{a_{\sigma\left(  1\right)  }}x_{\sigma\left(
2\right)  }^{a_{\sigma\left(  2\right)  }}\cdots x_{\sigma\left(  N\right)
}^{a_{\sigma\left(  N\right)  }}  &  =x_{1}^{a_{1}}x_{2}^{a_{2}}\cdots
x_{N}^{a_{N}}\ \ \ \ \ \ \ \ \ \ \left(  \text{since }\sigma\text{ is a
permutation of }\left[  N\right]  \right) \\
&  =x^{a}%
\end{align*}
(and no other monomials get transformed into $x^{a}$, since this substitution
is clearly bijective). Thus, the $x^{a}$-coefficient of $\sigma\cdot f$ is
precisely the $x^{a\circ\sigma}$-coefficient of $f$. That is, we have $\left[
x^{a}\right]  \left(  \sigma\cdot f\right)  =\left[  x^{a\circ\sigma}\right]
f$. This proves Lemma \ref{lem.symp.sort-a} \textbf{(a)}. \medskip

\textbf{(b)} $\Longrightarrow:$ Assume that $f$ is symmetric. Let
$a\in\mathbb{N}^{N}$ be an $N$-tuple. For any $\sigma\in S_{N}$, define an
$N$-tuple $a\circ\sigma$ as in part \textbf{(a)}. Now, $\operatorname*{sort}a$
is the $N$-tuple obtained by sorting the entries of $a$ in weakly decreasing
order; in particular, it is obtained by permuting the entries of $a$. That is,
$\operatorname*{sort}a=a\circ\sigma$ for some permutation $\sigma\in S_{N}$.
Consider this $\sigma$. From part \textbf{(a)}, we know that $\left[
x^{a}\right]  \left(  \sigma\cdot f\right)  =\left[  x^{a\circ\sigma}\right]
f$. Since $\sigma\cdot f=f$ (because $f$ is symmetric) and $a\circ
\sigma=\operatorname*{sort}a$, we can rewrite this as $\left[  x^{a}\right]
f=\left[  x^{\operatorname*{sort}a}\right]  f$. Thus, we have shown that all
$a\in\mathbb{N}^{N}$ satisfy $\left[  x^{a}\right]  f=\left[
x^{\operatorname*{sort}a}\right]  f$. This proves the \textquotedblleft%
$\Longrightarrow$\textquotedblright\ direction of Lemma \ref{lem.symp.sort-a}
\textbf{(b)}.

$\Longleftarrow:$ Assume that all $a\in\mathbb{N}^{N}$ satisfy
\begin{equation}
\left[  x^{a}\right]  f=\left[  x^{\operatorname*{sort}a}\right]  f.
\label{pf.lem.symp.sort-a.b.3}%
\end{equation}
We must show that $f$ is symmetric.

Let $\sigma\in S_{N}$. Let $a\in\mathbb{N}^{N}$ be an $N$-tuple. Then, the
$N$-tuple $a\circ\sigma$ (defined as in part \textbf{(a)}) is obtained by
permuting the entries of $a$; thus it has the same entries as $a$, just in a
different order. Consequently, the $N$-tuples $a\circ\sigma$ and $a$ become
identical if we sort them both into weakly decreasing order. In other words,
$\operatorname*{sort}\left(  a\circ\sigma\right)  =\operatorname*{sort}a$.
However, applying (\ref{pf.lem.symp.sort-a.b.3}) to $a\circ\sigma$ instead of
$a$, we find $\left[  x^{a\circ\sigma}\right]  f=\left[
x^{\operatorname*{sort}\left(  a\circ\sigma\right)  }\right]  f=\left[
x^{\operatorname*{sort}a}\right]  f$ (since $\operatorname*{sort}\left(
a\circ\sigma\right)  =\operatorname*{sort}a$). Comparing this with
(\ref{pf.lem.symp.sort-a.b.3}), we obtain $\left[  x^{a\circ\sigma}\right]
f=\left[  x^{a}\right]  f$. Meanwhile, Lemma \ref{lem.symp.sort-a}
\textbf{(a)} yields $\left[  x^{a}\right]  \left(  \sigma\cdot f\right)
=\left[  x^{a\circ\sigma}\right]  f=\left[  x^{a}\right]  f$ (since we just
proved $\left[  x^{a\circ\sigma}\right]  f=\left[  x^{a}\right]  f$).

Forget that we fixed $a$. We thus have shown that $\left[  x^{a}\right]
\left(  \sigma\cdot f\right)  =\left[  x^{a}\right]  f$ for each
$a\in\mathbb{N}^{N}$. In other words, each coefficient of the polynomial
$\sigma\cdot f$ equals the corresponding coefficient of $f$. Hence, these two
polynomials are equal. That is, $\sigma\cdot f=f$. We have proved this for
each $\sigma\in S_{N}$; that is, $f$ is symmetric. This proves the
\textquotedblleft$\Longleftarrow$\textquotedblright\ direction of Lemma
\ref{lem.symp.sort-a} \textbf{(b)}.
\end{proof}

\begin{definition}
\label{def.symp.monomial-symmetric} Let $\lambda$ be any $N$-partition. Then,
the $\lambda$\textbf{-th monomial symmetric polynomial} is defined to be%
\[
m_{\lambda}:=\sum_{\substack{a\in\mathbb{N}^{N};\\\operatorname*{sort}%
a=\lambda}}x^{a}\in\mathcal{S}.
\]

\end{definition}

\begin{example}
\label{exa.symp.monomial-symmetric} Let $N=3$. Then,%
\begin{align*}
m_{\left(  2,1,0\right)  }  &  =\sum_{\substack{a\in\mathbb{N}^{N}%
;\\\operatorname*{sort}a=\left(  2,1,0\right)  }}x^{a}\\
&  =x^{\left(  2,1,0\right)  }+x^{\left(  1,2,0\right)  }+x^{\left(
1,0,2\right)  }+x^{\left(  2,0,1\right)  }+x^{\left(  0,2,1\right)
}+x^{\left(  0,1,2\right)  }\\
&  =x^{2}y+xy^{2}+xz^{2}+x^{2}z+y^{2}z+yz^{2}%
\end{align*}
(again using the shorthands $x,y,z$ for $x_{1},x_{2},x_{3}$) and%
\begin{align*}
m_{\left(  3,2,1\right)  }  &  =x^{3}y^{2}z+x^{2}y^{3}z+x^{2}yz^{3}%
+x^{3}yz^{2}+xy^{3}z^{2}+xy^{2}z^{3}\\
&  =x^{3}y^{2}z+\left(  \text{all other }5\text{ permutations of this
monomial}\right)
\end{align*}
and%
\[
m_{\left(  2,2,1\right)  }=x^{2}y^{2}z+x^{2}yz^{2}+xy^{2}z^{2}%
\]
and%
\[
m_{\left(  2,2,2\right)  }=x^{2}y^{2}z^{2}.
\]

\end{example}

As we saw in this example, for each $N$-partition $\lambda=\left(  \lambda
_{1},\lambda_{2},\ldots,\lambda_{N}\right)  $, the polynomial $m_{\lambda}$ is
the sum of all monomials in the $S_{N}$-orbit of the single monomial
$x^{\lambda}=x_{1}^{\lambda_{1}}x_{2}^{\lambda_{2}}\cdots x_{N}^{\lambda_{N}}%
$. Thus, the $m_{\lambda}$ are also known as \textbf{orbit sums}. And
nowadays, they are often just called \textquotedblleft the $m$%
's\textquotedblright, since the notation $m_{\lambda}$ has become so standard
that everyone can be expected to understand this shorthand.

The definition of $m_{\lambda}$ shows that all coefficients of $m_{\lambda}$
are $0$ or $1$. More precisely: each $N$-partition $\lambda$ and each
$N$-tuple $a\in\mathbb{N}^{N}$ satisfy%
\begin{equation}
\left[  x^{a}\right]  m_{\lambda}=%
\begin{cases}
1, & \text{if }\operatorname*{sort}a=\lambda;\\
0, & \text{if }\operatorname*{sort}a\neq\lambda.
\end{cases}
\label{eq.m.xamla}%
\end{equation}
In particular, for each $N$-partition $\mu$, we have%
\begin{align}
\left[  x^{\mu}\right]  m_{\lambda}  &  =%
\begin{cases}
1, & \text{if }\operatorname*{sort}\mu=\lambda;\\
0, & \text{if }\operatorname*{sort}\mu\neq\lambda
\end{cases}
\nonumber\\
&  =%
\begin{cases}
1, & \text{if }\mu=\lambda;\\
0, & \text{if }\mu\neq\lambda
\end{cases}
\label{eq.m.xmula}%
\end{align}
(because $\operatorname*{sort}\mu=\mu$, since $\mu$ is already weakly decreasing).

It is also clear that each $a\in\mathbb{N}^{N}$ satisfies
\begin{equation}
\deg\left(  x^{a}\right)  =\left\vert a\right\vert =\left\vert
\operatorname*{sort}a\right\vert \label{eq.m.degxa}%
\end{equation}
(since $\operatorname*{sort}a$ has the same entries as $a$ up to order). Thus,
each monomial symmetric polynomial $m_{\lambda}$ is homogeneous of degree
$\left\vert \lambda\right\vert $.

Our symmetric polynomials $e_{n},h_{n},p_{n}$ are all easily rewritten in
terms of $m$'s:

\begin{proposition}
\label{prop.symp.ehp-as-m} \ 

\begin{enumerate}
\item[\textbf{(a)}] For each $n\in\left\{  0,1,\ldots,N\right\}  $, we have%
\[
e_{n}=m_{\left(  1,1,\ldots,1,0,0,\ldots,0\right)  },
\]
where $\left(  1,1,\ldots,1,0,0,\ldots,0\right)  $ is the $N$-tuple that
begins with $n$ many $1$'s and ends with $N-n$ many $0$'s. This $N$-tuple will
be written as $\left(  1^{n},0^{N-n}\right)  $ later on.

\item[\textbf{(b)}] For each $n\in\mathbb{N}$, we have%
\[
h_{n}=\sum_{\substack{\lambda\text{ is an }N\text{-partition;}\\\left\vert
\lambda\right\vert =n}}m_{\lambda}.
\]


\item[\textbf{(c)}] Assume that $N>0$. Then, for each $n\in\mathbb{N}$, we
have%
\[
p_{n}=m_{\left(  n,0,0,\ldots,0\right)  }.
\]
Under our identification of $N$-partitions with partitions of length $\leq N$,
we can (and will) simply write this as%
\[
p_{n}=m_{\left(  n\right)  }.
\]

\end{enumerate}
\end{proposition}

\begin{proof}
LTTR. (This means \textquotedblleft left to the reader\textquotedblright. In
this case, the proof is so easy that just starting at the definitions should suffice.)
\end{proof}

Next, we shall prove the following (fairly simple) theorem:

\begin{theorem}
[monomial basis of $\mathcal{S}$]\label{thm.symp.m-basis} \ 

\begin{enumerate}
\item[\textbf{(a)}] The family $\left(  m_{\lambda}\right)  _{\lambda\text{ is
an }N\text{-partition}}$ is a basis of the $K$-module $\mathcal{S}$.

\item[\textbf{(b)}] Each symmetric polynomial $f\in\mathcal{S}$ satisfies%
\[
f=\sum_{\substack{\lambda=\left(  \lambda_{1},\lambda_{2},\ldots,\lambda
_{N}\right)  \\\text{is an }N\text{-partition}}}\ \ \underbrace{\left(
\left[  x_{1}^{\lambda_{1}}x_{2}^{\lambda_{2}}\cdots x_{N}^{\lambda_{N}%
}\right]  f\right)  }_{\text{the }x_{1}^{\lambda_{1}}x_{2}^{\lambda_{2}}\cdots
x_{N}^{\lambda_{N}}\text{-coefficient of }f}m_{\lambda}.
\]
Or, to put this more compactly:%
\begin{equation}
f=\sum_{\lambda\text{ is an }N\text{-partition}}\left(  \left[  x^{\lambda
}\right]  f\right)  m_{\lambda}. \label{eq.thm.symp.m-basis.b.claim}%
\end{equation}


\item[\textbf{(c)}] Let $n\in\mathbb{N}$. Let%
\[
\mathcal{S}_{n}:=\left\{  \text{homogeneous symmetric polynomials }%
f\in\mathcal{P}\text{ of degree }n\right\}
\]
(where the zero polynomial $0$ counts as homogeneous of every degree). Then,
$\mathcal{S}_{n}$ is a $K$-submodule of $\mathcal{S}$.

\item[\textbf{(d)}] The family $\left(  m_{\lambda}\right)  _{\lambda\text{ is
an }N\text{-partition such that }\left\vert \lambda\right\vert =n}$ is a basis
of the $K$-module $\mathcal{S}_{n}$.
\end{enumerate}
\end{theorem}

\begin{proof}
\textbf{(b)} Let $f\in\mathcal{S}$. We must prove that
\begin{equation}
f=\sum_{\lambda\text{ is an }N\text{-partition}}\left(  \left[  x^{\lambda
}\right]  f\right)  m_{\lambda}. \label{pf.thm.symp.m-basis.goal1}%
\end{equation}


To do this, it suffices to show that%
\begin{equation}
\left[  x^{a}\right]  f=\left[  x^{a}\right]  \left(  \sum_{\lambda\text{ is
an }N\text{-partition}}\left(  \left[  x^{\lambda}\right]  f\right)
m_{\lambda}\right)  \label{pf.thm.symp.m-basis.goal2}%
\end{equation}
for all $a\in\mathbb{N}^{N}$ (because if two polynomials agree in all their
coefficients, then they must be equal). But $f$ is symmetric (since
$f\in\mathcal{S}$), and each $a\in\mathbb{N}^{N}$ satisfies%
\begin{align*}
&  \left[  x^{a}\right]  \left(  \sum_{\lambda\text{ is an }N\text{-partition}%
}\left(  \left[  x^{\lambda}\right]  f\right)  m_{\lambda}\right) \\
&  =\sum_{\lambda\text{ is an }N\text{-partition}}\left(  \left[  x^{\lambda
}\right]  f\right)  \underbrace{\left[  x^{a}\right]  m_{\lambda}%
}_{\substack{=%
\begin{cases}
1, & \text{if }\operatorname*{sort}a=\lambda;\\
0, & \text{if }\operatorname*{sort}a\neq\lambda
\end{cases}
\\\text{(by (\ref{eq.m.xamla}))}}}\\
&  =\sum_{\lambda\text{ is an }N\text{-partition}}\left(  \left[  x^{\lambda
}\right]  f\right)
\begin{cases}
1, & \text{if }\operatorname*{sort}a=\lambda;\\
0, & \text{if }\operatorname*{sort}a\neq\lambda
\end{cases}
\\
&  =\left[  x^{\operatorname*{sort}a}\right]  f\ \ \ \ \ \ \ \ \ \ \left(
\begin{array}
[c]{c}%
\text{since all addends of the sum are obviously }0\\
\text{except for the addend for }\lambda=\operatorname*{sort}a
\end{array}
\right) \\
&  =\left[  x^{a}\right]  f\ \ \ \ \ \ \ \ \ \ \left(  \text{by Lemma
\ref{lem.symp.sort-a} \textbf{(b)}, since }f\text{ is symmetric}\right)  .
\end{align*}
This proves (\ref{pf.thm.symp.m-basis.goal2}), and thus
(\ref{pf.thm.symp.m-basis.goal1}) follows. This completes the proof of part
\textbf{(b)}. \medskip

\textbf{(a)} The equality (\ref{eq.thm.symp.m-basis.b.claim}) from part
\textbf{(b)} shows that the family $\left(  m_{\lambda}\right)  _{\lambda
\text{ is an }N\text{-partition}}$ spans the $K$-module $\mathcal{S}$. It
remains to prove that this family is $K$-linearly independent.

Well, if a $K$-linear combination $\sum\limits_{\lambda\text{ is an
}N\text{-partition}}a_{\lambda}m_{\lambda}$ (with coefficients $a_{\lambda}\in
K$) equals $0$, then each $N$-partition $\mu$ satisfies%
\begin{align*}
\left[  x^{\mu}\right]  \left(  \sum\limits_{\lambda\text{ is an
}N\text{-partition}}a_{\lambda}m_{\lambda}\right)   &  =\sum\limits_{\lambda
\text{ is an }N\text{-partition}}a_{\lambda}\underbrace{\left[  x^{\mu
}\right]  m_{\lambda}}_{\substack{=%
\begin{cases}
1, & \text{if }\mu=\lambda;\\
0, & \text{if }\mu\neq\lambda
\end{cases}
\\\text{(by (\ref{eq.m.xmula}))}}}\\
&  =\sum\limits_{\lambda\text{ is an }N\text{-partition}}a_{\lambda}%
\begin{cases}
1, & \text{if }\mu=\lambda;\\
0, & \text{if }\mu\neq\lambda
\end{cases}
\\
&  =a_{\mu}%
\end{align*}
and therefore%
\[
a_{\mu}=\left[  x^{\mu}\right]  \underbrace{\left(  \sum\limits_{\lambda\text{
is an }N\text{-partition}}a_{\lambda}m_{\lambda}\right)  }_{=0}=0;
\]
that is, all coefficients $a_{\mu}$ of this linear combination $\sum
\limits_{\lambda\text{ is an }N\text{-partition}}a_{\lambda}m_{\lambda}$ are
$0$. This says precisely that the family $\left(  m_{\lambda}\right)
_{\lambda\text{ is an }N\text{-partition}}$ is $K$-linearly independent. The
proof of part \textbf{(a)} is thus complete. \medskip

\textbf{(c)} This is obvious: a sum of two homogeneous polynomials of degree
$n$ is again homogeneous of degree $n$, and the same applies to scaling.
\medskip

\textbf{(d)} This is analogous to part \textbf{(a)} but now, instead of the
equality (\ref{eq.thm.symp.m-basis.b.claim}), we need to show that each
$f\in\mathcal{S}_{n}$ satisfies%
\begin{equation}
f=\sum_{\substack{\lambda\text{ is an }N\text{-partition;}\\\left\vert
\lambda\right\vert =n}}\left(  \left[  x^{\lambda}\right]  f\right)
m_{\lambda}. \label{pf.thm.symp.m-basis.d.goal1}%
\end{equation}
But this equality can be easily derived from
(\ref{eq.thm.symp.m-basis.b.claim}): Let $\pi_{n}:\mathcal{P}\rightarrow
\mathcal{P}$ be the map that sends each polynomial $f\in\mathcal{P}$ to its
degree-$n$ homogeneous component (i.e., removes all monomials of degrees other
than $n$ from $f$). This is a $K$-linear map that fixes each monomial $x^{a}$
satisfying $\left\vert a\right\vert =n$ and annihilates all other monomials.

Now, let $f\in\mathcal{S}_{n}$. Then, $f$ is symmetric, so that $f\in
\mathcal{S}$, and therefore (\ref{eq.thm.symp.m-basis.b.claim}) applies. But
$f$ is furthermore homogeneous of degree $n$, so that $\pi_{n}\left(
f\right)  =f$. On the other hand, since $\pi_{n}$ is a $K$-linear map, we have%
\begin{align*}
&  \pi_{n}\left(  \sum_{\lambda\text{ is an }N\text{-partition}}\left(
\left[  x^{\lambda}\right]  f\right)  m_{\lambda}\right) \\
&  =\sum_{\lambda\text{ is an }N\text{-partition}}\left(  \left[  x^{\lambda
}\right]  f\right)  \underbrace{\pi_{n}\left(  m_{\lambda}\right)
}_{\substack{=%
\begin{cases}
m_{\lambda}, & \text{if }\left\vert \lambda\right\vert =n;\\
0, & \text{if }\left\vert \lambda\right\vert \neq n
\end{cases}
\\\text{(since }m_{\lambda}\text{ is homogeneous of degree }\left\vert
\lambda\right\vert \text{)}}}\\
&  =\sum_{\lambda\text{ is an }N\text{-partition}}\left(  \left[  x^{\lambda
}\right]  f\right)
\begin{cases}
m_{\lambda}, & \text{if }\left\vert \lambda\right\vert =n;\\
0, & \text{if }\left\vert \lambda\right\vert \neq n
\end{cases}
\\
&  =\sum_{\substack{\lambda\text{ is an }N\text{-partition;}\\\left\vert
\lambda\right\vert =n}}\left(  \left[  x^{\lambda}\right]  f\right)
m_{\lambda}%
\end{align*}
(here, we have removed all the addends with $\left\vert \lambda\right\vert
\neq n$ from our sum, since all these addends are $0$). Using
(\ref{eq.thm.symp.m-basis.b.claim}), we can rewrite this as
\[
\pi_{n}\left(  f\right)  =\sum_{\substack{\lambda\text{ is an }%
N\text{-partition;}\\\left\vert \lambda\right\vert =n}}\left(  \left[
x^{\lambda}\right]  f\right)  m_{\lambda}.
\]
In view of $\pi_{n}\left(  f\right)  =f$, this equality is equivalent to
(\ref{pf.thm.symp.m-basis.d.goal1}). Thus, we have proved
(\ref{pf.thm.symp.m-basis.d.goal1}). The proof of part \textbf{(d)} is
therefore complete.
\end{proof}

\begin{example}
\label{exa.symp.m-basis.tourn3}Let $N=3$. With the usual shorthands $x,y,z$
for the indeterminates $x_{1},x_{2},x_{3}$, we have%
\begin{align*}
\left(  x+y\right)  \left(  x+z\right)  \left(  y+z\right)   &
=\underbrace{x^{2}y+xy^{2}+x^{2}z+xz^{2}+y^{2}z+yz^{2}}_{=m_{\left(
2,1,0\right)  }}+\,2\underbrace{xyz}_{=m_{\left(  1,1,1\right)  }}\\
&  =\underbrace{m_{\left(  2,1,0\right)  }}_{=m_{\left(  2,1\right)  }%
}+\,2m_{\left(  1,1,1\right)  }.
\end{align*}
This can be generalized to higher $N$'s: expanding $\prod_{i<j}\left(
x_{i}+x_{j}\right)  $ in the basis $\left(  m_{\lambda}\right)  _{\lambda
\text{ is an }N\text{-partition}}$ gives a linear combination of $m_{\lambda}%
$'s whose coefficients count certain directed graphs (see a homework exercise).
\end{example}

\subsection{Functoriality}

We have been (and will be) working over a fixed commutative ring $K$. In
particular, the polynomial ring $\mathcal{P}$ and the symmetric polynomial
ring $\mathcal{S}$ were defined over $K$ (so that the coefficients of
polynomials are elements of $K$). We can also denote them by $\mathcal{P}_{K}$
and $\mathcal{S}_{K}$ to stress their dependence on $K$. Then we can compare
$\mathcal{P}_{K}$'s and $\mathcal{S}_{K}$'s for different $K$'s. The following
proposition is pretty obvious, but surprisingly useful:

\begin{proposition}
\label{prop.symp.functoriality}Let $K$ and $L$ be two commutative rings.

\begin{enumerate}
\item[\textbf{(a)}] If $K$ is a subring of $L$, then $\mathcal{P}_{K}$ becomes
a subring of $\mathcal{P}_{L}$, and $\mathcal{S}_{K}$ becomes a subring of
$\mathcal{S}_{L}$; moreover, we have $\mathcal{S}_{K}=\mathcal{S}_{L}%
\cap\mathcal{P}_{K}$ (that is, a symmetric polynomial over $K$ is the same as
a symmetric polynomial over $L$ whose coefficients belong to $K$).

\item[\textbf{(b)}] If $f:K\rightarrow L$ is a ring morphism, then $f$ induces
ring morphisms $\mathcal{P}_{K}\rightarrow\mathcal{P}_{L}$ (by applying $f$ to
each coefficient) and $\mathcal{S}_{K}\rightarrow\mathcal{S}_{L}$ (by
restricting the morphism $\mathcal{P}_{K}\rightarrow\mathcal{P}_{L}$ to
$\mathcal{S}_{K}$).
\end{enumerate}
\end{proposition}

\begin{proof}
LTTR. (All of this is essentially trivial.)
\end{proof}

\begin{example}
\label{exa.symp.functoriality.Q-and-Z3}\ \ 

\begin{enumerate}
\item[\textbf{(a)}] For $K=\mathbb{Z}$ and $L=\mathbb{Q}$, Proposition
\ref{prop.symp.functoriality} \textbf{(a)} is saying that polynomials over
$\mathbb{Z}$ are polynomials over $\mathbb{Q}$, and that the symmetric
polynomials over $\mathbb{Z}$ are precisely the symmetric polynomials over
$\mathbb{Q}$ whose coefficients belong to $\mathbb{Z}$.

\item[\textbf{(b)}] An example of a ring morphism that is not just an
inclusion is the projection $\pi:\mathbb{Z}\rightarrow\mathbb{Z}/3$. Thus,
Proposition \ref{prop.symp.functoriality} \textbf{(b)} shows that it induces a
ring morphism $\mathcal{P}_{\mathbb{Z}}\rightarrow\mathcal{P}_{\mathbb{Z}/3}$
and a ring morphism $\mathcal{S}_{\mathbb{Z}}\rightarrow\mathcal{S}%
_{\mathbb{Z}/3}$. These morphisms are known as \textquotedblleft reducing all
coefficients modulo $3$\textquotedblright.
\end{enumerate}
\end{example}

Proposition \ref{prop.symp.functoriality} is particularly useful because some
base rings like $\mathbb{Z}$ and $\mathbb{Q}$ are particularly easy to work
with, and then the proposition lets you transfer (some kinds of) knowledge
about $\mathcal{S}_{\mathbb{Z}}$ and $\mathcal{S}_{\mathbb{Q}}$ to the more
general setting of $\mathcal{S}_{K}$ for arbitrary $K$. (Recall that for any
ring $K$, there is a ring morphism $\mathbb{Z}\rightarrow K$.) Much of the
theory of symmetric polynomials/functions can be built over $\mathbb{Z}$ or
$\mathbb{Q}$, or occasionally over some larger rings such as the polynomial
ring $\mathbb{Q}\left[  t\right]  $ or the rational function field
$\mathbb{Q}\left(  t\right)  $.

\subsection{Schur polynomials}

We now come to a subtler and far more interesting family of symmetric polynomials.

\subsubsection{Alternants}

Here is one way to generate symmetric polynomials:

\begin{example}
\label{exa.schur.alt1}Let $N=3$ again. The Vandermonde determinant formula
\cite[Theorem 6.4.31 \textbf{(a)}]{21s} yields%
\[
\det\left(
\begin{array}
[c]{ccc}%
x^{2} & x & 1\\
y^{2} & y & 1\\
z^{2} & z & 1
\end{array}
\right)  =\left(  x-y\right)  \left(  x-z\right)  \left(  y-z\right)  .
\]


But what can we say about similar determinants where the powers are
\textquotedblleft raised\textquotedblright, such as%
\[
\det\left(
\begin{array}
[c]{ccc}%
x^{5} & x^{3} & 1\\
y^{5} & y^{3} & 1\\
z^{5} & z^{3} & 1
\end{array}
\right)  \ ?
\]
We can compute this determinant and factor it, thus obtaining%
\begin{equation}
\det\left(
\begin{array}
[c]{ccc}%
x^{5} & x^{3} & 1\\
y^{5} & y^{3} & 1\\
z^{5} & z^{3} & 1
\end{array}
\right)  =\left(  x-y\right)  \left(  x-z\right)  \left(  y-z\right)
q,\label{eq.exa.schur.alt1.det2}%
\end{equation}
where%
\begin{align*}
q &  =x^{2}y^{3}+x^{3}y^{2}+x^{2}z^{3}+x^{3}z^{2}+y^{2}z^{3}+y^{3}z^{2}\\
&  \ \ \ \ \ \ \ \ \ \ +xyz^{3}+xy^{3}z+x^{3}yz+2xy^{2}z^{2}+2x^{2}%
yz^{2}+2x^{2}y^{2}z.
\end{align*}
The presence of the three factors $y-z$, $x-z$ and $x-y$ here is not
surprising: The determinant in (\ref{eq.exa.schur.alt1.det2}) vanishes when we
put $y=z$, so it is divisible (as a polynomial over $\mathbb{Z}$) by $y-z$;
and similar arguments lead to it being divisible by $x-z$ and $x-y$. Since a
polynomial ring over $\mathbb{Z}$ is always a unique factorization domain,
this shows that the determinant is divisible by $\left(  x-y\right)  \left(
x-z\right)  \left(  y-z\right)  $. The remaining factor $q$ is easily seen to
be a symmetric polynomial in $x,y,z$ (this is easiest to see using Lemma
\ref{lem.symp.simple-transpositions}: if we swap $x$ with $y$, then both the
determinant in (\ref{eq.exa.schur.alt1.det2}) and the Vandermonde product
$\left(  x-y\right)  \left(  x-z\right)  \left(  y-z\right)  $ get multiplied
by $-1$, so that the quotient $q$ stays unchanged; likewise for any other swap
of two variables). But what is more surprising is that this quotient $q$ has
nonnegative integer coefficients (it is $m_{\left(  3,2,0\right)  }+m_{\left(
3,1,1\right)  }+2m_{\left(  2,2,1\right)  }$). Is this a fluke or does it generalize?
\end{example}

Let us try to generalize this:

\begin{definition}
[alternants and $\rho$]\label{def.schur.arho}\ \ 

\begin{enumerate}
\item[\textbf{(a)}] We let $\rho$ denote the $N$-tuple $\left(  N-1,N-2,\ldots
,1,0\right)  \in\mathbb{N}^{N}$. Thus, its $j$-th entry is $N-j$ for each
$j\in\left[  N\right]  $.

\item[\textbf{(b)}] For any $N$-tuple $\alpha=\left(  \alpha_{1},\alpha
_{2},\ldots,\alpha_{N}\right)  \in\mathbb{N}^{N}$, we define the $\alpha
$\textbf{-alternant}%
\[
a_{\alpha}:=\det\left(  \left(  x_{i}^{\alpha_{j}}\right)  _{i,j\in\left[
N\right]  }\right)  =\det\left(
\begin{array}
[c]{cccc}%
x_{1}^{\alpha_{1}} & x_{1}^{\alpha_{2}} & \cdots & x_{1}^{\alpha_{N}}\\
x_{2}^{\alpha_{1}} & x_{2}^{\alpha_{2}} & \cdots & x_{2}^{\alpha_{N}}\\
\vdots & \vdots & \ddots & \vdots\\
x_{N}^{\alpha_{1}} & x_{N}^{\alpha_{2}} & \cdots & x_{N}^{\alpha_{N}}%
\end{array}
\right)  \in\mathcal{P}.
\]

\end{enumerate}
\end{definition}

For example, $\det\left(
\begin{array}
[c]{ccc}%
x^{5} & x^{3} & 1\\
y^{5} & y^{3} & 1\\
z^{5} & z^{3} & 1
\end{array}
\right)  $ is $a_{\left(  5,3,0\right)  }$. Note that%
\begin{align*}
a_{\rho}  &  =\det\left(  \left(  x_{i}^{\rho_{j}}\right)  _{i,j\in\left[
N\right]  }\right)  \ \ \ \ \ \ \ \ \ \ \left(  \text{writing }\rho\text{ as
}\rho=\left(  \rho_{1},\rho_{2},\ldots,\rho_{N}\right)  \right) \\
&  =\det\left(  \left(  x_{i}^{N-j}\right)  _{i,j\in\left[  N\right]
}\right)  \ \ \ \ \ \ \ \ \ \ \left(  \text{since }\rho_{j}=N-j\text{ for all
}j\in\left[  N\right]  \right) \\
&  =\prod_{i<j}\left(  x_{i}-x_{j}\right)  \ \ \ \ \ \ \ \ \ \ \left(
\text{where }\prod_{i<j}\text{ means }\prod_{1\leq i<j\leq N}\right)
\end{align*}
(by the Vandermonde determinant formula \cite[Theorem 6.4.31 \textbf{(a)}%
]{21s}). Now we suspect:

\begin{conjecture}
\label{conj.schur.pos}For every $\alpha=\left(  \alpha_{1},\alpha_{2}%
,\ldots,\alpha_{N}\right)  \in\mathbb{N}^{N}$ satisfying $\alpha_{1}%
>\alpha_{2}>\cdots>\alpha_{N}$, the alternant $a_{\alpha}$ can be expressed as
$a_{\rho}$ times a symmetric polynomial with nonnegative integer coefficients.
\end{conjecture}

Note that the assumption $\alpha_{1}>\alpha_{2}>\cdots>\alpha_{N}$ is not much
of a restriction. Indeed, if two of the $N$ entries $\alpha_{1},\alpha
_{2},\ldots,\alpha_{N}$ are equal, then $a_{\alpha}$ is easily seen to be $0$
(since a determinant with two equal columns is $0$); in all remaining cases,
we can sort the $N$ entries $\alpha_{1},\alpha_{2},\ldots,\alpha_{N}$ in
strictly decreasing order by repeatedly swapping them, and each swap
multiplies $a_{\alpha}$ by $-1$. Thus, if we did not require $\alpha
_{1}>\alpha_{2}>\cdots>\alpha_{N}$ in Conjecture \ref{conj.schur.pos}, then we
would have to replace \textquotedblleft nonnegative\textquotedblright\ by
\textquotedblleft nonnegative or nonpositive, depending on the sign of the
permutation that sorts the $N$ numbers $\alpha_{1},\alpha_{2},\ldots
,\alpha_{N}$ in decreasing order\textquotedblright.

The \textquotedblleft nonnegative\textquotedblright\ part of Conjecture
\ref{conj.schur.pos} is far from easy. Try it! It was first proved by Carl
Kostka in an 1882 paper \cite{Kostka82} (relying on an older paper of his
\cite{Kostka77})\footnote{I find it peculiar that Kostka produced very few
papers apart from these; he essentially founded a field of mathematics and
left.} by finding a combinatorial interpretation of the quotient $a_{\alpha
}/a_{\rho}$. This is precisely what we shall do, though we will instead prove
it using a modern approach by Stembridge (2002, \cite{Stembr02}). The
quotients $a_{\alpha}/a_{\rho}$ in Conjecture \ref{conj.schur.pos} are known
as the \textbf{Schur polynomials}, and are perhaps the most interesting of the
classical bases of $\mathcal{S}$.

\begin{example}
\label{exa.schur.exa2}If $\alpha=\rho$, then the quotient $a_{\alpha}/a_{\rho
}$ is simply $1$. Boring.

The simplest case beyond that is the case when%
\[
\alpha=\left(  N,\ N-2,\ N-3,\ \ldots,\ 0\right)  .
\]
In this case, we have
\begin{align*}
a_{\alpha} &  =\det\left(
\begin{array}
[c]{ccccc}%
x_{1}^{N} & x_{1}^{N-2} & x_{1}^{N-3} & \cdots & x_{1}^{0}\\
x_{2}^{N} & x_{2}^{N-2} & x_{2}^{N-3} & \cdots & x_{2}^{0}\\
\vdots & \vdots & \vdots & \ddots & \vdots\\
x_{N}^{N} & x_{N}^{N-2} & x_{N}^{N-3} & \cdots & x_{N}^{0}%
\end{array}
\right)  \\
&  \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \left(
\begin{array}
[c]{c}%
\text{this is the Vandermonde determinant except that}\\
\text{its first column has exponent }N\text{ instead of }N-1
\end{array}
\right)  \\
&  =\prod_{i<j}\left(  x_{i}-x_{j}\right)  \cdot\underbrace{\left(  \text{some
homogeneous symmetric polynomial of degree }1\right)  }_{\text{i.e., a scalar
multiple of }x_{1}+x_{2}+\cdots+x_{N}}\\
&  \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \left(
\begin{array}
[c]{c}%
\text{using an argument similar to the one in}\\
\text{Example \ref{exa.schur.alt1} and using degree reasons}\\
\text{to nail down the degree -- exercise!}%
\end{array}
\right)  \\
&  =\text{a scalar multiple of }\prod_{i<j}\left(  x_{i}-x_{j}\right)
\cdot\left(  x_{1}+x_{2}+\cdots+x_{N}\right)  .
\end{align*}
By comparing coefficients of the monomial $x_{1}^{N}x_{2}^{N-2}x_{3}%
^{N-3}\cdots x_{N}^{0}$ on both sides, we can easily see that the scalar on
the right-hand side must be $1$, and thus we obtain%
\[
a_{\alpha}=\prod_{i<j}\left(  x_{i}-x_{j}\right)  \cdot\left(  x_{1}%
+x_{2}+\cdots+x_{N}\right)
\]
in this case.

What is the next interesting case?
\end{example}

\begin{thebibliography}{99999999}                                                                                         %


\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}

\bibitem[Kostka77]{Kostka77}\href{https://eudml.org/doc/148314}{Carl Kostka,
\textit{Ueber Borchardts Function}, Journal f\"{u}r die reine und angewandte
Mathematik \textbf{82} (1877), pp. 212--229.}

\bibitem[Kostka82]{Kostka82}\href{https://eudml.org/doc/148506}{Carl Kostka,
\textit{Ueber den Zusammenhang zwischen einigen Formen von symmetrischen
Functionen}, Journal f\"{u}r die reine und angewandte Mathematik \textbf{93}
(1882), pp. 89--123.}

\bibitem[Stembr02]{Stembr02}\href{https://doi.org/10.37236/1666}{John R.
Stembridge, \textit{A Concise Proof of the Littlewood-Richardson Rule},
Electronic Journal of Combinatorics \textbf{9} (2002), \#N5}
\end{thebibliography}


\end{document}