I'm making a research on Galois theory, and found something interesting regarding the ring of symmetric polynomials:
At least up to 5 variables, we can rewrite the elementary symmetric polynomials with powers of symmetric polynomials of degree 1.
For the formulation of the question, i will introduce the index set $I_q=\mathbb{N}_{<q}$, where $\mathbb{N}$ are the natural numbers, including zero.
First, we define the subset $S_k\subset I_q^k$ as
$$S_k=\{(i_1;\dots;i_k)\in I_q^k\mid i_1<i_2<\dots<i_k\}$$
This allows us to define the $k$-th elementary symmetric polynomial with $q$ variables $e_k=e_k(x_0;\dots;x_{q-1})$ as:
$$e_k=\sum_{(i_1;\dots;i_k)\in S_k}x_{i_1}\dots x_{i_k}$$
We also define the symmetric $p$-th powers of first degree polynomials $s_j^p=s_j^p(x_0;\dots;x_{q-1})$ to be:
$$s_j^p=\sum_{(i_1;\dots;i_j)\in S_j}(x_{i_1}+\dots+x_{i_j})^p$$
The question is then: can every elementary symmetric polynomial $e_k$ be written as a linear combination of these symmetric $k$-th powers of first degree polynomials $s_j^k$ for an arbitrary number of variables $q$?
As said earlier, i have made some progress, and found a pattern up to $q=5$ variables. I will leave my findings here:
[For $q=1$]
$$e_1=s_1^1$$
[For $q=2$]
$$\begin{align*}
e_1&=s_1^1+0s_2^1\\
e_2&=\frac{1}{2}(-s_1^2+s_2^2)
\end{align*}$$
[For $q=3$]
$$\begin{align*}
e_1&=s_1^1+0s_2^1+0s_3^1\\
e_2&=\frac{1}{2}(-s_1^2+0s_2^2+s_3^2)\\
e_3&=\frac{1}{6}(s_1^3-s_2^3+s_3^3)
\end{align*}$$
Continuing in a similar fashion, we can construct $q$-dimensional vector spaces $V_k$ over $\mathbb{Q}$, spanned by the set of symmetric $k$-th powers $G_k=\{s_1^k; s_2^k;\dots;s_q^k\}$.
The $k$-th elementary symmetric polynomial is then a vector $\vec{e_k}\in V_k$ in this setting. Note that these representations are not unique, e.g., for $q=2$, $\vec{e_1}=(1;0)=(0;1)$, and hence $G_k$ is not a basis for $V_k$. I'll add a few more values for $q$:
[For $q=4$]
$$\begin{align*}
\vec{e_1}&=(1;0;0;0)\\
\vec{e_2}&=\frac{1}{2}(-1;0;0;1)\\
\vec{e_3}&=\frac{1}{6}(4;2;-3;-1)\\
\vec{e_4}&=\frac{1}{24}(-1;-1;1;1)
\end{align*}$$
[For $q=5$]
$$\begin{align*}
\vec{e_1}&=(1;0;0;0;0)\\
\vec{e_2}&=\frac{1}{2}(-1;0;0;0;1)\\
\vec{e_3}&=\frac{1}{6}(-6;0;-1;3;0)\\
\vec{e_4}&=\frac{1}{24}(-5;2;-3;4;-1)\\
\vec{e_5}&=\frac{1}{120}(1;-1;1;-1;1)
\end{align*}$$
etc.
Best Answer
Yes.
For variables $x_1,\dots, x_k$, we have
$$ \sum_{J \subseteq \{1,\dots, k \} } (-1)^{ k- |J|} \left(\sum_{j \in J} x_j\right)^k = k! x_1 \dots x_k $$
so that $$ e_k = \sum_{(i_1; \dots; i_k )\in S} x_{i_1} \dots x_{s_k} = \frac{1}{k!} \sum_{(i_1; \dots; i_k )\in S} \sum_{J \subseteq \{i_1,\dots, i_k \}}(-1)^{ k- |J|} \left(\sum_{j \in J} x_{j} \right)^k $$
and each set $J$ occurs $\binom{ q-|J|}{k- |J|} $ times, so
$$ e_k = \sum_{ \substack{ J\subseteq I_q \\ |J| \leq k}} (-1)^{ k -|J|} \frac{ \binom{ q- |J|} {k-|J|}}{k!} \left(\sum_{j \in J} x_j\right)^k =\sum_{j=0}^k (-1)^{ k -j} \frac{ \binom{ q- j} {k-j}}{k!} s_j ^k.$$