Addendum: An alternate proof of the FTSFT

Last week I posted about the Fundamental Theorem of Symmetric Function Theory. Zarathustra Brady pointed me to the following alternate proof in Serge Lang’s book Algebra. While not as direct or useful in terms of changing basis from the $e_\lambda$`s to the $m_\lambda$`s, it is a nice, clean inductive proof that I thought was worth sharing:

Assume for contradiction that the $e_\lambda$`s do not form a basis of the space of symmetric functions. We have shown that they span the space, so there is a dependence relation: some nontrivial linear combination of $e_\lambda$`s, all necessarily of the same degree, is equal to zero. Among all such linear combinations, choose one (say $P$) that holds for the smallest possible number of variables $x_1,\ldots,x_n$. Furthermore, among the possible linear combinations for $n$ variables, choose $P$ to have minimal degree.

If the number of variables is $1$, then the only elementary symmetric functions are $x_1^k$ for some $k$, and so there is clearly no linear dependence relation. So, $n\ge 2$. Furthermore, if $P$ has degree $1$ as a polynomial in the $x_i$`s, then it can involve only $e_1$, and so it cannot be identically zero. So $P$ has degree at least $2$, in at least $2$ variables.

Now, if $P$ is divisible by $x_n$, then by symmetry it is divisible by each of the variables. So, it is divisible by $e_n$, and so we can divide the equation by $e_n$ and get a relation of smaller degree, contradicting our choice of $P$. Otherwise, if $P$ is not divisible by $x_n$, set $x_n=0$. Then we get another nontrivial relation among the $e_\lambda$ in the smaller number of variables $x_1,\ldots,x_n$, again contradicting the choice of $P$. QED!