In the book Polynomial Invariants of Finite Groups by Larry Smith, he proved the algebraic independence of elementary symmetric polynomials as follows:
Suppose $g(e_1,…,e_n) =0$ where $ g $ is not the zero polynomial, write $g$ as the sum of $cx_1^{a_1 - a_2} x_2^{a_2 -a_3}... x_n^{a_n}$ where $a_1 \geq ... \geq a_n$. Let the leading term of $g$ be $kx_1^{a_1 - a_2} ... x_n^{a_n}$.
Then the leading term of $g(e_1,…,e_n)$ is $kx_1^{a_1} ... x_n^{a_n}$ and hence k =0.
I don’t really understand the last part, it feels to me like the author just substitute the $e_i$’s into the leading term of g. If so, why does direct substitution of $e_i$’s into the leading term of $g$ work here?
Define an order of the monomials as follows: $x_1^{a_1} \cdots x_n^{a_n} > x_1^{b_1} \cdots x_n^{b_n}$ if and only $a_j + \cdots + a_n > b_j + \cdots + b_n$ for the first value of $j$ for which $a_j + \cdots + a_n \neq b_j + \cdots + b_n$. Call this the $S$-order. We would need to prove that this gives a well-defined order of course. This is well-defined, and there is a unique maximal element in any polynomial w.r.t the $S$-order.
With this definition if $x_1^{a_1 - a_2} \cdots x_n^{a_n}$ is the leading term of $g(x_1, \dots, x_n)$ w.r.t. the $S$-order, then $x_1^{a_1} \cdots x_n^{a_n}$ is the maximal term of $g(e_1, \dots, e_n)$ w.r.t. the lexicographical order ($L$-order). This is true simply because $a_j = (a_j - a_{j+1}) + \cdots + (a_{n-1} - a_n) + a_n$, and $x_1^{a_1} \cdots x_n^{a_n}$ is the unique maximal term of $e_1^{a_1 - a_2} \cdots e_n^{a_n}$ w.r.t the $L$-order.
More explicitly, let $x_1^{b_1} \cdots x_n^{b_n}$ be maximal w.r.t. the $L$ order. Assume there is a minimal $j$ such that $b_j > a_j$. $x_1^{b_1} \cdots x_n^{b_n}$ must occur in a polynomial $e_1^{b_1 - b_2} \cdots e_n^{b_n}$, where $x_1^{b_1 - b_2} \cdots x_n^{b_n}$ occurs in $g(x_1, \dots, x_n)$. But then $x_1^{b_1 - b_2} \cdots x_n^{b_n} > x_1^{a_1 - a_2} \cdots x_n^{a_n}$ w.r.t. the $S$-order. We conclude that $x_1^{b_1} \cdots x_n^{b_n} = x_1^{a_1} \cdots x_n^{a_n}$.