# Proving the uniqueness of the representation of the fundamental theorem on symmetric polynomials.

I am having trouble reconciling myself with a solution presented in the text Galois Theory by Harold M. Edwards. The question is presented here.

For context, the symmetric functions $$\sigma_1, \sigma_2, \dots, \sigma_n$$ in $$x_1, x_2, \dots, x_n$$ are understood to be the elementary symmetric polynomials $$\sigma_1 = x_1 + x_2 + \dots x_n \\ \sigma_2 = x_1 x_2 + x_1 x_3 + \dots + x_2 x_3 + \dots x_{n-1} x_n \\ \vdots \\ \sigma_n = x_1 x_2 \dots x_n.$$

The solution, also present in the book here, skims over some details which I have tried to prove and was unable to. My question pertains to the particular subcase for when $$F$$ can by written as $$G \cdot y_n^k$$ where $$k$$ is a nonzero integer and $$G$$ is a polynomial in the $$y$$'s that contains a nonzero term without a power of $$y_n$$. Here, why must it be so that $$G(\sigma_1, \sigma_2, \dots, \sigma_n)$$ must be the zero polynomial if $$F(\sigma_1, \sigma_2, \dots, \sigma_n)$$ is? We have that $$F(\sigma_1, \dots,\sigma_n) = G(\sigma_1, \dots, \sigma_n) \sigma_n^k \equiv 0$$ but $$\sigma_n$$ may be zero with $$G(\sigma_1, \dots, \sigma_n)$$ nonzero. Since $$\sigma_n$$ is zero when at least one of $$x_1, \dots, x_n$$ is zero, is this restriction enough to show that $$G(\sigma_1 \dots)$$, or equivalently in this case, the polynomial term of $$G(\sigma_1 \dots)$$ independent of $$\sigma_n$$ is zero? I am struggling to approach this, and beginning to wonder whether the solution is valid.

• Surely the $x_i$ are the polynomial symbols, they are not $0$ in the polynomial ring $K[x_1,\dots,x_n]$? Commented Dec 11, 2023 at 10:36
• @ancientmathematician shouldn't this comment be an answer? Commented Dec 11, 2023 at 10:47
• @ancientmathematician your comment I think settled this for me, this book attempting to be historical doesn't talk about fields/rings when talking about polynomials which also it defines somewhat vaguely and I was thinking about zero-ness by checking whether there existed any "values" that could be substituted with the symbols to give zero. Commented Dec 11, 2023 at 11:13
• I see that F in terms of the sigmas is the zero polynomial of the ring (I guess implied to be either $\mathbb{Q}[X]$ or $\mathbb{R}[X]$), then since $\sigma_n$ is not the zero polynomial and the ring has no zero divisors we must have that $G(\sigma_1,\dots)$ must be the zero polynomial. Commented Dec 11, 2023 at 11:23

The $$x_i$$ are the polynomial symbols, they are not $$0$$ in the polynomial ring $$K[x_1,\dots,x_n]$$.