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I am a little confused by the following exercise in Dummit and Foote, Exercise 14.6.31.

This is the setup of the exercise (not the statement to be proved):

Consider $f(x) = a_nx^n + \cdots + a_0$ and $g(x) = b_mx^m + \cdots + b_0$ as general polynomials and suppose the roots of $f(x)$ are $x_1,\dotsc,x_n$ and the roots of $g(x)$ are $y_1,\dotsc, y_m$. The coefficients of $f(x)$ are powers of $a_n$ times the elementary symmetric functions in $x_1,x_2,\dotsc,x_n$ and the coefficients of $g(x)$ are powers of $b_m$ times the elementary symmetric functions in $y_1,y_2,\dotsc,y_m.$

Earlier in the section, they give the definition

The general polynomial of degree $n$ is the polynomial $$(x - x_1)(x- x_2)\cdots (x-x_n)$$ whose roots are the indeterminates $x_1,x_2,\dotsc,x_n$.

I see how the general polynomial in this definition has coefficients which are elementary symmetric functions in $x_1,\dotsc,x_n$ (just by expanding the expression). I don't understand exactly what it means to consider $f(x)$ and $g(x)$ as general polynomials, since they don't seem to be general polynomials under this definition. I was guessing it meant to view the $a_i$ and $b_j$ as indeterminates and to view $f$ and $g$ as polynomials in $F(a_0,\dotsc,a_n,b_0,\dotsc,b_m)$ or something like that. But then, wouldn't $\,f$, for example, factor as $a_n(x - x_1)\cdots(x - x_n)$? Why would the coefficients of this polynomial involve powers of $a_n$ times the elementary symmetric functions in $x_1,\dotsc,x_n$ as opposed to just $a_n$ times the elementary symmetric functions?

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My interpretation of this is that to view these as general polynomials, we consider the coefficients as being polynomial functions of the roots $x_1,\dots,x_n$, which are indeterminant. So we write $f(x) = a_n(x - x_1)\cdots(x - x_n)$. In which case $a_n$ is a non-zero constant and $a_0,\dots,a_{n - 1}$ are symmetric polynomials in $x_1,\dots,x_n$. You are correct that we don't need "powers of $a_n$" but rather we multiply each elementary symmetric function by $a_n$.

The writing here could be better, it seems.

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  • $\begingroup$ Thank you. Just so I'm clear, you mean to view $a_i$ as an element of $F[x_1,\dotsc,x_n]$, namely, the $i$-th symmetric function in $x_1,\dots,x_n$, right? That seems like a sensible interpretation - hopefully the language they chose was a mistake! $\endgroup$ – user3075877 Jun 8 '18 at 3:50
  • $\begingroup$ Correct, yes @user3075877 $\endgroup$ – Trevor Gunn Jun 8 '18 at 4:41

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