While self studying algebra from Thomas Hungerford I have following question in understanding terminology on page 253 of textbook.

Consider this line of textbook

Let n be a positive integer, K an arbitrary field, E the subfield of symmetric rational functions in $ K(x_{1},..., x_{n} ) $ and $f_1, f_2,..., f_n \in E $ the elementary symmetric functions in $x_1 , x_2,..., x_n$ over K.

What does author mean by $f_1, f_2,..., f_n \in E $?

Does it mean that range of $f_{i}$'s belongs to E?

Clearly, it is a function in variables $x_{1} ,..., x_{n} $.


2 Answers 2


It means that the $f_i$'s them-self belong to $E$, not their range. The set $E$ contains all symmetric rational functions and hence the $n$ elementary symmetric functions $f_1, f_2, \dots, f_n$ are elements of $E$.

(Note that the elementary symmetric functions are called $e_i$ instead of $f_i$ in the Wikipedia article I linked.)

  • $\begingroup$ what does " over K " means in " Consider this line of textbook" in question asked? $\endgroup$
    – user775699
    Sep 24, 2020 at 10:11
  • 1
    $\begingroup$ It means that the coefficients in the symmetric polynomials are elements of $K$. $\endgroup$
    – Christoph
    Sep 24, 2020 at 10:52

Recall that \begin{align*} f_1(x_1,\dots,x_n)&=\sum_{i}x_i,\\ f_2(x_1,\dots,x_n)&=\sum_{i<j}x_ix_j,\\ &\ \ \vdots\\ f_n(x_1,\dots,x_n)&=x_1\cdots x_n. \end{align*} You can check that each of this function $f_i$ is symmetric, that is $$ f_i(x_{\sigma(1)},\dots,x_{\sigma(n)})=f_i(x_1,\dots,x_n) $$ for all $\sigma \in S_n$ and all $i=1,\dots,n$. So $f_1,\dots,f_n\in E$.


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