3.7 Elementary Symmetric Polynomial

Definition

Let $x_1,\cdots ,x_n$ be $n$ independent variables. Let $f_1=x_1+\cdots +x_n$, $f_2={\sum }_{1⩽i<j⩽n}{x}_i{x}_j$, $\cdots$, $f_n={\prod }_{i=1}^n{x}_i$.

Let $K$ be any field, and let $S_n$ act on $K(x_1,\cdots ,x_n)$ by

$$\sigma :f(x_1,\cdots ,x_n)/g(x_1,\cdots ,x_n)\mapsto f(x_{\sigma (1)},\cdots ,x_{\sigma (n)})/g(x_{\sigma (1)},\cdots ,x_{\sigma (n)}).$$

This action induces a map $S_n\to {\mathrm{Aut}}_K(K(x_1,\cdots ,x_n))$, which is an injective group homomorphism.

Let $E:=(K(x_1,\cdots ,x_n){)}^{S_n}$.

Proposition

The extension $K(x_1,\cdots ,x_n)/E$ is Galois, with the Galois group is isomorphic to $S_n$.

\begin{proof} By ^c7e515. \end{proof}

Lemma

Let $1⩽k⩽n-1$, and let $h_1,\cdots ,h_k$ be the elementary symmetric polynomials of $x_1,\cdots ,x_k$. Then each $h_j\in K[f_1,\cdots ,f_n,x_{k+1},\cdots ,x_n]$ for any $j⩽k$.

\begin{proof} If $k=n-1$, it is easy to verify $h_1,\cdots ,h_{n-1}\in K[f_1,\cdots ,f_n,x_n]$. In fact, $h_j=f_j-x_n{h}_{j-1}$ and $h_1=f_1-x_n$.

Now we prove the lemma by reverse induction. Suppose the argument holds for $k⩾r+1$. Now consider $k=r$. Let $g_1,\cdots ,g_{r+1}$ be elementary symmetric polynomials in $x_1,\cdots ,x_{r+1}$ and $h_1,\cdots ,h_r$ be elementary symmetric polynomials in $x_1,\cdots ,x_r$. By induction hypothesis, $g_1,\cdots ,g_{r+1}\in K[f_1,\cdots ,f_n,x_{r+1},\cdots ,x_n]$. Note that $h_1=g_1-x_{r+1}$ and $h_j=g_j-h_{j-1}\cdot x_{r+1}$. Now we finish the proof. \end{proof}

Theorem

We have $E=K(f_1,\cdots ,f_n)$.

\begin{proof} Let $F=K(f_1,\cdots ,f_n)\subseteq E$. Then $[K(x_1,\cdots ,x_n):F]⩾[K(x_1,\cdots ,x_n):E]=n!$ by ^nd8evl. Now it is suffices to show $[K(x_1,\cdots ,x_n):F]⩽n!$. Consider the chain

$$F\subseteq F(x_n)\subseteq F(x_n,x_{n-1})\subseteq \cdots \subseteq F(x_n,\cdots ,x_1)$$

and we claim that the chain satisfies $[F(x_m,\cdots ,x_n):F(x_{m+1},\cdots ,x_n)]⩽m$.

Let $g_n(T)=(T-x_1)\cdots (T-x_n)=T^n-f_1{T}^{n-1}+\cdots +(-1{)}^n{f}_n$. Then $x_n$ is a root of $g_n(T)$ yields that $[F(x_n):F]⩽n$. Let $g_k(T)=(T-x_1)\cdots (T-x_k)$. Then by ^d5a57f, $g_k(T)\in K[f_1,\cdots ,f_n,x_{k+1},\cdots ,x_n][T]\subseteq F(x_{k+1},\cdots ,x_n)[T]$ and so $[F(x_n,\cdots ,x_k):F(x_n,\cdots ,x_{k+1})]⩽k$. Now we finish the proof. \end{proof}

Remark. We also have $(K[x_1,\cdots ,x_n]{)}^{S_n}=K[f_1,\cdots ,f_n]$.

Summary. By this theorem, we prove that

$$\mathrm{Gal}(K(x_1,\cdots ,x_n)/K(f_1,\cdots ,f_n))\simeq S_n.$$