elementary symmetric polynomials are symmetric I'm trying to prove that elementary symmetric polynomials -- defined as usual --are indeed symmetric (i.e. invariant under the action of each $\sigma \in S_n$.) This seems almost trivial to me; in a commutative ring, you can simply reorder the permuted products and sums to get the same old $e_k$. I'm wondering what a more formal proof would look like.  
 A: One approach is to consider a permutation $\sigma:\{1,\ldots,n\}\rightarrow\{1,\ldots,n\}$ not just as a bijection of this set with itself, but more generally as inducing a bijection of the set $E_{n,k}=\{S\subset\{1,\ldots,n\}:|S|=k\}$ with itself, for each $1\leq k\leq n$: If $S=\{\ell_{1},\ldots,\ell_{k}\}\subset\{1,\ldots,n\}$, then define $\sigma_{k}(S)=\{\sigma(\ell_{1}),\ldots,\sigma(\ell_{k})\}$. Since $\sigma$ is a permutation, you know that $\sigma_{k}(S)\subset\{1,\ldots,n\}$, and $|\sigma_{k}(S)|=k$ since $\sigma$ is injective. Moreover, $\sigma_{k}$ is injective, since if $\sigma_{k}(S)=\sigma_{k}(S')$, $S$ and $S'$ must both contain $\sigma^{-1}(j)$ for each $j\in\sigma_{k}(S)$, of which there are $k$, and since $|S|=|S'|=k$, they must therefore have exactly the same elements, so $S=S'$. $\sigma_{k}$ is also surjective, since the set $\{j_{1},\ldots,j_{k}\}\subset\{1,\ldots,n\}$ is the image of $\{\sigma^{-1}(j_{1}),\ldots,\sigma^{-1}(j_{k})\}$ under $\sigma_{k}$.
Now, if we consider $s_{k}(x_{1},\ldots,x_{n})=\sum_{S\in E_{n,k}}\prod_{\ell\in S}x_{\ell}$, we see that, by making the change in order of summation $S'=\sigma_{k}(S)$, $S\in E_{n,k}$: $$s_{k}(x_{\sigma(1)},\ldots,x_{\sigma(n)})=\sum_{S\in E_{n,k}}\prod_{\ell\in S}x_{\sigma(\ell)}=\sum_{S\in E_{n,k}}\prod_{j\in\sigma_{k}(S)}x_{j}=\sum_{S'\in E_{n,k}}\prod_{j\in S'}x_{j}=s_{k}(x_{1},\ldots,x_{n}).$$ Since this can be done for any permutation $\sigma$, the proof is complete.
A: If $0 \leq k \leq n$, the coefficient of $t^k$ in the polynomial
$$f(x_1,\ldots,x_n,t) = (t - x_1)\cdots(t - x_n)$$
is the $(n - k)$th elementary symmetric polynomial, i.e.
$$f(x_1,\ldots,x_n,t) = \sum_{k=0}^{n}e_{n - k}(x_1,\ldots,x_n)t^k.$$
Let $\sigma \in S_n$. The map $f_{\sigma} : R[x_1,\ldots,x_n,t] \to R[x_1,\ldots,x_n,t]$ defined by
$$f_{\sigma}(f(x_1,\ldots,x_n,t)) = f(x_{\sigma(1)},\ldots,x_{\sigma(n)},t)$$
is a ring homomorphism.
Applying $f_{\sigma}$ to $f$, we get
$$f_{\sigma}(f) = \sum_{k=0}^{n}e_{n - k}(x_{\sigma(1)},\ldots,x_{\sigma(n)})t^k.$$
But we also have
$$f_{\sigma}(f) = (t - x_{\sigma(1)})\cdots(t - x_{\sigma(n)}) = (t - x_1)\cdots(t - x_n) = f.$$
Since $f$ is fixed by $f_{\sigma}$, the coefficients must be as well, i.e. the elementary symmetric polynomials are symmetric.
A: It suffices to show that they are invariant under transpositions. A proof of the following formula gives you the result. Fix. $1\leq i<j\leq n$ and $1\leq k\leq n$. Then
$$e_k(x_1,\ldots,x_n)=e_k(x_1,\ldots,\widehat{x_i},\ldots,\widehat{x_j},\ldots,x_n)+(x_i+x_j)e_{k-1}(x_1,\ldots,\widehat{x_i},\ldots,\widehat{x_j},\ldots,x_n)+x_ix_je_{k-2}(x_1,\ldots,\widehat{x_i},\ldots,\widehat{x_j},\ldots,x_n)$$
Each term is invariant under transposition of $i$ and $j$, hence so is the sum.
To prove the formula, simply distribute the $(1+x_it)(1+x_jt)$ factor in the generating function $\prod_{p=1}^n(1+x_pt)$.
