Show that the ring of polynomials $k[x_{1},...,x_{n}]$ is an integral extension of the ring of symmetric polynomials 
I want to prove that $A= k[x_{1},...,x_{n}]$, where $k$ is a field, is an integral extension of the subring $B$ consisting of the symmetric polynomials in $n$ variables.

It is sufficient by finding an expression with coefficients in B that nulls when evaluated the variables $x_{i}$ for every $i=1,...,n$. I tried it but I can’t find a suitable polynomial the satisfies this. Some help?
 A: Consider the polynomial $p_n=(t-x_1)\dots(t-x_n)\in A[t]$. Clearly any permutation of the $x_i$ will leave $p_n$ unchanged, so every coefficient of $p_n$ must lie in $B$, and hence $p_n\in B[t]$. On the other hand, $p_n(x_i)=0$ for any $i\in\{1,\dots,n\}$, so $p_n$ is the desired monic polynomial satisfied by each $x_i$.
A: This question may have interest for others visiting this site.
In fact the result is more general: If $H$ is any finite group acting on a commutative ring $A$ and if $A^H$ is the set of elements $a\in A$ with $ha=a$ for all $h\in H$
it follows $A^H \subseteq A$ is an integral extension of rings. Let $H:=\{h_1,..,h_n\}$
and let $a\in A$. Let
$f_a(t):=(t-h_1a)\cdots (t-h_na)\in A[T]$.
It follows
$hf_a(t)=(t-(hh_1)a)\cdots (t-(hh_n)a)=f_a(t)$
for any $h\in G$, since the element $h$ permutes the set $\{h_1,..,h_n\}$. Hence $f_a(t)\in A^H[t]$. The polynomial
$f_a(t)$ is monic and $f_a(a)=0$ since there is an $i$ with $h_ia=ea=a$. Hence $a$ is integral over $A^H$ for any $a\in A$.
Question: "It is sufficient to find an expression with coefficients in B that is zero when evaluated in the variables $x_i$ for every $i=1,...,n$. I tried it, but I can’t find a suitable polynomial that satisfies this. Some help?"
Answer: The polynomial $f_a(t)$ defined above gives a general answer to your question for an arbitrary finite group $H$ and commutative unital ring $A$.
Example: Let $S_n$ be the symmetric group on $n$ elements and let $k:=\mathbb{Z}$ be the ring of integers. Let
$S_n:=\{\sigma_1,..,\sigma_l\}$
and let $a(x_1,..,x_n)\in A:=\mathbb{Z}[x_1,..,x_n].$
Define
$f_a(t):=(t-\sigma_1a)\cdots (t-\sigma_la) \in B:=A^{S_n}[t]$.
It follows by the above argument that $f_a(t)$ is a monic polynomial and that $f_a(a)=0$. Hence the result is true over the ring of integers $\mathbb{Z}$:
The ring extension
$\mathbb{Z}[x_1,..,x_n]^{S_n} \subseteq \mathbb{Z}[x_1,..,x_n]$
is an integral extension.
