A neat way to prove that the set of symmetric polynomials with n variables is a subring of the ring of all polynomials with n variables So I am tasked to prove that the set of symmetric polynomials with n (n>2, let's say) variables is a subring of the ring of all polynomials with n variables.  
I know intuitively that if $A,B$ are such symmetric polymials then $A+B$, $AB$, and $-A$ are all closed respectively. Yet how do I word this proof neatly? Is there a good way to express symmetric polymials so I can do additions and multiplications and additive inverses without confusing the heck out of everyone? Thanks!
 A: For every fixed permutation of the $n$ variables, there is a corresponding ring automorphism of the ring of polynomials in those variables (just replace the variables by their image under the permutation, systematically wherever they occur in a polynomial). The set of fixed points of any automorphism$\sigma$ of a ring is clearly a subring (if $\sigma(x)=x$ and $\sigma(y)=y$ then clearly $\sigma(x+y)=x+y$ and $\sigma(xy)=xy$, and so forth). Finally the set of symmetric polynomials is just the intersection of these subrings taken over all permutations of the variables (or just over the adjacent transpositions is you want to be economical); the intersection of any collection of subrings is a subring.
A: A polynomial $f\in K[X_1,\dotsc,x_n]$ is symmetric iff
$$f(X_{\sigma(1)},\dotsc,X_{\sigma(n)})=f(X_1,\dotsc,X_n)$$
for all $\sigma\in S_n$.
So we have for example for symmetric $f$ and $g$ that
$$\begin{align}&(fg)(X_{\sigma(1)},\dotsc,X_{\sigma(n)})\\=&f(X_{\sigma(1)},\dotsc,X_{\sigma(n)})g(X_{\sigma(1)},\dotsc,X_{\sigma(n)})\\=&f(X_1,\dotsc,X_n)g(X_1,\dotsc,X_n)\\=&(fg)(X_1,\dotsc,X_n)\end{align}$$
and so $fg$ is also symmetric. Do the others in a similar way.
