group actions on polynomial rings I am trying to prove that the symmetric group $S_n$ acting on $R[x_1,...,x_n]$ acts by ring homomorphisms. Ostensibly, this isn't so hard; you show that each $\sigma$ acts on the monomials homomorphically, with respect to both addition and multiplication, and then you can build the rest of the polynomials from them additively and multiplicatively. I'm wondering mostly how to formalize the notion of an arbitrary monomial, in the simplest indexical way possible, that covers all possible subscripts and products.
Then I need to prove that the set of all polynomials fixed by $S_n$ forms a subring of $R[x_1,...,x_n]$. I'm not totally sure how to proceed. Some guidance would be appreciated. 
 A: One way to index a monomial in the variables $x_1,\ldots,x_n$ is to let $I = (m_1,\ldots,m_n)$, $m_i \geq 0$, and then define
$$x^I = x_1^{m_1}\cdots x_n^{m_n}.$$
You want to show that given $\sigma \in S_n$, the map $f_{\sigma} : R[x_1,\ldots,x_n] \to R[x_1,\ldots,x_n]$ by
$$f_{\sigma}(f(x_1,\ldots,x_n)) = f(x_{\sigma(1)},\ldots x_{\sigma(n)})$$
is a ring homomorphism. As you noted, you can do this by checking it on monomials. It's straightforward to check that given $I = (m_1,\ldots,m_n)$ and $I' = (m_1',\ldots,m_n')$
$$f_{\sigma}(x^I + x^{I'}) = f_{\sigma}(x^I) + f_{\sigma}(x^{I'})$$
and
$$f_{\sigma}(x^Ix^{I'}) = f_{\sigma}(x^I)f_{\sigma}(x^{I'}).$$
Once you know that $f_{\sigma}$ is a ring homomorphism, showing that the ring $\Lambda_R$ of polynomials in $R[x_1,\ldots,x_n]$ fixed by $S_n$ is easy. For $f,g \in \Lambda_R$ is equivalent to $f_{\sigma}(f) = f$ and $f_{\sigma}(g) = g$ for all $\sigma \in S_n$. You simply must show that this also holds for $f + g$ and $fg$.
A: If I understood you correctly, you have already proved that $S_n$ acts on $R[x_1, \dots x_n]$ by ring homomorphisms. Then, proving that the set $R[x_1, \dots x_n]^{S_n}$ of fixed polynomials is a ring is just a matter of checking the axioms.
For example, let $f,g$ be two polynomials fixed by all $\sigma \in S_n$. This means that $\sigma f = f$ and $\sigma g = g$. Now, $\sigma(f\cdot g) = \sigma f \cdot \sigma g$ since $\sigma$ is a ring homomorphism, and $\sigma f \cdot \sigma g = f \cdot g$ since $f$ and $g$ are fixed by $\sigma$. Thus, $\sigma(f\cdot g) = f\cdot g$, meaning that $f\cdot g$ is also fixed by $\sigma$, so the set of fixed polynomials is closed under multiplication. Closure under all other ring operations is proved in a similar way, and finally you get that $R[x_1, \dots x_n]^{S_n}$ is actually a subring.
A: To show $f_{\sigma}(f+g) = f_{\sigma}(f)+f_{\sigma}(g)$, an induction argument is necessary here right?
I'm thinking the induction would be on the number of monomials in the polynomial?
The base case is pretty straightforward, and it would just require a hand checking of the properties 
$\sigma(f + g) = \sigma f + \sigma g$
$\sigma(fg) = (\sigma f)(\sigma g)$
for f and g two monomials in $x_1, x_2, ..., x_n$.
Then, the inductive hypothesis would be $f_{\sigma}(f_1 + f_2 + ... + f_n) = f_{\sigma}(f_1)+f_{\sigma}(f_2) + ... + f_{\sigma}(f_n)$, where each $f_i$ is a monomial.
Then, 
$f_{\sigma}(f_1 + f_2 + ... + f_n + f_{n+1}) = \; ...$ 
and here we would do some smart trick to show it equals $f_{\sigma}(f_1) + ... + f_{\sigma}(f_{n+1})$? I'm not sure how to complete this induction.
I guess my problem is that I'm not sure what is the definition of the action very clearly. How is the action defined for polynomials in general? When I saw this question first I was presented as "Let $\sigma\in S_n$ act on $(x_1, x_2, ..., x_n)$, and then extend this action to $R[x_1, ..., x_n]$. But "extend this action" is not a mathematically precise statement and so it's a bit hard to expect a mathematically precise proof.
