Group of k-Algebra Automorphisms over a Polynomial Ring In Commutative Algebra with a View Toward Algebraic Geometry (which I'm clearly not ready for, but anyway...), there's the following example in section 1.3:

Let $S = k[x_1,...,x_r]$ by the polynomial ring, and let $\Sigma$ by
  the symmetric group of all permutations of $\{1,...,r\}$. The group
  $\Sigma$ acts on $S$ as follows: If $\sigma \in \Sigma$ and $f \in S$,
  we define
$$\sigma(f)(x_1,...,x_r) = f(x_{\sigma^{-1}(1)},\dots,x_{\sigma^{-1}(r)}).$$
The group $\Sigma$ acts as a group of $k$-algebra automorphisms of
  $S$.
Blah blah blah more stuff anjruu doesn't understand.

So, I get that $\sigma \in \Sigma$ is an automorphism from $S$ to $S$, since it is injective, and preserves the ring structure. 
I also get that $\Sigma$ is a group, since, well, it's based off of the group of permutations. 
What I don't get is how $\sigma \in \Sigma$ is a k-algebra. You need to be a module to be a k-algebra (or so I thought), but one of the properties of a module is scalar multiplication, and I don't understand how you would multiply $\sigma$ (what I was thinking of as the vector in this hypothetical module) with an element from $S$ (the scalars).
Can anyone enlighten me as to what's going on?
 A: It's not saying that each $\sigma$ is a $k$-algebra, but that each $\sigma$ is an automorphism of $k$-algebras.   That is, in addition to being a ring-homomorphism, each $\sigma$ is also $k$-linear.
A: Recall that the transpositions generate $S_n$, so we lose no generality by merely considering $k[x,y]$. Then $k$-linearity boils down to the fact that:
1.) If we switch independently $x$ and $y$ in two polynomials $f(x,y), g(x,y)$, and then add them, we get the same polynomial as if we added first, then switched $x$ and $y$.
2.) If we switch $x$ and $y$ in a polynomial $f(x,y)$, and then multiply every coefficient by $\alpha \in k$ , we obtain the same polynomial as we do if we multiply every coefficient by $\alpha$ first, and then swap $x$ and $y$.
It is clear that such a map $\sigma$ is bijective, so all we lack for $\sigma$ to be a $k$-algebra automorphism is to show $\sigma$ is multiplicative (which is similar to 1.) above).
In other words, all we are "really" doing with $\sigma$ is indexing "variable-relabeling" by a corresponding "finite-set-shuffling" of the indices of the variables (or indeterminates, more properly speaking). The discussion above indicates we can do this "two indeterminates at a time".
