Solution to Abstract Algebra by Dummit & Foote 3rd edition Chapter 2.2 Exercise 2.2.13
Let $n$ be a positive integer and let $R$ be the set of all polynomials with integer coefficients in the independent variables $x_1$, $x_2$, $\ldots$, $x_n$. I.e., elements of $R$ are formal sums $\sum_{i \in I} a_i \prod_{j=1}^n x_i^{r_{i,j}}$ for some finite set $I$, integers $a_i$, and nonnegative integers $r_{i,j}$.
For each $\sigma \in S_n$, define a mapping $\sigma \cdot$ by $$\sigma \cdot \sum_{i \in I} a_i \prod_{j=1}^n x_j^{r_{i,j}} = \sum_{i \in I} a_i \prod_{j=1}^n x_{\sigma(j)}^{r_{i,j}}.$$ Prove that this defined a left group action of $S_4$ on $R$.
Solution: Let $p = \sum_{i \in I} a_i \prod_{j=1}^n x_j^{r_{i,j}}$. We have \begin{align*}1 \cdot p =&\ 1 \cdot \sum_{i \in I} a_i \prod_{j=1}^n x_j^{r_{i,j}} = \sum_{i \in I} a_i \prod_{j=1}^n x_{\mathsf{id}(j)}^{r_{i,j}}\\ =&\ \sum_{i \in I} a_i \prod_{j=1}^n x_j^{r_{i,j}} = p.\end{align*}
Now let $\sigma$, $\tau \in S_n$. Then\begin{align*}\sigma \cdot (\tau \cdot p) =&\ \sigma \cdot (\tau \cdot \sum_{i \in I} a_i \prod_{j=1}^n x_j^{r_{i,j}})\\
=&\ \sigma \cdot \sum_{i \in I} a_i \prod_{j=1}^n x_{\tau(j)}^{r_{i,j}}\\
=&\ \sum_{i \in I} a_i \prod_{j=1}^n x_{\sigma(\tau(j))}^{r_{i,j}}\\
=&\ \sum_{i \in I} a_i \prod_{j=1}^n x_{(\sigma \circ \tau)(j)}^{r_{i,j}}\\
=&\ (\sigma \circ \tau) \cdot \sum_{i \in I} a_i \prod_{j=1}^n x_j^{r_{i,j}}\end{align*}Thus we have a group action of $S_n$ on $R$.
