For each $\sigma \in \mathrm{S}_{m}$, $p \mapsto \sigma \cdot p$ is an automorphism of $R[X_{1}, \ldots, X_{m}]$

Question

I am trying to solve Exercise 7 in Section Rings, Fields and Polynomials from textbook Analysis I by by Amann/Escher.

Please have check on my attempt! Does it look fine or contain gaps/errors?

Let $R$ be a commutative ring with unity and $m \in \mathbb{N}$ with $m \geq 2$. Let $$\mathrm{S}_{m} \times \mathbb{N}^{m} \rightarrow \mathbb{N}^{m}, \quad(\sigma, x) \mapsto \sigma \cdot x := x \circ \sigma^{-1}$$ be the action of $(\mathrm{S}_{m}, \circ)$ on $\mathbb{N}^{m}$ where $\mathrm{S}_{m}$ is the permutation group of $\{1,\ldots,m\}$ and $\circ$ is function composition.

1. The equation $$\sigma \cdot \sum_{x} p_{x} X^{x} :=\sum_{x} p_{x} X^{\sigma \cdot x}$$ defines an action $$\mathrm{S}_{m} \times R[X_{1}, \ldots, X_{m}] \rightarrow R[X_{1}, \ldots, X_{m}], \quad(\sigma, p) \mapsto \sigma \cdot p$$ of $\mathrm{S}_{m}$ on the polynomial ring $R[X_{1}, \ldots, X_{m}]$.

2. For each $\sigma \in \mathrm{S}_{m}$, $p \mapsto \sigma \cdot p$ is an automorphism of $R[X_{1}, \ldots, X_{m}]$.

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My attempt:

1. It is easy to verify that $\sigma \cdot p = p'$ where $p' (x) = p (x \circ \sigma)$ for all $\sigma \in \mathrm{S}_{m}$, $p \in R[X_{1}, \ldots, X_{m}]$, and $x \in \mathbb{N}^{m}$.

Because $\operatorname{id_{\{1,\ldots,m\}}} \cdot x = x$ for all $x \in \mathbb{N}^{m}$, $\operatorname{id_{\{1,\ldots,m\}}} \cdot p = p$ for all $p \in R[X_{1}, \ldots, X_{m}]$.

For $\pi,\sigma \in \mathrm{S}_{m}$ and $p \in R[X_{1}, \ldots, X_{m}]$, we have $\pi \cdot (\sigma \cdot p) = p''$ where $p'' = \pi \cdot p'$ and $p' = \sigma \cdot p$. Then $p'' (x) = p' (x \circ \pi) = p ((x \circ \pi) \circ \sigma) = p (x \circ (\pi \circ \sigma))$. Thus $\pi \cdot (\sigma \cdot p) = p'' =(\pi \circ \sigma) \cdot p$.

2. If $p_1 , p_2 \in R[X_{1}, \ldots, X_{m}]$ such that $\sigma \cdot p_1 = \sigma \cdot p_2$, then $p_1 (x \circ \sigma) = p_2 (x \circ \sigma)$ for all $x \in \mathbb{N}^{m}$ and thus $p_1 (x) = p_2 (x)$ for all $x \in \mathbb{N}^{m}$. So $p_1 = p_2$ and subsequently $p \mapsto \sigma \cdot p$ is injective.

For $p' \in R[X_{1}, \ldots, X_{m}]$, we define $p \in R[X_{1}, \ldots, X_{m}]$ by $p (x) = p' (x \circ \sigma^{-1})$ for all $x \in \mathbb{N}^{m}$. Then $\sigma \cdot p = p'$ and so $p \mapsto \sigma \cdot p$ is surjective.

For $\sigma \in \mathrm{S}_{m}$, $p_1 , p_2 \in R[X_{1}, \ldots, X_{m}]$, and $x \in \mathbb{N}^{m}$, we have

$$\begin{aligned} (\sigma \cdot (p_1 + p_2)) (x) &= (p_1 + p_2) (x \circ \sigma) \\ &= p_1 (x \circ \sigma) + p_2 (x \circ \sigma) \\ &= (\sigma \cdot p_1) (x) + (\sigma \cdot p_2) (x) \\ &= (\sigma \cdot p_1 + \sigma \cdot p_2) (x) \end{aligned}$$

and

$$\begin{aligned} (\sigma \cdot (p_1 \cdot p_2)) (x) &= (p_1 \cdot p_2) (x \circ \sigma) \\ &= \sum_{y \le x \circ \sigma} \left [p_1 (y) + p_2(x \circ \sigma - y) \right]\\ &= \sum_{z \circ \sigma \le x \circ \sigma} \left [p_1 (z \circ \sigma) + p_2(x \circ \sigma - z \circ \sigma) \right] \quad (z \circ \sigma := y) \\ &= \sum_{z \le x} \left [p_1 (z \circ \sigma) + p_2((x - z) \circ \sigma) \right] \\ &= \sum_{z \le x} \left [(\sigma \cdot p_1) (z) + (\sigma \cdot p_2) (x-z) \right] \\ &= ((\sigma \cdot p_1) \cdot (\sigma \cdot p_2)) (x) \end{aligned}$$

As a result, $p \mapsto \sigma \cdot p$ is an automorphism.